I have a question about Einstein’s general theory of relativity. Now, as I understand it, if you pick any point as a reference when measuring speed, every law of physics will work at that point. That means, if you are on earth looking at a space ship orbiting at near the speed of light, a meter stick will look as if it is 0.6 meters, time will slow down on the space ship and all that other stuff. If you were on the space ship a meter will appear to be a meter and time will seem the same. But when you look at earth, a meter will appear to be 0.6 meters and time will appear to slow down. I am also told, that if you synchronize watches, put one on earth and send the other on a ship in orbit at high speed, that when you land the ship and compare watches the watch on earth should be ahead of the one on the ship. So, you have two planets heading away from each other at near the speed of light. Now put a person on both planets. The person on planet X (isn’t that a cool letter) says his planet is the point of reference. He says the other planet is moving away from his at near the speed of light so it will have a time dilation where its clocks are moving slower. The person on planet Y says, no, no his planet is the point of reference and the other planet is moving away at near the speed of light. Now imagine that they synchronize a clock on each planet. They let so much time pass this way. Now, say a wormhole follows each planet. Planet X send planet Y the clock. Now both clocks are on the same planet, which clock is ahead in time? I asked my professor this question and can’t remember his answer. I asked my dad who is an engineer and as taken some physics and he refuses to answer my question adequately. Does anyone hear know enough physics to explain to me who’s clock is ahead (or neither)?
This is actually a question about special relativity, not general relativity. Typically, the rule is that whoever breaks the inertial frame is the one without the correct time (the one for which less time has passed) - however, in this situation depending on how your so called "wormhole" works, you could get some wacky behavior. See, the problem is, that you're jumping a huge spacial distance in, presumably zero time. This is basically like allowing your clock to move infinitely quickly. In the actual lorentz equation, there's an x term, too in time. In other words, if we have spacial distance separating two objects, we get a time difference due to that spacial gap, not just due to the velocity time dilation. Come to think of it, we haven't even defined how the clocks are synchronized. In whose frame are they synchronized, and at what time? edit: I think I can answer a better question than the one you asked that avoids the instantaneous jumping of a large distance. Say we have two planets moving toward eachother. From some frame we'll call "rest", the two planets are moving at equal speeds. If we have clocks on each of the planets, and we synchronize them from the rest frame at some time t_naut (with some distance x between them), when the planets pass eachother, the two clocks will read the same time. If we choose that exact moment to drop the clock off of one planet onto the other, the two clocks will continue to read the same time forever. If we synchronize our clocks from the frame of either one of the planets, when the two planets have that same distance x between them (measured from the "rest" frame so we don't change the location of our synchronization from last time), when the two planets pass, the clocks will read *different* times. Dropping one clock off onto the other planet will cause it to maintain that same time difference forever. Whose clock is ahead in this scenario depends only on whose frame we chose to do the synchronization in. The frame that we do the synchronization in will have the largest value for time when the two planets pass. This problem is still a bit tricky because it involves the synchronization of clocks that not only have different velocities but different locations as well, then measuring the time when they're brought to the same location and velocity.
Another way to describe it would be to have a third planet, slap bang in the middle, and use that as a reference. When the two clocks arrive there they will match each other perfectly, but they will be ahead of a third clock on that planet. I suspect that the clock will be ahead of the one on the planet it arrives on, but don't quote me on that.
Actually, it would work more like the second scenario I described. The only thing that really bugs me is whose frame the clock jumps instantaneously from one planet to the other in. If we have a "rest" frame, from which the two planets are moving the same, we synch the clocks in that frame, and instantaneously transport the clock from one planet to the other in that frame, the two clocks will read the same. If we synch the clocks from one of the planets frames, and transport the clocks instantaneously from the rest frame, the clock from the frame we used to do the synchronization will be behind (since the planets are moving away from eachother). If we synch the clocks from one of the planets frames and transport the clock instantaneously from that planet's frame, then the clock should be the same again, I think. The reason this gets tricky, though, is that if we try to do this problem from the "rest" frame, from a brief time, we have two clocks - one on each planet.
I think I follow most of what your saying, but I want to be sure. When you say frame, are you talking about a third point in space which we consider the speed to be zero?
Yes - the frame can be one planet, the other planet, or some other arbitrary point. We can consider any speed we want to be at rest, so I just picked three different ones that were useful for doing the problem. There is still something bothering me about this problem, though, so if I figure out that part of my answer was wrong, I'll post another message here.
Neither, they would both show the same time, assuming they synchronised at the same time initially, and were going at the same speed. Think about it, from each clock's point of view, the same time has passed.
It has to do with breaking of inertial frames, actually, and our definition of synchronization. Synchronization isn't the same in both frames. The frame in which their motion exhibits symmetry has a different definition of synchronization also.
