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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 $\Delta y' = \gamma(\Delta t - \frac{V\Delta x}{c^2})$ and so

$\Delta y' = -\gamma\frac{V\Delta x}{c^2}$. From this equation the car always observes B happening before A hence the negative sign.

I understand that if the car is in the middle B happens first because the car is going towards it and one could think of the light emmited by lightning reaching the car first. However suppose the car is to the left of A when both are emmited. The lorentz equation tells us that B stills happen before A. How is this possible? Isn't the car going towards A (also towards B) but A is closer so A should happen first ? Where did I go wrong

qubitz
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2 Answers2

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This is a common misconception that students can get due to over-emphasis on these “thought experiments”. The time that something happens is determined not purely by the time the light is received, but that time is corrected for the light travel time delay. So if I receive the light today from something 100 light years away then I do not think it occurred today, I know that it occurred 100 years ago.

So, yes, in your modified scenario you would receive the light from A first. But once you correct for the light travel time delay you will determine that B happened first. The center point is used because it makes the comparison easy, not because it is necessary.

Dale
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  • If the event was infinitely seperated. The event at B would still happen first then A how is that possible ? – qubitz Oct 22 '22 at 14:02
  • Infinity isn’t a place. You can take some specified finite distance and take the limit as it goes to infinity. The equation holds – Dale Oct 22 '22 at 14:44
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By placing car to the left of A is not good case for relativity of simultaneity because events are not equidistant. Question is when speed of light is constant for ground and car, then how they see things differently. If one says that car is moving to B, so light from B reach car first.

But from where it is seeing that car is moving, from ground. So one is calculating from ground that what car must observe. Because from car, it is at rest and ground is moving back. But light of lightning leaves points of events and independent of motion. So it must reach to the car in equal time from A and B.

There is no relativity of simultaneity from theory of relativity's postulates. So time is not different for different frames.

Neil Libertine
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