# Homework Help: Relativity Question

1. Nov 17, 2007

### joker_900

1. The problem statement, all variables and given/known data
An electron of energy 10GeV strikes a proton at rest. What is the velocity of the electron-proton CM system? What energy is available to produce new particles?

2. Relevant equations

3. The attempt at a solution
I've done the first part, but am stuck on the second. I was thinking that in the CM frame after the collision, as the sum of momenta is zero, if the max amount of energy possible went into new particles,

ECM = mc^2 + Mc^2 + E*

where m is the mass of an electron, M the mass of a proton and E* the energy going into new particles. I have ECM from the first part, so I can get E*. However I don't know how to transform this energy into energy in the lab frame.

I (hopefully) won't need a full solution, just some pointers on how to approach this please!!

2. Nov 17, 2007

### Shooting Star

Suppose that only one neutral particle with a non-zero rest mass m1 is produced. It’s also at rest in the CM frame. The max energy will be made available when the P and e are also at rest after collision, provided they are not destroyed. So, all three are travelling with v in lab frame after collision. The energy in CM frame of new particle is m1c^2, which is equal to your E*. If energy available in CM frame is m1c^2, then it must be m1*gamma*c^2 in lab frame. Express it as a function of initial electron energy Ee.

3. Nov 18, 2007

### joker_900

Thanks but I still can't get it. This is what I did after your help:

Conservation of energy in CM frame: ECM = mec^2 + mpc^2 + m1c^2

i.e. the initial energy in the CM frame equals the sum of the rest masses

Conservation of energy in lab frame: Ee + mpc^2 = Ee1 + Ep1 + En (En is energy of new particle in lab frame)

so Ee + mpc^2 = gamma(mec^2 + mpc^2 + m1c^2)
as they all have the same velocity they must have the same gamma in lab frame

so Ee + mpc^2 = gamma*ECM

This gives a gamma of about 2, and then doing

En = gamma*m1c^2 = gamma*(ECM - mec^2 - mpc^2)

gives the wrong answer (it should be 4.4 GeV)

4. Nov 18, 2007

### Shooting Star

Up to here, all seems to be fine. But gamma is dependent only on the velo of the CM frame. You should not have to find gamma(v) afterward. I presume that you can find the velo of the CM frame. Suppose it’s v.

In the lab frame,

Ee + Mpc^2 = (Mp + Me + m1)*gamma(v)*c^2.

From this, you can find m1*gamma(v)*c^2 = E_available in terms of the other quantities.

(Note that by CM frame here, I mean the centre of momentum frame, i.e. in which the total momentum vanish.)

5. Nov 18, 2007

### joker_900

Sorry I'm being really slow. Gamma changes after the collision right? So how do I find the new velocity of the CM frame in the lab frame? When I did this for the first part of the question, my method required knowing the masses of the electreon and proton and the energy of the electron in the lab frame. After the collision I don't know these?

6. Nov 18, 2007

### Shooting Star

The whole point is that for maximum energy to be available, all three particles should be at rest in the CM frame. The velocity of the CM frame does not change. So, you calculate that before the collision, and use it to find the reqd energy after the collision.

7. Nov 18, 2007

### joker_900

But I'm doing that and it's giving me the wrong answer :(

The velocity of the CM frame in the first part is 0.914c. So gamma = (1 - 0.914^2)^-0.5 = 2.46

Mpc^2 = 938 MeV Mec^2 = 0.512 Ee = 10000MeV

And putting this in i get m1*gamma*c^2 = 8625 MeV

!

8. Nov 18, 2007

### Shooting Star

Are you sure you've done the arithmetic correctly...? Anyway, try this formula for now, until I can verify your calc.

Energy available = sqrt([2Mp*c^2*Ee + (Mp*c^2)^2 + (Me*c^2)^2]