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While I'm learning general relativity, the definition of the distance really confuses me. For example, we observe the distance between the Earth and the Sun (usually by a transit of Venus), what does the distance mean? When we say the PSR1913-16's semi-major axis is 1,950,100 km, what does it mean?

In Schwarzschild space-time, we can be a stationary observer because the space-time is stationary. Then we have a clear definition of the simultaneity surface, so we can define the distance as the proper length in the simultaneity surface between two objects. (However,I'm not sure whether we can do so if we are not a stationary observer). It is the same as we do in flat space-time.

However, in general, the space-time isn't always stationary. We may even NOT have a time-orthogonal coordinate system and NOT have any Killing vector fields. So we may have NEITHER special coordinate systems NOR special vector field. We cannot give a suitable and unique definition of the simultaneity surface and the distance.

We have 3+1 formalism, but when the certain observer is given, it also gives arbitrary simultaneity surfaces so we have arbitrary definitions of the distance. I know little about this, so I'm not sure.

In short, I just want to know:
1. Given a certain observer(such as human in the earth),is there a suitable and unique definition of the distance in curved space-time in general relativity? If the answer is yes,what is it?
2. If the answer of the first question is no, what does the PSR1913-16's semi-major axis mean? What does the distance between the Earth and the Sun mean?

Thanks for your help.

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4 Answers 4

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The spatial distance between two points is, if you pick a foliation by spacelike hypersurfaces, the distance between point $A$ and $B$ is the integral along the shortest geodesic linking the two points with that induced metric. In other words,

$$d(A,B) = \min_i \left(\int_A^B \mathrm d\lambda \sqrt{h^{ab} \frac{\mathrm dx_i^a(\lambda)}{\mathrm d\lambda} \frac{\mathrm dx_i^b(\lambda)}{\mathrm d\lambda}}\right)$$

Where $h$ is the induced metric on the hypersurface, and $x_i$ obeys

$$\ddot x^a + {\Gamma^a}_{bc} \dot x^b \dot x^c = 0$$

and goes through points $A$ and $B$. If you are unable to get a foliation, I'm not quite sure if the spatial distance between two points is well defined, globally at least.

If you want to do it for a particular observer, it's probably a good idea to pick Fermi coordinates with the proper time of the observer for the foliation.

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  • $\begingroup$ What kind of foliation are you referring to? Hypersurface foliations are entirely arbitrary, they need not correspond to an observer. As we don't get concordant results for different observers you won't get a concordant result for different foliations. $\endgroup$
    – Moonraker
    Commented Jul 9, 2016 at 16:04
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    $\begingroup$ I'm not sure there is a way to define a unique distance between two spacelike separated points. I suspect it will depend on the foliation you define. $\endgroup$
    – Slereah
    Commented Jul 9, 2016 at 16:13
  • $\begingroup$ So what does the distance between the earth and the sun mean? Is it the answer after we pick Fermi coordinates with the proper time of the observer for the foliation? I'm not clear about it. $\endgroup$
    – lrh2000
    Commented Jul 10, 2016 at 0:38
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The answer to your first question is:

The distance in general relativity is given by;

$ ds^2 = g_{\mu\nu} dx^\mu dx^\nu $

In special relativity (minkowski spacetime) this reduces to;

$ds^2= dt^2 -dx^2-dy^2-dz^2$ $ = \eta_{\mu\nu} dx^\mu dx^\nu $

If you know the metric of spacetime (obtained by solving Einstein's field equations) then you can find the distance between 2 events using the above formulae

Hope that helps! :D

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  • $\begingroup$ In curved space-time,there may not be any time-orthogonal coordinate system,so I do not have a unique definition of simultaneity surfaces. "The distance between 2 events" is of course well defined,but how about "the distance beween two objects"(such as the sun and the earth)? $\endgroup$
    – lrh2000
    Commented Jul 10, 2016 at 0:38
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As far as I know, the answer to the first question is: No, there is no unique definition of a spacial distance in a curved space-time. Already in simple cases like the Friedmann metric there are several definitions of distance: https://en.wikipedia.org/wiki/Distance_measures_(cosmology) However, for small distances and small curvatures (which applies in the case of the Earth and Sun example) all the definitions give approximately the same result (the Minkowski one). As soon as the curvature or the distance (e.g. points next to a black hole or points far apart in a Friedmann universe respectively) the deviations of the different definitions become large.

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First of all, I suppose that you are referring to distance in space, and not to distance in spacetime (which is invariant as shown in the answer of user122066).

Special relativity reserves a clear answer what length is with regard to length contraction, and this principle may be adopted by general relativity including the Schwarzschild solution.

For the explanation, I begin with a simplified model where the worldline of Earth is stationary with regard to Sun (that means if both were belonging to the same reference frame). In this case, the proper distance is the distance between Earth and Sun measured by observers of the reference frame Earth-Sun. The proper distance is the longest distance that can be measured. Observers which are moving with respect to the reference frame Earth-Sun are observing shorter distances, because the observed distance is subject to Lorentz contraction.

However, Earth and Sun are not belonging to the same frame of reference. The work which has to be done in this case must start from the model above, by appropriate synchronization, also taking into account the exact question you want to answer, in particular if you are referring to a particle (or several particles) moving from Earth to Sun (and from Sun to Earth, for synchronization purposes), or if you are not referring to a particle. Particles (mass particles or massless particles such as light) are needed for the synchronization.

I think that this is the exhaustive definition of distance that we get from special (and general) relativity.


(Edit further to the comment of goodqt)

However,in general,there will be many frames of reference which cannot be synchronizable.

This is a highly interesting question. The question is concerning the problem of e.g. two stars in space at a far distance which are not related one with the other, perhaps even too far to send light signals from one to the other (one being not in the "observable universe" of the other).

According to the above explanations of my answer, there would be no distance! And also Slereah in the other answer comes to a similar conclusion:

"I'm not quite sure if the spatial distance between two points is well defined, globally at least."

This is an incredible conclusion, indeed, and I hope that there will be other answers for your question which will put light on this issue.

Personally, I can provide you with one solution, but with all reserves because there might be other solutions provided which are based on curved spacetime, this would be interesting to know.

My solution is the following:

We have to consider the character of curved spacetime as a model for gravity, that means that we can imagine gravity as curved spacetime, but curved spacetime is only a tool of physics which must not necessarily represent the whole structure of the universe.

The Schwarzschild solution provides us with a mean of the replacement of curved spacetime by gravitational time dilation. That means: Gravitational time dilation is equivalent to curved spacetime which is gravity.

You may get an entry to the equivalency issue by the answer of John Rennie to this question.

By consequence, the model of curved spacetime of the Schwarzschild metric may be replaced by an alternative model describing gravity as mere time dilation.

In this alternative model, we have eliminated curved spacetime, and we are coming back to absolute space. And in this fact lies the answer to your question: In a model with absolute space, distances are well-defined!

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    $\begingroup$ Downvoters why? $\endgroup$
    – Moonraker
    Commented Jul 9, 2016 at 16:08
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    $\begingroup$ my sincere apologies, my d/v was based on misreading the question at the end of the post, I expected your answer to contain an equation, when actually one was not required. An edit to your answer will allow me to reverse it. $\endgroup$
    – user108787
    Commented Jul 9, 2016 at 17:18
  • $\begingroup$ (I'm not sure whether I understood your meaning correctly.)However,in general,there will be many frames of reference which cannot be synchronizable. $\endgroup$
    – lrh2000
    Commented Jul 10, 2016 at 0:39
  • $\begingroup$ I'm not sure whether I understood your meaning correctly, @goodqt, but this is a great question! I added an edit to my answer. $\endgroup$
    – Moonraker
    Commented Jul 10, 2016 at 7:01

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