- #126

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"Simultaneously" in whose frame? Of course you could design things so that B will reset itself to read some time T "simultaneously" with A reading Tas simultaneity is defined in B's rest frame, but in that case A and B willnotread T simultaneously in A's rest frame. Likewise you could design things so they both read T simultaneously in A's rest frame, but then they wouldn't read T simultaneously in B's rest frame. So simultaneity hasn't been "ensured" in any absolute sense, only relative to some particular choice of reference frame. And since A and B are moving relative to one another, even if they both read T simultaneously in one of these frames, after that they will start to get progressively more out-of-sync because one is ticking slower than the other.

This is confusing, does "in its own reference frame" refer to A's frame or B's? Certainly when B receives the signal, it could figure out what count A has reached simultaneously with B receiving the signalusing the definition of simultaneity in B's frame, or it could figure out what count A has reached simultaneously with B receiving the signalusing the definition of simultaneity in A's frame.Which one are you talking about here, if either?

I don't get what you mean here. If B set sets itself to time T such that the event of B reading T is simultaneous with the event of A reading Tin A's frame, then they do not read T simultaneously in B's frame. Likewise if B sets itself so that they both read T simultaneouslyin B's frame, they do not read T simultaneously in A's frame. You seem to be suggesting it could be a "fact" inbothframes that they are both at the same count T the moment after B receives the first signal and resets itself, but that's impossible.

There is a problem with what you have said here.

Firstly, the answer to your question about what 'count' B is setting itself to was already explained here:

When counter B receives its signal, it will be able to do a relativistic calculation to find out the precise count that counter A has reached at the instant in its own reference frame that is simultaneous to the instant in B's reference frame for which it received the signal.

In other words, B is using the definition of simultaneity in A's reference frame to set itself.

Now, here is the problem with what you have said:

And since A and B are moving relative to one another, even if they both read T simultaneously in one of these frames, after that they will start to get progressively more out-of-sync because one is ticking slower than the other.

By making this statement, you are only repeating what I have already said. The fact that they get "progressively more out-of-sync" is the paradox!

Are you following me?

1. B calculates what the count is for A in A's reference frame.

2. B perceives A as going slower than itself.

3. When A reaches B, B sees that A is the same as itself.

Wait! no. What if A is not the same as B when they meet?

If A is not the same as B when they meet, then how could they be symmetrically relative?

That is the paradox.

I don't get what you mean here. If B set sets itself to time T such that the event of B reading T is simultaneous with the event of A reading Tin A's frame, then they do not read T simultaneously in B's frame.

Nothing is being "read". B is doing a calculation, which is not the same as "reading". "Reading" implies relativistic transformations brought about by spacial differences. Calculations are not dependent on location. Imagine that A and B are touching in space and that B is reading from A instantly. This would be the equivalent. Remember what you told me?

Yes, that's exactly right. Disagreements between frames about simultaneity only occur for events that happen at different spatial locations, if two events happen at the same timeandthe same position in one frame, then all other frames agree those events happened at the same time and position.

To understand why doing a relativistic calculation from B's frame is the same as if they were touching and exchanging the same information derived from the calculation, all you have to do is realize that the calculation is being done in B's frame, so the information that B derives about A's frame is not dependent on space or time differences (assuming the calculation is done instantly for B in B's frame. If the calculation is not done instantly, the same argument applies, but just with a slight deviation).

You seem to be suggesting it could be a "fact" inbothframes that they are both at the same count T the moment after B receives the first signal and resets itself

You are making an absolute statement. What "moment" are you referring to? The moment in A's frame or the moment in B's frame? I thought you didn't believe in absolutes? ;-) Joking.