# The vanishing of the covariant derivative of the metric tensor

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.
I think that is a very good intuitive of viewing it. ∇is measuring the gradient with respect to g. Therefore g has no gradient itself.

weirdoguy
Orodruin
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I think that is a very good intuitive of viewing it. ∇is measuring the gradient with respect to g. Therefore g has no gradient itself.
There is no such thing as a ”gradient with respect to g”. The connection defines a way of relating vectors in nearby tangent spaces. This can be done in such a way that parallel transport preserves angles or not. If it does, this is what metric compatibility means. This does not mean that it is a gradient with respect to the metric. There is also no unique metric compatible connection unless further constraints (such as no torsion) are imposed.

I agree that preserving the inner product is a nice and intuitive property and in many cases there is good reason to restrict oneself to metric compatibility. However, there is a priori nothing wrong with having a connection that is not metric compatible - it just might not be what we are looking for for our description of whatever we want to describe.

There is no such thing as a ”gradient with respect to g”. The connection defines a way of relating vectors in nearby tangent spaces. This can be done in such a way that parallel transport preserves angles or not. If it does, this is what metric compatibility means. This does not mean that it is a gradient with respect to the metric. There is also no unique metric compatible connection unless further constraints (such as no torsion) are imposed.

I agree that preserving the inner product is a nice and intuitive property and in many cases there is good reason to restrict oneself to metric compatibility. However, there is a priori nothing wrong with having a connection that is not metric compatible - it just might not be what we are looking for for our description of whatever we want to describe.
But the question was framed in the context of General Relativity, where the connection is a metric connection (plus torsion). The covariant derivative of the metric along any curve vanishes. (The metric is always parallel(ly?) transported onto itself.) In that sense the metric has a vanishing derivative and is covariantly "flat".

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But the question was framed in the context of General Relativity, where the connection is a metric connection (plus torsion). The covariant derivative of the metric along any curve vanishes. (The metric is always parallel(ly?) transported onto itself.) In that sense the metric has a vanishing derivative and is covariantly "flat".
The question is posed in the differential geometry forum. The only actual implication of GR is the literature the OP is using. The terminology is that the metric is parallel (meaning that the covariant derivative everywhere in all directions is zero).

Flatness is a geometric property of the connection, not the metric or any other tensor. Even when the connection is metric compatible, the space may not be flat (otherwise we would not talk about curved spacetime in GR). Let us not mix up these concepts.

TLDR: If the connection is metric compatible, the metric is parallel. If the space is flat, any parallel transport around a contractable loop returns the original tensor.

lavinia and weirdoguy
I suspect the question should have been posed in the Special and General Relativity forum. :-)

Actually this thread has been very helpful to me. I'm especially interested in following up on the method outlined in the Einstein paper, where one arrives at Christoffel symbols via the geodesic approach. Would anyone like to recommend a good source? I've never taken a course in the calculus of variations.

Actually this thread has been very helpful to me. I'm especially interested in following up on the method outlined in the Einstein paper, where one arrives at Christoffel symbols via the geodesic approach. Would anyone like to recommend a good source? I've never taken a course in the calculus of variations.
You could try Sean Carroll's GR lecture notes.
https://arxiv.org/abs/gr-qc/9712019Pages 66-67 (pdf version) has a good discussion of connections and metrics. It is not exactly what you're after, but it looks well explained and would be helpful.
PS I've come round to thinking that the best demonstration of ∇g=0 - in GR - is to note that it is trivially true in flat Minkowski space time, and there is always a locally Lorentzian frame in GR, so it has to hold everywhere in GR, as well.

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You could try Sean Carroll's GR lecture notes.
https://arxiv.org/abs/gr-qc/9712019
Thanks, Michael!
I've come round to thinking that the best demonstration of ∇g=0 - in GR - is to note that it is trivially true in flat Minkowski space time, and there is always a locally Lorentzian frame in GR, so it has to hold everywhere in GR, as well.
Exactly Schutz's approach!

Michael Price
lavinia
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Thanks, Michael!

Exactly Schutz's approach!
I don't know the Physics so this is a naive question.

The local Lorentz frame seems to be interpreted to say that the Space-Time metric can be represented in normal coordinates as diagonal ±1 with first partial derivatives equal to zero at a central point. But doesn't this assume that the connection and the metric are compatible? Why does one do a proof?

It also seems that the Christoffel symbols are assumed to be zero at the central point. Doesn't this require the affine connection to be torsion free? If not, I don't see how the Christoffel symbols can all vanish.

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Orodruin
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The local Lorentz frame seems to be interpreted to say that the Space-Time metric can be represented in normal coordinates as diagonal ±1 with first partial derivatives equal to zero at a central point. But doesn't this assume that the connection and the metric are compatible? Why does one do a proof?

It also seems that the Christoffel symbols are assumed to to be zero at the central point. Doesn't this require the affine connection to be torsion free? If not, I don't see how all of the Christoffel symbols can all vanish.
Sure. The usual assumption is that you have the Levi-Civita connection. You can also do something somewhat different and assume the connection and metric to be independent. In that case the Levi-Civita connection drops out from the stationary action principle (with some assumptions).

lavinia
I don't know the Physics so this is a naive question.

The local Lorentz frame seems to be interpreted to say that the Space-Time metric can be represented in normal coordinates as diagonal ±1 with first partial derivatives equal to zero at a central point. But doesn't this assume that the connection and the metric are compatible? Why does one do a proof?

It also seems that the Christoffel symbols are assumed to to be zero at the central point. Doesn't this require the affine connection to be torsion free? If not, I don't see how the Christoffel symbols can all vanish.
As a counter example: in spherical polars in a flat space and flat spacetime, the Christoffel symbols are not zero.

Orodruin
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