Wondering about clock transport

What SR?

 

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Emil, these have been long standing differences of opion.

From my perspective, special relativity has two major parts that apply.

One is as you see things, or how I think you do. Only clocks actually moving ever display a change in the way they record time.

The other is the one that results in the paradox. It has more to do with the fact that given just two clocks moving relative to one another, neither one can know whether they are moving or the other is moving. They both assume that they are at rest and the other clock must then be moving and running slow.

That is the basis of the time dilation paradox, which can only be reconciled when the two clocks are brought back together and compared side by side in the same frame of reference.

But there is another point that SR makes. Which is that there is no clock, meaning no frame of reference, that can with certainty say that in the universe they alone are not moving.

What SR says is that all clocks have to be considered as moving and only the difference in the motion between two clocks can be compared.

SR deals both with the actual affect that velocity has on the rate that a clock records time at and the perception that things look different from different frames of reference.

Time dilation paradoxes have been the subject of debate, well since shortly after Einstein published his 1905 paper. They will be debated long after we are gone.

 

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I hope not.

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You misunderstood my claim.
While I appreciate the effort you spent (is the integral really necessary for a constant velocity case?), it doesn't model what I said.

They're the same, Tach. I know - I wrote them both.
I'm sorry you misunderstood. It seems this has been a huge waste of time.

 

Tach,
I don't see how you can get this:

From this:

Or this:

Where do I state or imply that the clocks "travel a common distance (L) with the same speed (v)"?

 

"Now consider two synchronized clocks moving at velocity v together in frame S."

They are not the same. Maybe because 47 is very badly written, you did a much better job in writing post 66 since you took the trouble to cast it in a mathematical format. So, you can use the mathematical blueprint that I wrote up in conjunction with your problem statement from post 66 and find out the amount of clock desynchronization \(\Delta \tau=\tau_A-\tau_B\) in the original rest frame of the clocks.

 

That's the initial condition of Case 2.
So what did I say happens next?
"One clock remains moving at v, the other clock slowly separates at velocity v+dv until it is a particular distance L away."

Nowhere do I state or imply that "the clocks "travel a common distance (L) with the same speed (v)".

Post 66 is much more explicit, because I could see that there was a misunderstanding.
As I said, I'm sorry you misunderstood.

Starting with an integral seems like overkill.

Try this:

Let's see.

\(\begin{align} t &= \frac{L}{\Delta u + \Delta v} \\ \tau_A &= t\gamma(v + \Delta u) \\ \tau_B &= t\gamma(v - \Delta v) \\ \Delta \tau &= \tau_B - \tau_A \\ &= t(\gamma(v - \Delta v) + \gamma(v + \Delta u)) \\ &= \frac{L}{\Delta u + \Delta v}\left(\sqrt{1 - ((v-\Delta v)/c)^2} - \sqrt{1 - ((v+\Delta u)/c)^2}\right) \end{align}\)

I don't think there's anything stopping us from immediately taking one of the limits?
\(\begin{align} \lim_{\Delta v \rightarrow 0 \\ \Delta u \rightarrow 0} \left (\Delta \tau\right ) &= \lim_{\Delta v \rightarrow 0 \\ \Delta u \rightarrow 0} \ \frac{L}{\Delta u + \Delta v}\left(\sqrt{1 - ((v-\Delta v)/c)^2} - \sqrt{1 - ((v+\Delta u)/c)^2}\right) \\ &= \lim_{\Delta v \rightarrow 0} \ \frac{L}{\Delta v}\left(\sqrt{1 - ((v-\Delta v)/c)^2} - \sqrt{1 - (v/c)^2}\right) \\ &= -L \frac{d}{dv} \ \sqrt{1 - (v/c)^2} \mbox{(Definition of the derivative)}\\ &= \frac{Lv/c}{\sqrt{1 - (v/c)^2)} \end{align}\)

Now, let's apply that result to check the two cases described earlier:

In this case, v=0.
The synchronization difference is:
\(\Delta \tau = \frac{Lv/c}{\sqrt{1 - (v/c)^2)}\)
= zero, as expected.

The synchronization difference is:
\(\Delta \tau = \frac{Lv/c}{\sqrt{1 - (v/c)^2)}\)
which is non-zero, as expected.

 

Thanks for the pertinent reply, Pete. Actually, I do _not_ want to slowly separate the clocks; their job was over at the start; I was just using clock transport as an example; indeed, I tried to give a couple of other examples, but for some reason everyone more or less ignored them and focused intensely on clock transport.

Please allow me to slightly reword your above "agreement" conclusion:

What I really want everyone to agree on is this:
If two adjacent clocks read different times, then this conflicts with the reality that they can be absolutely synchronized (due to their proximity).

Whenever two frames are use to give some example [e.g., to invoke the relativistic transformation (often called the "Lorentz transformation")], then the two adjacent origin clocks at the start are always made to read the same time, usually zero.

pictorially:

two touching clocks
[0][0] OK

two touching clocks
[1][0] NOT OK

Here (at last) is my next step:

The following example is from a standard relativity text. It attempts to show Einstein's postulated (or assumed or stipulated) one-way light speed invariance:

Frame A moves to the right as Frame B moves leftward:
[clocks are in brackets] [S = light source]

Frame A
-------[0]--------------- -->
--------S~~>light
---<--[0]--------------------
Frame B

Frame A
------------[?]-----90m-------[300ns] -->
-------S-------------------------->light
[?]------------150m-----------[500ns]
Frame B

This is the only way to show one-way invariance on paper. However, it forces two adjacent clocks to read different times, and this, as we have seen, is a no-no.

We see that Einstein's proposed "one-way c invariance" conflicts with reality by improperly demanding that touching clocks read different times as they are "hit" by a light ray at absolutely the same time.

We know that at least one of the clocks must be wrong.

The problem is the lack of absolute clock synchronization in relativity theory.

In the above case, both of the touching clocks are wrong. Only truly synchronous clocks* can correctly record the light ray's travel time, which, in the given situation, happens to be 400ns. Thus, the distant clocks must read the _same_ time of 400ns when they are "hit" by the light ray. The A observer would measure light's one-way speed to be 90m/400ns = 0.225m/ns, a slower one-way light speed, whereas the B observer would find a one-way speed of 150m/400ns = 0.375m/ns, a somewhat faster one-way light speed. (*we are ignoring time dilation)

As we have seen, Einstein's assumed "c invariance" cannot happen even on paper because it causes clocks to conflict with reality. Upon being forced to "obtain" one-way c invariance, clocks will either improperly read the same start time when they are in fact started at absolutely different times (an example can be supplied), or they will improperly read different times when they should read the same time (as in the above simple textbook example).

Of course, all of this makes sense because everyone knows that asynchronous clocks are not properly related temporally, and special relativity's clocks are admittedly not absolutely synchronous.

What is the bottom line (or bottom lines)?

We all know that the one-way speed of light cannot be measured independently of clock synchronization, and we all know that the latter is conventional. But we have just discovered that the only proper convention is absolute or true synchronization - not Einstein's "synchronization" (which causes clocks to disagree with reality).

And even Einstein admitted that light's one-way speed will vary if the truly synchronous clocks of classical physics are used. //w w w . bartleby.com/173/7.h t m l

So what happens if we combine the extended Lorentzian theory (i.e., intrinsic clock slowing added to intrinsic length contraction) to the truly synchronous clocks of classical physics? Not much changes; instead of Einstein's very simple one-way speed equations of c - v and c + v, we have the only slightly less simple = c^2/(c+/-s), where s is the observer's absolute speed.

I highly recommend tossing aside the bad clocks (SR's asynchronous clocks), and quickly replacing them with the good clocks (truly synchronous clocks). But then, that's just my off-the-wall opinion.

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(If you wanna keep those asynchronous clocks, then who am I to force you to use good-uns!)

 

This is incorrect , already. The correct expression is:

\(\tau_A=\frac{t} {\gamma}\)

You might want to redo all the calculations.
The other objectionable part (not an error per se but objectionable nevertheless) is that you define the speeds and you calculate the gammas for \(v-\Delta u\) and \(v+\Delta v\) yet you define the separation time as \(t=\frac{L}{\Delta u +\Delta v}\) . Pick a frame and stick with it, don't jump in and out frames.

I got bored yesterday and I went ahead and I did all your cases as you described them, textually. If I misunderstood, it is because your description was inexact.

 

You set the bar for the discussion when you started the thread and titled it, "wondering about clock transport". And your first post...

And continued with...

Even if you need to set some background for what you are wanting to discuss, it is usually a good idea to state that up front. Threads like this tend to take on a life of their own fairly fast.

As you have noticed, your initial, questions (or statements at this point) were never fully addressed before the detail took over.


About this latest post...

The clocks in Frames A & B above are not in the same inertial frame of reference. They are not expected to read the same elapsed time except for that unique moment when they were set to zero.

So I don't understand what the rest of your argument is about. You could not put any properly functioning clock into that example and expect it to operate any differently.

 

Thanks, that was a typo in transcription.
I think it's correct from line 6 on.

I don't understand your objection, Tach. Where do you think I'm using a different reference frame?

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Because it of course impossible for it to be your fault.
I see you're still working on the misunderstanding that the clocks "travel a common distance (L) with the same speed (v)".
Why? Where do I state or imply that?

What do you not understand about this?

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.

Where does it say that the clocks "travel a common distance (L) with the same speed (v)"?

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.

Where does it say that the clocks "travel a common distance (L) with the same speed (v)"?

And you agree that the subsequent description was more explicit, so why are you sticking with the textual description which is apparently prone to being misunderstood?

 

I already explained it to you

\(t=\frac{L}{\Delta u + \Delta v}\) is calculated wrt the rest frame of the clocks, let's call it S.

\(\gamma(v+\Delta v)\) for example, uses the speed of one clock wrt a frame S' that moves at speed \(v\) wrt S.

Because you write badly.

We have been over this several times already.

We have been over this several times already as well.

Because it shows the importance of being accurate. It also shows that it is important to cast your statements in mathematical terms, natural language is not sufficient, especially when it is poorly written.

 

Hi timewarp,
I sincerely apologize for hijacking your thread on the sidetrack with Tach.

It conflicts with a statement that they are synchronized.
It doesn't say anything about whether they can be synchronized.

Yes

It's not a 'no-no', because they're not synchronized.

It causes them to conflict with the idea that they are absolutely synchronized.
But why do you imply that absolute synchronization is a required condition?

Correct.

You are assuming that absolute synchronization is part of reality.
Why?

The problem is that reality seems to conspire against us such that there is no way of telling whether two clocks separated in space are truly synchronous or not.

 

No, it's calculated wrt the frame in which the clocks are moving at \(v + \Delta u\) and \(v - \Delta v\) respectively.

Don't be an ass.
I've asked you several times to support your interpretation of what I said, but nowhere have you done so.
Now you're avoiding simple questions.
Why?
It seems painfully clear that you just misunderstood. Why not just say so, and move on?

So, your insistence on using a misinterpretation of a textual description instead of an explicit mathematical description is some sort of 'lesson'?

:bugeye:

 

How could it? \(\Delta u + \Delta v\) is the separation speed calculated in frame S.

Not really, it is you who has to live with the sloppy descriptions, not me.

 

Let me spell it out:

In a reference frame F(x,t):

• A is moving at dx/dt = v+du
• B is moving at dx/dt = v-dv
• At t = 0
  • A is at x=0
  • B is at x=0
• At t = L/(du+dv)
  • The distance between A and B is L

 

I don't know why you think that.

 

Have you read my explanation posted on the web? It is explained quite clearly.

 

Have you read the scenario?
It's quite clear that \(\Delta u\) and \(\Delta v\) are not defined in the clocks' rest frame:

You web 'explanation' is based on another botched reading of the posted scenario. Don't forget the importance of being accurate.