3Forces

IA Dynamics and Relativity

3.2 Motion in a potential
Given an arbitrary potential
V
(
x
), it is often difficult to completely solve the
equations of motion. However, just by looking at the graph of the potential, we
can usually get a qualitative understanding of the dynamics.
Example.
Consider
V
(
x
) =
m
(
x
3
3
x
). Note that this can be dimensionally
consistent even though we add up
x
3
and
3
x
, if we declare “3” to have dimension
L
2
.
We plot this as follows:
x
V
O
1 212
Suppose we release the particle from rest at
x
=
x
0
. Then
E
=
V
(
x
0
). We can
consider what happens to the particle for different values of x
0
.
x
0
= ±1: This is an equilibrium and the Particle stays there for all t.
1
< x
0
<
2: The particle does not have enough energy to escape the well.
So it oscillates back and forth in potential well.
x
0
< 1: The particle falls to x = −∞.
x
0
>
2: The particle has enough energy to overshoot the well and continues
to x = −∞.
x
0
= 2: This is a special case. Obviously, the particle goes towards
x
=
1.
But how long does it take, and what happens next? Here we have
E
= 2
m
.
We noted previously
t t
0
=
Z
dx
q
2
m
(E V (x))
.
Let x = 1 + ε(t). Then
2
m
(E V (x)) = 4 2(1 + ε)
3
+ 6(1 + ε)
= 6ε
2
2ε
3
.
So
t t
0
=
Z
ε
3
dε
0
6ε
2
2ε
3
We reach
x
=
1 when
ε
0. But for small
ε
, the integrand is approx-
imately
1
, which integrates to
ln ε −∞
as
ε
0. So
ε
= 0 is
achieved after infinite time, i.e. it takes infinite time to reach
ε
= 0, or
x = 1.
Equilibrium points
In reality, most of the time, particles are not flying around wildly doing crazy
stuff. Instead, they stay at (or near) some stable point, and only move very little
in a predictable manner. We call these points equilibrium points.
Definition
(Equilibrium point)
.
A particle is in equilibrium if it has no tendency
to move away. It will stay there for all time. Since
m¨x
=
V
0
(
x
), the equilibrium
points are the stationary points of the potential energy, i.e.
V
0
(x
0
) = 0.
Consider motion near an equilibrium point. We assume that the motion is
small and we can approximate
V
by a second-order Taylor expansion. Then we
can write V as
V (x) V (x
0
) +
1
2
V
00
(x
0
)(x x
0
)
2
.
Then the equation of motion is
m¨x = V
00
(x
0
)(x x
0
).
If
V
00
(
x
0
)
>
0, then this is of the form of the harmonic oscillator.
V
has a
local minimum at
x
0
, and we say the equilibrium point is stable. The particle
oscillates with angular frequency
ω =
r
V
00
(x
0
)
m
.
If
V
00
(
x
0
)
<
0, then
V
has a local maximum at
x
0
. In this case, the equilibrium
point is unstable, and the solution to the equation is
x x
0
Ae
γt
+ Be
γt
for
γ =
r
V
00
(x
0
)
m
.
For almost all initial conditions,
A 6
= 0 and the particle will diverge from the
equilibrium point, leading to a breakdown of the approximation.
If V
00
(x
0
) = 0, then further work is required to determine the outcome.
Example. Consider the simple pendulum.
m
`
d
θ
Suppose that the pendulum makes an angle
θ
with the vertical. Then the energy
is
E = T + V =
1
2
m`
2
˙
θ
2
mg` cos θ.
Therefore
V cos θ
. We have a stable equilibrium at
θ
= 0, and unstable
equilibrium at θ = π.
θ
V
π
π
mg`
mg`
If E > mg`, then
˙
θ never vanishes and the pendulum makes full circles.
If 0
< E < mg`
, then
˙
θ
vanishes at
θ
=
±θ
0
for some 0
< θ
0
< π
i.e.
E
=
mg` cos θ
0
. The pendulum oscillates back and forth. It takes a quarter
of a period to reach from
θ
= 0 to
θ
=
θ
0
. Using the previous general solution,
oscillation period P is given by
P
4
=
Z
θ
0
0
=
dθ
q
2E
m`
2
+
2g
`
cos θ
.
Since we know that E = mg` cos θ
0
, we know that
P
4
=
s
`
g
Z
θ
0
0
dδ
2 cos θ 2 cos θ
0
.
The integral is difficult to evaluate in general, but for small
θ
0
, we can use
cos θ 1
1
2
θ
2
. So
P 4
s
`
g
Z
θ
0
0
dθ
p
θ
2
0
θ
2
= 2π
s
`
g
and is independent of the amplitude
θ
0
. This is of course the result for the
harmonic oscillator.
Force and potential energy in three dimensions
Everything looks nice so far. However, in real life, the world has (at least) three
(spatial) dimensions. To work with multiple dimensions, we will have to promote
our quantities into vectors.
Consider a particle of mass
m
moving in 3D. The equation of motion is now
a vector equation
m
¨
r = F.
We’ll define the familiar quantities we’ve had.
Definition
(Kinetic energy)
.
We define the kinetic energy of the particle to be
T =
1
2
m|v|
2
=
1
2
m
˙
r ·
˙
r.
If we want to know how it varies with time, we obtain
dT
dt
= m
¨
r ·
˙
r = F ·
˙
r = F · v.
This is the power.
Definition
(Power)
.
The power is the rate at which work is done on a particle
by a force. It is given by
P = F · v.
Definition
(Work done)
.
The work done on a particle by a force is the change
in kinetic energy caused by the force. The work done on a particle moving from
r
1
= r(t
1
) to r
2
= r(t
2
) along a trajectory C is the line integral
W =
Z
C
F · dr =
Z
t
2
t
1
F ·
˙
r dt =
Z
t
2
t
1
P dt.
Usually, we are interested in forces that conserve energy. These are forces
which can be given a potential, and are known as conservative forces.
Definition
(Conservative force and potential energy)
.
A conservative force is a
force field F(r) that can be written in the form
F = −∇V.
V is the potential energy function.
Proposition. If F is conservative, then the energy
E = T + V
=
1
2
m|v|
2
+ V (r)
is conserved. For any particle moving under this force, the work done is equal to
the change in potential energy, and is independent of the path taken between
the end points. In particular, if we travel around a closed loop, no work is done.
Proof.
dE
dt
=
d
dt
1
2
m
˙
r ·
˙
r + V
= m
¨
r ·
˙
r +
V
x
i
dx
i
dt
= (m
¨
r + V ) ·
˙
r
= (m
¨
r F) ·
˙
r
= 0
So the energy is conserved. In this case, the work done is
W =
Z
C
F · dr =
Z
C
(V ) · dr = V (r
1
) V (r
2
).