Nope, the clocks traveled different distances, that much is clear.
...because they traveled the same distance .
...because they traveled different distances.
Is relativity of simultaneity measurable?
- Thread starter Pete
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Nope, the clocks traveled different distances, that much is clear.
...because they traveled the same distance .
...because they traveled different distances.
Excellent.
We agree that two events (A and B) are measured to happen simultaneously in one frame, and measured to happen non-simultaneously in another. You want to argue about why the simultaneity is relative, but it's irrelevant.
By definition, we have measured relativity of simultaneity.
We agree that two events (A and B) are measured to happen simultaneously in one frame, and measured to happen non-simultaneously in another. You want to argue about why the simultaneity is relative, but it's irrelevant.
By definition, we have measured relativity of simultaneity.
No, you haven't, you just measured different path lengths.
Quite the opposite, it explains why the setup of your gedank is invalid.
You just agreed that the clocks in one frame record the same time for events A and B, while the clocks in the other frame record different times for events A and B.
Do you agree that clocks measure time?
I guess he has to think about that one!
The answer to your question is given in posts 894,896.
No, he's got an offline life like the rest of us, which means he doesn't respond to every post within minutes.
Tach, the question was whether you agree that clocks measure time.
But, it was rhetorical. I know that you know.
Right.
I just had a flash of insight. Tach, are you saying that we can't compare the clocks in the non-simultaneous frame because by the time the second set of clocks align, the first set of clocks will no longer be aligned? If this is your objection, please realize that we can have the clocks stop ticking as soon as they pass each other, so they won't record the extra time spent moving apart after they become aligned.
according to me,this is not the way to say that events are non-simultaneous because clocks aren't synchronised.
Neddy Bate
Valued Senior Member
The ground clocks are synchronized in the ground frame. The ground clocks record the same time for two events. Therefore the events are simultaneous in the ground frame.
The rail clocks are synchronized in the rail frame. The rail clocks record different times for those same two events. Therefore the events are not simultaneous in the rail frame.
Comparing the results from the two frames shows that simultaneity is relative.
Does that help at all?
This is the third thread you have spawned from the original thread and you are nowhere closer to finding the answer to the original puzzle. The irony is that the original puzzle has a very simple solution if you don't insist on dumbing it down by removing the gravitational field.
To your question as to whether RoS is measurable, this is a still an open question, very hotly debated by professional physicists.
Suffice to say, that , in the framework of "SR Test Theories", there is no provision for testing RoS. There are tests for parameters $$a,b,d$$ and for the light speed anisotropy $$\frac{c}{c'}$$ but there are absolutely no tests for the parameter $$e$$, the one that "encodes" RoS. As a matter of fact, in all existent experiments, $$e$$ is assumed to be given depending on the assumed synchronization (internal vs. external).
The above explains why measuring RoS has proven an elusive as measuring OWLS. In fact, the two are inextricably connected, you cannot measure OWLS because it requires two synched clocks that turn out to be synched based on measuring .... OWLS.
Additionally, the above explains the conspicuous absence of any "RoS test" from the list of experimental tests of relativity. Pete would dearly want to put this on the account of insufficient technology (an argument that I debunked) , when, in reality, the insufficiency lies with the theoretical underpinnings of the experiment (see above).
This is the current state of affairs in mainstream physics. It also explains why your quest for putting together a (naive) way of measuring RoS has led to invalid setups.
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you are saying things theoretically.
Time dilation and length contraction is not observed.RoS has the same issue since it depends on selected frames.
Tach is right saying that RoS is not measurable
pete,i don't know exactly what your experiment is. But i do know that you are trying to prove that RoS is measureable.
Real fact is that it is not possible.
I hope you know that time dilation and length contraction are not observed.
RoS is related to one-way speed of light and we know that one-way speed of light is impossible to measure and measuring one-way speed of light is just like measuring some absolute frame of reference which is not possible.
I just said why RoS is not measurable.
I am in no condition to explain your experiment because i cannot understand anything about it at all.
Real fact is that it is not possible.
I hope you know that time dilation and length contraction are not observed.
RoS is related to one-way speed of light and we know that one-way speed of light is impossible to measure and measuring one-way speed of light is just like measuring some absolute frame of reference which is not possible.
I just said why RoS is not measurable.
I am in no condition to explain your experiment because i cannot understand anything about it at all.
False. There are tests for both time dilation and length contraction.
RoS eludes testing for reasons explained in post 95.
So, is Pete saying that the order in which the two events 1 and 2 are found to occur in will depend on the sign of x1− x2 or vx? It is only when the two events occur at the same point (i.e., x1 = x2) that the events will occur simultaneously in all frames of reference.
And is Tach simply saying that neither S nor S' is more physically correct than the other?
And is Tach simply saying that neither S nor S' is more physically correct than the other?
Post 95 explains why Pete is never going to put together a valid RoS test.