Wondering about clock transport

I can see now what the problem here really is - your usage of the English language is rather awkward. Perhaps it's not your native tongue.

 

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Hi timewarp. I apologize for Emil and Farsight - they mean well, but they can't help themselves when it comes to hijacking threads.

Correct. But GPS does have sychronization in a well defined reference frame.

Your approach is muddier than it needs to be. I suspect that you've had tangled discussions with nested misunderstandings on other forums, so now you're trying extra hard to get agreement on fundamentals first.


But anyway, it seems that the foundation that you want us to agree on is this:
If two clocks are touching and synchronized, then everyone agrees that they are synchronized.

Right?
So, what's your next step?


You want to slowly separate the clocks, right?
Note that if you do, they will not stay synchronized unless they were initially at rest.

 

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I deleted this post because Pete has a better approach to resolution...

Thanks Pete!

 

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This is false (or, at least, incorrectly phrased), they will stay synchronized if their speeds wrt the frame where they were initially at rest are the same.
You can say that they get de-synchronized by the same exact amount under the above conditions (but they remain in synch).

 

Tach, doesn't your example apply only if both clocks are moved?

I think Pete was referring to one stationary and one being moved slowly...?

At least that was what I understood the original question to involve.

But I have been having some bad luck in how I interpret things lately!

 

This is what separate means.

 

Hi Tach,
Yes, the clocks will remain synchronized with each other in a frame in which the magnitude of their velocities is equal.
But, my point is that in a different frame, they will not remain synchronized.


Also note that this thread is about arbitrarily slow separation. To spell it out:

Consider two synchronized clocks at rest together in frame S.
One clock remains at rest while the other is slowly moved at velocity dv to particular distance L away.
In frame S, the resulting difference in synchronization can be arbitrarily small with an arbitrarily small separation velocity.
If you take the limit as dv approaches zero, the resulting difference in synchronization is zero.


Now consider two synchronized clocks moving at velocity v together in frame S.
One clock remains moving at v, the other clock slowly separates at velocity v+dv until it is a particular distance L away. dv is parallel to v.
Now, the resulting difference in synchronization can not be made arbitrarily small.
If you take the limit as dv approaches zero, the resulting difference in synchronization is nonzero.

 

Sure, but this detail was not mentioned in your post, you just added it now.

Careful here, motion is relative, so how do you decide which clock "remains" at rest and which one is "slowly moved"?

Yes, this is the principle behind one of the many methods of clock synchronization via "slow" transport.

This can't be since the case you describe above is indistinguishable from the previous case: the relative speed between the two clocks is the same in both cases , dv.

 

What detail?
I said "Note that if you do, they will not stay synchronized unless they were initially at rest."
Ie in a frame other than that in which the clocks are initially at rest, they will not remain synchronized.

The one that has velocity zero in frame S remains at rest.
The one that has velocity dv in frame S is slowly moved.

Yes, it is essentially the same situation in a different reference frame.
And as we know, synchronization is frame dependent, so the result shouldn't be a surprise.

 

And I pointed out that your claim is false. The clocks stay synchronized wrt the frame they are initially at rest IF their speeds wrt this particular frame are the same.

You can look at clock B being "moved" at +dv wrt A or at clock A being moved at -dv wrt B. This is getting ridiculous.

 

Or if the separation speed is arbitrarily slow.
And my substantial claim, that they don't stay synchronized wrt any other frame, is correct.

Yes, of course you can.
But you asked how you can tell which clock "remains" at rest, a question I answered.
The clocks were initially at rest in frame S. Only one of them remains at rest in frame S.
Do you have a point, or is this an irrelevant sidetrack?

 

But you did not make this additional claim in the post I corrected you. This is precisely why I corrected you.

Yes, it is very relevant. I am going to upload a file that proves false your claim that there is a difference between the two cases. Stay tuned.

 

:bugeye:
I'm not sure where the misunderstanding is, but it's clear to me that:
"...they will not stay synchronized unless they were initially at rest."
is exactly equivalent to:
"they don't stay synchronized wrt any frame other than one in which they were initially at rest."

 

This is false. See here.

 

That is also false, I can impart the clocks such speeds wrt to the frame in which they were originally at rest, so they stay in synch wrt that frame or with respect to the frame "in which they were originally moving with speed v". It is just a matter of the speeds of the clocks wrt the frames of interest. While your statements may not necessarily be false, they are sloppy in the least.

 

You missed the part where we slowly separate the clocks.

Well, I'm not submitting for publication.

 

Doesn't matter, see here. You need to get your math straight.

 

I don't understand what I'm looking at. Can you walk me through the formalism?

 

It shows the calculation of the elapsed proper time on the two clocks considering the two cases the you outlined in post 47, it proves your assertion false. See also post 54, earlier.