What is the angular velocity of a satellite?

[vanhees71]

I think one must be a bit careful with the notion of "vorticity". In fluid dynamics you have on the one hand a kinematic vorticity tensor related to the displacement vectors of the congruence given by the worldlines of the fluid elements and the vorticity tensor (or better circulation tensor to distinguish it from the kinematic vorticity tensor), which is a two form defined from the enthalpy current, $w^{\mu}=h u^{\mu}$, where $h$ is the proper enthalpy density (i.e., the enthalpy density as measured in the local fluid-rest frame). The vorticity tensor then is defined by
$$\Omega_{\mu \nu}=\nabla_{\nu} w_{\mu}-\nabla_{\nu} w_{\mu}=\partial_{\nu} w_{\mu}-\partial_{\mu} w_{\nu}.$$
That's proportional to the kinematic vorticity tensor only for "dust", i.e., a pressureless fluid which is in free fall and the worldlines are thus geodesics.

Of course these quantities are in some sense frame dependent, i.e., it depends on the congruence of worldlines defined by the fluid. The distinct local frames where the physics of the tensor quantities is most easily interpreted are of course the local rest frames of the fluid, which thus is characterized by scalar quantities measured in the local rest frames of the fluid.

A very nice reference about fluid dynamics in GR is

L. Rezzolla and O. Zanotti, Relativistic hydrodynamics, Oxford University Press, Oxford (2013),
https://dx.doi.org/10.1093/acprof:oso/9780198528906.001.0001

 

[PeterDonis]

I think one must be a bit careful with the notion of "vorticity".

The notion I am referring to in this thread, which is the one arising from the kinematic decomposition of a congruence, is what you are calling "kinematic vorticity".

 

[binis]

So what?

The observer is watching his wristwatch running faster than a clock on the earth.Real?

 

[Ibix]

The observer is watching his wristwatch running faster than a clock on the earth.Real?

Sure.

I think your problem here is that you are trying to treat the Earth and Moon as symmetric and interchangeable. They aren't. So you should not be surprised that there are physical consequences to this lack of symmetry. Furthermore, you're trying to treat a GR problem using SR, which I suspect is causing headaches for those of us who know how that doesn't work.

Let's consider a pure SR case - a clock on the end of a piece of string swung in a circle around my head. I can easily adopt polar coordinates centered on me, and measure the orbital period of the clock. The clock could also adopt polar coordinates (for clarity, with the zero angle pointing in the same direction as mine) centered on itself. In this system it could measure the orbital period of my head, and would come out with a value that is lower than my value by a factor of $\gamma$.

You should not be surprised that the answers are not symmetric. One calculation is performed against an inertial frame and one against a non-inertial frame. The usual "everything's symmetric" rule of relativity only applies between inertial frames.

 

[Dale]

The observer is watching his wristwatch running faster than a clock on the earth.Real?

Yes, this has been experimentally demonstrated countless times with different types of experiments. This is how nature works.

 

[binis]

Yes, this has been experimentally demonstrated

Moon is the natural satellite of the earth. In the same way an observer on an artificial satellite is watching his watch running faster than a clock on the earth.True?

 

[binis]

Conclusively,time is running faster on the moon than the Earth but slower on some satellites. What about the ISS?

 

[Ibix]

Moon is the natural satellite of the earth. In the same way an observer on an artificial satellite is watching his watch running faster than a clock on the earth.True?

It depends on the altitude of the satellite because gravitational effects also affect clock rates. There is an altitude at which the two rate differences cancel, and which is faster depends on whether the satellite is above or below this altitude. That's why I suggested you consider a pure SR problem.

 

[Ibix]

In #42, @binis said: Conclusively,time is running faster on the moon than the Earth but slower on the satellites.

No. Time does not run fast or slow. There are just different paths through spacetime with different elapsed times along them. Thus if you find a way to compare tick rates of clocks (for example, standing beside one and watching the other through a telescope) you may find that they tick at different rates. But clocks in the same location may tick at different rates compared to each other, and different observers and different methods of observing and interpreting results may lead to different conclusions about which clock is ticking faster.

That said, there are symmetries in Schwarzschild spacetime that allow you to define a global time coordinate with a sensible relation to physically measurable times. In this spacetime, clocks at rest in this coordinate system will all regard other such "at rest" clocks as ticking slower than them if the other clock is below them and faster if they are above them. This is sometimes stated in popsci sources as "time runs slower near masses", but this is wildly unhelpful because it ignores all the caveats in my previous paragraph.

Just think of clocks as measuring the interval along a worldline, like an odometer measures distance along a spatial path. Use flocks of clocks to define a global notion of time (different flocks produce different such notions, of course). Compare clock rates. Don't think in terms of time running fast or slow.[/i][/i]

 

[Dale]

Moon is the natural satellite of the earth. In the same way an observer on an artificial satellite is watching his watch running faster than a clock on the earth.True?

Conclusively,time is running faster on the moon than the Earth but slower on some satellites. What about the ISS?

See this figure

At an altitude of about 3000 km is where the switch occurs. Above that altitude the time on a satellite runs faster than the time on Earth (gravitational time dilation is more important than kinematic time dilation). Below that altitude time on a satellite runs slower (kinematic time dilation is more important than gravitational). The ISS is below 3000 km and the moon is above 3000 km

 

[binis]

So the angular velocity vector between an Earth-fixed basis and a moon-fixed basis is $\boldsymbol{\omega}_1$

what is the magnitube of this vector?

 

[etotheipi]

I suppose you're asking about classical mechanics here? The two bases are related by a unique, orthogonal, time-dependent matrix $(\mathcal{R}_{ij})$ such that $\tilde{\mathbf{e}}_i = \mathcal{R}_{ij} \mathbf{e}_j$. Then take the time-derivative with respect to the frame $Oxyz$,$$\frac{\mathrm{d}\tilde{\mathbf{e}}_i}{\mathrm{d}t}\big|_{Oxyz} = \frac{\mathrm{d} \mathcal{R}_{ij}}{\mathrm{d}t} \mathbf{e}_j = \frac{\mathrm{d} \mathcal{R}_{ij}}{\mathrm{d}t} \mathcal{R}^{-1}_{jk} \tilde{\mathbf{e}}_k := \omega_{ik} \tilde{\mathbf{e}}_k$$and because $\mathcal{R}_{ij}$ is orthogonal we have $\mathcal{R}^{-1}_{jk} = \mathcal{R}_{kj}$ thus the tensor $\omega_{ik} := \dot{\mathcal{R}}_{ij} \mathcal{R}_{kj}$. You can easily check that $\omega_{ij}$ is antisymmetric (do it!), thus is reducible to a single index object via $\omega_i := \frac{1}{2} \epsilon_{ijk} \omega_{jk}$ which define the components of a vector $\boldsymbol{\omega}$. You can just take the Euclidean norm $||\boldsymbol{\omega}|| = \sqrt{\sum \omega_i^2}$ for the magnitude.

 

[Dale]

I suppose you're asking about classical mechanics here?

He is talking about general relativity and specifically two spatially separated observers in a curved spacetime.

 

[A.T.]

The ISS is below 3000 km and the moon is above 3000 km

One should add, that the Moon has non-negligible mass and thus it's own gravitational time dilation. So the time dilation graph for near Earth orbits of small objects doesn't strictly apply to the Moon.

 

[Dale]

One should add, that the Moon has non-negligible mass and thus it's own gravitational time dilation. So the time dilation graph for near Earth orbits of small objects doesn't strictly apply to the Moon.

Yes, that is a good point. I was neglecting the lunar gravitational time dilation (pure laziness, no justification).

 

[A.T.]

Yes, that is a good point. I was neglecting the lunar gravitational time dilation (pure laziness, no justification).

I don't think that it changes the qualitative answer, that a clock on the Moon runs faster than one on Earth. As the graph shows, the kinematic time dilation approaches zero, so the total time dilation is dominated by the gravitational potential. Near the Moon that gravitational potential drops by only a fraction of the Earth's potential well, so the clocks still run faster than on Earth.

 

[binis]

In #42, @binis said: Conclusively,time is running faster on the moon than the Earth but slower on the satellites.

Time does not run fast or slow.Don't think in terms of time running fast or slow.[/i][/i]

I like this view.Time cannot be extended nor shortened.I prefer to say "apparent" time dilation.

 

[A.T.]

I like this view.Time cannot be extended nor shortened.I prefer to say "apparent" time dilation.

Wordplay will not provide more understanding here. You have to learn how it relates to physical processes.

For example: If you take half of a radioactive rod to the Moon, leave it there for a long time, then bring it back, it will have decayed more than the other half that remained on Earth all the time. The same applies to any other process. So I don't know what is "apparent" about this.

 

[Nugatory]

I like this view.Time cannot be extended nor shortened.I prefer to say "apparent" time dilation.

Better would be to understand the role of relativity of simultaneity (in the simple case of velocity based time dilation and the more complicated case of the twin paradox) and of simultaneity conventions (in the curved spacetime of general relativity).

 

[binis]

If you take half of a radioactive rod to the Moon, leave it there for a long time, then bring it back, it will have decayed more than the other half that remained on Earth all the time.

Have this experiment done on the ISS or on a satellite below 3000 km?

The same applies to any other process.

It is known that gravity affects nuclear decay.Are there any examples for other processes?

 

[jbriggs444]

Have this experiment done on the ISS or on a satellite below 3000 km?

Clearly not. The measurement of nuclear decay rates makes for an insanely clumsy clock. You use it when no other alternative is available (e.g. carbon dating or potassium argon dating). If you want to measure tiny time dilation effects, you want to use good clocks.

For example, if you want to look at gravitational time dilation, something like Pound-Rebka is much more effective than putting one bucket of Uranium on the first floor and another on the second.

There is little point doing expensive and worthless experiments when there are expensive and worthwhile experiments left undone.

 

[valenumr]

The relevant invariant is the vorticity of the congruence of worldlines describing the object.

I think I understand what you mean, but I've thought long and hard in this, WRT Machian principal.

And I recently came to think on it more watching a starlink launch, where the vehicle had to induce spin to create momentum to deploy it's satellites.

Ultimately, I came to the conclusion that angular momentum is explicitly relative to the geodesic. An extended body is non-rotating for local observers if all parts are essentially parallel to the geodesic.

For example, considering a starship of anti-rotating rings along an axis, if energy were just applied to rotate the axis, the local observers would not experience artificial gravity.

However, if energy were applied to rotating the rings, the observes would experience artificial gravity.

Each case is totally distinguishable, merely by the fact that an extended (circular to keep it simple) body will have a velocity at it's extrema that is perpendicular to the motion along the geodesic.

Is that roughly correct? This has been bothering me.

 

[A.T.]

It is known that gravity affects nuclear decay.

It's not the strength of gravity, but the difference gravitational potential.

Are there any examples for other processes?

Yes most clocks use other processes.

 

[PeterDonis]

Ultimately, I came to the conclusion that angular momentum is explicitly relative to the geodesic.

Just to be clear, the "angular momentum" you are talking about is spin, not orbital. You are talking about an object spinning about its own axis, not revolving about some other object. But the OP of this thread was talking about the latter. The former came up in the course of discussion, and is what my post about vorticity that you quoted referred to.

An extended body is non-rotating for local observers if all parts are essentially parallel to the geodesic.

If you make "essentially parallel" mathematically precise, you will find that it means "zero vorticity".

Each case is totally distinguishable, merely by the fact that an extended (circular to keep it simple) body will have a velocity at it's extrema that is perpendicular to the motion along the geodesic.

Here you are talking about 3-velocity, not 4-velocity. Your intuitive idea here actually turns out not to work, for technical reasons that I'm not sure can really be explained at the "I" level. But the upshot is that, as I said above, if you work out the issues and make things precise, you will find that you come up with "nonzero vorticity" as the invariant, precise definition of "rotating" (in the sense of spin).

 

[valenumr]

Just to be clear, the "angular momentum" you are talking about is spin, not orbital. You are talking about an object spinning about its own axis, not revolving about some other object. But the OP of this thread was talking about the latter. The former came up in the course of discussion, and is what my post about vorticity that you quoted referred to.If you make "essentially parallel" mathematically precise, you will find that it means "zero vorticity".Here you are talking about 3-velocity, not 4-velocity. Your intuitive idea here actually turns out not to work, for technical reasons that I'm not sure can really be explained at the "I" level. But the upshot is that, as I said above, if you work out the issues and make things precise, you will find that you come up with "nonzero vorticity" as the invariant, precise definition of "rotating" (in the sense of spin).

Yes, about its own axis. I meant parallel WRT forward motion, and perpendicular "rotation" to that axis. I would be happy to have some pointers on the topic. Rotation and angular momentum are a very interesting point for me.

 

[PeterDonis]

I meant parallel WRT forward motion, and perpendicular "rotation" to that axis.

Yes, but as I noted, this turns out not to work as you state it. The problem is that, as I said in a previous post, "parallel" turns out to mean "zero vorticity" when you make it precise, "rotation" turns out to mean "nonzero vorticity" when you make it precise, and nonzero vorticity turns out to be incompatible with "perpendicular". In more technical language, nonzero vorticity means the congruence of worldlines that describes the object is not hypersurface orthogonal; that means it is impossible to define a family of spacelike "slices" of the object (each representing the object "at an instant of its time") that are perpendicular to the worldlines. (This observation has led, in the past, to such apparent "paradoxes" as the Ehrenfest paradox; I believe we have had some good past PF threads on that topic.)

I would be happy to have some pointers on the topic.

This Wikipedia article at least gives a start (note that it talks about all of the components of the kinematic decomposition, not just vorticity):

https://en.wikipedia.org/wiki/Congr...atical_decomposition_of_a_timelike_congruence

Also the references given at the end of that article are good sources.

 

[binis]

One calculation is performed against an inertial frame and one against a non-inertial frame.

Which one and why is the inertial frame?

 

[Ibix]

Which one and why is the inertial frame?

This was a scenario where I was whirling a ball on the end of a piece of string. One reference system was attached to my head and one to the ball. The system attached to my head is inertial because if I let go of a small body it stays at rest with respect to me while if the ball releases a small mass it will flycaway from it with a time varying distance growth.

This is a slight idealisation where I assume that I am very much more massive than the ball and am in zero g.

 

[Prishon]

Summary:: Angular speed ω is by definition the runned angle dθ per time dt elapsed: ω=dθ/dt. If the time elapsed in the center of the Earth is dt, the dilated time elapsed on satellite is dt′. What is the satellite's angular speed? Is it dθ/dt or dθ/dt′?

Angular velocity ω is by definition the runned angle dθ per time dt elapsed: ω=dθ/dt. If the time elapsed in the center of the Earth is dt, the dilated time elapsed on satellite is dt′. What is the satellite's angular velocity? Is it dθ/dt or dθ/dt′?

There is an additional general relativistic rotation.

 

[Ibix]

There is an additional general relativistic rotation.

What do you mean?

 

[Prishon]

What do you mean?

Well, that rotation is very tiny but if you parallel tranport a vector around the curved spacetime of the Earth it is rotated a tiny tiny bit. This doesn't happen in Newtonian spacetime.

 

[Ibix]

Well, that rotation is very tiny but if you parallel tranport a vector around the curved spacetime of the Earth it is rotated a tiny tiny bit. This doesn't happen in Newtonian spacetime.

Can I suggest that, given that we finally seem to have persuaded the OP to leave GR out of the problem while they continue to struggle with SR, introducing extra complexity to this problem is probably a mistake?

 

[Prishon]

Can I suggest that, given that we finally seem to have persuaded the OP to leave GR out of the problem while they continue to struggle with SR, introducing extra complexity to this problem is probably a mistake?

Yes you can.

 

[cianfa72]

However, the rate at which an object rotates (and whether it rotates rigidly) is not invariant. Not all inertial frames will agree about the rate of rotation of an unaccelerated object about its center of mass.

In the sentence above I took 'unaccelerated object' as amounting to the worldline of its center of mass is unaccelerated (i.e. the body's center of mass is in free fall - zero proper acceleration).

 

[PeterDonis]

the rate at which an object rotates (and whether it rotates rigidly) is not invariant.

The first of these statements is true, but the second is not. Rigid rotation, meaning whether or not the congruence of worldlines describing the object is Born rigid (has zero expansion and shear), is an invariant.