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.

  • 1
    $\begingroup$ 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. $\endgroup$ Commented Oct 12, 2013 at 15:43
  • $\begingroup$ I know this is old, but here is the fastest way I can think of: Imagine a clock (stationary in $K':(\tau,x',y',z')$) moving past a rod (stationary in $K:(t,x,y,z)$), all along the same axis. The clock moves past the two ends of the rod in a time interval $d \tau$ in $K'$, and we can find out how long this takes in $K$ from the invariant interval $ds = d \tau = \sqrt{1-\mathbf{v}^2} dt$, giving time dilation. The length of the rod in $K'$ is thus $L' = ||\mathbf{v}|| d \tau = (||\mathbf{v}|| d t) \sqrt{1-\mathbf{v}^2} = L_0 \sqrt{1-\mathbf{v}^2}$, giving length contraction. $\endgroup$
    – bolbteppa
    Commented Dec 11, 2023 at 6:29

2 Answers 2


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:

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


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

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

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

  • $\begingroup$ 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'? $\endgroup$
    – Thomas
    Commented Dec 29, 2018 at 10:59
  • $\begingroup$ >The interval between these events is therefore: 𝑠2=𝑐2𝑙2𝑣2−𝑙2 Why squared? $\endgroup$
    – m8labs
    Commented May 9, 2021 at 17:36

I would like to point few things out from the Lorentz transformation which make the above derivation for length contraction work and then we'll use that to derive length contraction. \begin{align} \tau &= \gamma\left(t-\frac{v}{c^2}x\right)\\ x' &= \gamma\left(x-vt\right) \end{align} Quickly here, we will evaluate the relevant differentials. \begin{align} d\tau &= \gamma\left(dt-\frac{v}{c^2}dx\right)\\ dx' &= \gamma\left(dx-vdt\right) \end{align}

For time dilation, we consider the spacetime interval given as:

\begin{align} -c^2dt^2+dx^2&=-c^2d\tau^2+dx'^2\\ -c^2dt^2+v^2dt^2 &=-c^2d\tau^2\\ d\tau &= \sqrt{1-\frac{v^2}{c^2}}dt \end{align}

here we used \begin{align} dx' &= \gamma\left(dx-vdt\right)\\ 0 &=\gamma\left(dx-vdt\right)\implies dx=vdt \end{align}

Similarly for Length contraction, we will need: \begin{align} d\tau &= \gamma\left(dt-\frac{v}{c^2}dx\right)\\ 0 &= \gamma\left(dt-\frac{v}{c^2}dx\right)\implies dt = \frac{v}{c^2}dx \end{align}

Let's us now evaluate it: \begin{align} dx'^2 &= -c^2dt^2+dx^2\\ dx'^2 &= \sqrt{1-\frac{v^2}{c^2}}dx \end{align} Here, we need to remind ourselves that, we set $d\tau=0$ in observer's frame to evaluate the length as measured by the observer. In the time dilation part, we set $dx'=0$ in object's frame to evaluate the time interval felt by it. Both calculations are happening in different frames of references.


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