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I have two particles, one of them is stationary, another one was speed $u$ in the lab's frame. I cna prove with Lorentz transformations that the speed of the zero momentum frame is

$$v=\frac{\gamma_u}{1+\gamma_u}u$$

I can also prove that if I have a particle with speed $u$, then the speed $u'$ of that particle observed from a different frame with speed $v$ (axes aligned with lab frame, moving along $x$-axis - ie standard Lorentz boost frame) is given by

$$u'=\frac{u-v}{1-uv/c^2}$$

Putting this two together, the particle which is stationary in the lab frame, has the same speed in the ZMF as the ZMF itself has in the lab frame (ie $|u'(u=0|=v$).

How could I see this intuitively, without much calculation?

Why the speed of a lab-frame stationary particle in ZMF is the speed of the ZMF in lab's frame?

zabop
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    Oh, as I wrote down the last sentence, I realized that: to go back to lab's frame, we need to inverse boost my ZMF speed in lab's frame. But if we approach it in the other way round: lab's frame is where the particle is stationary. In ZMF, it is not, so we need to inverse Lorentz boost by its speed to make it stationary. But that's just will give us lab's frame. So speed of particle in ZMF = speed of ZMF in lab's frame. If there is more to this story, would still appreciate answers. – zabop Apr 05 '21 at 21:31
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    That is perfectly fine at this stage of abstraction. Slightly more abstractly, you would be one step closer to proving that Lorentz transformations form a group, by proving that each element (boost) of the group has an inverse (inverse boost) that undoes the original and gets you back to the identity (zero boost) – Ninad Munshi Apr 05 '21 at 21:34

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