What is the angular velocity of a satellite?

[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.

 

[cianfa72]

I asked that because I'm often in trouble with the difference (if any) between physics and the mathematical model we use to represent it.

For example we were talking here about timelike and spacelike directions in spacetime. Are we really talking about physics or just of the mathematical model we use to represent it (namely spacelike and timelike vectors elements of the tangent space at each point of the Lorentzian manifold representing the spacetime) ?

I hope my doubt really makes sense...

 

[PeterDonis]

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.

For example we were talking here about timelike and spacelike directions in spacetime. Are we really talking about physics or just of the mathematical model we use to represent it

Both. Obviously the mathematical model has spacelike and timelike (and null) vectors in it, but those features of the model directly correspond to physical features of the world--for example, we measure spacelike intervals with rulers but we measure timelike intervals with clocks.

 

[cianfa72]

Both. Obviously the mathematical model has spacelike and timelike (and null) vectors in it, but those features of the model directly correspond to physical features of the world--for example, we measure spacelike intervals with rulers but we measure timelike intervals with clocks.

So, from a physical point of view a spacelike direction at a given event is obtained from a limiting procedure of a set of events spacelike separated w.r.t. the given event, while a timelike direction from a limiting procedure of a set of events timelike separated w.r.t. the given event.

 

[PeterDonis]

So, from a physical point of view a spacelike direction at a given event is obtained from a limiting procedure of a set of events spacelike separated w.r.t. the given event, while a timelike direction from a limiting procedure of a set of events timelike separated w.r.t. the given event.

Not really, because physically you can't perform such a procedure. You can't sit at a particular event in spacetime indefinitely while you extend smaller and smaller rulers from it or measure smaller and smaller clock intervals from it.

If you want to physically realize directions in spacetime, as opposed to intervals, then the direction in which your worldline points at a given event (which will depend on your state of motion at that event) is a timelike direction, and you can realize spacelike directions with gyroscopes. (Note that strictly speaking, you can't realize spacelike directions by pointing at distant objects like stars, because the light coming from those stars, which is what you're actually pointing at, is coming from a null direction, not a spacelike direction.)

 

[cianfa72]

If you want to physically realize directions in spacetime, as opposed to intervals, then the direction in which your worldline points at a given event (which will depend on your state of motion at that event) is a timelike direction, and you can realize spacelike directions with gyroscopes. (Note that strictly speaking, you can't realize spacelike directions by pointing at distant objects like stars, because the light coming from those stars, which is what you're actually pointing at, is coming from a null direction, not a spacelike direction.)

Let me use non technical language to describe my point of view to help intuition.

Just to 'visualize' spacetime events near an object's worldline consider a region of spacetime around it disseminated with exploding firecrackers. W.r.t. a given event on the object's worldline (say event A) we can split up events in the region around it (i.e. exploding firecrackers events) in two subset: the first one includes events can be 'reached' from (to) the given event A by massive objects or light rays (i.e. events inside the light-cone at event A); the second subset is the complement of the first one in that spacetime region.

The first subset defines events timelike separated w.r.t. the event A while the second subset events spacelike separated from it.

Now in order to physically define directions in spacetime at event A we can proceed as follows:

w.r.t. the first subset each path followed by a massive object through A with a different velocity defines a timelike direction in spacetime.

w.r.t. the events of the second subset we can further 'group' them based on axes locally defined in 'space' (e.g. gyroscope axes): events spatially aligned with a such axis are part of a group that -- in the limit of smaller and smaller region around the given event-- basically defines a spacelike direction in spacetime.

 

[PeterDonis]

The first subset defines events timelike separated w.r.t. the event A

Or lightlike separated, since you've included light rays in the definition.

w.r.t. the first subset each path followed by a massive object through A with a different velocity defines a timelike direction in spacetime.

Yes. And similarly, each path followed by a distinct light ray through A defines a lightlike direction in spacetime.

w.r.t. the events of the second subset we can further 'group' them based on axes locally defined in 'space' (e.g. gyroscope axes): events spatially aligned with a such axis are part of a group that -- in the limit of smaller and smaller region around the given event-- basically defines a spacelike direction in spacetime.

Yes.

 

[cianfa72]

Or lightlike separated, since you've included light rays in the definition.

Yes. And similarly, each path followed by a distinct light ray through A defines a lightlike direction in spacetime.

Sure, of course

 

[cianfa72]

I was thinking about this other topic: take two timelike separated events in the context of GR.

As definition of timelike separated events take the following: two events are timelike separated if there is at least a timelike path between them (it seems to me a reasonable definition).

My question is: Does that definition amount to say there is a timelike geodesic joining them ?

Thanks for you time !

 

[PeterDonis]

As definition of timelike separated events take the following: two events are timelike separated if there is at least a timelike path between them (it seems to me a reasonable definition).

The usual definition is that there must be a timelike geodesic connecting them. And similarly for null and spacelike separated. Requiring a geodesic in the definition is, AFAIK, most important for spacelike separation, since if we are allowed to use arbitrarily curved (non-geodesic) spacelike curves, we can connect any pair of points whatever.

 

[PeterDonis]

As definition of timelike separated events take the following: two events are timelike separated if there is at least a timelike path between them (it seems to me a reasonable definition).

My question is: Does that definition amount to say there is a timelike geodesic joining them ?

We can rephrase the question slightly so that it doesn't depend on a particular definition of "timelike separated": does the existence of any timelike curve between two events necessarily imply the existence of a timelike geodesic between those events?

I think the answer is yes for globally hyperbolic spacetimes, but I'm not sure it is yes for spacetimes that aren't. (For example, I'm not sure the answer is yes for Godel spacetime, which is not globally hyperbolic--it has closed timelike curves.)

 

[cianfa72]

Requiring a geodesic in the definition is, AFAIK, most important for spacelike separation, since if we are allowed to use arbitrarily curved (non-geodesic) spacelike curves, we can connect any pair of points whatever.

Why ? Take two timelike separated events: does it always exist an arbitrarily curved spacelike path connecting them ?

 

[PeterDonis]

Take two timelike separated events: does it always exist an arbitrarily curved spacelike path connecting them ?

Yes.

 

[cianfa72]

Yes.

Ah right, just to visualize it I made this sketch: events A and B are timelike separated yet there is a spacelike path (in black) connecting them (I believe there is no problem in smoothing out its acute angle)

 

[PeterDonis]

Ah right, just to visualize it I made this sketch: events A and B are timelike separated yet there is a spacelike path (in black) connecting them (I believe there is no problem in smoothing out its acute angle)

View attachment 288181

Not quite. You can't do it in just two spacetime dimensions, because when you smooth out the acute angle, you find that the path becomes timelike for some portion of that region.

If you add one more spatial dimension, however, you can have a helical path in spacetime that is always spacelike (no acute angle, just constant curvature) but connects two timelike separated (in the sense of having a timelike geodesic between them) events.

 

[PeterDonis]

Requiring a geodesic in the definition is, AFAIK, most important for spacelike separation

There is actually another reason for requiring a geodesic in the definition, which applies to any kind of separation. Given an event and a tangent vector at that event (and tangent vectors at a single event will always be definitely timelike, spacelike, or null), a unique geodesic is determined throughout the spacetime. But the uniqueness only holds for geodesics; there are an infinite number of non-geodesic curves that pass through the same event and have the same tangent vector at that event. So for the "separation" of two events to be well defined and unique, the definition has to require geodesics.