Relativity - Synchronization and slow transport

[RJBeery]

Useful is not the same as interesting. I can adjust my watch every morning to be in synch with the television's clock but I'm not going to start a forum thread about it.

What is interesting is that, GIVEN the existence of relativistic effects, we can effectively side step them when it comes to keeping a soon-to-be-remote clock in synch with a local one by moving it away very slowly. That's the entire point (or should be the entire point) of this thread.

 

[Tach]

Useful is not the same as interesting. I can adjust my watch every morning to be in synch with the television's clock but I'm not going to start a forum thread about it.

No one forces you to participate in the thread.

What is interesting is that, GIVEN the existence of relativistic effects, we can effectively side step them when it comes to keeping a soon-to-be-remote clock in synch with a local one by moving it away very slowly.

Actually, the math shows that you can't. You either get errors that you can't wish away or you get a ridiculously long transit time that renders the method useless. On the other hand, if you do things right, you can synchronize the clocks with no errors in a very short transit time. If you read the thread, you'd have learned that.

 

[RJBeery]

Tach: To your first point, the thread was started by Pete to specifically discuss slow transport synchronization, not other methods. Secondly, I would be surprised if anyone here with a basic understanding of SR thinks that absolute synchronization is possible if only one clock remains inertial. This is why I said "effectively side step" relativistic effects rather than claiming we can avoid them altogether. This is also why limits were being discussed...the more you want to avoid relativistic effects, the slower your transport velocity must be.

At some point the desynchronization is small enough that it remains less than our ability to measure it, so it is effectively gone...and this occurs at v > 0. It is most certainly a useful and interesting effect to discuss.

 

[Tach]

Tach: To your first point, the thread was started by Pete to specifically discuss slow transport synchronization, not other methods.

No, it was opened by Pete to defend a specific statement he made relative to slow clock transport. A statement that I have challenged.

Secondly, I would be surprised if anyone here with a basic understanding of SR thinks that absolute synchronization is possible if only one clock remains inertial.

So would I, who is claiming that? Or are you saying this because you don't understand what the thread is about?

This is why I said "effectively side step" relativistic effects rather than claiming we can avoid them altogether. This is also why limits were being discussed...the more you want to avoid relativistic effects, the slower your transport velocity must be.

...which has been proven to be a bad approach. Have you read the thread? Not yet?

At some point the desynchronization is small enough that it remains less than our ability to measure it,

So, what point would that be? How does it relate to the ridiculous increase in transport time? How do you strike the balance between minimizing the errors and reducing the transport time to non-infinite ? How do mainstream physicists deal with this problem?

so it is effectively gone...and this occurs at v > 0. It is most certainly a useful and interesting effect to discuss.

Problem is, you can't wish it away. So, you will need to find a practical way of solving it. talking in generalities doesn't solve the problem.

 

[RJBeery]

Well, we could get in to details about accuracy requirements and associated allowable transport velocities but this conversation seems to lack intellectual sincerity. I'll let you get back to convincing others how stupid they all are, I'm out.

 

[OnlyMe]

No, it was opened by Pete to defend a specific statement he made relative to slow clock transport. A statement that I have challenged.

It was my impression that Pete opened this thread, to continue a technical mathematical "debate" that began in a thread, which at the time appeared to have originated with a genuine lay question, about "slow clock transport". This discussion, in that thread was essentially high jacking the thread from the perspective of the thread's OP, as understood at the time.

While, much of this discussion has been of some interest, it does not appear from where I sit that you have been able to address one issue at a time. This does result in some, for at least one mostly silent observer.

I think the whole discussion or debate would have had far more clarity if the issue of slow clock transport had been addressed before the introduction of other synchronization issues and/or methods. So far the discussion seems so jumbled that it is difficult to separate some, of the separate issues.

BTW Pete's original "slow clock transport" description was completely consistent with and appropriate, within the context of the OP in that thread.

 

[Tach]

It was my impression that Pete opened this thread, to continue a technical mathematical "debate" that began in a thread, which at the time appeared to have originated with a genuine lay question, about "slow clock transport".

He opened the thread in order to concentrate on the debate between the two of us.

This discussion, in that thread was essentially high jacking the thread from the perspective of the thread's OP, as understood at the time.

Your understanding is wrong.

While, much of this discussion has been of some interest, it does not appear from where I sit that you have been able to address one issue at a time.

What appears to you is not really relevant. What really is, that is what is relevant.

I think the whole discussion or debate would have had far more clarity if the issue of slow clock transport had been addressed before the introduction of other synchronization issues and/or methods.

Clock transport (doesn't have to be slow, as shown in the thread) is a form of clock synchronization method.

 

[Pete]

Hi Tach,
Let's see what we do agree on:

We agree on the SR calculation for the clock descynchronization, as given in the first post of this thread:
$$\Delta \tau = \frac{L}{\Delta u + \Delta v}\left(\sqrt{1 - ((v-\Delta v)/c)^2} - \sqrt{1 - ((v+\Delta u)/c)^2}\right)$$
Yes?

We agree that for any v, L, $$\Delta v$$ and $$\Delta u$$:
$$\Delta \tau = 0$$ if and only if $$L(2v - \Delta v + \Delta u) = 0$$
Yes?

We agree that:
$$\lim_{\Delta v \rightarrow 0 \\ \Delta u \rightarrow 0} \left (\Delta \tau\right ) = \frac{\gamma Lv}{c^2}$$
Yes?

We disagree about whether the last expression is useful in practice.
Right?

We disagree on whether "slow separation" implies using that limit as an approximation.
Right?

In fact, we seem to disagree on whether slow has any meaning at all.
What do you think that "slow" implies, if anything, in the context it was originally used?

 

[Tach]

Hi Tach,
Let's see what we do agree on:

We agree on the SR calculation for the clock descynchronization, as given in the first post of this thread:
$$\Delta \tau = \frac{L}{\Delta u + \Delta v}\left(\sqrt{1 - ((v-\Delta v)/c)^2} - \sqrt{1 - ((v+\Delta u)/c)^2}\right)$$
Yes?

Yes.

We agree that for any v, L, $$\Delta v$$ and $$\Delta u$$:
$$\Delta \tau = 0$$ if and only if $$L(2v - \Delta v + \Delta u) = 0$$
Yes?

Yes

We agree that:
$$\lim_{\Delta v \rightarrow 0 \\ \Delta u \rightarrow 0} \left (\Delta \tau\right ) = \frac{\gamma Lv}{c^2}$$
Yes?

Yes

We disagree about whether the last expression is useful in practice.
Right?

Correct, it has only theoretical value.

We disagree on whether "slow separation" implies using that limit as an approximation.
Right?

Not quite. I disagree that "slow separation" has any meaningful use in practice.

In fact, we seem to disagree on whether slow has any meaning at all.
What do you think that "slow" implies, if anything, in the context it was originally used?

I answered a lot of questions, before I answer the last one I would like you to answer two questions I have asked you several times: why are you so fixated on slow clock transport? What does "slow" provide that makes it valuable in experimental physics

 

[Pete]

Not quite. I disagree that "slow separation" has any meaningful use in practice.

So can we agree that in the abstract, "slow separation" could reasonably imply using that limit.

I answered a lot of questions, before I answer the last one I would like you to answer two questions I have asked you several times: why are you so fixated on slow clock transport?

I'm focusing on slow clock transport in this discussion, because the claim you challenged is specifically about slow clock transport.

What does "slow" provide that makes it valuable in experimental physics

Firstly, the question isn't all that relevant - the context of the claim in question wasn't about experimental physics, but about synchronization under slow transport in the abstract.

Secondly, I'm sure you'll agree that limiting approximations are valuable in experimental physics.
Newtonian mechanics is a low-velocity approximation of SR.
SR is a flat space-time approximation of GR.
Classical physical theories are a macroscopic approximation of quantum theories.

For a concrete extreme example, consider the clock transport used in the OPERA experiment to measure certain timing delays. This is a high precision experiment, which implies much lower tolerance for approximating theories.

But even in that extreme case, if the clock was transported slowly enough, then no correction would need to be applied to the reading because the desynchronization error would be insignificant.

And that's an extreme case. For most lab experiments, it would be ludicrous to correct for the relativistic desynchronization involved in moving a clock from one side of the lab to the other.

 

[Tach]

So can we agree that in the abstract, "slow separation" could reasonably imply using that limit as an approximation?


I'm focusing on slow clock transport in this discussion, because the claim you challenged is specifically about slow clock transport.

This does not answer the question , why are you fixated on "slow" clock transport, especially in the context of it being useless in practice?

Firstly, the question isn't all that relevant - the context of the claim in question wasn't about experimental physics, but about synchronization under slow transport in the abstract.

Secondly, I'm sure you'll agree that limiting approximations are valuable in experimental physics.

This is not answering the question. We have seen, through an example provided by you, that the errors induced by the method cannot be neglected.

Newtonian mechanics is a low-velocity approximation of SR.

True but it fails miserably in explaining the advancement of Mercury perihelion. The same way slow clock transport , as used by you, in abstract would fail miserable the measurements in the Opera experiment.

For a concrete extreme example, consider the clock transport used in the OPERA experiment to measure certain timing delays. This is a high precision experiment, which implies much lower tolerance for approximating theories.

But even in that extreme case, if the clock was transported slowly enough, then no correction would need to be applied to the reading because the desynchronization error would be insignificant.

But you have seen that, through the example you worked yourself, the errors are of the order of 5ns for a measurement of the order of 60ns, so you falsified the above claim yourself, earlier in the thread.

And that's an extreme case. For most lab experiments, it would be ludicrous to correct for the relativistic desynchronization involved in moving a clock from one side of the lab to the other.

True. Yet you are still stuck answering how to solve the cases where the distances are hundreds of kms.

What I don't understand (and I still need a valid answer from you) is why are you so fixated on the slow clock transport, a hopelessly ill-defined approach, when I have shown you an exact method of synchronization?

 

[Pete]

This does not answer the question , why are you fixated on "slow" clock transport, especially in the context of it being useless in practice?

In this discussion, I'm focusing on what we're discussing.

This is not answering the question. We have seen, through an example provided by you, that the errors induced by the method cannot be neglected.

Wrong.
We can, of course, find situations in the errors of limiting approximations can't be neglected. That's why the more precise models exist.
But, there are also many situations in which the errors of limiting approximations can be neglected.

True

Excellent!
We agree that limiting approximations are not "useless in practice".
You do agree that newtonian mechanics is not "useless in practice", right?

but it fails miserably in explaining the advancement of Mercury perihelion.

Correct.
That is a context in which the limiting approximation is not appropriate.

The same way slow clock transport , as used by you, in abstract would fail miserable the measurements in the Opera experiment.

Wrong again.
The Opera experiment involved transporting a clock less than 10 km from the surface lab to the underground lab. The speed was not specified, but I think it's safe to assume that it was less than 300m/s.

The worked example, on the other hand, was for a clock transported 3000km at 300m/s.

But you have seen that, through the example you worked yourself, the errors are of the order of 5ns for a measurement of the order of 60ns, so you falsified the above claim yourself, earlier in the thread.

I still don't know how you're deriving that 60ns figure.
A clock transported 3000 km at 300m/s elapses 9999.999999995s instead of the limiting approximation value of 10000s.

True. Yet you are still stuck answering how to solve the cases where the distances are hundreds of kms.

I'm glad you agree that the limiting approximation is not always "useless in practice".
We use the more precise method when the situation demands, of course.
Just like we wouldn't use the SR equations if the situation was such that the GR solution was significantly different.

What I don't understand (and I still need a valid answer from you) is why are you so fixated on the slow clock transport, a hopelessly ill-defined approach, when I have shown you an exact method of synchronization?

To summarize:

• You did not show me an exact method. The exact SR solution is right there in the OP.
• It's not an exact method anyway. It's a limiting approximation that assumes flat spacetime and ignores quantum effects.
• I'm not fixated on slow clock transport, except in this thread which is about slow clock transport.

 

[Tach]

Wrong again.
The Opera experiment involved transporting a clock less than 10 km from the surface lab to the underground lab. The speed was not specified, but I think it's safe to assume that it was less than 300m/s.

The worked example, on the other hand, was for a clock transported 3000km at 300m/s.

...resulting into a 5 ns error. Pretty big for a practical case. You are still dodging the question. Why are you fixated on slow clock transport?

I still don't know how you're deriving that 60ns figure.
A clock transported 3000 km at 300m/s elapses 9999.999999995s instead of the limiting approximation value of 10000s.

That was the order of the measurement for Opera. And you are wrong about the distance, the distance was the whole 720 km, see reference [23] of the paper.

I'm glad you agree that the limiting approximation is not always "useless in practice".
We use the more precise method when the situation demands, of course.
Just like we wouldn't use the SR equations if the situation was such that the GR solution was significantly different.

You are still dodging my question, I answered all of yours , why can't you bring yourself to answer two of mine?

To summarize:

• You did not show me an exact method. The exact SR solution is right there in the OP.

• 
  
  This is absolutely false. The exact solution has been given to you in the link. Please check again, your solution is just an approximation.
  
  

  [*]It's not an exact method anyway. It's a limiting approximation that assumes flat spacetime and ignores quantum effects.

  
  ....and fails miserably at larger distances and/or at higher precision requirements (see the 3000km example)
  
  

  [*]I'm not fixated on slow clock transport, except in this thread which is about slow clock transport.

Why can't you be more open minded? Because it serves your purpose? Do you want me to show you another refutation of your claim?

 

[Tach]

Let's try again:

The expression $$\frac{L}{c^2} \frac{2v-\Delta v +\Delta u}{\sqrt{1 - ((v-\Delta v)/c)^2} +\sqrt{1 - ((v+\Delta u)/c)^2}$$ gives the exact desynchronization between clocks for transport at any speed, as viewed from an arbitrary frame. Why do you insist on nulling $$\Delta u$$ and $$\Delta v$$ when:

-they cannot be null
-nulling them adds imprecision
-nulling them results into infinite transport time, an unphysical situation

So, because you cannot make $$\Delta u=0$$ and $$\Delta v=0$$, you have to stick with the exact formula $$\frac{L}{c^2} \frac{2v-\Delta v +\Delta u}{\sqrt{1 - ((v-\Delta v)/c)^2} +\sqrt{1 - ((v+\Delta u)/c)^2}$$ meaning that your assertion that the clocks can only be synchronized when $$v=0$$ ("if two synchronized clocks are slowly separated, they will not stay synchronized unless they were initially at rest.")is false.

There is yet another way to falsify your claim. I have shown that the error for the slow clock transport is a function of the transport speed $$V$$:

$$\epsilon=\frac{L}{V}(1-\sqrt{1-(V/c)^2})=\frac{L}{V} \frac{V^2}{2c^2}$$

You can see from the above that the error halves every time you halve the speed but the transport time doubles. Obviously, you can NEVER have $$V->0$$ , because the transport time goes to infinity. This is true EVEN if the limit for the error exists (and is zero).

 

[Pete]

...resulting into a 5 ns error. Pretty big for a practical case.

Really? You can't think of a practical case in which a 5ns error in a 10,000s time span is acceptable?

You are still dodging the question. Why are you fixated on slow clock transport?

Question answered yet again:
This thread is about slow clock transport, so that's what I'm discussing.

That was the order of the measurement for Opera. And you are wrong about the distance, the distance was the whole 720 km, see reference [23] of the paper.

Ah, I thought you were talking about the other clock transports described in the paper.

The error in the measurement you describe was 2.3ns.
Have you found out how fast that clock was transported?
Do you agree that they could transport it slowly enough that the relativistic desynchronization would be insignificant?
How slowly would they need to transport it to make the error less than 0.1ns?
I haven't bothered to work it out, so I don't know if this particular situation is one in which the limiting approximation is appropriate.

You are still dodging my question, I answered all of yours , why can't you bring yourself to answer two of mine?

Answered have been provided, quite clearly.
Once more, just for you:

I am focusing on slow clock transport in this thread, because that's what the thread is about.

"Slow" in experimental physics, means slow enough that the error inherent in a low velocity approximation can be neglected for a particular measurement.

This is absolutely false.

The opening post gives the exact SR solution:
$$\Delta \tau = \frac{L}{\Delta u + \Delta v}\left(\sqrt{1 - ((v-\Delta v)/c)^2} - \sqrt{1 - ((v+\Delta u)/c)^2}\right)$$

....and fails miserably at larger distances and/or at higher precision requirements (see the 3000km example)

Correct. Like I said, you don't use a limiting approximation in an inappropriate situation (although I don't think that a 5ns synchronization difference after 10,000s is "failing miserably"!)

You can use a limiting approximation in appropriate situations.

Why can't you be more open minded? Because it serves your purpose? Do you want me to show you another refutation of your claim?

Getting personal?
Tach, this is really getting ridiculous.

We agree on the exact SR solution.
We agree that limiting approximations are not always appropriate.
We agree that limiting approximations are often useful.
The disputed claim in the opening post is very explicitly a claim about slow transport.

So why are you arguing?

 

[Pete]

Let's try again:

The expression $$\frac{L}{c^2} \frac{2v-\Delta v +\Delta u}{\sqrt{1 - ((v-\Delta v)/c)^2} +\sqrt{1 - ((v+\Delta u)/c)^2}$$ gives the exact desynchronization between clocks for transport at any speed, as viewed from an arbitrary frame.

Yes, that's an algebraic rearrangement of the expression I provided in the OP.
But like I said before, it's not exact except in an abstract SR model.

Why do you insist on nulling spacetime curvature when:

- it cannot be null
- nulling it adds imprecision

???

It's because the difference can be neglected in appropriate situations, right?

 

[Tach]

Really? You can't think of a practical case in which a 5ns error in a 10,000s time span is acceptable?

Not if the signal you are measuring has a 60 ns window.

Question answered yet again:
This thread is about slow clock transport, so that's what I'm discussing.

You are dodging the question.

Ah, I thought you were talking about the other clock transports described in the paper.

The error in the measurement you describe was 2.3ns.

Yes, I told you that earlier and you claimed that I gave you the wrong answer. Can you be so kind and retract your claim now?

Have you found out how fast that clock was transported?
Do you agree that they could transport it slowly enough that the relativistic desynchronization would be insignificant?
How slowly would they need to transport it to make the error less than 0.1ns?
I haven't bothered to work it out, so I don't know if this particular situation is one in which the limiting approximation is appropriate.

No more answers until you stop dodging and you give me an answer to my two questions.

Answered have been provided, quite clearly.
Once more, just for you:

I am focusing on slow clock transport in this thread, because that's what the thread is about.



"Slow" in experimental physics, means slow enough that the error inherent in a low velocity approximation can be neglected for a particular measurement.

So, you will stonewall, I see.

Correct. Like I said, you don't use a limiting approximation in an inappropriate situation (although I don't think that a 5ns synchronization difference after 10,000s is "failing miserably"!)

Wrong answer, the signal window you are trying to measure is 60 ns, so your error is a whopping 1/12.

We agree on the exact SR solution.
We agree that limiting approximations are not always appropriate.
We agree that limiting approximations are often useful.
The disputed claim in the opening post is very explicitly a claim about slow transport.

So why are you arguing?

Because you wouldn't give the honest answer, that you a clinging on "slow" because you are hoping that it justifies your claim. No matter, I have showed you another way that your claim is invalid, in the following post. One nice feature in dealing with your stonewalling has been that it gave me the opportunity to develop several totally different refutations.

 

[Tach]

Yes, that's an algebraic rearrangement of the expression I provided in the OP.
But like I said before, it's not exact except in an abstract SR model.

Why do you insist on nulling spacetime curvature when:

- it cannot be null
- nulling it adds imprecision

We are not talking about nulling the curvature, we are talking about the falsification of your claim, try staying on topic.

 

[Pete]

I think we're done.

 

[Tach]

In [thread=110633]another thread[/thread], a poster mentioned using slow transport to maintain clock synchronization.

[post=2843216]Post 42[/post]
I pointed out that if two synchronized clocks are slowly separated, they will not stay synchronized unless they were initially at rest.

Here is another disproof:

$$ \tau=\frac{L}{c^2} \frac{2v-\Delta v +\Delta u}{\sqrt{1 - ((v-\Delta v)/c)^2} + \sqrt{1 - ((v+\Delta u)/c)^2}$$

For $$v=0$$ per your claim:

$$ \frac{L}{c^2} \frac{-\Delta v +\Delta u}{\sqrt{1 - ((-\Delta v)/c)^2} + \sqrt{1 - ((\Delta u)/c)^2}} \ne 0$$

So, contrary to your claim, $$v=0$$ does not imply $$ \tau=0$$. Now, we are done.