Undergrad What I understand about Time Dilation

[PeterDonis]

Would an observer in an accelerating rocket not notice a clock at the front of the rocket going faster than at the rear

Yes, but that's because the front of the rocket is higher up in the "gravitational field" produced by the rocket's acceleration. It's not because of the acceleration itself (since the difference in clock rates will be present even if the acceleration is the same in the front and the rear).

So it is not the case that the velocity of the satellite going in a circular orbit creates an acceleration equal to the acceleration of gravity in an opposite direction to counter balance it, so that if the velocity were 0 the satellite would fall to the Earth, and if it were considerably greater than its orbit velocity it would leave orbit and go off into space?

Both of these "accelerations" are coordinate accelerations and depend on your choice of coordinates. Note that I said proper acceleration in my quote: that is the acceleration that you actually feel and measure with an accelerometer, and it is also the kind that the equivalence principle is talking about. A satellite in orbit feels no acceleration.

 

[name123]

Both of these "accelerations" are coordinate accelerations and depend on your choice of coordinates. Note that I said proper acceleration in my quote: that is the acceleration that you actually feel and measure with an accelerometer, and it is also the kind that the equivalence principle is talking about. A satellite in orbit feels no acceleration.

Ok, thanks.

 

[name123]

The General Relativity equation does not include a velocity term. Please look at the wiki link posted.

Thanks I had overlooked the part below.

I'm not sure I would use the words "relative", "correct", and "wrong" in the way you did. I would say the time dilation is relative (edit: better term provided: frame dependent), it's just not symmetrical if the situation isn't symmetrical. And nothing needs to be wrong with any of this as long as it is calculated correctly!

I assume with 2 satellites in free fall orbit at the same velocity and altitude but in opposite directions around a sufficiently large glass sphere, that they would observe each other's clocks to be going slower than their own (due to velocity time dilation), and also to observe a clock on the Sphere to be going slower than they would expect given gravity time dilation alone. The situation would be symmetrical for both of the orbiting satellites would it not?

Though presumably each time they pass each other they would notice that their clocks are actually the same, and that would be further confirmed when they are brought to the surface of sphere, where they would further notice that the clock on the sphere had actually been going faster than they would have expected given gravity time dilation alone. So they would notice a distinction between the appearance of time dilation which is relative, and the actual difference in the rate their clocks ticked.

 

[jartsa]

Would an observer in an accelerating rocket not notice a clock at the front of the rocket going faster than at the rear, as one might if the rocket were standing up on an object which had an appropriate gravitational field?

An inertial observer might say that the observer in the accelerating rocket gets fooled into thinking that the clock at the front of the rocket is going faster than the clock at the rear of the rocket by the way that light travels between the rear and the front.

So this is very "relative".If in an accelerating rocket a clock is moved next to another clock, then everyone must agree about the rate difference and the difference of the readings of the two clocks. So now the inertial observer must agree with the observer inside the rocket. So the two observers must agree that a front clock has ticked faster than a former rear clock that was at some time moved from the rear to the front.

So this is "absolute".

 

[PeterDonis]

With 2 satellites orbiting at the same velocity opposite ways around a sufficiently large glass sphere would observe each other's clocks to be going slower than their own,

Based on measurements over a period of time much shorter than one orbit, yes.

Though presumably each time they pass each other they would notice that their clocks are actually the same

Yes.

that would be further confirmed when they are brought to the surface of sphere, where they would further notice that the clock on the sphere had actually been going faster than they would have expected given gravity time dilation alone

No. Assuming the clock on the sphere is at rest on the sphere, it would have been going exactly as fast as gravity time dilation predicts. And faster than the clocks on the satellites (that is, if the satellites were both launched from rest on the sphere, made a bunch of orbits, and then came back and landed at rest on the sphere, the clock that stayed at rest on the sphere the whole time would show more elapsed time).

So they would notice a distinction between the appearance of time dilation which is relative, and the actual difference in the rate their clocks ticked.

Yes.

 

[name123]

An inertial observer might say that the observer in the accelerating rocket gets fooled into thinking that the clock at the front of the rocket is going faster than the clock at the rear of the rocket by the way that light travels between the rear and the front.

So this is very "relative".If in an accelerating rocket a clock is moved next to another clock, then everyone must agree about the rate difference and the difference of the readings of the two clocks. So now the inertial observer must agree with the observer inside the rocket. So the two observers must agree that a front clock has ticked faster than a former rear clock that was at some time moved from the rear to the front.

So this is "absolute".

So the truth of the situation would not be relative, the inertial observer would have initially been wrong.

 

[name123]

No. Assuming the clock on the sphere is at rest on the sphere, it would have been going exactly as fast as gravity time dilation predicts. And faster than the clocks on the satellites (that is, if the satellites were both launched from rest on the sphere, made a bunch of orbits, and then came back and landed at rest on the sphere, the clock that stayed at rest on the sphere the whole time would show more elapsed time).

But it would not be at rest from the satellites' frame of reference would it? I would have thought that the satellites would have thought it was going slower than gravity time dilation would predict, because they would also observe velocity time dilation. When they returned to the sphere though, they would notice that they were wrong about the velocity time dilation of the sphere clock, and that it was actually their clocks that had being going slower due to velocity time dilation. So that the while they had been correct on how much time the sphere clock had lost due to gravity time dilation, they were wrong about how much they thought it was losing to velocity time dilation, actually it had gained.

 

[Janus]

So once returned the proper time that has passed for ship C is less than the proper time that has passed for Earth. Its clocks have objectively "ticked" less than those on Earth, regardless of how it appeared to ship C. Not quite clear on how from C's perspective the Earth's clocks suddenly seem to have been ticking faster after all, and not have done 0.8 the amount of "ticks" that had happened on C. If you can explain that, that would be great, if not, it doesn't matter.

It's not that The Earth's clock "seem to be ticking fast after all"., It's that by C's measurement, the Earth clocks ran really fast during the short time C was under acceleration And ran slow the rest of the time.

But if for ship C its clocks have objectively ticked less than those on Earth's. Then its clock was ticking slower as it had appeared to the observer on Earth.

As far as C is concerned, it's clock only ticked slow relative to the Earth clock during the period it was under acceleration, during the rest of the time it ran faster.

Given that C's journey consisted of an outward leg, during which it's clocks were in synch with those of A and an inward leg during which it's clocks were in synch with those of B, then can one not conclude that like C the clocks on A and B were objectively ticking slower than those on Earth?

No. As measured by the Earth, the clocks on A, B and C all run slow at the same rate while C is cruising. There will be a short period (for our purposes we will treat is as being infinitesimally short) where C when from cruising speed to zero speed and back up to cruising speed, during which time C's clock rate varied between running slow and running at the same rate as the Earth clock.)
As measured by clock A, the Earth clocks always ran slow by a fixed rate, and the clock on C ran at the same rate as theirs for part of the trip and ran really slow (even slower than Earth's clocks) for the rest of the trip

For B, Earth's clocks ran slow at the same rate for the whole time, and C's clock ran really slow for the first part of the trip then ran at the same rate as theirs for the rest of the trip.
For C the Earth clock ran slow during the outbound leg, ran extremely fast during turnaround, and then ran slow again during the return trip.
So while everyone agrees as to what the respective time it is on Earth and on C's clock when they meet up again, they don't agree as to exactly how this came about.
And there is no objective way to say what really happened. In other words, each of these views of what happened during the course of the trip is just as valid as any other. You can't ever say that one clock "actually" ran fast or slow compared to another at any given point of the trip.

Supposing A goes out along side C, and passes B at the point C turns around, and as B passes A it sets its clock to the time on A's and comes back along side C. If one were to assume almost instantaneous acceleration and deceleration for C, then presumably its clock would be roughly in synch with A's on the outward journey and in synch with B's on the inward journey, and the time on B's, like C's would be less than the clock it passes on Earth as C returns. Unlike C, A and B would be in inertial frames.

Now I have to bring the third "leg" of Special Relativity into play, The Relativity of Simultaneity. In a nutshell, it means that events that are simultaneous in one frame, are not going to be simultaneous for another frame in motion with respect to the first frame, if the events are separated along the line of relative motion.

As an example. Imagine that we put a clock at the turn around point, that has been synchronized with the Earth clock according Earth's and the turn around point's frame. For C and A, when the clock on Earth reads 0, the clock at the turnaround already reads 0.36 yrs. For ship B, when the clock at the turn around reads 0, the clock on Earth already reads 0.36 yr. Thus for ship's A and C, when A leaves Earth at time 0, the clock at turn around reads 0.36 yr, and in the time it takes (0.8 yr by A and C's clocks) this clock runs at a rate of 0.8 and accumulates 0.64 yr to read 1 yr, while their own clocks read 0.8 yr.
Ship B is just passing the turnaround at this moment, and takes the reading of 0.8 yr onto its own clock. For him, it will also be 1 yr at the turnaround clock, but it will be 1.36 yrs at the Earth clock. He will continue on towards Earth, adding another 0.8 yr to his clock, while the Earth clock runs slow and accumulates 0.64y. His clock will read 1.6 yr and the Earth clock will read 2 yrs upon his arrival.

Note that according to ships A and C, when they reach the turnaround point and the clock there reads 1 yr, the Earth clock at that moment only reads 0.64 years, while according to B, when it passes ship A at the turnaround point, the Earth clock already read 1.36 yr. Relativity of Simultaneity in action. For Ships A and C, the events of the Earth clock reading 0.64 and the turnaround clock reading 1 yrs are simultaneous. But for ship B, they are not, for him the simultaneous events are the turnaround clock reading 1yr and the Earth clock reading 1.36 yr.

Unfortunately, when people first start learning about SR, they usually start with time dilation. The problem is that SR only really makes sense if you also include length contraction and the relativity of simultaneity. And of the three, I really think that the relativity of simultaneity should be the one you start with. Once you grasp this idea, the other two fall into place very easily.

 

[jartsa]

So the truth of the situation would not be relative, the inertial observer would have initially been wrong.

Well, the inertial observer says that moving the clock from the rear to the front caused the clock to (velocity) time dilate.

 

[PeterDonis]

So the truth of the situation would not be relative, the inertial observer would have initially been wrong.

No, not wrong, just talking about a different thing.
w.
Suppose the two satellites are just passing each other. By exchanging light signals as they pass, they can confirm that each of them sees the other's clock running slow. But by comparing their total elapsed time from the last time they passed to this time (i.e., over one complete orbit), they can also confirm that their elapsed times per complete orbit are the same. Both of these observations are perfectly "true"; they're just observations of different things.

it would not be at rest from the satellites' frame of reference would it?

The two satellites do not have a "frame of reference" in which they are both at rest while they are orbiting. And neither of the satellites' "frames of reference" work the same as a frame of reference in special relativity, or the same as the "frame of reference" of the clock at rest on the sphere. See below.

I would have thought that the satellites would have thought it was going slower than gravity time dilation would predict, because they would also observe velocity time dilation.

In the satellites' frames of reference (either one), there is no well-defined "gravity time dilation". One way to see this is via the equivalence principle: the satellites are in free fall, feeling no acceleration, so there is no "acceleration time dilation" in their frames. And by the EP, there is thus no "gravitational time dilation" in their frames either, because there is no "gravitational field". If an observer on the satellite releases an object, it doesn't fall; it floats right next to the observer.

By contrast, in the frame of the clock at rest on the sphere, there is a "gravitational field" and there is gravity time dilation, because the clock feels acceleration--and if an observer next to the clock releases an object, it falls.

The only useful concept of "gravity time dilation" that the satellites can use is the same one as the clock at rest on the sphere uses, and according to that concept, the clock at rest on the sphere is ticking at exactly the rate that "gravity time dilation" predicts.

 

[Janus]

But gravity or acceleration time dilation is objective regarding the slowing of the proper time. But velocity time dilation is supposed to be relative is it not? I have presumed that it is and that there are no objective results regarding the slowing of proper time for velocity time dilation. With the muon experiments are the muons not in a circular orbit, and thus undergoing acceleration?

For gravitational time dilation, there is some objectivity, If I place a clock higher in a gravity well it will tick slower than a lower one, and both clocks will agree.
But for acceleration, this only holds if both clocks are undergoing the same acceleration. Then the "lead" clock will run slower than the "trailing" clock.
However, if only one clock is accelerating, then this objectivity can go out the window.
If you You are accelerating away from me, I will measure your clock as running slow as a result of your velocity with respect to me at any given instant. Your acceleration only counts in that it is what changes the velocity.
However, you will measure my clock as running doubly slow. In part because of the increasing velocity between us and in part due to my position relative to you due to your acceleration and the magnitude of your acceleration.
If you were accelerating towards me, I still would only measure your clock as running slow due to velocity, but what you would measure for my clock would be a combination of running slow due to relative velocity and running fast due to the distance between us and the magnitude of you acceleration. Whether this results in net increase or decrease in my clock rate according to you depends on the exact parameters of the scenario.

 

[name123]

Given that C's journey consisted of an outward leg, during which it's clocks were in synch with those of A and an inward leg during which it's clocks were in synch with those of B, then can one not conclude that like C the clocks on A and B were objectively ticking slower than those on Earth?

No. As measured by the Earth, the clocks on A, B and C all run slow at the same rate while C is cruising. There will be a short period (for our purposes we will treat is as being infinitesimally short) where C when from cruising speed to zero speed and back up to cruising speed, during which time C's clock rate varied between running slow and running at the same rate as the Earth clock.)
As measured by clock A, the Earth clocks always ran slow by a fixed rate, and the clock on C ran at the same rate as theirs for part of the trip and ran really slow (even slower than Earth's clocks) for the rest of the trip

But as I also wrote:

Supposing A goes out along side C, and passes B at the point C turns around, and as B passes A it sets its clock to the time on A's and comes back along side C. If one were to assume almost instantaneous acceleration and deceleration for C, then presumably its clock would be roughly in synch with A's on the outward journey and in synch with B's on the inward journey, and the time on B's, like C's would be less than the clock it passes on Earth as C returns. Unlike C, A and B would be in inertial frames.

So if A's clock ticks are roughly in synch with C's on the outward journey (and then continue at the same rate) and B's clock ticks are roughly in synch with C's on the inward journey (and then continue at the same rate) then the amount of "ticks" on C's clock would be roughly equal to the amount A's clock ticked on the outward journey and the amount B's ticked on the inward journey. And since this is less than the clock ticked on Earth, why would it be wrong to conclude that like C the clocks on A and B were objectively ticking slower than those on Earth?

In another conversation another scenario was considered, one in which there are 2 satellites in free fall orbit at the same velocity and altitude but in opposite directions around a sufficiently large glass sphere. They would observe each other's clocks to be going slower than their own (due to velocity time dilation) but when pass each other again they would observe that actually the clock measurements were objectively the same. So while there relative observations were different (when they both observed the other's clock to be going slower), they are able to objectively tell that those observations were did not imply that the other's clock was actually going slower. So while I can understand that A would think the clock on Earth was running slow, given that is was in synch with C's and C's was found to be have objectively ran slower than that on Earth, then I am not sure why you think it would be wrong to conclude that A's was running slower also.

 

[russ_watters]

I assume with 2 satellites in free fall orbit at the same velocity and altitude but in opposite directions around a sufficiently large glass sphere, that they would observe each other's clocks to be going slower than their own (due to velocity time dilation)...

The situation would be symmetrical for both of the orbiting satellites would it not?

Yes. Please note that in orbit, half the time they are approaching each other and half the time they are moving apart. So the time dilation wrt each other oscillates around zero.

...and also to observe a clock on the Sphere to be going slower than they would expect given gravity time dilation alone.

No: Their acceleration means that they and the ground station are not doing similar things. Objectively, the satellite "knows" its clock is the slow one.

So they would notice a distinction between the appearance of time dilation which is relative, and the actual difference in the rate their clocks ticked.

No, the science works if they use it correctly. If it didn't, GPS would not work.

 

[name123]

The two satellites do not have a "frame of reference" in which they are both at rest while they are orbiting. And neither of the satellites' "frames of reference" work the same as a frame of reference in special relativity, or the same as the "frame of reference" of the clock at rest on the sphere. See below.

So one cannot use the special relativity equations considering one to be at rest to establish the velocity time dilation that they will observe?

No, not wrong, just talking about a different thing.
w.
Suppose the two satellites are just passing each other. By exchanging light signals as they pass, they can confirm that each of them sees the other's clock running slow. But by comparing their total elapsed time from the last time they passed to this time (i.e., over one complete orbit), they can also confirm that their elapsed times per complete orbit are the same. Both of these observations are perfectly "true"; they're just observations of different things.

It would be true what they experienced, but it wouldn't be true that they actually did observe the other's clock going slower, because it didn't go slower. They would pass each other twice per complete orbit, and their clocks would be the same at the other side also.

In the satellites' frames of reference (either one), there is no well-defined "gravity time dilation". One way to see this is via the equivalence principle: the satellites are in free fall, feeling no acceleration, so there is no "acceleration time dilation" in their frames. And by the EP, there is thus no "gravitational time dilation" in their frames either, because there is no "gravitational field". If an observer on the satellite releases an object, it doesn't fall; it floats right next to the observer.

By contrast, in the frame of the clock at rest on the sphere, there is a "gravitational field" and there is gravity time dilation, because the clock feels acceleration--and if an observer next to the clock releases an object, it falls.

The only useful concept of "gravity time dilation" that the satellites can use is the same one as the clock at rest on the sphere uses, and according to that concept, the clock at rest on the sphere is ticking at exactly the rate that "gravity time dilation" predicts.

The observers on the satellites could know the mass of the sphere, and thus how much gravity time dilation they would expect for the clock on it. Would they not also experience the velocity time dilation effect so experience the sphere clock going even slower than they would have expected given gravity time dilation alone?

If so then my point was that when they return to the sphere they would notice that rather than the sphere clock going slower than theirs due to velocity time dilation, it was theirs' going slower, and so the sphere clock would appear to have gone faster than they would have expected given gravity time dilation alone.

 

[name123]

Yes. Please note that in orbit, half the time they are approaching each other and half the time they are moving apart. So the time dilation wrt each other oscillates around zero.

But with special relativity wouldn't the other satellite's clock appear to be going slower whether it is coming towards it or going away?

No: Their acceleration means that they and the ground station are not doing similar things. Objectively, the satellite "knows" its clock is the slow one.

They aren't doing similar things, but would the satellite not observe the clock on the sphere to be going slower than it would expect simply because of gravitational time dilation, because would an observer on the satellite not observe a velocity time dilation effect?

No, the science works if they use it correctly. If it didn't, GPS would not work.

I wasn't suggesting that the science didn't work, I just thought that the adjustment for the GPS would take into account which clock was objectively going slower.

 

[Nugatory]

So one cannot use the special relativity equations considering one to be at rest to establish the velocity time dilation that they will observe?

The special relativity equations that you are thinking of can only be used in a local inertial frame (although if there were no gravity, the situation you find in most intro textbooks, "local" would extend out to infinity).

Extreme care is required when applying them in any other situation. Before you even try to take on the complexities of gravitational time dilation you should try to understand the Twin Paradox, as your counter-orbiting satellites are just a variation of that classic problem. Start with http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

But with special relativity wouldn't the other satellite's clock appear to be going slower whether it is coming towards it or going away?

What you SEE is the clock running slow when it is moving away from you,and running fast when it is moving away. This is the Doppler effect, described in that link above. You only realize that the other clock is running consistently slow relative to yours when you correct for the time that it took for light to make it from the clock to your eye; if the clock is one light-second away from you the time you SEE is what it read one second ago when the light started towards you.

 

[name123]

The special relativity equations that you are thinking of can only be used in a local inertial frame (although if there were no gravity, the situation you find in most intro textbooks, "local" would extend out to infinity).

Extreme care is required when applying them in any other situation. Before you even try to take on the complexities of gravitational time dilation you should try to understand the Twin Paradox, as your counter-orbiting satellites are just a variation of that classic problem. Start with http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

The twin paradox involves a turnaround, and accelerations etc., and an actual time difference when the clocks reunite. Here though there is no actual time difference when the clocks meet. The situation is symmetrical for both satellites, the situation is not symmetrical in the twin paradox.

What you SEE is the clock running slow when it is moving away from you,and running fast when it is moving away. This is the Doppler effect, described in that link above. You only realize that the other clock is running consistently slow relative to yours when you correct for the time that it took for light to make it from the clock to your eye; if the clock is one light-second away from you the time you SEE is what it read one second ago when the light started towards you.

But neither's clock was running slow when compared to the other's (each time they pass they can notice the same time has passed on each of their clocks). Even though when observing each other bounce torch beams off their mirrored ceilings they might assume the other's clock is running slower.

 

[PeterDonis]

So one cannot use the special relativity equations considering one to be at rest to establish the velocity time dilation that they will observe?

No, because special relativity only applies in flat spacetime, i.e., in the absence of gravitating masses. Here there is a gravitating mass present so spacetime is not flat.

(If spacetime were flat, two objects both in free fall could only meet once. So the fact that the two satellites, both in free fall, meet multiple times is enough to show that spacetime cannot be flat.)

They would pass each other twice per complete orbit, and their clocks would be the same at the other side also.

Yes, that's true, but it's also true that, while passing each other both times, they are in relative motion, and so each would see the other to have velocity time dilation.

The observers on the satellites could know the mass of the sphere, and thus how much gravity time dilation they would expect for the clock on it.

Yes, but this is using the "gravity time dilation" of the clock on the sphere. It is not using any concept of "gravity time dilation" that applies in the satellites' own frames.

Would they not also experience the velocity time dilation effect so experience the sphere clock going even slower than they would have expected given gravity time dilation alone?

No, because the "gravity time dilation" they are calculating doesn't apply in the satellites' own frames, as above, but the velocity time dilation is relative to the satellites' own frames. So there's no way to combine the two as you are describing.

 

[Nugatory]

The twin paradox involves a turnaround, and accelerations etc., and an actual time difference when the clocks reunite. Here though there is no actual time difference when the clocks meet. The situation is symmetrical for both satellites, the situation is not symmetrical in the twin paradox.

It's still a twin paradox situation, because it maintains the apparent paradox...

But neither's clock was running slow when compared to the other's (each time they pass they can notice the same time has passed on each of their clocks). Even though when observing each other bounce torch beams off their mirrored ceilings they might assume the other's clock is running slower.

Both correctly find that the other clock is running slower than theirs always, just as with the classic version of the twin paradox. In the classic paradox, the traveller is surprised to find that the Earth twin ends up older even though the Earth twin's clock was running slower throughout; here each satellite bserver is surprised to find that the twin on the other satellite has aged equally even though their clock was running slower for the entire time of separation. And the resolution is the same in both cases: the difference in clock rate (time dilation) does not lead to apparently obvious conclusion about the total elapsed time.

 

[russ_watters]

But with special relativity wouldn't the other satellite's clock appear to be going slower whether it is coming towards it or going away?

No, moving apart the other's clock appears slow and back together it appears to speed up(as someone else noted, this is more than just time dilation).

They aren't doing similar things, but would the satellite not observe the clock on the sphere to be going slower than it would expect simply because of gravitational time dilation, because would an observer on the satellite not observe a velocity time dilation effect?

Yes, sorry, I missed the switch to GR there.

I wasn't suggesting that the science didn't work, I just thought that the adjustment for the GPS would take into account which clock was objectively going slower.

I guess I'm not following, but what you said before appears to contain contradictions:

I assume with 2 satellites in free fall orbit at the same velocity and altitude but in opposite directions around a sufficiently large glass sphere, that they would observe each other's clocks to be going slower than their own...

Though presumably each time they pass each other they would notice that their clocks are actually the same...

So they would notice a distinction between the appearance of time dilation which is relative, and the actual difference in the rate their clocks ticked.

How can you "notice" something different from what you "observe"?

 

[name123]

No, because special relativity only applies in flat spacetime, i.e., in the absence of gravitating masses. Here there is a gravitating mass present so spacetime is not flat.

But in the Haele-Keating experiment did they not use the general relativity equations to work out the gravitational time dilation and the special relativity equations to work out the velocity time dilation? And when working out the velocity time dilation for the westbound direction, did they not do it from the plane's perspective?

(If spacetime were flat, two objects both in free fall could only meet once. So the fact that the two satellites, both in free fall, meet multiple times is enough to show that spacetime cannot be flat.)

I wasn't thinking that the spacetime was flat, only that scientists had used the special relativity equations for similar situations.

Yes, that's true, but it's also true that, while passing each other both times, they are in relative motion, and so each would see the other to have velocity time dilation.

So how could you work out what velocity time dilation they observed? Would it be much different from using the special relativity equations, because presumably if they were bouncing torch beams off mirrored ceilings there wouldn't be much difference from whether they were in flat space passing each other, or making an orbit around a huge sphere.

Yes, but this is using the "gravity time dilation" of the clock on the sphere. It is not using any concept of "gravity time dilation" that applies in the satellites' own frames.

Yes I was assuming they would work it out for the clock on the sphere.

No, because the "gravity time dilation" they are calculating doesn't apply in the satellites' own frames, as above, but the velocity time dilation is relative to the satellites' own frames. So there's no way to combine the two as you are describing.

I would expect them to be able calculate the gravity time dilation for the clock on the sphere the same as you could. And because the satellite is in relative motion to the sphere clock, also see it as having velocity time dilation.

 

[name123]

No, moving apart the other's clock appears slow and back together it appears to speed up(as someone else noted, this is more than just time dilation).

As Nugatory mentioned

What you SEE is the clock running slow when it is moving away from you,and running fast when it is moving away. This is the Doppler effect, described in that link above. You only realize that the other clock is running consistently slow relative to yours when you correct for the time that it took for light to make it from the clock to your eye; if the clock is one light-second away from you the time you SEE is what it read one second ago when the light started towards you.

Suggesting once you correct for the time that it took for the light to make it from the clock to your eye, that the other clock could be thought to be running consistently slow to yours compared to yours whether it is moving towards or away from you.

How can you "notice" something different from what you "observe"?

Well imagine when they first pass each other they set their clocks to 0, they then assume the other one's clock is running slower (maybe because of the time on their clock it took for a light beam to bounce off a mirrored ceiling or whatever), but when they pass again they notice that the same amount of time has passed on their clocks. They could even land on the sphere and bring the clocks together or whatever.

 

[russ_watters]

As Nugatory mentioned

Suggesting once you correct for the time that it took for the light to make it from the clock to your eye, that the other clock could be thought to be running consistently slow to yours compared to yours whether it is moving towards or away from you.

Clearly, it's not (and he did not mean to imply that). Symmetrical scenarios are symmetrical and also do not produce contradictions. The Twins Paradox describes how a non-symmetrical scenario works.

Well imagine when they first pass each other they set their clocks to 0, they then assume the other one's clock is running slower (maybe because of the time on their clock it took for a light beam to bounce off a mirrored ceiling or whatever), but when they pass again they notice that the same amount of time has passed on their clocks. They could even land on the sphere and bring the clocks together or whatever.

So that's calculating, not observing. But either way, you can calculate what you will observe and it will be accurate if you do it right. Otherwise, the theory would be of no value!

 

[PeterDonis]

in the Haele-Keating experiment did they not use the general relativity equations to work out the gravitational time dilation and the special relativity equations to work out the velocity time dilation?

No. They used one equation, derived using GR, that includes both gravitational time dilation and velocity time dilation. Pop science presentations that attribute velocity time dilation to "special relativity" are misleading; SR only applies if spacetime is flat, and as I've said before, if gravitating masses are present spacetime is not flat.

when working out the velocity time dilation for the westbound direction, did they not do it from the plane's perspective?

No. All of the calculations were done in a non-rotating frame in which the center of the Earth is at rest.

I wasn't thinking that the spacetime was flat, only that scientists had used the special
relativity equations for similar situations.

You can't use the SR equations if spacetime isn't flat, except within the confines of a single local inertial frame (as @Nugatory pointed out before). But no single local inertial frame can cover an entire orbit of either satellite, or even half an orbit (since, as you pointed out, the satellites meet every half orbit, yet they are both in free fall and free-fall objects in flat spacetime can only ever meet once). A single local inertial frame can only cover a very small part of one satellite's orbit.

So how could you work out what velocity time dilation they observed?

For the case I was speaking of, you use each satellite's local inertial frame to work out the velocity time dilation of the other.

I would expect them to be able calculate the gravity time dilation for the clock on the sphere the same as you could.

But, as I said before, that time dilation is not a time dilation in either satellite's local inertial frame, so you can't combine it with a velocity time dilation that's calculated in either satellite's local inertial frame. Such a calculation makes no sense. A calculation that combines gravitational time dilation and velocity time dilation only makes sense in a frame in which both are well-defined; the only such frame in your scenario is the frame of the clock on the sphere.

because the satellite is in relative motion to the sphere clock, also see it as having velocity time dilation.

More precisely: in either satellite's local inertial frame, the sphere clock is moving, so the satellite will see it as having velocity time dilation. But, for the reasons I've already given, there is no way to combine this with gravitational time dilation to get a meaningful comparison with anything.

 

[name123]

It's still a twin paradox situation, because it maintains the apparent paradox...

Both correctly find that the other clock is running slower than theirs always, just as with the classic version of the twin paradox. In the classic paradox, the traveller is surprised to find that the Earth twin ends up older even though the Earth twin's clock was running slower throughout; here each satellite bserver is surprised to find that the twin on the other satellite has aged equally even though their clock was running slower for the entire time of separation. And the resolution is the same in both cases: the difference in clock rate (time dilation) does not lead to apparently obvious conclusion about the total elapsed time.

What I cannot see with the two satellite situation is how you can claim that they both "correctly" find the other clock is running slower than theirs, when they can check and find that the same amount of time has elapsed. It seems to be a logical contradiction.

 

[Ibix]

What I cannot see with the two satellite situation is how you can claim that they both "correctly" find the other clock is running slower than theirs, when they can check and find that the same amount of time has elapsed. It seems to be a logical contradiction.

Because they can only check their clock rates unambiguously as they pass. At any other time they have to use some kind of complicated simultaneity criterion appropriate for their paths in a curved spacetime - and naive intuition from SR most definitely does not apply to this. The directly measured (with Doppler) rate will vary smoothly through the orbit, and the appropriate Doppler correction will also vary. The result will be a varying clock rate which will sum to the same elapsed time.

 

[PeterDonis]

What I cannot see with the two satellite situation is how you can claim that they both "correctly" find the other clock is running slower than theirs

Slower relative to their own local inertial frame. The qualifier is crucial.

It seems to be a logical contradiction.

No, it's just a reflection of the fact that each satellite has a different local inertial frame.

 

[name123]

Clearly, it's not (and he did not mean to imply that).

I think he did mean to state that, because he has also written:

Both correctly find that the other clock is running slower than theirs always, just as with the classic version of the twin paradox.

So I think he did mean to state that the other satellites clock is observed to be running slower.

Symmetrical scenarios are symmetrical and also do not produce contradictions. The Twins Paradox describes how a non-symmetrical scenario works.

Well given that they would both observe the other's clock to go slower than theirs and then find the same amount of time to have elapsed, how is that not a contradiction?

 

[PeterDonis]

given that they would both observe the other's clock to go slower than theirs and then find the same amount of time to have elapsed, how is that not a contradiction?

See my post #57.

 

[Nugatory]

What I cannot see with the two satellite situation is how you can claim that they both "correctly" find the other clock is running slower than theirs, when they can check and find that the same amount of time has elapsed. It seems to be a logical contradiction.

No worse than the logical contradiction that the traveller finds in the ordinary twin paradox: the Earth bound clock is slow for the entire journey, yet the Earth twin ends up more aged.

The explanation is the same in both cases: the actual time elapsed between two clock readings is the length of the path through spacetime that the clock followed and this quantity is unrelated to time dilation; the two clocks follow different paths through spacetime between meeting events, so there's no particular reason to expect the elapsed times to be the same. In the counter-orbiting satellite case we've arranged things so that the two paths happen to be of equal length, in the classic twin paradox we've arranged for them to be different lengths. But either way, both observers always find the other clock to be running slow.