69dodge,
The twins disagree about when the time difference accumulated. The stationary twin thinks it accumulated gradually over the entire trip: from his point of view, the travelling clock ran slowly because it was moving. But from the traveller's point of view, it is the Earth twin who was moving and whose clock therefore ran slowly. So how does he explain the fact that, when he returns, the Earth clock clearly shows more elapsed time than his, even though it was running slowly during the whole trip? He explains it by saying that it ran slowly during almost the whole trip; around the midpoint of his trip, however, while he accelerated towards Earth (parts 3 and 4), the Earth clock ran much faster than his, just as if it were at the top of a giant gravity well and his clock were at the bottom. Which, from his point of view, was precisely the case.
Sorry, but this is simply incorrect. The travelling twin will see the Earth's clock going very slow while he is moving away from the Earth, and very quickly while returning to the Earth. But the speed at which he sees the Earth's clock moving at any given time is strictly a function of their relative velocity at that time. It does not depend on acceleration at all.
Of course, most of the difference in the speed the Earth's clock seems to be moving will be due to the doppler affect. After correcting for that, the travelling twin will always calculate that the Earth's clock is running slow by a factor of sqrt(1-(v/c)^2), where v is whatever their relative velocity is
at that time.
Either way, he will not see the Earth clock suddenly speed up during the acceleration period. That just isn't how it works.
I don't see how it could accumulate only during the acceleration periods. If that were the case, the time difference would not even depend on the total trip time. It would depend only on the time spent accelerating. But of course, that isn't the case.
It depends also on the distance between the two twins during the acceleration, just as gravitational time dilation depends on the (vertical) distance between two clocks being compared.
This is incorrect. Nothing about either special relativity, or general relativity, suggests that the time dilation during periods of acceleration depends on the relative distance of the two objects. And in GR, the gravitational time dilation depends on the difference in gravitational potential, not on distance. In the case of planets, it just so happens that gravitational potential varies with distance.
And what about my example of the alien flying past the Earth? In that case, there is no acceleration. And yet the alien experiences less time travelling from the marker to the Earth than the Earth observer does.
The alien experiences less time starting from when it thinks it passed the marker, than the Earthling experiences starting from when he thinks the alien passed the marker. (They both agree about when the alien passed the Earth.) The alien still thinks the Earthling's clock is running slower than its own, however. But it also thinks that the Earthling started his stopwatch too early, before the alien actually passed the marker; that's why, in its opinion, the Earthling measured a longer period of time.
I know. That's the whole point. The twin flying back and forth between the Earth and his destination will, while he is moving, say that the twin back on Earth stopped his stopwatch at the wrong time (when the twin arrived at the destination), and likewise started it at the wrong time (when the twin heads back for Earth). Again, you have to recognize that we've got essentially two different experiments here.
Imagine there is a clock at the destination, which is (in the frame of the Earth and destination) synchronized with the clock on Earth. When the twin arrives at the destination, its clock will read (using the 4 light minutes and 0.8c example from before) 5 minutes later then the Earth clock read when he left. Let's assume that the acceleration period was very small (say 1 second).
When the twin leaves the Earth, he will see that the clock at the destination says -4 minutes. When he arrives, it will say +5 minutes. During the trip (which he says takes 3 minutes), he will see the clock advance 9 minutes. He will see the clock going 3 times its normal speed. He will not see it running slow all the way until the end, and then suddenly go through the rest of the 9 minutes just during deceleration. This is clear from the fact that even if he does not decelerate at all, and flies right past the destination clock, he will still see that it says +5 when he passes it! So the extra time
can't be elapsing during the deceleration time. My calculations for the flyby alien prove this, because if it were happening during the deceleration, the results for the flyby alien would have to be different.
For the first half of the experiment, we have 2 events.
Event one occurs at Earth, at the time when the twin leaves. [ ... ] Event two occurs at the destination, at the time when the twin arrives. [ ... ] the frame of the spaceship is exactly the frame in which the two events occur at the same location. That means that the time which passes in this frame must be smaller than the time between those events as measured from any other frame.
One could just as easily define an "event two" that occurs on Earth. Suppose the round trip takes two hours, as measured by an Earth clock. Then, let "event two" occur at the location of the Earth clock when it shows that one hour has elapsed. Now, the travelling twin ought to think that more than one hour passes for each one-way trip. Oops. (In other words, relatively moving inertial observers each think the other's clock runs slowly.)
You are forgetting about the simultineaty problem. If the Earth observer says that the instant just before the ship stops and turns around is simultaneous with event 2, then it necessarily follows that the observer on the ship will not, and vice-versa. Your experiment is thus rendered impossible, because the two observers will not be able to agree on what events mark the beginning and ending of the experiment.
In your experiment, we actually have 4 events.
Event 1) The Earth's clock says t=0.
Event 2) The spaceship leaves.
Event 3) The Earth's clock says t=2 hours.
Event 4) The spaceship stops.
Events 1 and 2 happen at the same location, so they are simultaneous in all frames. Thus they are effectively the same event.
Events 3 and 4 happen at different locations, so they are not simultaneous in all frames. If they are simultaneous in the Earth frame, then the twin on the spaceship will not agree that he turned around when the Earth clock said 2 hours. If the twin on the spaceship thinks they are simultaneous, then the Earth twin will say that the spaceship twin did not turn around at the right time.
The problem is the assumption that we can ignore the acceleration. We can't ignore it, because that's when all the interesting stuff happens; it's the only thing that breaks the symmetry between the two twins.
It's not. What breaks the symmetry is the fact that the events defining the beginning and ending of each part of the experiment happen at the same place for the spaceship, and at different places for the Earth. If you did design the experiment in such a way as to reverse this, you would get the opposite effect. What you have to consider is how such an experiment would be defined.
To see this better, imagine two huge rulers in space. One right next to the other like this ||. Now we could move ruler A up, until its middle is flush with the top end of ruler B, or we could move ruler B down, until its middle is flush with the bottom of ruler A. It
seems like these experiments should be equivalent, since when the middle of A is flush with the tip of B, the middle of B will be flush with the bottom of A. So there is no broken symmetry. The experiment looks the same to an observer on the middle of ruler A and to an observer on the middle of ruler B. By symmetry, they must experience the same duration of time, or there is a logical contradiction, a true paradox.
And yet we can think of the middle of B as being the Earth, and its top end being the destination, and the middle of A as being the spaceship. Then we've got the twin paradox, right? Likewise if we let the middle of A be the Earth, and the bottom of A be the destination, and the middle of B be the spaceship.
The problem is, you can't do it. If, from the POV of an observer in the middle of B (the Earth), the middle of A (the spaceship) arrives at the top of B (the destination) at the same time that the bottom of A arrives at the middle of B, then it will absolutely
not be the case that, from the POV of an observer at the middle of A, the middle of B arrives at the bottom of A at the same time that the top of B arrives at the middle of A.
I realize that is a bit confusing. Please try drawing it out on paper, and you will see what I mean. The point is that there is
no way to make the experiment symmetrical. If we define the beginning and end points of the experiment (in this case, each half of the experiment) in such a way that they happen at the same location for one observer, then the beginning and end points of the experiment cannot
possibly happen at the same location for the other observer. One of the two people is going to experience less time passing, and in any case, it is going to be the guy for whom both events happened at the same place. This would only create a paradox if it was possible to have the beginning and end of the experiment happen at the same place for both observers, but it isn't.
Each twin thinks the acceleration takes a small amount of his own time; the Earth twin thinks it also takes a small amount of spaceship time, but the spaceship twin thinks it takes a lot of Earth time.
The alien flyby example clearly shows that this is not the case. The alien, who has not decelerated yet at all, will see
exactly the same time on the clock at the destination that the twin who stops there will. The second before the spaceship stops (from the spaceship's POV), the clock at the destination will say 4 minutes 57 seconds (using the 4 light minute and 0.8c example I gave). After he stops, it will say 5 minutes. It is very easy to verify that this is the case.
Suppose the round trip takes two hours of Earth time and one hour of spaceship time, and suppose it starts at 10:00. Around the midpoint of the trip, as he is still moving away from Earth but right before he turns around, what does the spaceship twin think the time on Earth is? Well, his clock shows that 30 minutes have passed, but since---according to him---the Earth was moving and so its time slowed down, he says the Earth time is only 10:15.
Sure, but a clock which is synchronized with the Earth's (in the Earth's frame), which is at the turn around point, will say 11:00. It will not say 10:15 just before he stops, and then suddenly advance to 11:00 over the deceleration period.
Right after he turns around, as he is moving toward Earth, what does he think the time on Earth is? Well, he is going to reach Earth in 30 of his own minutes, which again correspond---according to him---to 15 Earth minutes. But when he does reach Earth, he will find that clocks there show 12:00. So, now they must be showing 11:45.
Yes, when he changes speed at a distance far from the Earth, his
calculation of what time it
currently is on the Earth changes. That is why it is useful to have a clock at the destination which is in the same frame as the clock on Earth. From the Earth frame, the two clocks are synchronized. From the frames moving relative to the Earth, they are not. While the ship is moving towards the destination, it sees the clock on at the destination as being 45 minutes ahead of the one on Earth. When the ship stops at the destination, it sees them as being synchronized again. Likewise going back it sees the clock on Earth as being 45 minutes ahead of the clock at the destination, but when it stops back on Earth they are again synchronized.
This may be where you are getting the idea that the time difference suddenly accumulates when the velocity changes. But it doesn't. All we have here are disagreements about what time it
currently is at some
distant location, due to relative velocity.
If the spaceship is looking back at a clock on Earth, he does not see the clock suddenly speed up and go through the missing 45 minutes as he decelerates. In fact, the time he sees pass on the Earth clock will be
less than what passes on his own during the deceleration period. All that happens is that his
calculation of
how long ago that light from Earth left, based on how far away the Earth is, has changed.
Just before he stops, he will look back at the clock on Earth. From my calculations, what he will see should be about 10:08. Note that from his POV, the earth is moving away from him at nearly light speed. So he is seeing where the Earth was some time ago. When he corrects for how far away the Earth was when this light was emitted, he will determine that
right now the Earth clock says 10:15.
When he stops, he will still see that the clock on Earth says 10:08. But now the Earth will appear to be much further away than before, so his calculations of how long ago the Earth clock said 10:08 will be different. Thus he now determines that the clock on Earth says 11:00.
And there you go. According to the twin on the spaceship, a full hour and a half of Earth time passed as he was turning around, even though the turnaround was practically instantaneous from his own point of view.
Wrong. During the turn around, his
calculation of what time it currently is on Earth will change by an hour and a half. But he will still only see a small amount of time pass on the Earth frame during this period.
Again, you have to remember that what he
sees happening on the Earth is largely due to the doppler effect. During his trip outward, he sees very little time pass on Earth (I figure about 8 minutes). During his trip back he sees the other hour and 52 minutes pass on Earth. He does not
see the actual rate at which time is passing on the Earth at all. He can only calculate it. And what he calculates depends on their relative speed. During his outward trip he calculates a passage of 15 minutes on Earth. Likewise during his return trip. He just can't add these two numbers together to get the total time, because his calculations of what time it was at the beginning of and ending of those periods don't match up.
Dr. Stupid
