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In the observer's frame the events A and B are simultanous ($\Delta t = 0$) and are separated by $\Delta x$

We can get the corresponding time between events in the car frame using the Lorentz Transformation $\Delta t' = \gamma(\Delta t - \frac{V\Delta x}{c^2})$ and so

$\Delta t' = -\gamma\frac{V\Delta x}{c^2}$. From this equation the car always observes B happening before A hence the negative sign.

Suppose that the car at t=0 is at A or before A. And suppose that B is almost infinitely seperated (millions of light years away). Even though A is super close to the car and B is super far away from the car the equation $\Delta t' = -\gamma\frac{V\Delta x}{c^2}$. Tells us that if $\Delta x$ is super-large then B happens waaay before A. How is that possible what is the intuition behind this ?

The question is why even tho car is going toward both A and B, B has to happen first not A ? Physically what is the intuition behind this

*I understand that this is not about light reaching the observer as some answers assume that I do

  • $\begingroup$ Events A and B are separated by the distance $\Delta x$ so $\Delta t=\frac{\Delta x}{c}$, the condition $\Delta t=0 $ is for events that happen in one same place i.e. $\Delta t=\frac{\Delta x}{c}=0$ $\endgroup$
    – The Tiler
    Oct 22, 2022 at 17:46
  • $\begingroup$ Could you please use consistent notation for coordinates? Is the $y'$ coordinate's axis parallel to that of $x$, and the car's velocity is co-axial with it, i.e. is your problem 1-dimensional? If so, rewrite everything in $x$ and $t$, and use primes for the car's frame. $\endgroup$ Oct 22, 2022 at 23:24
  • 1
    $\begingroup$ @TheTiler This is how you get nonsense with math. :) The A and B obviously don't happen at the same spatial location on the diagram. Consider the spatial separation of events from the observer, not from each other. $\endgroup$ Oct 22, 2022 at 23:47
  • $\begingroup$ woops I wrote y' but obviously meant t'. Sorry for that $\endgroup$
    – qubitz
    Oct 23, 2022 at 3:07

3 Answers 3


there is no inconsistency with what you have derived. With respect to a moving observer, the event B is happening at a much earlier time than event A.

But let me address what I think might be the confusion. The moving observer is not receiving the signal from B much earlier than it received the signal from A. Depending on where the observer is located, it might take light much shorter to travel from A to observer, than from B to observe. Here we are talking about an intelligent observer that knows about the speed of light, and determines the time at which the event B has occurred. So nothing paradoxical is going on, you can still see event A happening first - if you happen to be near it when the event occurs - and receive the signal from B at a later time, but realize that B must have happened much much earlier. It's like looking at the stars, knowing that what we are seeing is the past.

What is the mathematical intuition to this? Well, as a moving observer your spacetime is tilted with respect to an observer at rest. If you are familiar with Minkwoski diagrams, you might know that you can draw the coordinate frame of a moving observer with respect to another, as shown in the figure below. Now, this is greatly exaggerated, but you can imagine what would happen if you were moving slowly, the time axis of the observer ($t'$) would deviate slightly from the stationary one ($t$). The two events, A and B, happen both at $t = 0$ according to the stationary observer, but do not happen simultaneously for the moving observer. Specifically, A happens in the future and B happens in the past.

Minkowski coordinates

Now you ask what happens if you make the distance between the two events, from the first observer, very large. Then even the smallest deviations of the axis $t'$ from t would get accentuated. And that basically explains the paradox. The farther things are from us, the more little deviations in our coordinate system will matter.

I hope this helps

  • $\begingroup$ Yes sir I get that but physically why do event B happens first than A if observer is moving towards it ? car is moving towards both A and B so why should B happen first ? $\endgroup$
    – qubitz
    Oct 23, 2022 at 23:43
  • $\begingroup$ Because in a moving reference frame space and time are mixed in, things that are far away are also more in the past/future. Like look again at the diagram I sent in my answer, the line labeled x' is your time 0, but events A and B both fall on the actual horizontal x axis. So whichever event is farther to the right, the more they will be below your time = 0, thus in the past $\endgroup$ Oct 24, 2022 at 2:32
  • $\begingroup$ I understand that and can clearly see that from the diagram but that is a geometric explanation. Is there a physical explanation ? $\endgroup$
    – qubitz
    Oct 24, 2022 at 3:24
  • $\begingroup$ I made the modification to the figure, I think this should be bit clearer and more useful. As for @qubitz point, I am not sure I know what a physical explanation is in relativity, because geometry and coordinate systems are so intrinsically important to the subject that I think a geometric explanation is also a physical one. I should also point out that it doesn't matter where you are, whether you are to the left of both A and B and moving towards them, somewhere in the middle, or moving away from both of them. Your conception of the present is skewed, and so are both past and future. $\endgroup$ Oct 24, 2022 at 15:58

You need to think of the car frame as consisting of a set of meter sticks and synchronized clocks. When we say event B occurs at time t' in the cars frame, that's not necessarily the time in that frame at which the light reaches the observer in the car. That's because the light has to travel from point B to the car and the time that it reaches the car is not what is meant by the time t' in the car's frame.


we have $-x_{A}=x_{B}=x\;$ and $\;\;t'=\gamma\left(t-\frac{vx}{c^{2}}\right)$, the difference is $$\Delta t' =\gamma \,\frac{v}{c^{2}}\,\left(-x_{A}+x_{B}\right)=\gamma \,\frac{2vx}{c^{2}}$$

If we replace $x$ by $S= 2\pi R$ and $v= \omega R $, we find the formula of the Sagnac effect.


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