Lorentz transformation in-terms of hyperbolic rotation

March 14, 2019  657 words 4 mins read  Join the Discussion

Lorentz transformation is how an observer sees an event when he is moving on different points of spacetime. I want to note here, in special (or, general) relativity, our universe and the spacetime are interchangeable terms (For why, see this).  But, in reality, the same observer can’t be in different places at a time to see the same event. Thus, we need to have at least two observers. One owns his coordinate at $(x,t)$ and another at $(x’,t’)$ is moving with respect to the first observer $(x,t)$ with velocity $v$. We are supposing our observers to move in (1+1) dimensions because to make our proposal simple. (1+1) dimensions mean one time dimension and one space dimension. But, we can extend to a motion where we need to deal with all three space coordinates. I’m letting this job to you. Please let me know in the comment on how you could do it. So, the Lorentz transformation gives a relation between them as

$x’ = \gamma (x - vt)$, and

$t’ = \gamma (t - \frac{vx}{c^{2}})$

(say equation 1) where, $\gamma = \frac{1}{\sqrt{1- \frac{v^{2}}{c^{2}}}}$ is a Lorentz factor, and $c$ is the velocity of light.
You may aware or not that the matrix is another way of representing equations. So, in matrix form, we have our Lorentz transformation (say LT for short) as

$\begin{pmatrix}x’ \\ t’\end{pmatrix} = \begin{pmatrix}\gamma & -v\gamma\\ \frac{-v\gamma}{c^{2}} & \gamma\end{pmatrix}\begin{pmatrix}x \\ t\end{pmatrix}$

This means, if you multiply right-hand side matrices then, you can get the equation 1. If we suppose velocity of light $c$ to be 1 then,

$\begin{pmatrix}x’ \\ t’\end{pmatrix} = \begin{pmatrix}\gamma & -v\gamma\\ -v\gamma & \gamma\end{pmatrix}\begin{pmatrix}x \\ t\end{pmatrix}$.

If you are curious why I set $c = 1$ then, read this post. We want to see LT in terms of hyperbolic trigonometric functions so, we are interested in the matrix ($L$ say)

$L = \begin{pmatrix}\gamma & -v\gamma \\ -v\gamma & \gamma\end{pmatrix} = \begin{pmatrix} L_{11} & L_{12} \\ L_{21} & L_{22}\end{pmatrix}$.

So, if you do square of $L_{11}$ and $L_{12}$ (or, $L_{22}$ and $L_{21}$) and then subtract them so that you get one. i.e.

${L_{11}}^{2} - {L_{12}}^{2}$

$ = {\gamma}^{2} - {(-v\gamma)}^{2}$

$ = (\frac{1}{\sqrt{1- v^{2}}})^{2} - {(-v \times \frac{1}{\sqrt{1- v^{2}}})}^{2}$

$ = \frac{1}{1- v^{2}} - \frac{v^{2}}{1- v^{2}}$

$ = \frac{1- v^{2}}{1- v^{2}} = 1$

$\therefore {L_{11}}^{2} - {L_{12}}^{2} = {\gamma}^{2} - {(-v\gamma)}^{2} = 1$.

Similarly, ${L_{22}}^{2} - {L_{21}}^{2} = {\gamma}^{2} - {(-v\gamma)}^{2} = 1$.

This type of identity can be found in hyperbolic trigonometric function which I mean this

$cosh^{2}(\eta) - sinh^{2}(\eta) = 1$.
So, we set $cosh(\eta) = \gamma$ and $sinh(\eta) = -v\gamma$. Now, we put these in our matrix $L$ and thus

$\begin{pmatrix}x’ \\ t’ \end{pmatrix} = \begin{pmatrix}cosh(\eta) & sinh(\eta) \\ sinh(\eta) & cosh(\eta) \end{pmatrix}\begin{pmatrix}x \\ t \end{pmatrix}$.

This is a Lorentz transformation in terms of hyperbolic rotation.

Now, we may have two questions:

  1. Why do we make in terms of hyperbolic trigonometric functions? The answer is quite simple. Because we can have more mathematical tools (or, flexibility) when we working in hyperbolic trigonometric functions than in algebraic function so that we can quickly solve the problem.
  2. Why our $L$ is a hyperbolic rotation matrix? Let me make you clear, in the matrix $L$, we only have $v$ which manipulate our transformation. This $v$ is the velocity of second observer $(x’,t’)$ measured by first observer $(x,t)$. This means if second observer move in a velocity $v$ then, he will see all the points of spacetime which is seen by the first observer will shift (strictly speaking, linearly transform) towards the opposite direction of him (i.e. second observer). But, if we make our spacetime with hyperbolic coordinates then,  all the points rotate on the hyperbola along by some angle. So, the mathematics for this case is Lorentz transformation in terms of hyperbolic rotation.

If you are interested to know the visualization of Lorentz transformation, and LT in terms of hyperbolic rotation then, see this, and this.

Lastly, “Happy birthday to Prof. Einstein”.

Any feedback?

If you guys have some questions, comments, or suggestions then, please don't hesitate to shot me an email at [firstname][AT]physicslog.com or comment below.

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  • Damodar Rajbhandari
    Written by Damodar Rajbhandari, who is working on a PhD in the Mathematical Physics at the School of Mathematics & Statistics, University of Melbourne, Australia.
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