Download DYNAMICS AND RELATIVITY (PART II)

DYNAMICS AND RELATIVITY
(PART II)
Lecturer:
Dr. D.J. Miller
Room 535, Kelvin Building
[email protected]
Location:
257, Kelvin Building
No. of Lectures:
9
Recommended Text:
Young & Freedman, 11th Edition
Outline
5.
Work & Kinetic Energy
8.
Work
Work done by a varying force
Kinetic energy
Power
6.
Conservation of Energy
Potential Energy
Conservative and non-conservative forces
7.
Linear Momentum and Collisions
Momentum and impulse
Conservation of momentum
The centre-of-mass
Elastic and inelastic collisions
Angular Momentum and
Energy
Angular momentum and torque
Rotational energy
Analogy to linear quantities
9.
Special Theory of Relativity
Invariance of the physical laws
Relative nature of simultaneity
Time dilation
Length contraction
The Lorentz transformation
2
Question: A moving pendulum reaches the bottom of its swing. Which diagram
correctly represents the forces (in red) on the pendulum at this point?
A
B
C
D
The correct answer was B.
The acceleration is towards the pivot, therefore the force must be towards the pivot.
3
5. Work and kinetic energy
5.1 The definition of “work” in physics
Y&F: Ch 6.1
“Work” is done whenever a force is applied to move an object from one place to another.
If the force is F, and the displacement of the object is s, then the work done W is
given by
Note that both F and s are vectors, while W is a scalar. Recall the scalar product of two
vectors
θ
So we also have
4
Units:
Force, F :
Displacement, s :
Work, W :
Newtons [N]
metres [m]
Joules
[J]
1 Joule = 1 Newton metre
James Prescott Joule
1818-1889
1J=1Nm
Example: Consider a force of 5 N pushing an object 2 m in the same direction as the
direction of the force.
The work done is
5
If the motion is in a different direction from the force, we have to take the direction of the
vectors into account.
Example: What is the work done by gravity when a 5kg object moves 2m down a
frictionless 30o slope?
Force downwards is
θ
30o
Work done is
Newton’s second law
Displacement down the slope is
Angle between force and displacement is θ = 60o
g ≈ 9.8 ms-2
1 N = 1 kg m s-2
6
Work can also be negative
work is
positive
work is
negative
work is
zero
7
Question: A train is heading towards a bridge over a river. At the last minute the train
driver notices the bridge is out! He slams on the breaks and manages to stop the train
just before it goes over the edge…… If he is 100m away from the bridge when he puts
on the breaks and the brakes apply a force of 10,000N how much work does the
frictional force of the brakes do?
What is the work done by the frictional force of the brakes?
A: None
B: 1,000,000J
C: 50,000J
D: -1,000,000J
8
The correct answer was D: -1,000,000 J
The work done is
9
5.2 Work done by a varying force
Y&F: Ch 6.3
So far, the force applied has been constant, but this need not be the case. Let’s consider
a force which changes throughout the motion.
We can divide the displacement
such that
into lots of little (infinitesimal) displacements
Let the force on the object for a displacement
be
Then the work done on segment i is
The total work done is
Taking the limit
this turns into an integral:
10
Imagine a simpler case, where the motion is along only the x direction, starting at x1
and ending at x2.
Now
unit vector in
the x direction
⇒
and
component of F
in the x direction
The two shaded areas above are equal when ∆x → 0
11
Notice that when Fx is independent of the position, then we return to our old formula:
Alternatively
Example: An example of a force which varies with displacement is the force exerted by
a spring when you pull it.
The force applied by the spring is given by Hooke’s Law
k is known as the spring constant
12
Hooke’s Law tells us that the force required to stretch a spring
increases linearly with the distance that you pull it.
Robert Hooke
1635-1703
The work done in stretching the spring by an
amount X, is the area under the curve to the left.
Alternatively:
13
Example: Work done pushing a swing
Y&F: Ex 6.9
Let’s say we push a child in a swing. The child has a weight w, and
the length of the chain is R. Assuming that we push him very slowly,
what is the work done by each of the forces acting on the child?
“slowly” ⇒ all the forces are (approximately)
in equilibrium (i.e. sum to zero)
First we need to know how the forces vary with θ.
If T is the tension on the rope and F is the force with which we
push him, then
x-direction:
y-direction:
14
Now we can calculate the work done by each force moving from A (θ=0) to B (θ = θ0).
Remember that it is only the force in the direction ds which counts!
For the force F :
R
B
A
For the force w (gravity):
For the force T (tension): the angle between ds and T is always 90o, so
the work done by T is zero.
Notice that
and
since the total net force is zero, so can do no work!
15
5.3 Kinetic energy
Y&F: Ch 6.2
Consider an object of mass m, under a constant force F as it moves from x1 to x2.
Since F = ma, the object is accelerated. Its velocity changes from v1 to v2.
Displacement:
Velocities:
But Newton’s second Law tells us that
The work done by the force is
16
Definition: The Kinetic Energy of an object of mass m moving with velocity v is given by
It’s units are kg (ms-1)2 = N m = J
This should be familiar
from the last slide
Work-energy theorem: The work done on an object by the total net force is given by
the change of the object’s kinetic energy.
N.B. Notice that this is the work done by the total net force. In our swing example,
Wtot = 0 so the change in kinetic energy was zero. We could not have used this to
calculate the work done by the individual forces.
17
The work energy theorem is still true even if the force changes during the motion.
If we go back to our small steps ∆xi, then for step i the
work done is
change in kinetic energy for step i
The total work done is the sum over all the steps:
So the total work done is still equal to the change in kinetic energy.
18
Question: You throw a stone into a pile of nice soft mud. The mud exerts a constant
resistive force on the stone, stopping it in 3cm. How much faster do you have to throw
the stone to make it penetrate 12cm?
A: Twice as fast
B: Three times as fast
C: Four times as fast
D: Eight times as fast
E: Sixteen times as fast
The correct answer is….
A
The work done is force × distance, so we need 4 times as much energy.
But the energy is
so we only need twice as much speed to make 4 times the energy.
19
Example: work done by a pile-driver
Y&F: Ex 6.5
A pile-driver of mass m = 200kg drops a distance s12 = 3m
before hitting a beam. It drives the beam a distance s23 = 7.4cm
into the ground. If the guide-rails exert a constant frictional force
on the pile-driver of f = 60N, what is
a) the speed of the pile-driver, v, when it reaches point 2
b) the average force, F, the pile-driver exerts on the beam
a) Between points 1 and 2, the forces on the pile-driver are
friction and gravity:
Total net force
Total work done
Change in kinetic energy =
Work-energy theorem ⇒
20
b) Between points 2 and 3, there is an extra force on the pile-driver,
from the beam pushing it up. This is the force, F, that we need to
work out.
Total net force
Total work done
Change in kinetic energy =
Work-energy theorem
from a)
So
21
5.4 Power
Y&F: Ch 6.5
It is often useful to know how the work done by a force changes over time. We define
Power as the rate of change of work done with time:
Unit: Watt [W]
1 W = 1 Js-1
(Try not to confuse work W with Watts W!)
James Watt
1736-1819
The work done between time 0 and time T is given by
If the power is constant then total work done is
22
We can relate Power to Force and Velocity, just as we related Work to Force and
Displacement:
Recall that the work done for a constant force is
So the power is
Power = Force × velocity
23
Example: resistive forces on a car
A car has a maximum speed of 150km/h and produces a maximum power of 25kW.
What is the resistive force it feels at its maximum speed?
At its maximum speed, v, the car is not accelerating, so must be under zero net force.
i.e.
R
F
Work done by the engine is
So, the rate at which work is done against the resistance is the maximum power
output of the car.
number of seconds
in an hour
24
If the resistive force is constant and the car weighs 1000kg, what is its maximum speed
up a 30o slope?
F
N
R
w = mg
For no acceleration, there should be no force
Power output of the engine is
25
Question: Bob’s drives off in his new car and applying a (leisurely) constant
acceleration he does 0 to 30 mph in 10s. Later in the day, when trying to overtake,
he needs to accelerate from 30 to 60 mph in 5s. The power required for the overtake
is:
A: the same as
B: twice
C: four times
D: six times
E: eight times
… the power required for the acceleration from 0 to 30mph.
The correct answer is….
D
For the first acceleration,
with v = 30mph
For the overtake,
So we need 3 times the work done in half the time.
26
6. Conservation of Energy
6.1 Potential Energy
Y&F: Ch 7.1 & 7.2
Imagine an object of mass m, falling freely under gravity.
Let’s say it falls from a height y1 to a height y2.
The work done by the force of gravity is
and this causes an increase in the kinetic energy of the
object.
Where does this energy come from? The higher the object is at the start, the more
potential it has to gain energy when it falls.
We say that an object of mass m, at a height y, has gravitational potential energy,
given by
Notice that it doesn’t really matter where the point y = 0 is, since only changes in
potential energy matter.
27
In the example just shown, the object loses potential energy, but objects can also gain
potential energy by moving against the gravitational field.
If we apply an upwards force F to the object of mass m,
moving it from y1 to y2, we will be doing work
If we choose
then the object moves with zero
acceleration. Its velocity doesn’t change, so it has the same
kinetic energy as before, but it gains potential energy
The work done is stored as potential energy.
If F is larger than mg, then there will be a net
upwards force and the object will accelerate
upwards. The extra work done by the force will
contribute to the kinetic energy of the object.
28
Let’s return to our free-falling object (i.e.
)
Change in potential energy
Work done by gravitational force
Work-energy theorem
We define the total energy to be the sum of the kinetic and potential energies:
The total energy is conserved. It does not change.
When a force conserves the total energy, we say it is a conservative force.
29
Example: The maximum height of a projectile
Y&F: Ex 7.4
A projectile is fired from point 1 with an initial speed v0 at an initial angle α0. What is
the maximum height it reaches?
At point 1:
At point 2:
Conservation of energy
Therefore the maximum height is
Also, since there is no force in the x-direction, we know that the velocity in the x-direction
cannot change.
and we have
30
Example: Looping the loop!
1
h
Fc
If we don’t switch on the engine(!),
how high (in terms of R) does the
car have to start in order to loop-theloop successfully?
3
Fg
R
2
The crucial thing for survival is that the upward force should be bigger than the downward
force at point 3. We have two competing forces – the gravitational force Fg, pulling down,
and the centrifugal force Fc, pushing up.
Gravitational force at point 3 is:
we need
Centrifugal force at point 3 is:
31
So what condition on h gives
? Let’s use energy conservation again.
At point 1:
At point 2:
At point 3:
Conservation of energy
This tells us that
The requirement
then becomes
So we need a starting height
32
Question: A bottle is dropped from a balcony and hits the pavement
with a speed v. To double the speed of impact it would have to be
dropped from a balcony
A: twice as high
B: three times as high
C: four times as high
D: five times as high
E: six times as high
The correct answer is….
C
Conservation of energy,
So to double the speed we need 4 times the height
33
There are many other forms of potential energy too.
For example, a spring can contain elastic potential energy.
Remember Hooke’s Law?
So when we extend the spring from x1 to x2 we do work
The elastic potential energy stored in a spring is therefore given by
34
What happens if we take away our hand and let the spring pull the mass back?
Work done by the spring is
Work-energy theorem tells us that the work done is equal to the change in kinetic energy
So
Again, total energy (potential + kinetic) is conserved.
35
Potential energy is the area under the curve of
Compare this with gravity
Notice that the position of
O is no longer arbitrary
36
We can work in the opposite direction, and derive the force from the potential
In 1 dimension,
For our two examples:
More generally, in 3 dimensions,
37
The gravitational force exerted on a mass m by another mass M at a distance r, is
G is Newton’s constant
The potential energy is
constant
Choose
Isaac Newton
1643-1727
at
38
The potential we used before was an approximation which is valid when the change in
height is small
If
with y small
height above the Earth’s surface
radius of the Earth
“small” means
Then
with
Radius of the Earth = 6,380,000 m
Mass of Earth
= 5.97 × 1024 kg
39
Example: Escape Velocity of a satellite
A satellite of mass m is launched from the Earth’s surface with velocity v. What v do we
need, in order that the satellite escapes the Earth’s pull?
At launch
and
“Escape” means that the satellite reaches
with
So
Conservation of energy
40
Question: I want to mountain-bike a distance of 1km up a hill. I can choose to go straight
up the hill (1km) or I can choose to zigzag my way up, taking twice the distance (2km).
The average force which I must exert on the zigzag route is
A: 1/4
B: 1/3
C: 1/2
D: equal to
The correct answer is….
the average force I exert going straight up.
C
The work I need to do is the same, to provide the
gravitational potential energy (independent of the
path). But W=Fs so making s twice as long
means that I can make F half as big.
41
6.2 Conservative & non-conservative forces
Y&F: Ch 7.3
We have already seen two examples of a conservative force: gravity and a spring.
If the force is conservative, then we can convert potential energy into kinetic energy
and vice versa, with no energy lost.
potential
energy
kinetic
energy
potential energy
potential energy
kinetic energy
kinetic energy
Total mechanical energy (K +U) is conserved.
42
Imagine a conservative force F, which depends on position, acting on an object.
What is the work done by the force when an object moves from point A to point B?
Work done taking path 1 is
path 1
B
Work done taking path 2 is
A
path 2
For a conservative force we know the potential energy at both A and B, so the work
done must simply be
The work done depends only on the endpoints, not on the path taken.
43
This is really the definition of a conservative force. It is this definition which allows
us to define potential energy in the first place.
Imagine we now regard A → B → A
as the complete path
B
A
Work done by F is
and is independent of the path.
This means that the potential energy at point A (or B) is well defined and does not
change depending on the past history of the object.
44
A non-conservative force is a force where the work done by a force moving an
object from point A to point B depends on the path taken.
This is sometimes called a dissipative force.
A good example of a non-conservative force is friction.
The frictional force always opposes the motion so
always, so the work
done moving A → B → A is
B
A
The work done depends
on the path taken
⇒ we cannot define potential energy
The work done is usually dissipated as heat.
45
Y&F: Ex 7.12
Example: “Futon” fun
You are rearranging your furniture and want to move your 40kg “futon” 2.5m across
the room. The straight line path is blocked by a coffee table, so you need to move the
“futon” around the table as shown. How much work do you do moving the “futon”
around the table compared to the straight line path? (The coefficient of friction is 0.2)
Frictional force is
so
opposing the motion,
Work done moving the “futon” is
For the straight path:
For the path around the table:
46
Question: Two identical balls are rolled (from rest) down the slopes shown. Which
ball will be travelling faster when it reaches the other end?
1
2
A: Ball 1 will be faster.
B: Ball 2 will be faster.
C: The two balls will have the same speed.
The correct answer is….
C
The total energy is conserved, so
47
7. Linear momentum and Collisions
7.2 Momentum and impulse
Y&F: Ch 8.1
Definition: The momentum of an object is its mass × its velocity.
It is a vector quantity (it has magnitude and direction) usually denoted
It is measured in units of kg m s-1
Remember Newton’s second law
. Do you see a similarity?
If the mass is constant,
⇒ Force is the rate of change of momentum
48
Definition: The impulse applied to an object is the Force × the time the force is applied
This is also a vector quantity and is usually denoted
It is measured in units of Ns
e.g. A force of 3N acting on an object for 4s gives an impulse of 3N × 4s = 12Ns in the
same direction as the force.
If the force is constant, then
so
The impulse given to an object is equal to its change in momentum.
- this is the impulse-momentum theorem
49
This impulse-momentum theorem also holds when the force is not constant, but we
must slightly modify our definition of impulse.
Definition for a varying force:
Then
50
7.3 Conservation of momentum
Y&F: Ch 8.2
When two objects collide, the total momentum of the system is conserved.
This is a consequence of Newton’s third law: “For every action, there is an equal and
opposite reaction.”
m
If
But
m
is the force on the object with mass m, and
is the force on m then N3 tells us
so,
constant
momentum is conserved
51
Example: Recoil of a rifle
Y&F: Ex 8.4
A stationary rifle of mass mR = 3kg fires a bullet of mass mB = 5g at a velocity
vB = 300 ms-1 relative to the ground. If the rifle is allowed to recoil freely, what
is its recoil velocity? What are the kinetic energy and momenta of the bullet
and rifle just after firing?
momentum before = 0 (everything at rest)
momentum after
=
momentum before = momentum after ⇒
⇒
So, the velocity of the rifle in the x-direction is:
52
momentum before = momentum after = 0
Kinetic energy before
Kinetic energy after
2 things to notice:
The kinetic energy of the bullet is much larger
(velocity is more important to energy than mass since it is squared)
The kinetic energy is not conserved in this case, but momentum is.
(the energy comes from the explosive reaction which launched the bullet)
53
7.3 The centre-of-mass
Y&F: Ch 8.5
Conservation of momentum is true for more complicated systems too.
If there are no external forces on the system, then momentum is conserved.
the sum here is over all
the bodies in the system
constant
Definition: The centre-of-mass of a collection of n objects of mass
y
at positions
is
The centre-of mass will move with a velocity
x
54
Remember momentum
constant
, so
So as long as the masses don’t change,
momentum
conservation
constant
The velocity of the centre-of-mass is the same after the collision as it was before.
The centre-of-mass is also useful if there is a net force on the system and momentum
is not conserved.
The acceleration of the centre-of-mass is
55
The centre-of-mass moves like an object of mass
under a force
e.g. Consider an object breaking up in flight.
The motion of the centre-of-mass after the collision is the same is if the object had
stayed in one piece.
56
7.4 Elastic and inelastic collisions
Y&F: Ch 8.3 & 8.4
For the moment, lets assume there are no external forces.
In a collision total momentum is conserved, but total mechanical energy may not be.
Definition: An elastic collision is a collision in which total mechanical energy is conserved.
Example: A ball of mass 1 kg, travelling at 2 ms-1 collides elastically with another ball of
mass 2 kg travelling in the opposite direction at 1 ms-1. What are their velocities after the
collision?
Before:
After:
57
I have two equations, and two unknowns (
Solve for
and
: after some algebra (check it!)….
The + sign gives
(and
), so I can solve the system.
remember
) which is the trivial case of no collision
The - sign gives
and
58
Notice that I went about this more generally than I needed. In this case the momentum (in
the x-direction) is
We are in the centre-of-mass frame (the frame where the centre-of mass is at rest), so
59
Conservation of energy
So
and
.
As before, the + sign gives the trivial solution.
This is a general result for an elastic 2 → 2 collision in the centre-of-mass frame.
Even away from the centre-of-mass frame this result is useful.
The centre of mass velocity is
And the velocity
in a general frame is
velocity in centreof-mass frame
(similarly for
)
60
This collision was only in one dimension, but it could in principle be in more dimensions.
For example, imagine the collision of snooker/pool/billiard balls:
Momentum is a vector quantity, so it is conserved in both directions
61
Before the collision:
After the collision:
Momentum conservation
Kinetic energy conservation
⇒
now this is the magnitude
of the whole vector
rearrange
The velocity vectors form a right handed triangle, so the two balls fly off at 90o
62
Question: Two balls moving in a three dimensional space, collide are rebound in an
elastic collision. The balls are lit from above (from a light source very far away) and
cast shadows on a two-dimensional plane. Pretend the shadows have a mass equal to
their ‘parent’ ball. Does the collision of the shadows conserve
A: kinetic energy only
B: momentum only
C: both kinetic energy and momentum
D: neither kinetic energy nor momentum
63
The correct answer is….
B momentum only
momentum conservation: I could define the plane (with the shadows) as being z=0, then
momentum conservation for the shadow collision is the same as the x and y components
of momentum conservation for the ball collision.
mom. conservation for balls
⇒
mom. conservation for shadows
Kinetic energy:
For the shadows, the z-components are lost, so we can’t have kinetic energy conservation.
64
Definition: An inelastic collision is a collision in which total mechanical energy is not
conserved.
Example: A ballistic pendulum
A bullet of mass mB is fired with
velocity
at a wooden pendulum
of mass mA, as shown. It sticks in
the wood and the pendulum swings
up a vertical distance y. What was
the initial speed of the bullet?
velocity of block (with
bullet) immediately
after impact
momentum conservation
kinetic energy before collision is
kinetic energy after collision is
Kinetic energy is not conserved!
65
Now we need to calculate v, which we can do using energy conservation after the collision.
kinetic energy after collision is
potential energy after collision is
(this is just a definition of
where ‘0’ height is)
kinetic energy after “upswing” is
(the pendulum comes to rest)
potential energy after “upswing” is
Energy conservation
⇒
⇒
⇒
Putting this together, we have the velocity of the bullet:
66
Question: A 1kg lump of clay travelling at 1ms-1 smashes into another 1kg lump of
clay which is not moving. They stick together and become one 2kg lump. What is the
speed of the 2kg lump after the collision?
Before:
After:
A: 0 ms-1
B: 0.25 ms-1
C: 0.5 ms-1
D: 1 ms-1
E: 2 ms-1
1kg
1 ms-1
1kg
1kg
1kg
v ms-1
The correct answer is…. C
momentum is conserved (but not kinetic energy)
67
8. Angular momentum and energy
Y&F: Ch 10.4
8.1 Angular momentum and torque
Definition: The angular momentum of a point particle about O is given by
where
is the vector from O to the object and
is its momentum.
This is a vector product. A vector product of two vectors
and
is given by
is a vector perpendicular to the
plane of and , and is of length
68
The vector is perpendicular to both
(i.e. out of the slide)
and
O
For circular motion with angular velocity
magnitude of the angular momentum is
,
is at right angles to
so the
O
69
Definition: If a point particle with position vector relative to O, feels a force , then the
torque of the force (sometimes called the “moment of the force”)
on the particle with
respect to O is given by
The torque is a vector perpendicular to both
with magnitude
and
O
Torque is the rate of change of angular momentum with time:
In the absence of an applied torque, angular momentum is conserved.
70
Example: A plumber uses a long 80cm wench to unscrew a pipe fitting. If the angle
between the wench and the horizontal is 19o and he applies his full weight of 900N to
the wench, what is the torque on the pipe?
Y&F: Ex 10.1
Torque is into the page (by right hand rule) and is of magnitude
71
Y&F: Ch 10.6
An interesting example of torque and angular momentum is a gyroscope.
This is a spinning object balanced on a pivot.
The object is set to spin about its axis, and instead of falling down as you would
expect, it “precesses” about the pivot.
72
Consider first a gyroscope which is not spinning
Gravity pulls the centre-of-mass
down ( ), applying a torque
at right angles to both
and
.
The change in the angular
momentum in a time dt is
which is also at right angles to
both and .
With each dt, the changes in angular momentum add up, the angular
momentum gets bigger and the gyroscope falls more quickly.
(This is exactly what we naively expect.)
73
Now consider a gyroscope which is spinning
As before, Gravity pulls the
centre-of-mass down ( ),
applying a torque
at right angles to both
and
Now,
as before, but this
time the torque only changes the
direction of the angular momentum,
not its magnitude.
.
The angular momentum is always in the same direction as
doesn’t fall – it precesses.
so the gyroscope
74
For a gyroscope of mass m:
Downwards force on centre-of-mass
Torque
Change of angular momentum
So
The gyroscope precesses with an angular speed
75
Y&F: Ch 9.5
8.2 Rotational energy
Recall, a point particle moving with velocity
If it is moving in a circle of fixed radius r, then
has kinetic energy
.
so
Usually it is more useful to discuss the rotations of extended objects, for example, a
rotating disk. Imagine that the object is made up of lots of particles with mass
,
circulating at a radius with velocity
. Then the total energy is
This is the moment of inertia of the body
76
Definition: The Moment of Inertia of a system of particles about O is given by
It’s units are kg m2
where the particle i has mass mi and its distance from O is ri.
(Notice that this definition requires a specific axis of rotation – the moment of inertia
of a body is dependent on the body’s situation.)
The kinetic energy of a body rotating about an axis O with angular speed ω and
moment of inertia I is
77
Usually we will be interested in bodies that are continuous, rather than a discrete
number of point particles. Then the summation becomes an integral.
Example: The Moment of Inertia of a solid disk
O
Total Volume is
Total Mass is
Density is
First, lets think about a very thin annulus:
O
All segments of the annulus of equal mass
have the same moment of inertia (since
they are the same distance from O) so the
annulus has moment of inertia
volume
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To get the moment of inertia of the disk, we must
add up the moments of inertia for lots of annuli.
O
But recall
so the moment of inertia of the whole disk about the axis O is
(Notice that l dropped out.)
79
Question: What is the Moment of Inertia of a (non-thin) annulus about an axis
through its centre, perpendicular to its plane?
A:
O
B:
C:
(Hint: Think about the limits)
D:
The correct answer is…. B
If
I must get
If I increase
(a disk), so A and C are both wrong.
I must increase I because I am moving mass outwards ⇒ D is wrong
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More formally,
O
But now
so
Thus
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8.3 Analogy to linear quantities
You should have noticed parallels between this section and previous sections.
velocity
angular velocity
Momentum
Angular momentum
Force
Rate of change of momentum
Torque
Rate of change of angular momentum
Mass m
Moment of Inertia
Kinetic energy
Kinetic energy
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9. Special Theory of Relativity
9.1 Galilean relativity
Y&F: Ch 37.1
“Relativity” in physics tells us how to compare measurements made in different frames of
reference moving relative to one another.
Although there is a public perception that “relativity” belongs only to Einstein, classical
physics also contains relativity, called Galilean Relativity.
Consider two reference frames moving at a velocity u in the x-direction relative to each other.
y
y
x
O
ut
x
O
u
The coordinates are related by
a Galilean Transformation:
x
y = y
x
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More generally, to find the coordinates in a new frame O moving with a velocity
relative to the frame O:
i.e.
Everything else can be worked out from these. For example,
(We used this implicitly earlier in Section 7.)
Then the velocity is very large, this Galilean Transformation is wrong. We must
replace it with the Lorentz transformation.
We will use Einstein’s Theory of Special Relativity to derive this Lorentz transformation.
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Question: A spaceship is traveling at 0.5c relative to the bad guys, when it fires its
lasers. The laser light travels at speed c relative to the spaceship. What is the speed of
the laser with respect to the bad guys?
A.
B.
C.
D.
0.5c
1c
1.5c
2c
The correct answer is…
B
When the velocity between frames is very large, the Galilean
Transformation is wrong. We must replace it with the Lorentz
transformation.
We will use Einstein’s Theory of Special Relativity to derive this Lorentz
transformation, and we will see that the speed of light is c in all inertial frames.
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9.2 The Michelson-Morley Experiment
In 1864, Maxwell wrote down his theory of electromagnetism. This theory explains
light in the context of electric and magnetic fields, and predicts the speed of light in a
vacuum to be:
permittivity of
free space
permeability of
free space
So Maxwell showed that light travels at (roughly) 300,000kms-1. But relative to what?
By comparison, sound travels at roughly 330ms-1. Since sound is the vibration of molecules
in the medium (e.g. air) then this speed is relative to the medium itself.
If you shout into a wind of speed 20ms-1, then the sound will move ‘downwind’ at 350ms-1
and ‘upwind’ at only 310ms-1.
So what is light moving in?
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Scientists suggested that light moved through the “luminiferous aether”, some unknown
background medium, and the speed of light in Maxwell’s equations must be relative to
this.
In that case, then the speed of light should depend on the speed of the “aether wind” in
exactly the same way as the speed of sound depends on the normal wind’s speed.
“Aether
wind”
Earth
Sun
Even if the “aether wind” is constant with respect to the solar system we should find a
different speed of light in different directions, and at different times of the day and/or year.
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In 1887 Michelson & Morely set out to test this.
They split a coherent light beam (only one wavelength) into
two using a semi-transparent mirror, and let the two beams
travel in perpendicular directions before being bounced back
by mirrors (11m away).
Albert Michelson
(1852-1931)
mirror
semitransparent
mirror
coherent
light source
mirror
Edward Morely
(1838-1923)
The beams were recombined, and
if one has travelled faster than the
other they would see interference
fringes at the detector.
detector
They found no fringes, indicating that the split beams travelled at the same speed.
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9.2 Einstein’s postulates of Special Relativity
Y&F: Ch 37.1
E1: The laws of physics are the same in all inertial frames of reference.
E2: The speed of light in vacuum is the same in all inertial frames of reference
and is independent of the source.
An inertial reference frame is a reference frame which is not undergoing any acceleration.
The Newtonian mechanics (and the associated Galilean transformation) that we have seen
so far in this lecture course also obeys E1. For example,
only holds if the frame
is not accelerating, but holds for all non-accelerating (inertial) frames.
(Note that the Earth is not an inertial frame, since it is rotating. However, the acceleration
from the rotation is so small that we can pretend it is an inertial frame in most cases.)
So postulate 1 is not so surprising…..
…. it is E2 which throws us the surprises!
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9.3 Simultaneity is relative
Y&F: Ch 37.1
Einstein’s second postulates tells us that whether two events are simultaneous or not
depends on the reference frame of the observer.
Example: Imagine a train travelling at a speed u along the track. Observer O is
standing still on the platform while observer O is on the train, right in the middle of a
carriage.
O
A
B
A
B
u
Lightning strikes the carriage at
points A and B, when they are
adjacent to points A and B on
the ground.
O
Observer O observes the two lightning flashes at the same time. He marks the spot
where he was standing, measures distances OA and OB (the lightning left scorch
marks) and when he finds them equal concludes that the lightning stuck both ends of
simultaneouly.
In O’s frame, if the distance OA=OB, then the light from both bolts must have travelled
for time OA/c.
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Observer O must agree that the two light flashes reached observer O at the same time after all, this is a real event which really happened.
But the flashes do not reach observer O at the same time. He is travelling with speed u
away from observer O, and so O observes that the wavefront from bolt B reaches O first.
O
B
A
A
u
B
O
ut
Again, O must agree, but he disagrees on the interpretation.
O argues that since he is standing in the middle of the carriage, and the light from the
two bolts both travel at the same speed c in O’s reference frame (Einstein’s second
postulate) then the lightning bolts could not have been simultaneous.
Whether or not the lightning bolts were simultaneous
or not depends on the reference frame!
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Y&F: Ch 37.3
9.4 Time Dilation
We would like to use Einstein’s postulates to derive a new transformation between
coordinates in different inertial frames of reference.
d
mirror
O
source
d
O
O
O
u∆t
Returning to our train, imagine that O observes a light
from a source on the floor being projected vertically
upwards. It bounces off a mirror on the ceiling and
returns to the same position on the floor. He measures
the time the light takes for the round trip
u
u
Observer O measures the
round trip as time ∆t. But
since the source has
moved, he reckons the
light must travel further.
92
O reckons the distance travelled by the light is
d
so in his frame, the light must take time
Time is not universal!
It is reference frame dependent!
93
Definition: The proper time between two events is the time between the events in the
frame where the events happen at the same point in space.
In our example, the two events (departure and arrival of light at the source) occur at the
same point in space in O’s reference frame, so the proper time between the events is ∆t0.
Generally, the time between two events in a frame moving with relative (constant) speed
u with respect to the proper frame is
where
So clocks run slow if they are moving relative to you – this is known as time dilation.
Note:
photons take no time to travel
is ill-defined for
when
,
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The twin paradox
Imagine two twins, Bill and Ben. Ben decides to leave Earth in a spaceship which travels
at a fixed speed of u relative to the Earth, and travels a distance d (again relative to the
Earth) to a distant planet. Bill stays on the Earth during Ben’s journey.
Bill
Ben
u
In Bill’s reference frame the distance travelled is d, so Bill measures the time that Ben
takes to get there as
How long does Ben measure that the journey has taken?
95
In this example, we have two significant events:
• Ben’s departure from Earth
• Ben’s arrival at the new planet
In Ben’s reference frame, both of these events happen at the same space point (by
definition), so the time that Ben measures between these events is the proper time ∆t0 .
We know that the proper time interval is related to the (dilated) time interval in a
reference frame moving with speed u by our time dilation formula:
Note that the sign
of u doesn’t matter.
We know that the dilated time measured by Bill is
measured by Ben) between the events is:
so the proper time (as
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Now imagine that when Ben arrives at the planet, he immediately turns around and comes
back at the same speed (u).
Bill
Ben
u
Exactly the same reasoning as before can be applied and the time to get back is the
same as for the outward journey in both frames.
Therefore, the time elapsed for the total journey is:
In Bill’s frame:
In Ben’s frame:
When he returns, Ben is younger than Bill!
97
This seems paradoxical because what has distinguished Bill from Ben?
Relativity tells us that there is no ‘preferred’ reference frame – I could just have easily
considered Bill (and the entire planet Earth) as moving away from Ben at a speed u,
and when he returns, he would have been younger than Ben….or would he?
There is actually no paradox here, because there is something which tells us
that it was Ben who travelled.
Ben accelerated when he turned around (changing his velocity from u to –u)
– Ben was not in an inertial frame!
In fact, Ben knows this because he feels the g-force of the deceleration and
acceleration while turning round.
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Example: Elementary particles called muons are created in the upper atmosphere by
cosmic rays. If an average muon lives 1.56×10-6 s, will an average muon created at a
height of 2km and at a (downward) speed of 0.98c reach the ground before it decays?
Cosmic ray
In the Earth’s rest frame the muon lives a time
∆t before decaying, where
muon
0.98c
2km
Therefore an average muon travels a distance
Ground
before decaying.
The average muon reaches the ground, travelling 5 times further than non-relativistic
mechanics would otherwise predict.
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Y&F: Ch 37.4
9.4 Length Contraction
E2 gives us a really good way to measure distances. Since the speed of light is c in
every frame, we can time how long light takes to travel between two points and work
out the distance.
mirror
source
Back on the train, observer O decides to measure the
length of a box by using his source and mirror.
He measures a time ∆t0 for the light to reach the mirror
and return and concludes the box is of length
O
∆t0 is the proper time between the light’s emission and re-arrival.
O
O
O
u
Observer
O
watches
the
experiment and measures a time
∆t1 for the light to reach the mirror
and a time ∆t2 to return. These
times are not the same because
the train moves during the light’s
journey.
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On the journey to the mirror, the light travels a distance
But we know
u∆t1
distance travelled
by the train
so
l
length of
the box in
O’s frame
On the return journey, the light travels a distance
The total trip takes time
101
So in frame O’s frame,
, while in O’s frame
But we know the relation between ∆t and ∆t0 from our earlier example.
The length of the box measured by O is contracted compared to the length measured by O
Definition: The proper length between two points is the distance between the points in
the frame where the points are at rest.
In our example, the box is at rest in frame O, so its proper length is l0.
Generally, the distance between two points in a frame moving with relative (constant)
speed u with respect to the proper frame is
where γ was defined earlier.
So lengths are contracted if they are moving relative to you
– this is known as length contraction.
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The train in the tunnel paradox
A 25m long train approaches a 20m long tunnel at 0.6c. Some observers are standing
on the ground beside the tunnel.
The proper length of the train is l0=25m, but an
observer on the ground sees its length contracted to
So as the train passes through the tunnel, it the observers on the ground will see it
momentarily entirely enclosed.
The proper length of the tunnel is d0 = 20m, but the train driver see its length contracted to
So the train driver never sees the train enclosed completely in the tunnel.
We seem to have a paradox: How can the train be in the tunnel but not in the tunnel?
103
As before, this is not really a paradox if we define our expressions a little bit better.
What do we mean by “the train is in the tunnel”?
We mean that the front of the train is inside the tunnel and the back of the train is inside
the tunnel simultaneously.
But we saw earlier how simultaneity is relative. Two events which are simultaneous in
one frame may not be in another.
In our example, the two important events (front of train inside tunnel + back of train
inside tunnel) are simultaneous in the rest frame of the tunnel, but they are not
simultaneous in the rest frame of the train.
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Y&F: Ch 37.5
9.4 The Lorentz transformation
We are now ready to write down our replacement for the Galilean transformation
– the Lorentz transformation.
y
y
x
O
ut
x
O
u
x
y = y
x
x is a proper length in O’s frame. In O’s frame this distance is length contracted to
So
105
Alternatively, we could have compared the lengths in O’s frame:
So
from before
As an exercise you can show that
, so the transformation becomes:
106
Therefore the Lorentz transformation between the two frames, for a Lorentz “boost”
along the x-axis is:
This tells us how to relate the
coordinates of an event in one
frame (t,x,y,z) to the coordinates of
the same event in another frame
(t,x,y,z)
Notice:
•
The y and z components remain unchanged (as for the Galilean transformation
along the x-axis). We could find the general transformation by applying a rotation to
the above.
•
If u is small,
and
, so we return to the Galilean transformation
107
Also,
So the quantity
is Lorentz invariant.
doesn’t change with reference frame. We say that this
c.f. the length of a vector under a rotation
108
Velocity transformations
We can differentiate the Lorentz transformation equations to see how velocities transform.
(chain rule)
But
and
So
Notice that we return to the Galilean velocity addition rule when we take u → 0
Also, if
then
as required by E2.
109
Beware! Although the perpendicular (y and z) components don’t change under a boost
(in the x-direction), the perpendicular velocities do.
This is because a velocity is measuring a change in position with respect to time, and
our time coordinate changes with frame.
e.g.
Example: Two particles, A and B, approach each other with velocities vA=v and vB=–v
with respect to the ground. What is the velocity of B as measured in A’s rest frame?
A
B
110
Notice that the speed of B is always less than the speed of light in A’s frame.
1
If fact, our Lorentz transformation tells us that if an object has a speed
in any
frame, then it will have speed
in all other frames (which travel with relative speed
______ compared to the first frame).
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9.4 Relativistic energy and momentum
Y&F: Ch 37.7 & 37.8
Our classical definition of momentum was
. This was useful because it was
always conserved in collisions (elastic and inelastic).
Lets take a look at the previous example (particles A and B) imagine the particles
have the same mass m.
Total momentum in the ground’s frame =
Total momentum in the A’s rest frame =
Now imagine they collide, and stick together (inelastic) to form a particle c of mass 2m.
Boost to A’s rest frame:
But
This definition of momentum
cannot be conserved in all
frames!
112
Definition: The relativistic momentum of a particle of mass m, moving with velocity
is
This relativistic momentum is conserved in all frames (showing this from the velocity
transformation is a bit tedious).
Sometimes people redefine mass, rather than momentum.
This would leave the definition of momentum as we had before, i.e.
then mass becomes dependent on reference frame, which is very inconvenient.
, but
Please do not do this!
A non-linear relation between momentum and velocity is much better!
113
Since momentum is related to force and force is related to work and energy, we also
need a new definition of energy.
Definition: The relativistic energy of a particle of mass m, moving with velocity
(with no potential) is
Notice how similar this is to the new definition of momentum!
If
we can expand the denominator as a series in
Does this look familiar?
114
For
This is the
rest energy
of the object
This is the nonrelativistic
kinetic energy
of the object
Using our new definition of momentum to replace the velocity in
we find,
Notice that
is a Lorentz invariant?
What does this remind you of?
115
It is useful to think of time as another coordinate in the vector for the event.
Instead of having 3 components in a vector we have four
We collect these coordinates together as a four-vector
Position four-vector:
The “length” of this four vector is
, which is a Lorentz invariant.
We can do the same thing for momentum & energy.
Momentum four-vector:
The “length” of this four vector is
This transforms under a
Lorentz boost in the same
way as the position vector.
, which is also a Lorentz invariant.
116