# Kinetic Energy-Mass conversion

1. Dec 14, 2012

### ShamelessGit

1. The problem statement, all variables and given/known data
I'm trying to learn relativity on my own by doing practice problems on the internet. One of the ones I'm having difficulty with is one where you have to calculate the threshold kinetic energy necessary to create an anti-proton by smashing two protons together. I think you assume that one proton is at rest and the other smashes it, and then all 3 protons and the anti-proton move together at the same speed after the collision. You have to create a proton as well as an anti-proton in order to conserve charge.

2. Relevant equations

Kinetic energy = √(mc^2/(1-(v/c)^2)) - mc^2
Momentum = mv/√(1-(v/c)^2)

Rest energy = mc^2

Energy and momentum conserved

3. The attempt at a solution

I figured that I'd try to solve it the same way I'd try to solve for an elastic collision in classical physics, which is by setting up the conservation of energy and conservation of momentum and then by solving the system of equations.

I get:

K1 + M1 = K2 + M2

P1 = P2

P stands for momentum, K stands for kinetic energy, and M stands for mass energy.

This is the second time I've tried to work it this way. The first time I actually tried to solve it algebraically, which is a mess, and I found that I had set up an inconsistent system when I finally had rearranged everything and substituted. I figured I must have just set up the equation wrong and I tried again a few days later (today), but I noticed that this question is a part of a test that students would have to take and it is only worth 1 point, which is the same amount of points given to much easier questions, such as a simple Doppler-shift calculation, or simply writing down the Lorentz transformations (without having to use them). So I'm thinking that there must be a much easier solution to this problem? Also, are the equations I wrote down correct? I haven't actually tried to substitute anything this time because I can see that it will be an awful problem and probably not what the instructor was looking for.

2. Dec 14, 2012

### Ray Vickson

Such problems are almost always much easier if you go to the so-called "center of mass (CM) frame", in which the initial (and final) total momentum = 0. That is what physicists do: carry out the basics in the CM frame, then transform afterwards into other frames if needed.

3. Dec 14, 2012

### ShamelessGit

You're right! That does make it so much easier! Thanks!

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