Understanding Time Dilation in Einstein's Special Theory of Relativity

[Dale]

You state that fictitious forces are not 'Einstein'. Do you mean that they are not special theory?

Correct, the concept of fictitious forces predates Einstein by more than 100 years.

There is NO observer C in Einstein's section 4 STR depiction and the introduction of same is nothing more than a deliberate obfuscation!

Determinations made by a hypothetical observer C can have no effect whatsoever on what is taking place as far as A or B are concerned nor on the rates of operation of their clocks!

That is a bit of an extreme reaction. The whole point of relativity is that you can use any inertial reference frame you choose. The results will always be the same, as I explained.

Anyway, I only introduced C because you already refused to use non-inertial reference frames like A's. If you only admit B's reference frame (because A's is non-inertial and because no other inertial frame is explicitly mentioned) then it is hard to see what you are objecting to.

 

[cos]

Not if "physical truth" is defined to mean something that is true regardless of your choice of conventions about coordinate systems, and as I said before, this is how most physicists nowadays use the word "physical". If you want to use a different definition that's fine, it's just semantics, but I would ask that you spell out what you mean by physical. That's what DaleSpam asked you in post #90 above, and I also asked you about this in post #85 which you were responding to here:

You have allowed this thread to deteriorate into a totally in appropriate philosophical discussion.

When Einstein wrote that the equatorial clock 'must go more slowly' than a clock at one of the poles did he mean that the equatorial clock goes more slowly than a polar clock?

 

[Dale]

I made no suggestion that relativity does not work.

Then what is the discussion about?

 

[JesseM]

You have allowed this thread to deteriorate into a totally in appropriate philosophical discussion.

When Einstein wrote that the equatorial clock 'must go more slowly' than a clock at one of the poles did he mean that the equatorial clock goes more slowly than a polar clock?

As I said, I think he meant the average over an entire orbit, and I believe it would be true in all inertial frames that over a complete orbit an equatorial clock would tick less than a clock at the pole. Can we focus on the other situation Einstein discusses in section 4 where clock A and clock B are initially some distance apart, then A is briefly accelerated and afterwards moves inertially towards B? Do you assert that in this example A is "physically", "really", or "actually" ticking slower than B between the time it's accelerated and the time it reaches B, in spite of the fact that there are perfectly valid inertial frames where it is B that's ticking slower during this period of time? I just want to understand if you use words like "physically", "really" and "actually" to mean something that there is a single correct answer about, or if you just use these words to refer to the perspective of particular frames, so that you would be equally fine with saying that it is B that "physically", "really", and "actually" ticks slower than A in certain choices of frames.

 

[Dale]

It is not important what I think about the words 'physical' or 'real' but what Einstein meant by the words 'must go more slowly'!

I am of the opinion that he meant that clock A 'physically' or 'really' or 'actually' goes more slowly

The problem is that because you refuse to define "physical", "real", etc. I still don't know what you mean by that last. I cannot tell if we agree or disagree, and I don't know what words to use to clearly communicate my position back to you. It is, in fact, important what you think about those words because you are the one I am trying to communicate with. I even made it easy for you and suggested some definitions, all you have to do is say yes or no.

I think it is rather hypocritical that you accused me of "deliberate obfuscation" above.

 

[matheinste]

This doesn't answer the question but may be of some interest for those like me who have not seen it before.

http://www.sigmapisigma.org/radiations/2005/electrodynamics_fall05.pdf

"--------If there are two synchronously running clocks at A, and one of them is moved along a closed curve with constant velocity until it has returned to A, which takes, say, t sec, then,on its arrival at A, this clock will lag ½t(v R /c)2 sec [to lowest order in v R /c] behind the clock that has not been moved. From this we conclude that a balance-wheel clock located at the Earth’s equator must, under otherwise identical conditions, run more slowly by a very small amount than an absolutely identical clock located at one of the Earth’s poles.---------”

There are legion experimental demonstrations of time dilation, such as the ubiquitous muons-in-cosmic-rays example that appears in all the textbooks. When time could be measured to nanosecond precision fifty years after Einstein wrote these preceding lines, an experiment was done that recalled Einstein’s prediction explicitly:

Hay’s experiment as described J. Bronowski, The Ascent of Man, Little & Brown (1973), p. 255.

The experiment was done by a young man called H.J. Hay at Harwell. He imagined the Earth squashed flat into a plate, so that the North Pole is at the centre and the equator runs round the rim. He put a radio-active clock on the rim and another at the center of the plate and let it turn. The clocks measure time statistically by counting the number of radio-active atoms that decay. And sure enough, the clock at the rim of Hay’s plate keeps time more slowly than the clock at the centre. This goes on in every spinning plate, on every turntable. At this moment, in every revolving gramophone disc, the centre is ageing faster than the rim with every turn

Matheinste.

 

[DrGreg]

cos,

you have repeatedly been asked to explain what you mean by "physically", "really", or "actually". You might think these words are obvious and require no further explanation, but in relativity things that seemed obvious in Newtonian theory are no longer so. If you describe something as "physically slower" (for example), you need to explain what measurements or calculations you would perform to decide whether something is "physically slower" or not.

You often refer to a clock "ticking more slowly" but you fail to say slower than what and as measured by whom. In relativity these are not optional extras: different observers get different answers. It makes sense to assert "A ticks more slowly than B as measured by C". If we shorten this to "A ticks more slowly than B" this only makes sense (for instantaneous tick rates) in a context where the "as measured by C" is understood -- often the context is "as measured by B". To say "A ticks more slowly" makes no sense at all unless everyone implicitly understands what B and C are.

Note that in relativity it is possible for

"A ticks (instantaneously) more slowly than B as measured by B"

"B ticks (instantaneously) more slowly than A as measured by A"

to be simultaneously true. It's not a contradiction because A & B use different measurement procedures.

But "A actually ticks slower" (without mention of a B or C) means nothing. Can you give an unambiguous operational definition (what numbers you would measure or calculate) of what you think it means?

When Einstein wrote that the equatorial clock 'must go more slowly' than a clock at one of the poles did he mean that the equatorial clock goes more slowly than a polar clock?

He meant the equatorial clock ticks more slowly than a polar clock as measured by a polar clock.



============
Note: everything above applies to "instantaneous" clock rates. If you are talking about average clock rates where clocks A and B are initially together, separate and come back together again, everyone will agree which clock ticked fewer ticks than the other over the whole round-trip journey, but then at least one of the clocks must have accelerated (I'm assuming Special Relativity in the absence of gravity), so simple inertial frame analysis is not sufficient.

 

[phyti]

Jesse, post 81

Well, if you're just talking about average rate of ticking for a non-inertial clock between the times it departs from and returns to an inertial clock, then you aren't saying anything controversial if you say that the non-inertial clock has a slower average rate of ticking between these events, since this is true in all frames. I thought you were saying something more, Well, if you're just talking about average rate of ticking for a non-inertial clock between the times it departs from and returns to an inertial clock, then you aren't saying anything controversial if you say that the non-inertial clock has a slower average rate of ticking between these events, since this is true in all frames. I thought you were saying something more, that the clock A in his example in section 4 was objectively ticking slower than clock B during the after it was accelerated to the time it met clock B; that would be incorrect, but if you didn't mean to suggest this, please clarify. that would be incorrect, but if you didn't mean to suggest this, please clarify.

that the clock A in his example in section 4 was objectively ticking slower than clock B during the after it was accelerated to the time it met clock B

If it wasn't, when or where in the trip does the difference in time occur?

Please don't use the 1st postulate as a defense, it get's old fast!

 

[JesseM]

If it wasn't, when or where in the trip does the difference in time occur?

In a frame where the clock A that was accelerated was not ticking slower than B (because A and B initially had a nonzero speed and A's speed decreased after accelerating, from the perspective of this frame), the reason A was behind B when they met was because they were not synchronized in the first place (always remember the relativity of simultaneity!). In this frame B's time was always ahead of A's time by a constant amount prior to A accelerating, and after A accelerated its reading was "gaining on" B's reading, but not fast enough for it to surpass B's time by the time they met. If A was allowed to continue on at constant velocity past B after they crossed paths, then in this frame A's time would eventually surpass B's time.

Note that I gave a numerical example that worked like this back in post #64:

Suppose for example A and B are a distance of 60 light-seconds apart in the "stationary" frame K, and both are synchronized in this frame. Then if A is moved at 0.6c towards B at the moment when both clocks read a time of t=0, it will take 100 seconds in this frame for A to reach B, during which time A will only tick 80 seconds due to time dilation (the Lorentz factor being 1.25), so when A meets B, B will read t=100 seconds while A reads t=80 seconds.

Now consider things from the perspective of the inertial frame where A and B were initially moving at 0.6c and then A was accelerated to come to rest in this frame while B continued to move towards it at 0.6c. In this frame the clocks were not synchronized initially, so when A read t=0, B already read t=36 seconds according to this frame's definition of simultaneity. Then it takes 80 seconds in this frame for B to reach A (because the initial distance between them was 48 light-seconds in this frame due to length contraction, and 48 light-seconds/0.6c = 80 seconds), during which time B only ticks forward by 80/1.25 = 64 seconds due to time dilation, meaning B reads t=36 + 64 = 100 seconds when they meet, while A reads t=80 seconds when they meet. So you see that both frames make the same prediction about their respective times, even though in the first frame A was ticking slower while in the second frame B was ticking slower.

 

[cos]

When Einstein wrote that the equatorial clock 'must go more slowly' than a clock at one of the poles did he mean that the equatorial clock goes more slowly than a polar clock?

As I said, I think he meant the average over an entire orbit, and I believe it would be true in all inertial frames that over a complete orbit an equatorial clock would tick less than a clock at the pole.

On the basis that the equatorial clock does, on average over an entire orbit, 'go more slowly' than the polar clock I am of the impression that during this orbit the equatorial clock also 'goes more slowly' (i.e. ticks over at a slower rate) than the polar clock.

On the basis that, after one orbit, the equatorial clock is slower than (i.e lags behind) the polar clock then in my opinion it must, during that orbit, have ticked over at a slower rate than it would had it been located alongside the polar clock.

I think there is a distinction between the idea that the equatorial clock is slower than (i.e. lags behind) a polar clock after an entire orbit and that it 'goes more slowly' (i.e. ticks over at a slower rate) than the polar clock.

Can we focus on the other situation Einstein discusses in section 4 where clock A and clock B are initially some distance apart, then A is briefly accelerated and afterwards moves inertially towards B? Do you assert that in this example A is "physically", "really", or "actually" ticking slower than B between the time it's accelerated and the time it reaches B, in spite of the fact that there are perfectly valid inertial frames where it is B that's ticking slower during this period of time?

I agree that "...in this example A is "physically", "really", or "actually" ticking slower than B between the time it's accelerated and the time it reaches B." however I do not accept that "...there are perfectly valid inertial frames where it is B that's ticking slower during this period of time."

In my opinion B's rate of operation can in no way be affected by A's acceleration or deceleration or rate of uniform travel toward (or away from) B!

So when you say that it is B that's ticking slower during this period of time this is nothing more than a comparison of the calculated rate of operation of B to that of the rate of operation of clock A thus clock A 'is' ticking over at a faster rate than it was before it started accelerating however I am of the understanding that the concept of time contraction was, for Einstein, unacceptable.

Ergo, on that basis, A is not ticking over at a faster rate (time contraction) than it was before it started accelerating and on the basis that A's actions of any kind have no affect whatsoever on B's rate of operation A cannot be of the opinion that B is ticking over at a slower rate than it was before he started accelerating but can only conclude that B appears (according to his calculations) to be ticking over at a faster rate than it was previously.

In my opinion, which is probably controversial, clock A's instantaneous velocity can be substituted for v in the Lorentz transformations.

A is accelerating toward B and has attained a instantaneous velocity of s. He switches his rockets off and at that very instant is moving at the same (albeit, now) uniform velocity of s (i.e he is moving toward B at the same speed as he was at the very instant the rocket shut down).

I fail to see why the calculated rate of operation of clock A from B's point of view (or the calculated rate of clock B from A's point of view) will be any different if A has attained an instantaneous velocity of s or is moving toward B with a uniform velocity of s.

I just want to understand if you use words like "physically", "really" and "actually" to mean something that there is a single correct answer about, or if you just use these words to refer to the perspective of particular frames, so that you would be equally fine with saying that it is B that "physically", "really", and "actually" ticks slower than A in certain choices of frames.

Saying that B 'is' ticking slower (or faster) than A in certain choices of frames is NOT the same as saying that B ticks over at slower (or faster) rates than it did before A started moving (or accelerating). I reiterate - I am of the opinion that B's rate of operation is not affected by A's actions of any kind, it merely appears, from A's point of view (in accordance with his calculations), to change!

An astronaut has come to a stop at the end of his outward bound trip. He accelerates and attains an instantaneous velocity whereby a gamma factor of 400 000 is attained. He 'sees' (or 'determines' or 'calculates') the Earth clock ticking over at a fsater rate than his own clock by a factor of one second Earth time for each 400 000 seconds of his own time.

He flicks a switch extinguishing his rocket and at that very instant 'sees' (or 'determines' or 'calculates') that the Earth clock is no longer ticking over at that faster rate than his own clock but is immediately ticking over at a slower rate than his own clock by a factor of 400 000 seconds for each of his own seconds.

Is he not likely to ask himself what affect this would have on the clock (and on all of the planet's inhabitants) which instantaneously reverts from being 400 000 times faster than his clock to being 400 000 times slower?

Does he truly believe that this takes place because he flicked that switch?

I won't bother detailing the following but on the basis of the identical nature of time contraction and length contraction he would 'see' the planet instantaneously change from being shaped like a pancake of around 32 meters thick at the center (tapering to almost zero at its perimeter) to being more than 5 billion kilometers deep (in his direction of travel).

Would he not be of the opinion that this instantaneous and somewhat considerable change in that dimension could have a detrimental affect on the planet and a devastating effect on its inhabitants.

In fact, as his instantaneous velocity increases, so too 'does' the distance from him to the planet (as 'does' the rate of operation of the Earth clock) but when he flicks that switch the (seemingly increasing) distance to the planet instantaneously reverts to a close proximity to the plant - simply because he flicked a switch?

 

[cos]

It is not important what I think about the words 'physical' or 'real' but what Einstein meant by the words 'must go more slowly'!

I am of the opinion that he meant that clock A 'physically' or 'really' or 'actually' goes more slowly.

The problem is that because you refuse to define "physical", "real", etc. I still don't know what you mean by that last.

OK; what do you think Einstein meant by the phrase "...must go more slowly..."?

My opinion is that he was saying that a clock at the equator must tick over at a slower rate than it would if it were located at one of the poles or that it ticks over at a slower rate than a polar clock.

I cannot tell if we agree or disagree, and I don't know what words to use to clearly communicate my position back to you. It is, in fact, important what you think about those words because you are the one I am trying to communicate with. I even made it easy for you and suggested some definitions, all you have to do is say yes or no.

Your response to the question above might indicate whether or not we agree or disagree.

I think it is rather hypocritical that you accused me of "deliberate obfuscation" above.

The term 'hypocritical' implies that you are of the opinion that I have introduced 'deliberate obfuscation'.

Whilst my responses, or lack thereof, may have created obfuscation this was not my intention!

 

[cos]

cos,

you have repeatedly been asked to explain what you mean by "physically", "really", or "actually". You might think these words are obvious and require no further explanation, but in relativity things that seemed obvious in Newtonian theory are no longer so. If you describe something as "physically slower" (for example), you need to explain what measurements or calculations you would perform to decide whether something is "physically slower" or not.

When Einstein wrote in section 4 STR that an equatorial clock 'must go more slowly' he, I believe, related this to his equation .5tv^2/c^2 . having referred to that same equation in my postings I assumed that readers would automatically apply that same equation.

As to what measurements or calculations I would perform (as distinct from applying) I am of the opinion that Einstein did not explain what measurements or claculations he would perform (as distinct from applying) so perhaps you point should be directed to his depiction.

You often refer to a clock "ticking more slowly" but you fail to say slower than what and as measured by whom. In relativity these are not optional extras: different observers get different answers. It makes sense to assert "A ticks more slowly than B as measured by C". If we shorten this to "A ticks more slowly than B" this only makes sense (for instantaneous tick rates) in a context where the "as measured by C" is understood -- often the context is "as measured by B". To say "A ticks more slowly" makes no sense at all unless everyone implicitly understands what B and C are.

Perhaps you could provide a reference as to where I referred to a clock that is ticking more slowly but where I failed to say to what it is ticking more slowly than?

I do not care who makes the measurement. In my opinion a clock's intrinsic rate of operation will remain unchanged regardless of the point of view, or the determinations of, an observer.

Note that in relativity it is possible for

"A ticks (instantaneously) more slowly than B as measured by B"

"B ticks (instantaneously) more slowly than A as measured by A"

This may well be true of the previous sections of relativity however in section 4 Einstein points out that a clock at the equator must go more slowly than a clock at one of the poles and it is my understanding that from the point of view of observer B (at one of the poles) A does not 'tick more slowly' than his own clock as per you statement above ("A ticks (instantaneously) more slowly than B as measured by B") but faster!

to be simultaneously true. It's not a contradiction because A & B use different measurement procedures.

It matters not that A and B "...use different measurement procedures." Nothing that any of them 'measure' or 'calculate' or 'determine' or 'predict' will have any affect whatsoever on a clock's intrinsic rate of operation.

But "A actually ticks slower" (without mention of a B or C) means nothing. Can you give an unambiguous operational definition (what numbers you would measure or calculate) of what you think it means?
He meant the equatorial clock ticks more slowly than a polar clock as measured by a polar clock.

And as measured by the equatorial observer!

He determines that his clock is 'going more slowly' than the polar clock; he realizes that his having moved to the equator has had no affect whatsoever on the rate of operation of the polar clock which is still ticking over at the same rate as it was when he was at the same location.

In order to qualify that suggestion - neither of the observers can actually see what the other clock is doing but on the basis that the equatorial observer has read, and agrees with, Einstein's section 4 depiction he could agree with Einstein that his clock is 'going more slowly' than the polar clock and at a slower rate than it was before he moved to the equator.

Note: everything above applies to "instantaneous" clock rates. If you are talking about average clock rates where clocks A and B are initially together, separate and come back together again, everyone will agree which clock ticked fewer ticks than the other over the whole round-trip journey, but then at least one of the clocks must have accelerated (I'm assuming Special Relativity in the absence of gravity), so simple inertial frame analysis is not sufficient.

Most of the above applies to Einstein's section 4 STR depiction of equatorial and polar clocks where the the clocks are not separated and come back together again however observer A, as i pointed out in other message, could initially have been located at one of the poles where his clock was ideally synchronous with an identical clock. He moves to the equator then back to the pole (i.e. travels in a closed curve relative to the polar clock) whereupon he finds, as Einstein depicted in his description of a clock that moves in a closed curve relative to another clock, that his clock now lags behind that clock ergo he should be able to conclude, as Einstein pointed out, that his clock (progressively) went more slowly (i.e.ticked over at a slower rate) than the polar clock.

 

[Dale]

OK; what do you think Einstein meant by the phrase "...must go more slowly..."?

I think that he meant that A's proper time is slower than the coordinate time \left(\frac{dt}{d\tau}>1\right) in system K, the reference frame where A and B were initially at rest and synchronized. I believe that he understood that this is a frame-variant statement, which was the reason why he clearly identified frame K.

The term 'hypocritical' implies that you are of the opinion that I have introduced 'deliberate obfuscation'.

Whilst my responses, or lack thereof, may have created obfuscation this was not my intention!

I will accept this statement at face value and not impugn your motives. I would ask that you show me the courtesy of doing the same.

By the way, I would appreciate it if you would try not to underline so many words. I often go back through a thread looking for a link, and they are very hard to find with so many non-links in underlined font. Italics and bold are much preferable (as you have done in this quoted post).

 

[JesseM]

On the basis that the equatorial clock does, on average over an entire orbit, 'go more slowly' than the polar clock I am of the impression that during this orbit the equatorial clock also 'goes more slowly' (i.e. ticks over at a slower rate) than the polar clock.

Only on average over the whole orbit, not necessarily at every moment, depending what frame you choose. In certain frames there will be periods of time where the equatorial clock has a smaller speed than the polar clock, so during these periods of time the polar clock must be ticking slower in such a frame.

I agree that "...in this example A is "physically", "really", or "actually" ticking slower than B between the time it's accelerated and the time it reaches B."

You "agree"? I did not say that was what I thought.

however I do not accept that "...there are perfectly valid inertial frames where it is B that's ticking slower during this period of time."

So do you disagree with the math in the second paragraph in my example from post #64 below?

Suppose for example A and B are a distance of 60 light-seconds apart in the "stationary" frame K, and both are synchronized in this frame. Then if A is moved at 0.6c towards B at the moment when both clocks read a time of t=0, it will take 100 seconds in this frame for A to reach B, during which time A will only tick 80 seconds due to time dilation (the Lorentz factor being 1.25), so when A meets B, B will read t=100 seconds while A reads t=80 seconds.

Now consider things from the perspective of the inertial frame where A and B were initially moving at 0.6c and then A was accelerated to come to rest in this frame while B continued to move towards it at 0.6c. In this frame the clocks were not synchronized initially, so when A read t=0, B already read t=36 seconds according to this frame's definition of simultaneity. Then it takes 80 seconds in this frame for B to reach A (because the initial distance between them was 48 light-seconds in this frame due to length contraction, and 48 light-seconds/0.6c = 80 seconds), during which time B only ticks forward by 80/1.25 = 64 seconds due to time dilation, meaning B reads t=36 + 64 = 100 seconds when they meet, while A reads t=80 seconds when they meet. So you see that both frames make the same prediction about their respective times, even though in the first frame A was ticking slower while in the second frame B was ticking slower.

Unless you think the math is wrong in the second paragraph, the second paragraph is saying that in the frame where A and B were initially moving at 0.6c, it must be true that at the moment A accelerates (and we can assume the acceleration is instantaneously brief), A reads t=0 seconds and B reads t=36 seconds, and at the moment B and A meet, A reads 80 seconds and B reads 100 seconds. So B ticked forward by 64 seconds, while A ticked forward by 80 seconds, therefore B was ticking slower during this period, from the perspective of this frame. If you don't disagree with the math--and I can show you that these numbers follow directly from applying the Lorentz transformation to the scenario described in the first paragraph from the perspective of B's rest frame--then how can you say you disagree with the statement that "there are perfectly valid inertial frames where it is B that's ticking slower during this period of time"?

In my opinion B's rate of operation can in no way be affected by A's acceleration or deceleration or rate of uniform travel toward (or away from) B!

No one said anything about B's rate of operation being accelerated by A's acceleration! In the frame described by the second paragraph, B is always ticking at a rate of 0.8 ticks per second of coordinate time, both before and after A accelerates. Before A accelerates A is also ticking at 0.8 ticks per second of coordinate time, but then after accelerating A comes to rest in this frame, and is now ticking at 1 tick per second of coordinate time. Here are a few numbers to make this clear:

At coordinate time of t=-30 seconds in this frame, A reads -24 seconds, B reads 12 seconds

At coordinate time of t=-20 seconds in this frame, A reads -16 seconds, B reads 20 seconds

At coordinate time of t=-10 seconds in this frame, A reads -8 seconds, B reads 28 seconds

At coordinate time of t=0 seconds in this frame (the moment that A accelerates), A reads 0 seconds, B reads 36 seconds

At coordinate time of t=10 seconds in this frame, A reads 10 seconds, B reads 44 seconds

At coordinate time of t=20 seconds in this frame, A reads 20 seconds, B reads 52 seconds

At coordinate time of t=30 seconds in this frame, A reads 30 seconds, B reads 60 seconds

...So, you can see that for each interval of 10 seconds of coordinate time prior to A's acceleration, B advances forward by 8 seconds (from 20 to 28 seconds between t=-20 and t=-10 seconds, for example), and A also advances forward by 8 seconds (from -16 to -8 seconds between t=-20 and t=-10). On the other hand, for each interval of 10 seconds of coordinate time after A's acceleration, B still advances forward by the same amount of 8 seconds (from 44 to 52 seconds between t=10 and t=20, for example), while A now advances forward at the faster rate of 10 seconds (from 10 to 20 seconds between t=10 and t=20). So B's rate of ticking never changes in this frame, only A's rate of ticking changes, from ticking at the same rate as B before it accelerates to ticking faster than B afterwards. Again, do you disagree that these are the numbers we get if we apply the Lorentz transformation to the scenario as it was described in B's rest frame, or do you agree with the math but think that the description in B's rest frame describes what "really", "actually" happens while the description in this second frame is some sort of illusion?

So when you say that it is B that's ticking slower during this period of time this is nothing more than a comparison of the calculated rate of operation of B to that of the rate of operation of clock A thus clock A 'is' ticking over at a faster rate than it was before it started accelerating however I am of the understanding that the concept of time contraction was, for Einstein, unacceptable.

You still have never explained what "time contraction" means. You can see in the above scenario that even though A speeds up after accelerating, its rate of ticking can never be faster than the rate that coordinate time is passing--is that what you mean by time contraction? Or do you just think relativity forbids a clock's rate from ever speeding up at all after it changes velocities? If the latter, you're incorrect, if a clock changes velocities in such a way that its speed becomes smaller in a given frame, then its rate of ticking will get faster than what it was before changing velocities, from the perspective of that frame.

In my opinion, which is probably controversial, clock A's instantaneous velocity can be substituted for v in the Lorentz transformations.

Sure, if you want to figure out what things look like in the frame where A is instantaneously at rest at that instant (but in this frame A was not always at rest if it is accelerating; by definition an inertial frame must travel at the same constant velocity forever).

A is accelerating toward B and has attained a instantaneous velocity of s. He switches his rockets off and at that very instant is moving at the same (albeit, now) uniform velocity of s (i.e he is moving toward B at the same speed as he was at the very instant the rocket shut down).

Of course, did you think I had suggested otherwise?

I fail to see why the calculated rate of operation of clock A from B's point of view (or the calculated rate of clock B from A's point of view) will be any different if A has attained an instantaneous velocity of s or is moving toward B with a uniform velocity of s.

In SR we're only calculating things from the "point of view" of particular inertial coordinate systems, I don't know what it means to calculate things "from A's point of view" since A changes velocity at a certain point. If you want to calculate things from the point of view of the inertial frame where A is at rest after accelerating, that is exactly what I was doing in the second paragraph of my numerical example above.

Saying that B 'is' ticking slower (or faster) than A in certain choices of frames is NOT the same as saying that B ticks over at slower (or faster) rates than it did before A started moving (or accelerating).

As I point out above, I have never claimed this, so I don't know why you're making this point. When I said "...there are perfectly valid inertial frames where it is B that's ticking slower during this period of time", I thought it was fairly clear from the context that I meant B is ticking slower than A during this period of time, not that B is ticking slower than B was ticking prior to A's acceleration. Perhaps you misunderstood my meaning? If so, now that I have clarified, do you disagree that "there are perfectly valid inertial frames where it is B that's ticking slower than A during this period of time" (i.e. the period of time between A accelerating and A and B meeting)? If you don't disagree, then would you say that B is "physically", "really", and "actually" ticking slower than A during this period of time, from the perspective of these frames?

 

[cos]

OK; what do you think Einstein meant by the phrase "...must go more slowly..."?

I think that he meant that A's proper time is slower than the coordinate time \left(\frac{dt}{d\tau}>1\right) in system K, the reference frame where A and B were initially at rest and synchronized. I believe that he understood that this is a frame-variant statement, which was the reason why he clearly identified frame K.

His comment "...must go more slowly..." was not in relation to "...in system K, the reference frame where A and B were initially at rest and synchronized." but to his later reference to a clock at the equator which, he insisted 'must go more slowly' than a clock at one of the poles.

For the purpose of edification I submit his section 4 comments -

"If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by .5tv^2/c^2 (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B.

It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide.

If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the traveled clock on its arrival at A will be a .5tv^2/c^2 second slow. Thence we conclude that a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions."

I will accept this statement at face value and not impugn your motives. I would ask that you show me the courtesy of doing the same.

Noted - but I also trust that you will not impugn my motives in relation to any other of my postings and I will reciprocate

By the way, I would appreciate it if you would try not to underline so many words. I often go back through a thread looking for a link, and they are very hard to find with so many non-links in underlined font. Italics and bold are much preferable (as you have done in this quoted post).

"Sorry 'bout that Chief."

 

[cos]

On the basis that the equatorial clock does, on average over an entire orbit, 'go more slowly' than the polar clock I am of the impression that during this orbit the equatorial clock also 'goes more slowly' (i.e. ticks over at a slower rate) than the polar clock.

Only on average over the whole orbit, not necessarily at every moment, depending what frame you choose. In certain frames there will be periods of time where the equatorial clock has a smaller speed than the polar clock, so during these periods of time the polar clock must be ticking slower in such a frame.

On the basis that in my opinion this may well be the crux of the matter I make no apology for the fact that, although having read same, I have removed the remainder of your posting on the basis of my physical disabilities which severely restricts the time that I am able to devote to my responses.

"Depending what frame [is chosen]"?

There are, as I have previously stipulated, only two frames in this otherwise empty universe - that of the polar observer and that of the equatorial observer.

The planet could, as I have also previously stipulated, be replaced by a large rotating, transparent, massless sphere.

Apart from rotating, the sphere is not moving. From both points of view the equatorial clock is always moving at the same speed relative to the polar clock i.e. 1600K-h.

At no time does "...the equatorial clock [have] a smaller speed than the polar clock..."

 

[JesseM]

There are, as I have previously stipulated, only two frames in this otherwise empty universe - that of the polar observer and that of the equatorial observer.

First of all, frames are just coordinate systems. They don't depend on what objects happen to be present in the universe, you can certainly assign coordinates to events using a coordinate system where no object in the universe happens to be at rest. So, regardless of what objects happen to be present in the universe, you have an infinite number of different inertial frames you can use.

Second, in SR we usually stick to talking about inertial frames. The object at the equator does not remain at rest in any inertial frame, because it's moving non-inertially. You can think about non-inertial coordinate systems if you wish, but the usual rules of inertial frames, like light always moving at c or moving clocks running slow by a factor of \sqrt{1 - v^2/c^2}, no longer apply in non-inertial coordinate systems, so it's easier to just use inertial ones.

Finally, If you don't want to reply to my entire post that's fine, but could I request that you clarify what you meant when you said (speaking of the A-B thought experiment rather than the equatorial and polar clock thought experiment) that you do not agree with my statement that "...there are perfectly valid inertial frames where it is B that's ticking slower during this period of time"? Did you misunderstand what I meant by "it is B that's ticking slower", thinking I was saying that B's rate of ticking slowed down after A accelerated in some inertial frame? If so, please note that I didn't mean to imply any change in B's rate of ticking, I just meant that there are inertial frames where B is ticking slower than A after A accelerates, because in these frames A's rate of ticking speeds up after it accelerates (since its speed decreases in these frames) while B's rate of ticking remains unchanged. So with that clarification, would you still disagree that "there are perfectly valid inertial frames where it is B that's ticking slower than A during this period of time (the period after A accelerates)", or would you now agree with it? I'm not asking for a detailed answer here, just a simple "agree" or "disagree".

 

[cos]

So, regardless of what objects happen to be present in the universe, you have an infinite number of different inertial frames you can use.

On the basis of my presentation that the sphere is located in an otherwise empty universe thus a universe in which there is not an infinite number of frames and your continued insistence on an infinite number of frames - which, in my opinion, is a reprehensible attitude - this discussion is terminated.

 

[matheinste]

cos

Maybe for frames you mean rest frames.

In a universe empty except for ANY number of objects there are an infinite number of frames. This applies equally well to a universe containing only one object or our populated universe. However, there is at most only one frame in which any individual object can be at rest at any instant. So in the two clock scenario there are an infinite number of frames but for each clock there is only one frame in which it is at rest at any instant.

Matheinste

 

[Dale]

His comment "...must go more slowly..." was not in relation to "...in system K, the reference frame where A and B were initially at rest and synchronized." but to his later reference to a clock at the equator which, he insisted 'must go more slowly' than a clock at one of the poles.

Then I suggest that you misunderstood him. He postulated the equivalence of inertial reference frames and then he derived the relativity of simultaneity earlier in the paper. From then on he was repeatedly careful to identify the reference frame in which his analysis held. I don't know how you can read that work and come to any conclusion other than that he understood simultaneity, time dilation, and length contraction to be frame-variant effects.

Why do you think he repeatedly identified the reference frame in his later analysis if he believed that his results were frame-invariant?

 

[Al68]

During that trip he determines that B' is ticking over at a slower rate than his clock

yes.

whereupon he predicts that B' will resultantly lag behind his own clock

No, he would make no such prediction.

yet he arrives at that location to find that B' does not lag behind his clock but that his clock lags behind B'.

Which is what he would have predicted. Note that B' is only synchronized with the Earth clock in Earth's frame, not in the ship's frame. In Earth's frame B and B' both read zero when the ship leaves earth. In the ship's frame, B read zero when the ship left earth, but B' did not. B' runs faster than the ship's clock in the ship's frame during the initial acceleration at earth.

In his 1918 article (which I believe was merely an extension of his section 4 STR depictions) Einstein pointed out that it is ONLY the clock that experiences forces of acceleration (i.e. his section 4 clock A) that incurs a variation in it's rate of operation (a slower tick rate) NOT the unaccelerated inertial reference frame clock (i.e. his section 4 clock B).

He says no such thing in his 1918 paper.

The ship's clock has. as Einstein pointed out, accelerated thus it is, according to Einstein, the accelerated ship's clock that incurs time dilation - the Earth clock does not incur time contraction.

Again, he says no such thing in his 1918 paper.

The astronaut is moving at a velocity that generates a gamma factor of 400 000. He 'sees' or 'determines' that his clock is ticking over at the rate of 400 000 seconds for each of clock B' seconds (i.e. B is ticking over at a slower rate than his clock) but at the very moment that he puts his foot on the gas pedal to power up his retrorockets clock B stops
ticking over at that slower rate and instantaneously starts ticking over the faster rate of 400 000 seconds for each of his seconds

This is exactly the claim made by Einstein in his 1918 paper (except it would be much greater than 400,000 to one in your example during the turnaround). Referring to the ship's non-inertial rest frame: "the clock U1(earth clock), going at a velocity v, runs indeed at a slower pace than the resting clock U2(at rest with ship). However, this is more than compensated by a faster pace of U1(earth clock) during partial process 3 (turnaround acceleration)...The calculation shows that this speeding ahead constitutes exactly twice as much as the lagging behind during the partial processes 2 and 4. This consideration completely clears up the paradox that you brought up."

That clock instantaneously reverses its rate of operation from being 400 000 times slower than his clock to being 400 000 times faster?

No, nothing happens to his clock. The rate of Earth's clock is frame dependent and he is changing frames.

If you believe that I've got a bridge you might be interested in buying.

What I believe is irrelevant. This is what SR predicts and what Einstein claims in his 1918 paper.

I don't care in which frame the observations are made. In my opinion nothing that any frame 'observes' can physically affect the rate of operation of any clock!

Nothing physically happens to any clock. In fact the assumption is that each clock keeps proper time and is not affected by acceleration. And the rate of each clock is frame dependent. In other words the rate of a clock on Earth is different for different reference frames, although the clock doesn't change. This is the primary revelation Einstein made in 1905.

 

[JesseM]

On the basis of my presentation that the sphere is located in an otherwise empty universe thus a universe in which there is not an infinite number of frames and your continued insistence on an infinite number of frames - which, in my opinion, is a reprehensible attitude - this discussion is terminated.

Again, a "frame" is just a coordinate system, not something physical. Where did you get the idea that we are only allowed to assign x,y,z,t coordinates to events using a coordinate system where some physical object in the universe is at rest?

 

[phyti]

cos;

Irrelevant, I made no suggestion whatsoever that time dilation (as depicted by Einstein's section 4 STR comments) is not 'a real factor'!

I'm not saying you did. Those comments just reinforce that is is a real phenomena.
Particle physicists would not report things they did not observe, and gps satellites would not be corrected if they functioned accurately.

 

[cos]

In a universe empty except for ANY number of objects there are an infinite number of frames. This applies equally well to a universe containing only one object or our populated universe. However, there is at most only one frame in which any individual object can be at rest at any instant. So in the two clock scenario there are an infinite number of frames but for each clock there is only one frame in which it is at rest at any instant.

Matheinste

So is an observer standing alongside Einstein's section 4 equatorial clock entitled to realize that, as Einstein pointed out, his clock is 'going more slowly' (i.e. ticking over at a slower rate) than it would if he were at that pole?

If he were to be initially located at one of the poles and were to move to the equator would he be entitled to conclude that his clock is then ticking over at a slower rate than it was before he moved away from the pole?

 

[cos]

A168

For some reason I have been unable to respond to #121; when I hit the 'quote' button it brings up someone else's message however in an attempt to save time for both of us let's get back to basics.

In section 4 STR Einstein wrote -

"Thence we conclude that a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions."

I am of the opinion that by his comment "...must go more slowly..." he was saying that the equatorial clock ticks over at a slower rate than the polar clock.

Do you agree with my opinion?

 

[JesseM]

cos, you terminated the discussion with me because of my "reprehensible attitude" that one is free to use anyone of an infinite number of inertial frames regardless of how many physical objects are present in the universe, but you are continuing to talk to matheinste even though he argues exactly the same thing:

In a universe empty except for ANY number of objects there are an infinite number of frames. This applies equally well to a universe containing only one object or our populated universe.

I am sure if you asked anyone else on this thread they would agree that frames are just coordinate systems and you don't need any objects to be at rest in a frame in order to calculate what things look like from the perspective of that frame (and this perspective is as valid as the perspective of any other frame in SR)--even if you think this is incorrect somehow, given that it's a widespread opinion it's a point worth discussing, no? So, would you be willing to reconsider your termination of our previous discussion?

 

[cos]

cos, you terminated the discussion with me because of my "reprehensible attitude" that one is free to use anyone of an infinite number of inertial frames regardless of how many physical objects are present in the universe, but you are continuing to talk to matheinste even though he argues exactly the same thing:

I am responding to your posting on the basis that as far as i am concerned it is off thread.

I several times requested that you did not refer to a number of inertial frames however on the basis of your continued insistence upon doing so I terminated our discussion.

To the best of my knowledge i have not requested matheinste to 'cease and desist' however my attitude to our correspondence may well depend on his response to my recent message to him.

 

[cos]

Irrelevant, I made no suggestion whatsoever that time dilation (as depicted by Einstein's section 4 STR comments) is not 'a real factor'!

I'm not saying you did. Those comments just reinforce that is is a real phenomena.

You find me at a disadvantage - to what do you refer by the phrase "Those comments."?

Are you referring to Einstein's comments in section 4? If so, then do you agree with me that his statement that the equatorial clock "...must go more slowly..." (i.e. must tick over at a slower rate) than a clock at the equator means that it will tick over at a slower rate than it would if it were located at that pole - that this is a 'real phenomena'?

An observer is located at one of the poles; do you agree with me that if he moves to the equator he would be entitled to realize that his clock (although ticking over at it's 'normal' rate) would actually (really) be ticking over at a slower rate than it was before he left that location?

 

[JesseM]

I am responding to your posting on the basis that as far as i am concerned it is off thread.

But the question of which clock is ticking slower at a given moment (as opposed to average rate of ticking over the entire orbit) does depend on your choice of frame, so the fact that in relativity you are free to analyze things from the perspective of a frame where no physical object is at rest is relevant to what's being discussed on the thread, as far as I can tell. If you think it's incorrect that you can use a given frame even if there's no object in the universe at rest in that frame, that's a perception of yours that I think everyone else on this thread would disagree with, so it's worth discussing why you believe that you're right and everyone else is wrong.

I several times requested that you did not refer to a number of inertial frames however on the basis of your continued insistence upon doing so I terminated our discussion.

Again, all questions of "which clock is ticking slower at a given moment" depend on which frame you're using, and a lot of times you use language that makes it sound like you believe there's a single correct answer to questions about which clock is ticking slower, rather than a number of different possible answers depending on what frame you choose. You could avoid discussion of multiple frames if you modify your question to something like "which clock is ticking slower in the rest frame of the polar observer" (or 'in the rest frame of clock B' in the case of the other thought-experiment), and then no one would have reason to dispute your claim about which clock is ticking slower. But it is you who seem to be consciously refusing to qualify your statements about clock rates in a way that shows you understand the answer is specific to a particular choice of frame (why do you insist on phrasing your question in a way that doesn't refer to any specific frame?), and I think it's this refusal that leads many people here to think you are misunderstanding something about the SR...so, it's only natural that people will respond to you by pointing out that the question of which clock ticks slower at a given moment depends on what frame you choose, and that it is equally valid to analyze things from the perspective of any inertial frame. If you do already understand and accept this point, then just say so and no one will need to bring up the issue of multiple possible inertial frames again on this thread. If you don't understand/agree with this, then people will keep bringing up this point, because a failure to understand this is a major misunderstanding of SR.

 

[matheinste]

I am responding to your posting on the basis that as far as i am concerned it is off thread.

I several times requested that you did not refer to a number of inertial frames however on the basis of your continued insistence upon doing so I terminated our discussion.

To the best of my knowledge i have not requested matheinste to 'cease and desist' however my attitude to our correspondence may well depend on his response to my recent message to him.

Whether you talk to JesseM after your interpretation of his perfectly legitimate comment as "reprehensible" is entirely up to you. I do not wish to shoulder the responsibility for your action and so will not reply to your recent message in which you say ."---however my attitude to our correspondence may well depend on his response (matheinste's) to my recent message to him.----"

Matheinste.

 

[Dale]

do you agree with me that if he moves to the equator he would be entitled to realize that his clock (although ticking over at it's 'normal' rate) would actually (really) be ticking over at a slower rate than it was before he left that location?

Come on cos! I, and others, have asked you several times what do you mean by this? What does "actually (really)" mean to you? Can something be "actually (really)" true if it depends on the coordinate system? This is the crux of the entire thread!

 

[neopolitan]

An observer is located at one of the poles; do you agree with me that if he moves to the equator he would be entitled to realize that his clock (although ticking over at it's 'normal' rate) would actually (really) be ticking over at a slower rate than it was before he left that location?

I know you have terminated discussions with me, but perhaps someone else can pick up on this point.

The clocks do measure a rate. It's a pretty simple rate, one second per second, or one minute per minute, or if you have a more accurate clock, one microsecond per microsecond.

It seems that you are dividing time up into different types. If you want to measure a rate of "seconds at the equator per second at the pole", then you might get a much different answer to plain "seconds per second" which by default would be "seconds where I am, doing what I am doing per second where I am, doing what I am doing" which is a rate that would never change.

Is your argument that a clock at the equator runs at a rate of "seconds at the equator per second at the pole" which is "really" slower than the clock at pole which runs at "seconds at the pole per second at the pole" - and that this is "real" because the rate "seconds at the pole per second at the equator" is greater than the rate "seconds at the equator per second at the pole"?

Further, are you implying that when you have two frames which are inertial (A and B) where potential clock rates are:

seconds at A per second at A = seconds at B per second at B
and
seconds at A per second at B = seconds at B per second at A
where
seconds at A per second at A != (does not equal) seconds at B per second at A

then there is something illusory happening?

These are supposed to be clarifying questions. If you are not saying any of these things, it is helpful to know that you aren't.

cheers,

neopolitan

 

[phyti]

You find me at a disadvantage - to what do you refer by the phrase "Those comments."?

Are you referring to Einstein's comments in section 4? If so, then do you agree with me that his statement that the equatorial clock "...must go more slowly..." (i.e. must tick over at a slower rate) than a clock at the equator means that it will tick over at a slower rate than it would if it were located at that pole - that this is a 'real phenomena'?

An observer is located at one of the poles; do you agree with me that if he moves to the equator he would be entitled to realize that his clock (although ticking over at it's 'normal' rate) would actually (really) be ticking over at a slower rate than it was before he left that location?

Those comments were regarding particle accelerators and gps, actual cases of time dilation effects.

In the case where one clock moves away from the first, travels on a closed path, then rejoins it, and you are comparing the clocks in the same frame you started with, the difference in readings must be explained by the motion of the clock that traveled, which requires acceleration (+ and -) at the beginning and end of the trip. There is nothing else to be used as a cause.
This case is then depicted/extended to the Earth rotation example. The conclusion is the same.
The clock that takes the longest closed path, records the least amount of time.
The equatorial clock would run slower according to any Earth bound clock not on the equator.

Regarding another poster related to this:
Since the ratio of the Earth diameter to distance traveled in a 24 hr day is approx. 8/1500,
any other position on the Earth surface is always moving faster than the pole, in the sun reference frame. And as mentioned, in the Earth frame, it doesn't matter. The rotation is constant (disregarding small fluctuations).

 

[phyti]

Jesse,

Now consider things from the perspective of the inertial frame where A and B were initially moving at 0.6c and then A was accelerated to come to rest in this frame while B continued to move towards it at 0.6c. In this frame the clocks were not synchronized initially, so when A read t=0, B already read t=36 seconds according to this frame's definition of simultaneity.
Then it takes 80 seconds in this frame for B to reach A (because the initial distance between them was 48 light-seconds in this frame due to length contraction, and 48 light-seconds/0.6c = 80 seconds), during which time B only ticks forward by 80/1.25 = 64 seconds due to time dilation, meaning B reads t=36 + 64 = 100 seconds when they meet, while A reads t=80 seconds when they meet. So you see that both frames make the same prediction about their respective times, even though in the first frame A was ticking slower while in the second frame B was ticking slower.

The difference is 60(.6)(.8)/(1-.36) = 45 sec. The separation is 60 lsec. They cannot detect length change in their own frame, just as they can't detect their slower clocks.
A records arrival of B at 60/.6 = 100 sec. B records .8(100) =80 sec due to time dilation. Because of the shortened time, B thinks the distance is also short, i.e. .8(60) = 48 lsec.
(At this point why do you dilate A's time again, to 64 sec).

For A: t1=0, t2=100, elapsed time =100 sec
For B: t1=45, t2=125, elapsed time = 80 sec
Even though clock A lags clock B, B runs slower than A.
It's the elapsed times that are compared for unsynchronized clocks.

This scenario introduces a third rest frame with A and B moving at .6c. The original example involved two synchronized clocks with one moving in an arbitrary closed path to rejoin the other. The purpose is to demonstrate the clock taking the longer path records less time. They are not equivalent examples.

 

[phyti]

cos;

do you agree with me that if he moves to the equator he would be entitled to realize that his clock (although ticking over at it's 'normal' rate) would actually (really) be ticking over at a slower rate than it was before he left that location?

do you agree with me that if he moves to the equator he would be entitled to realize that his clock (although ticking at it's proper rate for the speed it has) would actually be ticking at a slower rate than it was at the pole?

My changes are in blue.
Is my translation of your quote correct?

 

[JesseM]

Suppose for example A and B are a distance of 60 light-seconds apart in the "stationary" frame K, and both are synchronized in this frame. Then if A is moved at 0.6c towards B at the moment when both clocks read a time of t=0, it will take 100 seconds in this frame for A to reach B, during which time A will only tick 80 seconds due to time dilation (the Lorentz factor being 1.25), so when A meets B, B will read t=100 seconds while A reads t=80 seconds.

Now consider things from the perspective of the inertial frame where A and B were initially moving at 0.6c and then A was accelerated to come to rest in this frame while B continued to move towards it at 0.6c. In this frame the clocks were not synchronized initially, so when A read t=0, B already read t=36 seconds according to this frame's definition of simultaneity. Then it takes 80 seconds in this frame for B to reach A (because the initial distance between them was 48 light-seconds in this frame due to length contraction, and 48 light-seconds/0.6c = 80 seconds), during which time B only ticks forward by 80/1.25 = 64 seconds due to time dilation, meaning B reads t=36 + 64 = 100 seconds when they meet, while A reads t=80 seconds when they meet. So you see that both frames make the same prediction about their respective times, even though in the first frame A was ticking slower while in the second frame B was ticking slower.

The difference is 60(.6)(.8)/(1-.36) = 45 sec.

Difference between what and what? Where are you getting that formula?

The separation is 60 lsec.

Separation between what and what?

They cannot detect length change in their own frame, just as they can't detect their slower clocks.

Not sure what you mean by "their own frame", or what you mean by "cannot detect length contraction". In the frame where A comes to rest after accelerating, the distance between A and B when A accelerates is certainly less than 60 light-seconds, an ruler-clock system at rest in this frame would measure the distance as 48 light-seconds.

A records arrival of B at 60/.6 = 100 sec. B records .8(100) =80 sec due to time dilation. Because of the shortened time, B thinks the distance is also short, i.e. .8(60) = 48 lsec.

You seem to have the notation backwards, B is the one who shows a time of 100 seconds upon meeting with A, A is the one who shows a time of 80 seconds when they meet. And the time has nothing to do with why the distance is 48 light-seconds in the frame where A is at rest after accelerating, 48 light-seconds is what would be measured by an actual ruler-clock system at rest in that frame; if you had a ruler with a rest length of 48 light-seconds which was at rest in that frame, with clocks at either end that are synchronized in that frame, then if one end of the ruler is passing next to A when the clock there reads T (at some time before A accelerates), that means that when the clock at the other end reads T at the moment it is passing next to B, showing that A and B are 48 light-seconds apart at time T in this frame.

(At this point why do you dilate A's time again, to 64 sec).

Again you seem to have the notation confused, it is B that only elapses 64 seconds between the event of A accelerating and the event of B meeting A, as measured in the frame where A is at rest and B is moving at 0.6c. The reason has to do with the relativity of simultaneity--do you understand this concept? Remember, in B's rest frame, B read t=0 at the moment that A accelerated (when we assume A read t=0 too, because A and B were initially synchronized in B's rest frame), which means that in the other frame where B is moving at 0.6c, B does not read t=0 at the moment A accelerated, instead it already reads 36 seconds. In general if two clocks are synchronized in their own rest frame and a distance L apart in this frame, then in a frame where the clocks are moving at speed v, they will be out-of-sync by vL/c^2. Here L=60 light-seconds (the initial distance between A and B in B's rest frame) and v=0.6c (B's velocity in the frame where B is moving at 0.6c), so in the frame where B is moving at 0.6c, A and B must be out-of-sync by (60)(0.6)/1 = 36 seconds, so B must already read t=36 seconds at the moment A reads t=0 seconds in this frame. And of course, you can also use the time dilation formula to show that in the frame where A is at rest after accelerating while B is moving at 0.6c, then if A elapses some amount of time T after the moment A accelerates, B must only elapse T * \sqrt{1 - 0.6^2} after the moment A accelerates in this frame. So, if A elapses 80 seconds between accelerating and meeting B, in this frame B must elapse 80*\sqrt{1 - 0.36} = 64 seconds between the time on B that's simultaneous with A accelerating (according to this frame's definition of simultaneity, that's the moment of B reading 36 seconds) and the time on B when it meets up with A (when it reads 100 seconds).

For A: t1=0, t2=100, elapsed time =100 sec
For B: t1=45, t2=125, elapsed time = 80 sec

I don't understand where these numbers are supposed to come from. I already stated the scenario in such a way that in B's rest frame, A and B both read t=0 at the moment that A accelerates, and they are a distance of 60 light-seconds apart in this frame at that moment. Do you disagree that with that assumption, if A is moving at 0.6c towards B after accelerating in this first frame, then it will take 100 seconds of coordinate time for A to catch up with B in this frame? Do you disagree that with A moving at 0.6c for 100 seconds it will only elapse 80 seconds in this first frame due to time dilation, and that since B remains at rest in this frame it will elapse 100 seconds, so A will read 80 seconds and B will read 100 seconds when they meet? Do you disagree that in the second frame where A is at rest after accelerating while B is moving at 0.6c, the event of A accelerating will be simultaneous with the event of B reading t=36 seconds?

Even though clock A lags clock B, B runs slower than A.
It's the elapsed times that are compared for unsynchronized clocks.

In what frame? In B's rest frame, where A was initially at rest too before accelerating, A and B were synchronized up until the moment A accelerated--that was how I defined the problem. In the frame where A was at rest after accelerating (and A and B were moving at 0.6c before A changed velocity), they were out-of-sync by 36 seconds until A accelerated.

This scenario introduces a third rest frame with A and B moving at .6c.

"Third"? I only mentioned two frames:
1) the frame where A and B were initially at rest, then after A accelerated it was moving at 0.6c while B remained at rest
2) the frame where A and B were initially moving at 0.6c, then after A accelerated it came to rest while B continued to move at 0.6c

What other frame are you thinking of?

The original example involved two synchronized clocks with one moving in an arbitrary closed path to rejoin the other.

Einstein assumed clock A and B were initially at rest with respect to one another and synchronized in their rest frame, then A was moved at constant velocity towards B; that's exactly what I assumed in my example too. Read what he wrote again:

From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by (1/2)tv^2/c^2 (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B.

Do you think my example differs in any way from this? If so, how?

 

[cos]

Are you referring to Einstein's comments in section 4? If so, then do you agree with me that his statement that the equatorial clock "...must go more slowly..." (i.e. must tick over at a slower rate) than a clock at the equator means that it will tick over at a slower rate than it would if it were located at that pole - that this is a 'real phenomena'?

An observer is located at one of the poles; do you agree with me that if he moves to the equator he would be entitled to realize that his clock (although ticking over at it's 'normal' rate) would actually (really) be ticking over at a slower rate than it was before he left that location?

In the case where one clock moves away from the first, travels on a closed path, then rejoins it, and you are comparing the clocks in the same frame you started with, the difference in readings must be explained by the motion of the clock that traveled, which requires acceleration (+ and -) at the beginning and end of the trip. There is nothing else to be used as a cause.

I appreciate that you have answered my question in relation to 'those comments' however my questions in relation to the equatorial clock's variation in it's rate of operation were not in relation to any eventual "...difference in readings..." nor any "cause" of that phenomenon.

This case is then depicted/extended to the Earth rotation example. The conclusion is the same.
The clock that takes the longest closed path, records the least amount of time.
The equatorial clock would run slower according to any Earth bound clock not on the equator.

This doesn't answer my question viz -

"An observer is located at one of the poles; do you agree with me that if he moves to the equator he would be entitled to realize that his clock (although ticking over at it's 'normal' rate) would actually (really) be ticking over at a slower rate than it was before he left that location?"

As an analogy - an observer is located on a mountain-top, he descends the mountain; is he entitled to be of the opinion that his clock is then ticking over at a slower rate than it was before he moved down the mountain?

I provide this as being an analogy on the basis of the principle of equivalence.

 

[cos]

do you agree with me that if he moves to the equator he would be entitled to realize that his clock (although ticking over at it's 'normal' rate) would actually (really) be ticking over at a slower rate than it was before he left that location?

do you agree with me that if he moves to the equator he would be entitled to realize that his clock (although ticking at it's proper rate for the speed it has) would actually be ticking at a slower rate than it was at the pole?

My changes are in blue.
Is my translation of your quote correct?

On the basis of some of the responses I have received then, with trepidation - yes.

 

[Al68]

A168

For some reason I have been unable to respond to #121; when I hit the 'quote' button it brings up someone else's message however in an attempt to save time for both of us let's get back to basics.

In section 4 STR Einstein wrote -

"Thence we conclude that a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions."

I am of the opinion that by his comment "...must go more slowly..." he was saying that the equatorial clock ticks over at a slower rate than the polar clock.

Do you agree with my opinion?

Yes, but in a relative sense, not in an absolute sense. Yes, in an absolute sense if you are referring to the proper time elapsed on each clock between two specified events, since proper time is not frame dependent.

It seems like you are referring to proper time when you use the terms "real" and "physical", since proper time is not frame dependent. And it seems you are using the term "illusion" to refer to coordinate time. If that's the case, then this whole misunderstanding can be cleared up by the simple statement that the reciprocal time dilation between clocks in relative motion refers to coordinate time, not proper time.

 

[cos]

In section 4 STR Einstein wrote -

"Thence we conclude that a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions."

I am of the opinion that by his comment "...must go more slowly..." he was saying that the equatorial clock ticks over at a slower rate than the polar clock.

Do you agree with my opinion?

Yes, but in a relative sense, not in an absolute sense. Yes, in an absolute sense if you are referring to the proper time elapsed on each clock between two specified events, since proper time is not frame dependent.

I am not referring to any time "...elapsed on each clock..." nor, in my opinion, was Einstein referring to any time "...elapsed on each clock..."

Einstein was, in my opinion, stating that the equatorial clock "..must go more slowly..." than the polar clock (i.e. that the equatorial clock is ticking over at a slower rate than the polar clock) on the basis that the equatorial clock is moving relative to the 'stationary' (in an otherwise empty universe) polar clock in the same way as the clocks aboard the aircraft in the Hafele-Keating experiment were 'going more slowly' (i.e. ticking over at a slower rate) than the laboratory clocks.

It seems like you are referring to proper time when you use the terms "real" and "physical", since proper time is not frame dependent. And it seems you are using the term "illusion" to refer to coordinate time.

I believe that to be a correct assumption. My interpretation of what is 'real' or 'physical' or 'actual' or 'normal' is what takes place in an observer's reference frame not what appears to be taking place from the point of view of a person who is moving relative to that reference frame i.e. a point of view that is frame dependent.

If that's the case, then this whole misunderstanding can be cleared up by the simple statement that the reciprocal time dilation between clocks in relative motion refers to coordinate time, not proper time.

There is NO reciprocal time dilation IN Einstein's section 4!

His equatorial clock 'goes more slowly' (i.e. ticks over at a slower rate) than the polar clock! The polar clock does not, reciprocally, 'go more slowly' than the equatorial clock!

The concept of 'reciprocal time dilation between clocks in relative motion' is sections 1 through 3 of STR not section 4!

The 'reciprocal time dilation between clocks in relative motion' applies to one inertial reference frame clock that is moving past another inertial reference frame clock. There is, in Einstein's depictions, only one clock (A) that is (having accelerated) moving whilst the other clocks (B), as Einstein pointed out, have 'remained at rest'.

As you pointed out - "...the reciprocal time dilation between clocks in relative motion refers to coordinate time, not proper time." So although the other clock appears on the basis of my determinations (which are based solely on the Lorentz transformations) to be ticking over at a slower rate than my clock it is not, in reality, physically ticking over at a slower rate than my clock however, in section 4 STR Einstein stipulated that clock A is 'going more slowly' than B ergo the proper time rate of A is not the same as the proper time rate of B.

Einstein's closed curve section 4 depiction could be applied to one observer (A) stationary alongside and some distance from B.

A accelerates and, continually firing his lateral rocket, moves in a closed curve around B and, having extinguished his main drive system, is then orbiting B at v.

From B's point of view A is moving (thus incurring time dilation) but from A's point of view B is not moving! (B could be spinning on it's axis thereby consistently presenting its face to A whereupon A determines that B is not moving whilst he, on the other hand experiencing g forces determines that he is centripetally accelerating).

B has, from A's point of view, remained at rest as Einstein stipulated he does.

 

[JesseM]

I believe that to be a correct assumption. My interpretation of what is 'real' or 'physical' or 'actual' or 'normal' is what takes place in an observer's reference frame not what appears to be taking place from the point of view of a person who is moving relative to that reference frame i.e. a point of view that is frame dependent.

Isn't this second person also "an observer", so isn't what they measure in fact "what takes place in an observer's reference frame"? (this point is independent of the other point that you refuse to discuss, namely that there is no need to have a physical observer actually at rest in a given frame in order to consider how things look from the perspective of that frame, and that in SR this perspective is just as valid as the perspective of any other frame).

There is NO reciprocal time dilation IN Einstein's section 4!

His equatorial clock 'goes more slowly' (i.e. ticks over at a slower rate) than the polar clock! The polar clock does not, reciprocally, 'go more slowly' than the equatorial clock!

As has been discussed, that's probably because he was talking about total time elapsed over an entire orbit. Unless Einstein wished to deny the selfsame theory he had set out in sections 1-3 (which would be a silly way to read him), of course he would not deny that at a given instant, the polar clock ticks more slowly than the equatorial clock in the instantaneous inertial rest frame of the equatorial clock at that instant.

Einstein's closed curve section 4 depiction could be applied to one observer (A) stationary alongside and some distance from B.

A accelerates and, continually firing his lateral rocket, moves in a closed curve around B and, having extinguished his main drive system, is then orbiting B at v.

In that section Einstein also discusses the simpler example where A and B are initially at rest with respect to one another, then A is moved at constant velocity (i.e. constant speed in a straight line rather than a curve) towards B. Why aren't you willing to consider things from the perspective of A's inertial rest frame during the phase where A is moving towards B? Is A not an inertial observer?

 

[Dale]

I believe that to be a correct assumption. My interpretation of what is 'real' or 'physical' or 'actual' or 'normal' is what takes place in an observer's reference frame not what appears to be taking place from the point of view of a person who is moving relative to that reference frame i.e. a point of view that is frame dependent.

So, do I understand this correctly? If A and B are in relative motion you would describe A in terms of A's rest frame and B in terms of B's rest frame and call that "real" etc. Any description of A from B's rest frame or B from A's rest frame would be "illusion" etc. Is that correct?

I assume this is correct in the case where A and B are both inertial observers, but what about non-inertial observers? You didn't seem to like non-inertial frames earlier, so would you hold to the same definition of "real" in the case that one or both of A and B are non-inertial?

 

[phyti]

Jesse; re: post 136

In your scenario, there is a C rest frame in which A and B, separated by 60 ls (light seconds), are moving to the right at .6c, with A in the lead. A changes to zero speed in the C frame ( change made arbitrarily brief and ignored). A's clock reads t. At a previous time, using the synchronization convention of sending a signal from half the distance between A and B to both, the B clock is ahead of the A clock by xbg = 45 sec, with x=60,b=.6, g=1.25.
The separation x is not 48 because when both were moving at .6c, they were equivalent to one object, therefore as long as each has the same velocity, their spacing is constant at 60 ls. That is their rest frame spacing, and is what they would measure at any other common speed. Only outside observers measure the separation between them differently.
(This is where you recite postulate 1.)
A's clock now reads t, B's clock reads t+45.
Now is when A and B see the spacing differently, because of the speed difference.
A sees B move to him in 60/.6 = 100 sec.
Because of time dilation, B's clock advances .8*100 = 80 sec.
A's clock now reads t+100, B's clock reads t+125.
The clock that experiences the least amount of time is not the one with the smallest reading, but the one with the longest path.

"Third"? I only mentioned two frames:

According to the relativity police, when both A and B move at .6c, it must be in reference to a specific frame! Follow your own rules!

In defense of cos, your responses are long and cluttered, with side excursions to things that aren't relevant to the original question, and are actually distracting. Post 1, was a simple example with two clocks at one (approx.) location, with one moving away and returning. The question was essentially, can Einstein's statement about the time difference be taken literally.
There was no question regarding other frames or what ifs. When people ask basic questions, they need answers in terms they can understand, not a course in 4-dimensional donut theory.

Referring to post 1:
If it isn't obvious that one clock is moving relatively to the other (no other clocks are mentioned), and the difference in time readings is attributed to the motion of the 'moving' clock, and Einstein is not known to lie about scientific experimentation, then there's a problem with comprehension, and getting meaning from the context of the writing.
If I offer you $1000, and deliver it in 100's, would you reject it just because you were expecting it in 50's?
I still recommend a good dictionary as your first information source.

 

[matheinste]

In the thread "Twin paradox negation" at the end of 2008 the original discussion deteriorated into the present discussion regarding section 4 clocks. It was eventually locked by jtbell in #220 with the words
------It looks like neither side is going to budge and the participants are simply getting testier and testier, so there is no useful purpose in continuing this discussion. -------

Matheinste.

 

[JesseM]

Jesse; re: post 136

In your scenario, there is a C rest frame in which A and B, separated by 60 ls (light seconds), are moving to the right at .6c, with A in the lead.

No, I specified that A and B had an initial separation of 60 ls in frame #1 where they were initially at rest before A accelerated (read post #64 again, where I said 'Suppose for example A and B are a distance of 60 light-seconds apart in the "stationary" frame K', which Einstein had defined as the frame where A and B were initially at rest). In frame #2 where A and B are initially moving at 0.6c, the initial separation between them is not 60 ls, it is 48 light-seconds, due to length contraction.

A changes to zero speed in the C frame ( change made arbitrarily brief and ignored). A's clock reads t. At a previous time, using the synchronization convention of sending a signal from half the distance between A and B to both, the B clock is ahead of the A clock by xbg = 45 sec, with x=60,b=.6, g=1.25.

No, if the two clocks were synchronized using the Einstein synchronization convention in the frame #1 where they were at rest, then in the frame where they are both moving at 0.6c, they will be out-of-sync by 36 seconds. Where do you get the idea that they will be out-of-sync by xbg? Maybe this relates to your misunderstanding about which frame they are 60 light-seconds apart in--it's true that if two clocks were a distance of x apart in the frame where they are moving at speed b, and the clocks are synchronized in their own rest frame, then in the frame where they're moving at speed b they'll be out-of-sync by xbg/c^2 (and we're using units where c=1 here). However, if two clocks are a distance of x apart in their own rest frame, and they are synchronized in their rest frame, then in a frame where they're moving at speed b they'll be out-of-sync by xb/c^2, and again I had specified that the 60 light-second separation was in their own rest frame. You can verify that xb/c^2 is the correct formula in this case using the Lorentz transformation. Suppose that in frame #1, A is at rest at position x=0 before accelerating, and B is at rest at position x=60. A accelerates at time t=0, and at that moment A reads 0 seconds and B reads 0 seconds. As long as each clock is at rest its reading matches with coordinate time; for example, at coordinate time t=-10, A reads -10 seconds and B reads -10 seconds. After A accelerates to 0.6c, its reading no longer matches with coordinate time in this frame, but B's continues to do so since it remains at rest; at t=10 seconds B reads 10 seconds, and at t=36 seconds B reads 36 seconds.

So, the event of B reading 36 seconds happens at position x=60, time t=36 in this frame, while the event of A reading 0 seconds (and instantaneously accelerating) happens at x=0, t=0 in this frame. Now we transform to frame #2 which is moving at 0.6c relative to frame #1; in this frame A and B were initially moving at 0.6c in the -x' direction, then A came to rest while B continued to move at the same speed and eventually caught up with A. If we know the coordinates x,t of an event in frame #1 and we want to know the coordinates x',t' of the same event in frame #2, then with gamma = 1.25, the Lorentz transformation equations are:

x' = 1.25 * (x - 0.6c*t)
t' = 1.25 * (t - 0.6c*x/c^2)

If you plug in x=0 and t=0 into this transformation, for the event of A reading 0 seconds and accelerating, you get x'=0 and t'=0 in frame #2. Now plug in the event of B reading 36 seconds, which has coordinates x=60 and t=36 in frame #1. This gives:

x' = 1.25 * (60 - 0.6*36) = 1.25 * (60 - 21.6) = 1.25 * 38.4 = 48
t' = 1.25 * (36 - 0.6*60) = 1.25 * (36 - 36) = 0

So, you can see that the event of B reading 36 seconds happens at t'=0 in this frame, and is thus simultaneous with the event of A reading 0 seconds which also happens at t'=0 in this frame. And you can also see that the spatial separation between A and B at this moment is 48 light-seconds in this frame.

We could also show that at any moment prior to A's acceleration, it's still true in this frame #2 that B is 36 seconds ahead and that the two clocks are 48 light-seconds apart. For example, consider the event of A reading -100 seconds, which in the unprimed frame #1 happens at x=0 and t=-100. In the primed frame #2 the coordinates of this event are:

x' = 1.25 * (0 - 0.6*-100) = 1.25 * (60) = 75
t' = 1.25 * (-100 - 0.6*0) = 1.25 * (-100) = -125

I claim that in frame #2, this event is simultaneous with the event of B reading -64 seconds. In the unprimed frame #1, B reads -64 seconds at x=60, t=-64, so in the primed frame #2 the coordinates of this event are:

x' = 1.25 * (60 - 0.6*-64) = 1.25 * (60 + 38.4) = 1.25 * (98.4) = 123
t' = 1.25 * (-64 - 0.6*60) = 1.25 * (-64 - 36) = 1.25 * (-100) = -125

So you can see that in frame #2 these events are indeed simultaneous, since they both happen at t'=-125. You can also see that the distance between A and B at this moment is 123 - 75 = 48 light-seconds, just as before.

The separation x is not 48 because when both were moving at .6c, they were equivalent to one object, therefore as long as each has the same velocity, their spacing is constant at 60 ls. That is their rest frame spacing, and is what they would measure at any other common speed. Only outside observers measure the separation between them differently.

The frame where both are moving at 0.6c is by definition not their "rest frame"! I specified that they had a separation of 60 light-seconds in the frame #1 where they were both initially at rest until A accelerated. This means that in the frame #2 where they were both moving at 0.6c until A accelerated, their separation is 48 light-seconds. Either you just misunderstood what frame the 60 light-second figure was supposed to refer to, or you are misunderstanding something more basic about the term "rest frame" and how length in the rest frame is related to length in other frames by the length contraction equation.

"Third"? I only mentioned two frames:
1) the frame where A and B were initially at rest, then after A accelerated it was moving at 0.6c while B remained at rest
2) the frame where A and B were initially moving at 0.6c, then after A accelerated it came to rest while B continued to move at 0.6c

According to the relativity police, when both A and B move at .6c, it must be in reference to a specific frame! Follow your own rules!

Of course it's in reference to a specific frame, the second of the two frames I mentioned--that's the primed frame #2 in the Lorentz transformation above, and also the one I put in bold in the quote from the previous post. So what is the third frame that you think is needed? Perhaps you are suggesting there needs to be a third object C which is at rest in frame #2...but this would be the same mistake cos made, in SR there is absolutely no need to have an object at rest in a given frame in order to analyze things from the perspective of that frame, a "frame" is just a coordinate system for assigning space and time coordinates to events. In any case, even if we do introduce an object C which is at rest in frame #2, there are still only two inertial frames to consider, because A and B would share the same inertial rest frame before A accelerated, and A and C would share the same inertial rest frame after A accelerated.

In defense of cos, your responses are long and cluttered, with side excursions to things that aren't relevant to the original question, and are actually distracting. Post 1, was a simple example with two clocks at one (approx.) location, with one moving away and returning. The question was essentially, can Einstein's statement about the time difference be taken literally.
There was no question regarding other frames or what ifs. When people ask basic questions, they need answers in terms they can understand, not a course in 4-dimensional donut theory.

But his whole question is about where there is a real physical truth about which clock is "actually" ticking slower at a given moment. I specifically mentioned that of course there was a real physical truth about which clock ticked more in total over the course of the two clocks departing and returning, but he made clear that he did not just want to talk about total time elapsed or average rate of ticking over the course of an extended trip, he wanted to talk about the relative rate of ticking at a single instant or a very brief time-interval. And if he doesn't understand that there is no single correct answer to the question of which clock is ticking slower at a given instant--that different inertial frames disagree about which is ticking slower at a given instant (because they disagree about which clock has a greater instantaneous velocity at that instant), and that all frames' perspectives are considered equally valid, and that to try to say there's a single correct answer is equivalent to introducing some notion of absolute time, which is the opposite of what relativity says. So, I think bringing up different frames is pretty critical to making sure he's not misunderstanding something very basic about SR.

If it isn't obvious that one clock is moving relatively to the other (no other clocks are mentioned), and the difference in time readings is attributed to the motion of the 'moving' clock, and Einstein is not known to lie about scientific experimentation, then there's a problem with comprehension, and getting meaning from the context of the writing.

Again, if you read cos' subsequent posts, it's clear he's not just talking about total time elapsed on each clock, he wants to talk about whether we can say one clock is ticking slower at any given moment. And the answer in relativity is "not in any frame-independent sense; different frames have different opinions about which clock is ticking slower at a given moment, and all inertial frames are considered equally valid in SR."

 

[Al68]

Einstein was, in my opinion, stating that the equatorial clock "..must go more slowly..." than the polar clock (i.e. that the equatorial clock is ticking over at a slower rate than the polar clock) on the basis that the equatorial clock is moving relative to the 'stationary' (in an otherwise empty universe) polar clock in the same way as the clocks aboard the aircraft in the Hafele-Keating experiment were 'going more slowly' (i.e. ticking over at a slower rate) than the laboratory clocks.

Right, assuming we ignore gravity.

I believe that to be a correct assumption. My interpretation of what is 'real' or 'physical' or 'actual' or 'normal' is what takes place in an observer's reference frame not what appears to be taking place from the point of view of a person who is moving relative to that reference frame i.e. a point of view that is frame dependent.

OK, you're simply using a definition of "real" that others on this forum do not. No problem.

His equatorial clock 'goes more slowly' (i.e. ticks over at a slower rate) than the polar clock! The polar clock does not, reciprocally, 'go more slowly' than the equatorial clock!

Right, because the equatorial clock does not stay in a single inertial frame. The standard time dilation equations can't be used in the non-inertial frame that the equatorial clock remains at rest in.

Einstein's closed curve section 4 depiction could be applied to one observer (A) stationary alongside and some distance from B.

A accelerates and, continually firing his lateral rocket, moves in a closed curve around B and, having extinguished his main drive system, is then orbiting B at v.

Not if we're still ignoring gravity. Without gravity, A would have to continuously fire his rocket in order to maintain a circular path around B.

From B's point of view A is moving (thus incurring time dilation) but from A's point of view B is not moving! (B could be spinning on it's axis thereby consistently presenting its face to A whereupon A determines that B is not moving whilst he, on the other hand experiencing g forces determines that he is centripetally accelerating).

B has, from A's point of view, remained at rest as Einstein stipulated he does.

Well, notice that in the non-inertial frame that A is at rest in while circling B, the relative velocity between the two clocks is zero. Both clocks are stationary with respect to this rotating reference frame.

This would be similar to a spinning wheel in deep space with negligible gravity, and a clock in the center, and a clock attached to the rim. If we're referring to the non-inertial reference frame in which the "rim" clock is at rest, then there is no relative motion between the clocks. Both clocks are at rest in this frame. The standard SR time dilation equations can't be used. This doesn't mean there is no time dilation, it just means that you can't use the standard SR equations to analyze non-inertial reference frames. We could use the gravitational time dilation equations to calculate the difference in the rates of each clock in this frame, and we'd get the same answer as we would by considering the center clock at rest in an inertial frame and the circling clock as in relative motion. That's no coincidence.

An (accelerating) observer at rest with and local to the circling clock would observe the center clock to run faster than his own, since he is accelerating toward the center clock continuously.

 

[cos]

Einstein was, in my opinion, stating that the equatorial clock "..must go more slowly..." than the polar clock (i.e. that the equatorial clock is ticking over at a slower rate than the polar clock) on the basis that the equatorial clock is moving relative to the 'stationary' (in an otherwise empty universe) polar clock in the same way as the clocks aboard the aircraft in the Hafele-Keating experiment were 'going more slowly' (i.e. ticking over at a slower rate) than the laboratory clocks.

Right...

Having made the comment "Right" can I take it that you agree that an equatorial clock is ticking over at a slower rate than a polar clock ("...under otherwise identical conditions.")?

This would be similar to a spinning wheel in deep space with negligible gravity, and a clock in the center, and a clock attached to the rim.

We have a large wheel in space that, initially, is not spinning; ignoring any effects of the wheel's mass a clock at the center would be ticking over at the same rate as an identical clock at the rim.

The wheel starts spinning. According to Einstein's section 4 depiction - the rim clock (A) that is now moving around the central clock (B) is then 'going more slowly' (i.e. ticking over at a slower rate) than B which has, according to Einstein, remained at rest.

An observer located at the center of the wheel sees clock A ticking over at a slower rate than his own clock (and, obviously, at a slower rate than it was before the wheel started spinning).

Having read and accepted Einstein's section 4 STR depiction of a clock that is made to move in a closed curve relative to another clock (and which will, as a result, incur time dilation relative to the 'at rest' clock) observer B realizes that clock A is ticking over at a slower rate than it was before the wheel started spinning due to the fact that it is now moving whilst he has remained at rest.

I am of the opinion that observer B could be aware of the fact that if he were to move to the rim of this wheel his own clock would then also be ticking over at a slower rate than it is whilst he remains at the center of the wheel; that the 'law' of physics that caused A to start ticking over at a faster rate than it was before the wheel started spinning will also apply to his clock.

An (accelerating) observer at rest with and local to the circling clock would observe the center clock to run faster than his own, since he is accelerating toward the center clock continuously.

I am of the opinion that he does not "observe the center clock to run faster than his own, [because] he is accelerating toward the center clock continuously." but he observes the central clock to be running faster than his own clock for the simple reason that it, having remained at rest, is physically ticking over at a faster rate than his own clock.

The observer I depicted moves to the rim thus then sees the clock at the center of the wheel ticking over at a faster rate than his own clock however on the basis that he can find no reason whatsoever as to why that clock would now be ticking over at a faster rate than it was before he moved to the rim of the wheel he can only(sensibly) conclude that his clock has slowed down in the same way as did clock A when the wheel started spinning.

He should be able to realize that the 'law' of physics that caused clock A to start ticking over at a slower rate when the wheel started spinning applies equally to his clock - his reference frame.

He is unable to carry out any experiment to confirm that his clock is ticking over at a slower rate than it was prior to his relocation - his heart-beat has also slowed down as have his mental processes however his intelligence has (presumably; hopefully) not been impaired.

 

[JesseM]

Having made the comment "Right" can I take it that you agree that an equatorial clock is ticking over at a slower rate than a polar clock ("...under otherwise identical conditions.")?

Note to Al68: keep in mind that cos seems to be trying to lead people into saying that the equatorial clock is ticking slower at every instant, not just over the course of an entire orbit, and not just visually when an observer next to one of the clocks looks at the other. I'm sure you'd agree that at any given moment, we can pick an inertial frame where the equatorial clock's instantaneous velocity is smaller than the polar clock's, and that it would therefore be the polar clock that is ticking slower at this moment in this frame (which is just as good as any other inertial frame).

 

[DrGreg]

cos

The complication in considering rotating frames is that this is an advanced topic in special relativity, halfway towards the mathematics of general relativity.

I'm going to modify your scenario slightly and consider a wheel rolling along a road. As always we ignore gravity.

Clock A is fixed to the wheel rim.
Clock B is fixed to the wheel centre.
Clock C is fixed to the road.

If the wheel is stationary, all 3 clocks agree that they are ticking at the same rate as each other. Now let the wheel roll.

We consider what is happening just at the moment that the wheel rolls over clock C in such a way that clocks A and C are momentarily at the same place.

From the point of view of inertial clock B, A and C are both ticking at the same rate as each other, both slower than B.

From the point of view of inertial clock C, A is ticking at the same rate as C, B is ticking slower than A and C.

From the point of view of accelerating clock A, C is ticking at the same rate as A, B is ticking faster than A and C.

Now how do you explain those 3 points of view in terms of "physically ticking", whatever that means?

 

[Al68]

Having made the comment "Right" can I take it that you agree that an equatorial clock is ticking over at a slower rate than a polar clock ("...under otherwise identical conditions.")?

Yes, in the rest frame of the polar clock, and in the accelerated frame of the equatorial clock, if we ignore gravity. Note again that this is not standard time dilation between inertial frames. Both clocks are stationary in the "rotating" reference frame. Relative velocity between the clocks is zero in this frame.

We have a large wheel in space that, initially, is not spinning; ignoring any effects of the wheel's mass a clock at the center would be ticking over at the same rate as an identical clock at the rim.

The wheel starts spinning. According to Einstein's section 4 depiction - the rim clock (A) that is now moving around the central clock (B) is then 'going more slowly' (i.e. ticking over at a slower rate) than B which has, according to Einstein, remained at rest.

That's right.

An observer located at the center of the wheel sees clock A ticking over at a slower rate than his own clock

Yes.

(and, obviously, at a slower rate than it was before the wheel started spinning).

Yes, in the center observers rest frame. But not for an observer stationary at clock A. In his non-inertial frame, clock A runs at the same rate it always did, and the center clock B started running at a faster rate when the wheel started spinning.

Having read and accepted Einstein's section 4 STR depiction of a clock that is made to move in a closed curve relative to another clock (and which will, as a result, incur time dilation relative to the 'at rest' clock) observer B realizes that clock A is ticking over at a slower rate than it was before the wheel started spinning due to the fact that it is now moving whilst he has remained at rest.

Yes, if you're referring to the inertial frame in which clock B is spinning, and clock A is in relative motion. If you're referring to the rotating reference frame in which both clocks are stationary after the wheel starts spinning, then the difference in clock rates is not due to relative motion, since there is none. In the rotating frame, the difference in clock rates is due to proper acceleration. This is simply a matter of which perspective you choose, the result is the same.

I am of the opinion that observer B could be aware of the fact that if he were to move to the rim of this wheel his own clock would then also be ticking over at a slower rate than it is whilst he remains at the center of the wheel; that the 'law' of physics that caused A to start ticking over at a faster rate than it was before the wheel started spinning will also apply to his clock.

No single clock changes its own rate in its own rest frame ever. If it does, then it doesn't qualify as a good time keeper in SR. That being said, if clock B were moved to the rim, it would run slow relative to a third clock "C" that was at the center and remained there.

I am of the opinion that he does not "observe the center clock to run faster than his own, [because] he is accelerating toward the center clock continuously." but he observes the central clock to be running faster than his own clock for the simple reason that it, having remained at rest, is physically ticking over at a faster rate than his own clock.

In the frame of clock A, both clocks are at rest. Both clocks are stationary in the accelerated frame of clock A.

The observer I depicted moves to the rim thus then sees the clock at the center of the wheel ticking over at a faster rate than his own clock however on the basis that he can find no reason whatsoever as to why that clock would now be ticking over at a faster rate than it was before he moved to the rim of the wheel he can only(sensibly) conclude that his clock has slowed down in the same way as did clock A when the wheel started spinning.

He has plenty of basis. He is no longer in an inertial frame. He can't use the standard time dilation equations in his frame. Light doesn't even travel at c in his accelerated reference frame. The standard lorentz transformations do not apply in this frame. He can prove to himself that he is not at rest in an inertial frame by the simple fact that if he releases his clock into freefall it will not stay near him. The clock must be accelerated to stay in the frame.

He should be able to realize that the 'law' of physics that caused clock A to start ticking over at a slower rate when the wheel started spinning applies equally to his clock - his reference frame.

Again, any clock that changes its own rate in its own frame does not qualify as a valid clock in SR.

He is unable to carry out any experiment to confirm that his clock is ticking over at a slower rate than it was prior to his relocation - his heart-beat has also slowed down as have his mental processes however his intelligence has (presumably; hopefully) not been impaired.

Well, his heartrate may very well be different under different circumstances. But if he has a watch that chimes every hour according to his clock A before the wheel started spinning, it will chime every hour according to clock A afterward as well.