The Twins Paradox: Exploring Time Dilations' Effects

In summary, the conversation discusses the differences between the time dilation effects in the GPS system and the common twins paradox scenario. While both involve time dilation, the GPS system has a real and verifiable time dilation that is asymmetric, while the twins paradox involves an undefined or unverifiable time dilation for the outgoing trip. The conversation also explores the impact of gravity and inertial frames on these scenarios.
  • #1
Buckethead
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TL;DR Summary
Twins paradox and GPS equivalent?
In a thread I started awhile back, in the common twins paradox scenario, it was indicated to me that the actual time (paraphrasing) on Earth for any given time in the ship is basically “undefined” (as it can’t be verified) and/or time dilated (ticking slower) for the trip out and then shifted to having advanced quickly for the trip back resulting in an older twin on Earth. In other words, it will be symmetrical (time dilation as seen from either twin) but not “real” or “verifiable” until the trip is complete in which case there in an asymmetry in rate of time resulting in the Earth twin being older (due to the frame change of the space ship).

However….In a GPS system, the time dilation is real and verifiable and is asymmetric. The orbiting satellite clock ticks more slowly, all the time, and this can be measured at any time. In addition from the point of view of the satellite the Earth clock is ticking faster and this also can be measured. Because of this the GPS clock is adjusted for the speed difference until the clocks both tick one second per second Earth time.

As a side note, both the satellite and (if you like) even the ground clock are inertial if you drop the ground clock during this experiment at the moment of dropping, although it’s not necessary since being non inertial or inertial for the ground clock has no effect on the rate of the clock when the clock is not moving (just dropped for example or sitting on the ground). Just trying to eliminate that possible objection).

So what I’d like to know is, if the GPS scenario and the twins paradox (for outgoing trip only) are equivalent, then the outgoing ship should see the Earth ticking faster, not slower, and the clock should be measurable or in the very least calculatable as it is in GPS. However it is not depicted this way. Why?
 
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  • #2
Height GR effect surpass moving satellite SR effect about double so GPS clock tick faster than our clock. Does it matter to your discussion?
 
  • #3
Buckethead said:
if the GPS scenario and the twins paradox (for outgoing trip only) are equivalent
They are not equivalent. During the outgoing trip, excluding initial acceleration and turnaround, both twins are in inertial frames. In the GPS case, the satellite is in an inertial frame but the ground observer is not. If the ground observer falls then maybe the scenario will be symmetric during the period of the fall, until significant air resistance builds up. But that period can only be very short - a second or two at most!
 
  • #4
anuttarasammyak said:
Height GR effect surpass moving satellite SR effect about double so GPS clock tick faster than our clock. Does it matter to your discussion?
I should have specified that I am ignoring gravity for the GPS system for simplification and because it is an additive effect that can be removed from the discussion.
 
  • #5
andrewkirk said:
They are not equivalent. During the outgoing trip, excluding initial acceleration and turnaround, both twins are in inertial frames. In the GPS case, the satellite is in an inertial frame but the ground observer is not. If the ground observer falls then maybe the scenario will be symmetric during the period of the fall, until significant air resistance builds up. But that period can only be very short - a second or two at most!
Yes, as I mentioned one can perform the GPS scenario with the ground clock in an inertial state simply by dropping the clock and before it accelerates. The time is short, but in principle it does not affect the experiment. In addition gravity effect can simply be subtracted if you want to leave the clock on the ground and the premise still holds. i.e. the clock discrepancies can be calculated and adjusted and without gravity the Earth based clock will tick faster, not slower as seen by the sattelite as predicted by SR.
 
  • #6
Buckethead said:
I should have specified that I am ignoring gravity for the GPS system for simplification and because it is an additive effect that can be removed from the discussion.

You can't just use part of the laws of physics. They all fit together. You can't just arbitrarily remove one and expect to get correct answers. If you want a scenario without gravity, you need to specify a scenario without gravity--a scenario that takes place in empty space, far away from all gravitating bodies. For example, one spaceship floating in space and a second ship using its rockets to orbit in a circle around the first one.
 
  • #7
Buckethead said:
In a thread I started awhile back, in the common twins paradox scenario

Please give a link. First, you shouldn't force others to have to look up the thread. Second, you should not be paraphrasing what other people said in the thread; you should be linking to it so everyone else can read what was said for themselves.
 
  • #8
PeterDonis said:
You can't just use part of the laws of physics. They all fit together. You can't just arbitrarily remove one and expect to get correct answers. If you want a scenario without gravity, you need to specify a scenario without gravity--a scenario that takes place in empty space, far away from all gravitating bodies. For example, one spaceship floating in space and a second ship using its rockets to orbit in a circle around the first one.
OK, let's leave gravity in since we have a similar situation with the twins. For GPS we have a ground based clock, non-inertial, in a gravitaitonal field. In the twins paradox we have the Earth bound twin with a ground based clock, non-inertial, in a gravitational field. To make the Satallite/ship more equivalent, we can start with a ship already up to speed, before passing Earth to synchronize his clock.
 
  • #9
PeterDonis said:
Please give a link. First, you shouldn't force others to have to look up the thread. Second, you should not be paraphrasing what other people said in the thread; you should be linking to it so everyone else can read what was said for themselves.
Fair enough. I was hoping that everyone would just think "Yea, ok it's undefined, I'm good with that".
 
  • #10
Buckethead said:
I should have specified that I am ignoring gravity for the GPS system for simplification and because it is an additive effect that can be removed from the discussion.
I am afraid whether your scenario goes with the observation that in order two to meet again at least one cannot stay still in a IFR.
 
  • #11
Buckethead said:
In the twins paradox we have the Earth bound twin with a ground based clock, non-inertial, in a gravitational field.

No. The standard twin paradox is set in flat spacetime--no gravity. If the term "Earth" appears, it is only used as a label for a point in space, not as a source of gravity. And the "Earthbound" twin is assumed to be in free fall--inertial--the whole time.
 
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  • #12
PeterDonis said:
No. The standard twin paradox is set in flat spacetime--no gravity. If the term "Earth" appears, it is only used as a label for a point in space, not as a source of gravity. And the "Earthbound" twin is assumed to be in free fall--inertial--the whole time.
No kidding!? I've never heard these stipulations implied or stated, but OK. Is the twin paradox a useful thought experiment if we allow for my scenario? i.e. Earthbound stuck to a real Earth. If we have real twins on a real planet Earth and one flys off at near light speed, will he return younger? If so, then let's keep that scenario.

Also with regard to my reference. I'm searching but having no luck (not very good at research). So I'll just simply ask instead of using a reference, Can the time on the Earthbound clock be calculated or determined to be ticking at a lower rate at any given time between the time the ship passes the Earth for synchronization and when it turns around? If the answer to this is no, then my original question needs to be re-stated as to why can it be calculated in the GPS scenario, but not in the twins scenario. If the answer is yes, then then why is it different (clock ticks slower) for the twins scenario, but from the frame of the satallite the Earth clock is ticking faster? If the answer is still yes but the Earth clock is ticking faster in the twins scenario, then why doesn't this violate SR symmetry (i.e. according to SR it should be ticking more slowly) ?
 
  • #13
Buckethead said:
In a thread I started awhile back, in the common twins paradox scenario, it was indicated to me that the actual time (paraphrasing) on Earth for any given time in the ship is basically “undefined” (as it can’t be verified) and/or time dilated (ticking slower) for the trip out and then shifted to having advanced quickly for the trip back resulting in an older twin on Earth.
This sounds like a mix between replies describing direct observation of the clock by telescope (apparent tick rate depends on direction of travel due to Doppler) and replies calculating the tick rate in an inertial frame (not actually "undefined", but a matter of your choice of simultaneity criterion).

The same is true of GPS satellites. If you monitor the signal sufficiently precisely you'll see Doppler as they orbit above you, but you need to choose a simultaneity criterion to work out what their clocks show "now". However, the presence of gravity supplies a natural simultaneity criterion. You aren't obligated to use it, but I suspect that most descriptions will do so.

As Peter says, references to what you've been reading would help. We can point out what's wrong, or where people are talking about different things, or where there are unstated assumptions.
 
  • #14
Buckethead said:
Fair enough. I was hoping that everyone would just think "Yea, ok it's undefined, I'm good with that".
If I guess correctly at the sort of thing you have in mind, it is "undefined within limits". And yes, I'm good with that.

The GPS satellites are never more than a few hundred light-milliseconds away from each other and from Earth-bound stations. This places a bound of a few hundred milliseconds plus or minus on the range of ambiguity in clock synchronization. Any "reasonable" standard of synchronization (e.g. Einstein synch with some arbitrarily selected standard of rest) must place the "actual" time at the remote clock somewhere within that range.
 
  • #15
Buckethead said:
No kidding!? I've never heard these stipulations implied or stated, but OK.
The usual twin scenario has significantly relatistic speeds (e.g. ##0.6c##), where time dilation and differential ageing are significant. By contrast, any gravitational effects would be negligible compared to the relative-velocity-based effects.
 
  • #16
Buckethead said:
Can the time on the Earthbound clock be calculated or determined to be ticking at a lower rate at any given time between the time the ship passes the Earth for synchronization and when it turns around?
The total elapsed proper time for each twin at the reunion event is what it is. No ambiguity.

However, the relative clock rate "at any given time" during the trip is not measurable without assuming a synchronization convention. Many conventions are possible. So the result is ambiguous.
 
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  • #17
Buckethead said:
Summary:: Twins paradox and GPS equivalent?

In a thread I started awhile back...

if the GPS scenario and the twins paradox (for outgoing trip only) are equivalent, then ... However it is not depicted this way. Why?
Buckethead said:
I should have specified that I am ignoring gravity for the GPS system
Buckethead said:
one can perform the GPS scenario with the ground clock in an inertial state simply by dropping the clock and before it accelerates
This is a terrible basis for a thread. First you allude to but do not properly reference the previous discussion. Then you want to know why the GPS is not depicted some way, but you want to modify the GPS to remove gravity and change the ground observer to a free falling observer.

How is anyone supposed to answer such a question? We don't know the previous discussion. We know that most GPS discussions involve gravity but not if the "not depicted this way" description was. Ditto for ground observers not in free fall.

I honestly have no clue how anyone is supposed to answer such a question. If we remove gravity and have the ground observer accelerate then how are we answering the question about why GPS is not depicted that way. If we answer the question about why GPS is not depicted that way, then we have to violate both of those conditions. Either way, how do we answer without giving the appearance of disagreement with some previous thread that may have had an entirely different basis of discussion?

I predict that this discussion will either leave you confused or everyone else frustrated or both.
 
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  • #18
Dale said:
This is a terrible basis for a thread. First you allude to but do not properly reference the previous discussion. Then you want to know why the GPS is not depicted some way, but you want to modify the GPS to remove gravity and change the ground observer to a free falling observer.
My apologize. Please see post #14 for corrections to my original question. I'll edit my original question to be more succint this afternoon when I get more time and address your issues.
 
  • #19
Buckethead said:
I'll edit my original question

Please don't. Post a correction instead. Changing the question almost 20 messages in is confusing. Some might even say antisocial.
 
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  • #20
Buckethead said:
No kidding!? I've never heard these stipulations implied or stated, but OK. Is the twin paradox a useful thought experiment if we allow for my scenario? i.e. Earthbound stuck to a real Earth. If we have real twins on a real planet Earth and one flys off at near light speed, will he return younger? If so, then let's keep that scenario.

Also with regard to my reference. I'm searching but having no luck (not very good at research). So I'll just simply ask instead of using a reference, Can the time on the Earthbound clock be calculated or determined to be ticking at a lower rate at any given time between the time the ship passes the Earth for synchronization and when it turns around? If the answer to this is no, then my original question needs to be re-stated as to why can it be calculated in the GPS scenario, but not in the twins scenario. If the answer is yes, then then why is it different (clock ticks slower) for the twins scenario, but from the frame of the satallite the Earth clock is ticking faster? If the answer is still yes but the Earth clock is ticking faster in the twins scenario, then why doesn't this violate SR symmetry (i.e. according to SR it should be ticking more slowly) ?
For the moment, we will assume no gravity. This means that the GPS satellite needs to have something else ( like a rocket engine) to provide the centripetal force needed to keep it circling the Earth. This also means that the GPS has a constant acceleration vector pointing at the Earth.
The twin flying past the Earth, is starting in an inertial frame, and since it and the Earth have a relative velocity with respect to each other, it measures the Earth clock as ticking slower than his own. This remains true for both outbound and return legs of the trip, as long as he is inertial.

However, at the end of the outbound trip, he needs to fire his engines to first reduce his relative velocity with respect to the Earth to zero, and then increase it again in order to start the return trip.
During which period, he also will have an acceleration vector that points towards the Earth.

Both the GPS clock and the twin while accelerating are in non-inertial frames. And measurements* made from non-inertial frames are subject to different rules than those made from inertial frames.

In addition to the time dilation measured due to relative motion between observer and clock, Its position with respect to the observer relative to the acceleration vector also has a determining effect. Clocks that are in the direction of the acceleration are measured to run fast, and clocks int he opposite direction run slow. The rate difference is determined by the magnitude of the acceleration and the distance as measured parallel to the acceleration vector. The greater the distance, the larger the tick rate difference.

So with the GPS satellite, the Earth observer measures the GPS clock as ticking slow because it has a relative motion with respect to the Earth.
The GPS satellite measures the Earth clock as ticking fast, because it is located some distance away in the direction of the acceleration, and the satellite is measuring from an non-inertial (rotating) frame. The two measurements agree with each other as the difference in ticks rates, but attribute it to different causes.

With the Twin, the Earth observer measure the twin's clock as running slow by a constant rate during both inbound and outbound legs and by a varying rate during the "turnaround" phase.

The twin measures the Earth clock as ticking slow during outbound leg.
Then once the acceleration phase starts an new factor is added. The long distance to Earth, and the magnitude of the acceleration tends to increase the measured tick rate. As the ship reduces its relative velocity with respect to the Earth, the time dilation factor decreases. The distance to the Earth continues to increase until the ship come to a rest with respect to it, so the factor causing the measurement of the Earth clock tick rate to be faster, increases. If it didn't start out doing so at the start of the acceleration period, The factor increasing the Earth clock tick rate will begin to win out. With the net result of the Earth clock more than making up the deficit in time it accumulated during the outbound trip.

Then the rocket begins to build speed with respect to the Earth, while decreasing the distance between them. The time dilation factor increases, and the factor due to measuring from an accelerating frame decreases. This will add even more to the amount of time that the Earth clock accumulates. Once the ship has gotten back up to its original relative speed to the Earth( but now heading towards it), the acceleration quits, the Twin is left with just the relative velocity time dilation factor affecting his measurement of the Earth clock.

The Earth clock runs slow until the twin and Earth meets up again. However the total amount the Earth clock gained during the acceleration phase of the trip supersedes the amount it loose during the inertial legs and the twin returns to find that the Earth Clock has accumulated more time than his own during their separation.
Again Both Earth and twin agree on the differences in accumulated times, but attribute it to different causes. The difference between this and the GPS example above, is that in the GPS example both observers agree that the ticks rate difference was constant at all times, while with the Twin example, the Earth observer says that the Twin clock ran slow the whole time (except for that brief moment when the twin was at rest with respect to the Earth), while the Twin will say that the Earth Clock went from running slow, to running fast, and then back to slow again.

* by "Measurement" I don't mean direct observation, but the determination made after factoring out light propagation delay effects (Doppler, etc.)
 
  • #21
Buckethead said:
I've never heard these stipulations implied or stated

Then you haven't read any standard twin paradox scenarios very carefully.

Buckethead said:
If we have real twins on a real planet Earth and one flys off at near light speed, will he return younger?

Yes. But as @PeroK pointed out, the effect of the Earth's gravity on the results is negligible. That's why the standard twin paradox scenarios do not even bother to include gravity and just assume flat spacetime everywhere.

Buckethead said:
why doesn't this violate SR symmetry

The "SR symmetry" you refer to only applies if both observers are always, for all time, at rest in one particular inertial frame (a different such frame for the two observers, of course, or they wouldn't be in relative motion).

Both the standard twin paradox and the "observer moving in a circle around another one that remains at the center" scenario violate this requirement. So there is no "SR symmetry" in either scenario.

In the standard twin paradox, the traveling twin is the one that is not at rest for all time in one particular inertial frame. The stay-at-home twin is.

In the "observer moving in a circle around another one that remains stationary" scenario, the "moving in a circle" observer is the one that is not at rest for all time in one particular inertial frame. The observer that remains at the center is.

In both these scenarios, the observer that is at rest for all time in one particular inertial frame is the one whose clock will show the most elapsed time between meetings of the two observers.

However, the analysis I just gave above is limited; it only applies in flat spacetime, and only if one observer is at rest for all time in one particular inertial frame and the other isn't.

The completely general method of analysis is to use geometry. In the standard twin paradox, the traveling twin's trajectory is two sides of a triangle in spacetime, and the stay-at-home twin's trajectory is the third side. A simple application of the spacetime version of the triangle inequality shows that the third side of this triangle has a length greater than the sum of the other two side lengths. (It's "greater than" instead of "less than" because of the minus sign in the spacetime interval formula.)

In the "moving in a circle" scenario, the trajectory of the observer moving in a circle is a helix in spacetime, and that of the observer at the center is a straight line. A somewhat more complicated calculation shows that the length of the helix is less than the length of the straight line. "Length" here is arc length along timelike curves, so it is the same as proper time, i.e., the time elapsed on the clock of an observer following that curve in spacetime. So "greater length" means "more elapsed time on the observer's clock".

So geometry gives the right answer for these cases; but it also allows you to analyze cases where the kind of simple analysis I gave at the start of this post can't even be applied--for example, cases where no observer is always at rest in anyone inertial frame, and cases where spacetime curvature--gravity--cannot be ignored.
 
  • #22
First off, thank you everyone for somehow managing to wade through my very messy and poorly written question. I was actually a little confused about what I wanted to ask and my question was a result of that. Regardless my "pseudo questions" resulted in replies from all of you that were helpful.

Janus said:
In addition to the time dilation measured due to relative motion between observer and clock, Its position with respect to the observer relative to the acceleration vector also has a determining effect. Clocks that are in the direction of the acceleration are measured to run fast, and clocks int he opposite direction run slow. The rate difference is determined by the magnitude of the acceleration and the distance as measured parallel to the acceleration vector. The greater the distance, the larger the tick rate difference.
PeterDonis said:
he "SR symmetry" you refer to only applies if both observers are always, for all time, at rest in one particular inertial frame (a different such frame for the two observers, of course, or they wouldn't be in relative motion).

Both the standard twin paradox and the "observer moving in a circle around another one that remains at the center" scenario violate this requirement. So there is no "SR symmetry" in either scenario.

In the standard twin paradox, the traveling twin is the one that is not at rest for all time in one particular inertial frame. The stay-at-home twin is.

In the "observer moving in a circle around another one that remains stationary" scenario, the "moving in a circle" observer is the one that is not at rest for all time in one particular inertial frame. The observer that remains at the center is.

In both these scenarios, the observer that is at rest for all time in one particular inertial frame is the one whose clock will show the most elapsed time between meetings of the two observers.

The part I didn't realize was that an orbiting satellite, even though inertial is not in an unchanging frame. I should have realized this since the "triplets paradox" (one outgoing ship and a second returning ship that stay inertial but pass each other to pass clock times) also suffer from this change in frame and therefore violate the symmetry. So that really gets to the heart of my question which was why does an outgoing inertial ship and an earthbound twin see each others clocks as running slowly whereas an inertial orbiting satellite sees a faster earthbound moving clock. The constant change in frame (of the satellite) is the culprit.

One thing that still confuses me is that (for example) in a triplet paradox, the change in frame between ship 1 (outgoing ship) and ship 2 (incoming ship) is an "unnatural" shift in frames, and by that I mean the ships are going in two different directions in space and in spacetime. In an orbiting satellite, the direction through spacetime remains constant and indeed is a straight line through spacetime, which it seems is similar to an inertial ship going in one direction in flat spacetime.

One thing I can see is that speed is limited for an inertial satellite in a given gravitational field. If the orbit is too fast, the orbit breaks, so a relativistic speed can never be achieved without going non-inertial (using retro jets for example). This means the gravity will always be a significant factor for an orbiting object. Is this the reason for the asymmetry? Or is it just that gravity in general without regard for the balance between speed and gravity screws with symmetry (and causing the frame change in and of itself) and my objection based on one way inertial direction through spacetime is invalid. It is gravity alone that is screwing with the symmetry. In other words, one way inertial motion through spacetime is irrelevant and the speed/gravity balance even though relevant is insignificant.
 
  • #23
Buckethead said:
So that really gets to the heart of my question which was why does an outgoing inertial ship and an earthbound twin see each others clocks as running slowly whereas an inertial orbiting satellite sees a faster earthbound moving clock. The constant change in frame (of the satellite) is the culprit.
Let spaceship and earthbound peoples speak what they want to. There is no contest because they will not meet again forever. On the other hand we are meeting with orbiting satellite periodically. We, Earth people and satellite pilots, have to get an agreement in comparing the clocks. Whose FR is a IFR is the point (ref post #10).
 
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  • #24
Buckethead said:
In an orbiting satellite, the direction through spacetime remains constant and indeed is a straight line through spacetime, which it seems is similar to an inertial ship going in one direction in flat spacetime.
Orbiting means going round like kids on merry-go-round. Its orbit is not a straight line but a circle. Not an inertial motion.
 
  • #25
Buckethead said:
an orbiting satellite, even though inertial is not in an unchanging frame.

Careful. There are at least two problems with putting it this way.

First, all objects are always "in" all "frames". Objects don't disappear from a frame just because they aren't at rest in that frame.

Second, the kind of "frame" you are talking about is, specifically, an inertial frame. There are also non-inertial frames.

A better way of putting what you are trying to say here is that an orbiting satellite, even though it is in free fall (zero proper acceleration), is not always at rest in a single inertial frame.

In fact, in curved spacetime, i.e., in the presence of gravitating bodies like the Earth, there is no such thing as a global inertial frame at all. There are only local inertial frames. More on that below.

Buckethead said:
The constant change in frame (of the satellite) is the culprit.

No, that can't be the case, because, as I just pointed out above, in the presence of a gravitating body like the Earth, nothing can be always at rest in a single inertial frame, because there aren't any global inertial frames at all. The orbiting satellite isn't always at rest in a single inertial frame, but neither is the Earth. So "change in frames" cannot be "the culprit" in this case.

The only type of case where not being always at rest in a single inertial frame is "the culprit" is cases like the standard twin paradox case as I described it earlier: flat spacetime, no gravitating masses anywhere, and the stay-at-home twin just floats freely in space while the traveling twin goes out, turns around, and comes back. The other case I described earlier, where one spaceship floats freely in empty space while a second spaceship uses its rockets to orbit in a circle around the first spaceship, is also of this type. But no other types of cases follow the simple rule that you are trying to use. In particular, the satellite orbiting the Earth is not the same as the two spaceships case I just described. It just isn't; and your attempts to treat it as if it were don't accomplish anything except to confuse the issue.

Buckethead said:
in a triplet paradox, the change in frame between ship 1 (outgoing ship) and ship 2 (incoming ship) is an "unnatural" shift in frames

This is a perfect example of why "shifting frames" is not a good way of looking at it. The spacetime geometry analysis I gave earlier solves this case easily: it's the same triangle in spacetime as the standard twin paradox, just with two different spaceships doing the outbound and inbound legs instead of one. And note that that analysis says nothing about "frames" at all.

Buckethead said:
In an orbiting satellite, the direction through spacetime remains constant and indeed is a straight line through spacetime, which it seems is similar to an inertial ship going in one direction in flat spacetime.

Only in the sense that both are geodesics. But that in itself isn't enough to make the cases work the same, because they are geodesics in different spacetime geometries. And the comparative elapsed proper times for different observers that meet up with each other multiple times will depend on the spacetime geometry, not just whether their particular worldlines are geodesics or not.

Buckethead said:
the gravity will always be a significant factor for an orbiting object. Is this the reason for the asymmetry?

No. You are failing to grasp the key point: other than the general spacetime geometry analysis I have described, there is no single rule you can apply that will explain the asymmetry in all cases. There just isn't. Looking for one is a waste of time. It isn't there.
 
  • #26
Buckethead said:
This means the gravity will always be a significant factor for an orbiting object. Is this the reason for the asymmetry? Or is it just that gravity in general without regard for the balance between speed and gravity screws with symmetry
I recommend you to forget about gravity at first. Say we are in a IFR. Satellite is in circular motion of tangential speed v. SR says the relation of its proper time ##\tau## to ours t is
[tex]t = \frac{\tau}{\sqrt{1-\frac{v^2}{c^2}}}...(1)[/tex]
This is the basic result. Then you may consider additional effects as you like e.g., mass, orbiting, spinning of the Earth, Earth radius, satellite's orbit radius or height, etc.

Though "additional" I said, in actual parameters of GPS Earth gravity effect by height difference almost minus doubles the effect of (1) so Earth clock goes slower (ref. post #2).
 
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  • #27
Buckethead said:
In an orbiting satellite, the direction through spacetime remains constant and indeed is a straight line through spacetime, which it seems is similar to an inertial ship going in one direction in flat spacetime.
The geodesic worldlines in curved spacetime are only locally equivalent to a straight line through flat spacetime. Globally you cannot ignore the overall spacetime geometry when computing accumulated proper time.

This can be visualized for 1D motion (radial fall), but not so good for orbits.
 
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  • #28
PeterDonis said:
A better way of putting what you are trying to say here is that an orbiting satellite, even though it is in free fall (zero proper acceleration), is not always at rest in a single inertial frame.

This is a key fact I had failed to grasp and this really helps. I have been thinking this whole time that something like an orbiting object following a geodesic through spacetime is at rest. (note to self: must remember not all IFR's are at rest!)

PeterDonis said:
Only in the sense that both are geodesics. But that in itself isn't enough to make the cases work the same, because they are geodesics in different spacetime geometries. And the comparative elapsed proper times for different observers that meet up with each other multiple times will depend on the spacetime geometry, not just whether their particular worldlines are geodesics or not.

This is another key point to remember. (note to self: IFR's can have different spacetime geometries!)

I was thinking about the simplest example of an orbiting satellite and two clocks, one of which is placed at the center of an ideal planet (perfectly round, even density) and the other on the satellite in a perfectly circular orbit.

In such a case can I assume that there is no relative velocity between the satellite and the fixed clock? And if so, can I then surmise that we can leave out effects of SR (since there is no relative motion) and simply view the situation as two clocks in different spacetime geometries (requiring only GR to solve) and further surmise from this that there is an asymmetry in clock rates? It seems like a natural conclusion but just want to be sure.

As a side question, is there a gravitational field at the center of this ideal planet. The field must be null, but probably not non-existant and as a result must be considered as having a different spacetime geometry than flat space.
 
  • #29
Buckethead said:
This is a key fact I had failed to grasp and this really helps. I have been thinking this whole time that something like an orbiting object following a geodesic through spacetime is at rest. (note to self: must remember not all IFR's are at rest!)
This is another key point to remember. (note to self: IFR's can have different spacetime geometries!)

I was thinking about the simplest example of an orbiting satellite and two clocks, one of which is placed at the center of an ideal planet (perfectly round, even density) and the other on the satellite in a perfectly circular orbit.

In such a case can I assume that there is no relative velocity between the satellite and the fixed clock? And if so, can I then surmise that we can leave out effects of SR (since there is no relative motion) and simply view the situation as two clocks in different spacetime geometries (requiring only GR to solve) and further surmise from this that there is an asymmetry in clock rates? It seems like a natural conclusion but just want to be sure.

As a side question, is there a gravitational field at the center of this ideal planet. The field must be null, but probably not non-existant and as a result must be considered as having a different spacetime geometry than flat space.
By jumping into GR before you have got the basics of SR you have got yourself in an almighty tangle here! What a confusion of ideas. I'm not sure where to start to try to disentangle that.
 
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  • #30
Buckethead said:
This is a key fact I had failed to grasp and this really helps. I have been thinking this whole time that something like an orbiting object following a geodesic through spacetime is at rest. (note to self: must remember not all IFR's are at rest!)
What does a claim that an inertial frame of reference is "at rest" even mean?

I think your best bet is to ignore General Relativity entirely and work on getting Special Relativity (or even Newtonian mechanics) firmly understood first.
 
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  • #31
Buckethead said:
note to self: must remember not all IFR's are at rest!)
Back up a bit, to “must learn what a frame is”. A frame, whether inertial or not, is a rule for assigning coordinates to events. Saying that a frame is or is not at rest makes no more sense than saying that the Pythagorean Theorem is at rest - it’s not a thing that moves.
 
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  • #32
PeroK said:
By jumping into GR before you have got the basics of SR you have got yourself in an almighty tangle here! What a confusion of ideas. I'm not sure where to start to try to disentangle that.

Maybe I'm just thinking of GR or rather relativity in a gravitational field as indicated by my example as being just another straightforward effect with regard to clock rates. For example in SR you can calculate time dilation using the Lorentz formula, very simple. Is it not also the case that given a pure example of two objects not moving relative to each other in a gravitational field an easily calculatable situation if you want to know how the two clocks vary over time? Or is it far far more complex than that?

jbriggs444 said:
What does a claim that an inertial frame of reference is "at rest" even mean?

From what I've learned in this thread, to me it means an IFR that is in flat spacetime. i.e. you can use the Lorentz transform to calculate differences in clock rates.

jbriggs444 said:
I think your best bet is to ignore General Relativity entirely and work on getting Special Relativity (or even Newtonian mechanics) firmly understood first.

Not sure my brain is that capable :)
 
  • #33
Buckethead said:
Is it not also the case that given a pure example of two objects not moving relative to each other in a gravitational field an easily calculatable situation if you want to know how the two clocks vary over time? Or is it far far more complex than that?
Before analysing one of these problems, I suggest you pick a coordinate system and stick with it. Part of the problem is that you jump between an undefined global coordinate system and local (inertial) coordinate systems.

The lack of global inertial reference frames in GR means you need an even more disciplined approach than you do in SR.
 
  • #34
Nugatory said:
Back up a bit, to “must learn what a frame is”. A frame, whether inertial or not, is a rule for assigning coordinates to events. Saying that a frame is or is not at rest makes no more sense than saying that the Pythagorean Theorem is at rest - it’s not a thing that moves.

As in "from Bob's frame of reference or Alice's frame of reference"? But what do you call something that is an IFR that is not changing direction? Can't you call that a frame at rest? What would you call it?
 
  • #35
PeroK said:
Before analysing one of these problems, I suggest you pick a coordinate system and stick with it. Part of the problem is that you jump between an undefined global coordinate system and local (inertial) coordinate systems.

The lack of global inertial reference frames in GR means you need an even more disciplined approach than you do in SR.

In a simple relative motion case I think of frames like this: From Bob's frame, Alice's clock is ticking more slowly. From Alice's frame, Bob's clock is ticking more slowly. So, yes, I guess I can see, in my example of a satellite orbiting a planet, I would have to do the same thing. From the planet bound clock, the orbiting clock does this, but from the orbiting clock, the planet bound clock does that.
 

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