# How do I derive the Lorentz contraction from the invariant interval?

While reviewing some basic special relativity, I stumbled upon this problem:

From the definition of the proper time: $$c^2d\tau^2=c^2dt^2-dx^2$$ I was able to derive the time dilation formula by using $$x=vt$$: $$c^2d\tau^2=c^2dt^2-v^2dt^2=c^2dt^2\left(1-\frac{v^2}{c^2}\right)\rightarrow d\tau = dt\sqrt{1-\frac{v^2}{c^2}}=dt/\gamma$$

Now, I would very much like to be able to derive the length contraction formula in a similar fashion, and feel strongly that this should be possible. The definition of the invariant interval is: $$ds^2=dx^2-c^2dt^2$$ using $$t=\frac{x}{v}$$ I tried: $$ds^2=dx^2-\frac{c^2}{v^2}dx^2=dx^2\left(1-\frac{c^2}{v^2}\right)\rightarrow ds=dx\sqrt{1-\frac{c^2}{v^2}}$$

This is where I'm stuck: I don't see how this can be converted to a Lorentz factor...

Any help that will allow me to arrive at the desired result $$ds=\gamma dx$$ would be very much appreciated.

• The cleanest way to do this is to generalize the $2-D$ rotation to a $2-D$ hyperboic rotation by subbing all of your sines and cosines for hyperbolic sines and cosines, and then making the definition $v/c = \tanh \phi$, and using that to eliminate your rotation angle. The results will just pop out. Oct 12, 2013 at 15:43

Suppose we have a rod of length $l$ at rest in the unprimed frame and we watch an observer in the primed frame speeding past: We'll take the origins in both frames to coincide when the observer in the primed frame passes the first end of the rod, so Event A is $(0, 0)$ in both frames.

In the unprimed frame the far end of the rod is at $x = l$, and we see the speeding observer pass it at $t = l/v$, so Event B is $(l/v, l)$. The interval between these events is therefore:

$$s^2 = \frac{c^2l^2}{v^2} - l^2$$

In the primed frame the stationary observer sees the rod, of length $l'$ coming towards him at speed $v$. The $x$ coordinate of both events is zero, and the time of Event B is $t = l'/v$, so the interval is:

$$s'^2 = \frac{c^2 l'^2}{v^2}$$

The intervals must be the same, $s^2 = s'^2$, so:

$$\frac{c^2 l'^2}{v^2} = \frac{c^2l^2}{v^2} - l^2$$

and a quick rearrangement gives:

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

$$l' = l \sqrt{1 - \frac{v^2}{c^2} } = \frac{l}{\gamma}$$

Response to comment:

To work out the time dilation you use a different pair of events. In the unprimed frame you have a clock, ticking with period $T$, stationary at the origin. So the events for the first and second ticks are $(0, 0)$ and $(T, 0)$. The interval $s^2 = c^2 T^2$.

As usual we choose the primed frame so the origins of the frames coincide, and the first tick is at $(0, 0)$. The second tick is at $t = T'$, and because the clock is moving at velocity $v$, the $x$ coordinate of the second tick is $x = vT'$ giving $(T', vT')$. The interval is therefore $s^2 = c^2T'^2 - v^2T'^2$.

As before, we set the intervals equal so:

$$c^2 T^2 = c^2T'^2 - v^2 T'^2$$

or:

$$T'^2 = T^2 \frac{c^2}{c^2 - v^2}$$

Now just divide the top and bottom of the RHS by $c^2$ and take the square root to get:

$$T' = T \frac{1}{\sqrt{1 - v^2/c^2}}$$

• Stupid question but why do you assume in the Prime frame that the length is l' and the speed is v? Why not length of l and speed of v'? Dec 29, 2018 at 10:59
• >The interval between these events is therefore: 𝑠2=𝑐2𝑙2𝑣2−𝑙2 Why squared? May 9, 2021 at 17:36