Download Part 1 - Mechanics and Thermodynamics

Physics I.
György Hárs
Gábor Dobos
2013.04.30
Contents
1 Kinematics of a particle - György Hárs
1.1 Rectilinear motion . . . . . . . . . . . .
1.1.1 Uniform Rectilinear Motion . . .
1.1.2 Uniformly Accelerated Rectilinear
1.1.3 Harmonic oscillatory motion . . .
1.2 Curvilinear motion . . . . . . . . . . . .
1.2.1 Projectile motion . . . . . . . . .
1.2.2 Circular motion . . . . . . . . . .
1.2.3 Areal velocity . . . . . . . . . . .
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Motion
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2 Dynamics of a Particle - György Hárs
2.1 Inertial system . . . . . . . . . . . . . . . . . .
2.2 The mass . . . . . . . . . . . . . . . . . . . . .
2.3 Linear momentum (p) . . . . . . . . . . . . . .
2.4 Equation of motion: . . . . . . . . . . . . . . . .
2.5 The concept of weight . . . . . . . . . . . . . .
2.6 The concept of work in physics . . . . . . . . .
2.7 Power . . . . . . . . . . . . . . . . . . . . . . .
2.8 Theorem of Work (Kinetic energy) . . . . . . .
2.9 Potential energy . . . . . . . . . . . . . . . . . .
2.10 Conservation of the mechanical energy . . . . .
2.11 Energy relations at harmonic oscillatory motion
2.12 Angular momentum . . . . . . . . . . . . . . . .
2.13 Torque . . . . . . . . . . . . . . . . . . . . . . .
2.14 Central force field . . . . . . . . . . . . . . . . .
3 Dynamics of system of particles - György Hárs
3.1 Momentum in system of particles . . . . . . . .
3.1.1 Collisions . . . . . . . . . . . . . . . . .
3.1.2 Missile motion . . . . . . . . . . . . . . .
3.2 Angular momentum in system of particles . . .
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3.3
3.2.1 The skew rotator . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.2 The pirouette dancer (The symmetrical rotator) . . . . . . . . . .
Discussion of the total kinetic energy in the system of particles . . . . .
4 Dynamics of rigid body - György Hárs
4.1 Moment of inertia . . . . . . . . . . . .
4.2 Equation of motion of the rigid body: .
4.2.1 Demonstration example 1. . . .
4.2.2 Demonstration example 2. . . .
4.3 Kinetic energy of the rigid body . . . .
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5 Non-inertial (accelerating) reference frames - György Hárs
5.1 Coordinate system with translational acceleration . . . . . . . . . . . . .
5.2 Coordinate system in uniform rotation . . . . . . . . . . . . . . . . . . .
5.2.1 Earth as a rotating coordinate system . . . . . . . . . . . . . . . .
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6 Oscillatory Motion - Gábor Dobos
6.1 The simple harmonic oscillator . . . . . . . . . . . . .
6.1.1 Complex representation of oscillatory motion .
6.1.2 Velocity and acceleration in oscillatory motion
6.2 Motion of a body attached to a spring . . . . . . . .
6.3 Simple pendulum . . . . . . . . . . . . . . . . . . . .
6.4 Energy in simple harmonic motion . . . . . . . . . .
6.5 Damped oscillator . . . . . . . . . . . . . . . . . . . .
6.6 Forced oscillations . . . . . . . . . . . . . . . . . . .
6.7 Superposition of simple harmonic oscillations . . . . .
6.7.1 Same frequency, same direction . . . . . . . .
6.7.2 Different frequency, same direction . . . . . .
6.7.3 Lissajous figures . . . . . . . . . . . . . . . . .
6.7.4 Fourier analysis . . . . . . . . . . . . . . . . .
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7 Waves - Gábor Dobos
7.1 Sine wave . . . . . . . . . . . . . . . .
7.2 Transverse wave on a string . . . . . .
7.3 Energy transport by mechanical waves
7.4 Group velocity . . . . . . . . . . . . .
7.5 Wave packets . . . . . . . . . . . . . .
7.6 Standing waves . . . . . . . . . . . . .
7.7 The Doppler Effect . . . . . . . . . . .
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8 First law of thermodynamics and related subjects
8.1 Ideal gas equation . . . . . . . . . . . . . . . . . . .
8.2 The internal energy of the gas (U ) . . . . . . . . .
8.3 The p-V diagram . . . . . . . . . . . . . . . . . . .
8.4 Expansion work of the gas . . . . . . . . . . . . . .
8.5 First law of thermodynamics . . . . . . . . . . . . .
8.5.1 Isochoric process . . . . . . . . . . . . . . .
8.5.2 Isobaric process . . . . . . . . . . . . . . . .
8.5.3 Isothermal process . . . . . . . . . . . . . .
8.5.4 Adiabatic process . . . . . . . . . . . . . . .
8.6 Summary of the molar heat capacitances . . . . . .
8.7 The Carnot cycle . . . . . . . . . . . . . . . . . . .
9 The
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
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György
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Hárs
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entropy and the second law of thermodynamics - György Hárs
The entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The isentropic process . . . . . . . . . . . . . . . . . . . . . . . . . . .
The microphysical meaning of entropy . . . . . . . . . . . . . . . . . .
Gay-Lussac experiment . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.4.1 Phenomenological approach . . . . . . . . . . . . . . . . . . . .
9.4.2 Statistical approach . . . . . . . . . . . . . . . . . . . . . . . . .
The Boltzmann equation . . . . . . . . . . . . . . . . . . . . . . . . . .
Approximate formula (ln n! ≈ n ln n − n) a sketch of proof: . . . . . . .
Equalization process . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.7.1 Equalization between gaseous components . . . . . . . . . . . .
9.7.2 Equalization of non-gaseous materials without phase transition .
9.7.3 Ice cubes in the water . . . . . . . . . . . . . . . . . . . . . . .
The second law of thermodynamics . . . . . . . . . . . . . . . . . . . .
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153
Introduction - György Hárs
Present work is the summary of the lectures held by the author at Budapest University of
Technology and Economics. Long verbal explanations are not involved in the text, only
some hints which make the reader to recall the lecture. Refer here the book: Alonso/Finn
Fundamental University Physics, Volume I where more details can be found.
Physical quantities are the product of a measuring number and the physical unit.
In contrast to mathematics, the accuracy or in other words the precision is always a
secondary parameter of each physical quantity. Accuracy is determined by the number
of valuable digits of the measuring number. Because of this 1500 m and 1.5 km are not
equivalent in terms of accuracy. They have 1 m and 100 m absolute errors respectively.
The often used term relative error is the ratio of the absolute error over the nominal
value. The smaller is the relative error the higher the accuracy of the measurement.
When making operations with physical quantities, remember that the result may not be
more accurate than the worst of the factors involved. For instance, when dividing 3.2165
m with 2.1 s to find the speed of some particle, the result 1.5316667 m/s is physically
incorrect. Correctly it may contain only two valuable digits, just like the time data, so
the correct result is 1.5 m/s.
The physical quantities are classified as fundamental quantities and derived quantities. The fundamental quantities and their units are defined by standard or in other
words etalon. The etalons are stored in relevant institute in Paris. The fundamental
quantities are the length, the time and the mass. The corresponding units are meter
(m), second (s) and kilogram (kg) respectively. These three fundamental quantities are
sufficient to build up the mechanics. The derived quantities are all other quantities which
are the result of some kind of mathematical operations. To describe electric phenomena
the fourth fundamental quantity has been introduced. This is ampere (A) the unit of
electric current. This will be used extensively in Physics 2, when dealing with electricity.
4
Chapter 1
Kinematics of a particle - György
Hárs
Kinematics deals with the description of motion, without any respect to the cause of the
motion. Strictly speaking there is no mass involved in the theory, so force and related
quantities do not show up. The fundamental quantities involved are the length and the
time only.
To describe the motion one needs a reference frame. Practically it is the Cartesian
coordinate system with x, y, z coordinates, and corresponding i, j, k unit vectors.
The particle is a physical model. This is a point like mass, so it lacks of any extension.
1.1
Rectilinear motion
(Egyenes vonalú mozgás)
The motion of the particle takes place in a straight line in rectilinear motion. This
means that the best mathematical description is one of the axes of the Cartesian coordinate system. So the position of the particle is described by x(t) function.
The velocity of the particle is the first derivative of the position function. The everyday concept of speed is the absolute value of the velocity vector. Therefore the speed is
always a nonnegative number, while the velocity can also be a negative number.
v(t) = lim
t=0
dx
∆x
=
∆t
dt
hmi
s
(1.1)
The opposite direction operation recovers the position time function from the velocity
0
vs. time function. Here x0 is the initial value of the position in t = 0 moment, t denotes
5
the integration parameter from zero to t time.
Zt
x(t) = x0 +
v(t, )dt,
(1.2)
0
The acceleration of the particle is the first derivative of the velocity vs. time function,
thus it is the second derivative of the position vs. time function.
dv
d2 x
∆v
=
= 2
a(t) = lim
t=0 ∆t
dt
dt
hmi
s2
(1.3)
The opposite direction operation recovers the velocity time function from the acceleration
0
vs. time function. Here v0 is the initial value of the position in t = 0 moment, t denotes
the integration parameter from zero to t time.
Zt
v(t) = v0 +
a(t, )dt,
(1.4)
0
1.1.1
Uniform Rectilinear Motion
(Egyenes vonalú egyenletes mozgás)
Here the acceleration of the particle is zero. The above formulas transform to the
following special cases. a = 0, v = v0 , x = x0 + vt.
Figure 1.1: Uniform Rectilinear Motion
1.1.2
Uniformly Accelerated Rectilinear Motion
(Egyenes vonalú egyenletesen gyorsuló mozgás)
Here the acceleration of the particle is constant. The above formulas transform to
the following special cases. a = const, v = v0 + at x = x0 + v0 t + a2 t2
Typical example is the free fall, where the acceleration is a = g = 9.81 m/s2 .
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Figure 1.2: Uniformly Accelerated Rectilinear Motion
1.1.3
Harmonic oscillatory motion
(Harmónikus rezgőmozgás)
The trajectory of the harmonic oscillation is straight line, so this is a special rectilinear
motion. First let us consider a particle in uniform circular motion.
Figure 1.3: Harmonic oscillatory motion
The two coordinates in the Cartesian coordinate system are as follows:
x = A cos(ωt + φ) y = A sin(ωt + φ)
(1.5)
If the uniform circular motion is projected to one of its coordinates, the motion of the
projected point is “harmonic oscillatory motion”. We choose the x coordinate.
x(t) = A cos(ωt + φ)
(1.6)
The displacement at oscillatory motion is called excursion. The sum in the parenthesis
is called the “phase”. The multiplier of time is called angular frequency, and additive
ϕ constant is the initial phase. The multiplier in front is called the “amplitude”. The
velocity of the oscillation is the derivative of the displacement function.
dx(t)
= v(t) = −Aω sin(ωt + φ)
dt
7
(1.7)
The multiplier of the trigonometric term is called the “velocity amplitude” (vmax ).
vmax = Aω
(1.8)
The acceleration is the derivative of the velocity:
dv(t)
= a(t) = −Aω 2 cos(ωt + φ)
dt
(1.9)
If one compares the displacement and the acceleration functions the relation below can
readily found:
a(t) = −ω 2 x(t)
(1.10)
Accordingly, the acceleration is always opposite phase position relative to the displacement.
In the kinematics of the harmonic oscillations it is very much helpful to go back to the
origin of the oscillatory motion and contemplate the phenomena as projected component
of a uniform circular motion. This way one gets rid of the trigonometric formalism and
the original problem could have a far easier geometric interpretation. Best example for
that if we want to find out the resultant oscillation of two identical frequency harmonic
oscillations with different amplitudes and different initial phases. In pure trigonometry
approach this is a tedious work, while in the circle diagram this is a simple geometry
problem, actually a cosine theorem application in the most ordinary case.
1.2
Curvilinear motion
(Görbervonalú mozgás)
The motion of the particle is described by an arbitrary r(t) vector scalar function,
where i, j, k are the unit vectors of the coordinate system.
r(t) = x(t)i + y(t)j + z(t)k
(1.11)
The velocity of the particle is the first derivative of the position function.
v(t) = lim
t=0
dx
dy
dz
∆r
=
i+ j+ k
∆t
dt
dt
dt
(1.12)
The velocity vector is tangential to the trajectory of the particle always.
The vector of acceleration is the derivative of the velocity vector. The vector of
acceleration can be decomposed as parallel and normal direction to the velocity.
∆v
dvx
dvy
dvz
d2 x
d2 y
d2 z
a(t) = lim
=
i+
j+
k= 2i+ 2j+ 2k
t=0 ∆t
dt
dt
dt
dt
dt
dt
8
(1.13)
Figure 1.4: Curvilinear motion
The parallel component of the acceleration (called tangential acceleration) is the
consequence of the variation in the absolute value of the velocity. In other words this is
caused by the variation of the speed. The normal component of the acceleration (called
centripetal acceleration) is the consequence of the change in the direction of the velocity
vector.
If one drives a car on the road, speeding up or slowing down causes the tangential
acceleration to be directed parallel or opposite with the velocity, respectively. By turning
the steering wheel, centripetal acceleration will emerge. The direction of the centripetal
acceleration points in the direction of the virtual center of the bend.
1.2.1
Projectile motion
(Hajı́tás)
In the model of the description the following conditions will be used:
Projectile is a particle,
Gravity field is homogeneous,
Rotation of the Earth, does not take part,
No drag due to air friction will be considered.
In real artillery situation the phenomenon is much more complex. This is far beyond
the present scope.
Figure 1.5: Projectile motion
9
The projectile is fired from the origin of the Cartesian coordinate system. The motion
is characterized by the initial velocity v0 and the angle of the velocity α relative to the
horizontal direction. The motion will take place in the vertical plane, which contains the
velocity vector. The motion is the superposition of a uniform horizontal rectilinear motion, a uniform vertical rectilinear motion and a free fall. Thus the velocity components
are as follows:
vx = v0 cos α
(1.14)
vy = v0 sin α − gt
(1.15)
The corresponding position coordinates are the integrated formulas with zero initial
condition.
Zt
0
0
vx (t )dt = v0 t cos α
(1.16)
g
0
0
vy (t )dt = v0 t sin α − t2
2
(1.17)
x=
0
Zt
y=
0
Two critical parameters are needed to find out. These are the height of the trajectory
(h) and the horizontal flight distance (d). First, the rise time should be calculated. The
rise time τrise is the time when the vertical velocity component vanishes. Accordingly
vy =0 condition should be met. From the equation the following results:
τrise =
v0 sin α
g
(1.18)
The height of the trajectory shows up as a vertical coordinate just in rise time moment.
g 2
h = y(t = τrise ) = v0 τrise sin α − τrise
2
(1.19)
By substituting the formula of τrise into the equation above, the height of the trajectory
results:
h=
v02 sin2 α g v02 sin2 α
−
g
2 g2
(1.20)
v02 sin2 α
2g
(1.21)
Accordingly:
h=
10
The rise and the fall part of the motion last the same duration, due to the symmetry of
the motion. Because of this, the total flight time of the motion is twice longer than the
rise time alone. The horizontal flight distance (d) can be calculated as the horizontal x
coordinate at double rise time moment.
v0 sin α
v02
d = x(t = 2τrise ) = v0 2
cos α = 2 sin α cos α
g
g
(1.22)
By using elementary trigonometry, the final formula of horizontal flight distance results:
d=
v02
sin 2α
2
(1.23)
This clearly shows that the projectile flies the furthest if the angle of the shot is 45
degrees.
1.2.2
Circular motion
(Körmozgás)
In circular motion, the particle moves on a circular plane trajectory. To describe the
position of the particle polar coordinates are used. The origin of the polar coordinate
system is the center of the motion. The only variable parameter is the angular position
ϕ(t) since the radial position is constant.
Figure 1.6: Circular motion
The derivative of the angular position is the angular velocity ω.
∆φ
dφ
1
ω(t) = lim
=
t=0 ∆t
dt
s
(1.24)
Up to this moment looks as if the angular velocity were a scalar number. But this is
not the case. The angular velocity is a vector in fact, because it should contain the
11
information about the rotational axis as well. By definition, the angular velocity vector
ω is as follows: The absolute value of the ω is the derivative of the angular position as
written above. The direction of the ω is perpendicular, or in other words, normal to the
plane of the rotation, and the direction results as a right hand screw rotation. This latter
means that by turning a usual right hand screw in the direction of the circular motion,
the screw will proceed in the direction of the ω vector. Just an example: If the circular
motion takes place in the plane of this paper and the rotation is going clockwise, the ω
will be directed into the paper. Counter clockwise rotation will obviously result in a ω
vector pointing upward, away from the paper.
With the help of ω vector number of calculation will be much easier to carry out. For
example finding out the velocity vector of the particle is as easy as that:
v =ω×r
(1.25)
This velocity vector is sometimes called “circumferential velocity” however this notation
is redundant, since the velocity vector is always tangential to the trajectory. The cross
product of vectors in mathematics has a clear definition. By turning the first factor ( ω)
into the second one (r) the corresponding turning direction defines the direction of the
velocity vector by the right hand screw rule. The absolute value of the velocity is the
product of the individual absolute values, multiplied with the sine of the angle between
the vectors.
Before going into further details, let us state three mathematical statements. Let
a(t) and b(t) are two time dependent vectors and λ(t) a time dependent scalar. Then
the following differentiation rules apply:
da(t)
db(t)
d
(a(t) × b(t)) =
× b(t) + a(t) ×
dt
dt
dt
(1.26)
d
da(t)
db(t)
(a(t)b(t)) =
b(t) + a(t)
dt
dt
dt
(1.27)
d
dλ(t)
da(t)
(λ(t)a(t)) =
a(t) + λ(t)
(1.28)
dt
dt
dt
These formulas make it possible to use the same differentiation rules among the vector
products, just like among the ordinary product functions. End this is true both the cross
product and the dot product operations. The proof of these rules, are quite straightforward. The vectors should be written by components, and the match of the two sides
should be verified.
Using the ω vector is a powerful means. This way the acceleration vector of the
particle can be determined with a relative ease.
a=
dv(t)
d
dω
dr
=
(ω × r) =
×r+ω×
dt
dt
dt
dt
12
(1.29)
The derivative of ω vector is called the vector of angular acceleration β. This is the
result of the variation in the angular velocity either due to spinning faster or slower or
by changing the axis of the rotation.
1
dω
=β
(1.30)
dt
s2
Last term is the derivative of the position vector. This is the velocity, which can be
written as above wit the help of ω vector. So ultimately the acceleration vector can be
summarized.
a = β × r + ω × (ω × r)
(1.31)
The above formula consists of two major terms. The first term is called tangential
acceleration. In case of plane motion, this is parallel or opposite to the velocity and
it is the consequence of speeding up or slowing down, as explained in the earlier part
of this chapter. The second term is called the centripetal or normal acceleration. This
component points toward the center of the rotation. The centripetal acceleration is the
consequence of the direction variation of the velocity vector. The absolute values of these
components can readily be expressed.
atan = βr
acpt = rω 2 =
v2
r
(1.32)
There are two special kinds of circular motion, the uniform and the uniformly accelerating
circular motion.
Uniform circular motion:
(Egyenletes körmozgás)
In here the angular velocity is constant. The angle or rotation can be expressed
accordingly:
φ(t) = ωt + φ0
(1.33)
Since the angular acceleration is zero, no tangential acceleration will emerge. However
there will be a constant magnitude centripetal acceleration, with an ever changing direction, pointing always to the center.
acpt = rω 2 =
13
v2
r
(1.34)
Uniformly accelerating circular motion
(Egyenletesen gyorsuló körmozgás)
In here the angular acceleration is constant. The corresponding formulas are analogous to that of uniformly accelerating rectilinear motion, explained earlier in this chapter.
ω(t) = βt + ω0
φ(t) =
β 2
t + ω0 t + φ0
2
(1.35)
(1.36)
The magnitude of the tangential acceleration is constant and parallel with the velocity
vector.
atan = βr = const
(1.37)
The magnitude of the centripetal component shows quadratic dependence in time.
acpt = rω 2 = r(βt + ω0 )2
1.2.3
(1.38)
Areal velocity
(Területi sebesség)
Figure 1.7: Areal velocity
Let us consider particle travelling on its trajectory. If one draws a line between the
origin of the coordinate system and the particle, this line is called the “radius vector”.
The vector of areal velocity is the ratio of the area swept by the radius vector over time.
The crosshatched triangle on the figure above is the absolute value of the infinitesimal
variation (dA) of the swept area vector.
1
dA = r × dr
2
14
(1.39)
1
dA = r × vdt
2
dA
1
= r×v
dt
2
m2
s
(1.40)
Areal velocity will be used in the study of planetary motion later in this book.
15
(1.41)
Chapter 2
Dynamics of a Particle - György
Hárs
(Tömegpont dinamikája)
Dynamics deals with the cause of motion. So in dynamics a new major quantity
shows up. This is the mass of the particle (m). The concept of force and other related
quantities will be treated as well. In this chapter only one piece of particle will be the
subject of the discussion, in the next chapter however the system of particles will be
treated.
2.1
Inertial system
In kinematics any kind of coordinate system could be used, there was no any restriction in
this respect. In dynamics however, a dedicated special coordinate system is used mostly.
This is called the inertial system. The inertial system is defined as a coordinate system
in which the law of inertia is true. The law of inertia or Newton’s first law says that the
motion state of a free particle is constant. This means that if it was standstill it stayed
standstill, if it was moving with a certain velocity vector, it continues its motion with
the same velocity. So the major role of Newton’s first law is the definition of the inertial
system. Other Newton’s laws use the inertial system as a frame of reference further on.
The best approximation of the inertial system is a free falling coordinate system. In
practice this can be a space craft orbiting the Earth, since the orbiting space craft is in
constant free fall.
The inertial systems are local. This means that the point of the experimentation and
its relative proximity belongs to a dedicated inertial system. An example explains this
statement: Imagine that we are on a huge spacecraft circularly orbiting the Earth, so
we are in inertial system. Now a small shuttle craft is ejected mechanically from the
spacecraft without any rocket engine operation. The shuttle craft also orbits the Earth
16
on a different trajectory and departs relatively far from the mother ship. Observing the
events from the inertial system of the mother ship the shuttle supposed to keep its original
ejection velocity and supposed to depart uniformly to the infinity. Much rather instead
the shuttle craft also orbits the Earth and after a half circle it returns to the mother ship
on its own. So the law of inertia is true in the close proximity of the experiment only. If
one goes too far the law of inertia looses validity.
On the surface of the Earth we are not in inertial system. Partly because we experience weight, which is the gravity force attracting the objects toward the center,
partly because the Earth is rotating, which rotation causes numerous other effects. Even
though in most cases phenomena on the face of our planet can be described in inertial
system, by ignoring the rotation related effects, and by considering the gravity a separate
interaction.
2.2
The mass
Mass is a dual face quantity. Mass plays role in the interaction with the gravity field.
This type of mass called gravitational mass and this is something like gravitational charge
in the Newton’s gravitational law.
F =G
m1 m2
r2
(2.1)
Here the m1 and m2 are the gravitational masses, r is the distance between the objects, F
is the resulting force and G is the gravitational constant (6.67x10−11 m3 kg−1 s−2 ). When
somebody measures the body weight with a bathroom scale he actually measures the
gravitational mass.
Other major feature is that the mass shows resistance against the accelerating effect. This resistance is characterized by the inertial mass. It has been discovered later
that these fundamentally different features can be related to the same origin, and so
the two types of mass are equivalent. Therefore the distinction between them became
unnecessary.
This equivalency makes the free falling objects drop with the same acceleration. The
gravity force is proportional with the gravitational mass, which force should be equal with
the acceleration times the inertial mass. So if the ratio of these masses were different,
then the free fall would happen with different acceleration for different materials. This
is harshly against the experience, so mass will be referred without any attribute later in
this book.
2.3
Linear momentum (p)
(Impulzus, mozgásmennyiség, lendület)
17
By definition the linear momentum is the product of the mass and the velocity.
Therefore linear momentum is a vector quantity.
kgm
(2.2)
p = mv
s
Newton’s second law: This law is the definition of force (F).
The force exerted to a particle is equal to the time derivative of the linear momentum.
The unit of force is Newton (N).
dp
kgm
F=
=N
(2.3)
dt
s2
Conclusion 1.
If the force equals to zero, then the linear momentum is constant. This is in agreement
with the law of inertia. However it is worth mentioning, that it is only true in inertial
system. Which means that on an accelerating train or in a spinning centrifuge it is not
valid.
Conclusion 2.
According to the fundamental theorem of calculus, the time integral of force results
in the variation of the linear momentum:
Zt2
p2 − p1 =
F(t)dt
(2.4)
t1
The right hand side is called impulse (erőlökés).
Conclusion 3.
The well-known form of the Newton second law can be readily expressed:
F=
d(mv)
dv
dp
=
=m
= ma
dt
dt
dt
(2.5)
Or briefly:
F = ma
(2.6)
Newton third law: (Action reaction principle)
When two particles interact, the force on one particle is equal value and opposite
direction to the force of the other particle.
18
Figure 2.1: Action reaction principle
2.4
Equation of motion:
(mozgásegyenlet)
The particle is affected by numerous forces. The sum of these forces, cause the
acceleration of the particle. This leads to a second order ordinary differential equation.
This is called the equation of motion:
X
Fi (r, t, v) = m
i
d2 r
dt2
(2.7)
In principle the forces may be the function of position, time and velocity.
Example 1 for the equation of motion :
Attenuated oscillation:
(csillapodó rezgés)
A particle is hanging on a spring in water in vertical position. The particle is deflected
to a higher position, and left alone to oscillate. Describe the motion by solving the
equation of motion. Ignore the buoyant force. The motion will take place in the vertical
line. The position is denoted y(t) which is positive upside direction.
The forces affecting the particle are as follows:
Fspring = −Dy
Fdrag = −k
dy
dt
Fgrav = −mg
(2.8)
Here D is the direction constant of the spring in N/m units, k is the drag coefficient and
g is 9.81 m/s 2 .Accordingly the equation of motion can be written:
−k
dy(t)
d2 y(t)
− Dy(t) − mg = m
dt
dt2
19
(2.9)
Ordering it to the form of a differential equation:
d2 y(t)
k dy(t) D
+ y(t) = g
+
2
dt
m dt
m
(2.10)
Let us introduce β for the attenuation coefficient with the following definition:
β=
k
2m
(2.11)
d2 y(t)
dy(t) D
+ y(t) = g
+ 2β
2
dt
dt
m
(2.12)
The mathematical method for solving this differential equation is beyond the scope of
this chapter. The solution below can be verified by substitution:
y(t) = Ae−βt cos ωt −
mg
D
(2.13)
Here A is the original value of the deflection, ω0 is called the Thomson angular frequency
and ω is the angular frequency of the attenuated oscillation with the following definitions:
r
q
D
ω = ω02 − β 2
(2.14)
ω0 =
m
Figure 2.2
20
Example 2 for the equation of motion: Conical pendulum
(kúpinga)
Figure 2.3: Conical pendulum
The conical pendulum circulates in horizontal plane with ω angular frequency. The
angle of the rope α relative to the vertical direction is the unknown parameter to be
determined. The coordinate system is an inertial system with horizontal and vertical
axes, with the particle in the origin. There are two forces affecting the particle, gravity
force (mg) and the tension of the rope (K). The equation of the motion is a vector
equation in two dimensions so two scalar equations are used.
← K sin α = macp
(2.15)
↓ mg − K cos α = 0
(2.16)
In addition the centripetal acceleration can be expressed readily:
acp = lω 2 sin α
21
(2.17)
After substitution K = mlω 2 results.
By means of this result the cosine of the angular position is determined:
g
cos α = 2
lω
2.5
(2.18)
The concept of weight
(A súly fogalma)
Let us place a bathroom scale on the floor of an elevator. The normal force (N ) is
displayed by the scale that is transferred to the object.
Figure 2.4: The concept of weight
The positive reference direction is pointing down. The following equation of motion
can be written:
↓ mg − N = ma
(2.19)
Here the acceleration of the elevator is denoted (a). Let us express the normal force
indicated by the scale:
N = m(g − a)
(2.20)
If the elevator does not accelerate (in most cases it is standstill) the scale shows the
force which is considered the weight of the object in general. (N =mg). This force is
just enough to compensate the gravity force, so the object does not accelerate. However,
when the elevator accelerates up or down, the indicated value is increased or decreased,
respectively. This also explains that in a freefalling coordinate system, where a = g, the
weight vanishes. Similarly zero gravity shows up on the orbiting spacecraft, which is also
in constant freefall.
22
2.6
The concept of work in physics
(A munka fogalma a fizikában)
The concept of work in general is very broad. Besides physics, it is used in economy,
also used as “spiritual work”. Concerning the physical concept, the amount of work is
not too much related, how much tiredness is suffered by the person who actually made
this work. For example, if somebody is standing with fifty kilogram sack on his back
for an hour without any motion, surely becomes very tired. Furthermore if this person
walks on a horizontal surface during this time, he gets tired even more. Physical work
has not been done in either case.
In high school the following definition was learnt. “The work equals the product of
force and the projected displacement”. This is obviously true, but only for homogeneous
force field and straight finite displacement. In equation: ∆W = F∆r. Here we used the
mathematical concept of dot product, which results in a scalar number, and the product
of the two absolute values is multiplied with the cosine of the angle.
In general case when the related force field F(r) is not homogeneous and the displacement is not straight, the above finite concept is not applicable. We have to introduce the
infinitesimal contribution of work (dW =F(r)dr). The amount of work made between
two positions is the sum or in other words integral of dW contributions. The physical
unit is Newton meter (Nm) which is called Joule (J).
Zr2
W =
F(r)dr
kgm2
J = Nm =
s2
(2.21)
r1
Figure 2.5: The concept of work
There is a special case when the force is the function of one variable F (x) only, and its
direction is parallel with the x direction. The above definition simplifies to the following:
Zx2
kgm2
(2.22)
W = F (x)dx
J = Nm =
s2
x1
23
Figure 2.6: The work is the area under the curve
In this special case the work done between two positions is displayed by the area
under the F (x) curve.
2.7
Power
(Teljesı́tmény)
The power (P ) is associated with the time needed to carry out a certain amount of
work. In mathematics, this is the time derivative of the work done. The physical unit is
Joule per second which is called Watt (W ).
d
kgm2
dW
=
(Fdr)
W =
(2.23)
P =
dt
dt
s3
Provided the force does not depend directly on time, the above formula can be transformed:
P =
dW
d
=
(Fdr) = Fv
dt
dt
(2.24)
So the instantaneous power is the dot product of the force and the actual velocity vector.
24
2.8
Theorem of Work (Kinetic energy)
Munkatétel (Mozgási energia)
Kinetic energy is the kind of energy which is associated with the mechanical motion of
some object. In high school the following simplified argument was presented to calculate
it:
Figure 2.7: Velocity vs. time function by the effect of constant force
A particle with mass (m) is affected by constant force. Initially the particle is standstill. The acceleration is constant, thus the v(t) graph is a sloppy line through the origin.
After (t) time passed, the displacement (s) shows up as the area under the v(t) curve.
Its shape is a right angle triangle.
s=
vt
2
(2.25)
v
t
(2.26)
The acceleration is the slope of the v(t) line.
a=
Let us multiply the above equation with the mass of the particle:
mv
t
ma =
(2.27)
The left hand side equals the force affecting the particle.
F =
mv
t
25
(2.28)
We also know that the work done in this simple case is:
W = Fs
(2.29)
So let us substitute the related formulas. Time cancels out:
1
mv vt
= mv 2
t 2
2
W =
(2.30)
This is the work done on the particle which generated the kinetic energy.
The above argument is not general enough, due to the simplified conditions used.
The general argument is presented below:
Let us start with Newton’s second law:
F=m
d2 r
dt2
(2.31)
The work done in general is as follows:
Zr2
W =
Fdr
(2.32)
d2 r
dr
dt2
(2.33)
r1
Substitute first to the second formula:
Zr2
W =
m
r1
Switch the limits of the integration to the related time moments t1 and t2 .
Zt2
W =m
(
d2 r dr
)dt
dt2 dt
(2.34)
t1
Take a closer look at the formulas in the parenthesis. In here the product of the first
and the second derivative of some function are present.
The following rule is known in mathematics:
d 1 2
df (x)
f (x) = f (x)
(2.35)
dx 2
dx
Using this formula for the last expression of work:
Zt2 " 2 #
Zt2 " 2 #
Zt2
d 1 dr
1 dr
m
dt = m d
=
W =m
d v2
dt 2 dt
2 dt
2
t1
t1
26
t1
(2.36)
By integrating the variations of the v2 , the total variation will be the result:
m
2
Zt2
1
1
d v2 = mv22 − mv12
2
2
(2.37)
t1
Zr2
W =
1
1
Fdr = mv22 − mv12 = Ekin2 − Ekin1
2
2
(2.38)
r1
Thus: The work done on a particle equals the variation of the kinetic energy. This is the
theorem of work.
Note there is no any restriction to the kind of force. So the force is not required to
be conservative, which concept will be presented later in this chapter. This can be even
sliding friction, drag or whatever other type of force.
The kinetic energy is accordingly:
1
Ekin = mv2
2
2.9
(2.39)
Potential energy
(Helyzeti energia)
Potential energy is the kind of energy which is associated with the position of some
object in a force field. Force field is a vector-vector function in which the force vector
F depends on the position vector r. In terms of mathematics the force field F(r) is
described as follows:
F(r) = X(x, y, z)i + Y (x, y, z)j + X(x, y, z)k
(2.40)
where i, j, k are the unit vectors of the coordinate system.
Take a particle and move it slowly in the F(r) force field from position 1 to position
2 on two alternative paths.
Let us calculate the amount of work done on each path. The force exerted to the
particle by my hand is just opposite of the force field -F(r). If it was not the case,
the particle would accelerate. The moving is thought to happen quasi-statically without
acceleration.
Let us calculate my work for the two alternate paths:
r

r

Z2
Z2
W1 =  (−F)dr
W2 =  (−F)dr
(2.41)
r1
r1
path1
27
path2
Figure 2.8: Integration on two alternative paths
In general case W1 and W2 are not equal. However, in some special cases they may
be equal for any two paths. Imagine that our force field is such, that W1 and W2 are
equal. In this case a closed loop path can be made which starts with path 1 and returns
to the starting point on path 2. Since the opposite direction passage turns W2 to its
negative, ultimately the closed loop path will result in zero value. That special force
field where the integral is zero for any closed loop is considered CONSERVATIVE force
field. In formula:
I
F(r)dr = 0
(2.42)
At conservative force field, one has to choose a reference point. All other destination
points can be characterized with the amount of the work done against the force field to
reach the destination point. This work is considered the potential energy (Epot ) of the
point relative to the reference point:
Zr
Epot (r) =
(−F(r, ))dr, = −
Zr
F(r, )dr,
(2.43)
ref
ref
The reference point can be chosen arbitrarily, however it is worth considering the practical
aspects of the problem.
Due to the fact that the reference point is arbitrary, the value of the potential energy
is also indefinite since direct physical meaning can only be associated to the variation of
the potential energy. In other words, the individual potential energy values of any two
28
points can be altered by changing the reference point, but the difference of the potential
energy values does not change.
Now the work done against the forces of the field between r1 and r2 points can be
expressed:
Zr2
Zr2
Zref
(−F(r))dr +
(−F(r))dr =
r1
r1
Zr1
(−F(r))dr = −
ref
Zr2
(−F(r))dr +
ref
(−F(r))dr
ref
(2.44)
The last two integrals are the potential energies of r1 and r2 points respectively.
Zr2
(−F(r))dr = Epot (r2 ) − Epot (r1 )
(2.45)
r1
2.10
Conservation of the mechanical energy
(Mechanikai energia megmaradása)
Mechanical energy consists of kinetic and potential energy by definition. Earlier in
this chapter the theorem of work was stated. Work done on a particle equals the variation
of its kinetic energy. In addition F(r) could be any kind of force.
Zr2
F(r)dr = Ekin2 − Ekin1
(2.46)
r1
Later the potential energy has been treated.
Zr2
(−F(r))dr = Epot (r2 ) − Epot (r1 )
(2.47)
r1
Let us switch the sign of the above equation:
Zr2
F(r)dr = Epot (r1 ) − Epot (r2 )
(2.48)
r1
At potential energy however conservative force field is required. This means that the so
called dissipative interactions are excluded, such as the sliding friction and the drag. Let
us make the right hand sides of the relevant equations equal.
Epot (r1 ) − Epot (r2 ) = Ekin2 − Ekin1
29
(2.49)
Ordering the equation:
Epot (r1 ) + Ekin1 = Epot (r2 ) + Ekin2
(2.50)
Using the conservation of mechanical energy requires conservative force, because this is
the more stringent condition.
Ultimately let us declare again clearly the conservation of mechanical energy: In
conservative system the sum of the kinetic and potential energy is constant in time.
Accordingly, these two types of energy transform to each other during the motion, but
the overall value is unchanged. In contrast to this when dissipative interaction emerges
in the system, the total mechanical energy gradually decreases by heat loss.
In this chapter the concept of work end energy have been used extensively. To improve
clarity, the following statement needs to be declared: Work is associated to some kind
of process or action. Energy on the other hand is associated to some kind of state of a
system, when not necessarily happens anything, but the capacitance to generate action
is present.
2.11
Energy relations at harmonic oscillatory motion
The equation of motion of the harmonic oscillation is as follows:
F (t) = −Dx(t)
(2.51)
Here Dis direction coefficient of the spring on which a particle with mass m oscillates.
In the chapter of kinematics the harmonic oscillatory motion has been introduced,
and the basic formulae have all been derived. The following relation was recovered:
a(t) = −ω 2 x(t)
(2.52)
ma(t) = −mω 2 x(t)
(2.53)
Let us multiply it with mass:
The left hand side of the equation is the force affecting the particle.
F (t) = −mω 2 x(t)
By comparing the two expressions of the force one can conclude as follows:
q
D
D = mω 2
ω= m
30
(2.54)
The harmonic oscillatory motion is a conservative process. This means that the total
mechanical energy (the sum of kinetic and the potential energy) should be constant.
1
1
(2.55)
Etot = Dx2 + mv 2 = const
2
2
Let us verify the above statement with the concrete formulas of displacement and velocity:
1
1
Etot = DA2 cos2 (ωt + φ) + mω 2 A2 sin2 (ωt + φ) = const
(2.56)
2
2
Now we can proceed on two alternate tracks by substituting the direction coefficient into
the equation and using the most basic trigonometric relation:
1
1
1
Etot = DA2 cos2 (ωt + φ) + DA2 sin2 (ωt + φ) = DA2
(2.57)
2
2
2
Or alternatively:
1
1
1
(2.58)
Etot = mω 2 A2 cos2 (ωt + φ) + mω 2 A2 sin2 (ωt + φ) = mω 2 A2
2
2
2
By using the velocity amplitude (vmax ) defined in the chapter of kinematics one can
conclude as follows:
1 2
Etot = mvmax
(2.59)
2
Ultimately we found two alternate formulae for the total mechanical energy. These
formulae prove that the process is truly conservative, and the total energy may show up
either as potential or kinetic energy. In amplitude position the total energy is stored
in the spring as potential (elastic) energy, at zero excursion position the total energy is
kinetic energy.
Figure 2.9: Energy relations of the oscillatory motion
In the figure above the energy relations are displayed. The motion takes place under
the solid horizontal line of total energy.
31
2.12
Angular momentum
(Impulzus nyomaték, perdület)
By definition the angular momentum of the particle is the cross product of the position
vector and the linear momentum.
kgm2
(2.60)
L=r×p
s
2.13
Torque
(Forgató nyomaték)
By definition the torque (M) is the cross product of the position vector and the force
affecting the particle.
kgm2
M=r×F
(2.61)
s2
Let us consider the situation when r and p and F are in the plane of the sheet. According
to the definition, both the angular momentum and the torque are normal to the sheet.
Figure 2.10: Both the angular momentum and the torque point into the paper
If the vectors depend on time, one can determine the derivative of the product:
dL
dr
dp
=
×p+r×
dt
dt
dt
Since
dr
dt
= v and
dp
dt
(2.62)
= F the above equation can be transformed:
dL
=v×p+r×F
dt
(2.63)
The first term on the right cancels out because v and p vectors are parallel. Therefore:
dL
=r×F
dt
32
(2.64)
The product on the right hand side is the torque. Ultimately one can conclude:
dL
=M
dt
(2.65)
In words: The time derivative of the angular momentum of some particle equals the
torque affecting this particle. (Obviously the reference point of both L and M must be
the same.)
This formula is analogous to that of Newton’s second law, expressed with the linear
momentum. By means of the fundamental theorem of calculus, this formula can be
integrated.
Zt2
L2 − L1 =
M(t)dt
(2.66)
t1
In words:
The variation of the angular momentums is the time integral of the torque affecting
the particle. This integral is called the angular impulse. (Nyomaték lökés)
2.14
Central force field
(Centrális erőtér)
If the force is collinear with the position vector and the magnitude depends on the
distance alone, then the force field is considered central force field:
F = k(r)r
(2.67)
here k is a scalar number which may depend only on the distance from the center.
As it has already been calculated:
dL
=r×F
dt
(2.68)
Let us substitute the central force field:
dL
= r × k(r)r
dt
(2.69)
The cross product is zero because of the collinear arrangement:
dL
=0
dt
(2.70)
Accordingly, in central force field the angular momentum is constant in time: (L= const)
It has the important conclusion. Planets, moons or spacecrafts which orbit their central
33
body in the space also move in the central force field of gravity. Therefore the angular
momentum referred to the central body is constant.
In the chapter of kinematics the concept of areal velocity was introduced in general.
Accordingly:
dA
1
= r×v
dt
2
(2.71)
On the other hand, the angular velocity is:
L = r × mv
(2.72)
By combining these two last equations:
dA
1
=
L = const
dt
2m
(2.73)
Ultimately the areal velocity is constant in the central force field.
Planetary motion: A meteorite is orbiting the sun on an ellipse trajectory. The
ellipse trajectory is the consequence of the Newton’s gravitational law. The constant
areal velocity will make the meteorite travel faster when close to the sun and slower
when it is far away. The crosshatched areas in the figure below are equal. So, the motion
is far not uniform.
Figure 2.11: Planetary motion with constant areal velocity
34
35
Chapter 3
Dynamics of system of particles György Hárs
(Tömegpont rendszer dinamikája)
3.1
Momentum in system of particles
The subject of analysis will be the system of particles. The system of particles in practice
may consist of several particles (mass points). Each of the particles may travel arbitrarily
in 3D space. The particles may exert force to each other (internal force) and may be
affected by forces originating in the environment (external force).
In mathematical calculations however it is worth reducing the number of particles
to two particles. This way, calculations become much easier without loosing generality.
The physical meaning behind the equations becomes even more apparent. At the end of
the argument the result will be stated in full generality for any number of particles.
Figure 3.1: System of particles
36
The center of mass is the weighted average of the position vectors.
m1 r1 + m2 r2
= rc
m1 + m2
(3.1)
Its time derivative is the velocity of the center of mass.
m1 v1 + m2 v2
= vc
m1 + m2
ptot =
X
pi = vc
i
X
mi
(3.2)
i
The numerator is the total momentum of the system of particles. So the total momentum
can be expressed as the product of the velocity of the center of mass multiplied by the
total mass.
Let us make one more time derivation:
m1 a1 + m2 a2
= ac
m1 + m2
(3.3)
Accordingly:
m1 a1 + m2 a2 = ac
X
mi
(3.4)
i
Now consider the Newton equation for m1 and m2 :
m1 a1 = F1 + F12
m2 a2 = F2 + F21
(3.5)
Internal forces show up with double subscript. By substituting the forces to the above
equation:
X
(F1 + F12 ) + (F2 + F21 ) = ac
mi
(3.6)
i
Here we have to take into account the fact, that the internal forces show up in pairs and
they are opposite of each other. F12 =-F21 . So they cancel out and only the external
forces remain.
X
X
Fiext = ac
mi
(3.7)
i
i
In words: The sum of the external forces accelerates the center of mass. Internal forces
do not affect the acceleration of the center of mass. This is the theorem of momentum.
If on the other hand the sum of the external forces is zero, the acceleration of the
center of mass becomes also zero, or in other words, the velocity of the center of mass is
constant. If the velocity of the center of mass is constant, then the total momentum of
the system of particles will also be constant.
37
So all together, let us state the conservation of momentum: In an isolated mechanical
system (in here the sum of the external forces is zero) the total momentum of the system
of particles is constant.
ptot = Const
(3.8)
This law can also be used in coordinate components. So if the system of particles is
mounted on a little rail cart, and external force parallel with the rail does not affect the
system, then that component of the total momentum will be constant which is parallel
with the rail. In terms of other directions no any law applies.
3.1.1
Collisions
(Ütközések)
At commonly happening collisions the conservation of momentum is valid because
the system of the two colliding particles represents an isolated mechanical system. There
are two specific types of collisions, the inelastic and elastic collision. The distinction is
based on the kinetic energy variation during the process.
Inelastic collisions:
(Rugalmatlan ütközés)
The two colliding particles get stuck together. The kinetic energy of the system
is partly dissipated. Substantial amount of heat can be generated. Let us write the
conservation of momentum:
m1 v1 + m2 v2 = (m1 + m2 )u
(3.9)
The velocity after collision (u) results:
m1 v1 + m2 v2
=u
m1 + m2
(3.10)
The “lost” mechanical energy, which has been dissipated to heat, is the difference of the
total kinetic energy before and after the collision:
2
1
1
1
m1 v1 + m2 v2
2
2
(3.11)
∆Eloss = mv1 + mv2 − (m1 + m2 )
2
2
2
m1 + m2
Elastic collision:
(Rugalmas ütközés)
38
Word “elastic” means that the mechanical energy is conserved. So in here both the
momentum and the kinetic energy are conserved. After collision the particles get separated with different velocities. The velocities before and after the collision are denoted
with v1 v2 and u1 u2 respectively. The conservation of momentum follows:
m1 v1 + m2 v2 = m1 u1 + m2 u2
(3.12)
The conservation of mechanical energy is also valid. Here the total mechanical energy is
kinetic energy since no potential energy is involved.
1
1
1
1
m1 v12 + m1 v22 = m1 u21 + m1 u22
2
2
2
2
(3.13)
Let us group terms with subscript 1 to the left and terms with subscript 2 to the right
hand side for two equations above.
m1 v1 − m1 u1 = m2 u2 − m2 v2
(3.14)
1
1
1
1
m1 v12 − m1 u21 = m2 u22 − m2 v22
2
2
2
2
(3.15)
Now factor out m1 and m2 from the equations, multiply the kinetic energy equation with
two and use the equivalency for the difference of squares:
m1 (v1 − u1 ) = m2 (u2 − v2 )
(3.16)
m1 (v1 − u1 )(v1 + u1 ) = m2 (u2 − v2 )(u2 + v2 )
(3.17)
Up to this point of discussion the 3D vector equations above are fully valid. Among dot
products, division operation is impossible. This is due to the fact that reverse direction
of the operation is ambiguous.
From this point, the mathematical argument is confined to the central collision only.
At central collision the velocities before the collision are parallel with the line between the
centers of the particle. This way the collision process takes place in a single line, and the
velocities before and after the collision will all be 1D vectors in the line of the collision.
The 1D vectors are practically plus, minus or zero numbers, and the dot product between
these vectors is basically product between real numbers. So from the above equations
the vector notation will be omitted. Accordingly any division can readily be carried out.
m1 (v1 − u1 ) = m2 (u2 − v2 )
(3.18)
m1 (v1 − u1 )(v1 + u1 ) = m2 (u2 − v2 )(u2 + v2 )
(3.19)
39
Let us divide the last equation with the former one:
v1 + u1 = v2 + u2
(3.20)
Now group the v terms to the left and u terms to the right hand side:
v1 − v2 = u2 − u1
(3.21)
m2 v1 − m2 v2 = m2 u2 − m2 u1
(3.22)
Multiply the equation with m2 :
The original equation for momentum conservation is simplified for 1D central collision:
m1 v1 + m2 v2 = m1 u1 + m2 u2
(3.23)
Let us subtract the former equitation from the last one. Here m2 u2 term will cancel out:
v1 (m1 − m2 ) + 2m2 v2 = u1 (m1 + m2 )
(3.24)
Thus u1 can be expressed:
v1
m1 − m2
2m2
+
v2 = u1
m1 + m2 m1 + m2
(3.25)
Due to symmetry, formula for u2 can be easily derived by switching the subscripts 1 and
2.
v2
2m1
m2 − m1
+
v1 = u2
m1 + m2 m1 + m2
(3.26)
The above formulas are not simple enough to provide plausible results. For this purpose
some special cases will be treated separately:
Discussion 1: What if m1 = m2 = m is the case.
Basically the masses are equal. Then the final result simplifies to:
v2 = u1 and v1 = u2
Accordingly the particles swap their velocities. If on the other hand one of the
particle had zero velocity originally (v1 =0), and the other particle slammed into it with
v2 velocity. Then :
v2 = u1 and u2 = 0
This means that the standing particle will start travelling with the velocity of the
moving particle, and the originally moving particle will stop.
40
Discussion 2: What if m2 is far larger than m1 .
−v1 + v2 = u1 and v2 = u2
This is the case when a ball bounces back from the face of the incoming bus. The
velocity of bus does not change (v2 = u2 ), and the velocity of ball is reflected plus the
speed of the buss is added.
Discussion 3: Billiard ball collision:
In this game, the balls are equal in mass, but the collisions are not necessarily central.
Consider the situation when one ball is standing and an equal weight ball collides to it
in a skew elastic collision. Let us go back to the original equation with vectors.
The momentum conservation for the present case:
mv = mu1 + mu2
(3.27)
The mechanical energy conservation for the present case:
1
1
1
mv12 = mu21 + mu22
2
2
2
(3.28)
After some obvious mathematical simplifications:
v = u1 + u2
v12 = u21 + u22
(3.29)
Figure 3.2: Billiard ball collision
First equation means that the vectors create a closed triangle. The second equation
shows that the created triangle is a right angle triangle, since the Pythagoras theorem is
true only then. As a summary, one can say that the balls travel 90 degree angle relative
to each after collision in billiard game.
Ballistic pendulum
(Ballisztikus inga)
This is a pendulum with some heavy sand bag on the end of some meter long rope.
The rope is hung on a high fix point, letting the pendulum swing. A simple indicator
mechanism shows the highest angular excursion.
41
Figure 3.3: Ballistic pendulum
The pendulum is left to get quiet and hang vertically. Then the gun is fired, the
bullet penetrates into the sandbag and get stuck in it. The pendulum starts to swing.
The first highest angular excursion is detected. From the above information, the speed
of the bullet can be found.
The whole process consists of two steps. In step 1 the bullet collides with the sandbag.
Up to this point, conservation momentum is valid but the mechanical energy is not
conserving quantity, due to the inelastic collision. After collision in step 2, there are no
more dissipative effects, so conservation of mechanical energy is true. The two relevant
equations are as follows:
mv = (m + M )u
(3.30)
1
(m + M )u2 = (m + M )gl(1 − cos φ)
2
(3.31)
Here m and M are the mass of the bullet and the sandbag respectively. The v and u are
the speed of the bullet and the speed of the sandbag respectively. The l is the length of
the rope and g is the gravity acceleration.
By eliminating u from the equations one can readily express the incoming speed of
the bullet:
m
v=u
(3.32)
m+M
m
1
(
v)2 = gl(1 − cos φ)
2 m+M
(3.33)
m+Mp
2gl(1 − cos φ)
m
(3.34)
v=
This is an excellent example how careful one must be. If wrongly the whole process is
assumed to be conservative, the resulting bullet speed will be some ten meters per second
which is roughly hundred times smaller than the real result.
42
3.1.2
Missile motion
(Rakéta mozgás)
Jet propulsion is the fundamental basis of the missile motion. This is based on the
conservation of momentum. If one tries to hold the garden hose when sprinkling the
garden, one will experience a recoil type force, which is pushing back. This force is
called “thrust”, and this drives the missiles, aircrafts and jet-skis.
Figure 3.4: Missile motion
The missile ejects mass in continuous flow with the ejection speed (u) relative to the
missile. The rate with which the mass is ejected is denoted (µ) and measured in kg/s.
The infinitesimal ejected momentum will provide the impulse to the missile:
udm = F dt
(3.35)
dm
=F
dt
(3.36)
uµ = F
(3.37)
u
So the product of the speed and ejection rate determines the thrust (F ). During the
missile motion the thrust is a constant force. As the missile progresses the overall mass
is continuously reduced by burning the fuel. The equation of motion is as follows:
F = m(t)a
(3.38)
Here m(t) is the reducing mass and m0 is the initial mass:
m(t) = (m0 − µt)
(3.39)
uµ = (m0 − µt)a
(3.40)
uµ
= a(t)
m0 − µt
(3.41)
The acceleration can be expressed:
43
In order to find out the velocity time function, the above formula needs to be integrated:
Zt
v(t) =
a(t, )dt, =
0
Zt
uµ
dt,
m0 − µt,
(3.42)
0
After the integration the velocity function is revealed:
v(t) = u ln
m0
m0 − µt
(3.43)
The final formula shows that approaching the m0 /µ time the speed grows to the infinity.
This value can not be reached since there must be a payload on the missile.
3.2
Angular momentum in system of particles
(Tömegpont rendszer impulzus nyomatéka)
Figure 3.5: Angular momentum in system of particles
Consider the total angular momentum of a system of particles.
L1 = r2 × m1 v1
L2 = r2 × m1 v2
(3.44)
This is the sum of the angular momentums of the individual particles:
L1 + L2 = Ltot
(3.45)
Check out the time derivative oft this equation:
dL1 dL2
dLtot
+
=
dt
dt
dt
44
(3.46)
In Chapter 2 it has been shown that the derivative of angular momentum is the torque
affecting the particle. Accordingly the above equation is transformed:
M1 + M2 =
dLtot
dt
(3.47)
Based on the definition of torque following formulas are true:
M1 = r1 × F1 + r1 × F12
M2 = r2 × F2 + r2 × F21
(3.48)
Let us substitute them to the equation above:
r1 × F1 + r1 × F12 + r2 × F2 + r2 × F21 =
Here we have to use that F12
principle.
dLtot
dt
(3.49)
= -F21 which is the consequence of action reaction
r1 × F1 + r1 × F12 + r2 × F2 − r2 × F12 =
dLtot
dt
(3.50)
Now regroup the left hand side:
r1 × F1 + r2 × F2 + (r1 − r2 ) × F12 =
dLtot
dt
(3.51)
If now one looks at the figure above, the fact is readily apparent that the r1 -r2 vector
and F12 force vector are collinear vectors, thus their cross product is zero. Therefore the
torques of the internal forces cancels out.
r1 × F1 + r2 × F2 =
dLtot
dt
(3.52)
The left hand side terms are all the torques of the external forces. So all together the
generalized statement is as follows:
X
Miext =
i
dLtot
dt
(3.53)
In words:
In system of particles, the time derivative of the total angular momentum is the sum
of the external torques. This statement is called the theorem of angular momentum.
Therefore the internal torques are ineffective in terms of total angular momentum.
If on the other hand the total external torque is zero, then the total angular momentum is constant. This is the conservation of angular momentum.
Ltot = Const
45
(3.54)
In summary: In a system of particles where the total external torque is zero, the total
angular momentum is constant, or in other words it is a conserving quantity.
This law can also be used in coordinate components. So if the system of particles
is mounted on a bearing, and external torque parallel with the axis of the bearing does
not affect the system, then that component of the angular momentum will be constant
which is parallel with the axis of the bearing. In terms of other directions no any law
applies.
3.2.1
The skew rotator
(Ferdeszögű forgás)
Consider the figure below. Two equal masses are placed on the ends of a weightless
rod. The center of mass is mounted on a vertical axis, which is rotating freely in two
bearings. The angle of the fixture is intentionally not ninety degrees, but a skew acute
angle. The system rotates with a uniform angular velocity. The job is to find out the
deviational torque which emerges, due to the rotation of asymmetric structure.
Figure 3.6: The skew rotator
The origin of the coordinate system is the center of mass. The coordinate system is
not rotating together with the mechanical structure and it is considered inertial system.
Gravity cancels out from the discussion, since the center of mass is supported by the
axis, and the gravity does not affect torque to the system. The mechanical setup is in
the plane of the figure. The two position vectors of the particles are (r) and (–r).
Momentums of the particles are p1 and p2 . They are normal to the paper sheet.
p1 = m(ω × r) p2 = −m(ω × r)
(3.55)
The corresponding L1 and L2 angular momentums are equal, because of the twice negative multiplication:
L1 = r × p1 = r × m(ω × r) L2 = −r × p2 = −r × m(ω × (−r))
46
(3.56)
So the total angular momentum is the sum of these two:
Ltot = 2r × m(ω × r)
Now we use the theorem of angular momentum:
X
dLtot
Miext =
dt
i
X
Miext =
i
d
dr
dr
(2r × m(ω × r)) = 2 × m(ω × r) + 2r × m(ω × )
dt
dt
dt
(3.57)
(3.58)
(3.59)
Now take it into consideration that the derivative of the position is the velocity which
can be expressed by means of angular velocity vector:
dr
=v =ω×r
dt
After substitution:
X
Miext = 2(ω × r) × m(ω × r) + 2r × m(ω × (ω × r))
(3.60)
(3.61)
i
The first term on the right hand side is zero, because this is a cross product of collinear
vectors. The final result comes up immediately.
X
Miext = 2r × m(ω × (ω × r))
(3.62)
i
After following the numerous cross products in terms of direction one can conclude, that
the torque is pointing out of the paper sheet. The direction of torque is rotating together
with the mechanical structure and always perpendicular to its plane. This is deviational
torque and this emerges because the angular momentum vector is constantly changing,
not in absolute value but in direction. This effect is very detrimental to the bearings
due to the load that it generates. There are cases however when such effect does not
show up. When the angular momentum vector is parallel with angular velocity vector
no deviational torque will emerge. These are called principal axes. In general there are
three perpendicular directions of principal axes.
A new interesting aspect:
Imagine that this experiment is carried out on a spacecraft orbiting the Earth. Suddenly the bearing and the mechanical axis disappear. How will the mass-rod-mass structure move after this?
Since no external torque affects the system, the angular momentum will be constant.
But now the angular velocity vector starts to go around on the surface of a virtual cone.
The symmetry axis of such virtual cone is just the angular momentum vector. This kind
of motion is called precession.
47
3.2.2
The pirouette dancer (The symmetrical rotator)
(A piruett táncos)
In conjunction with the previous section this section could be called as “symmetrical
rotator”. The setup fundamentally similar, the major difference is that the mass-rodmass system is positioned perpendicularly to the rotation axis.
Figure 3.7: Symmetrical rotator
p1 = m(ω × r) p2 = −m(ω × r)
(3.63)
Similarly to the previous section total angular momentum is:
Ltot = 2r × m(ω × r)
(3.64)
In present case however the position vectors and the angular velocity vector are normal
to each other. Therefore the direction of the angular momentum and the angular velocity
will be both parallel with the rotation axis. Since the direction of the vectors is clear, it
is enough to deal with the absolute values only.
Ltot = 2mr2 ω
(3.65)
Notice: The emerging 2mr 2 quantity is the so called the moment of inertia. Find details
in Chapter 4.
Let us see what happens when the pirouette dancer pulls his arms in. Then r2 ≤ r1
is the case.
The conserving quantity is the angular momentum.
2mr12 ω1 = 2mr22 ω2
48
(3.66)
From here:
ω1
r1
r2
2
= ω2
(3.67)
Accordingly, the dancer is spinning much faster.
Let us check out how the kinetic energy changed during the pirouette. Clearly the
difference between the final state and the initial state should be calculated.
1
1
∆Ekin = 2 m(r2 ω2 )2 − 2 m(r1 ω1 )2 = m(r22 ω22 − r12 ω12 )
2
2
(3.68)
Now let us substitute ω2:
∆Ekin = m r22 ω12
∆Ekin = mr12 ω12
r1
r2
4
r1
r2
!
− r12 ω12
2
(3.69)
!
−1
(3.70)
The kinetic energy increased. This only could happen due to the work done by the
dancer. The force of the dancer is internal force, so the kinetic energy of the system of
particles can be changed by internal forces too. In order to calculate the work done, the
first task is to find out the force function. The force against which the dancer pulls his
arm is the centrifugal force. The centrifugal force is an inertial force which emerges in
spinning coordinate system only. See details in chapter 5.
Let us find out the angular velocity as a function of the arbitrary position:
r 2
1
ω1
= ω(r)
(3.71)
r
The formula of centrifugal force in present case is:
4
1
2
2 r1
= mω12 r14 3
Fcf (r) = mrω = mrω1
r
r
(3.72)
The work carried out by the dancer is the integral of the dancer’s force, which is just
opposite of the centrifugal force.
Zr2
Wdancer = 2
(−Fcf )dr =
r1
−2mω12 r14
Zr2
r1
49
dr
r3
(3.73)
Now let us study the integral alone. Use the fundamental theorem of calculus:
Zr2
r1
r=r2
dr
1 1
1
1 −2
=
−
= − r
r3
2
2 r12 r22
r=r1
(3.74)
After substitution:
Wdancer = mω12 r14
1
1
− 2
2
r2 r1
= mω12 r12
r1
r2
2
!
−1
(3.75)
This result completely matches the growth of the kinetic energy calculated earlier, so the
increase of the kinetic energy is the consequence of the work done by the dancer.
3.3
Discussion of the total kinetic energy in the system of particles
In contrast to the earlier habit the system of particles will be treated in general up to n
pieces of particles.
Particle positions are defined by ri position vectors relative to the origin of the coordinate system. The center of mass has the position vector rc . Now, we introduce the center
of mass as a new coordinate origin. In this new coordinate system the corresponding
position vectors will be denoted ρi .
Figure 3.8: Kinetic energy in the system of particles
Therefore the following equation is true:
ri = rc + ρi
(3.76)
After derivation similar rule is found among the velocities:
v i = v c + υi
50
(3.77)
Here vi and vc are the velocity of i-th particle and the center of mass respectively in the
original coordinate system. . The υi is the velocity of the i-th particle in the center of
mass coordinate system.
Take a look at the total kinetic energy of the system in the laboratory coordinate
system:
X
X
n n n X
X
1
1
1
2
2
2
2
2
mi vi =
mi (vc + υi ) =
mi (vc + υi + 2vc υi )
Ekin =
2
2
2
i=1
i=1
i=1
(3.78)
Let us discuss the individual terms separately:
X
X
n n n
X
X
1
1
2
2
Ekin =
mi vc +
mi υi +
(mi vc υi )
2
2
i=1
i=1
i=1
In the first term the velocity of center of mass can be factored out:
!
n n
X
X
1
1
1
mi vc2 =
mi vc2 = mtot vc2
2
2 i=1
2
i=1
(3.79)
(3.80)
The first energy term is associated with the velocity of the center of mass in the laboratory
coordinate system. The second term is associated with the velocities relative to the center
of mass.
The third term gives zero result. The proof is as follows: From the third term vc can
be factored out:
n
X
(mi vc υi ) = vc
i=1
n
X
(mi υi ) = 0
(3.81)
i=1
The right hand side summa is the total momentum in the center of mass coordinate
system. The total momentum is the product of the total mass multiplied with the
velocity of the center of mass. But in the center of mass coordinate system the velocity
of center of mass is obviously zero. QED.
Accordingly the total kinetic energy of the system of particles can be divided into two
parts: Kinetic energy associated to the velocity of the center of mass in the laboratory
coordinate system, and kinetic energy associated to the velocities in the center of mass
coordinate system. In formulas:
n X
X
1
1
2
2
m i υi
(3.82)
Ekin = mtot vc +
2
2
i=1
The first term can be changed external force only, due to the theorem of momentum.
The second term can be changed by any type of forces.
51
Chapter 4
Dynamics of rigid body - György
Hárs
(Merev testek dinamikája)
4.1
Moment of inertia
(Tehetetlenségi nyomaték)
Up to this point only the motion of particles was discussed. The rigid bodies however
are extended objects. The kinematics of extended objects contains major distinctions
in terms of motion. Translation means that the points of the object travel the same
trajectory except for a shift, by which all the trajectories cover each other. The rotation
however contains circular trajectories with different radii.
Rigid body is a special system of particles. Here the positions of the particles are
fixed relative to each other. In addition the geometrical shape of the body is constant,
and it is independent of the mechanical load.
In present discussion a majorly simplified theory will be treated. The simplifications
are as follows:
ˆ The origin of the coordinate system will be the center of mass of the rigid body.
ˆ The rotation will take place around principal axis, thus the angular momentum
and the angular velocity vectors are parallel.
Let us consider the mass-rod-mass structure in chapter 3. The angular momentum in
general case can be expressed as follows:
Ltot = 2r × m(ω × r)
52
(4.1)
In present case however the position vectors and the angular velocity vector are normal
to each other. Therefore the direction of the angular momentum and the angular velocity
will be parallel. If two vectors are parallel then a scalar multiplier can be found between
them. (In the general case tensor describes the relation.)
Ltot = 2mr2 ω = θpair ω
(4.2)
The scalar multiplier is called the “moment of inertia”. This is denoted with the Greek
letter θ. So for a symmetrical pair of particles the moment of inertia is: θpair = 2mr2 .
For one single piece of particle this value is: θparticle = mr2 .
Let us calculate the moment of inertia for a large diameter thin wheel with the radius
r.
Z
θwheel = r2 dm
(4.3)
Since the radius is constant this can be factored out.
Z
2
θwheel = r
dm = mr2
(4.4)
Here m means the total mass of the wheel.
Now consider an arbitrary rotationally symmetric object. Imagine as this object
consisted of several wheels, each of them with dm mass.
Figure 4.1: Moment of inertia of a rotationally symmetric object
Z
θgeneral =
Z
dθ =
V
r2 dm
(4.5)
V
If the density of the object is constant then dm = ρdV . With this:
Z
Z
2
θgeneral = r dm = ρ r2 dV
V
V
53
(4.6)
By means of the above formula the moment of inertia for several symmetrical objects
can be calculated:
ˆ Tube: Here r = R =Const.
Figure 4.2: Tube
Since R is constant this can be factored out.
Z
Z
2
2
θtube = ρ R dV = ρR
dV = R2 ρV = mR2
V
(4.7)
V
ˆ Cylinder or disc:
Let us put together the cylinder from several tubes with increasing radii. Here dV =
2rπldris the case, where l is the length of the cylinder.
Figure 4.3: Cylinder
ZR
2
ZR
r 2rπldr = ρ2πl
θcylinder = ρ
0
0
54
r4
r dr = ρ2πl
4
3
r=R
=
r=0
π
ρlR4
2
(4.8)
On the other hand the mass of the cylinder is the product of its volume and density.
m = R2 πlρ
(4.9)
θcylinder
π ρlR4
R2
=
=
m
2 R2 πlρ
2
(4.10)
Divide them:
From here the moment of inertia is found:
1
θcylinder = mR2
2
(4.11)
ˆ Spherical shell. (Hollow sphere)
The radius is denoted R, the thickness of the shell is S.
Figure 4.4: Spherical shell
According to the general expression:
Z
r2 dV
(4.12)
(R cos φ)2 dV
(4.13)
θgeneral = ρ
V
The function to be integrated is: r = (R cos φ)2
φ= π2
Z
θshell = ρ
φ=− π2
55
The dV volume here is as follows: dV = (Rdφ)S(2πR cos φ)
φ= π2
φ= π2
Z
2
θshell = ρ
Z
4
(R cos φ) (Rdφ)S(2πR cos φ) =ρR S2π
φ=− π2
cos3 φdφ
(4.14)
φ=− π2
Let us deal with the integral for a while separately. The antiderivative is as follows:
π
Z− 2
− π2
φ= π2
4
1
=
cos φdφ =
(9 sin φ + sin 3φ)
12
3
φ=− π
3
(4.15)
2
By means of the integral value the moment of inertia can be expressed:
4
θshell = ρR4 S2π
3
(4.16)
On the other hand the total mass is the product of the density and the volume.
m = ρ4R2 πS
(4.17)
θshell
4 ρR4 S2π
2
=
= R2
2
m
3 ρ4R πS
3
(4.18)
2
θshell = mR2
3
(4.19)
Divide them.
The result is clear.
ˆ The bulky sphere
This can be composed of spherical shells with increasing radii. The infinitesimal volume
is the product of the surface of the inflating sphere and the infinitesimal thickness (dr):
dV = 4r2 πdr
ZR
θsphere = ρ
0
2 2
r dV = ρ
3
ZR
2 2 2
8π
r (4r πdr) = ρ
3
3
0
ZR
0
r=R
8π r5
8π
r dr = ρ
= ρ R5
3 5 r=0
15
4
(4.20)
On the other hand the mass of the sphere is the product of its volume and the density:
4R3 π
ρ
m=
3
56
(4.21)
Divide them:
θsphere
3
8π
2
= ρ R5 3
= R2
m
15 4R πρ
5
(4.22)
2
θsphere = mR2
5
(4.23)
Thus the result is:
ˆ Stick or rod.
The stick is put together from several particles with increasing radii. Early in this chapter
it was shown that in case of particles the moment of inertia is as follows: θparticle = mr2 .
In addition the infinitesimal volume is equal todV = Adr.
Figure 4.5: Moment of inertia for a rod
Z
2
Zl
r dV = ρ
θrod = ρ
V
2
Zl
r Adr = ρA
0
r3
r dr = ρA
3
2
0
r=l
r=0
3
l
1
= ρA
= ρAl3 (4.24)
3
3
The mass of the rod can also be expressed:
m = Alρ
(4.25)
1
θrod = ml2
3
(4.26)
Accordingly:
Now, we should find out how much the moment of inertia is through the center of mass.
A symmetrically rotating rod can be composed of two half size rods. Therefore half
length and half mass rod will be used and finally doubled:
1
1m l 2
θCenter = 2
( ) = ml2
(4.27)
32 2
12
57
ˆ Steiner’s theorem
If one calculates the difference between the moment of inertia of the rod with different
axes, an interesting relation can be revealed:
1
1
1
l
θrod − θcenter = ( − )ml2 = ml2 = m( )2
3 12
4
2
(4.28)
The difference is the total mass multiplied with the square of the distance between the
axes. The calculated result is a special case of a far more general law which is called the
Steiner’s theorem. This general law states that moment of inertia is smallest through
the axis of center of mass. In addition the moment of inertia for any parallel axes can
be calculated as follows:
θparallel = θcenter + mD2
(4.29)
where m and D are the mass and the distance between the axes respectively.
4.2
Equation of motion of the rigid body:
(Merev test mozgásegyenlete)
Let us take a look at the equation of motion of an extended rigid body. Two laws
will be used, such as the theorem of momentum
X
X
Fiext = ac
mi
(4.30)
i
i
and the theorem of angular momentum.
X
Miext =
i
dLtot
dt
(4.31)
In case of rigid body the theorem of momentum is written with the relevant quantities:
X
Fi = mtot ac
(4.32)
i
The theorem of angular momentum is written by means of the moment of inertia. Subscript c means that the moment of inertia is calculated to a principal axis through the
center of mass.
X
i
Mi =
d(θc ω)
d(ω)
= θc
= θc β
dt
dt
58
(4.33)
So all together the equation of motion consists of two vector equations:
X
Fi = mtot ac
(4.34)
i
X
Mi = θc β
(4.35)
i
They represent in principle six scalar equations, but in the most cases the system of equations contains only three equations. This is due to the fact that mechanical problems are
mostly 2D plane problems. This case the theorem of momentum contains two variables
x and y components in the plane of the sheet, and the theorem of angular momentum
contains z component normal to the sheet.
The time derivative of angular velocity is the angular acceleration β. In present
simplified situation the angular acceleration is always parallel with the angular velocity.
4.2.1
Demonstration example 1.
The yoyo Find the acceleration and the tension of the rope (rope force). The mass is
3 kg, g= 10m/s 2 the radii are R = 6 cm and r = 3cm
Figure 4.6: The yoyo
The equations of motion:
mg − K = ma
(4.36)
1
a
Kr = ( mR2 )( )
2
r
(4.37)
In the second equation the first factor is the moment of inertia for cylinder,
the second
1
R 2
factor is the angular acceleration. From the second equation K = 2 ma r comes out.
59
This and the first equation are added so K cancels out. The mathematical steps are as
follows:
"
2 #
1 R
ma
(4.38)
mg = 1 +
2 r
1
a=
1+
1
a=
1+
1
2
1
2
g
R 2
r
1
10 m
g= g=
2
6
3
3 s2
(4.39)
(4.40)
3
The rope force results as the acceleration is substituted to the expression of K.
2
2
R
R
mg
3 · 10
1
1
1
= m
K = ma
2 g =
2 =
2 N = 20N
R
r
1
2
r
2
r
1+ 2 r
1+2 R
1 + 2 63
(4.41)
4.2.2
Demonstration example 2.
Cylinder on the table Find the accelerations and the tension of the rope. Determine
the smallest possible static friction coefficient, which provides sliding free rolling for the
cylinder. m = 3 kg, M = 8 kg.
Figure 4.7: Cylinder on the table
Equation of motion for the block:
I. mg − K = ma
Equations of motion for the cylinder:
60
II. M g − N = 0
III. K + S = M a2
a
IV. KR − SR = ( 12 M R2 )( 2R
)
In addition there is a condition for the static friction force:
V. S ≤ µst N
IV. K − S = 41 M a
IV.+III. 2K = 34 M a
IV.+III. K = 38 M a
I.+IV.+III. mg = 38 M a + ma
I.+IV.+III. mg = ( 38 M + m)a
I.+IV.+III. a =
a=
m
3
M +m
8
g=
8m
g
3M +8m
8m
8·3
1
m
g=
g= g=5 2
3M + 8m
3·8+8·3
2
s
(4.42)
Let us find the tension of the rope:
IV.+III. K = 38 M a = 83 M 3M8m
g=
+8m
3mM
g
3M +8m
=
3·3·8
10
3·8+8·3
= 15N
Now determine the friction force:
-IV. −K + S = − 14 M a
IV.+III. K = 38 M a
(IV.+III).-IV. S = 83 M a − 41 M a = 18 M a =
The static friction coefficient is the last to deal with:
V. S ≤ µst N
II. M g − N = 0
61
mM
g
3M +8m
= 5N
II. M g = N
After substitution:
V. S ≤ µst M g
V.
V.
mM
g
3M +8m
m
3M +8m
=
≤ µst M g
3
3·8+8·3
=
1
16
=≤ µst
The summary of the solution is the following:
The acceleration of the block is a = 5 m/s 2 , The acceleration of the cylinder is a/2
= 2,5 m/s 2 . The tension of the rope is K = 15 N, the static friction force is S = 5 N.
The minimum required static friction coefficient is µst = 1/16 = 0.0625.
4.3
Kinetic energy of the rigid body
(Merev test mozgási energiája)
The motion of the rigid body represents kinetic energy. In chapter 3 the kinetic
energy of system of particles was discussed. Let us extend the validity of this discussion
to the rigid bodies.
n X
X
1
1
2
2
m i υi
(4.43)
Ekin = mtot vc +
2
2
i=1
The first term on the right hand side represents the kinetic energy of the rigid body
due to the speed of the center of mass. The second term is associated with the motion
relative to the center of mass. This is typically the rotation in case of rigid bodies,
therefore υi = ω × ri . Since they are normal to each other the absolute value of the
velocity is υi = ωri . Let us substitute to the formula of the kinetic energy.
n n
X
X
1
1 2X
1
1
2
2 2
2
mi ω ri = mtot vc + ω
mi ri2
(4.44)
Ekin = mtot vc +
2
2
2
2 i=1
i=1
The last summa term on the right hand side is the moment of inertia of the rigid body.
The total kinetic energy of a rigid body can be written as follows:
X
1
1
(4.45)
Ekin = mtot vc2 + θc ω 2
2
2
So the total kinetic energy consists of two terms. One of them is in conjunction with
the translation of the center of mass, the other one is related to the rotation around the
center of mass.
sectionCorrespondence between translation and rotation in the framework of the simplified model
62
Translation
Linear momentum
Force
Kinetic energy
Power
The corresponding
Linear momentum
Velocity
Force
Mass
Acceleration
p = mv
F = dp
= ma
dt
1
Ekin = 2 mv2
P = Fv
quantities
kg.m/s p
m/sv
kg.m/s 2 = N F
kg m
m/s2 a
63
Rotation
Angular momentum
Torque
Kinetic energy
Power
L = θC ω
M = dL
= θC β
dt
1
Ekin = 2 θC ω 2
P = Mω
L kg.m 2 /s
ω 1/s
M kg.m 2 /s2 =Nm
θC kg.m 2
β 1/s 2
Angular momentum
Angular velocity
Torque
Moment of inertia
Angular acceleration
Chapter 5
Non-inertial (accelerating) reference
frames - György Hárs
(Gyorsuló vonatkoztatási rendszerek)
So far dynamics has been treated in inertial systems. The inertial systems are equivalent, which means that the equations of physical laws show up in the same shape in
all inertial systems. There are practical aspects however which make the application
of the non-inertial reference frames necessary. In here the shape of equation of motion
changes relative to that in inertial system. New terms, the inertial forces will appear in
the equation of motion in addition to the real forces. The real forces are associated with
real interactions with other objects.
5.1
Coordinate system with translational acceleration
Figure 5.1: Coordinate system with translational acceleration
64
There are two coordinate systems. One of them is an inertial system, the other one is
a coordinate system with translational acceleration. The position vector of the origin of
the accelerating system is r0 . The position, the velocity and the acceleration in the two
systems are generated by consecutive derivations. The vectors in the accelerating system
are denoted with the subscript “rel” standing for relative.
r = r0 + rrel
(5.1)
v = v0 + vrel
(5.2)
a = a0 + arel
(5.3)
Let us multiply the last equation with the mass m.
ma = ma0 + marel
(5.4)
The left hand side of the equation is the sum of the real forces. Regroup the terms.
Fsum + (−ma0 ) = marel
(5.5)
The second term is called inertial force.
Finertial = −ma0
(5.6)
So altogether the sum of the real forces should be completed with the inertial force. This
way the equation of motion can formally be handled just like in the inertial system.
Example: A simple pendulum is hanging in a uniformly accelerating train. Find the
angle of the rope in stationary state relative to the vertical direction. The rope force is
denoted K the angle is ϕ.
Inertial system approach: In practice the observer is standing on the ground next to
the train.
↑ K cos φ − mg = 0
(5.7)
→ K sin φ = ma
(5.8)
65
Figure 5.2: Inertial system approach
The angle can be determined accordingly: tgφ = ag
Accelerating system approach: In practice the observer is on the train.
Figure 5.3: Accelerating system approach
66
↑ K cos φ − mg = 0
(5.9)
→ K sin φ + (−ma) = 0
(5.10)
The result is obviously the same as earlier. The point is that by choosing accelerating
system, the equation of motion becomes equilibrium equation, which easier to handle in
most cases.
5.2
Coordinate system in uniform rotation
All planets rotate thus this phenomenon is very frequent in nature. Let us have two
coordinate systems with common origin, the inertial system (x, y, z with the unit vectors
i, j, k) and the uniformly rotating system (p, q, s with the unit vectors e1 , e2 , e3 ). The
i, j, k vectors are constant in time since they belong to an inertial system, but the e1 ,
e2 , e3 vectors rotate together with the spinning system.
Figure 5.4: Coordinate system in uniform rotation
Consider an r vector in both coordinate systems:
r = xi + yj + zk = pe1 + qe2 + se3
(5.11)
Let us derivate it:
dr
dx
dy
dz
dp
de1
dq
de2
ds
de3
=
i + j + k = ( e1 + p
) + ( e2 + q
) + ( e3 + s
)
dt
dt
dt
dt
dt
dt
dt
dt
dt
dt
(5.12)
Regroup the right hand side:
dr
dp
dq
ds
de1
de2
de3
= ( e1 + e2 + e3 ) + (p
+q
+s
)
dt
dt
dt
dt
dt
dt
dt
67
(5.13)
The first three terms contain the derivative in the spinning system. Here it is worth
introducing a notation for the derivation in the spinning system. This will be similar to
the usual derivation symbol, but instead of d a Greek letter δ is used.
dp
dq
ds
δr
= e1 + e2 + e3
δt
dt
dt
dt
(5.14)
In the kinematics chapter it has been shown that the velocity due to spinning can be
written as follows:
de1
= ω × e1
dt
de2
= ω × e2
dt
de3
= ω × e3
dt
(5.15)
So altogether:
dr
δr
=
+ pω × e1 + qω × e2 + sω × e3
dt
δt
(5.16)
Let us factor out the angular velocity:
dr
δr
δr
=
+ ω × (pe1 + qe2 + se3 ) =
+ω×r
dt
δt
δt
(5.17)
So the rule of transformation between the derivatives in the inertial and in the spinning
system is as follows:
δr
dr
=
+ω×r
dt
δt
(5.18)
This has a direct consequence to the velocities: v = vrel + ω × r
If the above transformation rule is applied to the velocity vector then the relation
between the accelerations is recovered:
δv
dv
=
+ω×v
dt
δt
a=
(5.19)
δ
δvrel
δrrel
(vrel + ω × r) + ω × (vrel + ω × r) =
+ω×
+ ω × vrel + ω × (ω × r)
δt
δt
δt
(5.20)
The following notations are used: arel = δvδtrel vrel =
By means of which the acceleration is expressed:
δrrel
δt
a = arel + 2ω × vrel + ω × (ω × r)
(5.21)
Regroup the above formula and express the relative acceleration.
arel = a + 2vrel × ω + ω × (r × ω)
68
(5.22)
Multiply the equation with mass.
marel = ma + 2mvrel × ω + mω × (r × ω)
(5.23)
This equation has an important message. The equation of motion in a spinning system is
very much similar to that in the inertial system. The major difference is that additional
terms appear which are called the inertial forces. These forces are not exerted by some
other object, but they are due to the fact that the coordinate system is spinning. So
when solving problem in a spinning system these inertial force should be added to the
real forces. Take a closer look at these forces.
marel = F + FCoriolis + FCentrifugal
(5.24)
The second term is called Coriolis force. This force only appears if the particle is moving
relative to the spinning system. If we are on a spinning contraption in the amusement
park Coriolis force can be felt if we move our hand perpendicular direction to the axis
of spinning.
FCoriolis = 2mvrel × ω
(5.25)
Coriolis force does not make any work because its direction is normal to the relative
velocity. This only deflects the moving objects.
The third term is called the centrifugal force. This force affects also the stationary
objects in the spinning system and shows up in a turning car. Its direction points away
from the axis of rotation.
FCentrifugal = mω × (r × ω)
(5.26)
Therefore the absolute value of the centrifugal force depends on the distance from the
rotational axis and proportional with the square of the angular velocity.
FCentrif ugal = mrω 2
(5.27)
In technology it is a major limiting factor at manufacturing turbines and fast spinning
motors.
5.2.1
Earth as a rotating coordinate system
It is well known that planet Earth rotates with one day period. Accordingly inertial
forces appear. In most cases these are ignored, but in some special cases these forces
make major qualitative changes in the physical processes.
69
The effects of the centrifugal force
Figure 5.5: The effects of the centrifugal force on planet Earth
The gravity force and the centrifugal force affect the object together. The resulting
free fall acceleration will increase as one goes closer to the poles. At higher latitude we
are closer to the rotational axis therefore the centrifugal force is smaller. The vertical is
defined by the resulting direction of both effects. If planet Earth ceased to rotate strange
effect would follow. The water of the oceans would move to the poles, and the ocean
floor close to the equator would be a huge desert.
Let us calculate the effect of the rotation on the equator numerically: The acceleration
associated with the centrifugal effect can be calculated as follows:
acentrif ugal = Rω 2 = 6, 37 · 106
4π 2
−2 m
2 = 3, 4 · 10
s2
(24 · 3600)
(5.28)
The radius of Earth is denoted R which is 6370 km. The duration of a normal day shows
up in the denominator in seconds. If the acceleration value is compared to the nominal
70
value of the free fall acceleration we find that it is around one third of a percent.
3, 4 · 10−2
acentrif ugal
=
= 3, 4 · 10−3
g
9, 81
(5.29)
The effects of the Coriolis force
Figure 5.6: Decomposition of the omega vector on the northern hemisphere
Since the angular velocity is a vector, this vector can be decomposed to two components
as shown in the figure. The two resulting omega vector components are the horizontal
and vertical component. On the surface of the Earth one makes rotation around two axes
simultaneously. On the northern hemisphere there is an omega vector which is lying on
the ground horizontally and points to North, and another one which is standing out of
the ground vertically and points upside direction. Each of these rotations has different
Coriolis effects.
ω = ωv + ωh
FCoriolis = 2mvrel × ω = 2mvrel × (ωv + ωh ) = 2mvrel × ωv + 2mvrel × ωh
71
(5.30)
(5.31)
Let us study them separately.
The first and second terms on the right hand side are denoted as Coriolis 1 and 2
forces.
FC1 = 2mvrel × ωv
FC2 = 2mvrel × ωh
(5.32)
The effects of Coriolis 1 force Coriolis 1 force is associated with the vertical omega
vector.
ˆ The hurricanes
The first effect to be treated is the whirling motion of the hurricanes, which is a counter
clockwise (CCW) rotation on the northern hemisphere. Imagine that at certain spot
the air warms up more than in the surrounding places. Here the air lifts up and the
surrounding air will horizontally move to the place where the airlift occurred. This will
cause an air flow vector field, in which the horizontal velocity vectors will all point to
the place of the airlift in
cylindrically symmetric arrangement. In order to describe the situation quantitatively
a simplified mechanical model is introduced.
Consider a huge hypothetical frozen ocean on the northern hemisphere which is completely flat and the ice is free of any friction. We put down a particle on the ice (imagine
it weights some kilograms) and pull the particle with vrel speed by means of a weightless ideal thread through several kilometers toward the center. The initial and the final
positions of the particle are R1 and R2 respectively.
Figure 5.7: Elementary model for hurricane formation
The Coriolis 1 force is as follows:
FC1 = 2mvrel × ωv
(5.33)
The associated torque to the center can be written:
M = r × FC1 = 2mr × (vrel × ωv )
72
(5.34)
Now the following mathematical rule is used: a × (b × c) = b(ac) − c(ab)
By applying this rule to the case we find the following outcome:
M = 2m (vrel (r · ωv ) − ωv (r · vrel ))
(5.35)
The first term in the parenthesis cancels out since the factors in the dot product are
normal to each other.
M = −2mωv (r · vrel )
(5.36)
Now we use the integral form of the theorem of angular momentum. It declares that the
angular impulse equals the variation of the angular momentum.
Zt2
L2 − L1 =
M(t)dt
(5.37)
t1
For present case:
Zt2
L2 − L1 = −2mωv
Zt2
(r · vrel )dt = −2mωv
t1
r · (vrel dt)
(5.38)
t1
The integration by time is switched to position integral by using the fact: vrel dt = dr.
ZR2
ZR1
L2 − L1 = −2mωv r·dr = 2mωv r·dr
R1
(5.39)
R2
After integration we find the result:
L2 − L1 = mωv (R21 − R22 )
(5.40)
On the other hand it is known that the angular momentum can be expressed as the
product of the angular velocity and the moment of inertia. The angular velocity is
denoted as Ω.
L2 − L1 = Ω(mR22 )
(5.41)
Ω(mR22 ) = mωv (R21 − R22 )
(5.42)
The right hand sides are equal
Finally the angular velocity of the rotation can be expressed:
Ω = ωv (
R21
− 1)
R22
73
(5.43)
Back to the hypothetical experiment, it can be concluded that the mass on the thread
will rotate around the central position. The ratio of the initial and the final radii will
determine the angular velocity of the rotation.
The result above for the angular velocity can also be interpreted in the inertial system. In this approach the law to be used is the conservation of the angular momentum.
The result in the inertial system is very much similar to the actual formula except for
minus one in the parenthesis, which is missing in inertial system approach. In rotating coordinate system the angular velocity of the rotating coordinate system should be
subtracted.
When the speed of the hurricane is of interest this can readily be expressed:
R1 2
(5.44)
vhurricane = R2 Ω = ωv R2 ( ) − 1
R2
The angular frequency of the Earth’s rotation at 45 degree latitude is as follows:
1
2π
sin 45o = 5, 13 · 10−5
(5.45)
24 · 3600
s
Let us take some realistic numerical values. Assume that R1 =400 km and R2 =100 km.
Substitute these values to find out the wind speed in the hurricane.
ωv =
m
km
= 277
s
h
The result is realistic though the physical model was the most basic possible.
vhurricane = 5, 13 · 10−5 · 105 [16 − 1] = 77
(5.46)
ˆ The Foucault pendulum
The second effect to be treated is famous Foucault pendulum experiment. Leon Foucault
French physicist made the experiment in 1851. A 67 meter long pendulum was hung up
in the Panthéon in Paris. As the pendulum was swinging for several hours the plain of
the oscillation was gradually turning. The stick at the end of the pendulum was drawing
a rosette shape figure into the sand on the floor. The angular velocity of the turning
was measured to be 11.29 degrees per hour. Which value gave an exact match with the
actual vertical angular velocity ( ωv ) in Paris. Since Paris is at the 48.83 degree latitude,
the sine of this angle multiplies the total angular velocity of planet Earth.
11, 29o
15o
o
sin 48, 83 =
(5.47)
ωv =
hour
hour
Let us discuss the motion of the pendulum in the rotating coordinate system of planet
Earth. The problem is considered to be two dimensional. The pendulum is treated in
linear approximation. The equation of motion is as follows:
d2 r
dr
m 2 = −Dr + mrω 2v + 2m × ωv
dt
dt
74
(5.48)
At little angular excursions the motion can be approximated with a horizontal plane
and l is the length of the pendulum. On the right hand side
motion. Therefore D = mg
l
the first term is the returning force due to gravity, the second term is the centrifugal
force and the third one is the Coriolis force. After rearranging the equation the following
state is reached:
dr
d2 r
=0
+ (ω02 − ωv2 )r + 2ωv ×
2
dt
dt
(5.49)
Here ω02 = gl . This equation is difficult to handle due to the cross product. At this point
it is worth to consider the variable as a complex number rather than a two dimensional
vector. The complex calculation provides a straightforward means to rotate a complex
number. Let us switch to z as a complex variable. By this way the following equation is
recovered: Here j is the imaginary unit.
dz
d2 z
+ 2jωv + (ω02 − ωv2 )z = 0
2
dt
dt
(5.50)
This is an ordinary second order differential equation which is widely used. The solution
is looked for in the following form: z(t) = Aeλt . After substitution the equation is as
follows:
λ2 + 2jωv λ + ω02 − ωv2 = 0
By solving this equation the roots emerge:
p
−2jωv ± −4ωv2 − 4(ω02 − ωv2 )
= j(−ωv ± ω0 )
λ12 =
2
(5.51)
(5.52)
Ultimately there are two linearly independent solutions:
z2 (t) = Aej(−ωv −ω0 )t
z1 (t) = Aej(−ωv +ω0 )t
(5.53)
Any linear combination of the above solutions is a valid solution. For the sake of simplicity wee choose the average of the above:
1
ejω0 t + ejω0 t
z(t) = (z1 + z2 ) = Ae−jωv
= Ae−jωv cos(ω0 t)
2
2
(5.54)
The real and the imaginary parts give the solutions:
z(t) = A cos(ωv t) cos(ω0 t) − jA sin(ωv t) cos(ω0 t)
(5.55)
Finally the two coordinates of the motion can be written:
x(t) = A cos(ωv t) cos(ω0 t) y(t) = −A sin(ωv t) cos(ω0 t)
75
(5.56)
In inertial system approach the situation is completely different. The plane of the oscillation is constant and only planet Earth rotates under the experiment with ωv angular
velocity. This makes possible that basically a kinematical solution is given to this problem.
Figure 5.8: The Foucault pendulum in inertial system
The inertial coordinate system is characterized by the e1 , e2 unit vectors. The
rotating coordinate system, (which rotates together with Earth) is characterized by i, j
unit vectors.
The oscillation takes place along the e1 axis therefore it is written as follows:
r = Ae1 cos(ω0 t)
(5.57)
Looking from the rotating system the e1 unit vector seems turning to the negative ϕ
angle direction. Accordingly:
e1 = i cos(−φ) + j sin(−φ) = i cos φ − j sin φ
(5.58)
Let us substitute and take it into consideration that φ = ωv t.
r = A(i cos φ − j sin φ) cos(ω0 t) = A [i cos(ωv t) cos(ω0 t) − j sin(ωv t) cos(ω0 t)]
(5.59)
The two coordinates of the motion can be written accordingly:
x(t) = A cos(ωv t) cos(ω0 t) y(t) = −A sin(ωv t) cos(ω0 t)
(5.60)
This perfectly matches the result above, which was calculated in a more tedious way by
solving the differential equation of motion in the coordinate system of planet Earth.
The effects of Coriolis 2 force The Coriolis 2 force is associated with the horizontal
omega vector component.
FC2 = 2mvrel × ωh
76
(5.61)
Figure 5.9: The effects of horizontal omega vector
Dependent on the direction of the relative velocity two effects emerge.
If the relative velocity points vertically down, the FC2 force will point to the East.
This force will deflect the freefall from the perfect vertical direction slightly to the East.
Consider a tower with height denoted h. A particle is dropped with zero initial
velocity. The velocity of free fall is as follows: vrel = gt.The Coriolis acceleration in
present case is: aC2 = 2vrel ωh . After substitution the aC2 = 2gωh t formula results.
This acceleration should be integrated twice in order to find out the magnitude of the
deflection.
Zt
vC2 (t) =
2gωh t, dt, = 2gωh
t,2
2
t, =t
= gωh t2
(5.62)
1
= gωh t3
3
(5.63)
t, =0
0
The displacement is the integral of the above formula:
Zt
xC2 (t) =
gωh t,2 dt, = gωh
t,3
3
0
t, =t
t, =0
On the other hand the total duration of theqfreefall (denoted τ ) can be determined as
2h
follows: h = g2 τ 2 . From this formula τ =
results. The amount of the deflection
g
emerges by substituting τ into xC2 formula above.
1
1
xC2 (τ ) = gωh τ 3 = gωh
3
3
2h
g
32
= ∆C2
(5.64)
After arranging it to a decent form:
∆C2 =
2
ωh
3
77
r 3
2
h2
g
(5.65)
Let us take the constant values numerically and find out the value in parenthesis above.
The g equals 9,81 m/s 2 and angular velocity at 45 degree latitude is the following:
ωh =
2π
1
cos 45o = 5, 13 · 10−5
24 · 3600
s
(5.66)
After substitution the final formula shows up:
r
3
3
2
2
−5
5, 13 · 10
h 2 = 1, 54 · 10−5 h 2
∆C2 =
3
9, 81
(5.67)
If the height is 30 meters the deflection is 2.53 millimeters, when the height goes up
to 100 meters the deflection becomes 1,54 centimeters. These values are so small that
any air disturbance will causes much higher deflection, which makes this specific Coriolis
effect practically ignorable.
If the relative velocity points horizontally in East or West direction the FC2 force
will point to vertically up or down respectively. Consequently this effect will cause a
virtual decrease or increase in the weight of any object. Let us calculate the magnitude
of the effect. Assume that an aircraft travels 1008 km/h (280m/s) speed in either direction mentioned. The absolute value of the FC2 force is easy to calculate due to the
perpendicular position of the vectors in the cross product.
FC2 = 2mvrel ωh
(5.68)
The associated acceleration is aC2 = 2vrel ωh . Substitute the numerical values:
aC2 = 2vrel ωh = 2 · 280 · 5, 13 · 10−5 = 2, 87 · 10−2
m
s2
(5.69)
If one compares this to the nominal value of the freefall acceleration the following comes
out:
aC2
2, 87 · 10−2
=
= 2, 93 · 10−3
g
9, 81
(5.70)
Roughly one third of a percent reduction or increase in the weight of any object seems
negligible in most practical cases. And do not forget that the speed was high. At speeds
usual in ground transportation the effect is far more insignificant.
78
Chapter 6
Oscillatory Motion - Gábor Dobos
A body is doing oscillatory motion when it is moving periodically around an equilibrium
position. In mathematical terms this can be expressed in the following form:
r(t) = r(t + T )
(6.1)
where r is the position-vector of the body, t is the time and T is called the period of the
oscillation. (6.1) means that if the body is found in position r at a given t time, it will
return to the same position T time later. Any kind of motion that can be described by
an r(t) function which satisfies (6.1) is called and oscillation.
Oscillations and vibrations are one of the most common types of motion in nature. A
pendulum or a body attached to a spring is doing oscillatory motion. Musical instruments
and our vocal cords also utilise vibrations to create sounds. Most modern clocks measure
time by counting oscillations. From switching mode power supplies to radio transmitters
there is a wide variety of electronic devices that rely on the periodic motion of charged
particles. Atoms in solids are also vibrating around their equilibrium positions.
To understand all these phenomena, first we have to understand oscillations. We will
start our discussion of oscillatory motion with the simple harmonic oscillator, which is
the easiest to describe in mathematical terms. Then we will expand this simple model
to take into account other effects such as damping and external excitation. Finally we
will discuss how the superposition of oscillations can be used to describe any arbitrary
periodic motion.
6.1
The simple harmonic oscillator
A body is doing a simple harmonic motion if its displacement from the equilibrium
position can be described by a sinusoidal function:
x(t) = Acos(ωt + φ0 )
79
(6.2)
Figure 6.1: Displacement of a body doing simple harmonic oscillation as a function of
time
where:
ˆ x is the position (or displacement) of the body (Measured in meters)
ˆ A is called the amplitude of the oscillation. This defines the range in which the
body is moving. If the equilibrium point is in the origin, the body is oscillating
between x = A and x = −A. (Measured in meters)
ˆ The argument of the sinusoidal function (ωt+φ0 ) is called the phase of the oscillation. (Measured in radians)
ˆ At t = 0 the phase is equal to φ0 , thus it is called the initial phase. (Also measured
in radians)
ˆ The constant ω is called angular frequency (measured in radians/s, or simply 1/s),
and it is determined by the period of the oscillation. cos() is a periodic function:
it takes the same value when the phase is increased or decreased by 2π. Thus the
time required to change the phase by 2π is equal to the period of the oscillation.
In mathematical terms:
(ω(t + T ) + φ0 ) − (ωt + φ0 ) = 2π
(6.3)
ωT = 2π
(6.4)
ω=
80
2π
T
(6.5)
ˆ The number of oscillations in a unit of time is called the frequency of the oscillation.
It is usually measured in Hertz (Hz), which is one oscillation per second. Like the
angular frequency, it is determined by the period of the oscillation:
f=
6.1.1
ω
1
=
T
2π
(6.6)
Complex representation of oscillatory motion
An interesting feature of harmonic oscillations is their connection to circular motion.
Consider a body moving on a circular trajectory of radius A around the origin. When
the position of the body is projected to a straight line (such as the x or y axis) it can
be described by a sinusoidal function, thus the projection of the body is doing a simple
harmonic oscillation. The angular frequency of this oscillation is the same as the angular
velocity of the circular motion, and its amplitude is the radius of the circle.
Figure 6.2: The connection of circular and simple harmonic motion.
This connection can be used to describe harmonic oscillations in complex form. A
complex number may be represented by a vector in the complex plane. The coordinates
of this vector can be given either in Cartesian or in polar coordinates:
Ĉ = x + jy
(6.7)
Ĉ = Aejφ
(6.8)
√
where j = −1 is the imaginary unit, x and y are called the real and imaginary parts
of the complex number Ĉ (usually marked by Re(Ĉ) and Im(Ĉ)), while A and φ are its
magnitude and phase, respectively.
Consider a complex function Ĉ(t), whose magnitude is constant (A = const.), while
its phase is a linear function of time (φ = φ0 +ωt). The endpoint of the vector representing
Ĉ in the complex plane is moving on a circle around the origin. The real and imaginary
parts of Ĉ can be determined using Euler’s formula:
ejφ = cos(φ) + jsin(φ)
81
(6.9)
Using this, the real and imaginary parts of Ĉ are:
x = Acos(φ) = Acos(φ0 + ωt)
y = Asin(φ) = Asin(φ0 + ωt)
(6.10)
(6.11)
Since the real and imaginary parts are basically projections of Ĉ to the real and imaginary
axes, x and y are sinusoidal functions of time. This also means that a simple harmonic
oscillation can be described as the real part of such a “rotating” complex function. Such
functions are usually referred to as phase vectors or phasors. The main advantage of
such a representation is that the dependencies on A, ω and φ0 can be separated into
three independent factors:
jφ0 jωt
Aejφ = Aej(φ0 +ωt) = Ae
| {z } e
(6.12)
Â
As the first two terms on the right-hand side are time independent, they are usually
merged to a single constant, called complex amplitude (Â = Aejφ0 ), which describes both
the amplitude and the initial phase of the oscillation.
6.1.2
Velocity and acceleration in oscillatory motion
The velocity and the acceleration of the particle can be determined from (6.2):
dx
= −ωAsin(ωt + φ0 )
dt
d2 x
a = 2 = −ω 2 Acos(ωt + φ0 ) = −ω 2 x
dt
v=
(6.13)
(6.14)
Figure 6.3 shows the position, velocity and acceleration as a function of time. All of
these are sinusoidal functions, but in different phases. The velocity is late by a quarter
period with respect to the displacement, and the acceleration is also shifted by another
quarter period. This means, that in simple harmonic motion the acceleration is always
proportional and opposite to the displacement. This also gives us a hint about what kind
of systems may exhibit simple harmonic oscillations. As the sum of the forces acting on
the body is proportional to its acceleration there must be a retracting force, pulling the
body back towards the equilibrium with a strength proportional to its displacement.
6.2
Motion of a body attached to a spring
One of the simplest mechanical systems that exhibit simple harmonic oscillations consists
of a body attached to a fix point by a spring. Since the spring force is proportional to
82
Figure 6.3: Displacement, velocity and acceleration in simple harmonic motion
83
the elongation of the spring, this system satisfies the aforementioned criterion. Because
of its simplicity it is ideal to demonstrate how harmonic motion can be described in
mathematical terms. Using Newton’s law of motion:
F = ma
−Dx = m
(6.15)
d2 x
dx2
(6.16)
d2 x
m 2 + Dx = 0
dt
(6.17)
d2 x D
+ x=0
dt2
m
(6.18)
(6.18) is a homogenous second-order linear differential equation, which can be solved
by the ansatz (or trial function) x(t) = eλt . The values of λ can be determined by
substituting the trial function into the differential equation:
d2 λt D λt
e + e =0
dx2
m
λ2 eλt +
λ2 +
(6.19)
D λt
e =0
m
(6.20)
D
=0
m
(6.21)
r
λ = ±j
D
m
(6.22)
(6.21) is usually referred to as the characteristic polynomial. Since (6.21) has two different roots, (6.18) has two independent solutions. (These are the so called fundamental
solutions of the differential equation.)
x1 (t) = e
84
j
√D
m
t
(6.23)
−j
x2 (t) = e
√D
m
t
(6.24)
According to the superposition principle any solution of a linear differential equation
can be built up as a linear combination of its fundamental solutions.
√D
√D
x(t) = Ĉ1 ej m t + Ĉ2 e−j m t
(6.25)
where Ĉ1 and Ĉ2 are complex constants, that can be determined from the initial conditions. (The state of the system – such as the position and velocity at the start of the
experiment – is usually referred to as initial conditions.) But there is a further important
point to consider: the displacement of the object is a measurable quantity, thus it must
always be a real number. (There are no imaginary quantities in the real physical world.)
This means that only those solutions have valid physical meaning, where the imaginary
parts of the fundamental solutions cancel out each other for all moments of time. This
can be easily achieved by making the two constant multipliers the complex conjugates
of each other. This way the imaginary parts of the two terms on the right-hand side of
(6.25) are always the opposites of each other and their sum is a real number. The general
solution of (6.18) is:
√D
√D
x(t) = Ĉej m t + Ĉ ∗ e−j m t
(6.26a)
where Ĉ is determined by the initial conditions. Using Euler’s formula (6.26a) can be
transformed to:
r !
r !
D
D
x(t) = A1 sin
t + A2 cos
t
(6.26b)
m
m
Or by using trigonometric identities:
r
x(t) = Acos
D
t + φ0
m
!
(6.26c)
(6.26a), (6.26b) and (6.26c) are equivalent: they all give precisely the same result for all
values of t, and they can be transformed into each other using mathematical identities.
Comparing (6.26c) to (6.2) shows,
q that the body is doing a simple harmonic oscillation,
D
with angular frequency ω0 = m
. The amplitude (A) and initial phase (φ0 ) are determined by the initial conditions. (In fact A and φ0 can be expressed in terms of A1 and
A2 or in terms of Ĉ, that are also determined by the initial conditions.)
2π
An important feature of this system is that the period of oscillation (T =
=
ω0
r
m
2π
) is independent of the initial conditions. It is influenced only by the parameters of
D
85
the system itself (such as the mass of the body or the spring constant). If the amplitude
is increased, the body is going to move proportionally faster, and finish each cycle in
precisely the same amount of time. This means that such systems are ideal for building
clocks: as each cycle takes the same time regardless of the initial conditions, time can
be measured by counting cycles.
6.3
Simple pendulum
Another mechanical system, that exhibits similar oscillations, is the simple pendulum,
which is a bob suspended on a massless rope from a frictionless pivot. When it is displaced
from the vertical (equilibrium) position the weight of the bob pulls it back towards the
equilibrium. As the length of the rope is constant, the bob moves on a circular trajectory.
We can apply Newton’s second law to describe its tangential acceleration:
−mg sin(θ) = ml
d2 θ
dt2
(6.27)
Figure 6.4: A simple pendulum
The left-hand side of (6.27) is the tangential component of the weight of the bob.
(Ft = −mg sin(θ)) The minus sign indicates that it points towards the equilibrium
position. The right-hand side is the mass of the bob (m) multiplied by the tangential
86
acceleration (which can be derived from the angular acceleration, by multiplying it by
the length of the rope). Rearranging (6.27) gives:
d2 θ g
(6.28)
+ sin(θ) = 0
dt2
l
Although this is another second order homogenous differential equation, unlike (6.18) this
is not a linear differential equation, and it has no simple analytical solution. However
for small angles sin(θ) is very close to θ, therefore when the amplitude of the oscillation
is small we may approximate sin(θ) by θ, and (6.28) becomes:
d2 θ g
+ θ=0
dt2
l
(6.29)
D
g
(6.29) is identical to (6.18), with x replaced by θ and
replaced by . Therefore the
m
l
solution is also identical:
θ(t) = θ0 cos(ω0 t + φ0 )
(6.30)
where
r
ω0 =
g
l
(6.31)
and θ0 and φ0rare determined by the initial conditions. The period of the oscillation
2π
l
(T =
= 2π
) depends only on the length of the pendulum (l), and the gravitational
ω0
g
acceleration (g). As both of these parameters can be kept reasonably constant, pendulum
clocks were considered to be the most precise clocks from their invention in 1565 by
Christiaan Huygens until the invention of quartz clocks in 1927.
Note, that the calculation above is not an exact solution of (6.27), merely an approximation. It is true only for small angles. For higher amplitudes the difference between
sin(θ) and θ increases, and the approximation becomes less and less precise. It can be
shown, that at high amplitudes the behaviour of the pendulum becomes significantly
different from that of a harmonic oscillator, and the period of the oscillation is not
completely independent of the amplitude.
In fact the calculation above is never completly accurate. Even at small angles sin(θ)
and θ are not exactly the same, which means that the formulas above are never completly
precise. But for small angles the deviations are so small, that they are indistinguishable
from other errors.1 For example the length of the pendulum can be measured only
1
It must be noted however, that approximations cause so called systematic errors, while some measurement errors are random. When the experiment is repeated many times sytematic errors remain
the same, while random errors are different each and every time. This means that random errors tend
to average out when calculating the mean of a large number of measurement values, while systematic
errors cause a bias in the mean value, no matter how many measurements are made.
87
with a limited precision, and it may even change with temperature. The gravitational
acceleration may be different at different locations, the pivot is not completely frictionless,
and the viscosity of air may also influence the movement of the pendulum.
These disturbing effects may be decreased by careful design, but they can never be
completely eliminated. Physical measurements always have some level of error. If the
error which is caused by our approximations is considerably smaller than measurement
errors, it becomes undetectable: it is hidden by other disturbing effects.
The important lesson of this sub-chapter is that in many cases mathematical description of physical systems becomes too complicated, and no analytical solution can be
found. In such cases one may attempt to find an approximate solution by simplifying
the problem, like we did in case of the simple pendulum. But it is important to keep
it in mind that these kinds of calculations are not completely precise. They are applicable only under certain circumstances, when the error of the approximation becomes
negligible relative to other errors.
6.4
Energy in simple harmonic motion
Equation (6.14) gives the acceleration of a body doing a simple harmonic oscillation.
Using Newton’s law of motion we can calculate the force which is acting on the body
during harmonic oscillations:
F = ma = −mω 2 x
(6.32)
This is a conservative force, which means there is a potential associated with it. The
potential energy of the body can be obtained by integrating (6.32) .
F =−
Z
dEp
dx
Z
dEp =
(6.33)
Z
−F dx =
mω 2 xdx
(6.34)
An indefinite integral is only defined up to an additive constant. We may choose the
value of this constant, so that the potential energy is zero in the origin:
1
1
Ep = mω 2 x2 = mω 2 A2 cos2 (ωt + φ0 )
2
2
(6.35)
Besides its potential energy the body also has kinetic energy:
1
1
1
Ek = mv 2 = m(ωAsin(ωt + φ0 ))2 = mω 2 A2 sin2 (ωt + φ0 )
2
2
2
88
(6.36)
According to the Pythagoren theorem sin2 (φ) + cos2 (φ) = 1, thus:
1
Ek = mω 2 A2 (1 − cos2 (ωt + φ0 ))
2
(6.37)
The total energy of the system is:
1
E = Ep + Ek = mω 2 A2
2
(6.38)
which is a constant quantity. This is not surprising since there are no dissipative forces
in the system. As the body is moving, energy is transformed from one form to the other:
at the equilibrium the potential energy is zero, while the kinetic energy is maximal. At
the extrema the velocity and kinetic energy of the body becomes zero, and all of the
system’s energy is stored in the spring as potential energy.
6.5
Damped oscillator
So far we have ignored friction and the drag force of the medium in which the oscillator is
moving. But in practice some level of dissipation is inevitable. Therefore it is necessary
to expand our model of the simple harmonic oscillator to take this into account. (6.17)
can be modified by adding a new term representing a drag force, which is proportional
to the velocity of the body, but points to the opposite direction:
dx
d2 x
+k
+ Dx = 0
2
dt
dt
(6.39)
d2 x
k dx D
+
+ x=0
2
dt
m dt
m
(6.40)
m
We may use the same trial function as before:
x(t) = eλt
(6.41)
Substituting it into (6.40) gives:
λ2 +
λ1,2
k
D
λ+
=0
m
m
k
=−
±
2m
s
89
k
2m
(6.42)
2
−
D
m
(6.43)
r
D
is the angular frequency of the undamped oscillam
tor, which is usually referred to as the natural angular frequency. Introducing the new
k
constant β =
, (6.43) becomes:
2m
q
(6.44)
λ1,2 = −β ± β 2 − ω02
We already know that ω0 =
λ1,2 = −β ± j
q
ω02 − β 2
(6.45)
Depending on wether β is less than, equal or greater than ω0 we have three distinct cases:
ˆ ω0 > β (underdamped case):
√
−β+j ω02 −β 2 t
x(t) = Ĉe
+ Cˆ∗ e
−β−j
√
ω02 −β 2 t
√ 2 2
√ 2 2
x(t) = e−βt Ĉej ω0 −β t + Cˆ∗ e−j ω0 −β t
(6.47)
p
ω02 − β 2 , (6.47) becomes:
x(t) = e−βt Ĉejωt + Cˆ∗ e−jωt
introducing the new constant ω =
(6.46)
x(t) = e−βt Acos(ωt + φ0 )
(6.48)
(6.49)
In the underdamped case oscillations occur due to the imaginary component of
λ.
p It must be noted however that the angular frequency of these oscillations (ω =
ω02 − β 2 ) is lower than the natural angular frequency (ω0 ). The other main
difference between the undamped and the underdamped oscillator is the result of
the real part of λ, which is responsible for the e−βt factor. Damped oscillators lose
energy due to the dissipative drag force, thus the amplitude of the oscillation is
not constant but decreases exponentially with time.
ˆ ω0 = β (critical damping): Increasing the viscosity of the medium further decreases
the angular frequency of the oscillations. The angular frequency reaches zero when
β becomes equal to ω0 . In this case the second term on the right-hand side of (6.45)
is zero, and both roots are the same. (The solution is degenerated.) But a second
order linear differential equation must have two independent fundamental solutions.
The second fundamental solution can be found using the ansatz x(t) = teλt . The
general solution is the linear combination of the two fundamental solutions:
x(t) = (A + Bt)eλt
90
(6.50)
Figure 6.5: Underdamped oscillation
Figure 6.6: In the critically damped case even a single cycle takes infinitely long time
91
At this point the frequency of the oscillation decreases to zero, thus even a single
cycle would take an infinitely long time. This is the so called critically damped
oscillator.
ˆ ω0 < β (overdamped case): In the criticaly damped case, and at higher viscosities
λ has no imaginary part, thus no oscillations can occur. The system simply crawls
back toward the equilibrium position (without ever reaching it):
x(t) = Aeλ1 t + Beλ2 t
(6.51)
Figure 6.7: In the overdamped case there are no oscilations.
6.6
Forced oscillations
In most practical cases losses are inevitable, and vibrations continuously lose energy. For
an oscillator to be able to operate for an extended period of time these losses have to be
replenished. In other words a periodic external force has to act on the system to avoid
the decay of its vibration. This can be taken into account by expanding (6.39):
m
dx
d2 x
jωf t
+
k
+
Dx
=
F
Re
e
0
dt2
dt
92
(6.52)
where the right hand side represent the external force. Dividing both sides by m, and
using the constants β and ω0 again:
F0
d2 x
dx
2
jωf t
x
=
+
ω
Re
e
+
2β
0
dt2
dt
m
(6.53)
This is a second order inhomogeneous linear differential equation. The solution of such
equations is the sum of the solution of the homogeneous differential equation and a
particular solution of the inhomogeneous equation:
x(t) = xh (t) + xp (t)
(6.54)
where xh (t) is one of (6.49), (6.50) or (6.51), depending on the values of k and D. We
may look for xp (t)in the following form:
xp (t) = Âejωf t
(6.55)
This means that x(t) is the superposition of two oscillatory motions. In the p
beginning
(transient state) the system is oscillating like a damped oscillator with ω = ω02 − β 2
angular frequency. But the amplitude of this oscillation decreases exponentially due to
damping. On the other hand the system is forced to start vibrating with the frequency of
the external driving force. While the damped oscillation quickly dies out, the amplitude
of the forced oscillation increases until the power lost to damping becomes equal to the
power supplied to the system by the external driving force. The amplitude and phase of
the oscillation (both included in the complex amplitude) in this stationary state can be
determined by substituting (6.55) into (6.53) :
−Âωf2 ejωf t + 2jβ Âωf ejωf t + ω02 Âejωf t =
−Âωf2 + 2jβ Âωf + ω02 Â =
 =
F0 jωf t
e
m
F0
m
F0 /m
ω02 − ωf2 + 2jβωf
(6.56)
(6.57)
(6.58)
The magnitude of the complex amplitude is the amplitude of the oscillation:
A= q
F0 /m
2
ω02 − ωf2 + 4β 2 ωf2
93
(6.59)
Figure 6.8: The amplitude of the forced oscillation depends on the excitation frequency
The phase of the complex amplitude represents the phase difference between the
oscillation and the periodic external driving force:
∆φ = arctan
ω02 − ωf2
2βωf
(6.60)
As the system is linear, the amplitude of the oscillation in stationary state is proportional to F0 . A more interesting fact is that both the amplitude and phase-shift depends
not only on the parameters of the oscillator (such as the mass off the body, the spring
constant and the strength of damping), but also on the frequency of the external driving
force. Figure 6.8 shows the amplitude of the oscillation as a function of the driving
frequency for different dampings. When the damping is strong, the amplitude decreases
with increasing frequency. But as the damping decreases a pronounced peak appears on
the graph: the amplitude is maximal, when the denominator on the right-hand side of
(6.59) is minimal. This occurs when ωf reaches
ωA =
q
ω02 − 2β 2
(6.61)
This is called the angular frequency of amplitude resonance. The smaller the damping,
the more pronounced the resonance becomes and its frequency shifts towards the natural
frequency of the oscillator.
The velocity of the oscillator can be calculated by differentiating the displacement
with respect to time. In stationary state the velocity amplitude is:
94
v0 = ω f A = q
ωf F0 /m
2
ω02 − ωf2 + 4β 2 ωf2
v0 = r
ω02
ωf
F0 /m
2
− ωf + 4β 2
(6.62)
(6.63)
The velocity amplitude also depends on the frequency of the external driving force.
vo is maximal, when the denominator on the right-hand side of (6.63) is minimal. This
occurs when the quantity within the parenthesis becomes zero.
ω02
− ωf = 0
ωf
ωf = ω0
(6.64)
(6.65)
The velocity of the oscillator becomes maximal, when the frequency of the external
driving force is equal to the natural frequency of the oscillator. As the kinetic energy
also reaches its maximum at this frequency, the phenomenon is called energy resonance.
In forced oscillators both the amplitude and the kinetic energy increases until losses
due to damping become equal to the power supplied to the oscillator by the external
driving force. When the damping is weak this may result extremely strong vibrations,
giving the false impression that a well-tuned oscillator can generate energy. Many conmen try to utilise this false impression and claim that they are able to build perpetum
mobiles based on oscillators. Of course, this is not possible. It is important to understand, that oscillators cannot generate energy, only accumulate the power of the external
driving. (In fact, they are continuously losing energy to damping.) Although the amount
of accumulated energy can be very high, it is not generated by the oscillator. Without
the external driving force, vibrations decay exponentially.
6.7
Superposition of simple harmonic oscillations
As we have seen in in the previous section, different oscillations may get superimposed
on each other. In the following we will discuss some special cases:
6.7.1
Same frequency, same direction
Let us consider two oscillations in the x direction with the same frequency, but with
different amplitudes and different initial phases:
95
jωt
x1 (t) = Re Â1 e
x2 (t) = Re Â2 ejωt
(6.66)
(6.67)
The superposition of the two oscillations is:
x(t) = x1 (t) + x2 (t) = Re (Â1 + Â2 )e
jωt
= Re Âe
jωt
(6.68)
where  = Â1 + Â2 . In other words the superposition of the two oscillations is another
harmonic oscillation of the same frequency, with a complex amplitude, that is the sum
of the complex amplitudes of the two oscillations. The amplitude of this oscillation
(which is the magnitude of the complex amplitude) can be determined by using the law
of cosines:
Figure 6.9: Sum of the complex amplitudes
A2 = A21 + A22 + 2A1 A2 cos(φ1 − φ2 )
(6.69)
The initial phase is:
tg(φ) =
sin(φ)
A1 sin(φ1 ) + A2 sin(φ2 )
=
cos(φ)
A1 cos(φ1 ) + A2 cos(φ2 )
96
(6.70)
6.7.2
Different frequency, same direction
Let us consider two oscillations in the x direction with the same amplitude, but with
different frequencies:
x1 (t) = ARe ejω1 t
(6.71)
jω2 t
(6.72)
x2 (t) = ARe e
The superposition of the two oscillations is:
x(t) = x1 (t) + x2 (t) = Re A(ejω1 t + ejω2 t ) =
ω1 +ω2 i
h ω1 −ω2
ω1 −ω2
= Re A ej 2 t + e−j 2 t ej 2 t =
ω1 − ω2
ω1 + ω2
= 2Acos
t cos
t
2
2
(6.73)
(6.74)
(6.75)
Figure 6.10: The superposition of two oscillations with slightly different frequencies
creates a beat
ω1 + ω2
The superposition of the two oscillations is an oscillation (represented by the cos
t
2
term in (6.75)) whose frequency is the average of the frequencies ofthe two oscillations,
ω1 − ω2
with an amplitude modulated by a sinusoidal function (the cos
t term in
2
(6.75)). For example when two musical instruments emit similar sounds, we cannot detect them individually. Instead we hear a single tone with a fluctuating intensity. The
phenomenon is called beat.
The amplitude, of the oscillation will be maximal when the cos() term equals 1 or -1.
This means, that in each cycle of the modulation function we can detect two fluctuations.
Even though the frequency of the modulation function is half of the difference of the two
original frequencies, the frequency of the beat is the difference of the two frequencies.
97
6.7.3
Lissajous figures
It is also possible to describe the superposition of oscillations in different directions. The
simplest case is when two oscillations are perpendicular to each other. In this case we
may choose the x and y axes of our coordinate system to point in the direction of the
two oscillations. The trajectory of the oscillating particle (the so called Lissajous figure)
depends both on the frequencies of the two oscillations, and also on the difference of their
initial phases. (We will assume, that both oscillations have the same amplitude. If this
is not true, the Lissajous figure will be scaled in the x and y directions proportionally to
the respective amplitudes.)
Figure 6.11 represents typical Lissajous figures. In the first four cases the frequencies
of the two oscillations are identical. When the initial phases are equal to each other, the
x and y coordinate of the oscillating particle changes in unison and the particle moves
on a straight line. Since cos(φ + π) = −cos(φ), the particle also moves on a straight line,
π
when the phase difference is π. When the phase difference is , the coordinates of the
2
particle are:
x(t) = Acos(ωt)
(6.76)
π
y(t) = Acos(ωt + ) = Asin(ωt)
2
(6.77)
These are the horizontal and vertical components of circular motion, and consequently
the Lissajous figure becomes a circle. For any other arbitrary phase difference the trajectory of the oscillating particle is an ellipse, with an aspect ratio depending on the phase
difference.
Figure 6.11: Lissajous figures
When the ratio of the frequencies in the two directions is n:m, the particle finishes
n cycles in one direction while it does m cycles in the other direction. This makes the
Lissajous figures considerably more complicated. When the frequencies are the same,
the particle repeats the same trajectory in every cycle. When the horizontal and vertical
frequencies are different, the number of cycles it takes to close the Lissajous figure and
start repeating the same motion again is the least common multiple of n and m.
This also shows that Lissajous figures appear only when the ratio of frequencies is a
rational number. When the ratio is an irrational number there is no common multiple
98
other than infinity, which means that the trajectory is not a closed curve: although the
horizontal and vertical components of the motion are both periodic, their superposition
is not.
The ratio of frequencies can be determined by counting how many times the figure
touches the horizontal and vertical edges of the square that can be drawn around the
figure. Since the particle finishes n cycles in one direction while it does m cycles in
the other, it touches the edges perpendicular to the first direction 2n times, while it
touches the other edges 2m times. Consequently the ratio of the horizontal and vertical
frequencies is the inverse of the ratio of the number of cross-sections with the horizontal
and vertical edges. This technique can be used to measure unknown frequencies with a
two channel oscilloscope (that can plot the two signals against each other in the so called
X-Y mode) and a reference oscillator of known frequency.
6.7.4
Fourier analysis
During this chapter we have focused on oscillatory motions that can be described by
sinusoidal functions. But the definition of oscillatory motion is more general: it does
not require the motion to be harmonic, only to be periodic. Therefore it is necessary to
formulate a mathematical method to describe anharmonic oscillations. The problem is
that not all kinds of motion can be expressed in closed-form formulas. The best we can
do is to create a method that enables us to approximate any periodic motion at arbitrary
precision.
Our task is slightly similar to what we had to do to describe the simple pendulum.
Even though we cannot give a completely precise mathematical description, we may be
able to give an approximation that is so close to the actual behaviour of the system
that any deviations are unnoticable. Since we already have a basic understanding of
harmonic motion it is a sensible idea to approximate arbitrary periodic motions by the
superposition of sinusoidal functions.
Consider a periodic function f (t), with a period of T . In the zero order approximation we may try to calculate the time-average of f (t). Naturally, this is a very poor
approximation since a constant cannot describe how f (t) changes with time. We may
improve the approximation by adding a sinusoidal function with the same T period as
f (t) to the average value, and adjusting its phase and amplitude, to achieve the best
possible fit. Since f (t) is not a sinusoidal function we may not achieve a perfect fit, but
the difference between the arbitrary function and this simple harmonic approximation is
much smaller than the difference between f (t) and its average value.
If adding a single sinusoidal function has improved the approximation, we may try to
further improve it, by adding even more sinusoidal functions. Adding a second sinusoidal
component of the same frequency is pointless, since the sum of two sinusoidal functions
of equal frequency is another sinusoidal function of the same frequency with different
phase and amplitude. As we have already optimised the phase and amplitude of the
99
first sinusoidal component, adding another one with the same frequency will not improve
the approximation. Therefore the second sinusoidal component needs to have a different
frequency. But we also have to make sure, that the approximation has the same period as
f (t). This can be achieved by choosing the frequencies of further sinusoidal components
to be integer times the frequency of the first component. This way each sinusoidal
component will finish an integer number of cycles in T time, and their sum will have the
same period as f (t).
Figure 6.12: An arbitary periodic function (upper left) may be approximated by the
superposition of a series of sinusoidal functions
Adding each new sinusoidal component will slightly increase the quality of the approximation. By adding together a suitably high number of sinusoidal components we
may create an approximation of arbitrary precision.
f (t) = a0 + a1 cos(ωt + φ1 ) + a2 cos(2ωt + φ2 ) + a3 cos(3ωt + φ3 ) + ...
(6.78)
Ususally it is more efficient to represent all sinusoidal components in complex form:
f (t) = Re a0 + â1 ejωt + â2 e2jωt + â3 e3jωt + ...
(6.79)
To create an approximation of any arbitrary periodic function, all what we have to do, is
to determine the optimal values of a0 , â1 , â2 , . . . to achieve the best possible fit. These
100
values may be calculated by the formulas deduced by Joseph Fourier in the nineteenth
century:
Z
1 T
f (t)dt
(6.80)
a0 =
T 0
Z
1 T
ân =
f (t)enjωt dt
(6.81)
T 0
(6.82)
Using these formulas we may express any arbitrary periodic function as a sum of sinusoidal functions. It must be noted however that this is merely an approximation. The
more components included in the sum, the more similar it becomes to the original function, but to reach a perfect fit an infinite number of components would be required,
which is not possible in practice. When choosing the parameters of the approximation,
two criteria must be observed:
ˆ The frequency of the first sinusoidal component is determined by the period of the
oscillation. The longer the period, the lower the frequency needs to be.
ˆ Low frequency sinusoidal functions change very slowly. Therefore if the f (t) function changes rapidly, high frequency components are required to follow these rapid
changes. Since we have already chosen the frequency of the first sinusoidal component to set the period, this can be achieved only by increasing the number of
sinusoidal components. As a thumb rule the period of the highest frequency component should not be longer than the double of the required time resolution. Any
changes faster than that are going to be lost in the approximation.
101
Chapter 7
Waves - Gábor Dobos
Most people imagine waves as some sort of periodic disturbance which is moving in a
medium, such as the waves on the surface of a pond after a pebble is dropped into it.
It must be noted however that in physics the term is used in a much broader sense.
Electromagnetic waves can propagate in perfect vacuum without the presence of any
medium, shock waves are non-periodic, and standing waves are not moving in space.
The most general definition of a wave that includes the above examples is a disturbance
which can transport energy and momentum without any long range movement of matter.
Waves on the surface of water as well as sound waves, or waves in a string of a
musical instrument are called mechanical waves as they mechanically distort an elastic
medium. Electromagnetic waves consist of oscillations of electrical and magnetic fields
that can travel in space. In a plasma these fields can influence the movement of charged
particles forming different kinds of magnetohydrodynamic waves such as Alfvén waves or
magnetosonic waves. In some cases elementary particles also behave as waves. According to Einstein’s theory of general relativity even gravitational waves may be created
by distorting the fabric of the space-time continuum, although these have never been
experimentally detected due to measurement difficulties.
In this chapter we will discuss mechanical waves. These waves may be created by
suddenly distorting an elastic medium. Its surroundings will exert a restoring force on
the displaced portion of the medium, pulling it back towards the equilibrium position.
As we have seen in the previous chapter this may lead to vibrations. But in this case
vibrations are not limited to the displaced section. As its surroundings exert a force
upon it, it will also exert an opposite force on its surroundings. Due to this force,
the surrounding medium may also start to vibrate, and the disturbance may propagate
trough the medium.
An interesting feature of this type of motion is that although the disturbance may
travel large distances, the movement of a given piece of the medium is limited: it is oscillating around its own equilibrium position. The thunder of lightning can be heard from a
distance of several kilometres, but this does not mean that individual air molecules travel
102
this far. They collide with other particles, passing over their energy and momentum. It
is through these collisions that the disturbance reaches our ears, while the air as a whole
does not move.
The direction of the oscillation can be either parallel or perpendicular to the direction
into which the wave is traveling. Longitudinal waves occur when the direction of the
distortion is parallel to the propagation, like in case of sound waves in air. In other
cases the medium is oscillating at right angles to the direction into which the wave is
traveling. These are called transverse waves. It must be noted however that transverse
mechanical waves are not possible in all media since the restoring force must be parallel
to the displacement. In case of transverse waves this means that the restoring force has
to be perpendicular to the direction of propagation. In other words a shearing stress
must exist in the medium, and this is not possible in all materials. For example no
shearing stress may exist in gases, thus sound waves in air are always longitudinal waves.
As we are living in a three dimensional world there are two directions that are perpendicular to the direction of propagation. This means that in case of transverse waves the
oscillations at each point may be the superposition of two perpendicular oscillations. The
frequency of these perpendicular oscillations must be the same, but their initial phases
might be different. The situation is very similar to what we have discussed in section
6.7.3. The Lissajous figures created by the superposition of perpendicular oscillations
of the same frequency may form a straight line (when the phase difference is 0 or π), a
π
π
circle (when the phase difference is or − ) or an ellipse (in case of any other arbitrary
2
2
phase difference). Accordingly transverse waves might have linear polarisation, circular
polarisation or elliptical polarisation.
7.1
Sine wave
We have started the discussion of oscillatory motion with the simple harmonic oscillator,
as it was the easiest to describe in mathematical terms. For the same reason the first
type of wave we will discuss in more detail is the one dimensional sine wave, which can
be described by:
y(x, t) = Asin(kx − ωt)
(7.1)
where y is the displacement, A is the amplitude, ω is the angular frequency, and k is
called the wave number. If x is fixed, the displacement is a sinusoidal function of time.
In other words, each point of the medium is doing harmonic oscillations, with an initial
phase depending on the position. If we fix the time (like taking a snapshot of the wave)
one can see, that the displacement is also periodic in space. The spatial period of the
wave (the distance between two consecutive corresponding points – like tow crests or
103
Figure 7.1: A sine wave
zero crossings) is called the wavelength and it is usually marked by λ. While the period
of the oscillation is determined by the angular frequency (ω), the wavelength depends
on the wave number (k):
k=
2π
λ
(7.2)
The velocity of propagation can be determined by following the movement of a certain
point of the wave, like a crest or a trough. In other words we shall determine how x
should change as a function of time, to keep the argument of the sin() function (the
phase) constant:
kx − ωt = const.
dx
ω
=
vph =
dt
k
(7.3)
(7.4)
As it is the rate at which the phase of the wave propagates through space, it is called
the phase velocity.
In case of simple harmonic oscillations, the acceleration of the particle was proportional to the displacement. In case of a sine wave every point of the medium is doing a
simple harmonic oscillation, thus:
∂y
= −Aωcos(kx − ωt)
∂t
∂ 2y
= −Aω 2 sin(kx − ωt) = −ω 2 y(x, t)
2
∂t
104
(7.5)
(7.6)
As y(x,t) depends on two parameters, we may also calculate its differentials with
respect to the space coordinate (x):
∂y
= Akcos(kx − ωt)
∂x
∂ 2y
= −Ak 2 sin(kx − ωt) = −k 2 y(x, t)
∂x2
(7.7)
(7.8)
Comparing (7.6) and (7.8) gives:
∂ 2 y ω 2 ∂ 2 y
=
∂t2
k ∂x2
ω
Since we already know, that vph = is the phase velocity of the wave:
k
(7.9)
2
∂ 2y
2 ∂ y
=
v
(7.10)
ph
∂t2
∂x2
For sine waves the second differentials of the displacement with respect to time and
to the space coordinate are proportional to each other, and their ratio is the square of
the velocity of the wave. Systems that can be described by similar equations have the
potential to exhibit wave motion, thus (7.10) is called a wave equation
7.2
Transverse wave on a string
A string under tension is one of the simplest mechanical systems that can exhibit wave
motion. Imagine a thin wire with a linear mass density µ, fixed at one and, and pulled
with a constant F force at the other end. When a section of the string is displaced in
a perpendicular direction, the tension in the string pulls it back towards the original,
straight position. In other words it represents a restoring force, and we may expect the
displaced section to start oscillating. But the sections of the string are not separated
from each other. If one section starts oscillating, it will exert a periodic force on the
neighbouring sections, forcing them to start oscillating, too.
To give a mathematical description of the phenomenon, consider an infinitesimally
small section of the string. Both ends are pulled by the same F force, but in slightly
different directions. (If a transverse wave is traveling in the string, it is no longer straight,
but slightly curved.) Applying Newton’s law of motion in the horizontal direction:
∆m ax = F cos(θ + ∆θ) − F cos(θ)
µ ∆x ax = F (cos(θ + ∆θ) − cos(θ))
105
(7.11)
(7.12)
Figure 7.2: Forces acting on a section of the string
where ∆m is the mass of the section, ∆x is its length, and ax is its acceleration. At
small angles the cosine function changes very slowly. If θ << 1, cos(θ + ∆θ) ≈ cos(θ).
Thus the acceleration in the horizontal direction is zero. In the vertical direction:
µ ∆x ay = F (sin(θ + ∆θ) − sin(θ))
(7.13)
If the distortion is small (θ << 1) the sine function may be approximated by the tangent
of the same angle. As the tangent is basically the steepness of the curve, it may be
∂y
replaced by
:
∂x
sin(θ) ≈ tg(θ) =
∂y
∂x
Using this approximation (7.13) becomes:
∂y ∂y µ ∆x ay = F
−
∂x x+∆x ∂x x
∂y ∂y −
F ∂x x+∆x ∂x x
ay =
µ
∆x
(7.14)
(7.15)
(7.16)
If the section is infinitesimally small (∆x → 0):
F ∂ 2y
∂ 2y
=
∂t2
µ ∂x2
106
(7.17)
(7.17) is a wave equation, and shows, that if ther
distortion is small, transverse sine waves
F
are going to travel in the string with vph =
velocity. This may be verified by
µ
substituting (7.1) into (7.17):
−A ω 2 sin(kx − ωt) =
F
A k 2 sin(kx − ωt)
µ
(7.18)
As the equation must hold for all values of t and x:
F 2
k
µ
ω2 =
ω
=
k
vph =
7.3
(7.19)
s
F
µ
(7.20)
Energy transport by mechanical waves
When a sine wave is traveling in a medium, each point is doing a simple harmonic
oscillation. As we have seen in the previous chapter the energy of a harmonic oscillator
is:
1
E = mω 2 A2
2
(7.21)
where A is the amplitude of the oscillation, m is the mass of the oscillating body, and ω
is the angular frequency. Using this, the energy of the wave in an infinitesimally small
section of the medium is:
1
dE = dm ω 2 A2
2
(7.22)
where dm = µdl is the mass of the section, dl is its length, and µ is the linear mass
dl
density of the string. A wave travelling by v velocity needs dt =
time to travel dl
v
distance. Therefore:
1
dE = µ v dt ω 2 A2
2
The power transmitted by the wave is:
107
(7.23)
P =
1
dE
= µvω 2 A2
dt
2
(7.24)
The power transmitted by a wave traveling in a three dimensional medium may be
calculated in a similar manner. Consider a small dx · dy · dz section of the medium:
1
1
dE = dm ω 2 A2 = ρ dx dy dz ω 2 A2
2
2
1
dE = ρ v dt dy dz ω 2 A2
2
(7.25)
(7.26)
where ρ is the density of the medium. The power transmitted by the wave trough a unit
area of the medium is:
S=
1
dE
= ρ v ω 2 A2
dt dy dz
2
(7.27)
Human senses - such as hearing and vision - measure intensities on a logarithmic
scale. This is the reason why the intensity of certain types of waves is usually measured
W
not in 2 , but in other, special units, on a logarithmic scale. For example the intensity
m
level of sound is usually given in decibels (dB). By definition:
β = (10dB)lg
S
S0
(7.28)
W
where S0 = 10−12 2 . A 10dB increase in the noise level actually means that the power
m
of sound waves reaching our ears is increased by one order of magnitude. Our ears have
W
a very impressive dynamic range: 0dB (10−12 2 ) is usually the smallest sound level
m
W
that a human may detect. A whisper is around 20dB (10−10 2 ), while normal talking is
m
W
W
approximately 60dB (10−6 2 ). Our eras can tolerate noise levels up to 90dB (10−3 2 )
m
m
for an extended period of time, without any lasting damage, and only noises louder than
W
120dB (1 2 ) are immediately painful. Measuring a quantity on such a wide dynamic
m
range on a linear scale would be very difficult even with modern instruments. That is
why most biological systems measure such quantities on a logarithmic scale.
108
7.4
Group velocity
As we have seen in the previous chapter superposition of two oscillations of similar
frequencies creates a beat. A similar phenomenon can be observed with waves. Imagine
two waves travelling in the same medium with slightly different angular frequencies and
wave numbers. (This is actually a very realistic model. In practice it is impossible to
create a perfectly monochromatic wave: usually we have to deal with the superposition
of several waves with slightly different parameters.) The wave that we can observe is
going to be the superposition of the two waves:
y1 (x, t) = Asin(k1 x − ω1 t)
y2 (x, t) = Asin(k2 x − ω2 t)
y(x, t) = y1 (x, t) + y2 (x, t)
y(x, t) = Asin(k1 x − ω1 t) + Asin(k2 x − ω2 t)
using the trigonometric identity sinα + sinβ = 2cos
(7.29)
(7.30)
(7.31)
(7.32)
α−β
α+β
sin
:
2
2
(k1 + k2 )x − (ω1 + ω2 )t
(k1 − k2 )x − (ω1 − ω2 )t
sin
2
2
(k1 − k2 )x − (ω1 − ω2 )t
y(x, t) = 2Acos
sin(kx − ωt)
2
y(x, t) = 2Acos
(7.33)
(7.34)
Figure 7.3: Superposition of two sine waves of similar frequencies
where k is the average of k1 and k2 , and ω is the average of ω1 and ω2 . The superposition of the two waves gives a wave whose wave number and angular frequency
is the average of the wave numbers and angular frequencies of the two original waves
(this is called the carrier wave), but whose amplitude is not constant, but modulated
by a sinusoidal function. Since this modulation function also depends on both x and t,
its nodes (the positions where the amplitude is minimal) and antinodes (the positions
where the amplitude is maximal) are also going to move in the medium.
109
Two waves traveling in the same medium with similar angular frequencies are going
to have similar wave numbers and phase velocities, too. Since the angular frequency and
wave number of the carrier wave is the average of those of the two waves, its velocity is
also going to be similar.
vph
ω1 + ω2
ω
2
= =
k1 + k2
k
2
(7.35)
The groups formed by the modulation are going to travel with a different velocity:
vg =
ω1 − ω2
k1 − k2
(7.36)
If the frequencies and wave numbers of the waves are infinitesimally close to each other
vg =
ω1 − ω2
∂ω
=
k1 − k2
∂k
(7.37)
This is called the group velocity and it is a very important parameter not only in
physics, but also in the field of telecommunication. Information is usually transmitted
by the modulation of some sort of wave. The type of carrier waves may vary from
sound waves to electromagnetic waves, and different types of modulation techniques are
in use, but in most cases the velocity with which information can be transmitted from
one point to another is the same as the velocity of the groups formed by the modulation.
Even though electromagnetic waves travel at the speed of light, this does not mean that
information can be transmitted at the same velocity. For most media the group velocity
is lower than the phase velocity, therefore the speed of information transmission is also
lower.
It must be noted however that there are exceptions from this rule: the group velocity
is not necessarily the same as the “signal velocity”. The ω(k) function is called the
dispersion relationship. For most materials the slope of this function decreases for higher
frequencies, thus the group velocity is lower than the phase velocity. But there are some
exotic materials with a so called anomal dispersion relationship. In these materials the
group velocity may be higher than the phase velocity (or the speed of light). This does
not mean however that these materials enable faster than light information transmission.
As a thumb rule the signal velocity cannot be higher than the lesser of the phase and
group velocities.
110
7.5
Wave packets
Another interesting parameter that can be understood by considering the superposition
of waves is the amount of information that can be transmitted in a unit of time. Although
actual communication protocols are very complicated the rate of information transmission depends on the duration of pulses that can be formed by modulating the carrier
wave. (The shorter the pulses the more data can be transmitted in a unit of time.)
In section 6.7.4 we have used Fourier analysis to describe anharmonic oscillations as
the superposition of a large number of sinusoidal components. In a similar manner we
may attempt to construct short pulses from the superposition of sine waves. (This kind
of construct is usually referred to as a wave packet.) The problem is that Fourier analysis
is applicable only to periodic functions, and a pulse is not periodic. This problem may
be remedied by treating the non-periodic function describing the pulse as a periodic
function with an infinitely long period. (If the period is infinitely long, the function
never repeats itself...) As the frequency difference between Fourier components is the
inverse of the period, the Fourier components of an aperiodic function are infinitely close
to each other. In other words the components do not belong to discrete frequencies: they
form a continuous distribution. This distribution is called the Fourier transform of the
function.
Z ∞
ˆ
f (ν) = F (f (t)) =
f (t)e−2 π j t ν dt
(7.38)
−∞
where t is the time, ν is the frequency, and f (t) is the function whose Fourier transform
is fˆ(ν).
Let us consider the Fourier transform of a Gaussian-shaped pulse (In nature quantities
tend to have a so called normal distribution which is described by the Gaussian function.
Therefore it is reasonable to discuss Gaussian-shaped pulses. Of course, other pulse
shapes may also be described in a similar fashion.):
f (t) = √
2
1
− t
e 2(σt )2
2πσt
(7.39)
The duration of the pulse is determined by the constant σt , which is the full with at half
maximum of the Gaussian function. The Fourier transform of this function is:
Z ∞
2
1
− t 2 −2 π i t ν
ˆ
2(σ
t) e
√
f (ν) =
e
dt
(7.40)
2πσt
−∞
2 2 2
fˆ(ν) = e−2π σt ν
(7.41)
This function gives the amplitudes of the harmonic components whose superposition is
the σt wide Gaussian-shaped pulse. This is another Gaussian function whose full width
111
at half maximum is σν :
1
= 2π 2 σt2
2σν2
1
σν =
2πσt
(7.42)
(7.43)
It must be noted, that information is transmitted by the modulation of a carrier wave.
This means, that the components, that make up the pulse should centre around the
frequency of the carrier wave (ν0 ). In other words we shall shift the fˆ(ν) function by ν0 .
The pulse formed by these components is:
2
1
− t
f 0 (t) = F −1 fˆ(ν + ν0 ) = √
e 2(σt )2 e−2πjν0 t
2πσt
(7.44)
Of course this is not a pure Gaussian-shaped pulse, since it is modulated be the e−2πjν0 t
factor, which is basically a sinusoidal modulation with the frequency of the carrier wave.
Nonetheless the envelope of the pulse is still a Gaussian function whose full width at half
maximum satisfies:
σt σν =
1
2π
(7.45)
This means that in order to construct short pulses (and reach high data transmission
rates) we have to include Fourier components from a wide frequency range. The wider
the available frequency window, the shorter the pulses may become.
This explains why telecommunication corporations are willing to pay billions to obtain the right to use certain frequency windows. Although we usually say that - for
example - cell phones transmit data at 900 MHz frequency, in reality they use not just a
single frequency but a range of frequencies around 900 MHz. Each provider has its own
frequency window, for which it pays a concession fee to the government. The wider this
window is, the faster your mobile internet connection may become.
Based on this, we may also understand why the number of radio and television stations
is limited. Although in layman terms we usually say that a station transmits at a given
frequency, this refers to the frequency of the carrier wave. In reality the station uses a
range of frequencies around this frequency, to modulate the carrier wave. This is why the
frequencies of radio stations cannot be arbitrarily close to each other lest they interfere.
It must also be noted that (7.45) describes a much more basic principle, which has a
wide range of implications beyond telecommunication. Equations like (7.45) are usually
referred to as uncertainty relations. Although we may be able to form wave packets by
combining a large number of sinusoidal components, the length of these wave packets
cannot be infinitely short, and they always include a large number of different frequencies.
The dimensions of the wave packet in the time domain and in the frequency domain are
112
related to each other. The shorter the pulse is the wider range of frequencies it must
include. When we use only a narrow frequency range, the pulse is going to get longer.
It is easy to see, how a similar relation can be deuced between the length of a pulse
(its x dimension), and the range of wavenumber (k) components that must be combined
to form the wave packet:
1
(7.46)
2π
If we try to measure the position of the wave packet, the precision of our measurements
will be limited by σx . In a similar fashion σk limits the precision of wavenumber measurements. Equation (7.46) shows that the position and wave number of the wave packet
cannot be measured at arbitrarily high precision at the same time. The large number of
sinusoidal components that are required to form a short pulse makes wavenumber measurements uncertain. On the other hand, if we use only a narrow range of wavenumber
components σx increases and the pulse becomes blurred, which decreases the precision
of position measurements.
In general, where waves are concerned there are certain parameters that are connected
to each other. These parameters cannot be measured at the same time at arbitrarily high
precision. As we will see during the discussion of quantum mechanics, the quantum states
of physical systems are represented by wave functions, and quantum uncertainties are
closely related to the principles that we have discussed above.
σx σk =
7.6
Standing waves
Imagine two waves of the same frequency traveling in the same medium in opposite
directions. (Such situations may arise when the wave is reflected back from the end of
the medium.)
y1 (x, t) = Asin(kx − ωt)
y2 (x, t) = Asin(kx + ωt)
(7.47)
(7.48)
The superposition of the two waves is:
y(x, t) = y1 (x, t) + y2 (x, t)
y(x, t) = Asin(kx − ωt) + Asin(kx + ωt)
(7.49)
(7.50)
Using the same trigonometric identity as in the previous section:
y(x, t) = 2Asin(kx)cos(ωt)
113
(7.51)
The superposition of the two waves is a so called standing wave. Each point of
the medium is oscillating with the same frequency, but unlike other waves, in case of
standing waves, the initial phase is the same in every point of the medium. This means
that the wave is not moving. This is because the position and time dependencies of the
disturbance are separated into two terms. All points are oscillating in unison, with an
amplitude determined by the position. (The situation is slightly similar to that what
we have seen in section 7.4, but in this case the group velocity is zero: the groups are
not moving.) The positions where the amplitude of the oscillation is minimal are called
nodes, the ones, where the amplitude is maximal, are called anti-nodes.
Most musical instruments are designed to efficiently form certain types of standing
waves and reject others. For example in case of string instruments - such as guitars
and violins - the ends of the strings are held tight. This means that the standing wave
must have nodes at both ends, otherwise its energy dissipates away very quickly. When
the player uses the instrument many waves with different frequencies are created, but
only those may form stable standing waves, that have nodes at both ends. The distance
λ
between two neighbouring nodes is , therefore only those waves are stable, whose
2
half wavelength fits integer times into the length of the string. The wavelength (λ) is
determined by the wave number (k), which in turn depends on the velocity of the wave
(v) and its frequency (f ):
λ=
v
v
2π
= 2π =
k
ω
f
(7.52)
The so called principal mode is the standing wave with the lowest frequency that is
stable in the string. It has nodes at both ends, but there are no further nodes in between.
Therefore the wavelength is twice the length of the string:
λ0
2
λ0 = 2L
v
v
f0 =
=
λ0
2L
L=
(7.53)
(7.54)
(7.55)
This is called the fundamental frequency of the instrument. The first harmonic has
one extra node between the ends:
L = λ1
v
f1 =
L
114
(7.56)
(7.57)
The second harmonic has two nodes between the ends
λ2
2
2L
λ2 =
3
v
v
=3
f2 =
λ2
2L
L=3
(7.58)
(7.59)
(7.60)
And so on... In general:
2L
n
v
fn = n
2L
λn =
(7.61)
(7.62)
Figure 7.4: Standing waves in different types of resonators
Only waves with these frequencies form stable oscillations in the string, all other frequencies are rejected. The phenomenon is a type of resonance, and such instruments are
commonly referred to as resonators. The ratio of the intensities of the stable harmonics
(or overtones) depends on the design of the instrument, and it is characteristic to each
resonator. It is these intensity ratios that give each musical instrument its own unique
tone.
Other types of musical instruments have different resonators. For example a pan flute
consists of small tubes closed at one end, and open at the other end, so that the player
can blow air into it. The standing waves forming in such a resonator needs to have a
node at the closed end an anti-node at the open end. The distance between a node and
an anti-node is a quarter wavelength, thus:
115
λ0
4
λ0 = 4L
v
v
f0 =
=
λ0
4L
L=
(7.63)
(7.64)
(7.65)
The first overtone has an additional node between the two ends:
λ1
4
4L
λ1 =
3
v
v
=3
f1 =
λ1
4L
L=3
(7.66)
(7.67)
(7.68)
The second overtone has two nodes between the ends:
λ2
4
4L
λ2 =
5
v
v
f2 =
=5
λ2
4L
L=5
(7.69)
(7.70)
(7.71)
In general:
4L
2n + 1
v
fn = (2n + 1)
4L
λn =
(7.72)
(7.73)
Resonators open at both ends, have anti-nodes at both ends. The principal mode has
a single node in between:
λ0
2
λ0 = 2L
v
v
f0 =
=
λ0
2L
L=
116
(7.74)
(7.75)
(7.76)
The first harmonic has two nodes between the anti-nodes at the ends:
L = λ1
v
f1 =
L
(7.77)
(7.78)
The second harmonic:
λ2
2
2L
λ2 =
3
v
v
=3
f2 =
λ2
2L
L=3
(7.79)
(7.80)
(7.81)
In general:
2L
n
v
fn = n
2L
λn =
7.7
(7.82)
(7.83)
The Doppler Effect
When a racing car drives through the finish line the spectators on the stands will hear
a shift in the frequency in its sound: it seems to be higher when the car races towards
the spectators and drops quickly as it drives past. The phenomenon is called Doppler
Effect, and it is detected when the source or the observer is moving with respect to the
medium in which the wave is traveling.
To better understand the phenomenon imagine a stationary observer and a sound
source moving towards the observer at a given vs velocity. If the sound source would be
stationary, the wavelength (λ) of the sound waves could be calculated from the ratio of
v
the speed of sound (v) and the frequency (f ) emitted by the source (λ = ). But the
f
movement of the source compresses the waves and changes their wavelengths. This can
be understood by calculating the distance between two crests of the wave emitted by the
1
source. During one period (T = ) the crest of the wave emitted by the source travels
f
λ distance towards the observer. But in the same time the source is also moving, and it
vs
gets closer to the observer by vs T =
distance. Because of this, when the next crest of
f
117
the wave leaves the source, its distance from the previous one is not λ, but:
λ0 = λ − v s T =
v − vs
f
(7.84)
Due to this shift in the wavelength of the emitted sound, the observer is detecting a shift
in the frequency of the sound:
f0 =
v
v
=
f
λ0
v − vs
(7.85)
The Doppler Effect is also observed when the source is stationary, and the observer is
moving towards it. To determine the frequency detected by the observer we have to
calculate how many cycles of the wave reaches him or her in a unit of time. (You may
imagine this by following the crests of the sound waves. The detected frequency is the
number of crests reaching the observer in a unit of time.) Assume that the velocity of
λ
the wave is v and its wavelength is λ. The wave requires T = time to travel λ distance.
v
1
v
Therefore the frequency detected by a stationary observer is f =
= . (If a crest
T
λ
1
reaches the observer every T time, the frequency is f = .) But when the observer is
T
vo
oscillations
moving towards the source with a vo velocity it will encounter a further
λ
in the same unit of time. Therefore the frequency detected by the moving observer is
f 00 =
v + vo
v vo
+
=f
λ
λ
v
(7.86)
(7.85) and (7.86) can be combined into a single formula:
fd = f
v + vo
v − vs
(7.87)
where f is the frequency emitted by the source, fd is the frequency detected by the
observer, v is the speed of sound, vo is the velocity of the observer and vs is the velocity
of the source, respectively.
The phenomenon has widespread practical applications. Speed traps are important
tools in the hands of the police to enforce speed limits. Doppler radars may measure
wind speed. Modern ultrasonic imaging devices may utilise this principle to measure
blood flow in our veins. Even the expansion of the universe was detected by the red shift
in the spectrum of distant stars due to the Doppler Effect. (It must be noted however,
that the formula for electromagnetic waves is different form (7.87) due to relativistic
effects.)
118
Chapter 8
First law of thermodynamics and
related subjects - György Hárs
The subject of study is the thermodynamic system, which is separated from the environment by the boundary. The boundary does not allow material transport between the
environment and the system so the mass contained is constant. In general the boundary allows transfer of heat and mechanical work. The heat transfer may be inhibited
by insulation and so the system is called “thermally isolated” or in other words “adiabatic” system. The mechanical interaction may also be excluded by using rigid boundary,
which system is called “mechanically isolated”. If both thermal and mechanical isolation
is active then the system is considered “isolated” system.
The thermodynamic system contains gas in this chapter with definite primary state
parameters such as pressure (p), temperature (T ) and volume (V ). These parameters
are considered primary concepts with no further definition in the phenomenological discussion of the subject. The pressure and the temperature are intensive, driving force-like
parameters. The volume is extensive, quantity-like parameter. Microphysical substantiation of the state parameters is provided by the kinetic theory of gases.
8.1
Ideal gas equation
The ideal gas equation is an empirical law with units to be substituted as follows:
pV = N kT p [P a] V m3
N [1] T [K] k = 1.36 · 10−23 J/K
(8.1)
pV =
m
RT
M
p = nkT
p [P a]
p [P a]
V m3
n 1 m3
m [kg]
T [K]
k = 1.36 · 10−23 J/K
M [kg/mol]
119
T [K]
(8.2)
R = 8.3J/molK (8.3)
pV =
N
RT
A
p [P a]
V m3
N [1]
A = 6 · 1023 1/mol
T [K]
R = 8.3J/molK
(8.4)
One mol of material consists of 6 1023 pieces (Avogadro number A) of particles. This is
called the amount of substance. The molar mass (M ) is the mass of one mol substance,
which is equal with the same number of grams as the atomic mass number of the particle.
In System International the molar mass should be substituted in kilograms per mol. The
first and second version of the ideal gas law contain the Boltzmann constant (k), the
last two use the universal gas constant (R). The temperature is measured in (K) Kelvin
degrees which is the absolute temperature scale with the starting value at -273 degrees
Celsius. The following relations can be found between the constants: R = kA and
m
Nk = M
R
8.2
The internal energy of the gas (U )
The particles of the gas make chaotic motion in the container. This motion represents
kinetic energy which is considered the internal energy of the gas. The calculation is
carried out in the framework of the kinetic theory of gases with the following basic
assumptions.
ˆ The particles interact by elastic collisions with the chamber walls and with each
other.
ˆ No external potential is applied, so the total energy is kinetic.
ˆ Uniform spatial distribution is in the chamber.
ˆ There is no dedicated direction (isotropic structure).
Figure 8.1: Particles collide to the wall of the chamber
120
Particles bounce back from the wall of the chamber. Let us calculate the amount of
linear momentum transferred to the wall in a short dt time. The x direction is normal
to the wall, vx is the x component of the velocity. Half of those particles reach the wall
in dt time which are in vx .dt proximity of the wall, since the other half moves opposite
direction from the wall. Multiplying this thickness with the half of the density and the
area, the total colliding particles result. The mass of the particle is denoted µ. The
variation of momentum is 2µvx due to bouncing back with negative vx . So ultimately
the total momentum (dI ) transferred to the wall is as follows:
dI =
1
(vx dt · A · n) 2µvx
2
dI = A · nµvx2 · dt
(8.5)
(8.6)
The time derivative of the momentum is the force:
dI
= F = A · nµvx2
dt
(8.7)
The pressure is the ratio of the force over the area:
F
= p = nµ · vx2
A
(8.8)
p = nµ · vx2
(8.9)
p = nµ · vy2
(8.10)
p = nµ · vz2
(8.11)
This is true in all three directions.
Let us summarize. The root means square velocity is the Pythagorean sum of the component velocities:
2
3p = nµ · vrms
(8.12)
1
2
p = nµ · vrms
3
(8.13)
The pressure can be expressed:
The ideal gas equation can also be taken into consideration:
p = nkT
121
(8.14)
The right hand sides of the equations are equal:
1
2
nµ · vrms
= nkT
3
(8.15)
After some ordering the kinetic energy of the particle is revealed:
3
1
2
= kT
µ · vrms
2
2
(8.16)
In the argument above the kinetic energy of the particle was associated with three translational coordinates, which represent three thermodynamic freedom degrees. The conclusion demonstrates the principle of equipartition. According to this principle, all kinds
of energy storage capabilities contain equal amount of energy which is one half kT. In
reality the particles may also rotate in three normal directions, so maximum six thermodynamic freedom degrees are at disposal. If in general case the thermodynamic freedom
degree is denoted f the total kinetic energy stored in one particle is as follows:
εkin =
f
kT
2
(8.17)
If one considers N pieces of particles the value of the internal energy (U ) results.
U = N εkin =
f
kT · N
2
(8.18)
m
Let us use the relation expressed earlier: N k = M
R
With this the expression of the internal energy can be transformed:
U=
mf
RT
M2
(8.19)
Let us introduce the molar heat capacitance of constant volume with the following definition:
CV =
f
R
2
So ultimately the internal energy of the gas is revealed:
m
U=
CV T
M
(8.20)
(8.21)
The internal energy is a secondary state parameter of the gas, which means that it is
unambiguously generated from a primary state parameter (concretely from the temperature). The final parameter is T2 the initial is T1 . The variation of the internal energy
at finite and infinitesimal temperature variations is as follows:
m
m
∆U =
CV · (T2 − T1 ) dU =
CV · dT
(8.22)
M
M
122
8.3
The p-V diagram
The p-V diagram is a phase-plane on which each point represents a state of thermodynamic equilibrium. The horizontal axis is the volume the vertical is the pressure. Due
to the ideal gas law points of identical temperature are located on hyperbolas which
are called isotherms. The higher the temperature the further is the hyperbola from the
origin. The thermodynamic process which progresses reasonably slowly to reach equilibrium state at all times is called reversible process, provided no heat dissipation happened
due to friction. Therefore the process can be displayed by a solid line on the p-V diagram. Rapidly occurring irreversible processes (irreversible jumps) with finite variations
do not reach equilibrium during the process they reach equilibrium in the final state only.
State parameters such as p and T are defined for equilibrium, so these parameters are
non-existing during irreversible jump. A doted line from the starting point to the final
point shows the jump on the p-V plane.
8.4
Expansion work of the gas
If the volume of a thermodynamic system increases then the system carries out work on
the environment. This is called the work of the gas which is a positive number upon
expansion. The infinitesimal amount of work can be expressed as follows: dW = pdV If a
finite expansion process is concerned the work of the gas is the integral of the infinitesimal
contributions.
Figure 8.2: p-V diagram on work of expansion
ZV2
Wgas =
pdV
V1
123
(8.23)
The work of gas (Wgas ) can be viewed as the area under the curve in the p-V diagram.
The advantage is that the integral is positive or negative simultaneously with the work
of the gas.
8.5
First law of thermodynamics
The variation of internal energy can be caused by two effects. Either heat transfer (∆Q)
or external mechanical work (∆W ) can change the internal energy of the gas. The
transferred heat is obviously positive if the system accepted heat. The mechanical work
can either be external work, which is positive upon compression and work of the gas,
which is positive upon expansion. One has to be careful not to make confusion. From
practical point of view it is advised to insist on using the work of the gas.
The first law of thermodynamics is essentially the conservation of energy. Most
practical forms for infinitesimal and finite cases are as follows:
m
CV · dT + pdV
(8.24)
δQ =
M
m
∆Q = ∆U + ∆W =
CV · (T2 − T1 ) +
M
ZV2
pdV
(8.25)
V1
The above described form of the first law can be interpreted as follows: The transferred
heat (∆Q) is utilized for two purposes. Partly the internal energy (∆U ) of the system is
increased and partly the gas carried out mechanical work (∆W ) on the environment.
sectionThermodynamic processes
In this section reversible processes are discussed. Let us transform the ideal gas
equation in the following manner:
m
pV
=
R
T
M
(8.26)
If the amount of the gas does not change during a process the right hand side of the
above equation is constant.
pV
= Const
T
(8.27)
p2 V2
p1 V 1
=
T1
T2
(8.28)
Accordingly can be written:
This is called the united law of gases. Historically it is worth to mention the more specific
laws which date back to those scientists who discovered the actual laws.
124
8.5.1
Isochoric process
Here the volume is constant during the process. The volume cancels out from the united
law of gases:
p1
p2
=
T1
T2
(8.29)
This is the Gay-Lussac II. law.
The work of the gas is zero since there is no change in the volume (∆W =0 ). The
internal energy of the gas is increased by the total amount of the heat transferred.
Figure 8.3: p-V diagram of the isochoric process
∆Q = ∆U =
m
CV (T2 − T1 )
M
(8.30)
This equation explains the why the notation CV is used. In isochoric process this quantity
shows up as molar heat capacitance of the gas.
In any processes however the variation of the internal energy can be expressed by
means of CV with the above formula.
8.5.2
Isobaric process
Here the pressure is constant during the process. The pressure cancels out from the
united law of gases:
V1
V2
=
T1
T2
(8.31)
This is the Gay-Lussac I. law. (The name with hyphen corresponds to one person.)
125
The work of expansion is the mere product of the pressure and the change in the
volume.
∆W = p(V2 − V1 )
(8.32)
Let us use the ideal gas equation.
pV2 =
m
RT2
M
pV1 =
m
RT1
M
(8.33)
The difference of the above equations is as follows:
p(V2 − V1 ) =
m
R(T2 − T1 )
M
(8.34)
Now the first law can be used:
Figure 8.4: p-V diagram of the isobaric process
∆Q = ∆U + ∆W =
m
m
m
m
CV · (T2 − T1 ) + R(T2 − T1 ) =
(CV + R) · (T2 − T1 ) =
Cp · (T2 − T1 )
M
M
M
M
(8.35)
Accordingly the molar heat capacitance in case of isobaric process is denoted Cp .
Cp = CV + R
8.5.3
(8.36)
Isothermal process
Here the temperature is constant during the process. The temperature cancels out from
the united law of gases:
p1 V1 = p2 V2
126
(8.37)
This is the Boyle-Mariotte law. (The name with hyphen corresponds to two persons who
independently discovered this law.)
In here the internal energy is constant (∆U =0 ). The pressure is expressed from the
ideal gas equation.
p=
1 m
RT
V M
(8.38)
The work of the gas can be calculated by integrating the pressure between the border
volumes.
Figure 8.5: p-V diagram of the isothermal process
ZV2
∆W =
ZV2
pdV =
V1
m
1 m
RT · dV =
RT
V M
M
V1
ZV2
m
V2
dV
= RT ln( )
V
M
V1
(8.39)
V1
∆W =
m
V2
RT ln( )
M
V1
(8.40)
The first law looks like as it follows: ∆Q = ∆W
According to this result the total heat transferred has been turned to mechanical
work.
8.5.4
Adiabatic process
It needs to be emphasized that reversible adiabatic process what we are dealing with. The
word adiabatic on it own means a process without heat exchange with the environment
however the mechanical work can be transferred to the system. In contrast to this neither
heat nor mechanical work can be transferred to the system which is said to be isolated.
127
Since there is no heat exchange with the environment (δQ =0 ), the infinitesimal form
of the first law can be written as follows:
m
CV · dT + pdV = 0
(8.41)
M
Earlier the constant volume molar heat capacitance has already been introduced: CV =
f
R
2
mf
After substitution one can find: M
R · dT + pdV = 0
2
m f
R · dT + pdV = 0
M
2
(8.42)
On the other hand consider the temperature derivative of the ideal gas equation.
pV =
m
RT
M
d(pV ) m =
R
dT
M
(8.43)
(8.44)
The right hand side of the last equation can be recognized in the equation above in
parenthesis. Let us substitute it.
d(pV ) f
· dT + pdV = 0
dT 2
(8.45)
f
d(pV ) + pdV = 0
2
(8.46)
Here dT cancels out.
The variation of the product can be separated.
f
f
V dp + pdV + pdV = 0
2
2
(8.47)
f
f
− V dp = pdV + pdV
2
2
(8.48)
−f V dp = (f + 2)pdV
(8.49)
Here we introduce the adiabatic exponent denoted by the Greek letter (kappa κ).
−V dp =
f +2
pdV
f
128
f +2
=κ
f
(8.50)
−V dp = κ · pdV
(8.51)
This is a separable differential equation:
−
dp
dV
=κ·
p
V
(8.52)
Let us integrate from the initial value to the final value:
Zp2
−
(8.53)
p2
V2
) = κ ln( )
p1
V1
(8.54)
p1
− ln(
ZV2
dV
V
dp
=κ·
p
V1
p1
ln( ) = ln
p2
V2
V1
κ
(8.55)
Since the natural logarithm function is unambiguous so the arguments are equal.
κ
V2
p1
=
(8.56)
p2
V1
p1 V1κ = p2 V2κ
(8.57)
In general can be written the exponential equation of the adiabatic process:
pV κ = Const
Figure 8.6: p-V diagram of the adiabatic process
129
(8.58)
The equation of the adiabatic process can also be expressed by means of other primary
state parameters as well. Let us combine it with the universal gas law.
pV κ = Const
pV
= Const
T
(8.59)
After dividing the equations p cancels out:
T · V κ−1 = Const
(8.60)
Express the V from the second equation and substitute back to the first one:
Tκ
= Const
pκ−1
8.6
(8.61)
Summary of the molar heat capacitances
The molar heat capacitances have been used extensively at the discussion of the thermodynamic processes. The following parameters have been discussed: (R =8.3 J/molK )
The thermodynamic freedom degree is the number of the energy storing capabilities
each of which can contain kT kinetic energy, according to the equipartition principle. The
noble gases with mono-atomic molecule have three freedom degrees, which are the x, y, z
translational directions. The diatomic molecules such as the air and lot of others have
the three x, y, z translations and two additional rotational axes except for one direction
which connects the centers of the atoms. The exception is explained by the fact that
looking at the molecule in the central line it looks like a point without extension, thus
it lacks of moment of inertia. Higher number of atoms in the molecule generates six
thermodynamic freedom degrees since three translational and three rotational motions
are all available.
Number
atoms in
molecule
1
2
3 or more
of Thermodynamic Constant volthe freedom degree ume molar heat
f
capacitance
CV = f2 R
3
3
R
2
5
5
R
2
6
3R
130
Constant pres- Adiabatic exposure molar heat nent
f +2
capacitance =κ
f
f
Cp = 2 + 1 R
5
5
R
2
3
7
7
R
2
5
4
4R
3
8.7
The Carnot cycle
The Carnot cycle is a reversible cycle which consists of two isothermal and two adiabatic
processes. This is the classical example of the heat engine which is capable of generating
mechanical work from a certain part of the heat.
Figure 8.7: The Carnot cycle in p-V diagram
The corner points of the cycle are denoted A, B, C and D. The points A and B are
located on the T1 isotherm while the C and D points are located on T2 isotherm. The
T1 temperature is higher than T2 therefore T2 isotherm is closer to the origin of the p-V
plane. The process starts in point A.
ˆ AB section: (work output is positive in this section)
This is an isothermal expansion on T1 temperature. The heat intake from the heat bath
is ∆Q1 . The variation of the internal energy is zero (∆U =0 ) therefore the transferred
heat is fully converted to mechanical work ∆W1 .
∆W1 = ∆Q1 =
m
VB
RT1 ln( )
M
VA
(8.62)
ˆ BC section: (work output is positive in this section)
Adiabatic expansion (∆Q =0 ) as long as the temperature cools down to T2 temperature.
The gas carries out positive mechanical work on the expense of its own internal energy.
m
∆W2 = −∆U2 =
CV (T1 − T2 )
(8.63)
M
131
ˆ CD section: (work output is negative in this section)
Here an isothermal compression takes place on T2 temperature. The variation of the
internal energy is zero (∆U =0 ) therefore mechanical work ∆W3 is fully converted to
heat ∆Q2 which in turn has been transferred to the sink.
∆W3 = ∆Q2 =
VD
m
RT2 ln( )
M
VC
(8.64)
ˆ DA section: (work output is negative in this section)
Adiabatic compression (∆Q =0 ) as long as the temperature warms up to T1 temperature.
The mechanical work increased the internal energy of the gas.
∆W4 = −∆U2 =
m
CV (T2 − T1 )
M
(8.65)
The total amount of work carried out by the cycle ∆W is the sum of the mechanical
works of the processes above: Apparently ∆W2 and ∆W4 cancel out.
∆WCycle = ∆W1 + ∆W2 + ∆W3 + ∆W4 = ∆W1 + ∆W3
∆WCycle
VB
m
VD
R T1 ln( ) + T2 ln( )
=
M
VA
VC
(8.66)
(8.67)
The above formula can be simplified further by considering the fact that the BC and DA
section are adiabatic processes. According to the laws of adiabatic process the following
equations can be written: (See at the end of the section “adiabatic process”.)
T1 · VAκ−1 = T2 · VDκ−1
(8.68)
T1 · VBκ−1 = T2 · VCκ−1
(8.69)
After dividing the two equations the temperatures cancel out on both sides:
VD
VA
=
VB
VC
This can be substituted to the expression of the work of the cycle.
m
VB
VB
m
VB
∆WCycle =
R T1 ln( ) − T2 ln( ) =
R(T1 − T2 ) ln( )
M
VA
VA
M
VA
(8.70)
(8.71)
The work of the cycle can be viewed as the unusual rectangular area on the p − V plane.
If the direction of the circumference is clockwise then the cycle carries out a positive
work on the environment. In other words this is a heat engine.
132
The thermal efficiency of the heat engine (η) is defined as the ratio of the work of
the cycle (Wcycle ) over the heat intake from the heat bath (∆Q1 ). In other words this is
that percentage of the heat intake which has been converted to mechanical work during
the cycle. The rest of the heat intake necessarily goes to the sink.
Wcycle
η=
=
∆Q1
m
R(T1 − T2 ) ln( VVBA )
M
m
RT1 ln( VVBA )
M
=
T1 − T2
T1
(8.72)
Many terms cancel out.
η =1−
T2
T1
(8.73)
The work of the cycle can be expressed in an alternative way as well:
∆WCycle = ∆W1 + ∆W3 = ∆Q1 + ∆Q2
(8.74)
The efficiency is as follows:
η=
∆Q1 + ∆Q2
∆Q2
=1+
∆Q1
∆Q1
(8.75)
The two expressions of the efficiency are equal:
1+
T2
∆Q2
=1−
∆Q1
T1
(8.76)
After some ordering an important formula is the result.
∆Q1 ∆Q2
+
=0
T1
T2
(8.77)
The two terms of the above equation are called “reduced heat”, which are the ratio of the
heat and the actual temperature at which the heat transfer took place. This equation
declares the fact that the sum of the reduced heats for the Carnot cycle is zero. This has
far reaching consequences in connection with the concept of entropy.
133
Chapter 9
The entropy and the second law of
thermodynamics - György Hárs
At the end of the previous chapter an important relation has been revealed.
∆Q1 ∆Q2
+
=0
T1
T2
(9.1)
The two terms of the above equation are called “reduced heat”, which are the ratio of the
heat and the actual temperature at which the heat transfer took place. The “reduced
heat condition” declares the fact that the sum of the reduced heats for the Carnot cycle
is zero. This has far reaching consequences in connection with the concept of entropy.
9.1
The entropy
Consider the p-V phase plane. Let us draw in solid line numerous isotherms in equidistant
small temperature steps. Similarly draw in doted line numerous adiabatic curves with
small equidistant increments. Now an arbitrary reversible cycle is plotted on the top of
the isotherm and adiabatic grid.
The cycle on the figure can be approximated with little motions partly on the isotherm
and partly on the adiabatic curves. This way the arbitrary cycle is composed of several
tiny Carnot cycles. The reduced heat condition is valid for each of them. The reduced
heat intake at the high temperature side for a tiny Carnot cycle is equal to the reduced
heat egress at the low temperature side. Therefore the reduced heat contributions cancel
out by pairs, thus the total sum of the reduced heats is equal to zero for the whole
arbitrary cycle. If one makes the grid infinitely fine the sum of reduced heats is converted
134
to the closed loop integral. Accordingly this can be written:
I
dQ
=0
T
Arb.
cycle
(9.2)
Figure 9.1: Arbitrary cycle with grid
If the arbitrary closed loop integral is zero this is equivalent with the statement that
the integral between two points does not depend on the path of the integration.
Figure 9.2: Integration on two alternative paths
Let us break the closed loop with two points A and B. Point A will be considered
135
the starting point while B is the final point.
 B

 A

I
Z
Z
dQ 
dQ  dQ 
=
+
=0
T
T
T
A
B
P ath1
P ath2
Arb.
cycle
Integrating from B to A is the negative of the opposite direction integration.
 B

 B

Z
Z
dQ 
 dQ 
=
T
T
A
A
P ath1
(9.3)
(9.4)
P ath2
The integral depends on the initial and final points only.
Since the integral is independent of the path therefore the concept of entropy (S)
can be introduced. Variation of the entropy can be calculated on the most convenient
path, since the actual path is unimportant. First an isochoric process takes us to the
intermediate point (X) then an isothermal process reaches the final point (B).
Figure 9.3: Calculation of entropy variation
In isochoric process dQ = dU =
ZX
∆S1 =
m
C dT
M V
dQ
m
=
CV
T
M
A
therefore the following integral is evaluated
ZX
dT
m
TB
=
CV ln
T
M
TA
(9.5)
A
After this an isothermal process follows: dQ = pdV .
ZB
∆S2 =
X
dQ
=
T
ZB
X
136
pdV
=
T
ZB
X
p
dV
T
(9.6)
Here we use the ideal gas equation in the following form:
p
m 1
=
R
T
M V
ZB
∆S2 =
m 1
m
R dV =
R
M V
M
X
ZB
(9.7)
dV
m
VB
=
R ln
V
M
VA
(9.8)
X
So the total entropy variation is the sum of the two sub-processes:
SB − SA =
m
TB
m
VB
CV ln
+ R ln
M
TA M
VA
(9.9)
The entropy variation can be expressed by any two of the three (p, V, T ) primary state
parameters. We have made the (T, V ) version. The (p, V ) and the (p, T ) versions follow.
The ideal gas equation for the final and the initial states are as follows:
m
pB VB =
RTB
(9.10)
M
pA V A =
m
RTA
M
(9.11)
Let us divide them.
pB V B
TB
·
=
pA VA
TA
(9.12)
Take the natural logarithm of both sides:
ln
pB
VB
TB
+ ln
= ln
pA
VA
TA
(9.13)
The above equation can substituted into the formula of the entropy variation.
pB
VB
m
VB
m
VB
m
pB
m
CV ln
+ ln
+ R ln
=
(CV + R) ln
+ CV ln
SB − SA =
M
pA
VA
M
VA
M
VA M
pA
(9.14)
m
m
Accordingly: SB − SA = M
Cp ln VVBA + M
CV ln ppBA
Similarly ln VVBA = ln TTBA − ln ppBA
Now the above equation is substituted to the entropy variation:
SB − SA =
m
TB
m
TB
pB
m
TB
m
pB
CV ln
+ R(ln
− ln ) =
(CV + R) ln
− R ln
M
TA M
TA
pA
M
TA M
pA
(9.15)
137
m
m
Cp ln TTBA − M
R ln ppBA
Accordingly: SB − SA = M
Direct physical meaning can be associated to the variation of entropy between two
points of the p − V plane. By choosing a universal initial point as reference point, the
entropy function can be converted to a secondary state parameter. This time any point
on the p − V plane is characterized by a single entropy value. Practical choice can be
the following reference point. (T0 =273K, p0 =105 Pa, V0 = (m/M ) 2.266 10−2 m3 /mol).
9.2
The isentropic process
Let us find out where those points are located on the p − V plane, which have identical
entropy value. In other words we are looking for the isentropic curves. Here SA = SB is
the condition. The entropy variation is expressed as follows:
SB − SA =
VB
pB
m
m
Cp ln
+ CV ln
=0
M
VA M
pA
Cp ln
VB
pB
+ CV ln
=0
VA
pA
pA
Cp VB
ln
= ln
CV
VA
pB
(9.16)
(9.17)
(9.18)
Here we recall the adiabatic exponent (κ) which has already been introduced in chapter
8 as the ratio of the molar heat capacitances.
κ
VB
pA
=
(9.19)
VA
pB
pA VAκ = pB VBκ
(9.20)
This is a known formula, which was derived first at the discussion of the reversible
adiabatic process in the previous chapter. Now it has been proven that the isentropic
process is identical with the reversible adiabatic process.
9.3
The microphysical meaning of entropy
Let us make some elementary combinatorial calculations to reveal the fundamental roots
of entropy. We have two compartments (1 and 2) and four particles (N =4 ) with the
names a, b, c and d. From the point of view of micro states the particles are distinguishable, which means that a new micro state is generated if two particles replace each
138
other. However these micro states represent the same macro state since in macro state
only the number of the particles count. Accordingly, from point of view of macro states
the particles are indistinguishable. The concept of thermodynamic weight is associated
with the macro state and it is defined as the number of those micro states which generate
the given macro state. In the table below all the possible micro states are listed as well
as the corresponding thermodynamic weights.
Observable
macro state
A
B
C
D
E
Macro state charac- Names of particles
teristics
in compartment 1.
(N1 , N2 )
(Micro states)
4, 0
abcd
3, 1
bcd, acd, abd, abc
2, 2
ab, cd, ac, bd, ad, bc
1, 3
a,b,c,d
0, 4
-
Number of micro states
(Thermodynamic weight)
W = N1N!N! 2 !
1
4
6
4
1
In present case the thermodynamic weight (W ) can be expressed as follows:
W =
N!
N1 !N2 !
N = N1 + N2
(9.21)
In general case there are several elementary cells. The number of cells is denoted with k.
A series of numbers provide the characteristics of the macro state N1 , N2 , N3 , N4 , N5 .........Nk
where the numbers represent the numbers of particles in the elementary cells from 1 to
k. The corresponding thermodynamic weight is as follows:
W =
N!
N1 !N2 !N3 !.............Nk !
N = N1 + N2 + ......Nk
(9.22)
In terms of combinatorics the above formula is a permutation with repetition. In the
numerator the total number of permutation shows up provided all the particles are
distinguishable. In the denominator there is the divider that contains those cases which
differ in the order of particles within the cell. These cases do not constitute new macro
state since only the amount of particles in the cell counts.
9.4
Gay-Lussac experiment
The gas subjected to an isolated irreversible jump in this classical experiment. The
phenomena will be studied by both phenomenological and statistical thermodynamics.
139
9.4.1
Phenomenological approach
Consider a gas container which consists of two internal compartments with volumes V1
and V2 . Between the volumes there is a valve that can be operated remotely. The whole
container is thermally isolated from the environment. One of the volumes (V1 ) contains
gas with known parameters. Vacuum is in the other compartment (V2 ). Suddenly the
valve is opened the gas flows partly to the empty volume. During the gas flow a whistle
noise can be heard. Thermometers attached to both compartments show that V1 cooled
down and V2 warmed up by some extent. When the noise died out and some minutes
passed both thermometers show the initial temperature, in accordance with the first law
of thermodynamics. No work has been done by the gas no heat has been transferred to
the system so the total internal energy must have been conserved. This means that the
temperature is necessarily unchanged.
Though the temperature is identical with that in the initial state, this is far of being
an isothermal process. This is an isolated irreversible jump without work done. The
initial and the final states are in thermal equilibrium therefore drawing these two points
on the p − V plane is justified. However due to the irreversible jump it is not is not
allowed to draw any solid line much a rather a doted line from the initial to the final
point.
Figure 9.4: Gay-Lussac experiment on the p − V plane
Since the entropy is a state parameter the variation of entropy can be expressed
without any respect to the events which took the system from the initial to the final
state (here I did not use the word “process” intentionally). The known formula from the
140
previous page can readily be used with the condition that VA = V1 and VB = V1 + V2 .
SB − SA =
m
TB
m
VB
CV ln
+ R ln
M
TA M
VA
(9.23)
The first term cancels out due to the identical temperatures values.
SB − SA =
VB
m
R ln
M
VA
(9.24)
The final volume is obviously bigger than the initial therefore the entropy variation is
positive. The irreversible jump takes place necessarily to the right hand side direction
to a higher entropy value adiabatic curve. In addition the initial and final states are on
the same isotherm.
9.4.2
Statistical approach
In the figure below the gas container of the experiment is shown. Let us subdivide the
total VB volume to very small identical cells with the volume denoted with δ. The number
of cells (ωA and ωB ) comes out as follows:
ωA =
VA
δ
ωB =
VB
δ
(9.25)
Figure 9.5: Gay-Lussac experiment (statistical approach)
The actual emerging macro states carry the highest thermodynamic weight. In these
macro states all the cells contain the same amount of particles. Here N is the total
141
number of particles. The initial and final macro states are the following:
N N N N ,
,
...................
, 0|ωA +1 , 0|ωA +2 ................... 0|ωB −1 , 0|ωB (9.26)
ωA 1 ωA 2 ωA 3
ωA ωA
N N N N ,
,
...........................
ωB 1 ωB 2 ωB 3
ωB ωB
(9.27)
The corresponding thermodynamic weights can be readily written:
WA = h
N!
iωA
N
!
· [0!](ωB −ωA )
ωA
WB = h
N!
iωB
N
!
ωB
(9.28)
It is worth to be are aware of the fact that 0! =1. Let us calculate the ratio of the
thermodynamic weights.
h iωA
N
!
ωA
WB
= h iωB
(9.29)
N
WA
!
ωB
An approximate formula for the natural logarithm of the factorial will be used here:
ln n! ≈ n ln n − n
(9.30)
The relative error of this formula is tending to zero when n goes to the infinity. In present
case when n is extremely high the approximation is especially accurate. Sketch of proof
follows at the end of this section.
WB
N
N
N
N
N
N
= ωA
ln( ) −
− ωB
ln( ) −
(9.31)
ln
WA
ωA
ωA
ωA
ωB
ωB
ωB
WB
ln
=
WA
N
N
N
N
N ln( ) − N − N ln( ) − N = N ln( ) − ln( )
ωA
ωB
ωA
ωB
ln
WB
ωB
VB
= N ln
= N ln
WA
ωA
VA
(9.32)
(9.33)
Multiply the equation with the Boltzmann constant.
k ln
WB
VB
= N k ln
WA
VA
142
Nk =
m
R
M
(9.34)
k ln
m
VB
WB
=
R ln
WA
M
VA
(9.35)
Compare it with the variation of entropy stated earlier:
SB − SA =
m
VB
R ln
M
VA
(9.36)
The right hand sides match perfectly so the left hand sides are equal:
SB − SA = k ln
WB
WA
(9.37)
The above formula is called the Boltzmann equation. Now this has been proven for the
special case of the Gay-Lussac experiment QED.
9.5
The Boltzmann equation
The Boltzmann equation is a fundamentally significant equation which establishes the
bridge between phenomenological and the statistical approach of the thermodynamics.
Its importance is far greater than that it has just been proven for.
SB − SA = k ln
WB
WA
(9.38)
Here k is the Boltzmann constant (k = 1.36 · 10−23 J/K), WA and WB are the thermodynamic weights of the A and B macro states. The message of Boltzmann equation in
connection with the Gay-Lussac experiment is clear: The entropy increases according to
the phenomenological formula. Therefore the variation of entropy is positive so the logarithm value in the Boltzmann equation should also be positive. Accordingly the ratio of
thermodynamic weights is bigger than unit which means that far greater thermodynamic
weight belongs to the final macro state than to the initial macro state. The fact can be
generally stated that in the isolated system at irreversible jump the final equilibrium
state is located on a higher value isentropic curve on the p-V plane. The thermodynamic
weight of the final state is bigger than that of the initial. At irreversible jumps the system is tending to the state of highest possible thermodynamic weight which is the final
equilibrium macro state.
9.6
Approximate formula (ln n! ≈ n ln n − n) a sketch
of proof:
ln n! = ln 1 + ln 2 + ln 3 + ...... + ln n
143
(9.39)
The sum of logarithms of integers can be approximated with the following integral:
Zn
ln x · dx = [x ln x − x]x=n
x=1 = n ln n − n + 1 ≈ n ln n − n
(9.40)
1
Figure 9.6: Approximation of lnn! function. The crosshatched area is the absolute error
9.7
Equalization process
So far the subject of study was a single system consisting of gaseous particles. In equalization processes more thermodynamic systems take part, typically two systems will
be considered here. Initially these systems are in thermal equilibrium on their own.
Suddenly they are united and the equilibrium state of both systems vanishes. Violent
irreversible actions can proceed thus no equilibrium state parameters can be defined.
Strictly speaking temperature, pressure, internal energy and entropy are not existing
parameters in course of the events. After some time these actions are gradually relaxing
and the united system occupies its newly born equilibrium state parameters.
144
9.7.1
Equalization between gaseous components
Let us consider two separated gas systems. They contain the same type of gas with
the following equilibrium state parameters V1 , p1 , T1 and V2 , p2 , T2 .The gas systems are
united by removing the separating wall between them so the volumes are added. We
wonder the final state parameters and the variation of entropy.
The ideal gas equation will be used in the following form: pV = nRT Here n denotes
the number of moles.
Parameters
Volume
Internal energy
System1
n1 , p1 , V1 , T1
V1 = n1 R Tp11
U1 = n1 CV T1
System2
n2 , p2 , V2 , T2
V2 = n2 R Tp22
U2 = n2 CV T2
The internal energy, the volume, the number of moles and the entropy are added
together upon uniting the systems. The parameters without subscripts characterize the
united system.
U = U1 + U2 = CV (n1 T1 + n2 T2 ) = CV (n1 + n2 )T
(9.41)
The final temperature can be written accordingly:
n1 T1 + n2 T2
=T
n1 + n2
(9.42)
This is a weighted arithmetic mean in terms of mathematics where the weights are the
corresponding mole numbers.
Now the pressure of the united system needs to be found. This is expressed from the
ideal gas equation:
+n2 T2
(n1 + n2 )R n1nT11 +n
nRT
2
p=
=
V
R( np1 1T1 + np2 2T2 )
(9.43)
Two terms cancel out. The pressure finally can be expressed.
p=
n1 T1 + n2 T2
n1 T1
+ np2 2T2
p1
(9.44)
This is a weighted harmonic mean in terms of mathematics where the weights are the
corresponding nT products. For later use it is worth to express the T over p ratio in
145
the final state. This is a weighted arithmetic mean of the T /p ratios in the initial states
where the weights are the corresponding moles.
n1 Tp11 + n2 Tp22
T
=
p
n1 + n2
(9.45)
Now the entropy variations need to be dealt with. The term variation means the difference when the initial state parameters are subtracted from the final state parameters.
∆S1 = n1 Cp ln
T
p
− n1 R ln
T1
p1
∆S2 = n2 Cp ln
T
p
− n2 R ln
T2
p2
(9.46)
The total entropy variation of the equalization is the sum of the two variations above:
∆Seq = ∆S1 + ∆S2 = Cp (n1 ln
T
p
p
T
+ n2 ln ) − R(n1 ln + n2 ln )
T1
T2
p1
p2
T (n1 +n2 )
p(n1 +n2 )
∆Seq = Cp ln n1 n2 − R ln n1 n2 = (n1 + n2 ) Cp ln
T1 T2
p1 p2
T
p
− R ln
(n1 +n2 )
T1n1 T2n2
T
∆Seq = (n1 + n2 ) Cp ln (n +n p
+ R ln
1
2 ) T n1 T n2
1
2
p
(n1 +n2 )
pn1 1 pn2 2
(9.47)
p
p
(n1 +n2 )
pn1 1 pn2 2
(9.48)
!
p
(9.49)
The universal gas constant is factored out.
∆Seq = (n1 + n2 )R
Cp
ln
R
T
p
+ ln
(n1 +n2 )
T1n1 T2n2
p n n !
p1 1 p 2 2
p
(n1 +n2 )
(9.50)
For convenience reasons a new parameter is introduced here:
Cp
f
= +1=β
R
2
(9.51)
With this parameter the formula can be transformed:
T
∆Seq = (n1 + n2 )R β ln (n +n p
+ ln
1
2 ) T n1 T n2
1
2

∆Seq = (n1 + n2 )R ln
β
T
q
+ ln
(n1 +n2 )
βn1 βn2
T1 T2
146
p n n !
p1 1 p2 2
p
(9.52)
p n n 
p 1 1 p2 2 
p
(9.53)
(n1 +n2 )
(n1 +n2 )
!
Let us organize the fraction under a single logarithm.
!
r
p1 n1 p2 n2
Tβ
(n1 +n2 ) (
∆Seq = (n1 + n2 )R ln
) ·( β)
p
T1β
T2
(9.54)
The total variation of entropy due to the unification is finally revealed.


∆Seq = (n1 + n2 )R ln 
Tβ
p
(9.55)

q β
T1 n1
T2β n2
( p1 ) · ( p2 )
(n1 +n2 )
The newborn equilibrium parameters show up in the numerator while the initial parameters are in the denominator. The purpose is to prove the second law of thermodynamics
in mathematical precision. The second law states that the variation of entropy is always
positive at irreversible jumps of an isolated system.
Let us consider the numerator which is associated with the final state parameters. The
temperature over pressure has already been expressed above from the original formulae.
According to the well-known mathematical theorem weighted geometric mean is always not bigger than the weighted arithmetic mean.
s
T1
T2
β
n
+
n
1
2
T1
T
T2
T
p1
p2
= T (β−1) = T (β−1)
≥ T (β−1) (n1 +n2 ) ( )n1 ( )n2
(9.56)
p
p
n1 + n2
p1
p2
So replace the numerator with a smaller or equal quantity. This way the calculated
formula is inevitably diminished or unchanged.
q


T (β−1) (n1 +n2 ) ( Tp11 )n1 ( Tp22 )n2

q β
(9.57)
∆Seq ≥ (n1 + n2 )R ln 
T1 n1
T2β n2
(n1 +n2 )
( p1 ) · ( p2 )
Organize the fraction under a single root sign.
s
∆Seq ≥ (n1 + n2 )R ln T (β−1)
T1 p1 n1 T2 p2 n2
(
) (
)
p1 T1β
p2 T2β
(n1 +n2 )
!
(9.58)
Here the pressures cancel out.
s
∆Seq ≥ (n1 + n2 )R ln T (β−1)
(n1 +n2 )


∆Seq ≥ (n1 + n2 )R ln  (
1
(β−1)
T1
)n1 (
1
(β−1)
T2
!
)n2
(9.59)

T (β−1)

(β−1)  = (n1 + n2 )R(β − 1) ln
p
(n1 +n2 )
T1n1 · T1n1
T
p n
(n1 +n2 )
T1 1 · T1n1
(9.60)
147
!
Here f denotes the thermodynamic freedom degree which can be 3, 5 and 6.
!
f
f
T
β−1=
∆Seq ≥ (n1 + n2 )R( ) ln (n +n p
1
2 ) T n1 · T n1
2
2
1
1
Let us substitute the value of the final temperature to the numerator.
!
1
(n
T
+
n
T
)
1
1
2
2
f
2
p n
∆Seq ≥ (n1 + n2 )R( ) ln n1 +n
≥0
(n1 +n2 )
2
T1 1 · T1n1
(9.61)
(9.62)
In the numerator the weighted arithmetic mean of the initial temperatures is present.
In the denominator the weighted geometric mean of the same temperatures shows up
with the identical weights. Here again one has to refer to the mathematical theorem
of the inequality between the arithmetic geometric means. The numerator is surely not
smaller than the denominator so the fraction is not smaller than unit, which means
that the logarithm value is a non-negative number. So ultimately the second law of
thermodynamics has been proven for the concrete case of the isolated irreversible jump
at gas equalization. Q.E.D.
Figure 9.7: Isentropic curves are displayed
The curve of S1 + S2 entropy is jumped over by the isolated equalization
Fig. 9.7
148
Some special cases of the general entropy variation formula will be discussed here.
First the mole numbers of the initial components are equal. (n1 = n2 = n)


∆Seq = 2nR ln  q
Tβ
p
T1β
p1
·
T2β
p2
The initial temperatures can be equal too: (T1 = T2 )


∆Seq = 2nR ln  q
1
p
1
p1
·
1
p2
(9.63)

√
 = 2nR ln
p1 p2
p
(9.64)
The final pressure can be written for the actual conditions: This is the harmonic mean
of the initial pressures.
p=
n1 T1 + n2 T2
2
=
n1 T1
n2 T2
(1/p1 ) + (1/p2 )
+ p2
p1
This is substituted to the entropy variation above.
√
p1 p 2
∆Seq = 2nR ln
2
(1/p1 )+(1/p2 )
(9.65)
(9.66)
Here we should refer to the mathematical theorem that the harmonic mean is always
smaller or equal than the geometric mean. So the entropy variation is non-negative.
Finally the initial pressures are considered equal.


β
T
T
 = 2nRβ ln √
∆Seq = 2nR ln  q
(9.67)
T
·
T
β
β
1
2
T ·T
1
2
The final temperature can be written for the actual conditions:
T =
n1 T1 + n2 T2
1
= (T1 + T2 )
n1 + n2
2
f
∆Seq = 2nR( + 1) ln
2
1
2
(T1 + T2 )
√
T1 · T2
(9.68)
(9.69)
Here we should refer to the mathematical theorem that the arithmetic mean is always
greater or equal than the geometric mean. So the entropy variation is non-negative again.
149
9.7.2
Equalization of non-gaseous materials without phase transition
Heat equalization between liquid-solid and solid-solid components is studied mostly in
this point. In general the liquid-liquid equalization process is not part of this discussion
except for the case when both components are water for instance, when the mixing is
not associated with growth of entropy.
The product of mass and specific heat is the heat capacitance denoted capital C.
U1 = m1 c1 T1 = C1 T1
U2 = m2 c2 T2 = C2 T2
(9.70)
The internal energies are added together.
U = C1 T1 + C2 T2 = (C1 + C2 )T
(9.71)
The temperature after equalization is denoted T without subscript.
T =
C1
C2
C1 T1 + C2 T2
=
T1 +
T2
C1 + C2
C1 + C2
C1 + C2
(9.72)
Now the entropy variation is studied. The infinitesimal heat transfer can be used to
calculate the entropy variation. The actual temperature of the heat transfer is denoted
T∗ .
dQ = dU = C · dT∗
ZT
∆S1 =
dQ
=
T∗
T1
ZT
∆S2 =
T2
ZT
C1 · dT∗
= C1
T∗
T1
dQ
=
T∗
ZT
ZT
(9.73)
T
dT∗
= C1 ln
T∗
T1
(9.74)
T
dT∗
= C2 ln
T∗
T2
(9.75)
T1
C2 · dT∗
= C2
T∗
T2
ZT
T2
The total entropy variation of the equalization is the sum of the two components.
T
T
C1
T
C2
T
∆Seq = ∆S1 + ∆S2 = C1 ln
+ C2 ln
= (C1 + C2 )
ln
+
ln
T1
T2
C1 + C2 T1 C1 + C2 T2
(9.76)
After mathematical transformations this can be written:
!
( C C+C
( C C+C
1
2
)
)
1
2
1
2
T
T
∆Seq = (C1 + C2 ) ln
+ ln
T1
T2
150
(9.77)
The final result is as follows:


T
∆Seq = (C1 + C2 ) ln 
C1
( C +C
)
1
2
T1
C2
)
1 +C2
(C
· T2

Let us substitute the final temperature calculated earlier:


C2
C1
T
+
T
1
C1 +C2 2 
∆Seq = (C1 + C2 ) ln  C1 +CC21
≥0
C2
( C +C )
( C +C
)
1
2
1
2
T1
· T2
(9.78)
(9.79)
In the numerator of the fraction the weighted arithmetic mean of the initial temperatures
can be found, in the denominator the weighted geometric mean of the same temperatures
show up with the identical weights. Now the mathematical theorem of the inequality
between the arithmetic and geometric mean needs to be referred to. According this
theorem the arithmetic mean is never smaller than the geometric mean. Ultimately
the fraction is never smaller than unit which means that the entropy variation of the
equalization process is never negative. The second law of thermodynamics has been
proven for this case. Q.E.D.
The non-negative result of the entropy variation can also be understood by verifying
an infinitesimal entropy transfer from the warmer component to the colder one. The
warmer component carried out a dQ heat egress at T1 temperature while the colder
component took up the same dQ heat at a lower T2 temperature. The infinitesimal
entropy egress dS 1 and intake dS 2 can be written as follows:
T1 ≥ T2
dS1 =
−dQ
T1
dS2 =
dQ
T2
(9.80)
The total infinitesimal entropy production is the sum of the formulae above:
dSeq = dS1 + dS2 = dQ(
1
1
T1 − T2
− ) = dQ
≥0
T2 T1
T1 · T2
(9.81)
The difference in the numerator is obviously non-negative thus the original statement is
proven.
9.7.3
Ice cubes in the water
The next problem contains phase transition which is associated with latent heat of melting. This is a known parameter of any material. The actual latent heat (Qlat ) is calculated
as a product: (Qlat = mLmelt ). The entropy variation is the ratio of the latent heat over
melt
the melting point temperature in Kelvin degrees. ∆Smelt = mL
Tmelt
Frequent practice is to cool down the drinks by means of ice cubes.
151
We have a glass of water (m1 = 0.3 kg) at room temperature (T1 = 27 o C = 300K).
Three ice cubes (m2 = 0.03 kg, T 2 =-13 o C = 260K) are dropped into the water. After
the cubes melted we want to find the final temperature and to determine the variation
of the entropy of the system.
The chosen reference state is water at zero Celsius (T0 = 270K). The internal energies
(U1 , U2 )are calculated relative to the reference state.
U1 = m1 cw (T1 − T0 )
(9.82)
U2 = −m2 L − m2 cice (T0 − T2 )
(9.83)
The total internal energies of the initial components are equal with that of the final state.
U1 + U2 = m1 cw (T1 − T0 ) − m2 [L + cice (T0 − T2 )] = (m1 + m2 )cw (T − T0 )
After the system got into equilibrium the new temperature (T ) is formed.
m2
cice
L
m1
(T1 − T0 ) −
+
(T0 − T2 )
T = T0 +
m1 + m2
m1 + m2 cw
cw
(9.84)
(9.85)
The numerical constants are as follows:
cw = 4.187 · 103
J
kgK
cice = 2.108 · 103
cice
= 0.503
cw
J
kgK
L = 3.34 · 105
L
= 79.8K
cw
J
kg
(9.86)
(9.87)
After substitution the final temperature is revealed:
T = 273 +
300
30
27 −
[79.8 + 0.503 · 13] = 289.8K
330
330
(9.88)
The final temperature is 289.8 K which is 16.8 Celsius. So the drink is cooler by 10.2 o C.
The entropy variation of the water in the glass (∆S1 ) is negative.
∆S1 = m1 cw ln
T
T1
(9.89)
The entropy variation of the ice cubes (∆S2 ) is positive.
∆S2 = m2 cice ln
T0 m2 L
T
+
+ m2 cw ln .
T2
T0
T0
152
(9.90)
The entropy production of the equalization is the sum of the variations:
T0 m2 L
T
T
∆Seq = ∆S1 + ∆S2 = m2 cice ln
+
+ m1 cw ln m2 cw ln
(9.91)
T2
T0
T1
T0
After some transformations can be written as follows:
L
m1
T
m2
T
T0
+
+ cw (m1 + m2 )
ln
+
ln
∆Seq = m2 cice ln
T2 T0
m1 + m2 T1 m1 + m2 T0
(9.92)


T0
L
T
∆Seq = m2 cice ln
+
+ cw (m1 + m2 ) ln  ( m1 ) ( m2 ) 
(9.93)
T2 T0
T1 m1 +m2 · T2 m1 +m2
The numerical constants can be substituted:
273 3.34 · 105
289.8
J
∆Seq = 0.03(2.108 · 103 ln
+
) + 4.187 · 103 · 0.330 ln
= 7.13
1
10
)
(
)
(
260
273
K
300 11 · 260 11
(9.94)
The entropy variation of the equalization is positive as expected according to the second
law.
9.8
The second law of thermodynamics
In previous points the increase of entropy in isolated systems during irreversible events
has been demonstrated for several concrete cases. The second law of thermodynamics is
an empirically validated postulate. Increase of entropy in the isolated system is statistically substantiated by the Boltzmann equation by declaring that highest thermodynamic
weight belongs to the thermal equilibrium state.
The most accepted statement of the second law is as follows:
In isolated system (no heat and no work exchange with the environment) at naturally occurring irreversible events the total entropy production is positive.
In limit case the entropy production can be zero in reversible process.
To be isolated is a fundamentally important condition in the statement above. If
the system is not isolated, the entropy variation can be anything. This can be highly
negative too. By cooling and compressing the gas the entropy is dropping fast all the
way when it gets liquefied.
There are some related statements of the second law with the common feature of
denying the possibility of the followings: Perpetual motion machine of the second kind,
Heat engine operating cyclically with a single heat reservoir, Heat engine operating cyclically to convert heat to work in full extent, Heat that moves to the warmer object from
the colder one on its own. These statements are merely the consequences of the “most
accepted” version of the second law above.
153