I was watching this derivation of 'asynchronous' nature of clocks from which I got slightly confused.

Consider two reference frames moving relative to each other call them 'Red' and 'Blue' Frame. When the origin of blue Frame passes that of red frame a flash of light is emitted from the origin of both frames that is going to help the observers in the red and blue frame synchronise that frames 'clocks'. now we analyse the synchronisation process for the blue frame from the perspective of the red frame and of the blue frame.

Event 1: light emitted from origin of blue frame at $(t'_1)_0=0$

Event2: light arrives at $x'$ at $(t'_2)_0$ on origin clock

Blue Frame

in order to synchronise clock at $x'$ with that at origin must set $(t')_{x'}=(t'_2)_0=\dfrac{x'}{c}$

Red Frame

we will first calculate the time $(t_2)_0$ and then since the blue frame is moving we will use time dilation to find $(t'_2)_0$

$c(t_2)_0=v(t_2)_0+\dfrac{x'}{\gamma}$ the distance the light travels is the distance the blue frame travels in $(t_2)_0$ plus the distance from $x'=0$ to some generic $x'$ which is Lorentz contracted from our frame

$\implies (t_2)_0=\dfrac{x'}{\gamma(c-v)}$ since by time dilation $(t'_2)_0=\dfrac{(t_2)_0}{\gamma} \implies (t'_2)_0=\dfrac{x'}{\gamma^2(c-v)}=\dfrac{x'}{c}(1+\dfrac{v}{c})$

This tells us that from Red frame perspective using relativistic results we conclude that the clock at position 0 in blue frame is $\dfrac{x'}{c}(1+\dfrac{v}{c})$ at event 2 and so the observer at position $x'$ must set their clock to this value for that clock to be synchronised.

My confusion now arises: we have found that the time at $(t'_2)_0$ is not the same if we analyse this situation from the two reference frames, however shouldn't we have gotten the same value for this time? in our analysis from the Red frame we have used relativity to find a value for $(t'_2)_0$ that the observers in the blue frame will say that the time is at origin when event 2 happened however from the analysis of observers in the blue frame this time is different, thus it seems not to be consistent. i.e if now the blue frame stops and we were to look at the clock at origin for event 2 we would get a contradiction since observers in blue frame claim that $(t'_2)_0 = \dfrac{x'}{c}$ while observers in Red frame claim that $(t'_2)_0=\dfrac{x'}{c}(1+\dfrac{v}{c})$

So what is going on?

Thank You for any help

Example is from the course on special relativity from World Science U enter image description here

  • $\begingroup$ If x = x' = 0, the emitted light will be detected immediately by both frames. The problems with synchronism arise when light has some finite distance to run. $\endgroup$ Commented Dec 18, 2019 at 20:20

1 Answer 1


The relativity of simultaneity (an often forgotten feature of Einstein's train thought experiments) says that the synchronised blue clocks are not seen by the red observer to be synchronised.

If red tries to synchronise his clocks using calculations that assume the blue clocks are synchronised for him too, he will fail.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.