B Time dilation and Einstein's theorem

stevendaryl

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I'm sorry but we seem to be missing something here - Einstein's first postulate.

Let us look at this in its simplest view.
We have two frames, two inertial frames, in each of which there is a stationary clock. To an observer in frame A clock A is at rest and clock B, at rest in frame B , moving at v relative to A.
Two identical clocks in inertial frames.
Let us suppose they are light clocks.
When the light in clock A has travelled 1 light second to its mirror, it will measure 1 second to have passed and clock A will display (read) 1 second.
When the light in clock B has travelled 1 light second to its mirror, it will measure 1 second to have passed and will display (read) 1 second.
Frame A and Frame B are inertial frames and all we are measuring is how long it takes for light to travel 1 light second in each frame. the same time - 1 second measured from O when the clocks were co-located.
I am not saying that this is simultaneous because that judgement depends on the convention for simultaneity chosen; but I will say that they both happen at the end of equal intervals measured from event O.
Both clocks will read the same at turnround.
However, I am not saying that the travelling clock is not slow - it certainly is - but compared to what? Surely oit must be compared to A's measurement of the turnround time of T' that is the turnround time of B, in A's frame in which the light in clock B has travelled 1 light second to its mirror AND the distance that clock has travelled from clock A.
That is it will have travelledthe distance OT' or γT.
I would sort of like to help out, but I don't quite understand what is at issue. You have two rest frames A and B, and there are clocks at rest in each frame. So what exactly is the issue or question?
 

Ibix

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@Grimble - as @Dale notes, I explicitly use the Lorentz transforms which are derived from the two postulates. So no, I am not forgetting the first postulate.

Let's try this another way. The question is: what time is it on Earth when the rocket turns around?

In the Earth frame the answer is trivial - the rocket turns around at time ##T## because that's how we defined the experiment.

In the ship frame it's also trivial - it's ##T/\gamma##, which I called ##T'##, because that's what the ship's clock reads at turnaround.

But what are we to make of the fact that the frames disagree about the time of turnaround? What do Earth clocks show? Do they show ##T## or ##T'## or something else? The OP in this thread was trying to argue that this confusion proved that there was some kind of absolute time, I think. But he forgot about the relativity of simultaneity.

In the ship frame, the Earth is doing ##-v##. So at the same time as the ship turns round its coordinates are ##x'=-vT'##, ##t'=T'##. You can use the inverse Lorentz transforms to determine what an Earth frame clock reads at this event, which is ##T/\gamma^2##. And this is the point: "on Earth at the same time (as defined by the ship frame) as the turnaround" is an event a lot earlier on the Earth's worldline than "on Earth at the same time (as defined by the Earth frame) as the turnaround".

The slanted grey line in my Minkowski diagram connects all events that are "at the same time (as defined by the ship frame) as the turnaround". A horizontal line would connect all events that are "at the same time (as defined by the Earth frame) as the turnaround". They are different because simultaneity is relative. What "now" means depends on where you are and which frame you are using.
 
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All I am using is Newtonian mechanics PLUS Einstein's two postulates.
Those postulates tell us that the time for light to travel the distance OT in clock A in frame A, is equal to the time it takes light to travel the distance DT' in clock B in frame B.
Equal times for light to travel equal distances in inertial frames.
Simple mechanics...
 

jbriggs444

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All I am using is Newtonian mechanics PLUS Einstein's two postulates.
Those postulates tell us that the time for light to travel the distance OT in clock A in frame A, is equal to the time it takes light to travel the distance DT' in clock B in frame B.
Equal times for light to travel equal distances in inertial frames.
Simple mechanics...
Careful definition of terms would help. And a careful explanation of the point you are trying to reason toward.

Here, "O" denotes the event where the the near mirrors of the A and B light clocks are co-located and simultaneously started?

And, "T" denotes the event where light in A's light clock reaches its far mirror for the first half-tick?

And "T'" denotes the event where light in B's light clock reaches its far mirror for the first half-tick?

If so, what in the heck is D?
 

jbriggs444

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Newtonian mechanics is incompatible with Einstein’s second postulate.
I suspect (or hope) that @Grimble is only borrowing from Newtonian mechanics the notions of Euclidean 3-space and of inertial frames. In particular that, within a given inertial frame, an object subject to no external forces has constant velocity and a position function of the form ##\vec x=\vec{x_0} + \vec{v}t##.
 

Ibix

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If so, what in the heck is D?
I believe @Grimble is referencing my Minkowski diagrams from post #35. But I share your confusion over whether he is using the letters to reference events or places/times. Neither interpretation seems consistent with what he is writing at the moment.
 
Often it seems that beginners think they understand the relativity of simultaneity but actually do not. This is one reason I like David Morin's approach, which introduces the "rear clock ahead" principle early on. See also the "Andromeda Paradox," or this thought-provoking version of it (courtesy of Don Koks).
 

Mister T

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All I am using is Newtonian mechanics PLUS Einstein's two postulates.
But that's not the way the physics works. The consequences of Einstein's postulates don't agree with newtonian mechanics.
 

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