B Gravity vs Acceleration: The Impact on Clocks in Reference Frame R

[calinvass]

My question is, if gravity and acceleration are equivalent the what happens to a clock c1 , after a for example 1 year, if it is being accelerated by a force F, at g relative to a reference frame R, and what happens to another clock c2 , if it at rest relative to the same reference frame R, in a gravity field g=10 [mps/s] producing a force G=F?

What I understand is as the moving clock c1 time starts to slow down, the acceleration gets smaller and it an increased force is needed to maintain the acceleration g relative to the frame R. The same happens to the clock under G force. If we increase the force clock 1 time will go slower, if not they will show the same time. Is that correct ?

 

[Grinkle]

I was / am hoping to see someone who better knows what they are talking about respond, but in the meantime ...

I think if your question includes a third clock and you say that this third clock is in a reference frame that observes clock 1 accelerating and also observes clock 2 at rest but in a gravity well, then I think this third clock says that clocks 1 and 2 are ticking at the same rate.

I could be wrong.

 

[A.T.]

what happens to a clock c1 , after a for example 1 year, if it is being accelerated by a force F

https://en.wikipedia.org/wiki/Clock_hypothesis

 

[Stephanus]

My question is, if gravity and acceleration are equivalent the what happens to a clock c1 , after a for example 1 year, if it is being accelerated by a force F, at g relative to a reference frame R, and what happens to another clock c2 , if it at rest relative to the same reference frame R, in a gravity field g=10 [mps/s] producing a force G=F?

I think as @Grinkle has pointed out, you need the third clock.
Okay..., perhaps I can add.
To calculate the accelerated force, you use SR. While to calculate the clock in gravity field you would use GR. I don't know GR at all, but that's how to calculate it.

What I understand is as the moving clock c1 time starts to slow down, the acceleration gets smaller and it an increased force is needed to maintain the acceleration g relative to the frame R. The same happens to the clock under G force. If we increase the force clock 1 time will go slower, if not they will show the same time. Is that correct ?

Increased force? Ye-e-ess, but... Wouldn't it be faster than the speed of light if c1 is accelerated in someway that with respect to R frame it keeps constant acceleration?
Supposed c1 is accelerated relative to R 1 m /s². The speed of light is: 299,792,458m/second
So, after c1 speed is 299,792,456, (99.9999993328718% c) than 1 second later (according to whom? R?) then it will be 299,792,487 m/s. (99.9999996664359% c).
Let my try it with relativity addition formula.
So the speed difference wrt C1 would be ... 0.333 times the speed of light or 99,930,817 meter/seconds! I don't know if my calculator is accurate.
So c1 will be jerked (not accelerated) 99 thousands km/second wrt C1 to reach 299,792,487 m/s from 299,792,486m/s wrt R. I have no idea how much force to propel 1 kg mass that way. Of course you might want to calculate it yourself.

 

[PeterDonis]

if gravity and acceleration are equivalent

They are only equivalent locally, i.e., in a small patch of spacetime.

what happens to a clock c1 , after a for example 1 year, if it is being accelerated by a force F, at g relative to a reference frame R, and what happens to another clock c2 , if it at rest relative to the same reference frame R, in a gravity field g=10 [mps/s] producing a force G=F?

This thought experiment is not limited to a small patch of spacetime, so the results in the two cases don't have to be the same. And in this case you can't even really compare them. See below.

What I understand is as the moving clock c1 time starts to slow down, the acceleration gets smaller and it an increased force is needed to maintain the acceleration g relative to the frame R.

You are mixing up frame-dependent and frame-independent quantities. The frame-independent quantity is the acceleration that is felt by each clock; in order for the equivalence principle to apply at all, even in a small patch of spacetime, both have to be the same. So that is the condition of the problem: that the acceleration felt by both clocks, the one accelerating in free space and the one hovering at rest in a gravitational field. This acceleration--the precise term for it is proper acceleration--will not change at all throughout the problem--at least it shouldn't for the comparison you are trying to make.

For the clock accelerating in free space, if the acceleration it feels is constant, the acceleration it will appear to have relative to a fixed inertial frame--the precise term for this is "coordinate acceleration"--will decrease with time.

The same happens to the clock under G force.

Not really, because there is no "the same" in this case, because there is no fixed inertial frame we can use globally in the presence of gravity. The only thing we can compare between the two cases is the acceleration felt by the clocks--their proper acceleration.

We can't even compare "how much time elapses" for the two clocks, because they are in different spacetimes and there is no way to pick out "corresponding" events on their worldlines. This is just part of the fact I mentioned above, that the equivalence principle only covers small patches of spacetime--"small" meaning small enough that all these other issues don't arise.

 

[calinvass]

You are mixing up frame-dependent and frame-independent quantities. The frame-independent quantity is the acceleration that is felt by each clock; in order for the equivalence principle to apply at all, even in a small patch of spacetime, both have to be the same. So that is the condition of the problem: that the acceleration felt by both clocks, the one accelerating in free space and the one hovering at rest in a gravitational field. This acceleration--the precise term for it is proper acceleration--will not change at all throughout the problem--at least it shouldn't for the comparison you are trying to make.

Possibly.
From the conditions of the problem the proper acceleration for both clocks was supposed to be the same. An example in the real world would be a clock1 in a spacecraft , clock2 on Earth and reference frame at say in space at 10 light seconds from Earth and in the Sun reference frame . Relative speed between the sun an object on Earth is negligible.

For the clock accelerating in free space, if the acceleration it feels is constant, the acceleration it will appear to have relative to a fixed inertial frame--the precise term for this is "coordinate acceleration"--will decrease with time.

I also said that in order to maintain proper acceleration on clock1 you need to increase the force, perhaps it wasn't clear or I didn't understand your answer.
If the "coordinate acceleration " decreases it means it is similar to what happens on the clock2 reference frame.

 

[Ibix]

The problem as you described it holds coordinate acceleration constant while the proper acceleration (the acceleration felt by passengers in the rocket) grows without bound. It reaches infinity (obviously impossible to do) in under a year of the original rest frame's time. You want to hold proper acceleration constant.

However, what do you mean by time going slower? What are you comparing it to? You need a clock outside your rocket and far away from your planet to compare to. As soon as you do that, though, you run into obvious differences between the gravity and the rocket - for example the rocket observers will notice the external clock starting to move past them, while the planet observers will see it stationary at all times. This means that they aren't making a fair comparison - which is the start of the problems Peter mentioned.

The long and the short of it is that the equivalence principle is approximately true in some fairly limited circumstances. You are trying to operate outside those circumstances, so you need more sophisticated tools.

 

[PeterDonis]

An example in the real world would be a clock1 in a spacecraft , clock2 on Earth and reference frame at say in space at 10 light seconds from Earth

And how long will clock1 stay within this reference frame? Certainly not for a year. And 10 light seconds is much too large for the equivalence principle to apply; differences in the strength of the Earth's and probably even the Sun's gravity are observable over that range. In other words, your thought experiment is not limited to a small enough patch of spacetime to apply the EP.

I also said that in order to maintain proper acceleration on clock1 you need to increase the force

Not as measured by the person in the spacecraft . The "force" as calculated in a fixed inertial frame will increase, yes.

If the "coordinate acceleration " decreases it means it is similar to what happens on the clock2 reference frame.

I don't understand what you mean. In a reference frame fixed to the Earth, clock2 has zero coordinate acceleration; it is at rest and stays at rest. It has nonzero (and constant) proper acceleration.

 

[calinvass]

I don't understand what you mean. In a reference frame fixed to the Earth, clock2 has zero coordinate acceleration; it is at rest and stays at rest. It has nonzero (and constant) proper acceleration.

I meant to say that acceleration and gravity both produce time dilation.
For example, time for an object on Earth over a period of time of 1 year will be 0.0219 s less than time for an distant object, without taking into consideration Earth rotation.

 

[Stephanus]

I meant to say that acceleration constant velocity and gravity both produce time dilation.
For example, time for an object on Earth over a period of time of 1 year will be 0.0219 s less than time for an distant object, without taking into consideration Earth rotation.

Constant velocity and gravity both produce time dilation. If you accelerate, then you'll change frame. And time difference between velocity and gravity is very significant.

 

[calinvass]

Thanks, I get it now. It is also consistent with the way light travels as a wave. Under a gravity field similar to that produces by Earth EM waves are slightly affected. When a light source travels towards an object when reaching a certain speed, It gets blue-shifted, so the effect is significant and with acceleration it is progressive.

 

[PeterDonis]

I meant to say that acceleration and gravity both produce time dilation.

That's not a very precise way of putting it. The precise way of putting it is this: if we have two observers inside a rocket accelerating in flat spacetime ("accelerating" meaning "experiencing proper acceleration"), at rest relative to each other (as measured by them exchanging round-trip light signals and seeing that the round-trip travel time by each of their clocks remains constant), the observer at the bottom of the rocket will be time dilated (clock running slower) relative to the observer at the top (as measured by noting the elapsed time on both of their clocks between two successive round-trip light signals). Similarly, if we have two observers at rest in a gravitational field (and experiencing proper acceleration as required to hold them at rest), at rest relative to each other (same measurement method as above), the one lower down will be time dilated (clock running slower) relative to the one higher up (same measurement method as above).

Note that what causes the time dilation is not "acceleration" or "gravity" but difference in height between observers experiencing proper acceleration (where "height" just means "distance along the direction of proper acceleration"). Note also that the observers have to be at rest relative to each other; if they are moving relative to each other, things get more complicated (though in many cases there will still be a good approximation in which we can separate time dilation due to height from time dilation due to motion).

Note also that the above has a wider application than just a small patch of spacetime; that is, the above is not just a restatement of the equivalence principle. The EP makes a stronger claim: it says that locally--in a patch of spacetime small enough that both observers, to within the accuracy of measurement, experience the same proper acceleration--the same height difference produces the same time dilation in both cases (accelerating rocket in flat spacetime, at rest in a gravity field). But that claim no longer holds once we consider regions of spacetime large enough for differences in "field strength" (differences in the proper acceleration experienced by the two observers) to be measurable. In that case the same height difference no longer produces the same time dilation in both cases.

 

[calinvass]

Thanks, things are now very clear for me.

 

[David Lewis]

For example, time for an object on Earth over a period of time of 1 year will be 0.0219 s less than time for an distant object, without taking into consideration Earth rotation.

If my understanding is correct, time appears to pass more slowly on Earth from the viewpoint of the space object because gravitational field strength is weaker far from the Earth. And conversely, clocks seem to run fast for space objects from the perspective of Earth observers. Both clocks, if they are functioning correctly, actually tick at the same rate in reality. It's the difference in viewpoint that creates the illusion.

 

[PeterDonis]

Both clocks, if they are functioning correctly, actually tick at the same rate in reality.

This is not a good way to put it, because "in reality" implies that there is some "real" clock rate and the others are all "illusions" (your word). There is no such thing as the "real" clock rate. There is proper time along a particular world line, and there are measurements you can make to try to compare clock rates along different worldlines--for example, two observers both at rest relative to a large mass, but at different altitudes, can exchange round-trip light signals and measure the elapsed proper times on each of their clocks between two successive signals, and thereby determine that the clock at lower altitude is "running slow" compared to the higher one--because it shows less elapsed time between two successive signals. But all of these various observations and measurements are equally "real"; none of them has any privileged position, and none of them corresponds to how time flows "in reality".

 

[David Lewis]

[T]he clock at lower altitude is "running slow" compared to the higher one--because it shows less elapsed time between two successive signals.

Peter, thank you for correcting me. Would it be equally valid to say the the clock at the lower altitude is not running slow, but rather less time has passed?

 

[PeterDonis]

Would it be equally valid to say the the clock at the lower altitude is not running slow, but rather less time has passed?

The question is, less time has passed between what two events on the clock's worldline?

This is always the issue when trying to compare "clock rates" between spatially separated objects. What are the "corresponding" events on the two worldlines? In the standard twin paradox there is an easy answer: the two worldlines intersect at the start and end of the scenario, so those events are obvious choices for "corresponding". In other cases the choice can be more problematic; in the general case there is no single choice that is picked out by the scenario, all of them are arbitrary, which means that in the general case there is no way of comparing "clock rates" between spatially separated objects.

In the case under discussion, the fact that both objects are at rest, relative to each other and to the large mass, picks out a particular choice of "corresponding" events: basically they are the ones that are picked out by the natural simultaneity convention of the large mass's rest frame. With that choice, yes, we can say that the clock at the lower altitude has less time pass between a pair of corresponding events. But it's still good to be aware that that statement depends on a particular choice of simultaneity convention, i.e., of which pairs of events on the two worldlines are "corresponding".

 

[A.T.]

If my understanding is correct, time appears to pass more slowly on Earth from the viewpoint of the space object because gravitational field strength is weaker far from the Earth.

Gravitational time dilation depends on the potential, not the field strength. In the center of the Earth the field strength is zero, but the time dilation is maximal (lowest potential).

It's the difference in viewpoint that creates the illusion.

It's not an illusion, and it's not symmetrical like kinetic time dilation.

 

[A.T.]

Would it be equally valid to say the the clock at the lower altitude is not running slow, but rather less time has passed?

Time passed is what clocks measure.

 

[calinvass]

A clock in free fall under a gravitational field vs one hovering above the Earth is equivalent to one clock accelerating in respect to another.

 

[Ibix]

A clock in free fall under a gravitational field vs one hovering above the Earth is equivalent to one clock accelerating in respect to another.

If the two clocks are passing "close" to each other in the gravitational field and you mean to compare this to two clocks in flat spacetime, one moving inertially and one undergoing proper acceleration, then these two cases are the same, yes. Literally so, by the equivalence principle, although scare quotes are needed because "close" here depends on your measurement precision.

If that is the case, proper acceleration is a frame independent thing and you do not need to specify "in respect to". It's worth specifying the "proper" though, unless you are very sure that your audience will both be able to understand it from context and trust you to have used it correctly.

If you are meaning something else then it depends on what you actually meant.

 

[calinvass]

I meant to say a hypothetical constant gravitational field,so there is no change in gravitational filed during descent , but only gravity potential difference. This case works for a limited period of time, otherwise the object falling exceeds the speed of light. However, I don't understand acceleration as frame independent. It is something I've missed. Is it because no matter the current velocity (in one frame or another) , the acceleration remains constant by definition?

 

[PeterDonis]

I meant to say a hypothetical constant gravitational field

There is no such thing. More precisely, there is no solution of the Einstein Field Equation that has this property. The closest you can get is Rindler coordinates in flat Minkowski spacetime; but the "gravitational field" in that case is still not constant, it varies with "height" (but it doesn't vary "horizontally", i.e., its magnitude changes but not its direction). Also, the "field" only covers a portion of spacetime--there is a Rindler horizon beyond which observers that are "at rest" at a constant height in the "gravitational field" can never see.

This case works for a limited period of time, otherwise the object falling exceeds the speed of light.

In the case of Rindler coordinates, above, this is not a correct description: what happens is that a free-falling object falls below the Rindler horizon and can't be seen any longer by the static observers. But from the viewpoint of a global inertial frame, the free-falling object is just sitting at rest in flat Minkowski spacetime; it certainly isn't "exceeding the speed of light" or traveling on a spacelike worldline or anything like that.

I don't understand acceleration as frame independent.

Proper acceleration--the acceleration that is measured by an accelerometer and that you feel as weight--is frame independent. But coordinate acceleration--the second derivative of position with respect to time--is not. Carefully distinguishing between these two concepts is critical in understanding the topic under discussion.

 

[calinvass]

There is no such thing. More precisely, there is no solution of the Einstein Field Equation that has this property. The closest you can get is Rindler coordinates in flat Minkowski spacetime; but the "gravitational field" in that case is still not constant, it varies with "height" (but it doesn't vary "horizontally", i.e., its magnitude changes but not its direction). Also, the "field" only covers a portion of spacetime--there is a Rindler horizon beyond which observers that are "at rest" at a constant height in the "gravitational field" can never see.
In the case of Rindler coordinates, above, this is not a correct description: what happens is that a free-falling object falls below the Rindler horizon and can't be seen any longer by the static observers. But from the viewpoint of a global inertial frame, the free-falling object is just sitting at rest in flat Minkowski spacetime; it certainly isn't "exceeding the speed of light" or traveling on a spacelike worldline or anything like that.

That is why I said hypothetical and between some time limits. But, I, understand it is not possible to use it because you can't write any equations. It is possible to imagine it though.

 

[Nugatory]

That is why I said hypothetical and between some time limits. But, I, understand it is not possible to use it because you can't write any equations. It is possible to imagine it though.

The inability to write an equation is a very strong hint that the hypothetical you're imagining is not internally consistent. Starting with an internally inconsistent hypothesis pretty much guarantees confusing and illogical conclusions.

However, if by "within limits" you mean that you want to study the behavior of an object across a small enough region of spacetime that we can ignore any variations in the gravitational field and treat it as uniform... There's no difficulty writing equations that cover that situation, so we're good. (for example, it takes some seriously sensitive instruments to detect any non-uniformity in the Earth's gravitational field within a volume 100 meters on the side at the Earth's surface). A free-falling object within that volume will experience zero proper acceleration and coordinate acceleration determined by whatever coordinates you use.

 

[Ibix]

However, I don't understand acceleration as frame independent. It is something I've missed. Is it because no matter the current velocity (in one frame or another) , the acceleration remains constant by definition?

It's because you can detect proper acceleration in a "closed room" experiment. Basically you let something go. If it falls, your room is undergoing proper acceleration. And it is a frame-independent fact that you and the object were initially at relative rest and then later not, so proper acceleration is frame-independent.

Coordinate acceleration (the second derivative of your spatial coordinates with respect to your time coordinate) is not frame-independent by definition. For example, an object whose Rindler spatial coordinates do not change is undergoing constant proper acceleration, so has non-zero coordinate acceleration in a Minkowski frame.

 

[calinvass]

Thank you. I was confused because I had a wrong understanding of definitions. I've checked them again and it makes sense. Back to the original post, when I've started it, I thought time dilation occurs due to difference in gravitational field values not gravitational potential. In that case gravity wouldn't have been equivalent to acceleration. However, gravitational potential difference, alone, does not explain how time dilation occurs.

I've also found an explanation using space-time diagrams which didn't look very clear at first then I've plotted a graph of a constant acceleration and apparently, if you select any two points on x axis, you always get different velocities between them, which means you can apply simple Lorentz transformations. That is counterintuitive.
Then apply the equivalence principle.

 

[calinvass]

I've also found an explanation using space-time diagrams which didn't look very clear at first then I've plotted a graph of a constant acceleration and apparently, if you select any two points on x axis, you always get different velocities between them, which means you can apply simple Lorentz transformations. That is counterintuitive.
Then apply the equivalence principle.

That is actually wrong, meaning its not counterintuitive. We need acceleration curves separated by a distance on x-axis for two accelerating clocks.

 

[PeterDonis]

I've plotted a graph of a constant acceleration and apparently, if you select any two points on x axis, you always get different velocities between them, which means you can apply simple Lorentz transformations. That is counterintuitive.

That is actually wrong, meaning its not counterintuitive.

Can you clarify what you are doing here? You might want to also become familiar with these two different families of worldlines in Minkowski spacetime: the worldlines of Rindler observers (observers at rest in Rindler coordinates), and the worldlines of "Bell observers" (the worldlines that occur in the Bell Spaceship Paradox):

https://en.wikipedia.org/wiki/Rindler_coordinates#The_Rindler_observers

https://en.wikipedia.org/wiki/Bell's_spaceship_paradox

http://math.ucr.edu/home/baez/physics/Relativity/SR/BellSpaceships/spaceship_puzzle.html

PF also has an Insights article on the Bell Spaceship Paradox:

https://www.physicsforums.com/insights/what-is-the-bell-spaceship-paradox-and-how-is-it-resolved/

 

[calinvass]

Yes, I thought, clock rates difference between two points one above each other in the direction of motion within an accelerating object could be explained by Newtonian mechanics but the explanation is like in the Bell's spaceship. This seems to be essential for the EP.

 

[calinvass]

Thank you. This explanation at link https://www.physicsforums.com/insights/what-is-the-bell-spaceship-paradox-and-how-is-it-resolved/
Seems very clear up to a point.
The conditiom is, the clocks feel both the same proper acceleration and we set a refference frame F0, where they apear as accelerating at the same time.
We follow the acceleratio curve of the clock in the left hand side ( this is the lower or the rear clock in the direction of motion) then at any point we stop we can draw the coordinates (minkowski spacetime) of a reference frame featuring the left clock at rest FL. When you look on the x-axis to find the second clock, it will apear at point that is on the refference frame F0, further on the x-axis but also at a distance in time because the line of simulataneity (x) is tilted. On the FL reference frame the distance on the x-axis will also be at a greater.

This means the distance between them has increased in FL reference frame but not in F0 reference frame because of different simultaneity lines). Since FL is an accelerated reference frame, the distance between the clocks in this frame will be constantly increasing - constant speed . Now it is becoming a bit harder to follow. This will generate a reduced clock rate observed in the FL reference frame.
However in the F0 reference frame clocks should show the same rates so after the ship stops both clocks should show the same time because the coordinates are tilted always at the same angle. Now it seems to contradict the results in the twins paradox.

 

[PeterDonis]

FL is an accelerated reference frame

No, it isn't. It's an inertial frame in which the chosen spaceship (the one on the left) is momentarily at rest.

This will generate a reduced clock rate observed in the FL reference frame.

You have to be very careful when talking about clock rates and frames in cases like this. Here is a more careful way of stating what is going on:

Pick an event on the worldline of the left (or rear clock) while it is accelerating. Call that event L. We can pick two events on the worldline of the right (or front) clock which are simultaneous with L according to two different inertial frames: event R0, which is simultaneous with L according to frame F0, and event RL, which is simultaneous with L according to frame FL. These are different events.

Call the left clock's elapsed time at event L time T. At event R0, the right clock shows elapsed time equal to T. But at event RL, the right clock shows elapsed time greater than T.

Also, in frame F0, the distance between the two clocks always remains the same--so the distance between events L and R0 is the same as the distance between the clocks before they started accelerating. But the distance, in frame FL, between events L and RL is not the same--it is larger. (This is one way of seeing why the string stretches and eventually breaks.)

after the ship stops both clocks should show the same time

And they will, if they stop at the same time in frame F0. More precisely, if the left clock stops at some event SL, and the right clock stops at event SR0 which is simultaneous with event SL in frame F0, then the clocks will show the same time at those two events. And since the clocks will continue to move at the same speed (since they are no longer accelerating), they will remain synchronized according to frame F0.

But suppose that the left clock stops at some event SL, and the right clock stops at event SRL which is simultaneous with SL in frame FL. Then we have a more complicated situation. The time shown on the left clock at event SL will be less than the time shown on the right clock at event SRL. So in frame FL, the right clock will show more elapsed time than the left clock when they both stop accelerating.

But in frame F0, event SL happens before event SRL; so to see how the clocks end up in frame F0, we have to figure out what the left clock reads at event SLL, which is the event that is simultaneous with event SRL in frame F0. At event SLL, the left clock has been moving inertially for some time after stopping (i.e., in frame F0 the left clock stops accelerating first), so from the viewpoint of frame F0, it is "gaining time" relative to the right clock, which keeps accelerating. How things are when the right clock stops accelerating, at event SRL, will depend on the details, like the original distance between the clocks and how long the left clock accelerates.

Furthermore, in this second scenario, where the right clock stops accelerating at event SRL, the clocks are not moving at the same speed (in any inertial frame), so their synchronization will continue to change (unlike the first scenario, where the right clock stops at event SR0).

 

[calinvass]

Here is the diagram you've described

 

[PeterDonis]

Here is the diagram you've described

Not quite. Event SLL should be on the dotted line going up from event SL--remember, we assumed that the left clock stops accelerating at event L, so it moves inertially after that point, and the dotted line is the inertial worldline of the left clock after event L. Event SLL lies on that worldline.

 

[calinvass]

I was editing the diagram. However you've made some points I didn't take into account. I've judged the acceleration lines as helping curves not the actual path of the clocks. But the mistake was that the horizontal line from SRL should've intersected the dotted blue line not the helping acceleration path ( that path goes beyond c, so it clearly cannot represent the real path.

 

[calinvass]

 

[PeterDonis]

the mistake was that the horizontal line from SRL should've intersected the dotted blue line not the helping acceleration path

Yes.

that path goes beyond c

It shouldn't if you generated it correctly. You need to use the $\cosh$ and $\sinh$ functions.

 

[calinvass]

Yes, I understand. I've used constant Newtonian acceleration in flat spacetime but that is not the trajectory in frame F0.

 

[David Lewis]

It's not an illusion, and it's not symmetrical like kinetic time dilation.

Please explain what you mean by not symmetrical.

 

[Nugatory]

Please explain what you mean by not symmetrical.

Gravitational time dilation is asymmetrical in the sense that if B's clock is running slow relative to A's clock as observed by A, it is also running slow (or equivalently, A's clock is running faster) as observed by B. In physical terms: If B is deeper in a gravity well than A, and they exchange light signals from identically constructed (so the emission frequencies are the same) light sources B will receive a light signal that is blueshifted to a higher frequency than his own source, while A will receive a light signal that is redshifted to a lower frequency than his own source.

The speed-based time dilation is symmetrical: as observed by A, B's clock is running slow relative to A's clock but as observed by B, A's clock is running slow relative to B's clock. If they exchange light signals they will both receive either blue-shifted or red-shifted signals according to whether they are approaching one or another or moving apart.

 

[David Lewis]

Many thanks. Is it correct to say gravitational time dilation is asymmetric under interchange of observers?

 

[stevendaryl]

Many thanks. Is it correct to say gravitational time dilation is asymmetric under interchange of observers?

Yes, it's asymmetric.

But it's actually a little complicated as to what "gravitational time dilation" means. What's definitely true is that if you use a curvilinear, noninertial coordinate system, then in general the value of \frac{d \tau}{dt} (where \tau is the elapsed time on a standard clock, and t is coordinate time) will be position-dependent, as well as velocity-dependent, while in SR, using inertial Cartesian coordinates, it is only velocity-dependent.

 

[David Lewis]

So a satellite clock, having higher gravitational potential, would appear to tick faster than an Earth clock, and (assuming the satellite is moving with respect to the Earth clock) SR time dilation would make the satellite clock appear to tick slower than the Earth clock. I suppose you would add the two effects together to arrive at the net dT/dt.

 

[Nugatory]

So a satellite clock, having higher gravitational potential, would appear to tick faster than an Earth clock, and (assuming the satellite is moving with respect to the Earth clock) SR time dilation would make the satellite clock appear to tick slower than the Earth clock. I suppose you would add the two effects together to arrive at the net dT/dt.

Yes, and we have several older threads that show how to do this calculation. Getting it right is kinda important for the GPS system...

 

[calinvass]

There is also frame dragging effect thus the orbiting rotation direction or whether it is geostationary should influence the difference in tick rates.

 

[cianfa72]

if we have two observers inside a rocket accelerating in flat spacetime ("accelerating" meaning "experiencing proper acceleration"), at rest relative to each other (as measured by them exchanging round-trip light signals and seeing that the round-trip travel time by each of their clocks remains constant), the observer at the bottom of the rocket will be time dilated (clock running slower) relative to the observer at the top (as measured by noting the elapsed time on both of their clocks between two successive round-trip light signals).

Reading this old thread I would ask for a clarification: observer A sends a light signal reflected back from B and using his own clock measures the round-trip travel time; the observer B does the same. If the results of both measurements remain constant then we claim both observers are at rest each other.

Furthermore if in the above procedure each observer sends/encodes in the reflected light signal the reading of his own clock (when the signal is reflected back), the receiving observer can evaluate the difference between the encoded times and the readings of his own clock between the receipts of two successive round-trip light signals.

This way both observers will evaluate a non-zero difference (i.e. the observer at the bottom will be time dilated).

Is the above correct ? Thank you.

 

[PeroK]

Reading this old thread I would ask for a clarification: observer A sends a light signal reflected back from B and using his own clock measures the round-trip travel time; the observer B does the same. If the results of both measurements remain constant then we claim both observers are at rest each other.

Furthermore if in the above procedure each observer sends/encodes in the reflected light signal the reading of his own clock (when the signal is reflected back), the receiving observer can evaluate the difference between the encoded times and the readings of his own clock between the receipts of two successive round-trip light signals.

This way both observers will evaluate a non-zero difference (i.e. the observer at the bottom will be time dilated).

Is the above correct ? Thank you.

You could do it that way. Alternatively, each observer sends out a signal once per second, say. They each receive their own signals back at one second intervals. But, receive the other's signals with a longer or shorter time interval.

 

[cianfa72]

Alternatively, each observer sends out a signal once per second, say.

yes, once per second according their own clocks.

They each receive their own signals back at one second intervals. But, receive the other's signals with a longer or shorter time interval.

Again, as measured by their own clocks.

 

[jartsa]

Reading this old thread I would ask for a clarification: observer A sends a light signal reflected back from B and using his own clock measures the round-trip travel time; the observer B does the same. If the results of both measurements remain constant then we claim both observers are at rest each other.

Quite the opposite. All inertial observers say that there is some relative motion. They might call it "length contraction motion". I am guessing that "we" are inertial observers.

This way both observers will evaluate a non-zero difference (i.e. the observer at the bottom will be time dilated).

Yes, according to those non-inertial people the bottom people are time dilated. Different inertial observers have different opinions about the time dilation.

 

[DrGreg]

Reading this old thread I would ask for a clarification: observer A sends a light signal reflected back from B and using his own clock measures the round-trip travel time; the observer B does the same. If the results of both measurements remain constant then we claim both observers are at rest each other.

Quite the opposite. All inertial observers say that there is some relative motion. They might call it "length contraction motion". I am guessing that "we" are inertial observers

If "we" are inertial observers, you are right, but I suspect what cianfa72 meant to say was "each observer claims the other is at rest relative to themself", and, in that case, that is true.

EDIT: Just to clarify possibly ambiguous language:

• Observer A claims that Observer B is at rest relative to Observer A
• Observer B claims that Observer A is at rest relative to Observer B

Both these claims are true; they can be taken as a definition of what "at rest relative to a non-inertial observer" means.