1. Not finding help here? Sign up for a free 30min 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!

Relativistic dynamics

  1. Jul 31, 2007 #1
    1. The problem statement, all variables and given/known data
    Consider a particle of mass m traveling at speed v in the positive x direction. It splits into two pieces which travel in the x-y plane with velocities v1(vector)=v1cos(theta)x(hat) + v1sin(theta)y(hat) and v2(vector)=v2cos(phi)x(hat) - v2sin(phi)y(hat).
    a. Find the masses m1 and m2 of the two fragments.
    b. Transform to the center-of-mass frame and find the new velocities v1(vector) and v2(vector).


    2. Relevant equations



    3. The attempt at a solution
    I'm so lost as to where to start because it seems so arbitrary to say that the masses split into two pieces and then we need to determine the masses of each fragment. Right now I'm just looking for a conceptual explanation so i can wrap my head around it and at least start somewhere. Thanks in advance for the help.

    -Pat
     
  2. jcsd
  3. Jul 31, 2007 #2
    You must assume the angles [tex]\phi[/tex] and [tex]\theta[/tex] and, as far as I can see, also the velocities v1 and v2 as known for the problem to have a unique solution.

    The equation you're looking for is conservation of energy and momentum: [tex](E, \vec p) = (E_1, \vec p_1) + (E_2, \vec p_2)[/tex].

    Since I'm unfamiliar with the homework help rules in this forum I'll skip the rest of my post and copy it to some textfile for possible later use. Hope that helped you so far.

    Oh, one important triviality for number (b): The cms (geek-name for "center-of-mass") frame is defined by [tex]\vec p = 0[/tex].
     
  4. Jul 31, 2007 #3
    Thanks, that helps. Would you mind telling me how you made those vector notations?
     
  5. Jul 31, 2007 #4
    You can click on the images of the mathematical notations which pops up a box showing the source-code. You can also do a forum search for the terms "tex" or "latex" and see if there's some tutorial (I dunno if there is, but I'd expect it given that it's a very common question).
     
  6. Aug 2, 2007 #5
    I'm guessing that the conservation of energy and momentum is interrelated in this problem, so I'm not sure how to combine the two. In other words, I'm not sure how to express the mass in terms of both the conservation of energy and momentum.
     
  7. Aug 2, 2007 #6
    E² - (pc)² = (mc²)². Very important formula, btw.
     
  8. Aug 2, 2007 #7
    Is energy in this case some factor of [tex]\gamma[/tex] or is it kinetic energy? I'm referring to gamma as 1/[tex]\sqrt{1-u^{2}/c^{2}}[/tex].
     
  9. Aug 2, 2007 #8
    You have two ways to express the energy, one that uses the gamov-factor (that's the name of the gamma) and the version I gave in post #6 (i.e. via the momentum). The way I solved the problem I had to make use of both. It would probably help if you wrote how far you got so far and what you did. Otherwise responding involves a lot of guessing what you're talking about / thinking.
     
  10. Aug 2, 2007 #9
    So when you say that the energy is expressed in two ways then do you mean that they are equal to each other?
     
  11. Aug 2, 2007 #10

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    One way is to write

    [tex] E = \gamam m c^2 [/tex]

    the second way is
    [tex] E = \sqrt{p^2 c^2 + m^2 c^4 } [/tex]

    And yes, they are equivalent. Plug on in the other and solve for the magnitude of the three-momentum, p. You will find [itex] p = \gamma m v [/itex] as expected.



    By the way, I think you asked what E stands for here. It really is the rest mass energy plus the kinetic energy.


    By the way #2: maybe he best way to do this problem is by using four-momenta and conservation fo four-momentum. Have you seen that?
     
  12. Aug 2, 2007 #11
    im not sure what you mean by conservation of four-momentum.
     
  13. Aug 2, 2007 #12
    from your equations im getting:

    E=[tex]\gamma[/tex]mc[tex]^{2}[/tex]
    E=[tex]\sqrt{p^{2}c^{2}+m^{2}c^{4}}[/tex]
    [tex]\gamma[/tex]mc[tex]^{2}[/tex]=[tex]\sqrt{p^{2}c^{2}+m^{2}c^{4}}[/tex]
    take out c[tex]^{2}[/tex]
    [tex]\gamma[/tex]mc[tex]^{2}[/tex]=c[tex]\sqrt{p^{2}+m^{2}c^{2}}[/tex]
    cancel c
    [tex]\gamma[/tex]mc=[tex]\sqrt{p^{2}+m^{2}c^{2}}[/tex]

    Am i correct so far? and if so where should i go from there?
     
  14. Aug 3, 2007 #13

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Square both sides and isolate p. Now use the expression for gamma and do a bit of algebra. You will end up with [itex] p = \gamma m v [/itex].
     
  15. Aug 3, 2007 #14
    Ok, first, as promised, the full version of my original (1st, #2) post:
    The conservation of momentum gives you two equations for the two unknown quantities [tex] | \vec p_1 |, \ | \vec p_2 | [/tex]. Expressing the energy in the two different ways noted (once expressed by the momenta you now know, the other time by using the gamov-factors) then gives you two (hopefully) different equations involving the two unknown quantities (the two masses). That should be solvable for the masses, then. If you're still stuck I can also sketch the necessary mathematical manipulations. But given that you're taking a relativity course (where I'm assuming it is homework) it would be high time to learn and practice how to solve a system of equations for two unknown variabes - just in case that was/is/will be the problem.

    The whole thing reworded again:
    - Determine the momenta of the two particles from conservation of momentum.
    - Knowing the momenta and the gamov-factors, determine the masses from using conservation of energy (twice).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Relativistic dynamics
  1. Relativistic dynamics (Replies: 1)

  2. Relativistic dynamics. (Replies: 16)

Loading...