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What does general relativity say about the relative velocities of objects that are far away from one another? In particular:--

Can distant galaxies be moving away from us at speeds faster than $c$? Can cosmological redshifts be analyzed as Doppler shifts? Can I apply a Lorentz transformation in general relativity?

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What does general relativity say about the relative velocities of objects that are far away from one another?

Nothing. General relativity doesn't provide a uniquely defined way of measuring the velocity of objects that are far away from one another. For example, there is no well defined value for the velocity of one galaxy relative to another at cosmological distances. You can say it's some big number, but it's equally valid to say that they're both at rest, and the space between them is expanding. Neither verbal description is preferred over the other in GR. Only local velocities are uniquely defined in GR, not global ones.

Confusion on this point is at the root of many other problems in understanding GR:

Question: How can distant galaxies be moving away from us at more than the speed of light?

Answer: They don't have any well-defined velocity relative to us. The relativistic speed limit of c is a local one, not a global one, precisely because velocity isn't globally well defined.

Question: Does the edge of the observable universe occur at the place where the Hubble velocity relative to us equals c, so that the redshift approaches infinity?

Answer: No, because that velocity isn't uniquely defined. For one fairly popular definition of the velocity (based on distances measured by rulers at rest with respect to the Hubble flow), we can actually observe galaxies that are moving away from us at >c, and that always have been moving away from us at >c.[Davis 2004]

Question: A distant galaxy is moving away from us at 99% of the speed of light. That means it has a huge amount of kinetic energy, which is equivalent to a huge amount of mass. Does that mean that its gravitational attraction to our own galaxy is greatly enhanced?

Answer: No, because we could equally well describe it as being at rest relative to us. In addition, general relativity doesn't describe gravity as a force, it describes it as curvature of spacetime.

Question: How do I apply a Lorentz transformation in general relativity?

Answer: General relativity doesn't have global Lorentz transformations, and one way to see that it can't have them is that such a transformation would involve the relative velocities of distant objects. Such velocities are not uniquely defined.

Question: How much of a cosmological redshift is kinematic, and how much is gravitational?

Answer: The amount of kinematic redshift depends on the distant galaxy's velocity relative to us. That velocity isn't uniquely well defined, so you can say that the redshift is 100% kinematic, 100% gravitational, or anything in between.

Let's take a closer look at the final point, about kinematic versus gravitational redshifts. Suppose that a photon is observed after having traveled to earth from a distant galaxy G, and is found to be red-shifted. Alice, who likes expansion, will explain this by saying that while the photon was in flight, the space it occupied expanded, lengthening its wavelength. Betty, who dislikes expansion, wants to interpret it as a kinematic red shift, arising from the motion of galaxy G relative to the Milky Way Galaxy, M. If Alice and Betty's disagreement is to be decided as a matter of absolute truth, then we need some objective method for resolving an observed redshift into two terms, one kinematic and one gravitational. But this is only possible for a stationary spacetime, and cosmological spacetimes are not stationary. As an extreme example, suppose that Betty, in galaxy M, receives a photon without realizing that she lives in a closed universe, and the photon has made a circuit of the cosmos, having been emitted from her own galaxy in the distant past. If she insists on interpreting this as a kinematic red shift, the she must conclude that her galaxy M is moving at some extremely high velocity relative to itself. This is in fact not an impossible interpretation, if we say that M's high velocity is relative to itself in the past. An observer who sets up a frame of reference with its origin fixed at galaxy G will happily confirm that M has been accelerating over the eons. What this demonstrates is that we can split up a cosmological red shift into kinematic and gravitational parts in any way we like, depending on our choice of coordinate system.

For those with a more technical background in abstract math, the following description may be helpful. (The answer by knzhou does a nice job of explaining this in nontechnical terms.) Spacetime in GR is described as a semi-Riemannian space. A velocity vector is a vector in the tangent space at a particular point. Velocity vectors at different points belong to different tangent spaces, so they aren't directly comparable. To compare them, you need to parallel transport them to the same spot. If the spacetime is (approximately) flat, then you can do this, and you can say, for example, that the sun's velocity vector minus Vega's velocity vector is a certain value. But if the spacetime is not even approximately flat (e.g., at cosmological scales), then parallel transport is path-dependent, so the comparison becomes completely ambiguous.

Related: Why is the observable universe so big?

References

Davis and Lineweaver, Publications of the Astronomical Society of Australia, 21 (2004) 97, msowww.anu.edu.au/~charley/papers/DavisLineweaver04.pdf

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    $\begingroup$ Wait a second... I would hazard a guess that global conservation laws are not well defined as well and so they don't hold? $\endgroup$ – Joker_vD Apr 18 '18 at 9:44
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    $\begingroup$ Dear Ben Crowell, can you please tell me if the speed of the galaxy (moving away faster then c) is not defined in GR, then how can we define the speed in GR of the far away photon coming from that galaxy? That is not defined either in GR? $\endgroup$ – Árpád Szendrei Apr 18 '18 at 14:40
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    $\begingroup$ @ÁrpádSzendrei: The speed of a photon is like any other speed. It's well defined for a nearby observer, not defined for a distant observer. For example, a distant observer can say that a photon has zero speed as it approaches the event horizon of a black hole, and if they want to say that, there's nothing wrong with that, but it's not very meaningful. To a local observer, the photon moves at c. $\endgroup$ – Ben Crowell Apr 18 '18 at 15:19
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    $\begingroup$ I thought spacetime was approximately flat at cosmological scales. I thought there was a parameter that could be interpreted as the mean global curvature of the universe, and I thought observationally it was very close to zero. Am I confusing curvature of one choice of spatial slice with curvature of spacetime? $\endgroup$ – Mark Foskey Apr 18 '18 at 17:15
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    $\begingroup$ I've always wondered (as a thought experiment) what would happen if you had a billion light year long string tied between somewhere in the Solar System like a large asteroid, and a corresponding location in a distant galaxy. Obviously any forces transmitted would only propagate at the speed of light, but could it be used to answer this question? $\endgroup$ – Michael Apr 18 '18 at 18:10
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There's already a good answer, I'll just say it a slightly different way.

Consider two people running at the same speed on a road. Computing their relative speed is easy, we just do vector subtraction. If they're running the same direction, the relative speed is zero; if they're running opposite directions, the relative speed might be, say, twenty miles per hour.

If the two runners instead run on separate continents, it is much harder to define a relative speed. You could just compute their relative speed in three-dimensional space, but we don't want to leave the surface of the Earth, because in general relativity, you can't leave spacetime; there is no higher-dimensional space it's embedded in.

Staying on the surface, we could say they're going the "same way" if they're both running North, and that this means they have zero relative speed. But this is not self-consistent: if they eventually run up to the North pole and meet we will say they're going the same way when they clearly aren't. In fact, there is no self-consistent rule at all, because the surface of the Earth is curved; this problem is essentially what curvature is, mathematically.

In flat spacetime, i.e. the setting of special relativity, the curvature vanishes and you can define a consistent notion of "same way" by parallel transport. For example, if I spin around rapidly then in my frame the Sun is naively moving faster than the speed of light. But if you parallel transport the Sun's velocity to your position, to compute a relative velocity, you'll find that it's exactly what it would be if you weren't spinning, as expected. This is how special relativity handles relative velocities in its most general setting. In the case where the curvature is not negligible, there is no useful notion of "same way" and hence no relative velocity at all, as Ben Crowell explains. This isn't a problem, since we should only seek to define things we can measure locally.

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  • $\begingroup$ I see what you're saying with path-independence of parallel transport and how it depends on vanishing curvature, but connections aren't the only way to produce isomorphisms between tangent spaces. One thing I wonder about is spaces with trivial tangent bundle (Lie groups, say). There, we have a canonical way of identifying tangent spaces; would you say that we can compare velocities there? Or is parallel transporting vectors the only (physically?) acceptable way to do it? $\endgroup$ – Danu Apr 18 '18 at 7:19
  • $\begingroup$ @Danu : Doesn't this canonical identification rely on the 1-link (that is, a local copy) of the Cayley graph of the group -- i.e., on a global coordinate system? General covariance requires a great deal more agnosticism (or ignorance) about the frame of infinitesimal generators at various points in the group. $\endgroup$ – Eric Towers Apr 19 '18 at 3:24
  • $\begingroup$ @EricTowers I'm not sure what you are talking about; all one needs to do is pick a basis of the tangent space at a single point (e.g. the identity) and push it around using the differentials of left-translations. $\endgroup$ – Danu Apr 19 '18 at 7:42
  • $\begingroup$ @Danu : Picking a basis at one point is picking a simultaneous basis at all points by homogeneity. That is, you announce a global coordinate system as well as announcing that the intrinsic curvature of your space is constant everywhere. $\endgroup$ – Eric Towers Apr 19 '18 at 13:04
  • $\begingroup$ @EricTowers A global coordinate system? Not at all. Indeed, that would be a diffeomorphism onto an open subset of the Euclidean space of equal dimension, which is obviously not what I am talking about. (or indeed even possible). I am picking a basis of a vector space, which I then happen to be able to push around using homogeneity, trivializing the tangent bundle. Of course, I also not drawing any conclusions about the curvature of this manifold; this is a purely topological statement. I do not need any information regarding the Cayley graph of the group. $\endgroup$ – Danu Apr 19 '18 at 13:09
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Ben Crowell has answered this question already exhaustively. One aspect seems important. The core of the question whether space expands or distant galaxies are moving away lies in the understanding that these views depend on the chosen coordinates. So neither of these views - being not invariant - may be considered as a physical phenomenon. See Ned Wright's Frequently Asked Questions in Cosmology http://www.astro.ucla.edu/~wright/cosmology_faq.html

Here are some articles of cosmologists dealing with this issue:

A diatribe on expanding space https://arxiv.org/pdf/0809.4573.pdf

The kinematic origin of the cosmological redshift https://arxiv.org/abs/0808.1081

Expanding Space: the Root of all Evil? https://arxiv.org/abs/0707.0380

The kinematic component of the cosmological redshift https://arxiv.org/abs/0911.3536

Cosmological Redshift Interpreted as Gravitational Redshift http://www.ptep-online.com/2007/PP-09-06.PDF

There are more. In the end it seems a matter of taste what you prefer.

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  • $\begingroup$ This doesn't seem to be an answer, but an extended comment about a point made in a post. It's also mostly just links to other places, rather than an explanation. $\endgroup$ – Kyle Kanos Apr 19 '18 at 10:05
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    $\begingroup$ @KyleKanos it's not a full answer, but it adds useful information that could be lost in a comment. Timm could improve the answer by providing a sentence or two in summary of each link, but it's a worthwhile addition even without this. +1. $\endgroup$ – Reinstate Monica May 2 '18 at 5:16

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