Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Quick Special Relativity Q

  1. Jan 28, 2010 #1


    User Avatar

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

    Tevatron carries to beams of 1 TeV protons moving in opposite directions, so that for proton-proton collisions the COM frame is also the LAB frame. What proton energy must an accelerator provide to get the same COM energy when bombarding a stationary proton

    2. Relevant equations

    3. The attempt at a solution

    Tried working out the velocity of the two protons, and then using velocity addition to add these to see what it would be if one were at rest...hence I could work of the KE of the single proton..

    The problem is that the velocity is so close to the speed of light that my calculator won't allow me to do it this way..

    is there another way to approach it?
  2. jcsd
  3. Jan 28, 2010 #2


    User Avatar

    Any ideas?
  4. Jan 28, 2010 #3
    You don't need a calculator. 1 TeV is E in the COM frame. If you want to figure out what the energy of a collision is in a different frame, just boost to that frame. That means pick one proton and boost to its frame, so it's at rest, and then compute the energy of the other proton in that frame.
  5. Jan 28, 2010 #4
    I suggest that you do it algebraically, not with a calculator. Then you should be able to do it according to your original idea. The other benefit is that you see immediately from the formula how it would be for other values of the energy E ( = 2TeV in your case )

    Otherwise, I think it can be done without using a boost, like this:

    Assume that the one moving proton has a 4-momentum P1. The stationary proton's 4-momentum is denoted P2.

    The total 4-momentum is therefore P = P1 + P2.

    Now, the energy of two particle system in its rest frame (COM frame) is the same as the system's rest mass. This can be found by taking minus the square root of the square of P. This is analogous to the rest mass of a single particle, which satisfies P^2 = -M^2 where M is the proton rest mass.

    You would demand that this is equal to E:

    (P1 + P2)^2 = -E^2


    P1^2 + P2^2 + 2P1.P2 = -E^2

    Then I use:
    P1^2 = P2^2 = -M^2


    P1.P2 = -M*sqrt(M^2+p^2)

    where I used the components of the vectors P1 and P2 in the lab frame:
    P1 = (sqrt(M^2+p^2), p)
    P2 = (M, 0)

    I have restricted to one spatial dimension, and written a lowercase p for the spatial component of the momentum.

    The equation can then be reduces to:

    -2M^2 - 2Msqrt(M^2+p^2) = -E^2


    sqrt(M^2+p^2) = (E^2 - 2M^2) / 2M

    But the left hand side is actually the energy of the incoming proton that has 3-momentum p, so that would be the answer to the question.

    Please note that I may have made some errors here, so please check that you agree with everything yourself. Please also double-check signs. I have used the (-,+,+,+) signature. Maybe you use the opposite one. If so, you could correct for that.

    Last edited: Jan 28, 2010
  6. Jan 28, 2010 #5
    Now I'll check out the Forum Rules to see if there is a rule about helping too much...

    EDIT: I see that one should not provide a complete solution, so I guess I jumped the gun here (if my solution is correct, that is).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook