I Time dilation for the Earth's orbit around the Sun

[DanMP]

If we have 2 atomic clocks on Earth's orbit around the Sun, one on Earth's surface, at one pole, and the other on a spaceship, far from Earth, but traveling with the same speed around the Sun, the clocks would suffer the same kinematic time dilation or not?

I'm asking this because the clock on Earth would be stationary in the Earth's, co-moving, gravity well, while the spaceship clock would move (quite fast) through Sun's gravity well.

 

[pervect]

To measure time dilation, one needs a way of comparing clocks at two different locations. In special and general relativity, this process is not universal. This is called "the relativity of simultaneity". To define the process of clock comparison, it is sufficient to define a specific coordinate system, with "simultaneous" events having the same value of time coordinate. The natural coordinate system to use in this circumstance would be some version of Schwarzschild coordinates, but other choices are possible.

In addition, this particular problem requires General relativity, and not just special relativity. General relativity includes gravitational time dilation, and not just kinematic time dilation. General relativity does have formulae that allows one to compute time dilation if one specifies what one means by time dilation sufficiently well (for instance by specifying some particular coordinate system as above). However, one does not get the correct result by using the SR formulae for time dilation in this circumstance.

Given a choice of coordinate system, time dilation can conveniently be defined as the ratio of proper time to coordinate time for an observer following some specific path (worldline).

Using the metric associated with the chosen coordinate system, one can compute the proper time from the relation $d\tau^2 = g_{ij} dx^i dx^k$, where $\tau$ is the proper time, the $x^i$ are the choosen coordinates, and $g_{ij}$ are the metric coefficients associated with the particular coordinate choice.

This may seem slightly elaborate, but it's a safer way of doing the computating that makes it clear what assumptions are needed to answer the question. Sometimes assumptoins are not shared between the author of the question and the person who computes the answer, which can lead to confusion.

 

[PeroK]

If we have 2 atomic clocks on Earth's orbit around the Sun, one on Earth's surface, at one pole, and the other on a spaceship, far from Earth, but traveling with the same speed around the Sun, the clocks would suffer the same kinematic time dilation or not?

I'm asking this because the clock on Earth would be stationary in the Earth's, co-moving, gravity well, while the spaceship clock would move (quite fast) through Sun's gravity well.

In your example, you could take a clock at a large distance from the Solar system, at rest with respect to the Sun as a reference clock. There are then three factors affecting the time dilation of a clock in the Solar system (if we ignore other bodies in the Solar system and consider only the Sun and the Earth):

The Sun's gravitational potential; the Earth's gravitational potential; the speed of the clock relative to the Sun (and the distant reference clock). Therefore, the time dilation for a clock on the Earth's surface should be greater than a clock somewhere else on the Earth's orbit, as it has an extra factor. But, the "kinematic" component of that time dilation in this coordinate system would be the same for both clocks.

 

[DanMP]

this particular problem requires General relativity, and not just special relativity

I know that.

Given a choice of coordinate system, time dilation can conveniently be defined as the ratio of proper time to coordinate time for an observer following some specific path (worldline).

Using the metric associated with the chosen coordinate system, one can compute the proper time from the relation dτ2=gijdxidxkdτ2=gijdxidxkd\tau^2 = g_{ij} dx^i dx^k, where ττ\tau is the proper time, the xixix^i are the choosen coordinates, and gijgijg_{ij} are the metric coefficients associated with the particular coordinate choice.

This sounds very promising but unfortunately I don't have the skills to do the computations myself. I hoped that for the experts in this forum would be easy enough to do it and find the influence of the speed around the Sun on the clocks considered in the OP.

Sometimes assumptions are not shared between the author of the question and the person who computes the answer, which can lead to confusion.

What do you mean?

... the "kinematic" component of that time dilation in this coordinate system would be the same for both clocks.

I wouldn't be so sure, not without a proper computation, the kind that pervect suggested.

 

[Dale]

I hoped that for the experts in this forum would be easy enough to do it

To do it quantitatively is not easy. Your question has no known analytical solution so it would need to be done numerically.

I wouldn't be so sure, not without a proper computation, the kind that pervect suggested.

This annoys me. You know that it is too difficult for you. You ignorantly assume that it should be easy for others because they know more than you. But when they (whom you recognize know more than you) tell you the qualitative result of the computation you petulantly reject your assessment of their expertise and demand a full computation.

I agree fully with his assessment of the outcome based on my previous experience working with such problems, but I also recognize that calculating the result quantitatively will not be easy.

You have no rational basis for rejecting the assessment. You are asking (demanding) a lot of work for no compensation. The result of the calculation is a difficult to obtain piece of trivia for anyone besides yourself. But you are making no attempt to do any of it on your own.

If you really want this done why don’t you either put in the work yourself to learn the material and do the computation or pay someone for the large amount of work required to do it.

 

[DanMP]

To do it quantitatively is not easy. Your question has no known analytical solution so it would need to be done numerically.

If my question has no known analytical solution, maybe it would be a good exercise and even a material worth publishing ...

 

[jbriggs444]

If my question has no known analytical solution, maybe it would be a good exercise and even a material worth publishing ...

The word "trivia" in @Dale's post is relevant here. Numerical solutions are like grunt work. Time consuming but not noteworthy.

Lots of problems have no known analytical solution. They are the rule rather than the exception. For instance, in classical mechanics, the three body problem.

 

[DanMP]

You ignorantly assume that it should be easy for others because they know more than you.

Numerical solutions are like grunt work. Time consuming but not noteworthy.

I'm sorry for not knowing that it would be "grunt work" & "time consuming". I honestly believed that it should be fairly easy for experts.I don't quite understand why numerically? And if it has to be numerical, why not using a computer to do it?

On the other hand, with a circular orbit and constant speed would be easier?

 

[jbriggs444]

I don't quite understand why numerically? And if it has to be numerical, why not using a computer to do it?

Using a computer is what "numerical" means. More specifically, it tends to involve a differential equation solver using a method such as Runge Kutta.

 

[PeroK]

I'm sorry for not knowing that it would be "grunt work" & "time consuming". I honestly believed that it should be fairly easy for experts.I don't quite understand why numerically? And if it has to be numerical, why not using a computer to do it?

On the other hand, with a circular orbit and constant speed would be easier?

It's not too hard to see that for low speeds and weak gravity the time dilation is linear, in the sense that you can add "gravitational" and "kinematic" components. You can do this by considering four scenarios:

1) A clock at rest relative to the Sun.
2) A clock in the Earth's orbit relative to the Sun.
3) A clock on the Earth, with the Earth held at rest relative to the Sun.
4) A clock on the Earth, while the Earth orbits the Sun.

By considering these in relation to each other and in relation to a reference clock at rest and a long way from the Sun, you can see how to add the "components" of time dilation in each case. This was my argument anyway.

However, the Einstein Field Equations are non-linear, hence the solutions for two massive bodies don't, in general, superpose linearly. So, there is at least a small correction to the simple analysis above; even for low speed solar orbits.

Crunching that is not particularly interesting because a) we have a good approximation already; b) it doesn't tell us anything significant about GR; and, c) in general things like "kinematic" and "gravitational" time dilation are artefacts of the coordinates and have no physical significance in any case!

 

[Dale]

I don't quite understand why numerically?

There is not an analytical solution for the spacetime of a massive body orbiting another massive body. And unfortunately the EFE is nonlinear so you cannot simply add the two single body analytical solutions to get a valid overall solution. So it requires a numerical solution or some simplifying approximations.

 

[DanMP]

t's not too hard to see that for low speeds and weak gravity the time dilation is linear, in the sense that you can add "gravitational" and "kinematic" components.

You mean something like this?

However, the Einstein Field Equations are non-linear, hence the solutions for two massive bodies don't, in general, superpose linearly. So, there is at least a small correction to the simple analysis above; even for low speed solar orbits.

You said/implied that "at least a small correction" has to be expected. What if the "correction" is not small?

Crunching that is not particularly interesting because a) we have a good approximation already; b) it doesn't tell us anything significant about GR; and, c) in general things like "kinematic" and "gravitational" time dilation are artefacts of the coordinates and have no physical significance in any case!

You (all) are not interested to see how being static in a co-moving "gravity well" may affect the "kinematic component" of time dilation?

There is not an analytical solution for the spacetime of a massive body orbiting another massive body. And unfortunately the EFE is nonlinear so you cannot simply add the two single body analytical solutions to get a valid overall solution. So it requires a numerical solution or some simplifying approximations.

I suggested to consider the orbit a circle. If you know more "simplifying approximations", please post them.

 

[Dale]

I suggested to consider the orbit a circle. If you know more "simplifying approximations", please post them.

That wouldn’t simplify much. The big simplification would be to use the weak field approximation, or equivalently to use a linear approximation to the EFE.

 

[DanMP]

The big simplification would be to use the weak field approximation, or equivalently to use a linear approximation to the EFE.

I don't know if this would be appropriate.

Using a computer is what "numerical" means.

So why/how is this a "grunt work" if the computer does it?

 

[Dale]

I don't know if this would be appropriate.

Hence the need for a lot of work

So why/how is this a "grunt work" if the computer does it?

The computer won’t program itself

 

[Nugatory]

So why/how is this a "grunt work" if the computer does it?

It's a lot of work to get the differential equations and initial conditions into a form suitable for the computer; and then when the computer comes back with a number, you have only the same understanding of the physics involved that you started with.

 

[DanMP]

The computer won’t program itself

I thought that there are already made programs for EFE and/or problems like this.

If you really want this done why don’t you either put in the work yourself to learn the material and do the computation or pay someone for the large amount of work required to do it.

How much would it cost?

 

[Nugatory]

You said/implied that "at least a small correction" has to be expected. What if the "correction" is not small?

If the correction is not small, then the method @PeroK outlined isn't usable for this problem.

A more mundane example: We have two points on the surface of the earth, one five meters due east of a stake in the ground and the other five meters due west of the stake. You ask me to calculate the straight-line distance between them; I say that it's easy because distances just add, so it's ten meters. But in fact there is a small correction required because of the Earth's curvature - the two five-meter segments are actually two sides of a triangle and we're looking for the length of the third side. So is my add-the-lengths method an acceptable way of solving the problem? It depends on whether we care about the small correction - but it's there either way.

You (all) are not interested to see how being static in a co-moving "gravity well" may affect the "kinematic component" of time dilation?

That question is less interesting than it first appears, because"static" and "co-moving" are just (again, @PeroK) artifacts of the coordinate system.

Suppose that the clock in the gravity well is emitting a flash of light once every second so that we can watch it ticking from a distance by observing the arrival of the flashes. We will not, in general, receive the flashes at a rate of one per second; this will be a combination of time dilation and Doppler effect, and we can further divide the time dilation into a kinematic and a gravitational component. However, I can make this split come out pretty much any way that I please by choosing coordinates appropriately; all the real physics is in the ratio between the proper time elapsed between two emission events and the proper time elapsed between the two corresponding detection events.

 

[Dale]

I thought that there are already made programs for EFE and/or problems like this.

Not that I am aware of. Most labs doing this kind of research write their own code.

How much would it cost?

I would charge $60/hr. That is less than my time is worth, but I am not highly qualified so I cannot expect to charge full price.

 

[DanMP]

this will be a combination of time dilation and Doppler effect, and we can further divide the time dilation into a kinematic and a gravitational component. However, I can make this split come out pretty much any way that I please by choosing coordinates appropriately; all the real physics is in the ratio of proper time elapsed between two emission events and the proper time elapsed between the two corresponding detection events.

Consider the problem in another way: the spaceship clock starts from the Earth, where is synchronized with the other one, then flies off the Earth, going on orbit around the Sun, with the same speed as the Earth, but far from it (few millions km away). After a while (months, years), the spaceship returns and the reunited clocks are compared.

When I wrote "kinematic component" of time dilation I meant the influence of the orbital speed around the Sun in the calculation of the elapsed time between start and finish, for each clock.

I would charge $60/hr. That is less than my time is worth, but I am not highly qualified so I cannot expect to charge full price.

So I may expect more. Anyway, the next/big question is how many hours an expert would need to complete the task?

 

[Nugatory]

When I wrote "kinematic component" of time dilation I meant the influence of the orbital speed around the Sun in the calculation of the elapsed time between start and finish, for each clock.

The difference in elapsed time is the difference in the length of the world lines of the two clocks along their respective paths between the separation event and the reunion event. How much of this we attribute to the orbital speed and how much we attribute to gravitational effects is arbitrary.

 

[Ibix]

How much of this we attribute to the orbital speed and how much we attribute to gravitational effects is arbitrary.

Indeed. Isn't the point here that we either accept @PeroK's linear approximation, in which case the answer is trivial, or we don't, in which case the spacetime isn't static and the answer is a matter of personal preference?

 

[pervect]

Just for clarity, I'd advocate using the same linear approach that has been suggested. Doing a full-on GR computation just isn't sensible, the linear approximation is good enough and is widely used in any and all solar system problems.

The tricky part is that the problem is specified in a manner that makes it coordinate dependent, so one needs to choose a coordinate system. There are two main general philosophies here - geocentric and barycentric, i.e. earth-centered and sun-centered (actually the center of mass of the solar system, slightly different from sun-centered).

But it's unclear specifically which coordinate system to to recommend. PPN because it's in a lot of old textbooks? ICRS (or, for the Earth centered case, IERS) as being modern? What about BCRS and GCRS - the former of which I happen to have a line element for which would expedite the computation.

It would actually be better if the problem were specified in a manner that wasn't coordinate dependent, but the OP seems narrowly focused on doing things "their way" rather than learning new things.

I wouldn't actually expect much difference between any of the various coordinate choices by the way.

I''m also not feeling motivated to work through all the details - I don't actually think they'd help the OP, who in my opinion would be better off widening their focus. I am willing to comment a bit on the general approach that I'd use, which I've already done.

 

[DanMP]

... we either accept @PeroK's linear approximation, in which case the answer is trivial, or we don't, in which case the spacetime isn't static and the answer is a matter of personal preference?

By "spacetime isn't static" you mean linear frame dragging? It is possible? And it means that the clock on the Earth would not be affected at all by the orbital speed (of the Earth) around the Sun?

 

[DanMP]

It would actually be better if the problem were specified in a manner that wasn't coordinate dependent, but the OP seems narrowly focused on doing things "their way" rather than learning new things.

In a way it's true, I'm focused on learning if GR would predict/allow linear frame dragging.

 

[PeroK]

By "spacetime isn't static" you mean linear frame dragging? It is possible? And it means that the clock on the Earth would not be affected at all by the orbital speed (of the Earth) around the Sun?

A clock measures the spacetime distance along its worldline (which is equivalent to the proper time in its rest frame, if you prefer). That's a postulate of GR - or embedded in the theory in any case.

If you use the mathematical machinery of GR, then all you'll do is come up with this answer: that a specific clock on a specific worldline ... measured the spacetime distance along its worldline.

The equations of GR are not going to come up with something different, because that definition of a clock is embedded in the mathematics.

In particular, if a clock is moving in a particlar reference frame, you cannot ignore that motion and somehow get the "right" answer - by considering the distance along a different worldline, that the clock is not following?

 

[Dale]

In a way it's true, I'm focused on learning if GR would predict/allow linear frame dragging.

Linear frame dragging is predicted by GR. But it is not what you are thinking, it is not relevant to this scenario (I don’t mean it is present but too small to notice, I mean it doesn’t exist in this scenario at all)

 

[Ibix]

By "spacetime isn't static" you mean linear frame dragging?

No. I mean it doesn't have a timelike Killing vector field, which means that there's no way to pick a definition of "space" that doesn't change with time (in this case, because the orbiting planet loses energy through gravitational radiation and spirals into the star). That means that there's no non-arbitrary division of spacetime into space and time; hence no non-arbitrary definition of velocity; hence no non-arbitrary definition of kinematic time dilation.

If you are happy to accept PeroK's approximation then this doesn't matter (the Earth's kinetic energy is about 3×10³³J and its gravitational radiation emission is a few tens of Watts, if memory serves). If you aren't happy to accept it then there is no non-arbitrary answer to your question (although any sensible choice of coordinates will yield an answer close to the linear approximation).

 

[DanMP]

Linear frame dragging is predicted by GR. But it is not what you are thinking, it is not relevant to this scenario (I don’t mean it is present but too small to notice, I mean it doesn’t exist in this scenario at all)

Why? How do you know that linear frame dragging is not relevant to this scenario?

Do you know any experimental test/observation similar/related with this scenario that would confirm your claim?

 

[Dale]

Why? How do you know that linear frame dragging is not relevant to this scenario?

Linear frame dragging refers to the change in axes that happens as a gravitating object accelerates linearly near a test mass. In the solar system all of the gravitating objects are moving in free fall (geodesic), so their acceleration is 0.

Do you know any experimental test/observation similar/related with this scenario that would confirm your claim?

Sure, we experimentally observe that the sun and the planets are in free fall.

 

[DanMP]

Linear frame dragging refers to the change in axes that happens as a gravitating object accelerates linearly near a test mass.

Please provide a source/link for the above interpretation, because I searched and didn't find it ...

Sure, we experimentally observe that the sun and the planets are in free fall.

This is not what I meant. When I wrote "experimental test/observation similar/related with this scenario" I meant an experimental test similar with this:

the spaceship clock starts from the Earth, where is synchronized with the other one, then flies off the Earth, going on orbit around the Sun, with the same speed as the Earth, but far from it (few millions km away). After a while (months, years), the spaceship returns and the reunited clocks are compared.

and observations related/similar to such a scenario.

 

[Dale]

Please provide a source/link for the above interpretation, because I searched and didn't find it ...

See point 2 both before and after the derivation of equation 118 here:
https://en.m.wikisource.org/wiki/The_Meaning_of_Relativity/Lecture_4

 

[DanMP]

See point 2 both before and after the derivation of equation 118 here:
https://en.m.wikisource.org/wiki/The_Meaning_of_Relativity/Lecture_4

If I understood it correctly, it was about how a test body would be affected:

2. There is an inductive action of accelerated masses, of the same sign, upon the test body.

Now, I am interested in the effect of frame dragging on a clock with respect to timekeeping. In Wikipedia I see:

Rotational frame-dragging (the Lense–Thirring effect) appears in the general principle of relativity and similar theories in the vicinity of rotating massive objects. Under the Lense–Thirring effect, the frame of reference in which a clock ticks the fastest is one which is revolving around the object as viewed by a distant observer. This also means that light traveling in the direction of rotation of the object will move past the massive object faster than light moving against the rotation, as seen by a distant observer.

The text I colored in red seems to imply that a clock that revolves around a rotating massive object with the same angular speed (seen by a distant observer) as the dragged frame, would have no kinematic time dilation due to the rotation the distant observer sees. In other words, in the hypothetical case that the Sun is not rotating (no rotational frame dragging for the Sun) and the Earth is at the same/usual distance from the Sun, but stationary (in Sun's frame), and spinning very fast around its axis, a clock near Earth, co-moving with the rotational dragged frame would be faster than a clock at the same distance from the Earth but static in Sun's frame. How would GR explain that using worldlines?

 

[Dale]

If I understood it correctly, it was about how a test body would be affected:

Yes, a test body is any body, such as a clock, whose mass is small enough that it’s gravity is negligible.

In Wikipedia I see:

That is about rotational frame dragging, not linear frame dragging. Are you aware of that? I am not sure if you are deliberately changing topics now or if you think this is the same topic.

The text I colored in red seems to imply that a clock that revolves around a rotating massive object with the same angular speed (seen by a distant observer) as the dragged frame, would have no kinematic time dilation

I would have to work the math to be sure, but that sounds right. However, with the Earth surface clock at the pole it doesn’t matter in this scenario.

a clock near Earth, co-moving with the rotational dragged frame would be faster than a clock at the same distance from the Earth but static in Sun's frame. How would GR explain that using worldlines?

The rotating worldline is longer than the static worldline.

 

[pervect]

I'm not sure I understand the exact meaning of "linear frame dragging". In the weak field case, can we equate "linear frame dragging" with gravitomagnetism? WIki has a very specific definition of frame dragging that involves non-static stationary systems. Because a moving object isn't stationary unless the object is at rest, this would make fitting linear frame dragging into this definition problematic and most likely impossible, depending on exactly what the term means. However, in talking about frame dragging Wiki mentions "More generally, the subject of effects caused by mass–energy currents is known as gravitomagnetism, in analogy with classical electromagnetism". <<link>>

But I'm not sure I trust Wiki that much in this case, and I don't have a formal textbook definition at hand of what "frame dragging" and especially "linear frame dragging" actually means.

If we don't worry about the semantics of the meaning of "linear frame dragging", we can say that gravitomagnetic effects don't exist without mass-currents, and do exist when mass currents are present. As a consequence, we can conclude that the presence of gravitomagnetism (like the presence of time dilation) dependent on one's frame of reference. There will be no gravitomagnetic effects in a frame of reference at respect to a nearby massive body, but there will be gravitomagnetic effects in a frame of reference that is moving near a massive body.

We can even take the notion of gravitomagnetism out of the weak field into the strong field, by identifying the "gravitomagnetic field" as a particular part of the Riemann tensor, the so-called "magnetic part of the Rieemann".

This may be helpful, but if the OP has some other idea in mind of what the meaning of "linear frame dragging" is, it might not be relevant to the underlying question.

 

[sweet springs]

we have 2 atomic clocks on Earth's orbit around the Sun, one on Earth's surface, at one pole, and the other on a spaceship, far from Earth, but traveling with the same speed around the Sun, the clocks would suffer the same kinematic time dilation or not?

A simpler case of GPS satellite clock adjustment could help you. Against clock set on Earth’s surface, satellite clocks goes faster by GR gravity potential difference and goes slower by SR going round speed. Actual situation shows the former effect almost doubles the latter so satellite clock goes faster. We have to make periodical adjustment of satellite clock back.

 

[jbriggs444]

A simpler case of GPS satellite clock adjustment could help you. Against clock set on Earth’s surface, satellite clocks goes faster by GR gravity potential difference and goes slower by SR going round speed. Actual situation shows the former effect almost doubles the latter so satellite clock goes faster. We have to make periodical adjustment of satellite clock back.

There are periodic adjustments for other effects. The particular gross effect you mention was compensated for by design. The satellite clocks intentionally tick slowly so that the clocks in orbit and the clocks on the ground will all stay roughly synchronized in an Earth-centered inertial frame.

http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html

 

[DanMP]

That is about rotational frame dragging, not linear frame dragging. Are you aware of that? I am not sure if you are deliberately changing topics now or if you think this is the same topic.

Yes, it's about rotational frame dragging, but I think it is helpful in understanding how frame dragging in general may affect timekeeping, so in a way it is the same topic, as you will see soon.

I would have to work the math to be sure, but that sounds right. However, with the Earth surface clock at the pole it doesn’t matter in this scenario.

In this (second) scenario, both clock's are flying around the Earth, over the equator, one of them with the same speed (seen from Sun's frame) as the Earth's dragged frame (but this speed is now significant, not negligible as it is in reality) and the other with the speed needed to appear stationary from Sun's perspective (remember that the Earth is now stationary in Sun's frame).

The rotating worldline is longer than the static worldline.

This would mean that from Sun's frame/perspective the clock flying with the same speed as the Earth's dragged frame should tick slower than the other (static in Sun's frame), contradicting the red text in my quote from Wikipedia. What do you think about this contradiction?

 

[Dale]

In this (second) scenario, both clock's are flying around the Earth, over the equator, one of them with the same speed (seen from Sun's frame) as the Earth's dragged frame (but this speed is now significant, not negligible as it is in reality) and the other with the speed needed to appear stationary from Sun's perspective (remember that the Earth is now stationary in Sun's frame).

I don’t think that this scenario is self consistent. At least, I don’t see a way to translate it from English into math. There are not any “stationary” geodesics that the Earth can follow. Perhaps you intend for the Earth to be kept in place with some gigantic rocket thrusters?

This would mean that from Sun's frame/perspective the clock flying with the same speed as the Earth's dragged frame should tick slower than the other (static in Sun's frame), contradicting the red text in my quote from Wikipedia. What do you think about this contradiction?

It is not a contradiction, it is a mistake on your part. The longer worldline is the “faster” clock. The longer the worldline the more ticks between two given events, and more ticks is a faster clock.

 

[PeterDonis]

The text I colored in red seems to imply that a clock that revolves around a rotating massive object with the same angular speed (seen by a distant observer) as the dragged frame, would have no kinematic time dilation due to the rotation the distant observer sees.

It's more complicated than that.

First, consider the simpler case of a non-rotating, spherically symmetric massive body. Spacetime in the vacuum region outside this body is described by the Schwarzschild metric. This metric has a timelike Killing vector field--the term for this is "stationary"--and this KVF is hypersurface orthogonal--the term for this is "static". Note that in an earlier post, "has a timelike KVF" was called "static", but that's really not correct; the KVF also has to be hypersurface orthogonal, which basically means the gravitating mass can't be rotating. Since you're interested in the case where it is rotating, we have to be careful to distinguish "static" and "stationary".

Now suppose we have a test body in a free-fall circular orbit around the non-rotating massive body. If this body exchanges light signals with an observer at rest at infinity, the observer will calculate that the body's clock runs slow compared to the observer's clock. If we call this "time dilation", we can in the static case, for convenience, split it up into two pieces: gravitational and kinematic. The split is easily visualized: gravitational time dilation depends only on altitude above the object, and kinematic time dilation depends only on speed relative to the object. However, if we want to be rigorous, we have to define exactly what "height" and "speed" mean in terms of the known properties of the spacetime.

We do this as follows: we use the integral curves of the timelike KVF to define "points in space"--each curve corresponds to such a point (more precisely, to the worldline of such a point--you can think of each such curve as the worldline of a hypothetical observer if you like). Each point then has an altitude, defined as its areal radius $r = \sqrt{A / 4 \pi}$, where $A$ is the area of a 2-sphere centered on the massive body and containing the point. Gravitational time dilation then depends only on $r$.

We then define "speed" as the speed of an object relative to the points in space--i.e., relative to the hypothetical observers whose worldlines are the integral curves of the timelike KVF. So, for example, we can imagine a set of such observers that each occupy one of the points along the circular orbit of the test body described above, each of whom is measuring the speed of the test body as it passes by. Since the orbit is circular, all of these observers will measure the same speed. The kinematic time dilation of the body is then just the usual SR time dilation using this speed.

Now suppose the massive body is rotating. We assume that this means spacetime in the vacuum region outside the body is described by the Kerr metric (note that nobody has actually proven that this is the case, but it's a convenient assumption and should be at least reasonably accurate). The Kerr metric is stationary, but not static--it has a timelike KVF, but the KVF is not hypersurface orthogonal. Does this make a difference to the above, and if so, what?

It turns out that, as far as defining "points in space" are concerned, it makes no difference. The integral curves of the timelike KVF are the same as for the non-rotating case: they are worldlines of observers at rest relative to the center of mass of the rotating massive body. And each point in space still has an "altitude" which determines the gravitational time dilation at that point; the only difference is that now the "altitude" depends, not just on the areal radius $r$, but also on the colatitude $\theta$ (basically, $\theta = 0$ denotes the "north pole" of the body, $\theta = \pi / 2$ denotes the "equator", and $\theta = \pi$ denotes the "south pole"). But the key point is that these worldlines are not rotating: they are still at rest relative to the body's center of mass, and relative to the observer at infinity as well.

We can also still define "speed" as we did before--speed relative to observers at rest. So we still have well-defined notions of "altitude" and "speed". But we do have a problem: we can no longer split up the total time dilation into a "gravitational" piece that depends only on altitude and a "kinematic" piece that depends only on speed. And one way of looking at the reason why is the thing you pointed out from the Wikipedia article: the state of motion that has "minimum time dilation", for a given altitude relative to an observer at infinity, is rotating around the massive body (because of frame dragging), so it's not at rest. (Note that in the non-rotating case, the "at rest" state was also the state of "minimum time dilation" for its altitude--only gravitational, no kinematic.)

 

[PeroK]

This would mean that from Sun's frame/perspective the clock flying with the same speed as the Earth's dragged frame should tick slower than the other (static in Sun's frame), contradicting the red text in my quote from Wikipedia. What do you think about this contradiction?

The Wikipedia article relates to the relative time dilation of clocks orbiting the Earth only (as the rotating mass). If one clock is orbiting the Earth at just the right speed to match the "frame dragging", associated with the Earth's rotation, then it will be less time dilated than a clock at rest in the Earth's atmosphere.

This is completely unrelated to the time dilation of a clock at rest relative to the Sun, compared to one in orbit round the Sun - whether that clock is on the Earth and whether the Earth is rotating or not. The Wikipedia analysis has no second massive body about which the rotating mass is orbiting!

There are no contradictions here, only an increasingly elaborate scenario, with several competing factors. To an external observer, we have many factors:

Sun's mass
Angular momentum of the Sun
Earth's orbit round the Sun
Earth's mass
Angular momentum of the Earth
Motion of a clock relative to the Sun
Motion of a clock relative to the Earth

The question is which of these are numerically significant; and, how do you combine these factors (especially if not by linear combination). You don't get a contradiction if you look at only two of these factors and conclude that clock A is more time dilated than B; or, if you consider a different two factors and conlcude that clock A is less time dilated than B.

To compare the time dilation for two clocks you need to take all (seven) factors into account.

 

[DanMP]

Perhaps you intend for the Earth to be kept in place with some gigantic rocket thrusters?

Yes, something like that or not so gigantic if the Earth was orbiting the Sun at much greater distance.

It is not a contradiction, it is a mistake on your part. The longer worldline is the “faster” clock. The longer the worldline the more ticks between two given events, and more ticks is a faster clock.

Sorry for the mistake , you are right, but, on the other hand, if you remove the Earth (but leave the static clock there) and introduce a third clock, orbiting the Sun at the same distance as the static clock, this third clock would have (if I'm not mistaking again) a longer worldline than the static one, but would not be faster. So, if I'm not mistaking again, there is a contradiction.

 

[DanMP]

... the state of motion that has "minimum time dilation", for a given altitude relative to an observer at infinity, is rotating around the massive body (because of frame dragging), so it's not at rest.

So, if the state of motion that has "minimum time dilation" is rotating around the massive body (because of frame dragging), what is happening when the Earth is orbiting the Sun (back to the OP scenario)? If the frame with "minimum time dilation" can be dragged around the Earth, why it would be impossible to be also dragged along the orbit?

 

[PeterDonis]

if you remove the Earth (but leave the static clock there) and introduce a third clock, orbiting the Sun at the same distance as the static clock, this third clock would have (if I'm not mistaking again) a longer worldline than the static one, but would not be faster.

No. The orbiting clock would have a shorter worldline than the static clock between successive points where the two meet. And, correspondingly, the orbiting clock would run slower. Remember that this is spacetime, not space, and your ordinary intuitions about Euclidean geometry do not work. The orbiting worldline seems "longer" to your intuitions (because it spirals around rather than going straight up, so to speak), but your intuitions are wrong.

So, if the state of motion that has "minimum time dilation" is rotating around the massive body (because of frame dragging), what is happening when the Earth is orbiting the Sun (back to the OP scenario)?

If we consider the Sun as rotating (it is, but very slowly, much more slowly than the Earth), the "minimum time dilation" state of motion at the distance of the Earth's orbit will be rotating around it, but much, much slower than the Earth.

 

[Dale]

if you remove the Earth (but leave the static clock there) and introduce a third clock, orbiting the Sun at the same distance as the static clock, this third clock would have

I am sorry but all of these different scenarios are becoming frustrating to me. I think I am done with this discussion at this point. It feels like a moving target.

Best of luck in getting your question answered or at least in figuring out what your question is, but I am not enjoying participating any more.

 

[DanMP]

No. The orbiting clock would have a shorter worldline than the static clock between successive points where the two meet.

Ok, but

The rotating worldline is longer than the static worldline.

If we consider the Sun as rotating (it is, but very slowly, much more slowly than the Earth), the "minimum time dilation" state of motion at the distance of the Earth's orbit will be rotating around it, but much, much slower than the Earth.

This is not what I meant. When I said "If the frame with "minimum time dilation" can be dragged around the Earth, why it would be impossible to be also dragged along the orbit?" I meant some sort of linear frame dragging "done" by the Earth, not a rotational frame dragging due to Sun's rotation. If this "linear" frame dragging does not occur, it would mean that rotational frame dragging around the Earth would seem/be from the Sun's perspective as some kind of fast moving rotational deformation of the spacetime along Earth's orbit.

I am sorry but all of these different scenarios are becoming frustrating to me ...
... at least in figuring out what your question is ...

I'm sorry for these side-scenarios, but they were needed for a better understanding of the OP scenario/problem/question.

The Wikipedia analysis has no second massive body about which the rotating mass is orbiting!

There is such an analysis elsewhere? If not, don't you think that it would be something interesting, important for a better understanding of relativity and a new way to test it?By the way, I also asked about

experimental test/observation similar/related with this scenario

Regarding a possible test, it may be easier in the Earth/Moon system than in Sun/Earth one. Do we have an atomic clock on the moon?

 

[PeterDonis]

The rotating worldline is longer than the static worldline.

This is not correct. The rotating worldline is shorter; hence, the clock following the rotating worldline runs slower. More precisely, the static clock has more elapsed time between two successive meetings of the two clocks than the rotating clock does; hence, the rotating worldline is shorter between those two events than the static worldline is.

 

[PeterDonis]

Ok, but

See my post #47 just now.

When I said "If the frame with "minimum time dilation" can be dragged around the Earth, why it would be impossible to be also dragged along the orbit?" I meant some sort of linear frame dragging "done" by the Earth. If this "linear" frame dragging does not occur, it would mean that rotational frame dragging around the Earth would seem/be from the Sun's perspective as some kind of fast moving rotational deformation of the spacetime along Earth's orbit.

First, if we want to consider both the Sun and the Earth as having non-negligible mass and therefore causing non-negligible spacetime curvature, then we no longer have any exact solutions to use; there is no known exact solution for the case of two massive bodies. The main reason for this is that GR is nonlinear: you can't just add together two solutions and have another solution. Such cases, in general, have to be solved numerically. See further comments below.

For sufficiently weak fields, the nonlinearities should be small, so we could approximate a two-body solution by superposing solutions for the Sun alone and the Earth alone. This appears to be what you are intuitively doing. However, even then there are complications. First, linear frame dragging, as I understand it, requires the massive body doing the dragging to have nonzero proper acceleration. The Earth orbiting the Sun does not; it's in free fall. Second, even if I'm wrong in the previous sentence, linear frame dragging is a tiny, tiny effect, much smaller even than rotational frame dragging, so any such effect on the "minimum time dilation" state of motion for the Sun-Earth system would in all probability be unobservable with the accuracy of measurement we can achieve now or for the foreseeable future.

All that said, the "minimum time dilation" state of motion around the Sun, at the distance of the Earth, is not going to be anywhere near the Earth; it's going to be as far away from the Earth as possible (if we're just considering the Sun and the Earth), because the extra time dilation from the Earth's gravity well is much, much larger than any possible "frame dragging" effect.

don't you think that it would be something interesting, important for a better understanding of relativity and a new way to test it?

GR has already been tested for multi-body systems, using numerical solutions. The solar system and binary pulsar systems are two well known examples.

 

[Dale]

This is not correct. The rotating worldline is shorter;

This is one of the dangers of all of these random scenarios he has been throwing out.

The rotating worldline referred to above is one that is going around a rotating object at the right speed and direction to minimize time dilation. Due to rotational frame dragging this is not a stationary worldline, but one that slowly revolves around the central object. This scenario was identified in the Wikipedia article on frame dragging.

It is not a clock orbiting the sun at 1 AU. So my comment was correct, that worldline is longer, it was just referring to a scenario from a different random tangent. This is why I am done with this conversation.

 

[PeterDonis]

The rotating worldline referred to above is one that is going around a rotating object at the right speed and direction to minimize time dilation.

Ah, ok. You're right, all the different scenarios have gotten very garbled in this thread.