Infinitely separated objects in simultaneity

Question

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

Answer 1

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.

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

Answer 2

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.

Answer 3

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.