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Relativistic mass thought experiment/proof?

  1. Dec 29, 2011 #1
    Hey all,
    After reading some fairly elementary texts on special relativity, I understand length contraction and time dilation no problem, as it is well explained with examples of proofs, but I hate to say that relativistic mass is explained nowhere. I know this question has probably been asked before, but I will ask it anyway: Is there a fairly simple proof for relativistic mass, just as there is for length contraction and time dilation? Is it something to do with spacetime momentum four vector or something along those lines? I would really love to understand this, as it would allow me to understand special relativity more fully in order to read about general realtivity etc.
    Any answer would be gratefully appreciated.
    Thanks, Dom
     
  2. jcsd
  3. Dec 29, 2011 #2

    Bill_K

    User Avatar
    Science Advisor

    sorax123, In relativity the three-momentum is p = γmv where m is the rest mass. In olden times people tried to make this look like Newtonian mechanics by defining a relativistic mass M = γm, just so they could write it as p = Mv. Bad idea. It is not the actual mass and leads to a great deal of confusion. Suggest you avoid it like the plague.
     
  4. Dec 29, 2011 #3

    Dale

    Staff: Mentor

    I second Bill_K's suggestion. Relativistic mass is a generally deprecated concept. When modern physicists speak of "mass" they almost exclusively mean "invariant mass".
     
  5. Dec 29, 2011 #4
    Thanks for your additions guys, but for my piece of mind, is there any way to prove it, as it must have some weight behind it? I see some people use a right angled triangle with the hypotenuse being E=mc^2, and the other two being Ep=pc=mvc and Eo=moc^2, but how is this the case as where i have read the hypotenuse is mc, being the vector of the momentum four vector, the time part being γmc, and the space part being γmv, so how can this be?
    Any help to regain my sanity would be appreciated.
    D
     
  6. Dec 29, 2011 #5

    Dale

    Staff: Mentor

    Are you familiar with four-vectors?
     
  7. Dec 29, 2011 #6
    I am familiar with the momentum four vector, so I suppose I am :/
     
  8. Dec 29, 2011 #7

    Dale

    Staff: Mentor

    Then we have: [itex]\mathbf p = (E/c,p_x,p_y,p_z)[/itex]

    The invariant mass is then defined as: [itex]- c^2 m_{invariant}^2 = |\mathbf p|^2= -E^2/c^2 + p_x^2+p_y^2+p_z^2[/itex]

    The relativistic mass is defined as: [itex]m_{relativistic} = E/c^2[/itex]

    For a timelike particle we have [itex]p_i=m_{relativistic} v_i[/itex]

    So by substitution and solving you can obtain:
    [tex]m_{relativistic} = \frac{m_{invariant}}{\sqrt{1-v_x^2/c^2-v_y^2/c^2-v_z^2/c^2}}=\gamma m_{invariant}[/tex]
     
    Last edited: Dec 29, 2011
  9. Dec 29, 2011 #8
    I kind of get what you're saying in your reply, but I was wondering if there was a different way of getting it. The idea of the right angled triangle with the sides mentioned in my original post seems the most simple, so is there any way you could explain that?
    My brain says thanks
     
  10. Dec 29, 2011 #9

    Dale

    Staff: Mentor

    Well, (setting c=1 for convenience), from the above, you can always write:
    [itex]m_{relativistic}^2=m_{invariant}^2+p_x^2+p_y^2+p_z^2[/itex]
    which is the equation for the hypotenuse of a 4D "triangle". However, note that the space in which this "triangle" is NOT the usual Minkowski spacetime. The invariant mass is the "hypotenuse" of that "triangle".
     
  11. Dec 29, 2011 #10
    Aha I see, this seems to make sense more
     
  12. Dec 29, 2011 #11
    But still, out of curiosity, how on earth could you derive the expressions displayed in post 1? :S
     
  13. Dec 29, 2011 #12
  14. Dec 29, 2011 #13
    Thanks for your reply Dr, I am only a high school student, and haven;'t covered calculus :/, was wondering if there was a more elementary derivation?
     
  15. Dec 29, 2011 #14

    Dale

    Staff: Mentor

    There are no expressions displayed in post 1:
    Perhaps you meant to include something, but forgot.
     
  16. Dec 29, 2011 #15
    Apologies for that, stupid mistake on my behalf; I meant my second post regarding the right angled triangle :)
     
  17. Dec 29, 2011 #16
    That should be sufficient if you start with m(v):=mo·f(v) and forget about the energy.
     
  18. Dec 29, 2011 #17
    I originally tried to prove it as follows, but works only if time dilation is used, without length contraction, when length contraction is introduced the mass is deemed to be more;
    γmL/T=γm'L'/T'

    γmL/T=m'L/γT

    multiplying both sides by T gets us:

    γmL=m'L/γ

    multiplying by γ gets:

    γ2mL=m'L

    so:

    m'=γ2m


    I know there is something wrong here as is should be m'=γm, please correct my arithmetic, or let me know if it is possible to show relativistic mass in this form.
     
  19. Dec 29, 2011 #18
  20. Dec 29, 2011 #19

    Dale

    Staff: Mentor

    Just take the second equation in my post 7, multiply both sides by c², and then substitute in the third equation.
     
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