Physics 430: Lecture 10
Energy of Interaction
Dale E. Gary
NJIT Physics Department
4.9 Energy of Interaction of Two
Particles
Up to now we have been discussing the energy of a single particle. Now
let’s look at the energy of two particles.
 Imagine two isolated particles, alone in the universe (no external forces)
interacting with each other. Particle 2, at position r2 from the origin, exerts
a force F12 on particle 1, while the particle 1 at position r1 exerts an equal
and opposite force F21 = -F12 on particle 2. The distance between the two
particles is then r = r1 - r2 as shown in the figure.
 For definiteness, consider a gravitational interaction, so that, as usual,
Gm1m2
Gm1m2
ˆ
F12  r

r.
2
3
r
r
 From the definition of r and the vector difference of the
2 r = r1 - r2
two position vectors, we can rewrite this as:
1
Gm1m2
r1 - r2 .
F12  r2
3
r1 - r2
r1
 The fact that this depends only on the difference of the
two position vectors means that the force is
O
translationally invariant.

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Energy of Interaction of Two
Particles-2
This fact of translational invariance—the fact that the force of interaction
depends only on the distance between two particles, and not on their
absolution position, allows us to greatly simplify the discussion by choosing
a point, r2, say, as the origin (even though the position of particle 2 is
changing, i.e. the origin need not be a fixed point).
 With this choice, our earlier discussion of force on a single particle applies.
For example, if the force F12 on particle 1 is conservative, then

1  F12  0,
where the operator



 yˆ
 zˆ
x1
y1
z1
with respect to the coordinates r1 = (x1, y1, z1).
When the curl above is zero (i.e. the force is conservative) then we can
define a potential energy
F12  -1U (r1 ).
Note that U(r1) is defined throughout space, but the force at particle 1 is
the gradient of U(r1) evaluated at the point where particle 1 is, i.e. at r1.
1  xˆ


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Energy of Interaction of Two
Particles-3





The foregoing is specific to the case when particle 2 is at the origin, but we
can always translate to some other origin such that particle 2 is at position
r2, so that:
F12  -1U (r1 - r2 ).
The trick is that we do not have to modify the operator 1, because a
derivative like  / x1 is unchanged by adding a constant to x1.
We can then find the force on particle 2 by particle 1 as
F12  -1U (r1 - r2 )  -F21  2U (r1 - r2 ).
What this says is that we can consider a single potential energy U (r1 - r2 )
for the interaction between particles 1 and 2, and evaluate its gradient at
the position of each particle to find the force on each particle.
(Force on particle 1)  -1U 
.
(Force on particle 2)  -2U 
Now consider the two particles moving through space under their mutual
interaction, so that particle 1 moves through a distance dr1 under the force
F12 and particle 2 moves through a distance dr2 under the force F21.
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Total Energy of Interaction






By the work-KE theorem, there will be a change in kinetic energy of each of
the particles:
dT1  F12  dr1
and dT2  F21  dr2 .
We simply add these to find the total work done:
Wtot  F12  dr1  F21  dr2  F12  dr1 - F12  dr2 .
where the last equality is because again F21 = -F12. This can be rewritten
Wtot  dr1 - dr2   F12  d r1 - r2  - 1U r1 - r2 .
If we rename (r1 – r2) as r, we finally have:
Wtot  -dr  U (r)  -dU .
Let’s pause to appreciate this result. It says that if the force of interaction
between two particles is conservative, then we can define a potential
energy throughout space. As the particles move, the potential energy
changes, but the change is just the negative of the change in kinetic energy
that results from the interaction force.
Obviously, then, the total energy is conserved, i.e.
E  T  U  T1  T2  U  constant.
Notice that there are two kinetic energies, one for each particle, but only
one potential energy arising from the configuration of the particles.
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Elastic Collisions

Elastic collisions are ones that take place through conservative forces, so that no
energy is lost (to heat or other mechanism). A collision between a proton and
an electron, for example, occurs through the conservative electrostatic force.
Collisions between two billiard balls is largely elastic, because the balls are made
to act like stiff springs when they collide.
Example 4.8: An Equal-Mass Elastic Collision

Statement of the problem:


Consider an elastic collision between two particles of equal mass m1 = m2 = m, as
shown in the figure. Prove that if particle 2 is initially at rest then the angle between
the two outgoing velocities is q = 90o.
Solution:


Conservation of momentum gives: mv1  mv1  mv2 .
The fact that the collision is elastic means that
1
2

mv  12 mv1  12 mv
2
1
2
2
2
 v  v1  v .
2
1
2
2
2
1
Squaring the momentum equation, after eliminating m, gives:
2
v1
v  v1  2 v1  v2  v .
2
1

2
Comparing these, we see that
v1  v2  0.
2
2
1
2
v2
v1 and v2 are 90 degrees apart
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v1
q
4.10 Generalizing to Multiple
Particles
In the case of many particles (the text goes through the argument explicitly
for 4 particles, and then for N particles), the same thing holds.
 Taking the system of N particles interacting through conservative forces,
one can consider the “internal potential energy” of the system, U int, and
any “external potential energy”, U ext, imposed from outside.
int
ext
 The total potential energy is then U  U  U
   U ab   U aext

a b a
a
where the sums are over all of the particles and the b > a in the second of
the double sum ensures that we do not double count an interaction, e.g.
U12 and U21.
2
 The total kinetic energy is much simpler, just T   Ta   12 ma va .
a

a
One can calculate the force on particle a by Fa  -aU , meaning calculate
the gradient of U at the position of the particle a. The total energy
E = T + U is conserved.
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Rigid Bodies






The subtlety of
U  U int  U ext    U ab   U aext can be illustrated with
a
an example of rigid body rotation. a b a
For a macroscopic rigid body, the number of particles (atoms) may be vast.
However, because the body is rigid, the interaction forces between atoms is
constant, and the particles do not shift position appreciably to change the
internal potential energy.
Thus, the internal potential energy is constant and we can completely
ignore it.
However, if the body is NOT rigid, and the shape or configuration of atoms
changes, then there may be an appreciable change in internal kinetic
energy that DOES have to be taken into account.
An example is a star that may expand or contract during its lifetime. The
change in configuration of the internal layers of the star affect the potential
energy, and hence also the kinetic energy.
Since total energy is conserved, for example, an expanding star must
become cooler.
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Example 4.9: A Cylinder Rolling Down
an Incline

Statement of the problem:


h
A uniform rigid cylinder of mass M, radius R rolls without slipping down
a sloping track as shown in the figure. Use energy conservation to find
its speed v when it reaches a vertical height h below its point of release.
Solution:


The cylinder is a rigid body, so we may ignore its internal potential energy. The
external potential energy is U ext  Mgy , where y is the vertical position of its
CM measured from any convenient location.
The KE of a rotating (rolling) cylinder is T  12 Mv 2  12 Iw 2 , where I is the
moment of inertia, I = ½MR2, and w is the angular velocity of rolling. Because
the cylinder is rolling without slipping, w = v/R. Thus, the final kinetic energy
is
T  34 Mv 2 .

If the cylinder starts from rest, the change in kinetic energy is equal to minus the
change in potential energy
3
4

So the final speed is v 
4
3
Mv 2  Mgh,
gh .
October 3, 2008