LOG#009. Relativity of simultaneity.

Other striking consequence of Lorentz transformations and then, of the special theory of relativity arises when explore the concept of simultaneity. Accordingly to the postulates of relativity, and the structure of Lorentz transformations we can understand the following statement:

$\boxed{\mbox{Simultaneity is a relative concept. It depends on the inertial reference frame.}}$

What does it means? Not surprisingly, if two arbitrary events happening in space and time, $E_1, E_2$, take simultaneously in one inertial reference frame, they do NOT do so in any other inertial reference frame. Indeed, Einstein himself guessed an alternative definition of simultaneity:

“(…) Two events $E_1, E_2$ taking place at two different locations are said to be simultaneous if two spherical light waves, emitted with the events, meet each other at the center of the tie line connecting the locations of the events(…)”

A proof can be done using Lorentz transformations as follows. In certain frame S’, two events $E'_1$ and $E'_2$ are found to be simultaneous, i.e., they verify that $t'_1=t'_2$. Therefore, using the Lorentz transformations (in the case of parallel motion of S’ with respect to S without loss of generality) we get

$ct_1=\gamma (ct'_1+\beta x'_1)$

$ct_2=\gamma (ct'_2+\beta x'_2)$

and thus, since $t'_2=t'_1$, the substractiong produces

$c(t_2-t_1)=\gamma \beta (x'_2-x'_1)$

We can recast this result as:

$\boxed{\Delta t=t_2-t_1=\dfrac{\beta}{c}\gamma (x'_2-x'_1)=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\dfrac{v}{c^2}(x'_2-x'_1)}$

We can also derive this equation with a LIGHT CLOCK gedanken experiment. A light clock of proper length $L'=l'$ is at rest in the S’ frame. Its x’-axis moves at speed $v$ parallel to the x-axis in teh S-frame. We attach some mirrors (denoted by M) to the short ends of the clock, and light travels parallel to the direction of motion. See the next figure of this device:

At the initial time, we synchronize the clocks in S and S’, meaning that $t=t'=0$ when $x=x'=0$ and the light flash is emitted. In the S’-frame, light propagates in both directions at speed of light, and then, it reaches the mirrors at the ends at the same time

$t'=\dfrac{s'}{c}=\dfrac{L'}{2c}$

In the S-frame, by the other hand, the length of the light clock is contracted due to length contraction $L=\sqrt{1-\beta^2}L'$ but light still propagates at speed c. However, the left-hand mirror is moving toward the light at speed $v$. In the time interval $t_1$ required for light to reach the left mirror, light travels a distance

$ct_1=\dfrac{L}{2}-vt_1$

In the same way, the right-hand mirror is running away from the light flash. In a certain time $t_2$ required for light to reach it, light should travel a total length

$ct_2=\dfrac{L}{2}+vt_2$

Substracting both equations, we obtain

$t_2-t_1=\dfrac{L}{2(c-v)} - \dfrac{L}{2(c+v)}=\dfrac{L}{2}\dfrac{2v}{c^2-v^2}=L \dfrac{v}{c^2} \dfrac{1}{1-\dfrac{v^2}{c^2}}$

Using the contraction lenght result, i.e., using that $L=\dfrac{L'}{\gamma}=\sqrt{1-\dfrac{v^2}{c^2}}L'$, and that $L'=x'_2-x'_1$ in the S’-frame, we get the previous result

$\Delta t= t_2-t_1=\dfrac{1}{ \sqrt{1-\dfrac{v^2}{c^2}}}\dfrac{v}{c^2}(x'_2-x'_1)=\dfrac{\beta}{c}\gamma (x'_2-x'_1)$

Q.E.D.

Therefore, the meaning of this boxed formulae is straighforward:

If two events are simultaneous in the S’-frame are simultaneous in the S-frame in the case they do happen at the same position ($x'_2=x'_1$) and/or the relative velocity v between the two frames is zero ($v=0$).

Moreover, we have this interesting additional result:

The larger the spatial separation is between two simultaneous events in the S’-frame, and/or the higher the relative velocity is between the S and the S’ frame, the greater is the temporal separation of the events in the S-frame.