Detailed working out Lorentz contraction from the Lorentz transform

1. Apr 24, 2014

pervect

Staff Emeritus
We have a few posters struggling with this, I thought I'd post a step by step guide, to see if it would help. That seems easier than trying to untangle the confused threads we have. We'll see if it works...

Setup and notation:

We have a rocket, which has a front and a back.
We have a fixed reference frame S, and a moving frame S'.
The rocket is moving in frame S, and fixed in frame S'.

The mathematical representation of the path of the front of the rocket is a worldline, a line in space-time. The mathematical representation of the back of the rocket is another, different worldline.

Coordinates in frame S:
$t$ and $x$ (generic)
$t_f$ and $x_f$ the coordinates of some point on the worldline of the front of the rocket
$t_b$ and $x_b$, the coordinates of some point on the worldline of the back of the rocket
L the length of the rocket in S. L is equal to $x_f - x_b$ evaluated at $t_f = t_b$.

Coordinates in frame S'
$t^{'}$ and $x^{'}$ (generic)
$t^{'}_f$ and $x^{'}_f$ the coordinates of some point on the worldline of the front of the rocket
$t^{'}_b$ and $x^{'}_b$ the coordinates of some point on the worldline of the back of the rocket
$L^{'}$, the length of the rocket in S'. $L^{'}$ is equal to $x^{'}_f - x^{'}_b$ evaluated at $t^{'}_f = t^{'}_b$

Let the velocity of the rocket, and of the frame S', be v. Define $\beta = v/c$, the ratio of the velocity of the rocket to the speed of light.

step 1) Write the equation for the worldline of the back of the rocket in frame S:
answer: $x_b = v \, t_b$

step 2) Write the equation for the worldline of the front of the rocket in frame S:
answer: $x_f = v \, t_f + L$

This was derived from the fact that $x_f - x_b = L$ when $t_f = t_b$

step 3). Optional but highly recommended Draw a diagram / graph of these equations. Compare it to the numerous diagrams george has already drawn. Understand the concept of the "worldline".

How to draw a graph: I'll do this once, because it seems to me that people skip drawing the graphs, or even looking at them when someone else draws them. I'm not sure why - on the theory that the apparent laziness might hide underlying issues with understanding, I'll describe the process.

Label axes t and f. You'll see t going up the top of the page in many space-time diagrams, but if you're more comfortable drawing t left-right, feel free, it's your graph.

Select some specific value of velocity to draw the graph. It's recommended to use units of time and distance so that a graph of a light ray is a 45 degree line, for instance use units of seconds for time and light-seconds for distance.

Select a set of times, for instance t=(0,1,2,3,4,5)

Plug the value of time into the formula for distance, to compute a set of points.
$t_f=t_b=0$ gives $x_b = 0$ and $x_f = L$
$t_f=t_b=1$ gives $x_b = v$ and $x_f = L+v$
$t_f=t_b=2$ gives $x_b = 2v$ and $x_f = L+2v$

Plot the points on the graph. (I'm not sure how to explain this more clearly, I really hope people can manage to figure this out though!). Pair $t_f$ and $x_f$ to draw the graph of the worldline of the front of the rocket, pair $t_b$ and $x_b$ to draw the graph of the worldline of the rear of the rocket.

steps 4&5) We have the representation of the rocket in frame S. Now we want to find the representation in frame S'

One (graphical) approach would be to individually transform each of the points we just calculated to plot our graph in S to plot a simlar graph in S'.

Recall (or look up) the equations of the Lorentz transform.

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

The inverse transforms may also be handy:
$t = \gamma \left(t^{'} + v\, x^{'} /c^2 \right) \quad x = \gamma \left(x^{'} + v \, t^{'} \right)$

note that $\gamma$ is defined as
$$\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$$

so the point $t_f = 0$ and $x_f=0$ transforms to the point $t^{'}_f=0$ and $x^{'}_f=0$
and the point $t_f = 1, x_f = v$ transforms to the point $t^{'}=\gamma(1-v^2/c^2)=1/\gamma$ and $x^{'}=0$

We'd need to plot two lines go get the representation of the front and back of the rocket in s', one line would represent the front, the other the back.

One can also make the substitutions using the inverse transform into equations 1 and 2, i.e. substitute for x the equivalent expression $\gamma \left(x^{'} + v \, t^{'} \right)$ and for t the equivalent expression $\gamma \left(x^{'} + v \, t^{'} \right)$

One can use a hybrid approach if one realizes that the form of the worldline will be $x^{'}$ = constant (for instance by using the graphical approach) then using the Lorentz transform to find the value of the constant.

Any approach will give the results of step 4)

The worldline of the back of the rocket is given by $x^{'}_b = 0$ for all $t^{'}_b$.

To carry out step 5), we'll work out in detail the equation for the worldline of the front of the rocket using the second approach, the mathematical substitution of variables.

Given $x_f = v \, t_f + L$, to transform the variables from unprimed to primed
we substitute in the above $x_f = \gamma \left(x^{'}_f + v \, t^{'}_f \right)$ and $t_f = \gamma \left( t^{'}_f + v \, x^{'}_f / c^2 \right)$

Performing the above substitution and dividing by $\gamma$, we get

$x^{'}_f + v \, t^{'}_f = v \, \left( t^{'}_f + v \, x^{'}_f / c^2 \right) + L / \gamma$

Rearranging and cancelling terms we get

$x^{'}_f \left(1 - v^2/c^2 \right) = L / \gamma$

Because $\left(1 - v^2/c^2 \right) = 1 / \gamma^2$ we can write this as

$x^{'}_f = \gamma L$

Because $L^{'} = x^{'}_f - x^{'}_b$ we can write

$L' = \gamma L$

which is the desired result, which says that the rocket is longer in S' than it is in S by a factor of gamma, or conversely that the moving rocket is $1/\gamma$ times as long as the stationary rocket.

2. Apr 24, 2014

andromeda

Thanks,
This sets up some nice presentation standard and perhaps it should be pinned to the front part of the thread.
I will use your templates from now on If I have a next chance to present something of mathematical nature

I have some question:

Although this is correct mathematically, when you arrive at step 2 why you still use $t_f$?

The worldline is in my understanding another name for an explicit equation of motion of a point so common time coordinate t should be used then equation $x_f = v \, t + L$ represents the forward point moving in time that is the same as origin time.

3. Apr 24, 2014

pervect

Staff Emeritus
I chose to use $t_f$ and $t_b$ to insure there were no problems arising from the relativity of simultaneity. In frame S, one will choose $t_f = t_b = t$. In frame S', this relation will no longer be true, it will be replaced by the relation $t^{'}_f = t^{'}_b = t^{'}$. Using separate variable names for the front and back of the rocket essentially treats the front and back of the rocket as separate particles. This allows one to apply the appropriate frame-dependent clock synchronization explicitly as needed to calculate the length.

4. Apr 25, 2014

vanhees71

The calculation is of course correct, but isn't this made a bit too complicated? I'd simply put it in the following way. In the frame $\Sigma'$, where the rocket is at rest, the world lines of the front and the back are
$$x_b'(t')=0=\text{const}, \quad x_f'(t')=L'=\text{const}.$$
This defines the length of the rocket to be $L'$ if measured in its rest frame.

In the frame, where the rocket moves with velocity $v$ we have, according to the corresponding Lorentz transformation
$$t_b=\gamma t_b', \quad t_f=\gamma (t_f'+v L'/c^2), \quad x_b=\gamma v t_b', \quad x_f=\gamma (L'+v t_f').$$
Now you bring the ends of the rocket in coincidence with your ruler at rest simultaneously wrt. to the refrence frame $\Sigma$, where the rocket moves. The events of "reading off" the marks are thus determined as fulfilling
$$t_f-t_b=\gamma[(t_f'-t_b')+v L'/c^2] \stackrel{!}{=}0 \; \Rightarrow\; t_f'-t_b'=-\gamma v L'/c^2.$$
$$L=x_f-x_b=\gamma [L'+v (t_f-t_b)]=\gamma L' (1-v^2/c^2)=\frac{L'}{\gamma}.$$