Let us assume that there are two astronauts A and B who are floating in space. A sees B passing by and vice versa. A sends signals to B every minute. According to A since B is moving his clock will be slower. So B will receive the signals prior to the appointed minute. The same argument can be applied for B who will conclude A's clock is running slow. Who is right?
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$\begingroup$ 'he will receive signals prior to the appointed minute' - what exactly do you mean by this? Whether you take Special Relativity into account or not, you always need to account for the time it takes a signal to get from A to B. $\endgroup$– DJBunkAug 15, 2012 at 16:50
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$\begingroup$ I mean that say A is calling on the cell which sends waves at light spees and he says it is 15 minutes past 10. This is the time i mean by appointed second. $\endgroup$– SN77Aug 15, 2012 at 16:55
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1$\begingroup$ I have just now realized that B will receive the signal after the appointed minute. Actually there are two competing effects slowness of B's clock v/s the time travel of the signal. The Latter effect should outdo the prior $\endgroup$– SN77Aug 15, 2012 at 16:59
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$\begingroup$ @SN77: Your last comment is correct and the most important thing, but you must also include failure of simultaneity to understand everything. $\endgroup$– Ron MaimonAug 16, 2012 at 3:59
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2 Answers
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Both are right. Any moving clock is slower than a clock at rest, from the perspective of the frame at rest.
Maybe this simplified freehand graphic (apologies for its lack of precision) helps to see that both A and B feel the same about each other's time dilation:
Let's say that the red axis represents A and its proper time measured in minutes (first eight minutes are showed). Green axis and its numbers represents B observer.
Light or radio signals from A to B, represented in red oblique lines, are fired on a minute basis. Six of them are showed, that took six minutes of A proper time. However, these six signals from A to B take some eight minutes in B proper time. B concludes that A clock is slower. The same holds if we invert the situation (green lines from B to A). Well, almost the same (the last green line is intended to go from green 6 to red 8, blame my trembling fingers).
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$\begingroup$ How come both the persons say that their clocks are slower than the other and both be right? $\endgroup$– SN77Aug 15, 2012 at 17:01
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$\begingroup$ A says B is slower, and so does B about A. They are in different inertial frames, so both of them detect the time dilation of the other. $\endgroup$– CsourcesAug 15, 2012 at 17:08
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1$\begingroup$ Actually, you made me rethink this (as so often happens in answering questions of special relativity). I deleted my comment as such. $\endgroup$– mcFreidAug 15, 2012 at 17:09
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$\begingroup$ This is the right answer to the question asked. In the frame of A, B's clock runs slower. In the frame of B, A's clock runs slower. That's not a contradiction. In relativity, time measurements are reference-frame dependent, and we shouldn't expect observers in two different frames to agree. @MarkM's reply addresses the more complicated situation of the twin paradox, where you bring the two twins back together and then ask who's older. That's a good follow up, but goes further than the question that was actually asked. $\endgroup$ Aug 15, 2012 at 17:58
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The key is to remember that you need to pick a reference frame, and stick with it. So, let's say we are A.
We see B pass by us. Let's say, that B intends to travel 4 light years, turn around, and then come back. B accelerates away from us, reaches 80% of the speed of light, and begins his journey. According to us, how long will it take? If $d$ is the distance to his destination, then total time should be $t=\frac{2d}{v} = \frac{8 \,\text{lightyears}}{0.8\,c} = 10$ years. So, we say that 10 years will pass until B returns to us.
How long do we say B will say the journey will take? Multiply our time by the Lorentz factor, $$t'=\frac{t}{\sqrt{1-\frac {v^{2}}{c^{2}}}} = 6$$ B will only age 6 years when he returns, compared to the ten years we have aged.
How do B's calculations compare? Well, due to length contraction, the distance to his destination becomes our distance (4 light years) divided the Lorentz factor (since he sees everything moving by him, he says everything else is becoming length contracted.), which gives 2.4 light years. Since he is travelling there and back, the total distance traveled is 4.8 light years. Since his velocity is .8c, the total elapsed time is $t=\frac{d}{v}=\frac{4.8\, \text{lightyears}}{0.8\,c} = 6$ years.
So, when you include ALL effects (time dilation and length contraction), we can see the numbers work out such that B has measured 6 years, and A has measured 10.
So, what decides who elapses less? This is key - it's the one who had to accelerate. Whatever B used to get his journey started forced him to accelerate, so he was a non-inertial observer. So, he ended up elapsing less time. Similarly, a rocket leaving earth will elapse less time, since he was the one who had to accelerate away.
Now, you may ask, half way through the journey, what are the numbers? This is a meaningless question - since nothing can be transferred faster than light, either a signal must be sent (the same conclusion will hold), or they must meet up (as in our example). So, you can see, when the two observers meet up, they will find the one who accelerated elapsed less time. That's the whole origin pf the 'paradox' - they both can say the other's clock is slower. However, when you do the calculation (when they meet up to look at each other's clocks), you will find that B elapsed less.
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$\begingroup$ There are actually resolutions of the twin paradox using just special relativity, pretty much treating the acceleration as happening instantaneously. This works because there are three inertial frames involved (e.g. the earthbound twin, the frame of the traveling twin while he's moving away from Earth, and the frame of the traveling twin while returning to Earth). $\endgroup$ Aug 15, 2012 at 17:33
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$\begingroup$ In the returning frame, the outgoing twin experienced even more time dilation than the earthbound twin, so switching to this frame provides the necessary correction such that by the time he returns to Earth both twins agree on what their ages should be. $\endgroup$ Aug 15, 2012 at 17:33
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$\begingroup$ Mark, while certainly everything you've said is correct, I think you answered a different question than what was asked here. Then again, your answer is accepted... :-/ $\endgroup$– David ZAug 15, 2012 at 19:00