Wondering about clock transport

It isn't clear in what frame they are defined in your inaccurate description, it is quite clear that they are defined in S in my web solution. It says quite clearly in the text: "in the initial rest frame S, one clock moves at \(+\Delta u\), the other clock moves at \(- \Delta v\), therefore, the separation L is reached at \(t=\frac{L}{\Delta u +\Delta v}\)". The clocks move in opposite directions with infinitesimal speeds \(+\Delta u\) and \(- \Delta v\) , respectively, so there is a separation distance L at time \(t=\frac{L}{\Delta u +\Delta v}\). Quite clear.

 

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Tach's frame S is different. It's described in a document he linked a while ago:

 

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Ahem, no. Frame S is the initial rest frame of the clocks. Try reading the web file before you jump in. Ah, Pete already corrected you, beat me to it.

 

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....but JamesR doesn't read, he just trolls my posts so there is no way for him to know the facts. He just waits in the wings and pounces.

 

Feel free to ask for clarification on anything I say that isn't clear enough for you.
Given the statement "A moves at v + du", I don't see how you could possibly think that du is defined wrt A's rest frame, but we have it sorted now, so let's move on.

So what?
We are discussing your objection to my solution, remember?

Are you now clear that I'm working in a single reference frame?
Feel free to update your document to accurately reflect the scenarios it's supposed to be describing.

 

How about you tried to formulate your scenario in a precise manner? Define the frames, define the speeds. Start from the top.

 

Which bit is still not clear to you?

 

Since you never did a complete, clear , writeup , why don't you take the time to do it?

 

The writeup's there to read, Tach.
What do you find unclear?

 

You mean \(\frac{\gamma Lv}{c^2}\), right? What you wrote does not have dimensions of time.

Not necessarily. The clocks are synchronized from the perspective of a frame F that move with speed \(+v\) wrt your frame of reference.

No, it is \(\Delta \tau = \frac{Lv/c^2}{\sqrt{1 - (v/c)^2)}\). It should also have a sign that is dependent on the relative motion between the clocks and the frame of reference. The correct equation falls out straight from the Lorentz transforms. There is a lot of cleanup that you can do if you took the time to re-write your stuff from scratch. The way it is, it is a collection of posts with various mistakes.

 

That's right. Thanks.

What you say is correct, but I don't understand why you say "not necessarily".
If v=0, then \(\Delta \tau=0\), meaning the clocks are synchronized in the scenario reference frame, right?

That's right. Thanks.

That's right.
That can be built in by ensuring that v and L have a consistent sign convention.

I think the essential idea was clear enough way back at post 42:

This is an internet discussion forum.
If you expect publication quality rigor, you're in the wrong place.

 

Read the writeup I did, it is explained there.

No, it isn't. Think about the way you formulated the post, the clocks are "initially at rest" with respect to WHAT? It is this type of sloppy stuff that detracts from your posts.

 

That's cheap Tach. Pete's post back on page 3 of this thread was directed toward an explanation that did not require the kind of "rigor" you are demanding or would have been noticed by the original poster.

The questions and examples were clear within that context to those involved at the time.

And have you noticed that largely this obsession seems to have driven most others into the shadows if not completely away.

This thread began as a question about slow clock transport and a description of what that was and what effects it would have on the clocks.

Most of the original discussion was completely obscured by the next three pages, after you joined the discussion.

 

The statement works for any (x,t) inertial reference frame.
Think about it:

Choose any inertial (x,t) reference frame.
If the clocks were initially at rest in that frame, then they will stay synchronized in that frame on slow separation.
If the clocks were not at rest in that frame, then they will not stay synchronized in that frame on slow separation.

You can call it sloppy. I call it concise.
Remember, this is an internet discussion forum, not a journal.
Here, conciseness trumps rigor.

 

There is a fallacy in this type of reasoning. The desynchronization is predicated on \(L\) being non null. But \(L\) is a linear function of \(\Delta u\) and \(\Delta v\): \(L=(\Delta u +\Delta v )t\).

When \(\Delta u->0\) and \(\Delta v->0\) so does \(L->0\). So, the second sentence is not correct, you can't have it both ways.

 

What? The clocks are separated.

 

Not when \(\Delta u->0\) and \(\Delta v->0\). You can't have it both ways, your argument for the second sentence holds no water.

 

Lol and I was worried if I asked a question it could be dangerous.

The saga goes on.

 

To spell out the bleeding obvious for Tach (again):

The clocks are slowly separated by some given distance L.
The slower the separation, the longer it takes.
As the separation velocity approaches zero, the time taken approaches infinity.

This is ridiculous. 80-odd posts to explain a very simple sentence. Tach seems bent on deliberate misunderstanding. I don't know why, but I'm done explaining for him.

If anyone else doesn't get it and cares enough to ask about it, please do.

 

I probably should have been keeping an eye on this (unfortunately I've been spending my evenings playing Deus Ex - Human Revolution). Tach, could you please try to be a little less obtuse at times. You know full well this isn't the first time you've gotten responses like this from people, including myself.