Is relativity of simultaneity measurable?

The 1000km railcar between the two clocks is unnecessary. All we need is are two ends, moving at the same speed. But, a 1000km car would work too, if you can get the funding.

To be clear, in my setup both the \(S\) and the \(S'\) clocks are the same distance apart in \(S\).
This means that in \(S'\), the distance between the \(S'\) clocks is \(L\gamma\).

As Neddy said, that's not a flaw, it's the expected result.
In \(S'\), the clocks obviously don't line up at the same time - the events are not simultaneous. That's the whole point of the experiment.

The \(S'\) clocks measure the two events to be non-simultaneous.
The \(S\) clocks measure the events to be simultaneous.

The rail clock at B is measured by the \(S'\) clocks as being way past to the right of the corresponding ground clock at the time that the A clocks line up.
The rail clock at A is measured by the \(S'\) clocks as being still way off to the leftt of the corresponding ground clock at the time that the B clocks line up.

They are lined up at the same time in \(S\).
They are lined up at different times in \(S'\).

 

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Meaning that you can never line up both clocks in the pair. Meaning that you do not have a valid setup.

Meaning that you aren't measuring any RoS, you are measuring the effects of the clocks not lining up, the B clock in the railcar frame is way to the right of the B clock in the ground frame. What you are measuring is the effect of their spatial misalignment.
You still don't wonder why there is no such thing as a "RoS test" in this list?

 

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Of course we are.
Two events are measured as occurring at the same time in \(S\), and measured as occurring at different times in \(S'\).
That's relativity of simultaneity.

You mean not lining up at the same time, which is relativity of simultaneity.

 

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Maybe it's because it's impractical to set up synchronized clocks 1000km apart (and strictly speaking they should have vaccuum in between) moving at 100km/s.

 

The math shows that you are wrong, the clocks in S are synched up.
The clocks in S' show that they aren't, this is because the right end of the railcar went way past the corresponding marker on the ground.
So, you still have no thoughts about the total absence of any "RoS test" in the master list.

 

Let me give you a hint: you do not need 1000km train because the current measuring devices work at the femtosecond level, so , you could set it up with lengths that are in the 1m range. The real reason is that your setup is invalid.

 

Yes, this is exactly what relativity of simultaneity is!

 

You snipped the rest of the sentence: BECAUSE they are not lined up. What you are measuring (very badly) is the effect of spatial misalignment.
I note that you glossed over : "you do not need 1000km train because the current measuring devices work at the femtosecond level, so , you could set it up with lengths that are in the 1m range. The real reason is that your setup is invalid."

With the existent timing technology you could run an experiment as long as you satisfy the condition \(Lv=100 m^2/s\). You don't need large trains, you don't need outrageous speeds, so why isn't anybody doing a RoS test? So, once again, does the total absence of any "RoS test" from the list give you a clue? Still none?

 

At the femtosecond level, relativity of simultaneity would potentially be measurable with clocks moving at 10m/s, 10m apart.
But...

Can two clocks 10m apart be synchronized to the femtosecond?
Can the detectors reliably measure the time elapsed between synchronization and the event of interest to the femtosecond?
Can the clocks be accelerated to a precise constant speed (How precise? How constant? Feel free to do the error analysis), synchronized, then sent past the ground clocks so that they both pass within a femtosecond of each other?
Can the synchronization event be reliably detected by the two passing clocks? We'll need a detector that reliably activates in a window of 10 femtometers(!)

The engineering required is phenomenal.

 

And what is that effect? Relative simultaneity

 

Sure why not, the precision of cesium clocks definitely allows it.

Sure, no problem, and if you don't think so, increase the speed 10 fold, to 100 m/s. Maglevs go at 600km/h and have cars that are 30m long.

Well, when I objected to the fact that acceleration desynchronizes the clocks you replied that the synch is done after the acceleration phase. What's good for the goose, is good for the gander.

No, it cannot. This is painfully apparent in the gross limitations of your naive setup. So, again, what's good for the goose....

 

They couldn't because in \(S'\), distance between the \(S'\) clocks is \(L\gamma\)., remember?

 

Cite?

I don't see how that answers the question, which explicitly states that synchronization is done after acceleration.

Sorry, I meant the passing event.
I seriously doubt it can be reliably detected to the femtosecond, as you propose.

 

And the \(S'\) distance between the \(S\) clocks is \(L/\gamma\), which is why they are lined up at different times.
First the B clocks line up.
A short time later the A clocks line up.
(or the other way around, depending on which end is which)

 

...which means that the A clocks are never lined up at the same time as the B clocks in eiither frame. Thank you for making my point. Look, the setup is invalid, if it were, the technology to run such an experiment is available, experimentalists aren't stupid, they would have tried it. But the theoretical foundation is not there, you have to face it. This is what is precluding the experiment from happening, not the technology. The technology is available.

 

Google is your friend.

 

In \(S\), both the rail clocks and the ground clocks are \(L\) units apart, so the A clocks line up at the same time as the B clocks, as you showed in your equations.
In \(S'\), they line up at different times, as you showed in your equations, which demonstrates and measures relativity of simultaneity.
I don't see why this is so hard to acknowledge.
Two events. Same time measured in \(S\). Different times measured in \(S'\). QED

You're glossing over a lot of engineering. The practicality of this experiment is much more difficult than you seem to realise. Work through the details and see.
How are the clocks accelerated? How is their velocity maintained constant? How much error does a variation in clock velocity introduce?
How are the passing events detected?

You may also be interested in this:
Proposals for Two Satellite-Borne Experiments to Test Relativity of Simultaneity in Special Relativity

 

As you did. The way you describe it, it doesn't even begin to be realizable.

I am not the one who claims that the experiment is doable. The only difference is that you claim is not technically doable (this is false, did you see the part with Maglevs running at 167ms and having 30m cars? this is an \(Lv=5000\) that takes you out of the femtosecond range), I claim that it is technically doable (yet challenging) but that the theory of the experiment as stated by you is invalid.

Funny, I challenged you on this very issue when we started talking about the setup for detecting the "rod bending". You never answered.

Yes, I know this paper, it has lingered in arxiv since 2005, the author is a known fringer.

 

An example is clocks used for the GPS can make adjustments at least this small. 4.4453E-10s per s [tick]. .00000000044453 s/tick which equals the 38,407 nanosecond adjustment per day.

This link mentions the experimental uses associated with femtosecond time intervals.

http://whatis.techtarget.com/definition/femtosecond

 

Correct. I'm saying it's doable in principle; I'm skeptical of its practicality.

And yet own your mathematics supports my conclusion:

\(t_A - t_B = 0\)
\(t'_A - t'_B = -LV\gamma/c^2\)​

In \(S\), both the rail clocks and the ground clocks are \(L\) units apart, so the A clocks line up at the same time as the B clocks, as you showed in your equations.
In \(S'\), they line up at different times, as you showed in your equations, which demonstrates and measures relativity of simultaneity.
I don't see why this is so hard to acknowledge.
Two events. Same time measured in \(S\). Different times measured in \(S'\). QED