I Proper (and coordinate) times re the Twin paradox

[PeroK]

And how does that make a difference to the results?
Both A and B are moving away from C, either can be a positive or negative displacement depending on how the observer in C observes them...

It makes a difference to A and B. You have done a simple thought experiment:

At a certain time $t$ on C's clock, A's clock reads $t'$ and B's clock also reads $t'$. In any case, in C's frame A and B's clocks are synchronised.

Then you have made the intuitive assumption that A and B's clocks must be synchronised in each others frame.

This is a good thought experiment because:

If you understand the relativity of simultaneity, you will see the problem with this assumption.

If you don't understand the relativity of simultaneity, then you won't see the problem.

 

[Dale]

And how does that make a difference to the results?
Both A and B are moving away from C, either can be a positive or negative displacement depending on how the observer in C observes them...

If A is positive then B is negative and vice versa.

 

[Dale]

Could somebody please repeat what the question is?

Grimble does not have a well formed question that I can see and refuses to answer clarifying questions.

 

[Vitro]

Could somebody please repeat what the question is?

I'm not sure either, but I think he's making an argument that all clocks accumulate proper time at the same rate regardless of how they move relative to each other, and they only show different readings while in relative motion (because of the gamma factor) but if brought at rest in the same FoR they should always show the same (proper) time.

 

[PeterDonis]

Both A and B are moving away from C, either can be a positive or negative displacement depending on how the observer in C observes them...

Lorentz transformations don't work on displacements, they work on coordinates.

Here's my suggestion: pick an inertial frame, such as C's rest frame. Write down the coordinates of all of the events of interest in this frame, explicitly. Then write down the Lorentz transformation that goes from C's rest frame to A's rest frame. Then write down the (different!) Lorentz transformation that goes from C's rest frame to B's rest frame. Then transform the coordinates of all of the events of interest using each of these transformations, and write down the results.

Doing this will, first, help you clarify for yourself what the implications of your scenario are (I don't think you fully understand them), and second, help the rest of the posters in this thread understand what you are describing and what question you are asking.

If you are unable to complete the above exercise, then I strongly suggest closing this thread until you have taught yourself how to do so. Being able to do an exercise like the above is a basic skill in relativity, and if you don't have it, you shouldn't be posting an "I" level thread.

 

[sdkfz]

I'm not sure either, but I think he's making an argument that all clocks accumulate proper time at the same rate regardless of how they move relative to each other, and they only show different readings while in relative motion (because of the gamma factor) but if brought at rest in the same FoR they should always show the same (proper) time.

I believe you're correct. I think he wants "proper time" to be synonymous with "absolute time". (If it helps, I've been part of discussions (over many years, elsewhere) with Grimble where he strongly stuck to the claim that simultaneity is absolute, and the differing views of observers in relative motion is effectively just illusion. Any discussion with Grimble needs to account for his wanting to reject the "relative" part of "relativity", and to find something absolute underneath it all.)

 

[Herman Trivilino]

Ahhh... Now I think I understand the issue. 10 seconds of proper time for Twin A is the same as 10 seconds of proper time for Twin B. But the twins experience different amounts of proper time between departure and reunion. Theory predicts that they are different, and experiments have confirmed it. But if you don't understand the theory and refuse to accept the experimental results then you are left in a state of denial that can't be resolved.

 

[Grimble]

OK. Yes I have a problem with relativity. There seems to be something fundamental that is constantly glossed over, that we are expected to accept and believe in...

Introducing "Space and Time" Minkowski wrote

According to Lorentz every body in motion, shall suffer a contraction in the direction of its motion, namely at velocity v in the ratio [gamma]
This hypothesis sounds rather fantastical. For the contraction is not to be thought of as a consequence of resistances in the ether, but purely as a gift from above, as a condition accompanying the state of motion.

Now all that is just in my mind - I accept that. Everything should just fit seamlessly into a working model, yet each time I try something just doesn't line up. Every part works with every other but never all at once however one looks at it some part is left adrift.

I will try once more to show this by drawing an example; but first I will address Peter's suggestion

Here's my suggestion: pick an inertial frame, such as C's rest frame. Write down the coordinates of all of the events of interest in this frame, explicitly. Then write down the Lorentz transformation that goes from C's rest frame to A's rest frame. Then write down the (different!) Lorentz transformation that goes from C's rest frame to B's rest frame. Then transform the coordinates of all of the events of interest using each of these transformations, and write down the results.

C's frame. A is traveling with velocity -v, B is traveling with velocity v.
(t,-vt,0,0) event 1, A has traveled a distance -vt ,
(t,vt,0,0) event 2. B has traveled vt,
(t,0,0,0) event 3. C has remained at rest.

So for A,
t' = γ(t - (-v)(-vt)/c²
t' = γ(t - tv²/c²
t' = γt(1 - v²/c²)
t' = t/γ

And for B,
t' = γ(t - (v)(vt)/c²
t' = γ(t - tv²/c²
t' = γt(1 - v²/c²)
t' = t/γ

So as I see it ( and please explain where I am going wrong!) the proper time in A has the same duration as the proper time in B - as measured within each frame.

Everything is relative: A measures propertime on her clock, B measures proper time on his clock and C measures proper time on their clock and they each measure coordinate times on the other's clocks.

At its simplest we can take C out of Peter's thought experiment above leaving us with two events A and B separated with coordinates (t, 0 , 0 , 0) and (t, vt, 0, 0) in A's frame and (t, 0, 0, 0) and (t, -vt, 0, 0) in B's frame and calculating as above we have time t' = t/γ, for each coordinate time and of course t = γt' for the proper times as measured from either frame.

 

[stevendaryl]

C's frame. A is traveling with velocity -v, B is traveling with velocity v.
(t,-vt,0,0) event 1, A has traveled a distance -vt ,
(t,vt,0,0) event 2. B has traveled vt,
(t,0,0,0) event 3. C has remained at rest.

So for A
...
t' = t/γ

And for B,
...
t' = t/γ

So as I see it ( and please explain where I am going wrong!) the proper time in A has the same duration as the proper time in B - as measured within each frame.

Yes. Letting event 0 be the event where x=y=z=t=0, then what we conclude is:

1. The proper time between event 0 and event 1 is t/\gamma
2. The difference in coordinate times between events 0 and 1, according to A's coordinate system, is t/\gamma
3. The difference in coordinate times between events 0 and 1, according to C's coordinate system, is t
4. The proper time between event 0 and event 2 is t/\gamma
5. The difference in coordinate times between events 0 and 2, according to B's coordinate system, is t/\gamma
6. The difference in coordinate times between events 0 and 2, according to C's coordinate system, is t

Those are the facts. What is the question? Why do you think there is anything paradoxical or unclear about all this?

What we don't calculate (although we could) is:

• What is the coordinate time between events 0 and 1, according to B's coordinate system
• What is the coordinate time between events 0 and 2, according to A's coordinate system

The answer to both those questions is: \frac{\gamma' t}{\gamma} where \gamma' is the gamma factor computing using the relative speed between A and B (which will not be v but will be, using the velocity addition formula, \frac{2v}{1-\frac{v^2}{c^2}}).

So

• In A's coordinate system, event 1 takes place before event 2
• In B's coordinate system, event 2 takes place before event 1
• In C's coordinate system, the two events are simultaneous.

Please, please. Ask a question. If you just state facts, that's not a question. People will spend dozens of posts trying to guess what your point is. Why not actually say what your point is?

 

[Battlemage!]

Lorentz transformations don't work on displacements, they work on coordinates.

Here's my suggestion: pick an inertial frame, such as C's rest frame. Write down the coordinates of all of the events of interest in this frame, explicitly. Then write down the Lorentz transformation that goes from C's rest frame to A's rest frame. Then write down the (different!) Lorentz transformation that goes from C's rest frame to B's rest frame. Then transform the coordinates of all of the events of interest using each of these transformations, and write down the results.

Doing this will, first, help you clarify for yourself what the implications of your scenario are (I don't think you fully understand them), and second, help the rest of the posters in this thread understand what you are describing and what question you are asking.

If you are unable to complete the above exercise, then I strongly suggest closing this thread until you have taught yourself how to do so. Being able to do an exercise like the above is a basic skill in relativity, and if you don't have it, you shouldn't be posting an "I" level thread.

I know you're getting a lot thrown at you, but if you have time what do you think of this attempt to find velocities as measured by the observers and the consequential time measurement differences?

The coordinates of clock A are (x, t). The coordinates of clock C are (x', t'). The coordinates of clock B are (x'', t'')

Let v_A' = -v_A be the velocity of clock A relative to C (moving to the left in the negative x direction) and let v_B' be the velocity of clock B relative to C (moving to the right, in the positive x direction), with v_A' = -v_B' according to C.
If A is considered at rest, then C is moving at v_A according to A (to the right, in the positive x direction). Clock A knows that according to C, B moves at v_B' = -v_A' = v_A. So, we should be able to do a velocity transformation.

$$ v_B = \frac {v_A + v_B'}{1 + \frac{v_Av_B'}{c^2}} = \frac {v_A + v_A}{1 - \frac{v_A(v_A)}{c^2}} = \frac {2v_A}{1 - \frac{v_A^2}{c^2}}$$

So now we can calculate the time coordinate transformation from A to B, correct?

Then, the Lorentz factor from A to B would be: $γ_{AB} = \frac{1}{\sqrt{1 -\left( \frac {2v_A}{1 - \frac{v_A^2}{c^2}}\right)^2}c^{-2}}$
So, $t'' = γ_{AB} \left(t - \frac{v_B'x}{c^2}\right) = \frac{1}{\sqrt{1 -\left( \frac {2v_A}{1 - \frac{v_A^2}{c^2}}\right)^2}c^{-2}} \left(t - \frac{\left(\frac {2v_A}{1 - \frac{v_A^2}{c^2}}\right)x}{c^2}\right)$

And going the other way, $t = γ_{AB} \left(t'' + \frac{v_B'x''}{c^2}\right) = \frac{1}{\sqrt{1 -\left( \frac {2v_A}{1 - \frac{v_A^2}{c^2}}\right)^2}c^{-2}} \left(t '' + \frac{\left(\frac {2v_A}{1 - \frac{v_A^2}{c^2}}\right)x''}{c^2}\right)$So it looks to me that the only possible way for t'' to equal t according to observer A or B is if $\frac{\left(\frac {2v_A}{1 - \frac{v_A^2}{c^2}}\right)}{c^2} = 0$
How far off is that?

 

[Herman Trivilino]

C's frame. A is traveling with velocity -v, B is traveling with velocity v.
(t,-vt,0,0) event 1, A has traveled a distance -vt ,
(t,vt,0,0) event 2. B has traveled vt,
(t,0,0,0) event 3. C has remained at rest.

Your descriptions don't make clear the fact that events have a duration of zero time.

(t,-vt,0,0) event 1, A is located at the position -vt ,
(t,vt,0,0) event 2. B is located at the position vt,
(t,0,0,0) event 3. C is located at the position zero.

Is it you're assertion that these three events are simultaneous in C's frame?

[Edit: Note they are simultaneous in C's rest frame given my descriptions of the events.]

At its simplest we can take C out of Peter's thought experiment above leaving us with two events A and B separated with coordinates (t, 0 , 0 , 0) and (t, vt, 0, 0) in A's frame [...]

Well, now you've lost me. I thought the events were labeled 1, 2, and 3. Now you seem to be labeling them A and B.

Anyway, Events 1 and 2 are not simultaneous in A's frame if they were simultaneous in C's frame.

 

[PeterDonis]

I will address Peter's suggestion

Yes, you did; and stevendaryl gave the correct response, that yes, you have described a perfectly consistent scenario in accordance with relativity, and there is no issue with it at all. So what's the problem?

If the problem is that you simply can't believe that the scenario as you've described it is consistent, I'm sorry, we can't help you with that. Relativity is an experimental fact.

If you have some other question, then, as stevendaryl asked you, what is it? What is the specific issue you are having with the scenario as you described it? So far nobody has been able to figure out what that is.

the proper time in A has the same duration as the proper time in B - as measured within each frame.

Yes, that's correct. Proper time along a specific path in spacetime, which is what you are referring to here (proper time along A's path, and proper time along B's path) is an invariant; it's the same for all observers and in all frames. It's just geometry; it's no different than saying that the distance from New York to Washington, DC is the same no matter what coordinate grid we put on the Earth's surface.

Everything is relative: A measures propertime on her clock, B measures proper time on his clock and C measures proper time on their clock

Yes, that's correct; each clock measures proper time along its own path in spacetime.

they each measure coordinate times on the other's clocks

This is not quite correct. Nobody can directly measure times of events that are not on their paths through spacetime. What they can do is calculate what those times are, based on measurements they can make directly. So A and B and C can each calculate what the times registered on the others' clocks will be, and can relate that to coordinate times in their own frames.

 

[Dale]

So as I see it ( and please explain where I am going wrong!) the proper time in A has the same duration as the proper time in B - as measured within each frame.

The proper time between the origin and events 1 and 2 are the same in every frame.

 

[Ebeb]

Yes, but any clock in its own reference frame is at rest and only measures time passing. And, in Spacial Relativity which as I understand it is where the twin paradox may be considered, identical clocks can be placed in any frame and will keep identical time - identical clocks; identical laws of science; identical conditions...
so how do they measure different times?

Because of relativity of simultaneity.
When both clocks move relative to each other, the 'identical time' you refer to, f.ex event 5 seconds proper time clock A and event 5 seconds proper time on clock B are not occurring simultaneously anymore. When they move relative to each other you are not allowed to say that "When I tick 5 seconds per my inertial ref frame, the other clock also ticks 5 times per his inertial ref frame." It's a statement that doesn't make sense anymore.
Per clock A ref frame clock B ticks slower. It means that f.ex. when clock A shows 5 seconds proper time, clock B shows only 4 seconds of its proper time (for gamma factor 1.25 for 0.6c). For clock A, clock B ticks slower, not because clock B ticks slower in its own inertial ref frame, but because of relativity of simultaneity clock A ref frame and clock B ref frame take different directions in 4D spacetime. This slows the popping up of clock B proper time events per clock A ref frame (and slows the popping up of clock A proper time events per clock B frame).

The 5 seconds clock A proper time is also a clock A coordinate time for 4 seconds clock B proper time.

You know how to read and draw spacetime diagrams?

 

[Grimble]

Those are the facts. What is the question? Why do you think there is anything paradoxical or unclear about all this?

That is exactly what I think is happening; I am not saying there is anything wrong, I am first checking that what I think is happening is correct.

Your descriptions don't make clear the fact that events have a duration of zero time.

I am sorry I did not think that was necessary, as it is a fundamental fact that events are instants in time...

 

[Grimble]

I agree, this thread is going nowhere - my fault as you say for not knowing just what I am asking.

I suggest it be closed.

I have now seen just where my confusion lies and I will open a new thread shortly which will be specific about the concerns I have.

Thank you all once again for your efforts!

 

[Herman Trivilino]

I am sorry I did not think that was necessary, as it is a fundamental fact that events are instants in time...

When you write "C has remained at rest" you describe something happening for a period of time, not something happening at a particular time. It is indeed fundamental, and therefore necessary for a full understanding. The only reason I mention it is because it's a documented source of learner confusion that poses a stumbling block to learning.

We write $t$ to refer to a particular time or clock-reading. We write $\Delta t$ to refer to a duration of time, or a difference between two clock-readings. This all seems pedantic until you realize its importance.

 

[Grimble]

When you write "C has remained at rest" you describe something happening for a period of time, not something happening at a particular time. It is indeed fundamental, and therefore necessary for a full understanding. The only reason I mention it is because it's a documented source of learner confusion that poses a stumbling block to learning.

We write $t$ to refer to a particular time or clock-reading. We write $\Delta t$ to refer to a duration of time, or a difference between two clock-readings. This all seems pedantic until you realize its importance.

Hmmm...

What I wrote was

(t,0,0,0) event 3. C is located at the position zero.

then, surely, my specifying the coordinates (t,0,0,0) where 't' is a "...particular time or clock reading..."
is specifying a particular time - 't' ?
The phrase ""C has remained at rest" is only describing how c arrived at that event, just as I specified

A has traveled a distance -vt [...]
B has traveled vt,

indicating how A and B arrived at their specified events (by traveling there)

I do appreciate being educated about points where my thinking or writing is a bit imprecise or wooly but it is disheartening to receive criticisms that appear to be unjustified...

 

[Herman Trivilino]

What I wrote was

(t,0,0,0) event 3. C is located at the position zero.

In Post #98 what you wrote was

(t,0,0,0) event 3. C has remained at rest.

And in Post #101 I replied

(t,0,0,0) event 3. C is located at the position zero.

You have somehow attributed to yourself what I wrote. But the point here is that events last for a duration of zero time and to a lot of people this is a stumbling block to learning.

I do appreciate being educated about points where my thinking or writing is a bit imprecise or wooly but it is disheartening to receive criticisms that appear to be unjustified...

Even if a correction is made in error, as you seem to think happened in this case, it's not a criticism. It's just a correction. A lot of people take corrections as criticisms. It's an obstacle to learning things like math and especially physics.

 

[Dale]

then, surely, my specifying the coordinates (t,0,0,0) where 't' is a "...particular time or clock reading..."
is specifying a particular time - 't' ?

I think this is just a bit of notational confusion. Usually the symbol $t$ is a variable referring to all times. It seems that you meant it as a constant, which often would be written something like $t_0$ to distinguish it. The former would typically be interpreted as a worldline while the latter would typically be interpreted as an event.

I don't think this is a mistake, just a miscommunication.

 

[sdkfz]

Notation issues aside, Grimble, you basically seem to be setting up a symmetrical scenario and expressing surprise at some symmetrical results.

Why is that useful? (Especially when the basic symmetry between two observers is well known).

I'm curious, do you still assert the idea you presented here? ... https://forum.cosmoquest.org/showthread.php?101237-Ontological-concepts(-)-Special-relativity

(i.e. is your post #98 here trying to set up a situation as in point 13 in the link I gave above?)

 

[Grimble]

Take a light clock A, with the mirror 1 light second from the lamp, now add any number of imaginary identical inertial light clocks moving at different relative velocities.
Every one of our imaginary clocks will tick at the same rate as clock A, measuring the local time within that clocks frame of reference.
Yet, even though they all measure time at the same rate, relativity of simultaneity means that they are not synchronised for there is no observer who can measure simultaneity for all those clocks.

After 1 second measured on clock A, the light in clock A will reach the mirror. The proper time measured for clock A will be 1 second; yet that will only be measured by observer A. Observers with any other clock, moving relative to clock A, have to calculate the proper time in clock A from the coordinate time they measure with their clock.
And the same is true of course for any of our imaginary clocks.

Relativity of Simultaneity means that an observer can only measure simultaneity for clocks moving with the equal relative velocity to that observer using the Lorentz equation.

I have merely been trying to make an observation to test my understanding. That an observer permanently at the mid point of two moving clocks, having equal relative velocity to each of those clocks, must calculate the same coordinate time for each of those moving clocks.

 

[Dale]

I have merely been trying to make an observation to test my understanding. That an observer permanently at the mid point of two moving clocks, having equal relative velocity to each of those clocks, must calculate the same coordinate time for each of those moving clocks.

Yes.

 

[Ibix]

Take a light clock A, with the mirror 1 light second from the lamp, now add any number of imaginary identical inertial light clocks moving at different relative velocities.
Every one of our imaginary clocks will tick at the same rate as clock A, measuring the local time within that clocks frame of reference.

Yes, but note that this is an assumption of special relativity. It's completely supported by experiment, but the idea that identical devices behave in identical ways when viewed under identical circumstances is a manifestation of the principle of relativity, aka Einstein's first postulate.

Yet, even though they all measure time at the same rate, relativity of simultaneity means that they are not synchronised for there is no observer who can measure simultaneity for all those clocks.

Well, they can be synchronised at one time (typically t=0), so we can agree that (for example) they all emitted their first pulse simultaneously. But time dilation means that they will drift out of synch. And the relativity of simultaneity means that if the clocks weren't co-located at synchronisation different frames won't, in general, agree that they were ever all synched.

After 1 second measured on clock A, the light in clock A will reach the mirror. The proper time measured for clock A will be 1 second; yet that will only be measured by observer A. Observers with any other clock, moving relative to clock A, have to calculate the proper time in clock A from the coordinate time they measure with their clock.

I disagree slightly with @Dale here. Your clock, as described, ticks once every two seconds, when the light returns to the starting point. You can't measure half a tick without some other finer-grained clock. This is important because the coordinate time for a whole tick of a light clock depends only on the time dilation, while the coordinate time for half a tick depends on time dilation, your synchronisation convention, and the orientation of the clock with regard to its velocity. And you can't really talk about the proper time for half a tick because you're relating null-separated events. You can talk about the event half way between the emission of the pulse and its return, but whether or not this is simultaneous with the reflection event depends on your synchronisation convention.

If the above seems pedantic, it is, but for good reason. You are struggling with simultaneity and the devil is in the details.

That an observer permanently at the mid point of two moving clocks, having equal relative velocity to each of those clocks, must calculate the same coordinate time for each of those moving clocks.

First, where the observer is located is irrelevant as long as both clocks have equal and opposite velocities.

As you described it, the result depends on details you didn't specify. If you had talked about a complete out-and-back tick of the clock then I would agree with Dale - the ticks of the clocks will be simultaneous always (in this frame and no other!) if they ever were. However you talked about a half tick. In that case, the time of the half-tick depends on the orientation of the light clocks. If both of them emit the flashes from their rear ends (or front ends) simultaneously then the half ticks will be simultaneous. If they emit the pulses from their -x ends simultaneously then the half ticks won't be simultaneous.

 

[Dale]

I disagree slightly with @Dale here. Your clock, as described, ticks once every two seconds, when the light returns to the starting point. You can't measure half a tick without some other finer-grained clock.

Yes, that is a good point. Hopefully @Grimble will understand

 

[Herman Trivilino]

After 1 second measured on clock A, the light in clock A will reach the mirror.

You'd need two clocks to reach that conclusion. One at the lamp and one at the mirror.

The proper time measured for clock A will be 1 second; yet that will only be measured by observer A.

That won't be a proper time. Proper time is the time that elapses between two events that occur at the same place. One event is the light leaving the lamp, and the other event is the light reaching the mirror. The lamp and the mirror are not in the same place, not for Observer A and not for any observer because it would require the observer to move at the speed of light. This is what @Ibix referred to as "null separated" events.

And I'm not sure what you mean by "only be measured by observer A". Do you mean it's only a measured time (as opposed to some other kind of time) or do you mean only Observer A will measure it?

Observers with any other clock, moving relative to clock A, have to calculate the proper time in clock A from the coordinate time they measure with their clock.

Why would they have to do it that way? If Clock A measures a certain amount of proper time (and any single clock always measures proper time) they could just read Clock A.

 

[Grimble]

Thank you Gentle folk for guiding me through the 'mine field' of expressing thoughts in scientific terms! (Note: My thanks are genuine, not sarcasm as they could be read, due to the 'mine field' of semantics - which is another whole ball game!)

I will rewrite my post, taking note of your comments, in order to avoid getting lost in analysis of all those excellent points and focus on the points I was trying to express.

So, to try again Take a light clock A, with the mirror 0.5 light seconds from the lamp, now add any number of imaginary identical inertial light clocks moving at different relative velocities with all clocks beginning their 'ticks' simultaneously, at the same Spacetime Event, event E₀ that is (0,0,0,0) in each clock's frame.
Every one of our imaginary clocks will tick at the same rate as clock A, measuring the local time measured within that clocks frame of reference.

Yes, but note that this is an assumption of special relativity. It's completely supported by experiment, but the idea that identical devices behave in identical ways when viewed under identical circumstances is a manifestation of the principle of relativity, aka Einstein's first postulate.

Yes, I understand, but are not Einstein's postulates fundamental to everything in Relativity? I just wonder why you feel that needs saying.

Yet, even though they all measure time at the same rate and were all synchronised at E₀, relativity of simultaneity means that they cannot remain synchronised for there is no observer who can measure simultaneity for all those clocks. Synchronisation is determined by each observer in relation to his view of Spacetime.

But time dilation means that they will drift out of synch.

I thought time dilation was the affect of relative motion on the remote measurements of a moving clock, not that it physically changed the clock as an inertial clock's motion is only "in the eye of the beholder" - one cannot determine if an inertial clock is moving by examining the clock's frame - it can only be determined to be moving when examined from another frame.
After 1 second measured on clock A, the light in clock A will have returned from the mirror. The proper time measured for clock A will be 1 second which will be measured and displayed only by clock A. Observers with any other clock, moving relative to clock A, have to calculate the proper time in clock A from their measurements.

And I'm not sure what you mean by "only be measured by observer A". Do you mean it's only a measured time (as opposed to some other kind of time) or do you mean only Observer A will measure it?

My apologies, poor semantics, what I meant was

be measured only by observer A

Why would they have to do it that way? If Clock A measures a certain amount of proper time (and any single clock always measures proper time) they could just read Clock A.

Yes that is what I understand too, yet so often I have read of different observers reading different times on the same clock, or of a clock displaying different times to different observers...

And the same is true of course for any of our imaginary clocks.

Relativity of Simultaneity means that an observer can only measure simultaneity for clocks moving with the equal relative velocity to that observer using the Lorentz equation.

I have merely been trying to make an observation to test my understanding. That an observer permanently at the mid point of two moving clocks, having equal relative velocity to each of those clocks, must calculate the same coordinate time for each of those moving clocks.

 

[Herman Trivilino]

Yes, I understand, but are not Einstein's postulates fundamental to everything in Relativity? I just wonder why you feel that needs saying.

It usually doesn't, but it was a response to your assertion that all of the clocks in your thought experiment keep time correctly, something that we usually assume without asserting.

I thought time dilation was the affect of relative motion on the remote measurements of a moving clock,

It's the effect of relative motion on the local measurements (emphasis on the plural) of a moving clock. You need two events to measure an elapsed time. If the two events occur in the same place (in some frame) then the time that elapses in that frame is a proper time. In the rest frames of observers who move relative to those events the time that elapses will not be a proper time, for them those two events occur in different places. They will therefore need two clocks, one located (local!) at each of the two events. And the time that elapses in those frames will always be larger than the proper time.

not that it physically changed the clock as an inertial clock's motion is only "in the eye of the beholder"

I would word this differently: "not that it changed the clock as inertial motion is relative". This takes us back to the assumption that @Ibix mentioned and noted is a consequence of the Principle of Relativity.

so often I have read of different observers reading different times on the same clock, or of a clock displaying different times to different observers...

I can see then why you'd be confused. Such a thing is simply not true. You don't by any chance recall where you read that? Or what you read that led you to believe that?

 

[Grimble]

It's the effect of relative motion on the local measurements (emphasis on the plural) of a moving clock. You need two events to measure an elapsed time. If the two events occur in the same place (in some frame) then the time that elapses in that frame is a proper time. In the rest frames of observers who move relative to those events the time that elapses will not be a proper time, for them those two events occur in different places. They will therefore need two clocks, one located (local!) at each of the two events. And the time that elapses in those frames will always be larger than the proper time.

I am sorry for the confusion but it is due to the way that I have been thinking about this. I have been mentally 'labelling' measurements in the clock's frame 'local' and those in other frames, in which that clock is moving, as 'remote' - i.e. made from outside the clock's frame...
So let me rephrase, I understood that time dilation was the affect of relative motion on the measurements from another frame... .

I would word this differently: "not that it changed the clock as inertial motion is relative".

Yes, that is the scientific way to say it...

This takes us back to the assumption that @Ibix mentioned and noted is a consequence of the Principle of Relativity.

(But to what extent is it an assumption? For isn't it the very basis of scientific experimentation that if one repeats an experiment, under exactly the same conditions, one would expect the same results... ? That Scientific Laws are immutable... ? Is it an assumption that 'c' is a constant... ?)

I can see then why you'd be confused. Such a thing is simply not true. You don't by any chance recall where you read that? Or what you read that led you to believe that?

Not that I believe that, but it is something often implied, at least, when people talk of 'moving clock's slowing'; as if the clock will display a different time to a moving observer when, in fact, what Einstein wrote was

As judged from K, the clock is moving with the velocity v; as judged from this reference-body, the time which elapses between two strokes of the clock is not one second, but [...] a somewhat larger time.

i.e. that, for the stationary observer, the clock ticked more slowly - each tick took longer - so the same time was measured but measured to take longer.
I had trouble with this at first because it seems like nonsense until one understands that the proper time, i.e the time displayed by the clock, is only part of the time measured by the non-moving observer; for they have to also factor in the time for the clock to travel between the two events the stationary observer measures the clock to travel between. Then the whole thing falls into place and we see why the proper time is the invariant Spacetime Interval,
S = √(ct)² - x²
or, as x = vt
S = √(ct)² - (vt)²
S = t√ (1 - v²/c²)
S = t/γ
where S is the Spacetime Interval = τ
and t is the coordinate time.

 

[Herman Trivilino]

I am sorry for the confusion but it is due to the way that I have been thinking about this. I have been mentally 'labelling' measurements in the clock's frame 'local' and those in other frames, in which that clock is moving, as 'remote' - i.e. made from outside the clock's frame...

But there is no such thing as outside of a frame of reference. A frame of reference assigns coordinates to every point so there are no points outside.

(But to what extent is it an assumption? For isn't it the very basis of scientific experimentation that if one repeats an experiment, under exactly the same conditions, one would expect the same results... ? That Scientific Laws are immutable... ? Is it an assumption that 'c' is a constant... ?)

To a full extent. Yes, it is also an assumption that the speed of light is the same in all inertial frames.

We assume they are valid unless we find evidence to the contrary. And even then we often retain the laws for use within the limits of their validity, if they are useful to us.

Not that I believe that, but it is something often implied, at least, when people talk of 'moving clock's slowing'; as if the clock will display a different time to a moving observer

It's not implied. It's not as if the clock displays different times to different observers. It displays the same time to all observers, it's just that it doesn't in general display the same time as their own clocks.