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B Is this anywhere close for a derivation of P=γ(u)mu?

  1. Sep 11, 2016 #1
    Okay, this is going to sound weird. I know how to get relativistic energy using calculus and the assumption that momentum is γ(u)mu, but I haven't ever been able to do it without assuming P = γ(u)mu as a given. I wanted to see how close I could get to deriving relativistic momentum strictly using the Lorentz transformation equations and no assumptions about collisions. It ended up with some sort of weird frame mixing.

    Please let me know what you think if you have the time and inclination to dig through my (most likely) faulty reasoning.






    First of all, since I'm looking for just the momentum equation, rather than a transformation equation, I assumed the special case when v = u is always going to apply for the moving object itself (you're in S and you see an object moving; well in that object's frame, S', it's just at rest).

    Then I took the time transformation for the S' frame and did some basic algebra with that assumption (v = u):
    [tex] t' = γ(u)(t - \frac{u}{c^2}x)[/tex]
    [tex] t' = γ(u)(1 - \frac{u^2}{c^2})t[/tex]
    [tex] t' = γ(u)γ(u)^{-2}t[/tex]
    [tex] t' = \frac{t}{γ(u)}[/tex]

    This looked familiar so I kept going.

    I went back to S'. In general, u' = x'/t'. So, I multiplied u' by m and then replaced t' with the above:
    [tex]mu' = m\frac{x'}{t'}[/tex]
    [tex]mu' = m\frac{x'}{\frac{t}{γ(u)}}[/tex]
    [tex]mu' = γ(u)m\frac{x'}{t}[/tex]

    This looked almost right, but it's obviously mixing frames so it can't be. So here I just figure I'd replace x' with the Lorentz transformation for location along the x-coordinate and see what happened, but no longer explicitly assuming that v = u (since if v = u I get some weird thing that looks like it's moving me further away from P = γ(u)mu):
    [tex]mu' = γ(u)m\frac{γ(v)(x - vt)}{t}[/tex]

    Since this looked odd, I decided to cheat and let v approach 0. Why? I don't know why. But if I do, out pops the correct value:

    [tex]\lim_{v\rightarrow\ 0} γ(u)m\frac{γ(v)(x - vt)}{t} = γ(u)m\frac{x}{t} = γ(u)mu[/tex]

    So... the right side becomes the correct momentum formula. But the equation is
    [tex] \lim_{v\rightarrow\ 0}mu' = γ(u)mu [/tex]

    If I just declare that
    [tex]P = \lim_{v\rightarrow\ 0}mu' = γ(u)mu [/tex]
    then I win. But I can't think of any justification for that.

    So my question is, how much of this is complete b.s.? Is there anything of value from this little exercise or have I just wasted time learning something that is incorrect?

    Thanks.
     
  2. jcsd
  3. Sep 11, 2016 #2
    What do v and u represent here?
     
  4. Sep 11, 2016 #3
    v is the speed between S and S'. u is the speed of the projectile according to S.
     
  5. Sep 11, 2016 #4
    I haven't looked at what you've done in great detail (the limit-taking does seem pretty hand-wavey, as you say), but I think that you're trying to circumvent imposing the conservation principle on momentum for the general case of arbitrary relative speeds. The Tolman/Lewis thought experiment involving that elastic collision is compelling precisely because it prioritizes the conservation of momentum -- and really, that's why momentum is important in the first place.

    So I feel confident saying that whatever it is you're doing here, it's missing that key ingredient.
     
  6. Sep 11, 2016 #5

    Nugatory

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    Staff: Mentor

    This result becomes less interesting when you consider that it is true for any function ##\gamma## that has the property that $$\lim_{q\rightarrow{0}}\gamma(q)=1$$.

    I'm inclined to agree with @SiennaTheGr8 that you have to bring conservation of momentum in somewhere to get a solid derivation.
     
    Last edited: Sep 11, 2016
  7. Sep 11, 2016 #6
    That was actually why I decided to do it. I needed to get rid of the vt with an operation that also turned γ(v) into 1. I'm wondering if I reverse the prime and unprimed and do the same thing if I'll get the same result. If I did, would that in any way touch on conservation of momentum?




    Anyway, the reason I'm trying to do it without resorting to an elastic collision is because everything I've ever seen on it says "the old momentum isn't conserved in every inertial reference frame, but THIS one is", but no one ever explains where they derived it from. It sounds almost as if it was trial and error guess work until something worked (although P = γ(u)mu would certainly be an obvious guess). I figure, I can derive the Lorentz transformation by simply assuming c is the same in every frame, and from that I can derive the velocity transformation. So why can't I get momentum from that simply by starting with the transformation equations?

    So that's my new project these days: trying how to figure out how to derive relativistic momentum only from the two famous SR postulates.
     
  8. Sep 11, 2016 #7
    On the other hand, once one has established that ##E = \gamma E_0 = \gamma mc^2##, the most obvious educated guess for extending the definition of momentum to arbitrary speeds turns out to be the correct one. You already have ##\vec pc \approx E_0 \vec \beta## for ##|\vec \beta| \ll 1## (where ##\vec \beta = \vec v / c##), and by imposing the conservation principle you know that ##|\vec p|c## can grow arbitrarily large (unlike ##|\vec \beta|##). You're therefore looking for something to replace ##E_0## that reduces to ##E_0## in the classical limit and increases without bound as ##|\vec \beta| \rightarrow 1##. The all-important conserved quantity ##E## immediately suggests itself.

    But you said at the beginning that you don't know how to arrive at ##E = \gamma E_0 = \gamma mc^2## without first deriving ##\vec pc = E \vec \beta##. Try Einstein's original ##E_0 = mc^2## paper from 1905, which uses the relativistic Doppler effect: https://www.fourmilab.ch/etexts/einstein/E_mc2/www/
     
  9. Sep 11, 2016 #8
    But that's exactly what the Tolman/Lewis thought experiment does! See here: https://books.google.com/books?id=FrgVDAAAQBAJ&pg=PA76
     
  10. Sep 13, 2016 #9
    Hey I think I found exactly what I was lookong for.

    http://rsta.royalsocietypublishing.org/content/366/1871/1861


    Even cooler than its derivation of momentum, it gets the heart of special relativity without either appealing to arguments about the speed of light/Michelson Morley or discussions in four dimensional space time. Nothing but Newton and Maxwell. Very neat.
     
  11. Sep 13, 2016 #10
    Thanks for the link. Haven't read it in depth, but be sure to look at the supplementary PDF under the "Figures & Data" tab, if you haven't already. That's where they go through the math for the collision.
     
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