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There's a particle moving at a certain speed $v$ with respect to a reference frame $A$ through an axis I will call $x$. We thus know that for the frame of reference located at the particle itself, the relative velocity $u'$ of the particle is always $0$, no matter what $v$ is. All of this is trivial. $v$ can approach the speed of light $c$, but $u'$ will remain being $0$, it will approach $0$ as well.

Next, picture that there is another particle that has a parallel trajectory to the first one and the exact same velocity at all times. In this case, would $u'_2$ (the speed of the second particle with respect to the first one) be also $0$?

Now imagine that there is another particle that has a parallel trajectory to the first one, and at a certain time $t'=t''=0$ both particles have the same speed $v_0$ with respect to $A$ and are exactly at the same $x$ coordinate. The acceleration $a$ of both particles is exactly the same for every time $t'$ and $t''$ (of the first and second particle respectively), but it turns out that from $t'=t''=0$ up to a certain time $t'=\Delta t'$ the acceleration of the second particle is higher, and so for the rest of the trajectory there will be a displacement in the velocities, namely that $$v'-v=\Delta v=\Delta a\space \Delta t'$$ All of this with respect to $A$ (I suspect that the formula above might not be correct, in that case please do correct it). The relative velocity of the second particle with respect to the first one will then be $$u_2'=\frac{v'-v}{1-\frac{v'v}{c^2}}=c\frac{v'-v}{c-\frac{v'v}{c}}=c\frac{\Delta v}{c-\frac{v'^2}{c}+\frac{\Delta v \space v'}{c}}$$ So when $v'\to c$, $u'_2\to c$. $\Delta v$ can be as little as you want, as long as you make $\Delta t'$ small. It will, though, stay constant for the rest of the trajectory. What if $v'$ actually reaches $c$ and right after that, particle one keeps accelerating, so a time $\Delta t'$ after particle two reached speed $c$, particle one reaches the same speed. Now $u_2'$ is mathematically undetermined, but the limit will actually still be $c$ if we leave $v'=c$ constant. This confuses me a great deal mainly because the only difference now with the other hypothetical case of two particles always going at the same speed and accelerating together is a small displacement with respect to $A$ but an infinite displacement with respect to the reference frame of either particle (according to the Lorentz transforms). But in this other case, $u'_2$ will also always be $0$, right? The only difference between this case and the case of the velocity of the particle with respect to its own reference frame is a displacement in the axis perpendicular to $x$, which shouldn't change the first trivial result (or does it?). Does the "trivial" result even stand?

Please tell me where I'm wrong. Thank you.

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    $\begingroup$ No massive particle can reach the speed of light... and that's pretty much it. $\endgroup$
    – CuriousOne
    May 17 '16 at 22:24
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    $\begingroup$ Particles can not "reach" $c$, they either move under it if they have rest mass like electrons, or move at it, if they do not like photons. In the latter case there is no reference frame that can be attached to them (because the pseudometric degenerates along lightlike curves), so "relative to them" would be meaningless. Relativity formulas become mathematically meaningless at $v=c$, that some of them have limits is interesting but doesn't mean that those limits have any physical meaning. $\endgroup$
    – Conifold
    May 17 '16 at 22:42
  • $\begingroup$ @Conifold where can I learn about pseudometric degeneration along lightlike curves, then? $\endgroup$
    – sbs95
    May 18 '16 at 5:30
  • $\begingroup$ See explanation on Baez's blog math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/… and more discussion here physics.stackexchange.com/questions/29082/… $\endgroup$
    – Conifold
    May 19 '16 at 19:12
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If I'm reading the question (v1) right, you present a paradox in paragraph 2 (commonly called Bell's spaceship paradox) and then try to resolve it in the next few paragraphs. Your resolution doesn't make sense, as pointed out in the comments. The mistake is that $u_2'$ is not zero, by the relativity of simultaneity: different observers will disagree on the relative timing of events.

For example, suppose that in your frame, two particles have velocity $v$ and are separated by a distance $L$. Then you accelerate both particles to $v + \Delta v$ at the same time in your frame. If you plug in the Lorentz transformations, you'll find that in the particle frame, the particle in front gets accelerated $Lv/c^2$ earlier!

Therefore, whenever you're accelerating both particles "at the same time", the particle in front thinks the particle in behind is lagging, and getting farther away.

This effect can also be thought of in terms of length contraction. You've fixed the velocities of the particles so that they always act like a single object of length $L$ in your frame. Therefore, if the particles end up with a Lorentz factor of $\gamma$, the length between them in their frame will be $\gamma L$. So they must perceive $u_2' \neq 0$ during the acceleration.

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  • $\begingroup$ Thank you for the article. I didn't really understand any of this, but it's at least comforting to be absolutely certain that I forgot what I've learned about special relativity, if I ever really knew it in the first place. $\endgroup$
    – sbs95
    May 18 '16 at 5:27

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