Download 17 Lecture 17: Conservative forces in three dimensions

Mathematics for Physics 3: Dynamics and Differential Equations
17
81
Lecture 17: Conservative forces in three dimensions
So far we discussed conservative forces in one dimension, defining them as ones that can be obtained from a derivative
of some potential V (x). A conservative force acting along the x̂ axis is a force that can be written as:
dV (x)
.
dx
More generally, one can define conservative forces in three dimensions by introducing a potential function that
depends on all three coordinates V (x, y, z). We then have:
Fx = −
Fx = −
∂V (x, y, z)
,
∂x
Fy = −
∂V (x, y, z)
,
∂y
Fz = −
∂V (x, y, z)
,
∂z
(285)
where we denoted partial differentiation with respect to a given cordinate by ∂. This can be written in a vector form
as follows:
F = Fx x̂ + Fy ŷ + Fz ẑ = −
∂V (x, y, z)
∂V (x, y, z)
∂V (x, y, z)
x̂ −
ŷ −
ẑ ≡ −∇V (x, y, z) ,
∂x
∂y
∂z
(286)
where ∇, called the gradient, is a vector differential operator: it acts on a scalar function of the coordinates
V (x, y, z) and produces a vector. Each component of the vector is the slope in the respective direction. The direction
of ∇V (x, y, z) at a given point (x, y, z) is the direction along which the local change of the function V (x, y, z) is
maximal.
The simplest example is that of Galilean gravity where V (x, y, z) = mgz, where z is the vertical coordinate.
The gravitational potential increases as z increases, and it does not depend at all on the horizontal coordinates x
and y. The corresponding force is:
�
�
∂mgz
∂mgz
∂mgz
F = −∇V (x, y, z) = −∇mgz = −
x̂ +
ŷ +
ẑ = −mg [0x̂ + 0ŷ + 1ẑ] = −mgẑ
∂x
∂y
∂z
which is indeed in the direction −ẑ the direction along which the potential changes.
It is often convenient to denote the set of coordinates (x, y, z) by the vector r. We can think of the potential as
a function of the coordinate vector r. We defined a conservative force as one that can be written as a derivative of
a scalar function of the coordinates:
F = −∇V (r) .
(287)
Alternative (and equivalent) definitions for a conservative force are:
�b
• The work done by the force Wa→b ≡ a F · dr does not depend on the path from a to b.
• The curl of the force is zero: ∇ × F = 0.
Both these definitions will be discussed in detail in the next course, Mathematics for Physic 4.
Example: quadratic potential in 3 dimensions
Consider for example the following potential:
k
k
k
r · r = r2 = (x2 + y 2 + z 2 )
2
2
2
This is the three dimensional version of the harmonic oscillator we discussed in previous lectures.
The corresponding conservative force is:
�
�
�
�
k
k
k ∂(x2 + y 2 + z 2 )
∂(x2 + y 2 + z 2 )
∂(x2 + y 2 + z 2 )
2
2
2
F = −∇V (r) = −∇
r · r = − ∇(x + y + z ) = −
x̂ +
ŷ +
ẑ
2
2
2
∂x
∂y
∂z
�
�
k
= − 2xx̂ + 2y ŷ + 2z ẑ = −k(xx̂ + y ŷ + z ẑ) = −kr
2
V (r) =
This is of course the familiar linear restoring force, as in Hooke’s law, but now in three dimensions.
From the mathematical perspective we have learnt that
∇(r · r) = ∇(r2 ) = 2r
where r = |r|. Thus, by the chain rule we also have:
�
� 1
1
∇(r) = ∇ (r2 )1/2 = (r2 )−1/2 ∇(r2 ) = (r2 )−1/2 2r = r/|r| = r̂
2
2
Mathematics for Physics 3: Dynamics and Differential Equations
82
Conservation of energy
Recall that conservative forces in one dimension have the property that the total mechanical energy is conserved.
The same is true in three dimensions.
Let us first consider the time derivative of some function f of the coordinates (we assume that f only depends
on time through the position, not directly) via the chain rule:
d
dx ∂f (r) dy ∂f (r) dz ∂f (r)
f (r) =
+
+
= ṙ · ∇f
dt
dt ∂x
dt ∂y
dt ∂z
we see that it is neatly expressed in terms of the gradient of the function.
This applies in particular to the potential energy, whose time derivative is:
d
V (r) = ṙ · ∇V (r) = −ṙ · F
dt
which we identify as minus the work performed by the force per unite time (power). Recall that the work done over
an infinitesimally small distance dr is dW = dr · F . Dividing this work element by the corresponding time element
in which it is done, one obtains the power: dW/dt = ṙ · F .
Let us write now the second Newton law for a particle of mass m under a conservative force F = −∇V (r):
mr̈ = F
=⇒
mr̈ + ∇V (r) = 0
This is a vector equation. Forming a scalar product with the velocity vector ṙ we get:
mṙ · r̈ + ṙ · ∇V (r) = 0
� �� �
� �� �
d
d
1
dt V (r)
dt ( 2 mṙ·ṙ )
where we identify the first term as the time derivative of the kinetic energy and the second as the time derivative of
the potential energy. The conclusion is, as expected, that if the force is conservative, the time-derivative of the total
mechanical energy vanishes:
�
�
d 1 2
d
dE
mṙ + V (r) =
[T + V ] =
= 0.
(288)
dt 2
dt
dt
or, the total mechanical energy,
E=
1 2
mṙ + V (r)
2
is a constant of motion.
The two-body problem
We now turn to discuss the general problem of a two-body system. We assume an isolated system (i.e. one on which
no external forces apply) which is composed of just two bodies of masses m1 and m2 . Their positions in some inertial
frame is described by the two vectors
r1 = x1 x̂ + y1 ŷ + z1 ẑ ,
r2 = x2 x̂ + y2 ŷ + z2 ẑ .
The two bodies interact via some conservative force that depends on the relative coordinate r = r1 − r2 , namely
V = V (r1 − r2 ). We will show that without making any further assumptions about the nature of the potential we
can reduce the problem to the problem of a single particle.
We start with Newton law for the two bodies:
m1 r¨1 = −∇1 V (r1 − r2 ),
m2 r¨2 = −∇2 V (r1 − r2 )
(289)
where we denote by ∇i the gradient with respect to the coordinates of body i, namely,
∇1 =
∂
∂
∂
x̂ +
ŷ +
ẑ,
∂x1
∂y1
∂z1
∇2 =
∂
∂
∂
x̂ +
ŷ +
ẑ
∂x2
∂y2
∂z2
Therefore, given that V = V (r1 − r2 ), we get:
∇1 V (r1 − r2 ) = −∇2 V (r1 − r2 )
which is equivalent to the statement that
F 2→1 = −F 1→2
(290)
Mathematics for Physics 3: Dynamics and Differential Equations
83
namely that the force which body 2 exerts on body 1 is equal in magnitude and opposite in direction to the force
body 1 exerts on body 2. This is of course the third Newton law. We see that stating that the potential depends on
the relative coordinate r1 − r2 is not an extra assumption, but in fact a manifestation of the third Newton law.
Summing up the two equations of motion in (289) (II Newton law) together with (290) (the III Newton law) we
obtain:
m1 r¨1 + m2 r¨2 = −∇1 V (r1 − r2 ) − ∇2 V (r1 − r2 ) = 0 .
(291)
or, equivalently,
� dP
d�
m1 r˙1 + m2 r˙2 =
= 0.
dt
dt
where we identified the total momentum of the two-body system as
P ≡ m1 r˙1 + m2 r˙2 .
(292)
(293)
We therefore deduce that the total linear momentum (293) in an isolated system is a constant of motion.
Using the fact that the total momentum is conserved, we can simplify the problem. Instead of solving (289)
separately for the position of each of the two bodies, r1 (t) and r2 (t), momentum conservation calls for solving the
motion in terms of the centre of mass coordinate R(t) and the relative coordinate r(t), defined as follows:
R=
m1 r1 + m2 r2
,
m1 + m2
r = r1 − r2 .
(294)
This reasoning is based on the observation that the total momentum (293) is proportional to the derivative of the
centre-of-mass coordinate:
P = (m1 + m2 ) Ṙ
so the fact that P is conserves translates to the statement that the centre-of-mass coordinate is solved by
R(t) = R(0) +
P
t
m1 + m2
(295)
the centre-of-mass is moving in a straight line and in a constant speed, in agreement with the first Newton law (we
assumed that the system is isolated: no external forces act on the two bodies).
Recall that our original problem was to determine six coordinates as a function of time: x1 (t), y1 (t), z1 (t) and
x2 (t), y2 (t), z2 (t). We have already given the solution to three linear combinations of the latter in (295). We now
need to solve for the remaining three, that is to solve for the relative motion r = r1 − r2 . In order to express r1 and
r2 in terms of R and r let us first substitute r2 = r1 − r into the expression for R:
R=
m1 r1 + m2 (r1 − r)
m2
= r1 −
r
m1 + m2
m1 + m2
(296)
This implies:
m2
m1
r,
r2 = r1 − r = R −
r
(297)
m1 + m2
m1 + m2
Differentiating these expressions twice with respect to time, and using the fact that the centre of mass coordinate
does not accelerate: R̈ = 0 (as we established in (295) above) we obtain:
m2
m1
r̈1 =
r̈ ,
r̈2 = −
r̈ ,
(298)
m1 + m2
m1 + m2
r1 = R +
allowing us to convert the equations of motion in (289) into an equation of motion for the relative coordinate r(t).
For example, multiplying r̈1 by m1 and using m1 r¨1 = −∇1 V (r1 − r2 ) we get:
m1 m2
r̈ = m1 r̈1 = −∇1 V (r1 − r2 ) = −∇V (r)
m1 + m2
(299)
where in the last step we defined the gradient using the relative coordinate, namely given the relative coordinate
r = xx̂ + y ŷ + z ẑ, the corresponding gradient is
∇=
∂
∂
∂
x̂ +
ŷ +
ẑ .
∂x
∂y
∂z
We therefore obtained an equation of motion for the relative coordinate:
mr̈ = −∇V (r)
where the combination
(300)
m1 m2
m1 + m2
is called the reduced mass of the two-body system. Note that this is half of the harmonic mean of the two masses,
and it is always smaller than any of the two.
m≡