# Special Relativity Collision

1. May 15, 2010

### astoria

1. The problem statement, all variables and given/known data

A particle of rest mass M and total energy E collides with a particle of rest m at rest. Show that the sum E' of the total energies of the two particles in the frame at which their centre of mass is at rest is given by

(E')^2=(M^2+m^2)c^4+2Emc^2

3. The attempt at solution

I'll denote particle of rest mass M as having velocity v and particle of rest mass m as having velocity w and the relevant gammas as G(v), G(w)

E=Mc^2

Conservation of four momentum implies:

E'=G(v)Mc^2+G(w)mc^2

0=G(v)Mv+G(w)mw (where v and w are vectors)
G(v)Mv=-G(w)mw
(G(v)Mv)^2=(G(w)mw)^2 (taking modulus squared of both sides)
M^2(G(v)^2-1)=m^2(G(w)^2-1) (as G(u)^2=c^2(G(u)^2-1)

I can't see how to eliminate both G(v) and G(w) and get an answer for E', I need another equation.

2. May 15, 2010

### K29

hmmm. Thats not quite the way I'd do things (I get to an answer pretty close to the right one but not quite)

E (before the collision) = E' (after)
So use the fact that:
E = kinetic energy + rest mass.

Not sure why but I get the same solution as yours: in my final answer I have a $$\gamma M$$. In the first bracket. The 2Emc^2 part comes out right though. Maybe you can find if/where I made a mistake.

3. May 15, 2010

### astoria

I may be misunderstanding what you mean but I think what you are saying is that the energy is the sum of the Newtonian kinetic energy and mc^2 (i.e E=(mc^2+1/2mv^2))

This is only true if we take gamma as equal to 1+v^2/2c^2 and ignore the higher order terms.

4. May 15, 2010

### vela

Staff Emeritus
Try working in the original frame instead of the center-of-mass frame since E appears in your answer, and use E's and p's as your variables instead of writing them out in terms of gammas, masses, and velocities.

A very useful equation in these types of problems is E2-(pc)2=(mc2)2.

5. May 15, 2010

### astoria

Thanks, that came out fine :)

I've had a look at the official solutions (which are only outline) and it seems to do it a completely different way, using g(P+Q,V) etc

I've never seen this sort of problem done like that before, is it easier? unfortunately it's not comprehensively explained, how would I set it up?

6. May 15, 2010

### vela

Staff Emeritus
I have no idea. You'd have to explain it a bit more before I could tell what they did.

7. May 15, 2010

### astoria

Essentially it just says:

The four momenta of the particles are P=(E/c,p) and Q=m(c,0) in the given inertial frame. In centre of mass frame the four velocity V of the centre of mass decomposes as (c,0) and P+Q decomposes as (E'/c,0). Moreover g(V,V)=c^2 Therefore:

V=c(P+Q)/sqrt(g(P+Q,P+Q)) E'=g(V,P+Q)

Now use E'^2=(c^2)g(P+Q,P+Q)

8. May 15, 2010

### vela

Staff Emeritus
I'm not sure why the solution bothers working with the four-velocity. You can see directly from P+Q=(E'/c,0), that E'2=c2(P+Q)2. It's pretty much the same solution otherwise.

9. May 16, 2010

### astoria

Having read more solutions of this form, am I right to assume that if V,W are the four velocities in ICS 1 and V' W' are the four velocities of two particles in ICS 2g(V,W)=g(V',W')?

10. May 16, 2010

### vela

Staff Emeritus
Yes, because the product of two four-vectors is invariant.