# Homework Help: Relativistic Collission of Protons

1. Dec 1, 2011

### trevor51590

1. The problem statement, all variables and given/known data
A proton with M0 and energy 200 MeV collides with another proton at rest. How much kinetic energy does either proton have in the inertial frame in which they collide with equal and opposite momentum

2. Relevant equations
KE=1/2mv2
p=γmv
λ=1/√(1-v2/c2)
E=moc2γ

3. The attempt at a solution
Initially I figured I would need to solve for the velocity of the particle moving towards the rest particle. I use KE=1/2mv2 and get vim0 to be 0.65c

I also calculated my γ using my vi, and got 1.32

I understand that in all frames, the laws of physics hold, and as well that mass is going to be transferred from one particle to the other after the collision (energy). I am not sure how I should set up a diagram to have it make sense to me though - the wording of the problem is throwing me off a bit.

My mind tells me that if one proton a hits rest proton b, some energy from proton a will be transferred in the form of mass to proton b. Proton b will then start moving in the positive direction with momentum p. Proton a will bounce back with momentum -p, but the mass will be less and the velocity higher.

Am I on the right track at all? Or am I misinterpreting the question all together?

Thanks!

2. Dec 1, 2011

### Simon Bridge

The problem wants you to do the calculation in the frame were the two protons have equal and opposite momenta. This is usually the center-of-momentum frame - which has some nice properties.

The numbers you are given are the "lab frame".

I'll give you some notes:

You don't transfer mass in collisions - you transfer momentum and energy. The "mass increase" idea comes from thinking the gamma in the momentum and energy formula is applied to the rest-mass when it is more usefully thought of as applying to the whole thing as the kinetic energy contribution.

It can be helpful to think in terms of energies instead of masses:
$E_{tot}=E_0+E_K = \gamma E_0$ where $E_0=m_0c^2$

So: $E_K=(\gamma -1)E_0$

Or - in terms of momentum:
$E_{tot}^2 = (\gamma E_0)^2 + (pc)^2$

It can help to express speed as a proportion of c:
$pc=\gamma E_0 \beta : \beta =v/c$

All right - that should get you started.

Working out the relative speed to get gamma etc will get you there in the end, but you don't need to.
You need to decide if the 200MeV is kinetic or total energy.

From the energy relations you get gamma.
Then you need to make an equation that gives you kinetic energy in terms of momentum.
You know what the momentum of each proton has to be because of the reference frame.

Last edited: Dec 1, 2011
3. Dec 6, 2011

### trevor51590

I genuinely appreciate your help

After solving through, I got an Etot=1138 MeV, and a γ of 1.21

From there I solved for β using the E2tot equation and got 0.07c

V'1=V1-Vcm

Because v2=0, Vcm=0.035c

Therefore:
V'1=0.035c
V'2=-0.035c

Look okay?

Thanks again!

4. Dec 6, 2011

### trevor51590

I realized this part is incorrect because it is not relativistic.

Would this be the formula to use? I know that the speed of light is invariant under a Lorentz translation, and that leads to conservation of momentum, I'm having a hard time putting it together though

5. Dec 6, 2011

### Simon Bridge

A competent teacher will not mark your answers, but your method and how you understand the subject. So there is no point commenting on your answers. That said, it's not like there is no information ... you speeds look awfully low.

Note:
the relativistic equations always work - even in the non-relativistic case.
the total energy minus rest-mass energy is the kinetic energy - this is correct. Try expressing your m in terms of E/c2 - for a proton that is 938.3MeV//c2 so E0 is easy from there.

This should also tell you that the 200MeV is kinetic energy.

(Represent the classical KE equation in terms of rest-mass-energy and beta, then find out the speed ... it is relativistic?)

The total energy must be 1138.3MeV =γE0 so:
γ = 1138.3/938.3 = 1.213 (4sig.fig.)

Note: I got an equation wrong. Under "In terms of momentum" the first term should not have a gamma.

$E_{tot}^2 = E_0^2 + (pc)^2$

My apologies if this confused you. I do try hard to stop people from relying on me for answers though :) If I do this right, you no longer need me.

anyway - you can get beta from gamma, allowing you to get the speed the com frame is moving in the lab frame (the "stationary" proton frame). This gets you a new gamma, from which you get the momenta and kinetic energies in the com frame.

It can be tricky keeping all the equations in their right places. Lynchpin:$$E_{tot}=\gamma E_0=E_K+E_0=\sqrt{E_0^2 + (pc)^2}$$

For instance - see if you can derive the following:

$$p = \sqrt{\gamma^2 - 1} \frac{E_0}{c}$$

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