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 2−1−2

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

/ε

, 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

).