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I just watched this video regarding block universe: https://www.youtube.com/watch?v=wwSzpaTHyS8&t=676s and it provoked the following thought experiment:

Let's assume two observers, O1 and O2, at the same location in space, L1. There is a relativistic speed difference between the two, and they are both observing the same location, L2, one lightyear away.

According to relativity of simultaneity, the relativistic speed difference means that they perceive different timeframes, and as such have contrasting perceptions about the current "now" at L2. For the observer O1, it is currently T1, for the observer O2, it is T2. Let's assume that T2 = T1 + half a year.

Due to the speed of light, when O1 observes L2, it observes events transpiring at T1 minus one year, while O2 observes events transpiring at T2 minus one year = T1 minus half a year.

Now, O2 transmits its observations of L2 to O1. Since they are in the same location, this transmission can occur with no delay. This allows O1 to access information about events transpiring at a location one lightyear away only half a year after they took place.

What's wrong with this picture?

EDIT: A perhaps a bit more tangible version:

Imagine the sun goes supernova right now. Due to the finite speed of light, an observer standing on the surface of Earth will not know about this event for another 8 minutes.

4 minutes later, an alien spacecraft is passing by Earth at relativistic speeds. In that spacecrafts frame of reference, the "now" of the sun is shifted 4 minutes forward relative to the observer standing on the surface of Earth.

From the alien spacecrafts point of view, the sun went supernova 8 minutes ago. Since it's at the same location as Earth (8 lightminutes away), it would now observe the sun going supernova. If it transmits this information to Earth during its fly-by, an observer on Earth would know about the supernova right now, 4 minutes sooner than the information would ordinarily be able to arrive.

Accepted answer:

I have accepted benrg's answer, which, if I interpret it correctly, dictates that in the alien spacecrafts frame of reference, the distance between Earth and the sun is increased to 12 lightminutes, so it still won't detect the supernova for another 4 minutes (disregarding that the spacecraft will have moved closer by then).

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    – Buzz ♦
    Commented Jun 3 at 0:36

2 Answers 2

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when O1 observes L2, it observes events transpiring at T1 minus one year, while O2 observes events transpiring at T2 minus one year

This part is incorrect. The distance is different in the two frames (length contraction), and moreover the emitter is moving in at least one of the two frames, so it isn't in the same place when it emits the light.

If you work it out correctly, you'll still find that your two observers see events at different time and position coordinates, but they're the same events. They're just assigned different coordinates in the two frames, and the coordinates are mapped to each other by the Lorentz transformation.

An easier way to see that they see the same thing is that they are not really seeing L2, but rather light at their own location that enters the pupil of their eye, and since they're at the same location they see the same light.

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  • $\begingroup$ Thank you! I updated my post with a more tangible illustration and my interpretation of the implications of your answer. Did I get it right? $\endgroup$
    – Uffe Poul Hansen
    Commented Jun 2 at 21:39
  • $\begingroup$ @UffePoulHansen That's the right idea, but the numbers (12 light minutes and 4 minutes) are wrong because, as you said, you disregarded the ship's speed relative to the earth and sun. The easiest way to get the right answer is to use the fact that $Δx^2-(cΔt)^2 = Δx'^2-(cΔt')^2$, and plug in Δx = 8 light minutes, Δt = 4 minutes, Δt' = 8 minutes. $\endgroup$
    – benrg
    Commented Jun 3 at 1:09
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There is simply no way for a light signal in vacuum to go from a source to a repeater to a receiver faster than for it to go directly from the source to the receiver.

By the triangle inequality the distance between the source and the repeater plus the distance between the repeater and the receiver must be at least as large as the distance between the source and the receiver. This holds in every inertial frame. And in every inertial frame the speed of light is $c$. So the time for the indirect path is always equal to or greater than for the direct path.

In more geometrical terms, light travels on a geodesic. Any small deviation from the geodesic path will increase the time.

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  • $\begingroup$ What is the repeater in the OP's scenario? $\endgroup$
    – PM 2Ring
    Commented Jun 2 at 17:05
  • $\begingroup$ @PM2Ring it is O2 $\endgroup$
    – Dale
    Commented Jun 2 at 17:56
  • $\begingroup$ At the time of the thought experiment, O1 and O2 are both located at L1. O1 and L1 are at rest relative to each other, and to L2. $\endgroup$
    – PM 2Ring
    Commented Jun 2 at 18:23
  • $\begingroup$ @PM2Ring yes. The triangle inequality holds, it is just a degenerate case $\endgroup$
    – Dale
    Commented Jun 2 at 20:02

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