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Let's say, Observer A is at rest and Observer B is in a spaceship moving at a significant fraction of speed of light. B synchronized his atomic clock with A's when his spaceship just passed A. enter image description here

Now, some time later a star becomes a supernova and A and B observe (record the time coordinate) of this event supernova. The exchange messages to compare the time they recorded with each other and they see that they measured different times for the supernova event.

Cesium's transition frequency (9,192,631,770) does not change, for A and B. So, how did the measurements change?

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  • $\begingroup$ "When they meet," - wait, how could they meet? Did the spaceship decelerate to a stop and then accelerate back towards A? $\endgroup$
    – Alfred Centauri
    Commented Nov 2, 2018 at 15:35
  • $\begingroup$ No when the supernova exploded, they noted down the time on a paper. Whatever the clock does afterwards due to deceleration and acceleration is irrelevant. A and B compare the time they noted on the paper. $\endgroup$
    – Krutik Desai
    Commented Nov 2, 2018 at 15:37
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    $\begingroup$ Got it, thanks for clarifying that. One last question, do A and B write down the time they see the supernova explosion or do they write down the time they observe the supernova explosion? $\endgroup$
    – Alfred Centauri
    Commented Nov 2, 2018 at 15:38
  • $\begingroup$ When the supernova occurs, they note down the time. The supernova is just a placeholder for any event. $\endgroup$
    – Krutik Desai
    Commented Nov 2, 2018 at 16:08
  • $\begingroup$ Krutik, if they write down the time they see the supernova, then time of flight of the light from the supernova to their eyes must be taken into account. This is different from observation in SR where to observe is to 'record the spacetime coordinates of an event in an inertial coordinate system in which one is at rest'. $\endgroup$
    – Alfred Centauri
    Commented Nov 2, 2018 at 16:30

2 Answers 2

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A sees that the calcium clock of B runs slower, and viceverza. But the problem is not symmetric. First, the supernova is moving relative to one of the observers. That observer will see that the distance betweem him and the supernova is smaller than what the stationary observer will see. Second, at least one of the observers has to accelerate if they are going to meet again. The acceleration changes things in such a way that the cesium clock has marked less ticks for the moving/accelerating observer. To him, the cesium clock of the stationary observer was running slow when he moved at constant speed, and then accelerated when he was decelerating.

Case with no acceleration: Asumming that the star explodes when the distances to both is the same according to A, then A will see that B marks a larger time than him. Even if to him B's clock is running slower, he see that B marks the explosion after it happens. To A both events, the explosion and B writing the time of the explosion, are not simultaneous. B will also see A's clock as running slower than him, and will also see that the explosion happens before A marks the time. But to B the explosion happens when the star is closer to A than to him.

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  • $\begingroup$ I clarified the question in a comment above. The acceleration and deceleration is not considered. Also I understand that supernova is moving relative to the B but I meant it as an "event" that both A and B see so as to note down their time. So I intended that relative motion to be neglected. Sorry for being unclear. $\endgroup$
    – Krutik Desai
    Commented Nov 2, 2018 at 15:44
  • $\begingroup$ But they cannot meet unless one of them decelerates. And the answer of what time each measures depends on the position of the supernova. And there is not only time dilation but also lenght contraction, so the moving guy will see the distances along the line of movement as shorter. $\endgroup$
    – user65081
    Commented Nov 2, 2018 at 16:06
  • $\begingroup$ And also as Alfred Centauri asked, are they recording the time the light reached them or the time they calculate the explosion happened? $\endgroup$
    – user65081
    Commented Nov 2, 2018 at 16:10
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    $\begingroup$ I only ask the question as a part of special relativity. And the moment the supernova occurs, both observers note the time (the supernova is just a placeholder for any event). They note down the time on paper. The clock may change depending on the deceleration afterwards but the observers compare the time on paper. $\endgroup$
    – Krutik Desai
    Commented Nov 2, 2018 at 16:13
  • $\begingroup$ @AaronStevens yes, that is why I added the second paragraph to the answer, it does not involve any acceleration $\endgroup$
    – user65081
    Commented Nov 2, 2018 at 22:09
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Cesium's transition frequency (9,192,631,770) does not change, for A and B. So, how did the measurements change?

The measurements didn’t change, they were just measuring different things. If I measure how many meters away London is from me and you measure how many meters away London is from you then even though our meter sticks do not change we will get different measurements. This is because the distance to London depends on the reference point, and our two measurements were using different reference points.

Similarly, the time of the supernova depends on the reference velocity and the two measurements were using different reference velocities. Nothing changed about the measuring devices, they were just used to measure different things.

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