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I was thinking about the relative speed of an observation reference frame and an object which has been accelerated to a speed close to the speed of light. I'm by no mean an expert and the last physics class I took was more than 20 years ago so my question could be silly... If we accelerate a particle, let's say an electron, to 99.99% of the speed of light and then we start moving in the opposite direction and reach about the 0.01% of the speed of light in the opposite direction, by the original point of observation, using a reasonable amount of energy, we should be dilating our time, so the time in the original frame of reference should pass faster then the time in our moving frame, that means that a relative velocity observed from the original frame should increase from ours. Doesn't that mean that we will observe the $e$ passing $c $?

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There are at least two misunderstandings in your argument. The most fundamental one is that you tried to compare times in two different reference frames, but you can't do that in relativity. For example, if Bob zooms off at $\frac{1}{2}c$ to the right, and Alice stays where she is, then both of them will see "time dilation" when they look at the other person. Suppose that they are both carrying clocks. Both of them will think that the other person's clock is running slower than their own.

This comes into your scenario because I think you are conflating time dilation observed from the particle moving at $0.01\% c$ and time dilation from the rest frame.

The second misunderstanding here is that observers will not only see time dilation, but also space dilation, so this also makes figuring out the relative velocities a little bit harder.

Doing this calculation properly leads to the velocity-addition formula. Suppose that you have $v_1=0.9999c$ and $v_2=0.0001c$. Then the velocities that these two particles will perceive each other to be moving at (if they could perceive things and take measurements!) would be $$v_\text{rel} = \frac{v_1+v_2}{1+\frac{v_1 v_2}{c^2}} = \frac{c}{1+0.9999 \cdot 0.0001} \approx 0.99990002 c $$

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  • $\begingroup$ Shouldn't it be $$v_{\text{rel}}=\frac{v_1-v_2}{1-\frac{v_1v_2}{c^2}}=\frac{0.99c-(-0.01c)}{1-\frac{(0.99c)(-0.01c)}{c^2}}\approx0.99990002c$$? The OP seems to have $v_1$ and $v_2$ both relative to the same frame. $\endgroup$ – Aaron Stevens Jul 29 at 17:14
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    $\begingroup$ @AaronStevens I think that the fact that your answer is algebraically equivalent to mine is a hint that we are just labeling the frames differently. Call the particle moving to the right at $99.99\%c$ (from the OP's perspective) particle A and the particle moving to the left at $0.0001\%c$ particle B. Let's also use the notation of the associated Wikipedia article to distinguish the frames. Let $u$ be the velocity of particle A viewed from the frame of particle B. Let $u'$ be the velocity of A viewed from the OP's perspective. ... $\endgroup$ – sasquires Jul 29 at 19:30
  • $\begingroup$ ... Finally, let $v$ be the boost required to go from B's frame to the OP's frame. With these choices and the equation written as it is on Wikipedia, we get the result that I wrote (with the same signs). If you label the frames differently, you need different signs. (But I deliberately avoided getting into frame-labeling issues in the answer, since the OP said it has been 20 years since his/her last physics class and the main question was conceptual.) $\endgroup$ – sasquires Jul 29 at 19:31
  • $\begingroup$ You're right, it is just a difference of which velocity you report relative to which frame. However, I think for future readers we should strive to stay consistent with typical sign conventions. Or at least specify when we are going against them. Based on what the OP has said, it seems like $v_1$ and $v_2$ should have different signs. $\endgroup$ – Aaron Stevens Jul 29 at 21:17
  • $\begingroup$ @AlfredCentauri I'm not against that convention in general. I was just speaking in terms of what the OP seems to be going for. We are given the speed $v_1$ of a particle relative to the "starting frame" and the speed $v_2$ of us relative to the "starting frame". We want to find the speed $v_{rel}$ of the particle relative to us. Therefore it makes sense to me to use the $v_{rel}=v_1-v_2$ convention here. $\endgroup$ – Aaron Stevens Jul 29 at 23:28

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