# Lorentz Transformation of four-velocity using the transformation matrix

Lets assume we have a two reference Frames $$O$$ and $$O'$$. $$O$$ is at rest and $$O'$$ is moving towards $$O$$ with velocity $$v$$ along the $$x$$-axis. Now we look at an object in which is moving with veloctiy $$u$$ along the $$z$$-axis in $$O$$. I have to find the veloctiy of the object in the frame of reference $$O'$$. So wat i have tried is following: The object is has a four-velocity $$u^{\mu}=(c\gamma(u), 0, 0, u\gamma(u))^T$$ in $$O$$. Now I multiply it with the Lorentz-Transformation Matrix $$\Lambda_{\nu}^{\mu}$$: $$u^{\prime\mu} = \Lambda_{\nu}^{\mu} u^{\mu} = \begin{pmatrix} \gamma(v) & -\beta_x\gamma(v) & 0 & 0 \\ -\beta_x\gamma(v) & \gamma(v) & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} c\gamma\\ 0\\ 0\\ u\gamma \end{pmatrix} = \begin{pmatrix} c\gamma(v)\gamma(u)\\ -\beta_x\gamma(v)c\gamma(u)\\ 0\\ u\gamma \end{pmatrix}= \begin{pmatrix} c\gamma(v)\gamma(u)\\ -v\gamma(v)\gamma(u)\\ 0\\ u\gamma(v) \end{pmatrix}$$ But in the solution (in which they use 3-vectors) they get $$\vec{u}'=(-v, 0, u/\gamma)$$, which makes more sense and I know how this result is achieved. However I'm trying to understand the whole four-vectors thing. If we ignore the first index of four vector, shouldn't the solution be the same? I know it's a really basic problem and that I'm missing something obvious, but I have been looking around for hours and couln't find anything useful. I'd really appreciate your help.

• If we ignore the first index of four vector, shouldn't the solution be the same? Yes, but the spatial components of the 4-velocity are not the 3-velocity. Dec 8, 2020 at 18:46

There are two issues to address here.

First, you made a minor mistake in your Lorentz transformation. It should be:

$$u^{\prime\mu} = \Lambda_{\nu}^{\mu} u^{\mu} = \begin{pmatrix} \gamma(v) & -\beta_x\gamma(v) & 0 & 0 \\ -\beta_x\gamma(v) & \gamma(v) & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} c\gamma(u)\\ 0\\ 0\\ u\gamma(u) \end{pmatrix} = \begin{pmatrix} c\gamma(v)\gamma(u)\\ -v\gamma(v)\gamma(u)\\ 0\\ u\gamma(u) \end{pmatrix}.$$

Note the last component of the result is $$\gamma(u)$$, not $$\gamma(v)$$.

Second, now that you have the four-velocity you need to obtain the three-velocity. Accounting for the correction above, the four-velocity you found is:

$$u^{\prime\mu} = \begin{pmatrix} c\gamma(v)\gamma(u)\\ -v\gamma(v)\gamma(u)\\ 0\\ u\gamma(u) \end{pmatrix}$$

and the general relationship between the four-velocity and three-velocity is given by the following formula

$$u'^{\mu} = \begin{pmatrix}c \, \gamma(u') \\ u_x' \, \gamma(u') \\ u_y' \, \gamma(u') \\ u_z' \, \gamma(u') \end{pmatrix}.$$

Comparing the first component in the two equations we see that

$$\gamma(u') = \gamma(v) \gamma(u)$$

and so we find the components of the three-velocity $$u'$$

$$u_x' = -v \\ u_y' = 0 \\ u_z' = \frac{u}{\gamma(v)}.$$

• thank you very much, that helps a lot. Never thought that I would get help so quickly here, really appreciate it. Dec 8, 2020 at 20:07