Undergrad What is the angular velocity of a satellite?

[binis]

TL;DR

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′?

 

[haushofer]

As you are obliged to ask in relativity theory: according to whom?

 

[PeroK]

What is the satellite's angular velocity? Is it dθ/dt or dθ/dt′?

Why not both? Both might be useful quantities.

 

[Dale]

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′?

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′?

That is a relative quantity. That means that its value depends on the coordinate system chosen. Roughly speaking, if you choose the coordinate system $(t,r,\theta,\phi)$ then the angular velocity will be $d\theta/dt$. If you choose the coordinate system $(t’,r,\theta,\phi)$ then the angular velocity will be $d\theta/dt’$. Either choice is valid.

Any time you have a relative quantity you must specify the coordinate system you are using. Otherwise it does not have a unique definite value.

 

[binis]

Any time you have a relative quantity you must specify the coordinate system you are using.

Let's stay in the local (ECI) frame.I think I am getting it wrong,in this frame,there is not dt',there is not time dilation?Where the dt' is appear?

 

[Dale]

Let's stay in the local (ECI) frame.I think I am getting it wrong,in this frame,there is not dt',there is not time dilation?Where the dt' is appear?

Ok, so in your OP you listed $dt$ as the time at the center of the Earth and $dt’$ as the time at the satellite. But the time coordinate in the ECI is not either of those. The ECI uses time as at the surface of the Earth for hypothetical clocks that are not fixed to the earth, but stay in place as the Earth rotates under it. So if we use $dT$ to denote the time in the ECI $(T,r,\theta,\phi)$ coordinates then the angular velocity in the ECI would be $d\theta/dT$

 

[PeterDonis]

The ECI uses time as at the surface of the Earth for hypothetical clocks that are not fixed to the earth, but stay in place as the Earth rotates under it.

Actually, I don't think even this is quite correct. As I understand it, the ECI uses hypothetical points in space not fixed to the earth, but staying in place as the Earth rotates underneath them, to fix the simultaneity convention and the directions of the spatial axes. But the clock rate in the ECI is the rate of clocks at rest on the geoid of the rotating Earth, which will not be the same (it will be slightly slower) as the clock rate of actual "inertial" clocks sitting at rest in the ECI at the altitude of the surface of the Earth. (I put "inertial" in quotes because of course such hypothetical "at rest in the ECI" clocks would have to have nonzero proper acceleration in order to maintain altitude, and their clock rates would be affected by the Earth's gravitational potential.)

 

[Dale]

Actually, I don't think even this is quite correct.

You could be correct. I am going from memory. So then a clock attached to the Earth at the level of the geoid would not be at rest in the ECI, but its proper time would match the ECI coordinate time.

 

[PeterDonis]

a clock attached to the Earth at the level of the geoid would not be at rest in the ECI, but its proper time would match the ECI coordinate time.

That is my understanding, yes. Unfortunately I can't find any good sources online at the moment for this; none of the references I can find talk about the time coordinate at all, they only talk about the spatial coordinates.

 

[Motore]

Possibly relevant:
https://www.itu.int/dms_pubrec/itu-r/rec/tf/R-REC-TF.2018-0-201208-W!PDF-E.pdf

 

[vanhees71]

For GPS I found this a good reference:

http://www.nbmg.unr.edu/staff/pdfs/Blewitt_Encyclopedia_of_Geodesy.html

 

[binis]

That is a relative quantity. That means that its value depends on the coordinate system chosen. Roughly speaking, if you choose the coordinate system $(t,r,\theta,\phi)$ then the angular velocity will be $d\theta/dt$. If you choose the coordinate system $(t’,r,\theta,\phi)$ then the angular velocity will be $d\theta/dt’$. Either choice is valid.

But only relative velocity is physically meaningful. We can choose to interpret either satellite or Earth to be at rest.Relative velocity ω is dθ/dt = dθ/dt' => dt=dt' => no time dilation ?

 

[jbriggs444]

But only relative velocity is physically meaningful. We can choose to interpret either satellite or Earth to be at rest.Relative velocity ω is dθ/dt = dθ/dt' => dt=dt' => no time dilation ?

If we are working within the framework of special relativity, the frame in which the satellite is at rest is not inertial. That matters.

When using a non-inertial frame of reference, things get... tricky. You can't always make the coordinates line up.

 

[Dale]

But only relative velocity is physically meaningful. We can choose to interpret either satellite or Earth to be at rest.Relative velocity ω is dθ/dt = dθ/dt' => dt=dt' => no time dilation ?

No, this assertion fails. First, it fails because angular velocity is not the same as velocity so it doesn’t follow the same rules. Second, the physically meaningful angular velocity is invariant not relative, and is measured by local gyroscopes. Third, there is no requirement that relative rotation be reciprocal. Fourth, when spacetime is curved even relative linear velocity is not physically meaningful for spatially separated objects like in this scenario.

If I had to pick one reason as the most important I would pick the second. The idea that only relative angular velocity is physically meaningful is very wrong. Angular velocity is not a relative quantity, it is invariant and is easily measured locally with a gyroscope without reference to any other object. That is an essential and fundamental difference between angular velocity and linear velocity.

 

[jbriggs444]

Angular velocity is not a relative quantity, it is invariant and is easily measured locally with a gyroscope without reference to any other object.

I have one small quibble with this.

Whether an object is rotating or not is an invariant fact of the matter. It either rotates or does not. All inertial frames will agree. 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.

For instance, the rotation rate of the second hand on an analog wristwatch cannot be a relativistic invariant if time dilation is a thing.

 

[etotheipi]

Wait, I thought @Dale refers to the endomorphism $\mathbf{\Omega} = \mathbf{\Omega}_{\text{FW}} + \mathbf{\Omega}_{\text{rot}}$ where $\mathbf{\Omega}_{\text{FW}}$ depends only on invariant 4-acceleration $\mathbf{a}$ of the observer and $\mathbf{\Omega}_{\text{rot}}(\mathbf{v}) = \boldsymbol{\omega} \times_{\mathbf{u}} \mathbf{v}$ depends only on the invariant 4-rotation $\boldsymbol{\omega}$ of the local frame, i.e. such that $d\mathbf{v}/d\tau = \mathbf{\Omega}(\mathbf{v})$. Also $\mathbf{a}$ and $\boldsymbol{\omega}$ can both be determined by the observer himself with local experiments.

 

[Dale]

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.

This is exactly analogous to coordinate acceleration and proper acceleration. What you are talking about above is the coordinate rotation. What I am talking about is the “proper” rotation. That is invariant because it is the rotation measured by gyroscopes carried by the object. All frames must agree on the outcome of such a measurement.

By the way, does anyone know the correct term for “proper” angular velocity? I don’t think that I have seen such a term used before.

 

[etotheipi]

By the way, does anyone know the correct term for “proper” angular velocity? I don’t think that I have seen such a term used before.

I think the formalism is this, although I hope someone will correct/improve: given a vector field $\mathbf{v}(\tau) = v^{\mu} \mathbf{e}_{\mu}(\tau)$ fixed with respect to an observer $\mathscr{O}$ [i.e. $v^{\mu}$ are independent of $\tau$], then one may define an endomorphism $\mathbf{\Omega}$ by$$\frac{d\mathbf{v}}{d\tau} = \mathbf{\Omega}(\mathbf{v}) = \underbrace{c(\mathbf{a} \cdot \mathbf{v})\mathbf{u} - c(\mathbf{u} \cdot \mathbf{v})\mathbf{a}}_{\mathbf{\Omega}_{\mathrm{FW}}(\mathbf{v})} + \underbrace{\boldsymbol{\epsilon}(\mathbf{u}, \boldsymbol{\omega}, \mathbf{v}, \, \boldsymbol{\cdot} \, )}_{\mathbf{\Omega}_{\mathrm{rot}}(\mathbf{v})}$$where $\mathbf{u} = d\mathscr{O}/d\tau$ is the 4-velocity of $\mathscr{O}$, $\mathbf{a}$ is the 4-acceleration of $\mathscr{O}$ and $\boldsymbol{\omega}$ is the 4-rotation of $\mathscr{O}$'s local frame $\{ \mathbf{e}_{\mu} \}$.

 

[Dale]

I think the formalism is this, although I hope someone will correct/improve: given a vector field $\mathbf{v}(\tau) = v^{\mu} \mathbf{e}_{\mu}(\tau)$ fixed with respect to an observer $\mathscr{O}$ [i.e. $v^{\mu}$ are independent of $\tau$], then one may define an endomorphism $\mathbf{\Omega}$ by$$\frac{d\mathbf{v}}{d\tau} = \mathbf{\Omega}(\mathbf{v}) = \underbrace{c(\mathbf{a} \cdot \mathbf{v})\mathbf{u} - c(\mathbf{u} \cdot \mathbf{v})\mathbf{a}}_{\mathbf{\Omega}_{\mathrm{FW}}(\mathbf{v})} + \underbrace{\boldsymbol{\epsilon}(\mathbf{u}, \boldsymbol{\omega}, \mathbf{v}, \, \boldsymbol{\cdot} \, )}_{\mathbf{\Omega}_{\mathrm{rot}}(\mathbf{v})}$$where $\mathbf{u} = d\mathscr{L}_{\mathscr{O}}/d\tau$ is the 4-velocity of $\mathscr{O}$, $\mathbf{a}$ is the 4-acceleration of $\mathscr{O}$ and $\boldsymbol{\omega}$ is the 4-rotation of $\mathscr{O}$'s local frame $\{ \mathbf{e}_{\mu} \}$.

Yes. Mathematically, that seems correct (@vanhees71 do you spot any problems I missed), but is there a specific term for that? I am not aware of it.

 

[Nugatory]

But only relative velocity is physically meaningful. We can choose to interpret either satellite or Earth to be at rest.Relative velocity ω is dθ/dt = dθ/dt' => dt=dt' => no time dilation ?

You could make the same argument for two clocks moving inertially (that is, in a straight line with no rotation, angular velocity, centripetal force, or any other things that make circular motion more tricky). It might be worth working out for yourself why the argument fails in that case before taking on the more complicated orbiting satellite case.

 

[vanhees71]

That looks plausible to me, but I must say I've never thought about "angular velocity" in relativity. If I understand it right, what @etotheipi defines as $\boldsymbol{\omega}$ quantifies the additional change of a vector along an observer's timelike worldline with time which goes beyond the Fermi-Walker transport, and this precisely describes the rotation of this vector relative to a Fermi-Walker (i.e., rotation-free) transported tetrad of the observer, and that's of course just describing an "angular velocity" in close analogy to how you define the angular velocity in Newtonian rigid-body dynamics.

 

[weirdoguy]

But only relative velocity is physically meaningful.

You've already been told, in the previous thread you were participating in, that this is not true.

 

[binis]

the physically meaningful angular velocity is invariant not relative, and is measured by local gyroscopes. Angular velocity is not a relative quantity, it is invariant and is easily measured locally with a gyroscope without reference to any other object.

An observer on the moon's surface sees the Earth revolve around the moon.

 

[Dale]

An observer on the moon's surface sees the Earth revolve around the moon.

So what? That is not the earth’s physically meaningful angular velocity.

Please address the points I raised before proceeding.

 

[PeterDonis]

does anyone know the correct term for “proper” angular velocity?

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

 

[Dale]

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

Of course. I cannot believe I never made the mental connection between vorticity and what I was mentally calling proper rotation. It is obvious in retrospect.

 

[PeterDonis]

Of course.

Note, though, that this sense of "rotation" corresponds to "spin", not "orbiting". For example, in the case of the Earth, it describes the Earth's rotation about its axis, but not its orbiting about the Sun. In the case of the Moon, this can get confusing because its (sidereal) rotation about the Earth has the same angular velocity as its rotation about its axis, because of tidal locking. But the vorticity of the congruence of worldlines that describes the Moon still only describes the latter, not the former.

 

[PeterDonis]

In the case of the Moon, this can get confusing because its (sidereal) rotation about the Earth has the same angular velocity as its rotation about its axis, because of tidal locking.

I should probably also point out that there are other relativistic effects--Thomas precession, de Sitter precession, and Lense-Thirring precession--that affect the relationship between "spin" and "orbital" rotation. These are all very small for bodies like the Moon, Earth, and Sun, so they are usually ignored.

 

[Dale]

Note, though, that this sense of "rotation" corresponds to "spin", not "orbiting".

Yes, but that is intended since it is the only invariant angular velocity. Orbital angular velocity is not invariant.

In the case of the Moon, this can get confusing because its (sidereal) rotation about the Earth has the same angular velocity as its rotation about its axis, because of tidal locking.

That brings up a good point. Even in Newtonian physics with no time dilation the concept of relative angular velocity is ambiguous. If you have a planet which spins once per day and a moon which spins once per week and orbits once per month, then what is the relative angular velocity?

 

[PeterDonis]

Orbital angular velocity is not invariant.

In the general case, no, because there is no general invariant notion of "coming back to the same place" or "going through 360 degrees of angle" or any other concept that could be used to define what one "orbit" means.

There are some special cases where I think an invariant notion of "one orbit" could be defined, but they're very special--I think one would need the spacetime to be stationary and spherically symmetric, or at least axially symmetric with the plane of the orbit being an "equatorial plane" of the axial symmetry. And any such system could at most have one "primary" and one "satellite"--a system like the Solar System, where we have satellites orbiting planets orbiting the Sun, would not meet the requirements.