# Length contraction , Time dilation and Lorentz Transformation: Pokemon edition

“Most of Special Relatvity is like playing a game of – He said, She said.”

Inspired by the recent Pokemon movie trailer, this post takes you on a journey with Ash and Pikachu  as they find out about the founding principles of special relativity – Length contraction, Time dilation  and Lorentz Transformation.

## Time dilation

Let’s consider a scenario where Pikachu is on a  car moving with a velocity $v$ and Ash is standing the ground.

There is a Pokeball on the floor of the car. It is constantly emitting light straight toward the ceiling, Let us try to understand how Ash and Pikachu would see this photon of light from the time it leaves the floor to the ceiling and back, in their respective frame of references:

How pikachu describes the event:

Pikachu only sees the photon bouncing up and down and therefore:

$2L = c t^{'}$

$4L^2 = c^2 t^{'2} \rightarrow (1)$

How Ash describes the event:

But when the trajectory of the light is viewed from Ash’s frame of reference it is not a straight line at all and therefore, according to Ash:

$2 \sqrt{L^{2} + \frac{v^{2}t^{2}}{4}} = c t$

$4 L^{2} + v^{2}t^{' 2} = c^{2} t^{2}$

$4 L^{2} = (c^{2} - v^2) t^{2} \rightarrow (2)$

Since both these descriptions of what happened are true and have to agree, we must therefore from (1) and (2) have that,

$t = \gamma t^{'}$ where $\gamma = \frac{1}{\sqrt{1- \frac{v^2}{c^2}}}$

This ensures that both the event descriptions are satisfied. The implication being that Pikachu who is on a cart moving with some velocity has to be experiencing time slower than someone who is not moving at all (Ash).

## Length Contraction

Consider the same scenario as before except that this time the Pokeball is emitting light along the direction of motion and back.

How Pikachu describes the event:

If the time measured by Pikachu is slower than Ash from the previous analysis, then we expect that the distance traveled by the photon in Pikachu’s frame of reference will not be $L$. Instead let’s call that distance $L^{'}$. Therefore,

$2L^{'} = c t^{'}$

$\frac{2L^{'}}{c} = t^{'} \rightarrow (3)$

How Ash describes the event:

According to Ash, the photon was initially moving along the direction of motion and then started moving against the direction of motion and therefore the total time taken would be the addition of the time taken in both the cases i.e

$\frac{L}{c+v} + \frac{L}{c-v} = t$

$\frac{2L}{1 - \frac{v^{2}}{c^{2}}} = c t$

$\frac{2L}{c} = \left( 1 - \frac{v^{2}}{c^{2}} \right) t$

$\frac{2L}{c} = \frac{1}{\gamma^2} t$

But, we found from our previous analysis that $t = \gamma t^{'}$ and

$\frac{2L^{'}}{c} = t^{'}$. Plugging these back into the previous equation gives us:

$L = \frac{L^{'}}{\gamma}$

The length of the cart as observed by Ash ($L$) is shorter than that observed by Pikachu ($L^{'}$). This implies that if Pikachu is holding a 1-meter stick,  then Ash would see the 1-meter stick to be shorter than 1-meter. And in general any object that is held parallel to the direction of motion would appear ‘squished’

## Lorentz Transformation

Now consider the final scenario where at $t=0$, the pokeball is at a distance $x$ from both of them and emits a photon of light. Pikachu decides to move towards the pokeball with a velocity $v$ to catch it before Ash does.

This pulse of light would reach Pikachu when he is at a distance $x^{'}$ from the Pokeball and would reach Ash at a distance $x$ since he is not moving.

Length contraction according to Ash:

Having  understood the concept of Length contraction, Ash would say:

$x^{'} = \gamma \left( x - vt \right) \ \ \rightarrow (4)$

Length contraction according to Pikachu:

But Pikachu also understands Length Contraction and thinks that he is at rest and it is Ash who is receding backwards. And therefore:

$x = \gamma \left( x^{'} + vt^{'} \right)$

We must understand that both are valid representations of  what is happening and must agree. Therefore plugging in (4) into the above equation and performing some simple algebraic manipulations we get,

$x = \gamma \left( \gamma \left( x - vt \right) + vt^{'} \right)$

$\frac{x}{\gamma^{2}} = \left( x - vt \right) + \frac{vt^{'}}{\gamma}$

$vt - \frac{v t^{'}}{\gamma} = x \left( 1 - \frac{1}{\gamma^2} \right)$

$vt - \frac{v t^{'}}{\gamma} = \frac{v^{2} x}{c^{2}}$

$t - \frac{t^{'}}{\gamma} = \frac{v x}{c^{2}}$

$t^{'} = \gamma \left( t - \frac{v x}{c^{2}} \right) \ \ \rightarrow (5)$

(4) and (5) are known as Lorentz Transform equations  and relate the distances and time measured in Ash’s reference frame with that measured in Pikachu’s frame.

$x^{'} = \gamma \left( x - vt \right)$

$t^{'} = \gamma \left( t - \frac{v x}{c^{2}} \right)$

One can start off in Ash’s frame of reference and perform a “Lorentz Boost” to see how things are happening in Pikachu’s frame of reference. Often it is convenient to write them in a matrix form:

$\begin{bmatrix} ct^{'} \\ x^{'} \end{bmatrix} = \begin{bmatrix} \gamma & -\frac{v}{c} \gamma \\ - \frac{v}{c} \gamma & \gamma \end{bmatrix} \begin{bmatrix} ct \\ x \end{bmatrix}$

In this post we have constrained Pikachu to move only in the x-direction but this need not be the case. A fun little exercise would be to extend this analysis to Lorentz boosts in 3-dimensions.

It is often more enlightening to see these relations when applied to problems and my personal recommendation would be the book – ‘Spacetime physics by Edwin F. Taylor’. Thank you for joining Ash and Pikachu on this journey. Hope you learned something new. Have fun!

# Cooking up a Lorentz invariant Lagrangian

Let’s consider a scalar field, say temperature of a rod varying with time i.e  $T(x,t)$. (something like the following)

We will take this setup and put it on a really fast train moving at a constant velocity $v$ (also known as performing a ‘Lorentz boost’).

Now the temperature of the bar in this new frame of reference is given by $T^{'}(x^{'}, t^{'})$ where,

$x^{'}(x,t) = \gamma \left( x - vt \right)$

$t^{'}(x,t) = \gamma \left( t - \frac{v x}{c^{2}} \right)$

Visualizing the temperature distribution of the rod under a Length contraction.

Temperature is a scalar field and therefore irrespective of which frame of reference you are on, the temperature at each point on the rod will remain the same on both the frames i.e

$T^{'}(x^{'}, t^{'})= T(x, t)$

Therefore we can say that Temperature (a scalar field) is Lorentz invariant. Now what other quantities can we make from T that would also be Lorentz invariant ?

Is $\nabla . T^{'}(x^{'}, t^{'})= \nabla . T(x, t) ?$

Well, let’s give it a try:

$\frac{\partial T}{\partial x} = \frac{\partial T^{'}}{\partial x^{'}} \frac{\partial x^{'}}{\partial x} + \frac{\partial T}{\partial t^{'}} \frac{\partial t^{'}}{\partial x}$

$\frac{\partial T}{\partial x} = \frac{\partial T^{'}}{\partial x^{'}} \gamma - \frac{\partial T}{\partial t^{'}} \gamma v$

$\frac{\partial T}{\partial x} = \gamma \left( \frac{\partial T^{'}}{\partial x^{'}} - \frac{\partial T}{\partial t^{'}} v \right)$

——–

$\frac{\partial T}{\partial t} = \frac{\partial T^{'}}{\partial x^{'}} \frac{\partial x^{'}}{\partial t} + \frac{\partial T^{'}}{\partial t^{'}} \frac{\partial t^{'}}{\partial t}$

$\frac{\partial T}{\partial t} = - \frac{\partial T^{'}}{\partial x^{'}} \gamma v + \frac{\partial T^{'}}{\partial t^{'}} \gamma$

$\frac{\partial T}{\partial t} = \gamma \left( - \frac{\partial T^{'}}{\partial x^{'}} v + \frac{\partial T^{'}}{\partial t^{'}} \right)$

Clearly, **

$\frac{\partial T}{\partial x} + \frac{\partial T}{\partial t} \neq \frac{\partial T^{'}}{\partial x^{'}} + \frac{\partial T^{'}}{\partial t^{'}}$

But just for fun let’s just square the terms and see if we can churn something out of that:

$\left( \frac{\partial T}{\partial x} \right)^{2} = \gamma^{2} \left( \frac{\partial T^{'}}{\partial x^{'}} - \frac{\partial T^{'}}{\partial t^{'}} v \right)^{2}$

$\left( \frac{\partial T}{\partial t} \right)^{2} = \gamma^{2} \left( - \frac{\partial T^{'}}{\partial x^{'}} v + \frac{\partial T^{'}}{\partial t^{'}} \right) ^{2}$

We immediately notice that:

$\left( \frac{\partial T}{\partial t} \right)^{2} - \left( \frac{\partial T}{\partial x} \right)^{2} = \gamma^{2} \left[ \left( - \frac{\partial T^{'}}{\partial x^{'}} v + \frac{\partial T^{'}}{\partial t^{'}} \right) ^{2} - \left( \frac{\partial T^{'}}{\partial x^{'}} - \frac{\partial T^{'}}{\partial t^{'}} v \right)^{2} \right]$

$\left( \frac{\partial T}{\partial t} \right)^{2} - \left( \frac{\partial T}{\partial x} \right)^{2} = \left( \frac{\partial T^{'}}{\partial t^{'}} \right)^{2} - \left( \frac{\partial T^{'}}{\partial x^{'}} \right)^{2}$

Therefore in addition to realizing that $T$ is Lorentz invariant, we have also found another quantity that is also Lorentz invariant. This quantity is also written as $\partial_{\mu} T \partial^{\mu} T$ .

** There is a very important reason why this quantity did not work out.  This post was inspired in part by Micheal Brown’s answer on stackexchange . I request the interested reader to check that post for a detailed explanation.