I The vanishing of the covariant derivative of the metric tensor

snoopies622

Summary
Axiom or theorem, part two.
I brought up this subject here about a decade ago so this time I'll try to be more specific to avoid redundancy.

In chapter five of Bernard F. Schutz's A First Course In General Relativity, he arrives at the conclusion that in flat space the covariant derivative of the metric tensor is zero. This can be done very simply, it seems to me. Then in chapter six he argues that since Riemannian manifolds are locally flat, this relation holds there as well, and says no more about it — evidentally satisfied with this reasoning.

I have read elsewhere, however, (including here) that $\nabla g = 0$ should be taken as an axiom instead since without it funny things happen, like the lengths of vectors and the angles between them changing when they're parallel transported, which quite reasonably seems like something to avoid.

So what I'm wondering is . . what are the premises of flat space that make $\nabla g = 0$ true, such that there one does not have to bother positing it as a separate axiom? Obviously in Cartesian coordinates the coefficients of $ds^2 = dx^2 + dy^2 + . . .$ do not change as we move around a flat manifold, and equally obviously the basis vectors parallel transport, so the Christoffel symbols are all zero. But maybe there's something deeper to it? I'm just troubled by the idea that one can derive $\nabla g = 0$ in flat space but outside of flat space — or at least outside of Riemannian manifolds — one should posit it as an axiom instead.

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Orodruin

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Summary: Axiom or theorem, part two.

Then in chapter six he argues that since Riemannian manifolds are locally flat, this relation holds there as well, and says no more about it — evidentally satisfied with this reasoning.
This depends on what meaning you give to "Riemannian manifold". If you just give it it should have a metric, then it is not true. If you do not but also include that the connection should be the Levi-Civita connection, then it is true by assumption as the Levi-Civita connection is the unique torsion-free and metric compatible affine connection.

snoopies622

Not sure I'm completely following. If I say that $\nabla g=0$ is true in a plane or on the surface of a sphere (describing the metrics), would that be correct by itself or is it not really correct without also specifying the Levi-Civita connection?

Orodruin

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You can have a manifold with a metric and introduce a connection that is not metric compatible (although I am not sure why you would want to). You can also have metric compatible connections that are not the Levi-Civita connection.

There are examples of metric compatible connections that are flat (globally), but not torsion free. Being flat is a property of the connection.

Michael Price

In the first order Palatini formalism you write down the action in terms of g and Γ and allow them to vary independently. The equations of motion then establish the relation between g and Γ where Γ is the Christoffel symbol, which ensures that ∇g = 0. In the second order formalism you just have to establish the relation by fiat.
This assumes the torsion is zero, but easy to extend to cover non-zero torsion as well.

snoopies622

Being flat is a property of the connection.
Now that's very interesting since Schutz goes about it in the opposite direction, perhaps because his book is for beginners. Must think about this more deeply.

snoopies622

Okay, here's a slightly different version of my original question:

Suppose I say . .

"Let's take a plane, and put a Cartesian coordinate system on it. The metric is $ds^2 = dx^2 + dy^2$. Obviously the coefficients of this metric (1,0,0,1) don't change as we move around the plane, and just as obviously $e_x$ and $e_y$ stay the same length and point in the same direction as we move those around as well. Therefore the Christoffel symbols are all zero and the covariant derivative of the metric is zero."

Would you then say, "well no, you've actually made unstated assumptions about your connection. These unstated assumptions are . . "?

Michael Price

Looking back at my Uni GR notes, I see a derivation ∇g=0 from the geodesic equation. If they sneaked in some hidden assumptions, I can't see them.

snoopies622

Yes I believe that's how Einstein does it in his original paper, too. More than one way to cook a chicken! :)

• Michael Price

Orodruin

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Okay, here's a slightly different version of my original question:

Suppose I say . .

"Let's take a plane, and put a Cartesian coordinate system on it. The metric is $ds^2 = dx^2 + dy^2$. Obviously the coefficients of this metric (1,0,0,1) don't change as we move around the plane, and just as obviously $e_x$ and $e_y$ stay the same length and point in the same direction as we move those around as well. Therefore the Christoffel symbols are all zero and the covariant derivative of the metric is zero."

Would you then say, "well no, you've actually made unstated assumptions about your connection. These unstated assumptions are . . "?
First of all, components not changing is very different from a field "not changing". Whether components change or not is just a matter of differentiating numbers. Whether a field (such as a basis vector field) changes from point to point is a question that is only answered by the connection. By saying that $\vec e_x$ and $\vec e_y$ "do not change" as you move around the plane, you have made the assumption that you have the standard connection on the plane.

strangerep

"Let's take a plane, and put a Cartesian coordinate system on it. The metric $ds^2 = dx^2 + dy^2$
To minimize confusion, refer to $ds^2$ as the "(invariant) line element". A "metric" is actually a (bilinear) mapping from a pair of vectors to a number, (at a given point on the manifold).

Obviously the coefficients of this metric (1,0,0,1) don't change as we move around the plane,
That's only true for particular choices of coordinates (you're using Cartesian coordinates here). But in plane polar coordinates the line element is $ds^2 = dr^2 + r^2 d\theta^2$, so the metric is $(1,r^2)$.

The Christoffel symbols are all zero in Cartesian coords, but not all zero in plane polar. Nevertheless, the covariant derivative of the metric is a tensor, hence if it is zero in one coordinate systems, it is zero in all coordinate systems.

Then, in General Relativity (based on Riemannian geometry), one assumes that the laws of physics "here, today" are not fundamentally different from the laws of physics "there", "yesterday", or "tomorrow". Hence, at every point in spacetime, it is possible to find a coordinate system in which the metric components have the familiar constant-component form, at least in an infinitesimal neighbourhood of that point.

Hence the covariant derivative of the metric (being a tensor) is also zero "there", or "yesterday", or "tomorrow".

Hence we may assume $\nabla_\lambda g_{\mu\nu} = 0$ everywhere and everywhen.

From that equation (expressed in terms of general connection coefficients), one can (by cyclic permutation of indices) derive [Exercise!] that the connection coefficients must equal the Christoffel symbol coefficients (if one also assumes zero torsion).

Summary: if we specify that $\nabla_\lambda g_{\mu\nu} = 0$ and zero torsion, then the connection coefficients are the Christoffel symbols.

HTH.

• dextercioby

haushofer

To add: that summary also follows from the fact that the connection appears algebraically with an invertible tensor in the eqn. for metric compatibility, and simply counting the number of equations and the number of independent components of the torsion-free connection: 1/2*D*(D+1).

A similar question by the way can be asked for the vielbein postulate. Different authors motivate this differently.

snoopies622

strangerep your explanation seems to very closely follow Schutz's approach, and like it, it does not posit a particular connection on the plane as a separate hypothesis. So I'm still wondering if such a hypothesis is necessary. Now I'm re-reading section 9 of Einstein's 1916 paper The Foundations of the General Theory of Relativity, where he arrives at Christoffel symbols by starting with a geodesic, as Michael Price suggested.

• Michael Price

Orodruin

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Now I'm re-reading section 9 of Einstein's 1916 paper The Foundations of the General Theory of Relativity, where he arrives at Christoffel symbols by starting with a geodesic, as Michael Price suggested.
Note that this has other assumptions, ie, that the geodesics of the connection coincide with the paths extremizing distances and zero torsion.

Michael Price

Note that this has other assumptions, ie, that the geodesics of the connection coincide with the paths extremizing distances and zero torsion.
The geodesic route only sets the symmetric part of the connection to the Christoffel symbols. The antisymmetric torsion part of the connection is left undefined.

Orodruin

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The geodesic route only sets the symmetric part of the connection to the Christoffel symbols. The antisymmetric torsion part of the connection is left undefined.
Not if, as I clearly stated, torsion is assumed to be zero.

Michael Price

Not if, as I clearly stated, torsion is assumed to be zero.
Yes, but we don't have to assume the torsion is zero for this derivation to work. (Einstein assumed it, we don't have to.)

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Orodruin

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Yes, but we don't have to assume the torsion is zero for this derivation to work. (Einstein assumed it, we don't have too.)
To find the symmetric components no. To find the anti-symmetric components, yes, unless you add some additional requirement that is essentially equivalent on top.

Michael Price

To find the torsion you have to move to the spin connection, which is outside the scope of this thread.

snoopies622

By saying that $\vec e_x$ and $\vec e_y$ "do not change" as you move around the plane, you have made the assumption that you have the standard connection on the plane.
Personally I find it a bit strange that this requires an additional hypothesis. Given what the Cartesian coordinate system on a plane looks like, how could the basis vectors not parallel transport? Isn't the parallelism of the coordinate lines built into it? Given two points very close to each other, their x=constant lines, for instance, go on forever and never touch.

Michael Price

According to the equivalence principle you can always find a locally Lorenzian, free falling frame through any point. In that frame you must have ∇g=0 where g is the Minkowski metric. Transforming back into the general coordinate frame we will still have ∇g=0.

Orodruin

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According to the equivalence principle you can always find a locally Lorenzian, free falling frame through any point. In that frame you must have ∇g=0 where g is the Minkowski metric. Transforming back into the general coordinate frame we will still have ∇g=0.
This is true in GR. This is the differential geometry forum, even if the OP is using a GR textbook.

Personally I find it a bit strange that this requires an additional hypothesis. Given what the Cartesian coordinate system on a plane looks like, how could the basis vectors not parallel transport? Isn't the parallelism of the coordinate lines built into it? Given two points very close to each other, their x=constant lines, for instance, go on forever and never touch.
In an affine space (such as Euclidean space), there is a built in notion of identification of tangent spaces at different points. This supplies a predetermined connection and thereby a sense of parallellism. However, if you disregard this notion and just look at the coordinates and the metric, it is perfectly possible to introduce other connections.

snoopies622

Thanks, Orodruin. Looks like I'll have to take out a differential geometry book or two. None of my GR books go into this stuff with any depth.

Orodruin

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Thanks, Orodruin. Looks like I'll have to take out a differential geometry book or two. None of my GR books go into this stuff with any depth.
Well, that depends on what you need it for. If you just need it for learning GR, you can probably skip a lot of things for the time being and concentrate on the things necessary for GR.

snoopies622

I read in another forum (unfortunately I no longer remember where) the following idea: An intuitive justification for having $\nabla g=0$ is, if you move a meter stick around a room and the meter stick grows or shrinks as you move, it will still measure one meter in length if what you're using to measure it is the meter stick itself.

• Michael Price

"The vanishing of the covariant derivative of the metric tensor"

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