0
$\begingroup$

The twin paradox states that if you have one twin on earth and one twin is sent in a round trip journey to space in a rocket close to the speed of light, when the twin sent in the rocket returns to earth they will have aged less than their twin on earth.

Now let's call the twin on earth Twin A and the twin sent to space Twin B. During the time when B is accelerating to turn around, it will no longer be in an inertial frame and it will see A's clock moving rapidly. However, let's assume A has a light clock, so when B will see A's clock moving rapidly B will observe A's clock moving faster than speed of light. Moreover, the sequence of events of a car moving on earth will also be observed to pass rapidly by B and B might see the car moving faster than speed of light.

So my question is that although B will observe the events happening on earth rapidly during the acceleration, but doesn't that mean some events of moving objects will be happening rapidly such that the objects are moving faster than speed of light relative to B hence violating the second postulate of special relativity?

$\endgroup$
2
$\begingroup$

In B's frame, A's clock says (say) noon. Then B quickly turns around, and in his new frame, A's clock says (say) 3PM. The change of frames can happen arbitrarily fast.

This is exactly the same phenomenon as any other change of frames. Stand at sunset so that in your frame, the sun is 93 million miles in front of you. Now turn 180 degrees so that in your new frame, the sun is 93 million miles behind you. Did you just "see" the sun move 186 million miles in an instant ---- or did you simply change frames very fast?

$\endgroup$
  • $\begingroup$ Dear WillO. After your comments, I edited, but just a few comments on your answer. You are trying to explain a phenomenon with SR, and it is not the real thing. " The change of frames can happen arbitrarily fast." So this would mean that the ship turns around in a jiffy. Sorry, that is not how it works in real life. And no, you cannot turn around instantly. Nothing moves instantly. $\endgroup$ – Árpád Szendrei Apr 14 at 16:10
  • $\begingroup$ the spaceship needs to decelerate, use energy, and then turn around and accelerate again. This needs time. During this time (that you disregard) happens the real thing. The two twins start moving in the time dimension with different speeds, they age differently. And this is the only period when the twin on the spaceship could see the other twin's clocks' parts, and other things on Earth possibly move faster then c. If you try SR, then during constant speed travel, both twins could see the other one's clock move faster. Constant speed is symmetrically relative. Acceleration is absolute. $\endgroup$ – Árpád Szendrei Apr 14 at 16:14
  • $\begingroup$ @ÁrpádSzendrei : No, I did not disregard the time in which the ship is accelerating; I said that the time could be very short (which is needed to be consistent with the OP's question, where he wants $B$ to see a very fast change in the time on $A$'s clock). The ship travels out (in a straight line, for simplicity), then takes some time to turn around, then travels in (in another straight line). To answer the OP's question, all that matters are the initial and final frames, not what happens in between. (Just as what happens during the turning has no bearing on the sun example.) $\endgroup$ – WillO Apr 14 at 16:57
  • $\begingroup$ I believe this answer is misleading,as this is not how this happens in real life. He is asking about what the traveling twin sees from the spaceship on Earth. What he is asking is, can the twin in the spaceship see things move faster (then c) on Earth during the turn (acceleration)? The real answer is yes, and this is caused by GR time dilation. $\endgroup$ – Árpád Szendrei Apr 14 at 17:08
  • $\begingroup$ Thank you for your answer WillO. So the way I see your answer is that if we show how the angle of B's slice in the Minkowski diagram changes during the acceleration, over a long distance the slice swipes through events much faster than speed of light due to transitioning between inertial frames. And displacement and time between two events which were observed by B in two different inertial frames during his transition would be inadequate to determine speed because the coordinate system of the two inertial frames are not the same. Please correct me if I am wrong. $\endgroup$ – Salar Khan Apr 14 at 21:32
0
$\begingroup$

as you indicated, he's not in an inertial frame so postulates of relativity don't apply to him.

$\endgroup$
-2
$\begingroup$

First of all, after the correct comments, you are asking (or mentioning) about two main things:

  1. why is one twin getting older. since this is not your main question, I will not go into this

  2. why can the twin on the spaceship see things move faster then light on earth.

I believe your main question is 2., why does the twin in the spaceship see things move faster than c on Earth, an why is SR violated. You can already feel that SR is not the right thing for this phenomenon. That is in part why GR was developed, GR describes gravitational time dilation, and that is what to our current knowledge best describes the events you are asking about. Basically in short version, SR states that nothing can move faster then c, when measured locally in vacuum. Your measurement is not local. You are trying to measure speed from far away (from a different gravitational zone).

If you try to explain this with SR, you will realize that both twins could symmetrically say that the other's clock moves faster. Which one is right? Constant speed is symmetrically relative. You need acceleration. Acceleration is absolute. That is why you need GR. The problem with the SR explanation is too, that it disregards the period of turnaround, and how it works in real life. Turning around with a spaceship needs time and energy. It needs acceleration, and that is equal to a gravitational zone according to the equivalence principle. That is the specific period during which the traveling twin will see the other's clock tick faster, and the twin on Earth will see the other's clock tick slower. SR cannot give you that with a real life explanation (or at least it is much more complicated).

You are on the right track, just need a few things to clarify:

  1. you are correct, this is due to gravitational time dilation (that is where you ask about the "during acceleration" part of the journey)

  2. because of the equivalence principle, the accelerating ship has the same effect on the clock as a gravitational zone (slowing down the clock relatively)

  3. light travels at speed c in vacuum when measured locally

  4. you are right, you can measure a different speed then c when you measure the speed of light from a far away observer's view

You can measure a speed different from c even when measured locally in a medium (not vacuum), but your case is about the nonlocal measurement.

The main thing is, that you need to measure the speed of light from a far away observer's view, and the observer needs to be in a region of space that has a different stress-energy (gravitational zone) then where the light is actually traveling.

There are two main cases:

  1. Let's say that you are measuring the speed of light as it passes next to the Sun, and you are measuring it from Earth. This is the Shapiro delay, and you will measure a speed less then c, because of the Sun's stronger (relative) stress-energy, and Earth's weaker stress-energy (relative to the Sun). The clock on earth ticks faster, and you divide the path of light with a time that is more (relatively), so you get a smaller speed. The path is longer too, because of gravity, but let's disregard that.

  2. And yes, contrary to popular belief, it is possible to measure a speed more then c when measured from far away. When we say the maximum speed is c, we mean when measured locally in vacuum. If you would measure the speed of light as it passes next to Earth, and the observer is at the Sun, you would measure a speed more then c. How is it possible? It is because your clock at the Sun ticks slower, and you will divide the path of light with a smaller amount of time (relatively), so you get a speed more then c. The path is a little bit longer because of gravity, but let's disregard that (actually the time component is more dominant so the path length won't matter in this case).

So basically yes, the answer to your question is if you look from the spaceship as it accelerates, you could see things move faster then c on Earth. This is because you are measuring speed non-locally, from far away. SR only states that if you measure speeds locally, then you cannot measure speeds faster then c (in vacuum). So SR is not violated.

Please see here why you would prefer GR to explain this:

How is the classical twin paradox resolved?

$\endgroup$
  • $\begingroup$ Despite the title, the question is strictly about special relativity. Gravitational time dilation has nothing to do with it. $\endgroup$ – WillO Apr 14 at 4:30
  • $\begingroup$ @WillO sorry but that is not true. SR cannot explain why (OK it can, but that is not the real thing) one twin gets older, because constant speed is symetrically relative, Both twins could say the other got older. But, you are right in that he is asking why SR is violated. I will edit to explain. He is specifically asking about "during the acceleration", so he understands that the change in the twins' speed in the time dimension gets altered during the acceleration, which is equivalent to a gravitational zone, that is the equivalence principle. $\endgroup$ – Árpád Szendrei Apr 14 at 15:54
  • $\begingroup$ After your edit, this is still very misleading. Of course you don't need GR to calculate the integral of $dt$ along a non-straight path. SR is perfectly adequate to deal with this question. $\endgroup$ – WillO Apr 14 at 16:51
  • $\begingroup$ "If you try to explain this with SR, you will realize that both twins could symmetrically say that the other's clock moves faster." This is incorrect. The stay-at-home twin is in a single inertial reference frame the whole time, but the traveling twin switches frames during his turn-around. Only the stay-at-home's reference frame can be used in SR. $\endgroup$ – D. Halsey Apr 14 at 18:35
  • $\begingroup$ Thank you for your answer Árpád Szendrei. So your saying the problem with my measurement is that it is globally rather than locally which misleadingly makes me assume that v = Δx / Δt when I should be doing v = dx / dt for events near me. And it is this speed v that can never exceed the speed of light. Am I correct? $\endgroup$ – Salar Khan Apr 14 at 21:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.