Not Understanding Time Dilation - mk 2.

[cmb]

OK, so what is it?

I've provided figures from someone else. I'll stick with them 'til there is a wiser head that uses that equation properly and modifies the +45,900 ns/day gravity effect/-7,200 ns/day kinematic effect.

I regret I cannot follow this because I'm asking straight questions to aid my understanding... and then I get an answer to a different question.

 

[Dale]

I regret I cannot follow this

I know. That is why I suggested multiple times that you avoid this question until you have the right background. You need to learn SR before jumping into GR. Avoid scenarios involving gravity for now.

 

[pervect]

I second Dale's the motion. If you don't understand SR first, you won't get anywhere with GR at all.

One way to avoid GR with something close to your original question is to place the experiment on a small asteroid, rather than Earth.

This avoids gravity, at the minor expense of the hovering spaceship having to do a powered orbit.

However, this is still not the simplest case to understand, the simplest and standard textbook case does not involve any accelerating clocks.

 

[cmb]

I know. That is why I suggested multiple times that you avoid this question until you have the right background. You need to learn SR before jumping into GR. Avoid scenarios involving gravity for now.

Sorry, but I am not following it because I am not getting any direct answers. I'm sure it would help if I knew more, but I want to work with some numbers first because to me that is easier to make 'real sense' out of this specific paradox than these space-time plots.

I do appreciate you trying, but the way I see it is that when we look up at a GPS satellite it is running Xns/day fast due to a lower gravity field, and Yns/day slow due to kinematic effects. If we were to look down from the satellite, we would see clocks running Xns/day slow due to a higher gravity field, and Yns/day slow due to kinematic effects.

I don't see why that is a wrong understanding, nor anything in your equation that says it is wrong, so I'm unable to progress this dialogue where there is no direct response to that essential question. The answer is yes or no, and maybe we can take it from there once there's a definite answer to whether that last paragraph is correct.

 

[Dale]

If we were to look down from the satellite, we would see clocks running Xns/day slow due to a higher gravity field, and Yns/day slow due to kinematic effects.

The problem is that this separation into gravitational and kinematical effects can only be done where the gravitational field is weak and not changing over time. In the astronaut's inertial frame the gravitational field changes over time, so you cannot decompose it like that. All you can do is evaluate the integral that I provided, which is guaranteed to agree with the previous overall calculation regardless of what coordinate system you use.

We are not trying to be evasive. I have, in fact, provided the answer, even though I knew you would not understand it. You simply need to learn SR first, you are going to be unsuccessful with the GR-first approach.

 

[Dale]

One way to avoid GR with something close to your original question is to place the experiment on a small asteroid, rather than Earth.

This avoids gravity, at the minor expense of the hovering spaceship having to do a powered orbit.

This is a good suggestion.

cmb, would you be interested in this approach?

 

[PAllen]

I undertand the question, but there again there is something about that question that makes me hesitate. One clock must be sped up relative to the other so that they stay in sync. I'm finding it tough to fathom that you can slow one OR other, whichever, and still end up with a 'consistent' time frame where both match up.

This get's at a key misunderstanding. I think both Dalespam and Pervect have answered it, but I'll try a more direct answer.

IF you want all clocks to be synchronized for a ground observer (obviously what you want for a GPS system for use by people on the ground), you speed up the orbital clocks. However, from the point of view of any orbital clock, it is not synchronized relative to a ground clock. In fact, the adjustment of the orbital clock increased de-synchronization from the point of view of the orbital clock.

If you want all clocks synchronized from the point of view of a given orbital clock, you have to speed up the ground clock. You would also have to make complex, time varying adjustments to other orbital clocks for them all to synchronized from the point of view of a given orbital clock.

I am guessing you are thinking, on some level, that synchronization is objective. Instead, it is observer dependent.

 

[cmb]

This get's at a key misunderstanding. I think both Dalespam and Pervect have answered it, but I'll try a more direct answer.

IF you want all clocks to be synchronized for a ground observer (obviously what you want for a GPS system for use by people on the ground), you speed up the orbital clocks. However, from the point of view of any orbital clock, it is not synchronized relative to a ground clock. In fact, the adjustment of the orbital clock increased de-synchronization from the point of view of the orbital clock.

If you want all clocks synchronized from the point of view of a given orbital clock, you have to speed up the ground clock. You would also have to make complex, time varying adjustments to other orbital clocks for them all to synchronized from the point of view of a given orbital clock.

I am guessing you are thinking, on some level, that synchronization is objective. Instead, it is observer dependent.

I don't think there is any misunderstanding. I have not said anything against this at all, and recognise it as the 'simple' description you'd give a first-timer at SR.

But I am taking it one step further than this. I am asking what happens when you bring those mismatched clocks back into the same gravitational and inertial frame, put them on the same table together, and look again. Now it most definitely IS objective. There is no longer any observer-dependence. So if you agree (as per my previous comments) that indeed the two clocks de-synchronise differently, my question remains - does the act of bringing them back together somehow re-synchronise them?

 

[PAllen]

I don't think there is any misunderstanding. I have not said anything against this at all, and recognise it as the 'simple' description you'd give a first-timer at SR.

But I am taking it one step further than this. I am asking what happens when you bring those mismatched clocks back into the same gravitational and inertial frame, put them on the same table together, and look again. Now it most definitely IS objective. There is no longer any observer-dependence. So if you agree (as per my previous comments) that indeed the two clocks de-synchronise differently, my question remains - does the act of bringing them back together somehow re-synchronise them?

No. If you bring an adjusted orbital clock back to Earth in some smooth way, without changing any adjustments made to it, it will have an accumulated offset and a faster rate relative to the unadjusted ground clock.

 

[cmb]

No. If you bring an adjusted orbital clock back to Earth in some smooth way, without changing any adjustments made to it, it will have an accumulated offset and a faster rate relative to the unadjusted ground clock.

...Yet for someone who has been watching this clock from the ground all the time it was up there, it had looked like it was on time all that while?

 

[ghwellsjr]

You cannot "watch" an orbiting clock from the ground and agree that it has the correct time on it. Two people at various locations on the ground would see the clock with different times on it because they would have different light travel times. The only reason GPS can be used as an accurate time keeper for everybody on Earth at the same time is because it also performs the function of positioning so the receivers know how to compensate for the light travel time. It's a very complex process and it's awesome that you can buy GPS receivers for less than $100 that tell you the time to within a fraction of a microsecond. When you consider that light travels 1000 feet per microsecond, this is quite remarkable.

In any case, once two clocks have become desynchronized, they do not automatically resynchronize to the same time just because they are colocated. Their rates will become the same so that whatever difference they have between them will remain forever, but that is not what we mean by synchronized. They also have to display the same time and have the same rate to be synchronized.

 

[cmb]

You cannot "watch" an orbiting clock from the ground and agree that it has the correct time on it. Two people at various locations on the ground would see the clock with different times on it because they would have different light travel times.

This is a thought experiment, I can set up any conditions I like: I send a satellite into a perfect circle at a constant distance from my perfectly round earth, and I take a signal from it each time it is directly overhead. I know the distance and can adjust for ToF of the signal perfectly.

If I adjust the clock I send into orbit such that it always matches my clock on Earth when it comes overhead, then the clocks remain perfectly matched - for however long the satellite stays up there. However, for someone floating with the clock they observe the clocks increasingly desynchronise.

Just before the clock begins its decent back to earth, on the ground I am reading that the clocks are perfectly synchronised, because that is how I have set them up in my thought experiment. But to the guy floating with the clock, they look desynchronised because they have always been increasing their desynchronisation. For me, I monitor the clock coming back down to Earth and would I notice that they become more desynchronised as the clock gets closer to me, and if so by how much? That act of the clock coming back to me surely cannot change the time on the clock by a variable amount, according to how long the satellite has been up there. So what do I see the radio signal, from the clock as it is coming back down to earth, telling me? Do I see the time signal slowing down considerably for the period of its decent, so that, by the time it gets to me, it matches the desynchronised time the guy floating with it saw and expects to see when he gets down here?

If so, why would the rate of 'correction' during the same descent path be different depending on whether I bring the satellite down after one month compared with 10 years?

If not, then the clocks will read the same time and the guys who was with the clocks gets real confused?

 

[PAllen]

...Yet for someone who has been watching this clock from the ground all the time it was up there, it had looked like it was on time all that while?

Yes, while it was in orbit. It would start going out of synch, from ground observer's point of view, when its motion and position in gravity well changed. If it looked in synch for some position (in gravity well) and state of motion, why would one expect it to remain so when these things change?

 

[cmb]

If it looked in synch for some position (in gravity well) and state of motion, why would one expect it to remain so when these things change?

Please see my post which hit the board at the same time as yours.

I have no problem with the synch looking different once it is returned to earth, but why would it be a variable amount of synch according to what has happened in the previous years?

If you visited me in my control shack and we watched my satellite clock just before its return to earth, you'll say 'when that gets back here, it'll be desyncronised because its changed its gravtiy field, etc.' and I say 'Yup, sure will. But, tell me, by how much will it be desynchronised once I bring it back down this particular flght path?'

Y'see, at that moment in time, the 'actual' (or 'latent', if you will) desychronisation is independent of how long it has been up there, but this is inconsistent with the observations of the guy floating along with it who has watched the desynch build up.

 

[ghwellsjr]

This is a thought experiment, I can set up any conditions I like: I send a satellite into a perfect circle at a constant distance from my perfectly round earth, and I take a signal from it each time it is directly overhead. I know the distance and can adjust for ToF of the signal perfectly.

If I adjust the clock I send into orbit such that it always matches my clock on Earth when it comes overhead, then the clocks remain perfectly matched - for however long the satellite stays up there. However, for someone floating with the clock they observe the clocks increasingly desynchronise.

Just before the clock begins its decent back to earth, on the ground I am reading that the clocks are perfectly synchronised, because that is how I have set them up in my thought experiment. But to the guy floating with the clock, they look desynchronised because they have always been increasing their desynchronisation. For me, I monitor the clock coming back down to Earth and would I notice that they become more desynchronised as the clock gets closer to me, and if so by how much? That act of the clock coming back to me surely cannot change the time on the clock by a variable amount, according to how long the satellite has been up there. So what do I see the radio signal, from the clock as it is coming back down to earth, telling me? Do I see the time signal slowing down considerably for the period of its decent, so that, by the time it gets to me, it matches the desynchronised time the guy floating with it saw and expects to see when he gets down here?

If so, why would the rate of 'correction' during the same descent path be different depending on whether I bring the satellite down after one month compared with 10 years?

If not, then the clocks will read the same time and the guys who was with the clocks gets real confused?

It is possible to construct a pair of clocks as you described so that they both keep exactly the same time on them all the time and in synchronization if you are willing to calculate out the ToF. In general, the orbiting clock will be time dilated because of its speed and the Earth clock will be time dilated because of gravity. If the amount of time dilation is the same then you don't have to do anything to make one run faster than the other. Then when you bring the orbiting clock down to the surface of your earth, it might be possible that they could remain in sync. If the amount of time dilation is different, then you would have to make one of the clocks run faster to be able to make them keep the same time. Then when you bring them together, they will start going out of sync.

I hope this is correct, I may have overlooked something. And remember, this is on a fictitious Earth of your choosing that may be a different size, mass, spin rate, than our real earth.

 

[PAllen]

I have no problem with the synch looking different once it is returned to earth, but why would it be a variable amount of synch according to what has happened in the previous years?

I don't think this would happen with such an adjusted clock. The time reading deviation would only begin accumulating from when its motion from what it was specifically adjusted for.

 

[cmb]

I don't think this would happen with such an adjusted clock. The time reading deviation would only begin accumulating from when its motion from what it was specifically adjusted for.

But this is the whole point of my question. The kinematic effects* would mean that from the satellite you'd see the Earth clock falling behind but also the satellite clock has already been adjusted to run fast aswell. Why would a satellite observer not see a kinematic desynch?


*[we take it as read that the Gravity effects are known and compensated - they are not observer dependent insofaras the observers know what g field they're in]

 

[PAllen]

But this is the whole point of my question. The kinematic effects* would mean that from the satellite you'd see the Earth clock falling behind but also the satellite clock has already been adjusted to run fast aswell. Why would a satellite observer not see a kinematic desynch?


*[we take it as read that the Gravity effects are known and compensated - they are not observer dependent insofaras the observers know what g field they're in]

The satellite observer and a funny adjusted clock are two different things. The satellite observer would see an accumulating difference between the funny clock and a normal clock on the satellite (one brought from the ground without any adjustments). If the satellite had a funny clock and a regular clock, then when the satellite was brought down:

- the regular clock would now be going at the same rate as the ground clock, but with a time reading difference proportional to the orbit time.

- the funny clock would now be going fast compared to the ground clock, with a time difference accumulating only from when the satellite's motion changed from what the funny clock was adjusted for.

 

[ghwellsjr]

If I adjust the clock I send into orbit such that it always matches my clock on Earth when it comes overhead, then the clocks remain perfectly matched - for however long the satellite stays up there. However, for someone floating with the clock they observe the clocks increasingly desynchronise.

I wonder why you think an observer traveling with the orbiting clock would think that it increasingly desynchronises with the ground clock. I'll bet you're thinking that when two clocks are traveling inertially with respect to each other (no gravity) then they each observe that they are running slower than the other and therefore increasingly desynchronize, right? And so now you're thinking that the same thing happens (neglecting gravity and GR) with an orbiting clock and a stationary clock on the surface of earth, is that what you're thinking?

If so, then you have overlooked the fact that the orbiting clock is not inertial, it is always accelerating toward the Earth so in this case (neglecting rotation of the surface of the Earth where the "stationary" clock is), one clock is inertial and the other is not. This means that the orbiting clock will be time dilated while the Earth clock is not and so everytime the orbiting clock is overhead of the Earth clock, they will each "see" that the time dilation is occurring on just the orbiting clock.

An observer traveling with the orbiting clock will not see the two clocks desynchronising if they have been adjusted so that they are keeping the same time. In fact no observer will see them desynchronising.

Does this help?

 

[Dale]

But this is the whole point of my question. The kinematic effects* would mean that from the satellite you'd see the Earth clock falling behind but also the satellite clock has already been adjusted to run fast aswell. Why would a satellite observer not see a kinematic desynch?


*[we take it as read that the Gravity effects are known and compensated - they are not observer dependent insofaras the observers know what g field they're in]

As I said before, you cannot separate the kinematic and gravitational effects in a time varying field like in the satellite frame.

 

[Dale]

One other thing to note, the separation of time dilation into kinematic and gravitational components is only an approximation, even in a non-time varying gravitational field. It is an approximation that works reasonably well for the GPS satellites, but not in general. The exact formula is not separable. See equations 3 and 4 at:
http://en.wikipedia.org/wiki/Time_dilation#Time_dilation_due_to_gravitation_and_motion_together

Generally you need to evaluate the integral I gave above in your coordinate system of choice.

 

[cmb]

One other thing to note, the separation of time dilation into kinematic and gravitational components is only an approximation, even in a non-time varying gravitational field. It is an approximation that works reasonably well for the GPS satellites, but not in general. The exact formula is not separable. See equations 3 and 4 at:
http://en.wikipedia.org/wiki/Time_dilation#Time_dilation_due_to_gravitation_and_motion_together

I'm looking at equation 4, in which the approximation separates out the terms, and it says underneath "For applications near the Earth this approximation will introduce an error on the order of .76 nanoseconds per century." In my examples, I'm discussing 7,200ns per DAY. So the approximation is entirely satisfactory and represents an effect less than one millionth of the effect I am discussing.

But thanks for directing me to that, as I now know that arguing that my scenario fails, on the basis that it is unreasonable to separate the terms, is not a legitimate argument.

I feel like I am asking reasonable questions, and I'm being thrown curve-balls back to dissuade me from enquiring into the detail. Well, thanks to that equation I can now see that the 'combined-effect-disargument' of gravitational and kinematic effects in my example is a red-herring.

 

[PAllen]

I'm looking at equation 4, in which the approximation separates out the terms, and it says underneath "For applications near the Earth this approximation will introduce an error on the order of .76 nanoseconds per century." In my examples, I'm discussing 7,200ns per DAY. So the approximation is entirely satisfactory and represents an effect less than one millionth of the effect I am discussing.

But thanks for directing me to that, as I now know that arguing that my scenario fails, on the basis that it is unreasonable to separate the terms, is not a legitimate argument.

I feel like I am asking reasonable questions, and I'm being thrown curve-balls back to dissuade me from enquiring into the detail. Well, thanks to that equation I can now see that the 'combined-effect-disargument' of gravitational and kinematic effects in my example is a red-herring.

Maybe it would help to describe exactly what you are asking. Since every responder has (in my opinion) tried to be helpful, your feeling that your questions are not being answered suggests a complete disconnect between what you think you are asking and what others understand of your questions. Try a fresh, complete, precise as possible, statement of what you are asking.

From my point of view, it seems that what I would see as direct answers produce responses: not what I was asking, misleading, etc.

One thing I can tell you is that outside of the specific application of GPS, no papers or books on GR I've read ever bother to separate kinematic and gravitational effects. It is not worth it, as straighforward integrations of the line element give you your measured quantities for arbitrary situations.

 

[cmb]

Maybe it would help to describe exactly what you are asking. Since every responder has (in my opinion) tried to be helpful, your feeling that your questions are not being answered suggests a complete disconnect between what you think you are asking and what others understand of your questions. Try a fresh, complete, precise as possible, statement of what you are asking.

Yes, I was thinking on trying to frame it as a different example. Seems I have introduced what are considered too many variables.

Indeed, I do appreciate all the comments so far. I am happy to say again; thanks for your time responding. I recognise there is no thought to try to dismiss my question, but that it looks like it is not a very obvious paradox. After all, if it were obvious it'd have already been discussed many times!

I'll chew it over and see if I can put it into a different form away from gravity or orbits.

Bear with me..Thanks again!

 

[ghwellsjr]

Maybe it would help to describe exactly what you are asking.

I think he did that in post #42 and I responded to him in post #45 and #49 but he has not reacted to my posts.

I'm guessing that he thinks the orbiting clock is like the traveler in the Twin Paradox whose age difference is indeterminate until he comes back to Earth and then all of a sudden there's this vast age difference. So even if the orbiting clock were adjusted so that it looks like it's keeping time with the Earth clock, when it comes down from orbit, he thinks, it will all of a sudden have this huge time shift.

 

[Dale]

I'm looking at equation 4, in which the approximation separates out the terms, and it says underneath "For applications near the Earth this approximation will introduce an error on the order of .76 nanoseconds per century." In my examples, I'm discussing 7,200ns per DAY. So the approximation is entirely satisfactory and represents an effect less than one millionth of the effect I am discussing.

Yes, which is why the approximation is used in GPS.

But thanks for directing me to that, as I now know that arguing that my scenario fails, on the basis that it is unreasonable to separate the terms, is not a legitimate argument.

I feel like I am asking reasonable questions, and I'm being thrown curve-balls back to dissuade me from enquiring into the detail. Well, thanks to that equation I can now see that the 'combined-effect-disargument' of gravitational and kinematic effects in my example is a red-herring.

Apparently you didn't read the very next sentence where it says "Unfortunately equation (4) is inconsistent with special relativity". Regardless of the accuracy in one frame (ground frame), if the equation is inconsistent with special relativity then you can get big errors in another frame (astronaut frame).

This is not a red-herring, you only feel like you are asking reasonable questions because you are unfamiliar with the material and seem to be completely unwilling to listen to the good advice and responses that you have received from multiple well-informed people.

 

[Dale]

cmb, pay attention to this comment:

One thing I can tell you is that outside of the specific application of GPS, no papers or books on GR I've read ever bother to separate kinematic and gravitational effects. It is not worth it, as straighforward integrations of the line element give you your measured quantities for arbitrary situations.

 

[cmb]

This is not a red-herring, you only feel like you are asking reasonable questions because you are unfamiliar with the material

Yes, that is a possibility and I'll see if I can provide a 'refreshed scenario'. I thought I'd been clear before, but it seems not. Maybe in doing so I will see my error and realize the issue for myself, if not I'll post it.

...and seem to be completely unwilling to listen to the good advice and responses that you have received from multiple well-informed people.

That's a bit unfair. What makes you say that if I disagree my point has been answered it means I am not listening? It might mean I am wrong and that I simply don't understand my own question, but I assure you I am reading every word to see if it answers my questions.

Better you say "..and it seems that you are prepared to be completely wrong on something just to see if there is something new to find out, even though everyone else is saying something different". I'd agree with that!

It is a non-sequitur to argue someone is not listening to you because they don't agree with you.

 

[cmb]

Apparently you didn't read the very next sentence where it says "Unfortunately equation (4) is inconsistent with special relativity". Regardless of the accuracy in one frame (ground frame), if the equation is inconsistent with special relativity then you can get big errors in another frame (astronaut frame).

OK, message heard! I'll see if I can reframe it away from any gravity.

 

[ghwellsjr]

OK, message heard! I'll see if I can reframe it away from any gravity.

Before you do that, would you please respond to my comments regarding your previous scenario in post #42. My comments are in posts #45, #49 and #55.