# Is relativity of simultaneity measurable?

Discussion in 'Physics & Math' started by Pete, May 8, 2013.

1. ### TachBannedBanned

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$S'$ IS the lab frame, the clocks don't line up in $S'$, you don't know your own setup?

3. ### OnlyMeValued Senior Member

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Tach, in my earlier post I was referring to a single event.., a single lightning strike and that two observers separated by either distance or some relative velocity would measure that single event differently. That is do to RoS, or at least the result of the same underlying assumtions, that leads to RoS in the classical hypothetical.., but it was not the clasic example imvolbing "two" separated events.

This reconstruction clearly began with two events simutaneous in the emnpbankment frame, and a comparrison of how the two events were recorded to have occurred by two observers, separated by some relative velocity (sufficient to meet the demands of time device accuracy).

The fact is even if the two evens are not simutaneous in their common frame, that is they have no common frame, how the timing of those events is recorded by observers separated by ether distance or relative velocity, is an example of RoS... Though it may not always serve as experimental proof. Ideally events which are established as simutaneous in a common frame, best serve the purpose experimentally. That does not mean that all other cases where time of light delays in measurement are present, are not examples of RoS.., or the same underlying mechanisms.

5. ### PeteIt's not rocket surgeryModerator

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My mistake, I am sorry. I've been using the natural convention of $S$ for the ground frame and $S'$ for the car frame. I note that you introduced the opposite convention (it's not part of my setup).

So let's go back to where this misunderstanding originated and clarify.

There are four clocks. Let us label the ground clocks $Ag$ and $Bg$, and the rail clocks $Ar$ and $Br$.
In the below calculations, you seem to use $A$ to refer to the event of $Ar$ and $Ag$ aligning, and $B$ to refer to the event of $Br$ and $Bg$ aligning.

I notice you've used the convention that $v$ is the velocity of $S$ relative to $S'$, rather than the other way round. Any particular reason?
I'm guessing that:
$t_A =$ the time in $S$ at which $Ag$ and $Ar$ align.
$t_B =$ the time in $S$ at which $Bg$ and $Br$ align.
So, you've set the clocks to align simultaneously in the car frame.
This is the opposite to what was specified in the setup, but let's run with it anyway.

And these appears to be is the positions at which the clocks align:
$x_A =$ the position in $S$ at which $Ag$ and $Ar$ align.
$x_B =$ the position in $S$ at which $Bg$ and $Br$ align.

So the clocks do not line up simultaneously in the $S'$, as shown by the times $t'_A$ and $t'_B$ record on the $S'$ clocks $Ag$ and $Bg$. This is enough to measure relativity of simultaneity - the events A and B are simultaneous in $S$, and not simultaneous in $S'$.

In $S$, the clocks align simultaneously.
In $S'$, the clocks do not align simultaneously, with a measured time difference.

Obviously, this means that in $S'$ the distance between $Ar$ and $Br$ is not the same as the distance between $Ag$ and $Bg$. If it was the same, then the would align simultaneously, which would mean that the events A and B were simultaneous in both $S$ and $S'$.

The whole point of the experiment is that the clocks are aligned in $S$ at a particular instant, and always misaligned in $S'$

7. ### TachBannedBanned

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Well, you should know your own description.

Evidently.

You can use either convention as long as you are consistent, which is what I have been.

Evidently. This means that the car frame measures a Ros of....zero.

Yes.

The only part that is true is that you are measuring zero Ros in frame $S$ (by the way you constructed the scenario) but you are measuring the difference in traveled distance in frame $S'$ since the clocks do not align at the $B$ endpoint.

Correct, you are measuring the fact that the clocks have traveled different distances in the two frames.

...meaning that you measuring the results of traveling different distances, thank you for agreeing with my calculations and with my original point. The setup you created is not a valid setup for measuring RoS.
At this point, you have to ask yourself again, why there is no "RoS test" in the list of tests for SR? All the other SR effects have their own tests, even length contraction is testable, why not RoS? Not because it isn't a valid prediction of SR, not because the technology does not exist (I have already shown you that it exists).

8. ### PeteIt's not rocket surgeryModerator

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I do. You introduced an odd convention, and you also changed the setup.

You defined two events to be simultaneous in the car frame.
So we measure relativity of simultaneity in the lab frame.
This is the opposite to my original setup, but it works just as well.

You set it up so that A and B are simultaneous in $S$.
We measure relativity of simultaneity in $S'$ to be:
$t'_B-t'_A=\gamma vL/c^2$
, because, as you say, in $S'$ the B clocks do not align at the same time as the A clocks align.

We have described any travel distances, only positions and times of alignment.

Wrong. Your own equations show that:
A and B are set up to be simultaneous in S. Do you disagree?
A and B are measured as nonsimultaneous in S'. Do you disagree?

9. ### TachBannedBanned

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Ok, since you want it this way, I'll "check in $S$":
You arranged for the railcar to have a different distance between the clocks than the platform
$x'_B-x'_A=\gamma L$

$x_A=\gamma(x'_A-vt'_A)=-\gamma vt'_A$
$x_B=\gamma(x'_B-vt'_B)=\gamma^2 L-\gamma vt'_B$

You made $x_B-x_A=L$ (in the lab frame, the platform), so:

$L=\gamma^2 L-\gamma v(t'_B-t'_A)$

The above is true if and only if:

$t'_B-t'_A=\frac{\gamma Lv}{c^2}$

Let's see what happens in the platform frame, the frame where you are doing the measurements:

$t_A=\gamma(t'_A-\frac{vx'_A}{c^2})$
$t_B=\gamma(t'_B-\frac{vx'_B}{c^2})$

so:
$t_B-t_A=\gamma(t'_B-t'_A)-\frac{\gamma v}{c^2}(x'_B-x'_A)=\gamma \frac{\gamma Lv}{c^2}-\frac{\gamma^2 Lv}{c^2}=0$

So, your experiment designed to measure Ros in the lab frame , $S$, measures a RoS of ....zero.

At this point, you have to ask yourself again, why there is no "RoS test" in the list of tests for SR? All the other SR effects have their own tests, even length contraction is testable, why not RoS? Not because it isn't a valid prediction of SR, not because the technology does not exist (I have already shown you that it exists). Can you stop pretending that this thorn doesn't exist and start reflecting in earnest on the question?

10. ### PeteIt's not rocket surgeryModerator

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I also arranged for the clocks to align simultaneously in the the lab frame, and measured RoS in the car frame. But you went and changed that.
Wrong. You seem to be confused about in which frame the events are set to be simultaneous, and in which frame the events are measured. It could be done either way:
If we set up the experiment so that A and B are simultaneous in the lab frame (my original setup, and the equations in your last post), we measure RoS in the railcar frame.
If we set up the experiment so that A and B are simultaneous in the railcar frame (your equations in [post=3068567]post 827[/post]), we measure RoS in the lab frame.

As I said, I am not convinced that it's feasible. If you want to prove me wrong, then feel free to work out all the engineering details.

11. ### TachBannedBanned

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No, I didn't , I pointed out your misconceptions in either frame.

False, you are measuring the spatial misalignment at the right hand endpoint.

False, as well, you seem to forget that you rigged the proper length of the railcar to be $x'_B=\gamma L$ in frame $S'$ resulting into measuring a null value for the Ros in frame $S$. I have just showed you that, you demanded the calculation and I showed you the exact math <shrug>.

This is a cop-out, I disproved each one of your contentions on technology.

I proved you wrong by proving that your gedank is invalid.

12. ### PeteIt's not rocket surgeryModerator

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To measure the spatial misalignment, we would use a ruler.
To measure the temporal misalignment, we use clocks.

No, you're still confused about in which frame RoS is measured.
In the original setup, the lab-frame distance between the lab-frame clocks is $L$, and we 'rigged' the clocks to align simultaneously in the lab-frame (just like Einstein 'rigged' the lightning strikes to occur simultaneously in the platform frame).
This means that the the lab-frame distance between the railcar clocks is L, and the proper distance between the railcar clocks is $\gamma L$.
In that situation (the original setup), we then measure RoS using the railcar clocks.

In the setup I described, we set the clocks to align simultaneously in the lab frame, and measure RoS in the railcar frame.

13. ### TachBannedBanned

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Correction: in order to measure temporal misalignment , the clocks need to be co-located, otherwise, you can't compare them. The clocks are not co-located at the right end.

14. ### PeteIt's not rocket surgeryModerator

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To measure temporal misalignment, the clocks need to be synchronized.

We're comparing the A railcar clock with the B railcar clock (which is why they need to be synchronized in the railcar frame), and comparing the A ground clock with the B ground clock (which need to be synchronized in the lab frame).

They are co-located for an instant as they pass each other. If you really wanted to compare the two B clocks, you could do it then.

15. ### Fednis48Registered Senior Member

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I'm a little confused. Is your position that spatially separated clocks in the same frame cannot be compared? Or is it that there's no way to make the train-frame clocks align with their ground-frame counterparts?

16. ### TachBannedBanned

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"Then" the clocks at A are misaligned , so you cannot compare them. What you are measuring is spatial separation, no matter how you twist and turn .

17. ### PeteIt's not rocket surgeryModerator

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Of course you can, you can compare the A clocks when they are aligned. Although I don't know why you want to, since it's not part of the experiment.
We could measure the spatial separation. But for that, we would use a ruler.

Tach, this is simple, and I don't know why you're avoiding it. Look at [post=3068567]your maths in post 827[/post].
We set two events to occur simultaneously in one frame (call it $S$, it doesn't matter whether it's the railcar frame or the ground frame):
Do you agree that this sets the events to be simultaneous in $S$?
Now, measure the time of those events in $S'$:
Do you agree that the measured time between the events in $S'$ is $\gamma vL/c^2$?

Enough evasion, Tach. Answer those two questions and be done.

18. ### TachBannedBanned

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Then the ends at B are misaligned. Look , Pete, I did the math for your scenarios as viewed from either $S$ or $S'$ and I proved you wrong in either case, at this point, why don't you do the math, who knows, you may realize your error?

19. ### PeteIt's not rocket surgeryModerator

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What, you mean that the A's don't align simultaneously with the B's?

You seem to be avoiding the simple questions from the previous post:

Do you agree that this sets the events A and B to be simultaneous in $S$?

Do you agree that the measured time between the events in $S'$ is $\gamma vL/c^2$?

20. ### TachBannedBanned

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But you claim to be measuring time (RoS). In reality you are measuring the fact that the clocks traveled different distances. I pointed out this error in your thinking several times.

So, the conclusion of my post 827 points out your error:

"So, your scenario has a flaw, the clocks on the rail car do not line up with the clocks on the ground, the clock at B in the railcar is measured by the platform as being way past to the right of the corresponding ground clock. No wonder the ground clock has accumulated the extra time, the railcar clock has passed it by quite a bit. The larger the relative speed, the larger the missalignment, both spatial and temporal "

21. ### PeteIt's not rocket surgeryModerator

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Yes, time is what clocks measure. We are recording measurements made by clocks.

Yes, the distance between the clocks is different in the two frames. We could measure that fact, using rulers.
The time between when the clocks align is also different, and we can measure this using clocks.

No wonder the ground clock has accumulated the extra time.
That extra time is the measurement of the relativity of simultaneity.
Yes, the clocks are misaligned spatially. That's length contraction, which we can measure using rulers (and synchronized clocks).
Yes, the clocks are misaligned temporally. That's relativity of simultaneity, which we can measure using synchronized clocks.

Why are you ignoring the simple questions in the previous post?

22. ### TachBannedBanned

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...and this is why the clocks show different times, what you are measuring is the difference in traveled time due to difference in traveled distance.

23. ### PeteIt's not rocket surgeryModerator

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Well, the causality isn't well defined. You could equally say that the temporal misalignment causes the spatial misalignment.
But it doesn't matter - we're interested in the temporal misalignment, whatever it's cause, so that's what we're measuring.
No, we're measuring the times on the clocks.

The clocks in one frame record the same time for events A and B.
The clocks in the other frame record different times for events A and B.

Agreed?