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Homework Help: Special Relativity Collision

  1. May 15, 2010 #1
    1. The problem statement, all variables and given/known data

    A particle of rest mass M and total energy E collides with a particle of rest m at rest. Show that the sum E' of the total energies of the two particles in the frame at which their centre of mass is at rest is given by

    (E')^2=(M^2+m^2)c^4+2Emc^2

    3. The attempt at solution

    I'll denote particle of rest mass M as having velocity v and particle of rest mass m as having velocity w and the relevant gammas as G(v), G(w)

    E=Mc^2

    Conservation of four momentum implies:

    E'=G(v)Mc^2+G(w)mc^2

    0=G(v)Mv+G(w)mw (where v and w are vectors)
    G(v)Mv=-G(w)mw
    (G(v)Mv)^2=(G(w)mw)^2 (taking modulus squared of both sides)
    M^2(G(v)^2-1)=m^2(G(w)^2-1) (as G(u)^2=c^2(G(u)^2-1)

    I can't see how to eliminate both G(v) and G(w) and get an answer for E', I need another equation.
     
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  3. May 15, 2010 #2

    K29

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    hmmm. Thats not quite the way I'd do things (I get to an answer pretty close to the right one but not quite)

    E (before the collision) = E' (after)
    So use the fact that:
    E = kinetic energy + rest mass.

    Not sure why but I get the same solution as yours: in my final answer I have a [tex]\gamma M[/tex]. In the first bracket. The 2Emc^2 part comes out right though. Maybe you can find if/where I made a mistake.
     
  4. May 15, 2010 #3
    I may be misunderstanding what you mean but I think what you are saying is that the energy is the sum of the Newtonian kinetic energy and mc^2 (i.e E=(mc^2+1/2mv^2))

    This is only true if we take gamma as equal to 1+v^2/2c^2 and ignore the higher order terms.
     
  5. May 15, 2010 #4

    vela

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    Try working in the original frame instead of the center-of-mass frame since E appears in your answer, and use E's and p's as your variables instead of writing them out in terms of gammas, masses, and velocities.

    A very useful equation in these types of problems is E2-(pc)2=(mc2)2.
     
  6. May 15, 2010 #5
    Thanks, that came out fine :)

    I've had a look at the official solutions (which are only outline) and it seems to do it a completely different way, using g(P+Q,V) etc

    I've never seen this sort of problem done like that before, is it easier? unfortunately it's not comprehensively explained, how would I set it up?
     
  7. May 15, 2010 #6

    vela

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    I have no idea. You'd have to explain it a bit more before I could tell what they did.
     
  8. May 15, 2010 #7
    Essentially it just says:

    The four momenta of the particles are P=(E/c,p) and Q=m(c,0) in the given inertial frame. In centre of mass frame the four velocity V of the centre of mass decomposes as (c,0) and P+Q decomposes as (E'/c,0). Moreover g(V,V)=c^2 Therefore:

    V=c(P+Q)/sqrt(g(P+Q,P+Q)) E'=g(V,P+Q)

    Now use E'^2=(c^2)g(P+Q,P+Q)
     
  9. May 15, 2010 #8

    vela

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    I'm not sure why the solution bothers working with the four-velocity. You can see directly from P+Q=(E'/c,0), that E'2=c2(P+Q)2. It's pretty much the same solution otherwise.
     
  10. May 16, 2010 #9
    Having read more solutions of this form, am I right to assume that if V,W are the four velocities in ICS 1 and V' W' are the four velocities of two particles in ICS 2g(V,W)=g(V',W')?
     
  11. May 16, 2010 #10

    vela

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    Yes, because the product of two four-vectors is invariant.
     
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