Physical reasons for having a metric-compatible affine connection?

[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.

 

[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 :)

 

[nearlynothing]

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 :)

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?

 

[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?

 

[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.

 

[nearlynothing]

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?

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.

 

[nearlynothing]

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.

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

 

[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.

 

[WannabeNewton]

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

See appendix E of Wald.

 

[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!

 

[ChrisVer]

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

 

[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.

 

[TrickyDicky]

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

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.

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.

 

[TrickyDicky]

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.

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.

 

[pervect]

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.

 

[nearlynothing]

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.

I can't 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.

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!

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.

 

[atyy]

The physical reason is that it works. But here are some references exploring theories with other connections.
http://arxiv.org/abs/gr-qc/0604006
http://arxiv.org/abs/1007.3937
http://arxiv.org/abs/1008.0171

 

[nearlynothing]

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


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.

What about theories where a connection with torsion is used? there the two types of geodesics don't coincide.

 

[pervect]

What about theories where a connection with torsion is used? there the two types of geodesics don't coincide.

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

The concept of a geodesic can be formulated in two ways. One is the definition
given by Wald in Eq. (3.3.1): a “straightest possible line” whose tangent
vector is parallel propagated along itself. The other is the source of the name
(as I understand it): a “shortest possible path” between any two of its points
(or more generally, an “extremal length path” between them). In the pres-
ence of torsion, these two concepts need no longer be equivalent. For the
sake of consistency with Wald’s definition, we will take the term “geodesic”
to imply the first meaning but not necessarily the second.

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

With parallel transport understood, the next logical step is to discuss geodesics. A
geodesic is the curved-space generalization of the notion of a “straight line” in Euclidean
space. We all know what a straight line is: it’s the path of shortest distance between
two points. But there is an equally good definition — a straight line is a path which
parallel transports its own tangent vector. On a manifold with an arbitrary (not necessarily
Christoffel) connection, these two concepts do not quite coincide, and we should discuss
them separately.

In the same document:

Thus, on a manifold with metric, extremals of the length functional are curves which parallel transport their tangent vector with respect to the Christoffel connection associated with that metric. It doesn’t matter if there is any other connection defined on the same manifold. Of course, in GR the Christoffel connection is the only one which is used, so the two notions are the same.

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

 

[pervect]

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.

 

[Geometry_dude]

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.

I'm not sure whether one can say that. First of all, ascribing paths to quantum objects is always a dangerous business due to the Heisenberg uncertainty relation and even if you ignore this or find some kind of way around this, you are assuming here that no force acts on the atom, since parallel transport with respect to a metric connection is just a mathematical machinery to describe inertial motion in the GR sense. It appears that I was not very clear when desribing why the connection has to be metric or that the physical explanation I gave was insuffucient. The reasoning is as follows: Gravitational motion is just inertial motion, inertial motion is motion on straight lines, that is geodesics in the sense of proper time functional extremizing curves. A metric connection gives you auto-parallels that are geodesics.

Thus if you want to use your connection to describe (purely) gravitational motion as auto-parallels, your connection has to be metric.

Torsion is only relevant for parallel transport, it has no relevance for the shape of geodesics as can be seen by the following formula for auto-parallels
$\ddot x ^k + \Gamma^k{}_{(ij)} \, \dot x^i \, \dot x^j = 0$
You need it, for example, when you have a frame attached to your auto-parallel and you want to know what happens to that frame during the motion. For this you need to parallel transport the frame vectors.

I can't 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.

Yes, but do you have any physical reason to use this coordinate system? Also, not every manifold admits a flat connection. A necessary and sufficient condition would be that the tangent bundle is trivial, which is quite a strong property.

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.

Yes, but then you wouldn't describe inertial motion anymore, see above. In that case, pervect has given you a condition for it to still make sense physically:

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.

It would be interesting to figure out whether the reverse implication holds (probably not) and when a connection has this property.

 

[TrickyDicky]

I'm not sure whether one can say that. First of all, ascribing paths to quantum objects is always a dangerous business due to the Heisenberg uncertainty relation and even if you ignore this or find some kind of way around this, you are assuming here that no force acts on the atom, since parallel transport with respect to a metric connection is just a mathematical machinery to describe inertial motion in the GR sense.

Well, what Haushofer says was the objection that Einstein made to Weyl's 1918 unified field theory, that didn't use a metric connection: "If in nature length and time would depend on the pre-history of the measuring instrument, then no uniquely defined frequencies of the spectral lines of a chemical element could exist, i.e., the frequencies would depend on the location of the emitter." (my bold)

http://relativity.livingreviews.org/open?pubNo=lrr-2004-2&page=articlesu9.html

Of course this was 8 years before the modern QM conception and Heisenberg uncertainty. But for the moment it is maybe better to keep things classical and leave for later quantum implications on paths.

My point in the previous post is that the inherent path-dependence of GR for distant locations due to curvature would seem to indicate that Einstein's physical objection to Weyl's theory applies also to his own theory.

The following exchange from the same livingreviews page seems to imply that both Weyl and Einstein were aware of this for GR except in the special static case:

" Weyl answered Einstein’s comment to his paper in a “reply of the author” affixed to it. He doubted that it had been shown that a clock, if violently moved around, measures proper time ∫ds. Only in a static gravitational field, and in the absence of electromagnetic fields, does this hold:

“The most plausible assumption that can be made for a clock resting in a static field is this: that it measure the integral of the ds normed in this way [i.e., as in Einstein’s theory]; the task remains, in my theory as well as in Einstein’s, to derive this fact by a dynamics carried through explicitly.”69View original Quote ([395], p. 479)

Einstein saw the problem, then unsolved within his general relativity, that Weyl alluded to, i.e., to give a theory of clocks and rulers within general relativity. Presumably, such a theory would have to include microphysics. In a letter to his former student Walter Dällenbach, he wrote (after 15 June 1918):

“[Weyl] would say that clocks and rulers must appear as solutions; they do not occur in the foundation of the theory. But I find: If the ds, as measured by a clock (or a ruler), is something independent of pre-history, construction and the material, then this invariant as such must also play a fundamental role in theory. Yet, if the manner in which nature really behaves would be otherwise, then spectral lines and well-defined chemical elements would not exist."

 

[stevendaryl]

Well, what Haushofer says was the objection that Einstein made to Weyl's 1918 unified field theory, that didn't use a metric connection: "If in nature length and time would depend on the pre-history of the measuring instrument, then no uniquely defined frequencies of the spectral lines of a chemical element could exist, i.e., the frequencies would depend on the location of the emitter." (my bold)

http://relativity.livingreviews.org/open?pubNo=lrr-2004-2&page=articlesu9.html

I don't know any of the details of Weyl's theory, but I had the impression (which might be wrong) that it lacked an inherent standard of length. The weird possibility that this allowed was that an object could start off one size, make some complicated journey through space, and end up in the same location bigger or smaller than it started. If I'm right about that, then Einstein's comment would certainly make sense, because there wouldn't be a unique notion of the "size" of a hydrogen atom--different hydrogen atoms would have different sizes and correspondingly different spectra.

It's kind of interesting that prior to the discovery of quantum mechanics, nothing in physics (as far as I know) offered any explanation for why objects had definite sizes. Ultimately, the scale of physical objects is determined by constants such as the Bohr radius, which involves \hbar.

 

[TrickyDicky]

I don't know any of the details of Weyl's theory, but I had the impression (which might be wrong) that it lacked an inherent standard of length. The weird possibility that this allowed was that an object could start off one size, make some complicated journey through space, and end up in the same location bigger or smaller than it started. If I'm right about that, then Einstein's comment would certainly make sense, because there wouldn't be a unique notion of the "size" of a hydrogen atom--different hydrogen atoms would have different sizes and correspondingly different spectra.

It's kind of interesting that prior to the discovery of quantum mechanics, nothing in physics (as far as I know) offered any explanation for why objects had definite sizes. Ultimately, the scale of physical objects is determined by constants such as the Bohr radius, which involves \hbar.

Yes, it lacked such length standard, what I'm saying is that in general GR itself also lacks it(unlike SR), in non-static cases(i.e. proper length problem in FRW geometry) due to curvature effects on parallel transport path dependence.
Then again at the time Einstein made these considerations he thought that GR implied that only static solutions were physically valid, it wasn't til 1925 with Friedmanns results that he realized it wasn't so.

Also note that QM doesn't really explain it either, the \hbar constant is not given a physical explanation, it is just introduced by the SE formalism from Planck's ad hoc addition of it to quantize energy and solve the blackbody problem.

 

[TrickyDicky]

The example of Weyl's theory with non-metric connection highlights the basic reason for having a metric connection as GR does, while there is still the issue of path-dependence and the absence of an inherent standard vector length one can still mitigate this by imposing by hand a standard clock(basically assuming that in distant points in the universe elements have the same spectrum as here, which it is basically a form of the cosmological and copernican principle) and by virtue of the affine parametrization of the worldlines any particular standard Einsteinian clock admits a proper time parametrisation independent of its worldline.
However in the case of Weyl's torsionless non-metric connection the characteristic frequencies of atomic clocks depend not only on their actual spacetime locations but also on their past worldlines and therefore the spectral lines would be blurred, which is not what we observe.
So with a metric connection one gets to be able to introduce an ideal or standard set of clocks and rods in a reliable way.

 

[robphy]

Here my comment from an old (2007) thread "Why torsion free metric compatible connection ?"
https://www.physicsforums.com/showthread.php?t=199500

One view is that
"[torsion-free] metric compatibility" means that the metric tensor field carries all of the information of the geometry of spacetime...
Physically, this means that the metric tensor field determines the motion of free particles (the geodesic structure) and the propagation of light (the conformal structure..and causal structure).

This might be a useful resource:
http://relativity.livingreviews.org/open?pubNo=lrr-2004-2&page=articlesu3.html

(I think the document above was already linked by an earlier poster.)

also possibly useful:
http://www.google.com/search?q=ehlers+pirani+schild

 

[Geometry_dude]

Well, the proper time functional is really just the natural distance functional in Lorentzian geometry. It measures the length ('length'='time' for c=1) of causal, that means physical, curves. Of course this depends on the curve you're measuring.
Either way, I really don't see the relation to atomic spectra here, while I agree that they should depend on the geometry of spacetime in principle.

I personally do not believe Weyl's conceptual model to be a fruitful way of unifying gravity with electromagnetism, but that is just an opinion.

 

[stevendaryl]

Well, the proper time functional is really just the natural distance functional in Lorentzian geometry. It measures the length ('length'='time' for c=1) of causal, that means physical, curves. Of course this depends on the curve you're measuring.
Either way, I really don't see the relation to atomic spectra here, while I agree that they should depend on the geometry of spacetime in principle.

I personally do not believe Weyl's conceptual model to be a fruitful way of unifying gravity with electromagnetism, but that is just an opinion.

It's not just a matter of opinion. Weyl's theory is contradicted by the data.

 

[gentsagree]

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.

What happens if we vary with respect to an arbitrary connection with torsion? I assume from your writing that it won't yield metric compatibility, and this confuses me.

On a heuristic level (I haven't done the calculation yet, just curious): S. Jensen (pdf given above) writes

<<Requiring that parallel transport leave inner products invariant essentially means that the vectors’ lengths and angles relative to one another must be unchanged from point to point. But this does not specify anything about “global” rotations of the tangent space during parallel transport: that is the physical meaning of torsion.>>

So, if torsion does not affect metric compatibility, shouldn't the calculation of varying the Palatini action be independent of torsion/no torsion?