x^\mu = \left(\begin{array}{c} ct \\ x \\ y \\ z \end{array}\right)

The Lorentz transform can be written in matrix form, using

`\beta=\frac{u}{c}`

and `\gamma=\frac{1}{\sqrt{1-\beta^2}}`

:

\mathbf{L}=\left(\begin{array}{cccc}

\gamma & -\beta\gamma & 0 & 0 \\

-\beta\gamma & \gamma & 0 & 0 \\

0 & 0 & 1 & 0 \\

0 & 0 & 0 & 1 \end{array} \right)

Now, one can write a Lorentz transform in the x-direction from a frame

*S*to a frame

*S'*moving with velocity

*u*relative to

*S*as:

x^{\prime\nu} = \mathbf{L}^\nu_\mu x^\mu

Where I've used the Einstein summation convention, in which one sums over repeated indices... in this case from 0 to 3:

\begin{align*}

x^{\prime 0} = \mathbf{L}^0_0 x^0 + \mathbf{L}^0_1 x^1 + \mathbf{L}^0_2 x^2 + \mathbf{L}^0_3 x^3 \\

x^{\prime 1} = \mathbf{L}^1_0 x^0 + \mathbf{L}^1_1 x^1 + \mathbf{L}^1_2 x^2 + \mathbf{L}^1_3 x^3 \\

x^{\prime 2} = \mathbf{L}^2_0 x^0 + \mathbf{L}^2_1 x^1 + \mathbf{L}^2_2 x^2 + \mathbf{L}^2_3 x^3 \\

x^{\prime 3} = \mathbf{L}^3_0 x^0 + \mathbf{L}^3_1 x^1 + \mathbf{L}^3_2 x^2 + \mathbf{L}^3_3 x^3

\end{align*}

Since we're dealing with a Lorentz transform in the

*x*direction, we'll ignore the 2 and 3 indices; they correspond to the y and z directions. This leaves the 0 (time) index, and the 1 (x-direction) index to deal with:

\begin{align*}

x^{\prime 0} = \gamma ct -\beta\gamma x\\

x^{\prime 1} = -\beta\gamma ct + \gamma x

\end{align*}

Expanding

`\beta`

gives:

\begin{align*}

x^{\prime 0} = \gamma\left(ct - \frac{ux}{c}\right) = c\gamma\left(t - \frac{ux}{c^2} \right)\\

x^{\prime 1} = \gamma(x - ut)

\end{align*}

From the Lorentz transforms I discussed in another earlier post:

\begin{align*}

t^\prime = \gamma\left(t - \frac{ux}{c^2}\right) \\

x^\prime = \gamma(x - ut)

\end{align*}

It should therefore be clear to see that our transformed four-vector,

`x^{\prime \nu}`

has the following form:

x^{\prime \nu} = \left(\begin{array}{c} ct^\prime \\ x^\prime \\ y \\ z \end{array}\right)

We have, therefore, managed to transform a four-vector and equate its components to the expected forms for the Lorentz transformation of x and t coordinates, bearing in mind that the zeroth component of a four-vector is not just time t, but ct. It is then a simple case of dividing the 1st component by the 0th, having removed the c from component 0:

\begin{align*}

v^\prime = \frac{x^\prime}{t^\prime} = \frac{vt - ut}{t-\frac{uvt}{c^2}} \\

= \frac{t(v-u)}{t(1-\frac{uv}{c^2})} \\

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

\end{align*}

But there's also another way to approach this problem. We can define a four-velocity

`u^\mu = \gamma \left(\begin{array}{c}c\\v\end{array}\right)`

. This can be transformed in exactly the same way, using the **L**matrix:

\begin{align*}

u^{\prime \nu} = \mathbf{L}^\nu_\mu u^\mu \\

= \gamma\left(

\begin{array}{c}

\gamma c - \beta\gamma v \\

-\beta\gamma c + \gamma v

\end{array}\right) \\

= \gamma\left(\begin{array}{c}

\gamma c\left(1-\frac{uv}{c^2}\right) \\

\gamma(v-u)\end{array}\right)

\end{align*}

Now, dividing the "space-like" component by the "time-like" one and multiplying by

*c*, i.e.

`c\frac{u^{\prime 1}}{u^{\prime 0}}`

gives:

\begin{align*}

v^\prime = \frac{\gamma(v-u)}{\gamma\left(1-\frac{uv}{c^2}\right)} \\

v^\prime = \frac{v-u}{1-\frac{uv}{c^2}}

\end{align*}

We have recovered the expression for relativistic velocity transformations!

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