# What is one full orbit

1. Apr 26, 2007

### MeJennifer

I was doing some thinking (I know, some might feel that is perhaps not such a good idea ) about classical, i.e. non quantum based, clocks and was confronted by the question if in general relativity there is such thing as a frame independent notion of a full orbit.

Let's get a bit more specific to avoid misunderstandings or irrelevant matters.

Suppose we have two point masses that are in orbit with each other. The masses are significant, they are not so-called "test particles" and their masses are not identical. We also assume that this system is not influenced by far away gravitational fields.

Now, can we construct a frame invariant definition of a full orbit? Any orbit is ok and with or without rotation of the point masses.

I am trying to think of one I fail to see how.

Last edited: Apr 26, 2007
2. Apr 26, 2007

### Wallace

I think I see what you are getting at. You can construct a rotating frame in which the masses are not moving at all, and hence the oribit is not frame independent. In order to describe the orbits you need a reference point external to the two point masses in order to define their motion with respect to, if you made a rotating frame in which the particles were still you would then see that the frame rotated wrt this third reference point and hence you could still describe the orbits. I think this is essentially a Mach's Principle question.

As long as you have an external distant mass (and hence observer) to define motion with respect to then you could describe the orbits, but without some other point to reference I think you may be right in suggesting this becomes impossible. Again I think this is the essence of Mach's Principle.

3. Apr 26, 2007

### MeJennifer

Thanks Wallace.

I was thinking more in terms of the "time gap" and gravitational Thomas precession.

From the perspective of each mass there is a different view of when a full orbit was completed. The question is, is there an observer independent way of defining when one full orbit what completed. In other words it this a unique event.

Last edited: Apr 26, 2007
4. Apr 26, 2007

### Dick

You don't need a specific external observer to define rotation for your system. What you need is that the spacetime containing the system is 'asymptotically flat'.

5. Apr 26, 2007

### Wallace

But you need something to define the precession with respect to. Isn't that the problem? I seem to have misunderstood your question I think. What are the impediments you are thinking of that prevent 'one orbit' being defined? Mach's principle came straight to my mind when I read your question and I can't think of anything else?

How does asymptotic flatness help you define rotation? Don't you still need a fixed reference point?

6. Apr 26, 2007

### Dick

If your spacetime is asymptotically minkowski any infinitely distant observer is a fixed point. It's a 'fixed star'. That's not very machian. But then the Kerr-Newmann metric is asymptotically flat. The rotation is embedded in 'something'.

7. Apr 26, 2007

### Wallace

But the distant observer is the key part though right? Without that you can't pin the edges of the space-time down. If there is no distant fixed point for the observers in the rotating frame to observe then they cannot know they are rotating.

I would think that you do need to embed the rotation in something, but that space alone is insufficient to be that something, since it is unobservable. I'll stop on this line of thought though as we don't need another 'is space-time real' thread!

8. Apr 27, 2007

### MeJennifer

Ok let's define it a bit more detailed and simpler.

Two non-rotating point masses A and B (for the sake of argument A having twice as much mass as B) are in orbit with each other. No other masses exist in this universe and the masses are not test particles. We are interested in a full GR understanding not a weak field solution.

Is there an observer independent formulation of B making one complete revolution around A?

In other words could such a configuration ever work as a clock?

To me that seems to be impossible since I cannot think of any way how we can associate a complete revolution with one unique spacetime event.

Ok, so then how does that help in defining a complete revolution of say B around A in an observer independent way?

Last edited: Apr 27, 2007
9. Apr 27, 2007

### cesiumfrog

MeJennifer, since A (for example) can individually confirm (by centripetal force measurements) that he is non-rotating, but can also observe (by receiving light pulses from B) that B is wandering around in the "sky" (relative to A's coordinate axes) with an almost fixed period (changing only due to the continuous slow radiation of gravitational waves), so A can certainly use the configuration as a clock (without reference to the distant background stars, nor anything else except A and B).

Likewise, B can use the motion of A (in B's sky) as a clock. This will have almost the same period, with a difference attributable to the different "time dilation" factor according to how many Scharschild radii B's surface is at, compared to A's surface (I'm presuming observers are on the surface, and A and B are spherical masses).

Similarly, an independent third observer can use the motion as a clock. The period will be again different, owing to the time dilation of the combined frame, the extent of the frame dragging effect, etc. But everybody will be happy using the system as a clock.

10. Apr 27, 2007

### Wallace

I think this is the crux of my problem with this. How does A define their co-ordinate axes when there is no other observable points to fix those axes?

11. Apr 27, 2007

### MeJennifer

Censiumfrog,

the trouble I have is that I think we need (to associate) a unique spacetime event for each tick of the clock in order to have an obsever independent clock. How do you get around that or do you think it is not neccesary. And isn't the (gravitational) Thomas precession going to avoid you from having an observer independent definition of what a full revolution actually means?

12. Apr 27, 2007

### robphy

It seems to me that the correct formulation of the problem posed
will, at some point [at least in discussing the test-particle limit],
have to make contact with the issues discussed in this paper:

On Relative Orbital Rotation in Relativity Theory
by
David B. Malament
http://www.lps.uci.edu/home/fac-staff/faculty/malament/papers/RelOrbitalRotation.pdf
http://philsci-archive.pitt.edu/archive/00000117/00/RelOrbitalRotation.pdf [Broken]

One might want to see if issues discussed in this paper can be used as a starting point for the problem posed.

Last edited by a moderator: May 2, 2017
13. Apr 27, 2007

### cesiumfrog

It should be trivial? Mathematically, parallel transport, or physically, by means of a simple gyroscope. Of course these axes wouldn't be perfectly fixed to distant stars, due to frame dragging effects.

I don't think Mach's principle comes into GR at this level. In GR an observer at any point on the manifold can measure intrinsic acceleration and angular velocity; but if we choose to believe Mach's principle it might be very awkward to define a universe consisting of just two objects "orbiting one another" with nothing else.

I don't understand what you're getting at.

Trivially, A or B can mark a space-time event (the tick of the clock, perhaps by flashing a light) at every "completion" (noting the start is chosen arbitrarilly) of a period, and they should also agree on this (though they may disagree on how many seconds it takes). Likewise, they can mark a separate sequence of space-time events for the ticking of their personal atomic clocks.

Now, further observers located at varying distances will surely experience differing degrees of frame dragging, and so disagree on what fraction of the period our first sequence of tick-events corresponds to. So what? They'll also disagree on the frequency of those atomic clocks. Why is all this a problem? These observers are no doubt clever enough to calcalute how to correct for the gravitational effects, if they wish.. but this seems not to satisfy you.

Isn't it one of the most fundamental principles of relativity theory, that there is no such thing as absolute time? But then, it seems like absolute time is exactly what you're looking for in your "observer independent" clock? What figures?

14. Apr 27, 2007

### Garth

In GR there isn't any Thomas precession (the masses are 'free falling' along their geodesic paths), there is however the geodetic precession.

The expression for the geodetic precession includes a space-curvature component (the part that includes $\gamma$) and a time dilation component.

$$\Omega = \nu \times (\gamma + \frac{1}{2})\nabla U$$

The time dilation component will account for extra time needed to make up 'the full orbit'.

Like Wallace I see this your OP question as essentially Machian in nature.

Garth

Last edited: Apr 27, 2007
15. Apr 27, 2007

### Dick

The Einstein equations are differential equations. You can't solve differential equations without boundary conditions. If you don't have boundary conditions this debate will go on forever. Probably will even with boundary conditions. Who am I kidding? I knew I should have stayed with the homework forums.

Last edited: Apr 27, 2007
16. Apr 27, 2007

### Wallace

Isn't that pretty much the point of Mach's principle?

But yeah I'm sure it's much more satisfying to hang around the homework forums and actually achieve something with a discussion. Although the open ended and never ending type discussion around these parts are not completely without merit, since you get to exercise your brain and physics skills. It's more the journey rather than the destination whereas I guess the homework forums are all about the destination!

17. Apr 27, 2007

### Dick

True, true. And I suppose that could be considered a version of Mach's principle. My point is that in the case of two orbiting bodies the presence of distant observers is built into the boundary conditions. It wouldn't have to be - I'm sure others are possible. The former just seems more like the world we live in.

18. Apr 27, 2007

### pervect

Staff Emeritus
I think this is a good observation. With two isolated orbiting bodies, it is standard to assume asymptotic flatness at infinity. This is a boundary condition. The need for a boundary condition is rather similar to the need for boundary conditions for solving E&M.

The definition of angular momentum in GR that I've seen involves this assumption of asymptotic flatness, for instance the one offered in Held, "General Relativity and Gravitation" (section 3 by Jeffrey Winicour). I think that defining angular momentum also solves the original problem.

Possible nit - this is the "ADM-like" definition of angular momentum, does the Komar-like defintion also require asymptotic flatness?

In crude and possibly inaccurate terms, this background at asymptotic infinity can be thought of as defining a sort of "inertial frame at infinity".

So I think the observation that the boundary conditions are the issue is a good one.

19. Apr 27, 2007

### George Jones

Staff Emeritus
http://www.science.psu.edu/alert/Bruegmann5-2004.htm" [Broken] may be of interest.

Last edited by a moderator: May 2, 2017
20. Apr 27, 2007

### Chris Hillman

Thomas precession in gtr?

Some of this discussion seems to involve posters who are in my "ignore list", but the following caught my eye:

Pedantic quibble: I think Garth misspoke; no doubt he meant to say something like this:

Thomas precession is an effect which arises when a small object moves in a circular orbit by virtue of some nongravitational force. (Think of a stone whirled around by some schoolchild at the end of a string.) In any region of spacetime having neglible gravitational field, it can be treated using only the formalism of str. The effect certainly makes sense in gtr too, it just isn't often considered as a separate precession effect in regions of strong curvature.

In Minkowski spacetime one could in principle try to treat a planet in a circular orbit around the Sun as a stone whirled at the end of a string, but this would not yield the correct precession (as measured in solar system tests). However, one can employ the weak-field formalism of "linearized gtr" to analyze precession effects of a small spinning object orbiting a rotating massive object, and then one can decompose the result into Thomas, de Sitter, and Lense-Thirring terms, but of course the Thomas term vanishes whenever no non-gravitational forces act on the orbiting object! See Ciufolini and Wheeler, Gravitation and Inertia for such an analysis.

Last edited: Apr 27, 2007