# Physical reasons for having a metric-compatible affine connection?

1. Jul 14, 2014

### nearlynothing

So as the title says, what are the physical reasons behind requiring the connection between tangent vector spaces to be metric-compatible?

My guess is that this is desired from wanting different points in space-time to be "equivalent", in the sense that if any two vectors at a point are the same at another point, their inner product is the same, meaning that if a physical quantity is the same on different tangent spaces, their properties are still the same (no way of distinguishing one point on the manifold from the other).

I'm not used to talk in proper mathematical language so i apologize if the argument is not properly presented.

2. Jul 14, 2014

### haushofer

One reason is that metric compatibility can be seen as the general-covariantization of the fact that the partial derivative of the minkowski metric vanishes :)

3. Jul 14, 2014

### nearlynothing

I see what you mean, but then the reason that the connection in minkowski space-time is metric compatible relies on observation alone? or is there any pure theoretical reasons that tell us this should be so?

4. Jul 14, 2014

### Matterwave

A connection which is non-metric compatible would have some undesirable properties. The most important example is then that two vectors which are parallel transported would not maintain a constant inner product as your OP suggested. This would suggest that two vectors which are orthogonal, for example, even though they are transported parallel along a curve will not remain orthogonal. In what sense then are we still "parallel transporting" vectors?

5. Jul 14, 2014

### WannabeNewton

A variation of the Palatini action with respect to an arbitrary torsion free connection will yield metric compatibility so it is naturally built into GR given Einstein's equations. We don't need to put it in by hand.

6. Jul 14, 2014

### nearlynothing

I was also thinking along the lines of: Given a vector field that is covariantly constant, if the metric were not covariantly constant then the vector's norm would have different values just by virtue of being evaluated at different points. So this would imply equal vectors at different points have different properties just for being at different points.

7. Jul 14, 2014

### nearlynothing

Hey WannabeNewton, i'm not familiarized with the Palatini action, but i will make some time to check it out, thanks :)

Last edited: Jul 14, 2014
8. Jul 14, 2014

### haushofer

Parallel transport indeed preserves vector lengths. Physically, this means that e.g. the spectrum of an atom does not depend on the history (path) of the atom. It is the same all along the path.

In the palatini formalism you treat both metric and connection as independent variables. In Sugra you often take this formalism, as it makes (susy) variations of the action easier; you then only need to vary wrt th vielbein.

9. Jul 14, 2014

### WannabeNewton

See appendix E of Wald.

10. Jul 15, 2014

### Geometry_dude

We choose the connection $\nabla$ in GR to be metric, for the simple reason that this is the natural choice of a connection on a pseudo-Riemannian manifold $(\mathcal Q,g)$. Why?

Well, first ask yourself the following question: Why do we need a connection in the first place? The mathematician would tell you now that on a generic manifold $\mathcal Q$ there is nothing that tells you that a tangent vector at $q \in \mathcal Q$ is "the same" as a tangent vector at $q'$. "The same" means that we have chosen an isomorphism between the tangent spaces.
Of course, you could take a chart $(U, \kappa)$ and simply say that the vector
$$X_q = X^i \, \partial_i \upharpoonright_q$$
at $q$ is "the same" as the vector
$$X_{q'} = X^i \, \partial_i \upharpoonright_{q'} \, ,$$
but this would depend on a particular choice of chart and thus be quite arbitrary as well as local. Without any reference to any other geometric structure, you would take a connection $\nabla$ (get the name now?), find an auto-parallel $\gamma$ with tangent $\dot \gamma$
$$\nabla_{\dot \gamma} \dot \gamma = 0 \, ,$$
that connects $q$ and $q'$ and then find $X_{q'}$ by parallel transporting $X_q$ along $\gamma$
$$\nabla_{\dot \gamma}{X} = 0 \, .$$

So which $\nabla$ should we take?
Well, there already is a global structure on our manifold, namely the metric $g$! Can we use this somehow to get said isomorphism? The answer is, of course, yes and you can derive a condition on $\nabla$ that relates it with $g$ within the theory of G-structures. The condition is
$$X(g(Y,Z)) = g (\nabla_X Y, Z) + g (Y,\nabla_X Z)$$
for any vector fields $X,Y,Z.$ If this condition holds, we call $\nabla$ a metric connection or compatible with the metric $g$. This does not uniquely determine $\nabla$, but a computation in local coordinates will show you that it is enough to uniquely determine the auto-parallels $\gamma$. That same computation will tell you that the $\gamma$s are just plain geodesics!
In turn, this means that the condition for $\nabla$ to be "compatible with the metric $g$" is required to make sure that if there exists a geodesic $\gamma$ between $\gamma(0)=q$ and $q'=\gamma(s)$ and $X_q= \dot \gamma(0)$, then $X_{q'}= \dot \gamma(s)$. Alternatively, to put it in physical terms: If your connection is not metric and you compute its curvature, then that will have nothing to do with gravity!

11. Jul 15, 2014

### ChrisVer

May I ask then, what if there is no such path/curve $\gamma$ connecting $q,q'$?

12. Jul 15, 2014

### Geometry_dude

That's a good question. Obviously, the construction fails and hence you will have trouble finding a canonical isomorphism between the two tangent spaces. In general parallel transport is highly path-dependent, so you cannot just use another curve and it might even be that there is more than one auto-parallel connecting $q$ and $q'$. The example of the $2$-sphere with the standard metric and Levi-Civita connection is very illustrative.

13. Jul 15, 2014

### TrickyDicky

As an example of the case where unlike the 2-sphere you don't have this path-dependence there's the negatively curved pseudosphere surface.

14. Jul 15, 2014

### TrickyDicky

How does the path-dependence of GR manage to keep this physical feature of our universe?
In other words, it is kind of counter-intuitive, that we would need a metric compatible connection to allow us to preserve vector lengths and therefore spectra of atoms that are path(history) independent and curvature identified with gravity as geometry-dude said only to find that in GR there is no path-indepence in general.
It seems we can use the GR metric connection in general only to preserve vector lengths of infinitesimally distant points.

15. Jul 15, 2014

### pervect

Staff Emeritus
A slightly different take on this quesiton, not fully worked out.

The notion of parallel transport defines geodesics (via parallel transporting a vector along a curve, and the metric defines distances. We know that a metric-compatible connection with no torsion defines geodesics which minimze distance. I believe that if we use a different connection (still torsion free for the moment), geodesics would no longer minimize distances.

My argument for this is uses the geodesic equations (written in some suitable coordinate system) as a base. If we change the connection coefficients / christoffel symbols, we'd change the differential equation, which implies that we change the solution for geodesics. But the metric remains the same, so the only way to keep geodesics as distance minimizing curves is to stick with the metric compatible connection.

So the basic idea is that we want geodesics to minimze distance, just as straight lines used to, as part of our notion of what geodesics should be and to give them some physical significance.

16. Jul 15, 2014

### nearlynothing

I cant see how this is coordinate dependent, it only tells me that the manifold is flat, and that the connection reduces to an identity in these particular coordinates.

I understand this as saying that it should be metric compatible because we want the geodesics defined by the connection to be the same as the ones defined by the metric, but how does this solve the problem? we could still choose not to have that.

17. Jul 15, 2014

### atyy

Last edited: Jul 15, 2014
18. Jul 15, 2014

### nearlynothing

What about theories where a connection with torsion is used? there the two types of geodesics dont coincide.

19. Jul 15, 2014

### pervect

Staff Emeritus
Yes, see for instance http://www.slimy.com/~steuard/teaching/tutorials/GRtorsion.pdf

For the other half, the difference between the two definitions when a non-metric compatible connection is used, see http://preposterousuniverse.com/grnotes/grnotes-three.pdf [Broken]

In the same document:

Another few fun factoid about torsion - on a flat manifold with no torsion and the Leva Civita connection, parallelograms close perfectly. With no torsion, the parallelograms fail to close in terms of order 3 in a series expansion of the length of the side of the parallelogram. With torsion the parallelograms fail to close in terms of order 2. (This is a paraphrase from memory from Penrose's "Road to Reality").

Sean Caroll has a paper on the observational consequence of the presence of torsion (which I stumbled across and haven't read). http://arxiv.org/abs/gr-qc/9403058

Last edited by a moderator: May 6, 2017
20. Jul 16, 2014

### pervect

Staff Emeritus
Another thing that's worth noting is that the metric compatible connection implies that a geodesic that starts out as timelike will remain timelike, similarly for spacelike and null. From the reading above, this might not necessarily follow if a different non-metric compatible connection was used.