# How can we test space-time for pseudo-Riemannian-ness?

1. Jun 30, 2011

### fr.jurain

Hi all,

One thing I found puzzling about several intros to GR is that, after a summary exposition of absolute diff. calculus, without further ado they posit a Levi-Civita connection, then derive the Einstein field eq. As if all manifolds had to be pseudo-Riemann, then.

Do they really have to, though? I tried to look a bit deeper into the matter and my provisional, & somewhat dismaying, conclusion is that yes, by a weird spin-off of the Equivalence principle they sort of have to. At least those posing as space-time models, that is.
That is, it could pretty well be that in practice we can't test whether our space-time continuum is pseudo-Riemann like it says in the papers, or not.
Did I miss s/thing? If so, what would be a practical test of pseudo-Riemannian-ness?

Here's how I understand the modelling of space-time in GR:
1) we need it 4-dim because of how we experience space & time,
2) we want a smooth manifold because the laws of Nature come out to us as differential eq's,
3) we want a tangent (1, 3)-metric tensor at every 4-point because again, it's our experience that lengths & velocity of light can be measured locally, and that SR holds locally,
4) we need a connection because the associated covariant derivative (CD) is what'll make the laws of Nature "generally covariant".
However, the only motivation for, in addition, requiring the CD of the metric to vanish everywhere, i. e. the manifold to be pseudo-Riemann, seems to be mathematical coziness, not physical necessity.

Now, imagine a (classical, i.e. non quantum) universe, whose physicists were lucky enough that their home world verified (1)-(4) exactly; only they all, like one man, picked the same wrong connection when putting it into theoretical shape.
I think their resulting GR will be mostly OK; they'll have inertia/fictitious forces, curvature-based gravitation, EP, local SR, general covariance, all there & accounted for. It's not even certain their EFE would look awkward.
Of course, they can't fail to notice discrepancies; e. g., test particles which we (who watch their universe from outside) know to be free-falling, will not follow their notion of a geodesic. Not enough to let them revamp their connection though, as they're still free to blame these discrepancies on a physical cause: one of the as yet unmodelled, but not fictitious, "fundamental forces of Nature"; because, as the difference of 2 CD's, the cause in question has the tensor nature.

In other words: there is no (practical) way to sort out "fundamental forces of Nature not described by GR" from "bits we dispensed with adding to the expression of the connection"; every bit of what we describe one way we could just as consistently describe the other way.
This is true as long as the theory comes out with the disclaimer "not completely unified yet".

Hence the odd conclusion: as long as we refrain from unification, i. e. just trying to account for inertia, the EP, and gravitation, then "everything goes"; that is, every connection is roughly as good as any other.
Or is it? Is there a physical necessity to have the CD of the metric vanish? Any light appreciated.

Best regards,
François Jurain.

2. Jun 30, 2011

### atyy

The left side of the EFE is picked so that the stress-energy is covariantly conserved. Could one do this with a different connection?

3. Jun 30, 2011

### bcrowell

Staff Emeritus
If I take a non-covariant derivative of a tensor, $\partial_aT_{bc}$, it can be nonzero for two reasons: because T is varying, or because the metric is varying. The derivative can vanish for one choice of coordinates but not for another.

Now substitute g, the metric, for T above. The non-covariant derivative of the metric can be nonzero for two reasons: because the metric is varying ... or because the metric is varying. In other words, it seems to me that there is no physically meaningful way to say whether the metric is varying, because you can't separate physical changes from changes that exist only because of a choice of coordinates. So for this reason it seems to me that we need the covariant derivative, and the covariant derivative must vanish when evaluated on the metric.

BTW, there are formulations of GR, such as the Ashtekar formulation, in which the signature can be different on different parts of the manifold.

4. Jun 30, 2011

### martinbn

I might have misunderstood your post, but pseudo-Riemannian is not related to the choice of connection. The defining property is having a metric (pseudo-Riemannian). The choice of the Levi-Civita connection is convenient, and it may come also for historical reasons. But the difference with any other connection is a tensor, so if another plays a role in the physics it can be represented as the chosen one plus a tensor field. The tensor field would be interpreted as additional field.

5. Jun 30, 2011

### atyy

6. Jun 30, 2011

### Staff: Mentor

My understanding seems to be a little different from yours. To my understanding any test for SR is a test for pseudo-Riemannian-ness.

I think that the question of the connection is independent from the question of pseudo-Riemannian-ness. We want a metric compatible connection because we want the physical laws which determine if something is 1 second in duration to be the same everywhere.

7. Jun 30, 2011

### TrickyDicky

Yes, gravity defined as curvature demands it.
Maybe if you think of situations where you don't want a Levi-civita connection like for instance in the gauge theories of particle physics that don't deal with gravity, you'll understand better why the Levi-civita connection is the only choice in GR where you need to relate different points each with its particular vector spaces in a globally curved manifold, while the vector spaces in gauge theories are independently added to the manifold. They don't need to preserve any metric since the vector space is just added to the manifold structure so they use a different connection.
But in GR you really need a metric, because of the intrinsic curvature of the manifold, and once you have a metric the connection naturally should be compatible with that metric.
I'm curious too about why you relate the need of metric-compatibility with the fact that the signature in relativity is not definite-positive. Apparently these are facts independent of each other as has been pointed out above.

8. Jul 1, 2011

### fr.jurain

Hi all,

thanks for your thoughtful responses.

Sorry, I could help you understand by using the same language as everybody else, to begin with. If you replace "pseudo-Riemannian" everywhere in my OP with "equipped with the Levi-Civita connection induced by the metric" then you'll get my meaning. No wonder if you're not sure you got me right.

I'd say of course it could, possibly at the cost of inflicting the equation with an ungainly look. It's quite clear the way martinbn wrote it:
It's quite clear to me only a covariant derivative will do. It's also clear GR practitioners have more than a fondness for conserving metrics, what puzzles me is that they don't seem physically compelled (by experiments, I mean) to do so.

Your understanding seems to be martinbn's, see above. Yes, to posit a tangent metric everywhere is to express beliefs like "MM-experiments also yield null-results around Alpha Centauri".

I see. The motivation is to share a common unit length (or a common unit time-interval, same thing) all over the universe.

Interesting, it looks like an essentially different motivation from DaleSpam's. Is it?

OK, we do have a tangent metric. Naturally the connection *could* be metric-cancelling, yet I fail to see why it *should*. E. g. we could still write down a field equation. What breaks down?

It might be that your understanding of "pseudo-Riemannian" is the same martinbn's, and just about everybody else's; see above. It's hard to tell where the emphasis is on a written text; I certainly had no intent to put it on "pseudo". I see no relation between signature and metric-compatibility, rather the former seems to be backed by experiments whereas the latter is not.

Anyway, thanks again for your enlightening responses. It seems I can prune all of my OP down to:
* suppose you & I sail to Alpha Centauri by 2 different routes, dropping a buoy at every mile along our ways with a 1-meter long iridiated-platinum ruler attached (with which to calibrate the next one); shall we find a closure error when we rendez-vous at last?
* if there is none, is this how the Alpha-Centaurian meter is the same as ours?
* until we go there & do that, what practical tests can we actually carry out?
* And if we can't, when do we care if some theories predict non-null results for these tests?

Best regards,
François Jurain.

9. Jul 1, 2011

### TrickyDicky

Not really,the difficulty transporting vectors in a curved manifold is precisely what the covariant derivative tries to solve and why it is so important to share a common unit length all over in a curved spacetime setting. This wouldn't be necessary in a flat manifold where the tangent vector space is isomorphic to the flat spacetime.
The principle of Equivalence, I should have mentioned it.

The latter is backed by experiments confirming the (weak) equivalence principle.

10. Jul 1, 2011

### atyy

OK, so we write the EFE in terms of a different connection and an additional field. Would the new field be considered gravitational or matter?

Some possibly useful articles on "principles of gravity":

Sotiriou, http://arxiv.org/abs/0707.2748" [Broken] (try the discussion on p19)

Hinterbichler, "[URL [Broken] Aspects of Massive Gravity
[/url] (says that the idea that gravity is a massless spin-2 particle is much more constraining than any classical "derivations")

Bekaert et al, http://arxiv.org/abs/1007.0435" [Broken] (contains the Weinberg "low-energy theorem")

Last edited by a moderator: May 5, 2017
11. Jul 1, 2011

### fr.jurain

Ah thanks, although I think I missed a step. My read of the above is you're saying weak EP => normal coordinates are available => connection is Levi-Civita; are you? Seems to me normal coordinates are enough to account for the weak EP, by making geodesics look straight and run along at constant velocity, at any prescribed 4-point e (d^2M/dt^2 = 0 at e); then any torsion-free connection will do the trick. If in addition we needed the metric to look stationary (so add $\partial_ig_{jk}$ = 0, also at e), then only Levi-Civita would do; need we, though?

Well, if I'm still with you, I'll say my OP was prompted in part by the impression you put it into this or that bin according to personal religion and taste. Guess I'll look into it again after digesting TrickyDicky's response. Thanks for the refs, might take some days to digest them too.

Best regards,
François Jurain.

12. Jul 1, 2011

### TrickyDicky

Normal coordinates can be used both locally (minkowskian) and globally (LC connection) in any (Pseudo)riemannian manifold (because inertial frames of SR don't exist globally in a curved manifold).

13. Jul 1, 2011

### TrickyDicky

Since the Equivalence principle has very different formulations, most of them not totally satisfactory due to use of vague or not well defined terms, maybe the LC connection is better understood directly as the implementation in GR of the empirical observation (Eotvos experiments, etc) of the equal falling of all bodies (universality of free-fall), thru the covariant derivative. In this way this observation is generalized in the theory.

14. Jul 3, 2011

### atyy

One part of the EP usually involves the use of Riemann normal coordinates, in which the Christoffel symbols can be set to zero at a point, eg. Eq 8.8, 8.9 and the discussion following in http://arxiv.org/abs/1102.0529

Matter is usually thought of something that has a stress-energy tensor field and is minimally coupled to the metric. The Sotiriou review linked to above discusses some theories of gravity in which matter and geometry seem interchangeable, eg. scalar-tensor theories in the Einstein and Jordan frames on p13, where the EEP in the Einstein frame is "empty" since there is always a force acting on massive particles, while the EEP is satisfied in the Jordan frame.

Last edited: Jul 3, 2011
15. Jul 4, 2011

### fr.jurain

Yes any connection will do, on my own I eventually figured as much with http://arxiv.org/abs/gr-qc/9806062v1; strong EP actually. Seems you even get a wee bit more if the connection is torsion-free: they vanish in a small domain around 4-point e. So perfectly-normal-looking-inertialness all the way between the 8 corners of the usual elevator car, not only at the center. I guess "small" is so the theory can get away in case you managed to somehow stuff a whole black hole inside the car.

16. Jul 4, 2011

### fr.jurain

Sorry I intended to click "preview". Here's to the 2nd part:

So it's more like "you can drop it into whichever bin without breaking down the math" than like "you may drop it into whichever bin you please today". Your ref seems to say, "what if photons run straight from A to B while electrons mind the curb", and to deserve more than this passing glance. I'll have a closer look, then maybe start a few new threads.

Back to my original question: it seems in fact I was requesting a practical test for the Levi-Civita-ity of the particular connection the Lord was good enough to bestow on us. The posts of these last days pointed to the existence of a common unit length. The direct test would be to carry 2 identical standards from A to B by 2 different routes; the null-result is if both standards agree upon arrival. Given that, for precision, A & B may need to be either a few LY apart or in a few centuries from now, what else can we do?

Thanks for your help anyway.

Best regards,
François Jurain.

17. Jul 4, 2011

### atyy

If one doesn't use the Levi-Civita connection, are the coordinates in which the connection components vanish at a point also coordinates in which the metric is Minkowski at that point?

18. Jul 4, 2011

### TrickyDicky

This is confusingly stated, in the linked paper they refer to any "linear connection" in the sense of a connection on a vector bundle and specified by a covariant derivative, so it is not just any connection, but actually by just adding the torsion-free requisite, the one we have been referring all along as Levi-Civita connection.

19. Jul 4, 2011

### fr.jurain

Yes, although not necessarily the standard one, i. e. one such that the tensor is Diag(-1, +1, +1, +1). That's how I read his paper at least, which was not in depth.

You're right of course. Only thing I can use in a connection is the associated covariant derivative; I would be clearer if I just wrote so.
Ah, this is where we've been disagreeing.
The way I understand it, torsion-free *LINEAR* connections are exactly those that differ from a LC one by a (1 index up, 2 down)-tensor field if the field in question is symmetric in the down indices; whereas you allow no difference.
Am I deluded in believing that there are torsion-free linear connections which can't be LC?

20. Jul 4, 2011

### atyy

I think I'm using wrong terminology. By Minkowski did you mean that the signature is (1,3)? I meant that it should have standard form diag(-1,1,1,1), which is what I've usually understood for the EP.

21. Jul 4, 2011

### TrickyDicky

Here's how I understand it: the paper you linked states the existence (and uniqueness) of normal coordinates for symmetric linear connections, and the presence of normal coordinates seem to imply the preservation of the metric, then the fundamental theorem of Riemannian geometry states that on any Riemannian manifold (or pseudo-Riemannian manifold) there is a unique symmetric connection that preserves the metric.

22. Jul 5, 2011

### fr.jurain

The Wolfram folk http://mathworld.wolfram.com/MinkowskiMetric.html are consistent with your meaning, only they don't distinguish between the tensor and the array of its components in a given basis; what I dub "to be Minkowski" they dub "to be equivalent to the Minkowski metric" i. e. to have sig (-1, 3) as you guessed. Again I may be at odds with the usual practice, especially among physicists.
Others frown upon the use of the word "metric" (because of the topological undertones), and only allow "Minkowski space", a cute gig step around the core of your question.
I won't pretend I answered it by the way. I believe that, at least in the case where the conn. coeffs are 0'ed at point e only, it is always possible to additionally bring the metric to the form diag(-1,1,1,1) at e; and that it's enough for EEP (which, on 2nd reading, is what the paper I ref'd seems to call SEP, just in case things weren't muddy enough already).

23. Jul 5, 2011

### fr.jurain

So do I.

OK.

X marks the snag...
Do we agree at least on an aside: even if the Chr. symbols formulas ({ab,c}=(Gac,b + Gcb,a-Gab,c)/2 etc.) hadn't been invented yet, only that the conn. coeffs are symmetrical in the down indices, still you could prove 0 torsion?
More on the main contention tonight (which is your afternoon I guess).

24. Jul 5, 2011

### fr.jurain

A bit sooner than I'd expected actually.

Here's an example of a symmetric connection which, if I'm not mistaken, cannot derive from a metric.
Start with R^4 as the manifold, Diag(-1, +1, +1, +1) as the metric, and the cov. der. of the corresponding LC connection, the "Minkowskian connection"; less pedantically known as the plain directional derivative.
Now, switch to polar coordinates t, r, u, z (i. e. x = r*cos(u), y = r*sin(u)), remove the plane r = 0 as a precaution, and to the coefficients of the connection add the following tensor H, with components given in the tangent basis:
Htu^t = Hut^t = exp(A(r)) where A is some non-constant, finite, function of r only,
Htt^u = exp(B(r)), also non-constant & finite (-exp(B(r)) is OK, too, maybe even B = A),
all other Hab^c = 0.
The "perturbed" connection is still symmetric, so it has normal coordinates if we trust Iliev (or else just plain Wikipedia http://en.wikipedia.org/wiki/Normal_coordinates), yet I found there's no way it could derive from a metric tensor. Try it yourself by all means, there are too many ways I could get it wrong.

25. Jul 5, 2011

### fr.jurain

Confirmed, it could even be 0. I found a simple test: the symmetry of the Ricci tensor in its 2 indices. In the present case Rtu != Rut, with the indicated choice of coordinates and Htu^t = exp(A(r)).

Although it does not, I may not have been perfectly clear about this: if you can BOTH make the coordinates normal at any prescribed 4-point e, and SIMULTANEOUSLY, preserve the metric at the same e, then yes, I believe the connection is LC, i. e. the metric is actually preserved everywhere.
By "preserving the metric at e" I mean canceling the cov. derivative of the metric tensor (also at e), or equivalently since the coordinates are normal, its partial derivatives (again, at e).