# Find a reference frame where momenta of electron and proton are equal

1. Feb 7, 2012

### khfrekek92

1. The problem statement, all variables and given/known data
The electron is traveling at a speed of β=.9999999, γ=1957, with mass mc^2=.51099 MeV.
The Proton is traveling at a speed of β=.9, γ=2.29, with mass mc^2=938.27 MeV.
They are heading in opposite directions, directly towards each other on the x-axis

Find the reference frame where their momenta are equivalent, and find the total energy in the frame, and the total kinetic energy.

2. Relevant equations
see below

3. The attempt at a solution
I'm not exactly sure if I'm going about this right but I have started a lorentz transform with the momentum 4-vectors of both the electron and proton. Then I took the second component of the resultant vectors which would be equal to γp, then I set them equal to each other and tried to solve for a velocity. Am I going about his right? All the work is shown in the attached jpeg.

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2. Feb 8, 2012

### BruceW

I don't think your working is correct. For the electron, you've got its original momentum 4-vector in terms of $\gamma_e$ and ue, and then you do a Lorentz transform on that, but you use the same $\gamma_e$, so the momentum 4-vector you get on the right hand side is just the 4-vector for an electron at rest. (You can check by cancelling down).

You should be able to answer the question by using the same Lorentz transform on the electron and proton (to get them both into the new reference frame). But you need to use a new gamma and beta. Also, you mentioned making the spatial momenta of the electron and proton equal. But I don't think this is what the question is asking for. I think the question is asking for the spatial momenta to be equal and opposite.

One last thing: you should be able to work out the answer by using a Lorentz transform of the electron and proton 4-momenta. But you can make things easier on yourself if you think about what you are trying to find in the new reference frame, you'll see that although transforming the individual 4-momenta is the most straightforward thing to do, it is not the easiest calculation to get the answer.

3. Feb 8, 2012

### khfrekek92

Oh! Yeah I have no idea what I was doing.. But I fixed my mistake and now no matter what I do I get that the speed of the reference frame is 1.000116893*c! That is definitely not possible....
And I'm not sure what you mean about making it easier, I have to find the frame where the mass*velocity is the same for the proton as it is for the electron.. hmmm..

4. Feb 9, 2012

### BruceW

a speed greater than c? yeah, there must be something going wrong... maybe you're trying to do the transform of reference frame in the wrong direction or something like that?
About making it easier: (hint) you can do addition or subtraction on 4-vectors you already have, to get another 4-vector.

5. Feb 9, 2012

### khfrekek92

I didn't even think of that!! So all I have to do is sum the fourvectors and do the lorentz transform on that vector to get the the zero momentum frame?! Then use the invariant to get the speed?!

6. Feb 9, 2012

### BruceW

Yep, just add the two fourvectors. Its easy to forget this 'trick', but it often makes questions easier.