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.(adsbygoogle = window.adsbygoogle || []).push({});

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.

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

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