I What is the angular velocity of a satellite?

[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.

 

[cianfa72]

How 'proper angular velocity' as in post#17 is related to the vorticity of the congruence of worldlines describing the object ?

 

[PeterDonis]

How 'proper angular velocity' is related to the vorticity of the congruence of worldlines describing the object ?

The magnitude of the vorticity is the proper angular velocity.

 

[ergospherical]

Vorticity is a 2-form $\boldsymbol{\Omega} = \mathrm{d} \mathbf{u}$ and can be turned into a vector by Hodging and then taking the metric dual, $\boldsymbol{\omega} = (\star \boldsymbol{\Omega})^{\sharp} = \nabla \times \mathbf{u}^{\sharp}$.

The magnitude of the vorticity is the proper angular velocity.

Also the angular velocity of a material element is half the vorticity.

 

[cianfa72]

Related to this topic I would like to ask: in the context of SR and GR what is the proper velocity of an object from a (spacetime) geometrical point of view ?

I'm aware of object's proper acceleration (as measured by an accelerometer attached on it) corresponds in the mathematical model of spacetime (Lorentzian manifold) to the path curvature of the worldline representing it.

What about proper velocity (if any ) ? Thanks.

 

[DrGreg]

Proper velocity is the 3-vector $$ \left( \frac{\text{d} x^1}{\text{d} \tau} , \frac{\text{d} x^2}{\text{d} \tau}, \frac{\text{d} x^3}{\text{d} \tau} \right) $$ which is the spatial component of the 4-velocity. It is also known as "celerity", a name I much prefer because "proper velocity" doesn't really behave like other "proper" things in relativity. As it's a 3-vector, not a 4-vector, it is coordinate-dependent and so doesn't really have a geometrical spacetime interpretation.

To get a geometrical interpretation related to velocity, think of the angle between two intersecting worldlines, the worldline of the observer and the worldline of the object being measured (or, strictly, hyperbolic angle). It's called "rapidity". The speed of one relative to the other is the "slope" of one worldline relative to the other worldline.

 

[cianfa72]

Proper velocity is the 3-vector $$ \left( \frac{\text{d} x^1}{\text{d} \tau} , \frac{\text{d} x^2}{\text{d} \tau}, \frac{\text{d} x^3}{\text{d} \tau} \right) $$ which is the spatial component of the 4-velocity. It is also known as "celerity", a name I much prefer because "proper velocity" doesn't really behave like other "proper" things in relativity. As it's a 3-vector, not a 4-vector, it is coordinate-dependent and so doesn't really have a geometrical spacetime interpretation.

From above the four-velocity, instead, as a 4-vector should be geometrically well defined -- namely the tangent vector to the object worldline (tangent vector as an element of the tangent space defined at each point along the object worldline).

So the claim that velocity is not frame-invariant does not apply to four-velocity, I believe...

 

[PeterDonis]

So the claim that velocity is not frame-invariant does not apply to four-velocity, I believe...

That's correct; the 4-velocity of a given object at a given point is an invariant.

 

[cianfa72]

That's correct; the 4-velocity of a given object at a given point is an invariant.

As for the object's proper acceleration (physically measured by an accelerometer attached to it) there is the analogue for the object's 4-velocity (since we said it is frame-independent - i.e. each frame assign to the given object at a given point the same 4-velocity vector) ?

 

[PeterDonis]

As for the object's proper acceleration (physically measured by an accelerometer attached to it) there is the analogue for the object's 4-velocity (since we said it is frame-independent - i.e. each frame assign to the given object at a given point the same 4-velocity vector) ?

If you are asking if there is a way to directly measure an object's 4-velocity, the way an accelerometer can directly measure the object's 4-acceleration, first we need to ask what, exactly, we measure with an accelerometer.

The simplest kind of accelerometer (such as a bathroom scale) just measures the magnitude of an object's proper acceleration. The magnitude of an object's 4-velocity is always $1$, by definition (or $c$ if you're using conventional units), so there's nothing to measure in that respect.

More complicated accelerometers (like the ones used in inertial navigation systems) include gyroscopes to measure the direction of proper acceleration. But this "direction" is a direction in space and requires a reference system. We can do the same with velocity, but in both cases the direct measurement we make is not quite the same as the direction in spacetime of the corresponding 4-vector (4-acceleration or 4-velocity). We have to calculate that direction in spacetime from the measurement data if that's what we want to know. So in this sense we don't directly measure 4-acceleration (the full 4-vector); we measure things from which we can indirectly calculate 4-acceleration. We can do the same for 4-velocity.

 

[cianfa72]

The simplest kind of accelerometer (such as a bathroom scale) just measures the magnitude of an object's proper acceleration. The magnitude of an object's 4-velocity is always $1$, by definition (or $c$ if you're using conventional units), so there's nothing to measure in that respect.

ok, got it thanks.

More complicated accelerometers (like the ones used in inertial navigation systems) include gyroscopes to measure the direction of proper acceleration. But this "direction" is a direction in space and requires a reference system.

Can you elaborate this point please ?

We have to calculate that direction in spacetime from the measurement data if that's what we want to know. So in this sense we don't directly measure 4-acceleration (the full 4-vector); we measure things from which we can indirectly calculate 4-acceleration. We can do the same for 4-velocity.

How can we actually 'rebuild' the full 4-acceleration vector (or the full 4-velocity vector) from those measurements ?

 

[PeterDonis]

Can you elaborate this point please ?

Consider, say, an inertial navigation system for an airplane on Earth. It will have gyroscopes oriented North-South, East-West, and up-down. The direction it assigns to the instantaneous velocity of the plane will be with respect to those axes; it is a direction in space, where "space" is defined by those axes.

How can we actually 'rebuild' the full 4-acceleration vector (or the full 4-velocity vector) from those measurements ?

You would need a clock traveling along the same worldline as the accelerometer (or inertial navigation system) to provide the proper time. Then you would have to use the clock and the spatial axes of your reference system to construct what amount to Fermi normal coordinates centered on the chosen worldline. You can then treat the spatial 3-vectors from your measurements as the "space" components of 4-vectors, with your clock providing the "time" component, in those coordinates.

 

[vanhees71]

In other words, a smartphone will do. It has a clock as well as an accelerometer

 

[PeterDonis]

In other words, a smartphone will do. It has a clock as well as an accelerometer

If you add the app to do all the 4-vector calculations, sure.

 

[vanhees71]

Well, yes, I only think that neither the clock nor the accelerometer are accurate enough to resolve the relativistic effects. Nevertheless it does pretty well for all kinds of experiments, but that's a bit off-topic in this thread:

https://phyphox.org/

 

[PeterDonis]

I only think that neither the clock nor the accelerometer are accurate enough to resolve the relativistic effects

With current smartphones, yes, I would agree. But someday someone might sell one that has an atomic clock and a high precision inertial navigation system in it.

 

[vanhees71]

Well, maybe then I'll finally buy one!

 

[cianfa72]

Consider, say, an inertial navigation system for an airplane on Earth. It will have gyroscopes oriented North-South, East-West, and up-down.

Sorry for the stupid question: the inertial navigation system you were talking of is actually onboard on the airplane I guess...

 

[PeterDonis]

the inertial navigation system you were talking of is actually onboard on the airplane I guess...

Yes.

 

[cianfa72]

You would need a clock traveling along the same worldline as the accelerometer (or inertial navigation system) to provide the proper time. Then you would have to use the clock and the spatial axes of your reference system to construct what amount to Fermi normal coordinates centered on the chosen worldline. You can then treat the spatial 3-vectors from your measurements as the "space" components of 4-vectors, with your clock providing the "time" component, in those coordinates.

So basically the inertial navigation system (with its gyroscopes oriented North-South, East-West, and up-down, plus a clock attached to it) physically 'implements' the Fermi-Walker transport of the tetrad defined by its four axes (one timelike and three spacelike).

 

[Ibix]

So basically the inertial navigation system (with its gyroscopes oriented North-South, East-West, and up-down, plus a clock attached to it) physically 'implements' the Fermi-Walker transport of the tetrad defined by its four axes (one timelike and three spacelike).

Yes.