- #1
Thadriel
- 21
- 7
- TL;DR Summary
- (no "master clock" at rest with the center of the disk)
Sorry I know this is a common one. I'm sure you've all had to answer this a thousand times, so I apologize, but I'm hoping to find a comment that lights the spark to get my brain to grasp what seems to not make sense.
Obviously, a third observer who is at rest with respect to the disk will see that the clock on the outside has a much faster velocity than a clock on the interior of the disk, so clearly the outside clock will show that it has measured less time.
But that's one question. What about looking at it from the perspective of the two clocks on the disk as they look at each other, and not bringing the "at rest" third clock into the conversation at all?*
These two clocks do not move with respect to each other during the rotation, so presumably only light signal delay would cause differences in what they see of each other's clocks during the rotation. Which means neither should measure the other as having slow or fast clock.
For simplicity, you could just put clock A at the center and clock B at the outside. As they the disk rotates, the two clocks remain at rest with respect to each other. I believe that, as the disk is moving, they should see no difference unrelated to light signal delay of their clocks. (not discussion differential aging yet).
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If that is the case, how to we reconcile that the outer clock will have experience less time when the two clocks are brought together and compared locally? The only thing I can think of is that they will disagree on when the disk stops rotating. But how? They are both at rest with respect to each other according to each other's perspective. It's really easy to see the symmetry break in the standard twin paradox, because one must return. But with this example, we're free to move either clock and leave the other alone.
Somewhere in this example there has to be something that breaks the apparent symmetry between these two clocks according to both reference frames of the two clocks (as opposed to a third "master clock."). Because otherwise, if you remove all the matter in the universe except these two clocks, they have no way to distinguish each other from one another other than accelerometers. And acceleration is not the cause of time dilation (unless maybe it's a "gravitational field" caused by the acceleration, and then we're bringing GR into the problem).
*Or, is this what they mean when they say in special relativity you have to treat acceleration "very carefully?" As in, "don't you go examining acceleration without using an outside clock and measuring rod that are not accelerating." Which would mean, special relativity can deal with acceleration, but only if the reference frame used for examination is inertial.
But that's one question. What about looking at it from the perspective of the two clocks on the disk as they look at each other, and not bringing the "at rest" third clock into the conversation at all?*
These two clocks do not move with respect to each other during the rotation, so presumably only light signal delay would cause differences in what they see of each other's clocks during the rotation. Which means neither should measure the other as having slow or fast clock.
For simplicity, you could just put clock A at the center and clock B at the outside. As they the disk rotates, the two clocks remain at rest with respect to each other. I believe that, as the disk is moving, they should see no difference unrelated to light signal delay of their clocks. (not discussion differential aging yet).
.
If that is the case, how to we reconcile that the outer clock will have experience less time when the two clocks are brought together and compared locally? The only thing I can think of is that they will disagree on when the disk stops rotating. But how? They are both at rest with respect to each other according to each other's perspective. It's really easy to see the symmetry break in the standard twin paradox, because one must return. But with this example, we're free to move either clock and leave the other alone.
Somewhere in this example there has to be something that breaks the apparent symmetry between these two clocks according to both reference frames of the two clocks (as opposed to a third "master clock."). Because otherwise, if you remove all the matter in the universe except these two clocks, they have no way to distinguish each other from one another other than accelerometers. And acceleration is not the cause of time dilation (unless maybe it's a "gravitational field" caused by the acceleration, and then we're bringing GR into the problem).
*Or, is this what they mean when they say in special relativity you have to treat acceleration "very carefully?" As in, "don't you go examining acceleration without using an outside clock and measuring rod that are not accelerating." Which would mean, special relativity can deal with acceleration, but only if the reference frame used for examination is inertial.