# Centre of Mass frame of colliding molecules

1. Mar 10, 2008

### jbunten

I'm not sure this is the right place to post this question but here goes:

In feynman vol.1 39-9 there is a situation where two molecules are about to collide in a CM frame, the frame has velocity Vcm and the two molecules have respective velocities v1 and v2, then there is no correlation between Vcm and the relative velocity, w (w=v1-v2) "coming in". what does Feynman mean by "coming in"?

I find this a very difficult argument to understand, specifically, is w really relative velocity *before* the collision? in that case how can it be independent of Vcm, surely if one of the molecules is very heavy and very fast and the other slow and light, then Vcm is moving very fast in roughly the same direction of the heavy molecule and a correlation exists.

If someone could look this up I'd appreciate it as I've been trying to get my head around it for ages, many thanks.

2. Mar 11, 2008

### pkleinod

My copy of Feynman, Leighton, and Sands (1963) has the
discussion on pages 39-8 and 39-9. Maybe you have a more
recent version, in which case some of what I say may not
be appropriate.

The presentation in my copy is rather confusing and the
diagram (Fig. 39-3) is incorrect: there are two different
vectors with the same label v1 and two
different vectors with label v2. In
addition, the caption says "...viewed in the CM system",
whereas the text defines the vectors v to be in the
laboratory system.

In what follows, capital letters refer to vectors in
the laboratory system and lower case letters refer
to the centre-of-mass (CM) coordinate system or to the
coordinates of one particle relative to the other. I am
going to consider the two-body problem where there are
no external forces and where the internal forces are
central forces obeying Newton's third law.

Let Ri and Vi be the position and
velocity vectors respectively of particle i in the
laboratory system and let mi denote the mass of particle i.
Introduce two new coordinates:

R = (m1R1+m2R2)/M
r = R1 - R2
M = m1 + m2

The inverse transformation is

R1 = r + m2r/M
R2 = r - m1r/M

There are similar equations for the velocities, obtainable
from the above by differentiating wrt time. In the absence
of external forces, the transformation achieves a separation
of the two-body problem into two single-particle problems:

M dV/dt = 0
m dr/dt = F1i
m = m1m2/M

where F1i is the force exerted by
particle 2 on particle 1. and m is the reduced mass.
You can see from this that the
centre-of-mass travels in a straight line with constant
velocity, regardless of the force acting between the
particles. That is what is meant by "uncorrelated", nothing
more. When Feynman says "coming in" he means the relative
velocity before the collision.

To see what happens after the collision, you must see two
things: the linear momentum is MV both before and
after the collision, and the kinetic energy is

T = 1/2(MV2 + mv2)

For an elastic collision, the first condition will
be satisfied if the velocities of
particles 1 and 2 after the collision as seen from the
centre-of-mass are in opposite
directions and the second condition will be satisfied if
the magnitude of v is the same before and after
the collision. This means that the CM velocity vector
for particle 1 must lie on a sphere centred at the end
of the vector V and the CM velocity vector for
particle 2 must be in the opposite direction and
ending on a second sphere centred at the end of V.

Draw vectors V1 and V2
with their tails placed in the origin of the laboratory
system. Draw vector v. The vector V has its
tail in the laboratory origin and its head somewhere on the
vector v---the exact position depends upon the masses.
Using the end of this vector as a new origin, draw a circle
of radius m2v/M showing all possible
velocities of particle 1 after the collision. Draw another
circle of radius m1v/M showing the
possible velocities of particle 2 after the collision.

Hope this is of some help. You have to stare at the diagram
a while.

3. Mar 13, 2008

### jbunten

Thanks for your reply and sorry for taking a while to answer, that was a really good and detailed explanation. I think my confusion arose from the poor definition of "uncorrelated" and the confusing diagram. Once I drew the diagram as you said it made perfect sense.

Many thanks.