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If a particle moves with velocity $(u_x,u_y,u_z)$ in $S$ and $S'$ moves velocity $v$ relative to $S$, what s the relation between $\gamma_{u'}$ ($u'$ speed of the particule in $S'$) and $\gamma_u, \gamma_v$?

I know the solution is $\gamma_u' =(1-\frac{u_xv}{c^2})\gamma_u\gamma_v$ but I've been unable to prove it.

Qmechanic
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Gabriel
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  • @robphy Lorentz equations didn't seem to work. It looks like $\frac{1}{(\gamma_u \gamma _v)^2}=\frac{(1-u_xv/c^2)^2}{\gamma_u^2}$ reasoning from the solution, but I cannot get to the first equality from there. – Gabriel Apr 18 '17 at 14:10
  • Have you tried matrix multiplication of two boosts along the x-axis? Alternatively, do you know about rapidities? I replied to a similar question recently in https://physics.stackexchange.com/questions/326272/applying-time-dilation-twice – robphy Apr 18 '17 at 14:21
  • @robphy "multiplication of two boos along the x-axis" could you explain that further? – Gabriel Apr 18 '17 at 14:26
  • A boost is the technical term for the usual Lorentz Transformations, which relates coordinates in two frames of reference. (Technically speaking, rotations in space are also Lorentz Transformations. So, I am just being specific.) Rather than write a pair of equations to transform t and x, one can write it in matrix form using a 2-by-2 matrix multiplying a vector. In this form, it's easier to do a sequence of transformations by multiplying matrices: http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html#c2 – robphy Apr 18 '17 at 14:45
  • So difficult and so simple simultaneously. See my answer therein : Lorentz transformation of velocity 4-vector, EDIT B, EDIT C. – Frobenius Jan 11 '18 at 08:36

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So a Lorentz boost (in this case in the $x$-direction by speed $v$) has the structure$$\gamma_v\begin{bmatrix} 1&-v/c&0&0\\ -v/c&1&0&0\\ 0&0&1&0\\ 0&0&0&1\end{bmatrix}.$$

Everyone agrees on 4-vectors in spacetime, though the components change by Lorentz transforms. One such 4-vector is the 4-velocity of the particle, which in $S$ is given by the column vector $\gamma_u\cdot[c,~u_x,~u_y,~u_z].$ This is a standard form for a 4-velocity and if you have not yet memorized it, you probably should, 4-velocities always look like $\gamma_v\cdot[c,\vec v]$ where $\vec v$ is their velocity in those coordinates; their 4-D magnitude according to the Minkowski metric is always $c.$

Now please abandon your hesitation and just do the transform; after transformation the 4-velocity must be $\gamma_{u'}\cdot[c, \vec u']$ so we can get $\gamma_{u'}$ from the very first component from the expression $$\gamma_{u'}\begin{bmatrix}c\\u'_x\\u'_y\\u'_z\end{bmatrix} = \gamma_v\begin{bmatrix} 1&-v/c&0&0\\ -v/c&1&0&0\\ 0&0&1&0\\ 0&0&0&1\end{bmatrix} \cdot \gamma_u \begin{bmatrix}c\\u_x\\u_y\\u_z\end{bmatrix}.$$ So just do the transform and you will see that the very first component gives you the equation that you seek.

CR Drost
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