1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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: 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.
     
  2. jcsd
  3. May 15, 2010 #2

    K29

    User Avatar

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Yes, because the product of two four-vectors is invariant.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook