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Derivation of Relativistic Momentum WITHOUT using Relativistic Mass?

  1. Aug 18, 2011 #1
    Derivation of Relativistic Momentum WITHOUT using Relativistic Mass???

    Does anyone know a way to derive relativistic momentum without falling back on the concept of relativistic mass?

    Also, if it is not already part of the derivation given, does anyone know how to show that relativistic momentum is conserved in elastic collisions?
     
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  3. Aug 18, 2011 #2

    jtbell

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    Re: Derivation of Relativistic Momentum WITHOUT using Relativistic Mass???

    I've always seen it the other way around: first you show that mv (where m is the classical mass) is not conserved in relativistic collisions, after taking time dilation etc. into account, but [itex]\gamma mv[/itex] is conserved; and then defining "relativistic mass" as [itex]\gamma m[/itex].

    I don't know any Web sites offhand that use this treatment because I learned it from paper-and-ink books a long time ago, e.g. Beiser's modern physics textbook which I taught out of for several years. Or was it cuneiform tablets?...
     
  4. Aug 18, 2011 #3
    Re: Derivation of Relativistic Momentum WITHOUT using Relativistic Mass???

    That's exactly the method I've been trying. In another thread I started, "Help with Relativistic Collision" (which so far no one has responded to), I consider an elastic collision. It's easy to show that mv is not conserved but so far I have not had success in showing that [itex]\gamma[/itex]mv is. Do you know how to show that it is?
     
  5. Aug 18, 2011 #4

    BruceW

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    Re: Derivation of Relativistic Momentum WITHOUT using Relativistic Mass???

    search 'four-velocity' on wikipeida, and it gives an explanation for the four-velocity. From there, you just multiply by the rest mass to get the four-momentum.
    The 0'th component of the four-momentum is the relativistic energy and the 1,2,3 components of the four-momentum are the relativistic momentum.
     
  6. Aug 18, 2011 #5

    pervect

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    Re: Derivation of Relativistic Momentum WITHOUT using Relativistic Mass???

    Try Goldstein, "Classical Mechanics", if you get get a hold of it.
     
  7. Aug 18, 2011 #6
    Re: Derivation of Relativistic Momentum WITHOUT using Relativistic Mass???

    and once you have four momentum how do you prove conservation?
     
  8. Aug 18, 2011 #7

    bcrowell

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    Re: Derivation of Relativistic Momentum WITHOUT using Relativistic Mass???

    In 1+1 dimensions, the answer to the OP is pretty trivial. Let m be invariant mass, [itex]m_r=\gamma m [/itex] relativistic mass. If you have a derivation that's written in terms of [itex]m_r[/itex], substitute [itex]\gamma m[/itex] for it everywhere, and you have a derivation without using relativistic mass.

    The generalization from 1+1 to 3+1 dimensions also seems pretty trivial to me. If you've established [itex]m^2=E^2-p^2[/itex] and conservation of (E,p) in 1+1 dimensions, is there really any mystery about how to generalize p from a 1-vector to a 3-vector?
     
    Last edited: Aug 18, 2011
  9. Aug 18, 2011 #8

    BruceW

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    Re: Derivation of Relativistic Momentum WITHOUT using Relativistic Mass???

    I guess relativistic momentum is conserved for the same reason that classical momentum is conserved in the classical limit.
    In the classical limit, we say momentum is conserved because the laws of physics are invariant under a change in the position of the origin of the coordinate system.
    Maybe relativistic momentum is conserved for the same reason?
     
  10. Aug 19, 2011 #9
    Re: Derivation of Relativistic Momentum WITHOUT using Relativistic Mass???

    I seem to be having trouble getting the velocity of the mass to match up with the velocity referred to in the gamma function (1-v2/c2)-1/2. Let me take you through my thinking. This is a simple derivation of relativistic mass using the classic elastic collision example http://faculty.physics.tamu.edu/adair/phys222/chapt2/Special Relativity Part II a.pdf . On the last step the author gets to the point where she can say

    mouy=m'uy/[itex]\gamma[/itex]

    It seems innocent enough at first glance, but going back through the author's reasoning, it looks like that [itex]\gamma[/itex] is a function of v (the velocity between the two frames) as opposed to uy, which is what we would want for a consistent momentum formula.
     
  11. Aug 19, 2011 #10
    Re: Derivation of Relativistic Momentum WITHOUT using Relativistic Mass???

    Perhaps this is less confusing: http://en.wikibooks.org/wiki/Special_Relativity:_Dynamics
     
  12. Aug 19, 2011 #11

    DrGreg

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    Re: Derivation of Relativistic Momentum WITHOUT using Relativistic Mass???

    I don't know if my own thread Derivation of momentum & energy formulas is the sort of thing you are looking for?
     
  13. Aug 19, 2011 #12
    Re: Derivation of Relativistic Momentum WITHOUT using Relativistic Mass???

    Besides analyzing collisions, I've seen a derivation of the relativistic mass that goes something like this:

    Take gravity to be a constant, downward acceleration. There is a long, straight pole parallel to the ground (and perpendicular to the gravitational acceleration) which is held up by another pole extending from the ground to its center of mass. In the reference frame in which the center of mass isn't moving, two equal masses suspended from the pole move in opposite directions with speed u. Clearly the net torque is always zero at any point in time. Here is an illustration of the system:

    [PLAIN]http://k.min.us/icl8xvvNg.jpg [Broken]

    Now analyze the system from the reference frame in which the left mass is at rest. Now the system looks like this:

    [PLAIN]http://k.min.us/ibipRGSAQ.jpg [Broken]

    From relativistic velocity addition, we know that (working in units where c=1):

    [tex]u'=\frac{2u}{1+u^2}[/tex]

    This velocity is less than what would be expected in a Newtonian system, yet (as we already know) the net torque must still be zero. We therefore conclude that the mass mu' must be greater than m0 to compensate.

    From the fact that [itex]\tau_{net} =0[/itex]:

    [tex]m_0gut = m_{u'}g(u'-u)t[/tex]

    [tex]\frac{m_0}{m_{u'}} = \frac{u'}{u}-1[/tex]

    Now if you get everything in terms of u' (solve the first equation for u and sub it into the above equation) you can simplify it down to:

    [tex]\frac{m_0}{m_{u'}} = \sqrt{1-u'^2}[/tex]
     
    Last edited by a moderator: May 5, 2017
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