Consider two inertial frames of reference, $S$ and $S'$, sharing the $x$-axis. $S'$ is travelling at constant velocity $\vec{v}=v\hat{x}$ from S and at $t=t'=0$, the origins of the two frames coincide.

In frame $S'$, a particle is travelling at constant velocity $\vec{u'}=u'\hat{x}$. At $t'=0$, the particle is at the origin of $S'$.

(a) Calculate the velocity 4-vector $\eta^{\mu}{'}$ of the particle in the frame $S'$.

(b) Using a suitable Lorentz transformation, determine the velocity 4-vector $\eta^{\mu}$ of the particle in the frame S.

(c) Using the component $\eta^{0}$, calculate $u$, the velocity of the particle as observed from frame S, thus deriving the relativistic velocity addition formula:



(a) This part is fairly straightforward, I get:


by differentiating $x^{\mu}{'}$ with respect to proper time and applying the chain rule.

(b) This is where I am no longer sure of my work. I decide to use the transformation matrix

$\Lambda^{-1}= \left[ {\begin{array}{ccccc} \gamma & \beta \gamma & 0 & 0 \\ \beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} } \right] $

with positive signs as this situation is viewed as $S$ travelling away from $S'$ at -v$\hat{x}$ (or, I believe it is the inverse of the standard transformation matrix). We define also $\beta = \frac{v}{c}$.

Then, using [ ] to indicate matrices, I carry out the transformation $[\eta]=\Lambda^{-1} [\eta]'$ and get, via matrix multiplication, the following, where the gamma indices indicate which velocity gamma is a function of:

$\eta^{\mu}=\gamma_v \gamma_{u'} \left( {\begin{array}{c} c+v \\ \frac{vu'}{c}+u'\\ 0\\ 0\\ \end{array} } \right) $

I'm not really sure if this is correct. I have attempted (c) with this solution, equating $\eta^{0}$ from above to $\eta^{0}=\gamma_u c$ but the velocity addition formula didn't drop out.

I appreciate any corrections, hints and guidance! Thanks.


Hints :

  1. If you intend to use the matrix $\:\Lambda^{-1}\:$ given in $\:\S(\bf b)\:$ then you must swap the spatial and temporal components of $\:\eta^{\mu}{'}\:$ given in $\:\S(\bf a)$.
  2. It's not necessary to work with 3+1-dimensions.Work in 1+1-dimensions.
  3. If it's constraint to work with 4 or 2-vectors, then OK. But the exercise could be solved easily directly by the 1+1-Lorentz transformation.
  4. The way you try to solve your exercise might demand the difficult case of determining the relation between the $\:\gamma-$factors : $\:\gamma_{\upsilon},\gamma_{u},\gamma_{u'}$.


1+1-Lorentz Transformation $\:S-\!\!\!-\!\!\!\longrightarrow S'\:$
\begin{align} \text{Spatial} : & \quad x'_{1}=\gamma_{\upsilon}\left(x_{1}-\beta x_{0}\right) \tag{01a}\\ \text{Temporal} : & \quad x'_{0}=\gamma_{\upsilon}\left(x_{0}-\beta x_{1}\right) \tag{01b} \end{align} so \begin{equation} \begin{bmatrix} x'_{1}\vphantom{\dfrac12}\\ x'_{0}\vphantom{\dfrac12} \end{bmatrix} =\Lambda \begin{bmatrix} x_{1}\vphantom{\dfrac12}\\ x_{0}\vphantom{\dfrac12} \end{bmatrix} \tag{02} \end{equation} where \begin{equation} \Lambda= \begin{bmatrix} \hphantom{\boldsymbol{-}\beta }\gamma_{\upsilon} & \boldsymbol{-}\beta \gamma_{\upsilon}\vphantom{\dfrac12}\\ \boldsymbol{-}\beta\gamma_{\upsilon} & \hphantom{\boldsymbol{-}\beta }\gamma_{\upsilon}\vphantom{\dfrac12} \end{bmatrix} \tag{03} \end{equation} and \begin{equation} \begin{bmatrix} x_{1}\vphantom{\dfrac12}\\ x_{0}\vphantom{\dfrac12} \end{bmatrix} =\Lambda^{-1} \begin{bmatrix} x'_{1}\vphantom{\dfrac12}\\ x'_{0}\vphantom{\dfrac12} \end{bmatrix} \tag{04} \end{equation} where \begin{equation} \Lambda^{-1}= \begin{bmatrix} \hphantom{\beta }\gamma_{\upsilon} & \beta \gamma_{\upsilon}\vphantom{\dfrac12}\\ \beta\gamma_{\upsilon} & \hphantom{\beta }\gamma_{\upsilon}\vphantom{\dfrac12} \end{bmatrix} \tag{05} \end{equation} In all above equations we have as first component the spatial with subscript *$_{1}$ and as second component the temporal with subscript *$_{0}$.

So for the 2-velocity \begin{equation} \boldsymbol{\eta'}= \begin{bmatrix} \eta'_{1}\vphantom{\dfrac12}\\ \eta'_{0}\vphantom{\dfrac12} \end{bmatrix} =\gamma_{u'} \begin{bmatrix} u'\vphantom{\dfrac12}\\ c\vphantom{\dfrac12} \end{bmatrix} \ne \underbrace{ \gamma_{u'} \begin{bmatrix} c\vphantom{\dfrac12}\\ u'\vphantom{\dfrac12} \end{bmatrix} = \begin{bmatrix} \eta'_{0}\vphantom{\dfrac12}\\ \eta'_{1}\vphantom{\dfrac12} \end{bmatrix}}_{\textbf{yours in}\:\boldsymbol{\S(a)} } \tag{06} \end{equation}


Difficult to prove but you must try. If you succeed to prove them I suggest to post them as an answer to your own question herein : \begin{equation} \dfrac{\gamma_{u'}}{\gamma_{u}}=\gamma_{\upsilon}\left(1-\dfrac{\upsilon\, u}{c^{2}}\right) \tag{07} \end{equation} or for the inverse transformation \begin{equation} \dfrac{\gamma_{u}}{\gamma_{u'}}=\gamma_{\upsilon}\left(1+\dfrac{\upsilon\, u'}{c^{2}}\right) \tag{08} \end{equation}


Strange or not the easy way to prove equation (07), or equivalently equation (08), is not algebraic. Instead of many pages of algebra I like very much the following game with the $\:\mathrm dt'$s (I post the proof since this doesn't answer your question)

So, let a third system $\:S''\:$ attached to the particle. This system is moving with velocity $\:u'\:$ with respect to $\:S'\:$ and with velocity $\:u\:$ with respect to $\:S$. In system $\:S''\:$ the time $\:t''\:$ is the proper time $\:\tau\:$. In this system let two events 1 and 2, the particle at rest ($\:\mathrm dx''=x''_{2}-x''_{1}=0\:$) at time moments $\:\mathrm t''_{1},\mathrm t''_{2}\:$ apart by the infinitesimal (proper) time interval $\:\mathrm dt''=t''_{2}-t''_{1}=\mathrm d\tau\:$.

From the Lorentz Transformation $\:S''\longleftarrow\!\!\!-\!\!\!-\!\!\!\longrightarrow S'\:$ \begin{equation} \dfrac{\mathrm dt'}{\:\mathrm dt''}=\dfrac{\,\mathrm dt'}{\mathrm d\tau}=\gamma_{u'} \tag{09} \end{equation} and from the Lorentz Transformation $\:S''\longleftarrow\!\!\!-\!\!\!-\!\!\!\longrightarrow S\:$ \begin{equation} \dfrac{\mathrm dt}{\:\:\mathrm dt''}=\dfrac{\mathrm dt}{\mathrm d\tau}=\gamma_{u} \tag{10} \end{equation} But from the Lorentz Transformation $\:S'\longleftarrow\!\!\!-\!\!\!-\!\!\!\longrightarrow S\:$ we have \begin{equation} \dfrac{\,\mathrm dt'}{\mathrm dt}=\gamma_{\upsilon}\left(1-\dfrac{\upsilon\, u}{c^{2}}\right) \tag{11} \end{equation} Dividing (09) by (10) and using (11) we have (07).

  • $\begingroup$ Thanks for the response! 1. I don't quite follow the reasoning, why is that the case? 2. True, thanks! 3. Would this involve deriving and applying the velocity addition formula in part (b)? I have thought about this but since I am asked to verify it in part (c), so I presume that I should not use it in part (b). 4. Correct, I have tried to do this but cannot see an easy way out. $\endgroup$ – Samalama Jan 9 '18 at 12:12
  • $\begingroup$ Thanks, all clear now! I actually realised that I also made a basic matrix multiplication error previously ... Certainly that didn't help either. I now have the relations as shown in edit B, but after a page or two of algebra I cannot see how this might work. See my answer for what I did. $\endgroup$ – Samalama Jan 9 '18 at 15:05
  • $\begingroup$ Re edit C, that's very neat! It would have taken me time to get that. Really appreciate your help and detail - thanks! $\endgroup$ – Samalama Jan 10 '18 at 13:47

For part (c), after playing around with the transformed 4-vector a little, I found that the velocity addition formula neatly drops out by dividing the spatial 4-velocity by the its temporal component:

$$ \frac{\eta^1}{\eta^0}=\frac{\gamma_{u}u}{\gamma_{u}c}=\frac{\gamma_v \gamma_{u'}(v+u')}{\gamma_v \gamma_{u'}(c+\frac{vu'}{c})}$$

All gamma factors neatly cancel and multiplying by $c$ yields the velocity addition formula. I'm not sure if this is the method as suggested by the question but it is much simpler than trying to equate the three gammas.


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