# Moving objects in Stationary and Static Spacetimes

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1. Jun 17, 2014

### latentcorpse

Hi,

I understand the definition of stationary as existence of a timelike Killing vector field and as a result the fields (metric plus anything else e.g. gauge fields etc) cannot have any time dependence.

Static is a stronger constraint that satisfies the above plus requires the Killing vector fields to be orthogonal to the spacelike hypersurfaces. This orthogonality tells us g_{tx}=0 and consequently, there can be no time-space components of the metric.

That's all fine.

My question is: can objects move in a) static spacetimes b) stationary spacetimes

According to post #2 in https://www.physicsforums.com/showthread.php?t=593850, in a static spacetime, if we make a measurement at t1 and again at t2, along the observer's world line, nothing will have changed i.e. the curvature of spacetime (metric), the fields e.g. gauge fields etc are all unchanged. Does this mean that any objects in the system just sit around and do nothing as time evolves? Surely not otherwise, why would we study static spacetimes - they would be insanely boring!

What about stationary spacetimes? Here we can have dt dx components of the metric eg in the Kerr black hole metric. This black hole is rotating and so surely there is time dependence on something? I would have guessed there should be time dependence in the metric due to such a rotation but that appears to be wrong since the metric components aren't functions of t!

This is annoying because based on this understanding, nothing would ever move and based on experimental observations of cars on the road outside my bedroom window, that is not true! Of course....we probably don't live in a static/stationary spacetime!

2. Jun 17, 2014

### Bill_K

The only things that can move are test particles, i.e. objects whose energy is so small that in an approximation their presence and motion does not affect the spacetime geometry.

3. Jun 17, 2014

### latentcorpse

Thanks. A few follow up questions:

1, This is the answer for both static and stationary?

2, So test particle is something with very low mass that we put in basically to see what happens?

3, What about the situation of the charged Reissner Nordstrom black hole. Could we find some charged particle that would move? I guess to avoid affecting the spacetime geometry it would also need to have low mass and so again would be considered a test particle, right?

4, What would happen if you put something really massive in? The geometry would be deformed so that the spacetime was no longer static/stationary?

4. Jun 17, 2014

### bcrowell

Staff Emeritus
No, this is wrong. For example, the field of the rotating earth is stationary, but the presence and motion of the matter does affect the spacetime geometry.

A static example is Einstein's static universe, in which the "dust" particles (galaxies) are moving, and are conceived of as a perfect, continuous fluid.

5. Jun 17, 2014

### WannabeNewton

No. Only quantities constructed purely out of the metric will be "time-independent".

The rotation of the source is time-independent so there's no reason the space-time should be time-dependent in the coordinates adapted to the source rotation.

6. Jun 17, 2014

### WannabeNewton

I think you mean a perfect fluid that isn't dust because dust cannot source a static space-time, unless you mean "dust" in a sense other than what it connotes in GR.

7. Jun 17, 2014

### bcrowell

Staff Emeritus
Einstein's static universe was a model in which he used the cosmological constant to balance out gravity and produce a static spacetime.

8. Jun 17, 2014

### latentcorpse

So we can have time dependent motion in both static and stationary?

Are there any differences between the motion of particles in static and stationary?

Say I want to write down the Lagrangian describing some theory in my static/stationary spacetime and I want some time dependent field that will describe the time dependent motion of matter - what would such a term look like.

Thanks.

9. Jun 17, 2014

### bcrowell

Staff Emeritus
Depends on what you mean by time-dependent. An atom in the earth has motion that is time-dependent, but in the nonrotating frame based on the earth's center of mass, the motion is time-independent at any given point in space.

Static is basically stationary+nonrotating.

10. Jun 17, 2014

### WannabeNewton

Relative to the rigid frame fixed to the distant stars yes. You are talking about motion on space-like hypersurfaces so the foliations are not unique. The foliation defined by the time-like Killing field itself is what I mean by "rigid frame fixed to the distant stars". Why wouldn't there be? Again the time-like Killing field Lie transports all metric defined quantities along its flow but not all quantities in a stationary or static space-time constructed purely from the metric.

This is a bit of a vague question but one obvious difference is frame dragging.

11. Jun 17, 2014

### pervect

Staff Emeritus
Having a timelike Killing vector just means that there exists a metric that describes the space-time where the coefficients are independent of coordinate time. Welll, one also needs to specify that the time coordinate is actually timelike, I suppose.

Note, though, that just because such a metric exists for the space-time doesn't mean you have to use it. So any given metric for a spacetime with a timelike killing vector may not have the property of being independent of coordinate time. But it will be diffeomorphic to some metric that has this property.

Note that the description of this property in terms of Killing vectors is coordinate independent, hence why it is preferrred to the "no time dependence of the metric" description, which isn't coordinate independent.

A logical consequence of the definitions is that such a system with a timelike Killing vector may be either static or stationary. The difference between the two is a matter of rotation, which can be described in tensor terems as the "vorticity tensor".

IF you consider an idealized Earth, there exists a time-independent metric (or equivalenlty, the metric of the idealized Earth in any coordinate system has timelike Killing vectors), howver, because the Earth rotates this metric is stationary and not static. The failure of the hypersurface orthogonality condition is the mathematics behind why you can't Einstein-synchronize clocks around the Earth's equator.

Hopefully you are already familiar with why rotating objects do not allow Einstein clock synchronization around them. If not, it's a bit hard to explain with complete technical accuracy in a short post, but you would benefit from studying the Haefele Keating experiment and the Sagnac effect here on the Earth.

12. Jun 17, 2014

### WannabeNewton

Thanks!

13. Jun 17, 2014

### latentcorpse

Thanks but still a few points I don't understand:

1, Why is the foliation not unique? Maybe I have just misunderstood you. Did you mean that I can pick any coordinate system (t,x,...) I like but there is only one coordinate system (the one where the foliation is defined by the KVF flow i.e. the one "fixed with respect to distant stars") where the time indepent motion becomes manifest?

2, What does it mean to be parallel transported along the KVF flow? Does this mean that in the rigid frame fixed to distant stars, there is no time dependence and so measurements at t1 and t2 will yield the same results?

3, A field strength term in a Lagrangian will look like F^{\mu \nu} F_{\mu \nu} with the contraction made by the metric. Does this mean that the gauge fields get parallel transported along the KVF flow?

14. Jun 17, 2014

### latentcorpse

Thanks. Just to pick up on the Einstein-synchronisation point:

Is it true this is due to the fact that two clocks in a rotating frame need to be synchronised by a chain of intermediate clocks and this depends on the path chosen. If so, can you explain how to see this?

Secondly, I would think this effect would apply everywhere EXCEPT the equator! Since on the equator there is only one path to choose - follow the equator! Clearly I'm missing something here.

15. Jun 17, 2014

### WannabeNewton

There is no unique way to foliate space-time into space-like hypersurfaces. Different choices will result in different spatial trajectories for particles which as Ben noted may or may not have "time" dependence, where "time" refers to the parameter characterizing the foliation.

It's Lie transport not parallel transport so you have to be a bit careful there. If a quantity is Lie transported by the time-like Killing field then yes it means exactly what you stated.

Yes but again it's Lie transport not parallel transport. Anyways, this has to be true otherwise the gauge field, through its coupling to the gravitational field, would introduce time dependent terms in the Einstein equations.

16. Jun 17, 2014

### WannabeNewton

The clock synchronization paths taken for the local Einstein synchronization gauge connection are not curves in physical space like the equator. They correspond to simultaneity curves in the quotient manifold obtained by taking the space-time and identifying all points that lie on the same orbit of the time-like Killing field.

17. Jun 17, 2014

### pervect

Staff Emeritus
Suppose you have 1000 clocks spaced around the equator, in order so that clocks with nearby numbers are close, ie. clock 1 is near to 2, 2 is near to 3, etc. Also clock 1 is next to clock 1000 because the equator is a circle Then you start pair-wise Einstein synchronizing them, clock 1 to 2, clock 2 to 3, clock 3 to 4, etc. When you 've worked your way around the equator, you finally Einstein synchronize clocks 999 and 1000. Clock 1000 is right next to clock 1, but they won't be Einstein synchronized, because of the Earth's rotation and the effects of special relativity. I've said "The Earth" here, but the same result applies in flat space-time with a rotating disk. You sometimes need to account for GR effects on the Earth (gravitational time dilation, for instance) but the effect occurs in special relativity too, and may be easier to analyze there.

You can resync clock 1 to clock 1000, but then it will no longer be Einstein synchronized to clock 2.
There will always be a "gap" in the synchronization.

Last edited: Jun 17, 2014
18. Jun 18, 2014

### latentcorpse

So can you give me an example of a term in the Lagrangian that would allow time dependence?

19. Jun 18, 2014

### latentcorpse

I understand the idea of this quotient manifold but don't see the connection between that and the equator. Taking such a quotient allows us to pick a particular spatial slice of our spacetime - this manifold will include the equator but surely it will also include other space as well?

20. Jun 18, 2014

### latentcorpse

So the synchronisation gap is due to SR effects eg time dilation. That seems fair enough. But what's the connection to lack of hypersurface orthogonality? Is it just that lack of orthogonality means the spacetime is stationary and therefore allowed to rotate which introduces these SR effects? Or is it something deeper?

Also, if I set up a similar series of clocks on a circular path above the equator, why would this not happen? What's special about the equator here?

Last edited: Jun 18, 2014
21. Jun 18, 2014

### WannabeNewton

22. Jun 18, 2014

### latentcorpse

Yes I have read through it but I'm afraid the connection between the path and the quotient manifold is not clear to me. Is it to do with post #3 that you make.

You pick a frame where the clocks are at rest (equivalent to quotient manifold) and set up the clocks on a path $\gamma$ and say the action is transitive if integral of dt is zero. This would clearly require g_{0i}=0 for spatially separated clocks i.e. a non-rotating spacetime. Or is it a non-rotating frame? I guess a frame but then wouldn't this mean clock synchronisation is frame DEPENDENT?

Also, why does synchronisation require $g_{00} dt = -g_{0i} dx^i$?

Thanks.

23. Jun 18, 2014

### pervect

Staff Emeritus
Relativity of simultaneity is actually the SR effect of interest.

It's 100% identical to hypersurface orthognoality - it's an operational definition of hypersurface orthogonality.

A spatial surface (not necessarily orthogonal) is a surface of constant time. An orthogonal hypersurface is, operationally, a surface of constant time as defined by Einstein clock synchronization convention. Assume that you're given a bunch of worldlines (each of which you can imagine is carrying a clock). In this case, these worldlines, or clocks, are just those clocks or worldlines that are co-rotating with the Earth. If you can synchronize all the clocks via Einstein clock synchronization (you may have to adjust the rate of some clocks in general, this is allowed but not necessary for this example when we are restricting our points on the equator), then the set of all events "at the same time" is the orthogonal space-like hypersurface that's orthogonal to the worldlines of the clocks.

But we've just seen that there isn't any such surface (at least not a closed surface), because there isn't any such synchronization possible.

You get the same effect, the equator isn't special.

24. Jun 18, 2014

### WannabeNewton

Yes it is a locally non-rotating frame. This means that, given the Lorentz frame (field) $e_{\alpha}$, with $e_0 \propto \xi$, where $\xi$ is a time-like Killing field, we must have $F_{e_0}e_{\alpha} = 0$ where $F_{e_0}$ is the Fermi-Walker derivative. This condition is equivalent, for time-like Killing fields, to $\xi_{[\alpha}\nabla_{\beta}\xi_{\gamma]} = 0$. This latter condition, through Frobenius' theorem, is equivalent to the condition that $\xi$ is hypersurface orthogonal to a one-parameter family of space-like hypersurfaces i.e. a space-like foliation of space-time.

However the quotient manifold is not the frame; here is how the quotient manifold is defined. Let $\sim$ be the equivalence relation between $p,q\in M$ given by $p\sim q$ if $p,q$ belong to the same orbit of $\xi$. Then the quotient manifold representing the space on which the local Einstein simultaneity curves for the congruence are defined is given by $M / \sim$. Here $M$ is the space-time.

Clock synchronization is always frame dependent. This is true even for inertial frames in SR.

That is only true for local Einstein synchronization. To see this, consider two integral curves (or orbits) of $\xi$ that are infinitesimally close to one another and let $p,q$ be events from the respective curves that are infinitesimally separated. Then we can apply SR in the immediate vicinity of $p$, which will include $q$, and use the fact from SR that two events $p,q$ are Einstein simultaneous if the space-time displacement vector $\eta$ joining them is orthogonal to $\xi$ at $p$. Now choose a coordinate system $\{x^{\hat{\mu}}\}$ adapted to $\xi$ so that $\xi = \partial_\hat{t}$ hence $\xi_{\hat{\mu}} = g_{\hat{\mu} \hat{t}}$; this is called Fermi-Normal coordinates. In this coordinate system $\eta^{\hat{\mu}} = dx^{\hat{\mu}}$ and the condition for local Einstein simultaneity becomes $0 = dx^{\hat{\mu}}g_{\hat{\mu} \hat{t}} = d\hat{t} g_{\hat{t}\hat{t}} + dx^{\hat{i}} g_{\hat{t}\hat{i}}$. I'm using hats to indicate that the Fermi-normal system comes from a physical frame.

You can think of $\xi_{\hat{\mu}}$ as defining a gauge connection on the bundle associated with the quotient manifold $M / \sim$. If $\xi_{[\alpha}\nabla_{\beta}\xi_{\gamma]} = 0$ then local Einstein synchronization can be propagated under the gauge connection defined by $\xi_{\hat{\mu}}$ along any two curves in $M / \sim$ that have the same endpoints and the result will be the same. This amounts to having a transitive Einstein synchronization convention for this congruence.

Last edited: Jun 18, 2014