SR-time dilation vs GR-time dilation on rotating Earth

[cianfa72]

TL;DR Summary

About SR vs GR time dilation effect on rotating Earth

Reading this old thread does-special-and-general-relativity-affect-aging-of-apollo-astronauts I came with some question.

To me isn't clear how SR-time dilation by rotation of the Earth is compensated by GR-time dilation effect.

 

[Dale]

Personally, I would not consider the SR effect to be separate from the GR effect. SR is part of GR. Similarly, I would not consider the gravitational potential to be separate from the centrifugal potential.

 

[Sagittarius A-Star]

To me isn't clear how SR-time dilation by rotation of the Earth is compensated by GR-time dilation effect.

I described it in other words here:
https://www.physicsforums.com/threads/gravity-at-earths-poles-vs-equator.1065794/post-7121118

A clock at the equator at sea level is in a higher gravitational potential than a clock at a pole at sea level - with respect to the (non-rotating) ECI frame. In this frame, the clock at the equator is moving (SR-time-dilation). In the (rotating) ECEF frame, the clock at the equator is at rest, but the gravitational potential is modified by the centrifugal force (also SR-time-dilation).

In 1905, when Einstein couldn't consider GR-effects, he wrote:

From this, we conclude that a clock placed at the equator must be slower by a very small amount than a similarly constructed clock which is placed at the pole, all other conditions being identical.

Source (see end of §4):
https://en.wikisource.org/wiki/On_the_Electrodynamics_of_Moving_Bodies_(1920_edition)

 

[Vanadium 50]

I would argue "please reopen this thread just for me" is, well, sounds like a privileged place in the universe, and cosmology tells us thee is no such thing.

More importantly, why don't you calculate it yourself? If you don't know how, reread the thousands of posts PF members have made explaining it to you.

 

[cianfa72]

I described it in other words here:
https://www.physicsforums.com/threads/gravity-at-earths-poles-vs-equator.1065794/post-7121118

A clock at the equator at sea level is in a higher gravitational potential than a clock at a pole at sea level - with respect to the (non-rotating) ECI frame. In this frame, the clock at the equator is moving (SR-time-dilation). In the (rotating) ECEF frame, the clock at the equator is at rest, but the gravitational potential is modified by the centrifugal force (also SR-time-dilation).

In ECI frame the center of the Earth is at rest. One can assume the Earth's center is free-falling. Does an object at fixed $(x,y,z)$ in it in free-fall as well?

 

[PeterDonis]

isn't clear how SR-time dilation by rotation of the Earth is compensated by GR-time dilation effect.

In general there is not a well-defined split between "SR time dilation" and "GR time dilation".

In a stationary spacetime, you can use the timelike Killing vector field to separate the two, but in that case the "compensation" is perfectly clear, and, as @Vanadium 50 has already pointed out, there are plenty of previous threads that discuss how it works.

 

[PeterDonis]

In ECI frame the center of the Earth is at rest. One can assume the Earth's center is free-falling. Does an object at fixed $(x,y,z)$ in it in free-fall as well?

Can't you compute its proper acceleration and see?

In fact, since in the post you were responding to the objects were at rest on the surface of the rotating Earth, it's even simpler: is an object at rest on the surface of the rotating Earth in free fall?

 

[PeterDonis]

Moderator's note: Thread level changed to "I".

 

[cianfa72]

Can't you compute its proper acceleration and see?

In fact, since in the post you were responding to the objects were at rest on the surface of the rotating Earth, it's even simpler: is an object at rest on the surface of the rotating Earth in free fall?

I was interested in the proper acceleration of an object at rest in ECI frame not on the rotating Earth.

If one considers a rotating gravitating body like Earth, then the relevant spacetime is Kerr spacetime which is stationary but not static (btw I think this model actually neglects the effect of Universe mass distribution). Since spacetime around the gravitating body isn't flat, ECI frame can't be inertial (only the Earth's center at rest in it is free-falling i.e. has zero proper acceleration).

Therefore I believe objects at rest in ECI frame aren't in free-fall in general.

So the qualification Inertial in ECI="Earth Centered Inertial" refers just to the Earth's center?

 

[Dale]

I was interested in the proper acceleration of an object at rest in ECI frame not on the rotating Earth.

The proper acceleration of an object at rest in the ECI is non-zero except at the center. A better name would have been ECNR (earth centered non-rotating)

 

[Vanadium 50]

At the risk of repeating not just myself, why don't you calculate this?

 

[cianfa72]

At the risk of repeating not just myself, why don't you calculate this?

You mean calculating the proper acceleration of an object at rest at $(x,y,z)$ ECI coordinates. To do this we can assume either non-rotating or rotating Earth. That means Schwarzschild vs Kerr spacetime respectively.

In the former I think ECI $(x,y,z)$ are actually the "corresponding" of Schwarzschild coordinates $(r,\theta, \phi)$.

Does it make sense to do such an assumption and calculate accordingly ?

 

[Vanadium 50]

It;s your question. Don't ask me what I mean. What do you mean? (And if you don't know, why are you wasting everyone's time?)

 

[Dale]

Does it make sense to do such an assumption and calculate accordingly ?

Yes. And it would be instructive to do it for both and compare the difference

 

[cianfa72]

It;s your question. Don't ask me what I mean. What do you mean? (And if you don't know, why are you wasting everyone's time?)

Sorry, it isn't my intention to wast your time. I was simply asking whether that kind of scenarios make sense or not.

 

[Vanadium 50]

Stop trying to wriggle out of it. If you are interested in the results of the calculation, do the calculation. Or, better still, as @Dale says, do it both ways and compare.

 

[Vanadium 50]

And it would be instructive to do it for both and compare the difference

It would, bur probably calculationally, not physically. Gravitational time dilation is a ppb effect, so perturbations will be ppt, and because they tend to pull in oppiste directions (alternatively, surfaces are nearly equipotential - same physics, different words) maybe even ppq.

These are tiny, tiny effects. If you want to learn how to calculate tiny, tiny effects, doing the calculation is instructive. If you just want other people to do the calculation for you because you told them to, I fail to see the point.

 

[PeterDonis]

I was interested in the proper acceleration of an object at rest in ECI frame not on the rotating Earth.

That question is just as simple: what would be the proper acceleration of an object at rest on a non-rotating Earth?

 

[PeterDonis]

If one considers a rotating gravitating body like Earth, then the relevant spacetime is Kerr spacetime

Actually, no, we don't know that this is the case. There is no analogue of Birkhoff's theorem, which is what tells us that the vacuum surrounding a non-rotating spherical planet must be Schwarzschild, for the rotating case.

To put it another way, for exact spherical symmetry all multipole moments higher than monopole (mass) must be zero, whether there is matter present or not; but for the rotating case, Kerr spacetime has only two nonzero moments, mass and angular momentum, but a rotating planet can have higher multipole moments nonzero also (at a minimum, rotating planets like the Earth will have a nonzero quadrupole moment). These higher multipole moments must also be nonzero in the vacuum region around the rotating planet (measurements by orbiting satellites have measured the effects on the vacuum surrounding the Earth of the Earth's quadrupole moment), so that spacetime geometry can't be Kerr.

 

[PeterDonis]

I believe objects at rest in ECI frame aren't in free-fall in general.

You don't have to "believe". You can calculate. Or, if all you need is to answer the zero/nonzero question, you can apply simple physical arguments to the question I posed in post #18.

 

[cianfa72]

That question is just as simple: what would be the proper acceleration of an object at rest on a non-rotating Earth?

Ok, given the mapping of ECI $x,y,z$ coordinates to spherical coordinates $r,\theta, \phi$, one must evaluate the proper acceleration of timelike worldlines described by fixed corresponding Schwarzschild coordinates $(r,\theta,\phi)$ and varying coordinate time $t$, i.e. said $u$ the worldline's 4-velocity we need $$a^{\mu} = u ^{\nu}\nabla_{\nu} u^{\mu}$$
The computation gives that the only nonzero component is $$a^{r} = \frac {GM} {r^2}$$
Note that, given the physical meaning of Schwarzschild $r$ coordinate, a point at ECI coordinates $(0,0,z)$ has not proper length $z$ from the center of the Earth (i.e. the spacelike geodesic length joining the worldline described by $(r,0,\phi)$ and the worldline through the center of the Earth evaluated on any spacelike hypersurface of constant coordinate time $t$ is always the same since spacetime is stationary, however its value is not $z$).

 

[Sagittarius A-Star]

In ECI frame the center of the Earth is at rest. One can assume the Earth's center is free-falling. Does an object at fixed $(x,y,z)$ in it in free-fall as well?

No, as others mentioned.
It maybe helpful to look, how Haefele and Keating used the ECI frame for an approximate calculation of their twin"paradox" experiment.

Around-the-World Atomic Clocks:
Predicted Relativistic Time Gains
...
Nevertheless, the relative timekeeping behavior of terrestrial clocks can be evaluated by reference to hypothetical coordinate clocks of an underlying nonrotating (inertial) space (6).
For this purpose, consider a view of the (rotating) earth as it would be perceived by an inertial observer looking down on the North Pole from a great distance. A clock that is stationary on the surface at the equator has a speed $R\Omega$ relative to nonrotating space, and hence runs slow relative to hypothetical coordinate clocks of this space in the ratio $1 - R^2\Omega^2/2c^2$, where $R$ is the earth's radius and $\Omega$ its angular speed
...
General relativity predicts another effect that (for weak gravitational fields) is proportional to the difference in the gravitational potential for the flying and ground reference clocks.
...

Source:
https://web.archive.org/web/2017033...psu.edu/rq9/HOW/Atomic_Clocks_Predictions.pdf

 

[cianfa72]

If my post #21 looks correct, next step is the rotating Earth case. One can reasonably assume Kerr metric to make the calculation.

Kerr metric in BL-coordinates employs oblate spheroidal coordinates $(r,\theta,\phi)$. As in the non-rotating case (i.e. Schwarzschild spacetime), does a timelike worldline described by fixed ECI $(x,y,z)$ or the corresponding oblate $(r,\theta, \phi)$ and varying coordinate $t$ represent a non-rotating observer w.r.t. the rotating Earth's center ?

 

[A.T.]

To me isn't clear how SR-time dilation by rotation of the Earth is compensated by GR-time dilation effect.

In the rotating rest frame of the Earth, the shape of the surface tends towards an equipotential surface with respect to the effective potential (graviational + centrifugal, see geoid). Since in this frame there is no kinetic time dilation for objects at rest on the surface, and the gravitational time dilation depends on the difference in potential, the tick rates of clocks on that equipotential surface are the same.

In the non-rotating frame the centrifugal potential contribution to the gravitational time dilation turns into kinetic time dilation from tangential velocity. But the tick rates must still be the same on the geoid, so the two effects (graviational and kinetic) must cancel here.

 

[cianfa72]

In the non-rotating frame the centrifugal potential contribution to the gravitational time dilation turns into kinetic time dilation from tangential velocity. But the tick rates must still be the same on the geoid, so the to effects (graviational and kinetic) must cancel here.

As far as I can understand, you mean that effects of kinetic and gravitational time dilation are actually opposite, hence to give the same tick rate on any clock at rest on the geoid, they must "compensate" somehow I would say.

 

[A.T.]

As far as I can understand, you mean that effects of kinetic and gravitational time dilation are actually opposite, hence to give the same tick rate on any clock at rest on the geoid, they must "compensate" somehow I would say.

In the non-rotating frame the geoid is not an equipotential surface anymore, because there is no centrifugal potential in that frame. So based on potential alone, the clocks on it would not tick at the same rate. But the kinetic time dilation from tangential motion accounts for that discrepancy, and combined with the potential it results with equal clock rates on the geoid.

 

[Vanadium 50]

As far as I can understand,

If you want to understand, do the calculation.

 

[cianfa72]

Btw, when one talks of "rest frame of geoid" or in general "rest frame of something" it is not univocally defined. Indeed, starting from a "rest chart/coordinates", any one-to-one transformation of the timelike coordinate into another one that doesn't involve spacelike coordinates also defines a valid "rest frame/chart" as well.

 

[PeterDonis]

One can reasonably assume Kerr metric to make the calculation.

Only as an approximation. Whether it's a good enough approximation depends on the specific case, for example how large a quadrupole moment the planet has and how close to the planet the object you are calculating for is. See my post #19.

 

[PeterDonis]

when one talks of "rest frame of geoid" or in general "rest frame of something" it is not univocally defined.

The worldlines of the geoid are uniquely defined. That only leaves this:

any one-to-one transformation of the timelike coordinate into another one that doesn't involve spacelike coordinates

But such a transformation does nothing but change the "tick rate" of coordinate time in the frame. But the actual tick rate of clocks on the geoid is also uniquely defined, so if we use that to define the "tick rate" of coordinate time, then the rest frame of the geoid is uniquely defined.

Note that the "rest frame of the geoid" as I have just defined it is not the ECI frame. It is the ECER ("Earth centered Earth rotating") frame. In other words, it's the natural frame we use in our everyday lives on the rotating Earth.

 

[cianfa72]

The worldlines of the geoid are uniquely defined.

But such a transformation does nothing but change the "tick rate" of coordinate time in the frame. But the actual tick rate of clocks on the geoid is also uniquely defined, so if we use that to define the "tick rate" of coordinate time, then the rest frame of the geoid is uniquely defined.

Ok yes, you mean use the proper time along the geoid's timelike worldlines to define the "tick rate" of the coordinate time $t$ employed.

Regarding the Kerr spacetime model, I still have this doubt: does a worldline described in BL-coordinates by fixed values of oblate spheroidal coordinates $(r,\theta, \phi)$ and varying $t$ represent an observer at rest in ECI frame ?

 

[Vanadium 50]

To me isn't clear how SR-time dilation by rotation of the Earth is compensated by GR-time dilation effect.

Of course it's not. Because you have hijacked your own thread.

You got some very good advice when this thread was 2 or 3% of its present age. You declined to take it. That's your choice, but it will not lead to understanding. Just frustration on everybody's part, including your own.

 

[PeterDonis]

you mean use the proper time along the geoid's timelike worldlines to define the "tick rate" of the coordinate time $t$ employed.

Yes.

does a worldline described in BL-coordinates by fixed values of oblate spheroidal coordinates $(r,\theta, \phi)$ and varying $t$ represent an observer at rest in ECI frame ?

The ECI frame doesn't use Kerr spacetime as its background to begin with, so this question is meaningless.

A meaningful question would be whether a worldline with fixed spatial BL coordinates in Kerr spacetime is at rest relative to an observer at rest at infinity. The answer to this should be simple to read off from the Kerr metric and its limit as $r \to \infty$.

 

[cianfa72]

The ECI frame doesn't use Kerr spacetime as its background to begin with, so this question is meaningless.

Ok, so the background spacetime for ECI frame is Schwarzschild spacetime.

A meaningful question would be whether a worldline with fixed spatial BL coordinates in Kerr spacetime is at rest relative to an observer at rest at infinity. The answer to this should be simple to read off from the Kerr metric and its limit as $r \to \infty$.

In the limit $r \to \infty$ Kerr metric reduces to flat metric in polar coordinates. To answer the question posed, I believe one can use light beams exchanged from each other by verifying that the elapsed/proper time (along any of the two observer worldlines) between sending and receiving the beam back does not change over time.

Do you mean the above can be directly derived from the form of the Kerr metric and its limit as $r$ goes to infinity ?

 

[PeterDonis]

the background spacetime for ECI frame is Schwarzschild spacetime.

No, that's not quite correct either. The gravitational potential that is used in the ECI frame includes the effects of Earth's quadrupole moment, so it recognizes that the Earth is not precisely spherical. This affects the definition of the geoid.

In the limit $r \to \infty$ Kerr metric reduces to flat metric in polar coordinates.

Not quite, no. The rotation parameter $a$ does not vanish at infinity, so the coordinates cannot be standard polar coordinates.

I believe one can use light beams exchanged from each other by verifying that the elapsed/proper time (along any of the two observer worldlines) between sending and receiving the beam back does not change over time.

Yes.

Do you mean the above can be directly derived from the form of the Kerr metric and its limit as $r$ goes to infinity ?

Yes.