## Tuesday, 15 April 2008

### Four-Vector Relativistic Velocity Transformations

Following on from my earlier post, I decided to write about the relativistic velocity transformations using a four-vector approach. We can define a space-time four-vector (or position-time) as follows:
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!