SR-time dilation vs GR-time dilation on 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.

 

[cianfa72]

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.

Ok, so ECI coordinates are not simply the "inverse mapping" of spatial Schwarzschild coordinates $(r,\theta,\phi)$ of Schwarzschild spacetime.

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

Ah ok, the Kerr metric is asymptotically flat however the spatial BL-coordinates at infinity are not the standard polar coordinates.

 

[PeterDonis]

so ECI coordinates are not simply the "inverse mapping" of spatial Schwarzschild coordinates $(r,\theta,\phi)$ of Schwarzschild spacetime.

That's correct.

the Kerr metric is asymptotically flat however the spatial BL-coordinates at infinity are not the standard polar coordinates.

That's correct.

 

[cianfa72]

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 $r \to \infty$.

About this point I reasoned as follows: since the Kerr metric is stationary it follows that light beam paths $ds^2=0$ with fixed $(\theta, \phi)$ are described in BL-coordinates by ${dr} / {dt}$ that doesn't depend on coordinate time $t$ along the path. Hence light beam paths emitted at different coordinate time from an observer at fixed $(r,\theta, \phi)$ -- included at $r \to \infty$ -- are actually "congruent" in BL-chart. Therefore the amount of coordinate time between sending and receiving back a light beam is always the same. Since the metric doesn't depend on $t$ the amount of proper time along the observer worldline doesn't change as well.

 

[PeterDonis]

I reasoned as follows

Yes, this is correct.

 

[cianfa72]

Yes, this is correct.

Ok, therefore the above reasoning holds true in any stationary spacetime.

 

[PeterDonis]

Ok, therefore the above reasoning holds true in any stationary spacetime.

As you stated it, it only holds for radial trajectories in a stationary spacetime in which spherical spatial coordinates can be adopted.

However, it should be straightforward to see that the argument can be generalized so as to apply to any round-trip light signals between two fixed integral curves of the timelike Killing vector field.

 

[cianfa72]

However, it should be straightforward to see that the argument can be generalized so as to apply to any round-trip light signals between two fixed integral curves of the timelike Killing vector field.

Yes, since in a stationary spacetime there are coordinates such that integral curves of timelike KVF are described by fixed spatial (spacelike) coordinates and varying coordinate time $t$. Therefore, fixed two of them, one can apply the above argument to get the result.

 

[PeterDonis]

fixed two of them, one can apply the above argument to get the result.

You don't even have to use coordinates to make the argument. Just the fact that you have two fixed integral curves of the timelike KVF is enough.

 

[cianfa72]

You don't even have to use coordinates to make the argument. Just the fact that you have two fixed integral curves of the timelike KVF is enough.

Ok, the point is that light cones from any point on fixed integral curves of timelike KVF are congruent each other (when "trasported" along the flow of timelike KVF).

 

[PeterDonis]

light cones from any point on fixed integral curves of timelike KVF are congruent each other (when "trasported" along the flow of timelike KVF).

Yes.

 

[cianfa72]

Coming back to ECI frame: we said it is not the "inverse mapping" of spherical coordinates of Schwarzschild spacetime in Schwarzschild coordinates. So which is their significance ?

In other word: which is the "rule" for assigning ECI coordinates to a point ?

 

[PeterDonis]

which is the "rule" for assigning ECI coordinates to a point ?

https://en.wikipedia.org/wiki/Earth-centered_inertial

 

[cianfa72]

The Wikipedia link claims that for ECI frame Cartesian coordinates can be used. However my doubt is: since spacetime is not flat, $z$ coordinate of a point along the Earth's rotating axis is not the Euclidean length from the Earth's center up to that point P along $z$ axis. In other words Euclidean geometry doesn't apply to such Cartesian coordinates.

 

[PeterDonis]

The Wikipedia link claims that for ECI frame Cartesian coordinates can be used.

Yes.

However my doubt is: since spacetime is not flat

Cartesian coordinates do not require spacetime to be flat. Cartesian coordinates does not mean the metric of a spacelike hypersurface of constant time must be Euclidean, any more than spherical coordinates means $r$ must be the physical distance from the center.

 

[cianfa72]

Cartesian coordinates do not require spacetime to be flat. Cartesian coordinates does not mean the metric of a spacelike hypersurface of constant time must be Euclidean, any more than spherical coordinates means $r$ must be the physical distance from the center.

Ok, so the key feature of Cartesian coordinates is that they are composed by mutually orthogonal geodesic coordinate lines regardless of the fact that coordinate values may not be the same as the physical distance measured along any of them from the center.

 

[PeterDonis]

the key feature of Cartesian coordinates is that they are composed by mutually orthogonal geodesic coordinate lines regardless of the fact that coordinate values may not be the same as the physical distance measured along any of them from the center.

Yes.

 

[cianfa72]

The weird thing is that, for example in 2D cartesian coordinates, a geodesic starting orthogonal to the $x$ axis and another starting orthogonal to the $y$ axis might not intersect orthogonally (i.e. forming 4 congruent angles).

 

[PeterDonis]

in 2D cartesian coordinates

On what manifold? A flat Euclidean plane? Or something else?

 

[cianfa72]

On what manifold? A flat Euclidean plane? Or something else?

I was talking about a generic 2D surface (or better a non-Euclidean 3D space), like the spacelike hypersurface of constant coordinate time $t$ discussed in post #49.

 

[PeterDonis]

I was talking about a generic 2D surface (or better a non-Euclidean 3D space)

Cartesian coordinates are generally used for cases where the space is very close to Euclidean. That is true for the spacelike surfaces of constant time that you refer to.

 

[cianfa72]

Cartesian coordinates are generally used for cases where the space is very close to Euclidean. That is true for the spacelike surfaces of constant time that you refer to.

So, as in Schwarzschild spacetime in Schwarzschild coordinates the coordinate $r$ represents the square root of nested 2-spheres's area divided by $4\pi$ (and it is not the radial distance from the center along the spacelike geodesic integral curves of $\partial_r$), which is the significance of Cartesian coordinates (e.g. $x$) in ECI frame ?

 

[PeterDonis]

which is the significance of Cartesian coordinates (e.g. $x$) in ECI frame ?

Have you read the article I referenced? What does it say?

 

[cianfa72]

Have you read the article I referenced? What does it say?

Yes, the Wikipedia link says that the orientation of ECI frame is defined by the ecliptic and the Earth's rotation axis (i.e. the fundamental plane is the ecliptic).

The location of an object in space can be defined in terms of right ascension and declination which are measured from the vernal equinox and the celestial equator.

Then it claims:

Locations of objects in space can also be represented using Cartesian coordinates in an ECI frame.

However, it seems to me, there is not a definition of Cartesian coordinates applied to ECI frame. Are they just the "inverse mapping" of right ascension and declination in spherical coordinates ?

 

[PeterDonis]

Are they just the "inverse mapping" of right ascension and declination in spherical coordinates ?

That in itself is not sufficient; you would also need $r$ in spherical coordinates. The vernal equinox and the celestial equator define the orientation of the Cartesian axes.

 

[cianfa72]

That in itself is not sufficient; you would also need $r$ in spherical coordinates. The vernal equinox and the celestial equator define the orientation of the Cartesian axes.

Yes of course. For any 2-sphere of given $r$ of where the point to be located is, one gets the Cartesian coordinates by "inverse mapping" its right ascension and declination.

So the question is: what is the $r$ coordinate ? I don't think it is the $r$ coordinate of Schwarzschild coordinates in Schwarzschild spacetime (i.e. it doesn't have the same significance/feature as "reduced circumference").