# Relative Velocity

1. Aug 24, 2006

### Oxymoron

What is everyone's views on relative velocity is General Relativity? Is it useless? Is it useful, if we make specific choices about how we parallel transport? Is there a definite relative velocity? Is there relative velocity at all? Or is there a better one than all the others?

2. Aug 24, 2006

### pervect

Staff Emeritus
Relative velocity can always be defined if two objects are at the same point in space, or close enough together that space-time is still locally flat.

Relative velocity to a distant object isn't uniquely defined in general, though one sees specific cases where defintions have been imposed, for instance Hubble's law. These defintions have been the source of a fair amount of confusion, however :-(.

If one has a a static and/or stationary metric, it makes sense to talk about one's velocity relative to the metric. This often comes about when one has a single massive body, in which case one can use the velocity relative to the metric to define the velocity relative to the massive body. However, this would probably be a case where one should explain the details of what one means by 'velocity'. One way of describing this is that given a stationary metric, one can define a body "at rest" relative to the metric at any point in space-time. (The body at rest relative to the metric will experience constant metric coefficients - to be ultra-formal, one can describe this body at rest as following the orbit of a time-like Killing vector associated with the stationary metric - all stationary metrics have by defintion time-like Killing vectors).

One can then measure one's velocity relative to this body "at rest", definiing a meaningful concept of "velocity" in any stationary metric.

3. Aug 26, 2006

### Oxymoron

Well, that sounds like a fair way to do it.

What is this "stationary metric" you mention? Is the Schwarzschild metric tensor one of these? So are you saying that one way to define relative velocity between two observers, is to treat one observer as the massive, spherically symmetric object and then use the stationary metric "constructed" from that object?

This is good, I like it. At least, it is a start - something for me to investigate! :)

4. Aug 26, 2006

### pervect

Staff Emeritus
One way of describing a stationary metric is that it is a metric where none of the metric coefficients are functions of time. This is a coordinate dependent defintion.

Stationary metrics include the Schwarzschild metric and the Kerr metric, for example.

The Schwarzwschild metric meets some stronger conditions, so its static as well as stationary.

This is discused in Wald BTW, which is where I'm getting the defintions from.

The coordinate independent definition is a little nicer, but may be less accessible. A stationary metric is any metric with a time-like Killing vector.

I'm not sure if you've run into Killing vectors before, they are quite useful and important so I'll try to explain.

A Killing vector is present whenever you have a one parameter group of isometries. A timelike Killing vector means you have a time-like one paramete group of isometries, i.e. a one parameter time translation symmetry of the metric.

Because a Killing vector is a vector field, it's not the one parameter group of isometries itself, it's the "generator" of the group.

This is also discussed in Wald.

In the special but most convenient case where none of the metric coefficients are functions of time, $$\xi^a = (1,0,0,0)$$ is a Killing vector. Killing vector's satisfy Killing's equation, $$\nabla_a \xi_b + \nabla_b \xi_a = 0$$.

5. Aug 27, 2006

### Oxymoron

Wow, I've never had it explained like that to me before - very enlightening, thankyou.

Some questions though. Whenever I think of a one-parameter group of transformations I think of the flow of water. More precisely, I think of a collection of maps where any one of these maps defines how a particle in the flow of the water moves. Is this a safe way to think? I guess whenever you have a one-parameter group of transformationsyou need a vector field, which generates the group of transformations, is this right?

So in most cases one will have a manifold one which there is defined a smooth vector field which generates a one-parameter group of transformations. My question is: can this group of transformations be thought of as a "velocity field" or is that something completely different?

An invariant metric tensor, g, is one which does not change under any sort of transformation. (should I be careful to state what kind of transformation Im talking about here? I mean, aren't all tensors invariant under a coordinate transformation?) So a transformation which leaves the metric tensor invariant is called an isometry right?

So now, instead of talking about one-parameter groups of transformations (O-PGTs) we can talk about those which leave the metric invariant (i.e. the isometries). But, I said earlier than all O-PGTs are "generated" by a vector field. Are you saying now that a one-parameter group of isometries (O-PGIs) are in fact generated by a Killing vector field? I mean, vector fields which generate O-PGIs are specially called Killing vector fields?

I heard somewhere that a Killing vector field has the effect of "dragging" a metric into itself. If this is an apt statement could you possibly clarify it for me?

Last edited: Aug 27, 2006
6. Aug 28, 2006

### Oxymoron

I suppose this is because of how the Lie derivative affects the metric tensor, i.e. L g = 0.

7. Aug 28, 2006

### pervect

Staff Emeritus
Yep, that's exactly the defintion of a Killing vector. It's a vector field that generates a OPG of transformations which leaves the metric invariant, i.e the generator of an OPG of isometeries.

Killing vectors can be space-like, or time-like. It turns out that time-like Killing vectors generate a conserved energy via Noether's theorem. This conserved energy can either be a conserved quantity for a partricle orbit, or can be used as an actual defintion of system energy (system mass) for systems which have stationary metrics. In the later it is case it's known as the "Komar mass" of the system.

Space-like Killing vectors similarly generate conserved momenta.

I think the "dragging" idea comes from the fluid flow analogy you mentioned, doesn't it? Personally I tend to use a rotating sphere as my mental model for a OPG of isometries (rotations leave the metric invariant), and interpret the Killing vector field as the velocity field of the points on the surface of the sphere.

8. Aug 29, 2006

### Oxymoron

Ok, let us (well...me...) just concentrate on the Schwarzschild metric. The first thing I think of after hearing this metric is "spherical symmetry". The Schwarzschild metric is stationary if there is a coordinate system in which the metric is time-independent, right? And is it correct that the vector field, X, is Killing because the metric is stationary? I mean, if I have a stationary metric, then there is a coordinate system in which the metric is time-independnt

$$\frac{\partial g_{ab}}{\partial x^0} = 0$$

then define a vector field $X^a = \delta^a_0$ in the special coordinate system, then

$$\mathcal{L}_X g_{ab} = X^c g_{ab,\,c} + X^c{}_{,\,b}g_{ac} + X^c{}_{,\,a}g_{bc} = \delta^c_0 g_{ab,\,c} = g_{ab,\,0} = 0$$

and since $\mathcal{L}_Xg_{ab}$ is a tensor, if it vanishes in this special coordinate system, it must vanish in all coordinate systems. Hence X is a Killing vector field.

Conversely, if I have a timelike Killing vector field X, then there exists a coordinate system which is adapted to the Killing vector field and then

$$\mathcal{L}_X g_{ab} = g_{ab,\,0}$$

and therefore the metric is stationary. So A spacetime is stationary if and only if it admits a timelike Killing vector field.

Then A spacetime is spherically symmetric if and only if it admits exactly 3 linearly independent spacelike Killing vector fields whose orbits are closed and which satisfy

$$[X^1,X^2] = X^3,\quad[X^2,X^3] = X^1,\quad[X^3,X^1] = X^2$$

Is this what you mean when you said

Is a line of longitude on the Earth an example of one of these orbits?

Then solving Einstein's field equations in a vacuum, assuming spherical symmetry, you get the Schwarzschild solution (which is a metric) which is stationary and the coordinate system used is adaptable to a Killing vector field $X^a = \delta^a_0$ since

$$X_a = g_{ab}X^b = g_{ab}\delta^b_0 = g_{0a} = g_{00}\delta^0_a = (1-\frac{2m}{r},0,0,0)$$

From this, is it possible for me to say that the Killing vector field X is hypersurface-orthogonal?

I mean, if I now have a hypersurface-orthogonal Killing vector field then surely the vorticity tensor of the timelike congruence vanishes. But basically that means that a spherically symmetric spacetime with the Schwarzschild metric admits an irrotational timelike Killing vector field.

So if I have a spherically symmetric spacetime (i.e. one with a singular massive object at the origin) in which I can assume that it is asymptotically flat then the isometry group contains a subgroup isomorphic to the rotation group $\mathcal{SO}(3)$, and the orbits of this group are 2-dimensional spheres.

The isometries are generated by the Killing vector field which implies, among other things, that the Schwarzschild metric is invariant under rotations.

Last edited: Aug 29, 2006
9. Aug 29, 2006

### pervect

Staff Emeritus
The Schwarzschild metric has a spatial surface orthogonal to the Killing vector fields, but the Kerr metric doesn't. (I think you already worked this out from your remarks about vorticity).

Metric with a time-like Killing vector and hypersurface orthogonality are called "static".

10. Aug 29, 2006

### Oxymoron

In my efforts to find a suitable way of working out relative velocity in a spherically symmetric spacetime with the Schwarzschild metric, I came across the following lemma:

Let X be a Killing vector field on a smooth manifold M, and let C be a geodesic in M. Then X, restricted to C, is a Jacobi Field and g(C',X) is constant along C.

Now, I had always thought that to have any sense of relative velocity I would need to be able to parallel transport tangent vectors from one geodesic to another, but the way in which I do this has always seemed imprecise. I figured I may need some knowledge of Jacobi fields or something to work it out. Does anyone reckon that this lemma is useful in my quest, or not?

Whenever I think of Jacobi fields I think of a two-parameter group of transformations. Is this accurate?

Because I was thinking, the way in which I could do this is to do the following: If I have a geodesic (say an observer moving through an empty universe with one spherically symmetric static mass), call it C, then surely I must be able to vary the geodesic by a One-Parameter Group of geodesics. In other words, given the structure of my universe just stated, and given a Killing vector field X, defined along a geodesic C, I must be able to define a variation of C having a variation vector given by the Killing vector field itself!

Last edited: Aug 29, 2006
11. Aug 29, 2006

Staff Emeritus
Notice the lemma requires you to have a a Killing vector field, and only a few very special spacetime have them.

12. Aug 29, 2006

### Oxymoron

Right, that is why I am only considering spherically symmetric spacetime at the moment, which allow me to have Killing vector fields (...I hope, at least!). I am aware that relative velocity, in general, will be impossible to define. Instead Im working in the few places where I hope I will find a nice way to define relative velocity, and hopefully work out if there is a corresponding way to view Frequency shifting there as well.

Basically, once I have worked out relative velocity I will then attempt to derive a frequency shifting formula which hopefully will be analogous to the Doppler shift in Minkowski spacetime.

So far, Pervect has guided me to the Spherically Symmetric Spacetime with a Stationary Schwarzschild Metric which he has told me has a notion of relative velocity with respect to the metric. Hopefully in the next couple of days I will understand what this means! :)

Last edited: Aug 29, 2006
13. Aug 29, 2006

### robphy

You may find this interesting

AIP Conference Proceedings -- June 19, 2006 -- Volume 841, pp. 633-634
A CENTURY OF RELATIVITY PHYSICS: ERE 2005; XXVIII Spanish Relativity Meeting

On the concept of relative velocity
V. J. Bolós

Last edited by a moderator: May 2, 2017
14. Aug 29, 2006

### pervect

Staff Emeritus
I was just re-reading that section of Wald, and Wald only claims that an object that is following an orbit of the Killing field is "staying in place" in a static space-time.

I was originally thinking this should work in a general stationary metric, but I may have to think about the matter more, I might be missing something. I also realized that I have to think about circular orbits in the Schwarzschild metric some more as far as the notion of "staying in place" goes.

15. Aug 31, 2006

### Oxymoron

I believe I will definately need the space-time to be static.

16. Aug 31, 2006

### pervect

Staff Emeritus
The issue that your remarks got me thinking about, more clearly expressed, is this:

Let's take the Schwarzschild metric, and use coordinates (t,r,theta,phi) for definiteness.

Then (1,0,0,0) is a timelike Killing vector. (0,0,0,1) is a spacelike killing vector. Their weighted sum is also a Killing vector - it could be either timelike or spacelike, depending.

The issue is that while (1,0,0,0) is certainly an orbit of the Killing field, I don't think it's unique. (1,0,0,alpha), where alpha is some constant would also seem to me to be an orbit of the Killing field.

I'm not quite sure how to resolve this. I don't think Wald goes into any more detail than saying that a stationary object follows an orbit of the Killing field. It's true, but it doesn't seem to be a complete defintion of a stationary object. I've been meaning to do more reading, but I've been busy on other projects.

In words, any circular orbit is an isometry. Thus while a particle hovering at constant (r,theta,phi) isn't moving "relative to the metric", because it's always at a place of constant metric coefficients, neither is a particle in a circular orbit.

So the defintion I was offering, based on Wald, doesn't seem to actually be a complete coordinate independent description of a "stationary" object :-(.

17. Aug 31, 2006

### Oxymoron

A circular orbit is an isometry, but is it correct that it is not a geodesic? Since keeping a fixed distance from the mass requires constant acceleration?

From what I understand any observer in a static space-time (with the Schwarzschild metric) can measure a relative velocity with respect to the metric. But then does this mean that we can work out the relative velocity between two observers so long as at least one of them is travelling parallel with the Killing vector field?

18. Sep 1, 2006

### pervect

Staff Emeritus
We certainly know how to define relative motion in the Schwarzschild coordinate system, and more physically via a frame field

http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity

(Actually, I don't know for sure if you grok the notion of frame fields or not, I suppose that's another issue. They are extremely useful IMO though, and the way I'd actually go about calculating the velocity of an object in the Schwarzschild metric).

The issue currently in my mind is how we can describe relative motion in a coordinate independent manner, i.e. without referring to the Schwarzschild coordinates specifically.

I thought that Wald's defintion of following an orbit of the Killing field would work, but I have some strong doubts now.

Note that the transformations I was talking about earlier would represent everything from "hovering", to a "powered orbit", to a pure circular orbit (which would be a geodesic). I believe they all qualify as 'following an orbit of a Killing field'. So this is a necessary condition, but I don't think its sufficient without some additional constraints.

19. Sep 3, 2006

### Oxymoron

Well, isn't that all we need? All I need is a way of determining motion of an object in a static spacetime with respect to an observer travelling along the flow of the Killing vector field. But if your Frame field's are the key to this then I must understand them first.

I briefly read through the wikipedia article and it was too hard to understand. Perhaps you could give a small recount of what you know about these frame fields and how they allow us to determine relative motion.

I've decided that I will insist upon working in a static spacetime with the Schwarzschild metric. By definition, a static spacetime admits a timelike hypersurface-orthogonal Killing vector field. Let me explain what I think of this definition (it could be wrong!):

If an observer, call him A, is travelling along (or parallel to, I guess?) an integral curve (or flow line) of the Killing field, which, I assume, he is allowed to do, and suppose there is a second observer, call him B, travelling, possibly accelerating, along his own worldline. Then since the spacetime is static, the Killing field has a orthogonal hypersurface which I understand to be a hypersurface which is orthogonal to each and every integral curve, i.e. orthogonal to the worldline of A.

Therefore this surface is spacelike (?) and connects A and B. Then is this not a way of parallel transport? Could I simply parallel transport the velocity vector of A along the orthogonal hypersurface to B and proceed to compare? I could imagine then that I would be comparing the velocities of A and B at events that are not causally connected?!

Last edited: Sep 3, 2006
20. Sep 3, 2006

### pervect

Staff Emeritus
OK, while from a purist POV I would like to see it done in a coordinate independent manner, but I do agree that it's more important to be able to do it at all, first - one can worry about the finer details later.

Frame fields are a very importnat tool in GR, but they tend to be a bit confusing. I'm not aware of any particularly easy to understand treatment of them.

The way I think of them is this. Suppose you are at a location. In some small region, the space-time is, locally, Lorentzian and Minkowskian.

Because space-time is Minkowskian, you can set up local coordinates that reflect this, and will be well behave the way you are used to. You want your coordinates to be orthonormal, by your local rulers and local clocks.

The easiest way to do this is to define an orthonormal basis of 1-forms. A 1 form is just a map from a vector to a scalar.

Defining 4 orthonormal basis one-forms defines 4 local coordinates, by defintion.

Because space-time is nearly flat, one can think of coordinate displacements as vectors, non-rigorously. (I don't know a rigorous way to approach the issue).

So the one-forms maps the vectors, which are essentially equal to small displacements, into local coordinates. Making the one-forms orthonormal means that the coordinates are also orthonormal.

If you use tools like GRTensor, the mathematical manipulations are done for you by the computer. This is an area where I got lazy - I probably should have studied the Newman-Penrose formalism more (this is where the stuff is explained), but instead I let the computer do all the dirty work.

If you go digging around for Newman-penrose, you should be able to find out more.

Armed with this knowledge, you might want to look at the Wiki article again, it works out some examples.

If you look at a textbook treatment, look for "orthonormal basis of one-forms" and "hats" over the vectors.

This is good for other things other than velocity. For instance, it's often useful to know what the stress-energy tensor is in a "local" frame-field, rather than the density per unit change in coordinate. It's a bit more "physical".