# I When will metric compatibility hold/not hold?

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1. Aug 10, 2017

### Ron19932017

Hi everyone,

I am reading Sean Carroll's note on gr and he mentioned metric compatibility.
When ∇g=0 we say the metric is compatible.

However from another online material, the lecturer argues ∇ of a tensor is still a tensor,
and given that ∇g vanish in locally flat coordinate and this is a tensorial equation, therefore it vanishes in any other coordinate. That gives us ∇g always = 0.

I guess the contradiction comes from some implicit use about ∇g=0 vanish in locally flat coordinate but I am not sure what exactly is it. The first derivative in local coordinate vanishes, but I am not sure if the connection symbol vanishes too. I mean in locally flat coordinate the metric is cartesian like, but does that immediately imply the connection is also cartesian like (=0)?

Does anyone know why is there a contradiction? Sean Carroll and Schutz did not talk much about non-torsion-free cases so I really don't know what is going on.

2. Aug 10, 2017

### Staff: Mentor

There isn't one. The statement that $\nabla g$ vanishes in locally flat coordinates assumes that you are using the $\nabla$ operator that is metric compatible. There are other possible ways of defining a $\nabla$ operator for which $\nabla g$ would not vanish, even in locally flat coordinates.

3. Aug 12, 2017

### dextercioby

One more point for the Op: zero torsion is generally unrelated to zero nonmetricity tensor. In GR the two conditions are taken as valid and henceforth one has an unique way to link the connection to the metric tensor.

4. Aug 12, 2017

### Ibix

Something I'm not clear on - what happens if you do pick a different definition of $\nabla$? You're presumably giving yourself a degree of freedom that's not present in vanilla GR by allowing yourself a free choice of connection or metric-compatibility. So are you defining a family of more general theories of spacetime containing GR as a special case, or are you just giving yourself more ways to describe the same thing?

I suspect the latter from how little comment Carroll makes over picking the metric-compatible torsion-free case, but I'm not certain.

5. Aug 12, 2017

### vanhees71

In standard GR you assume a pseudo-Riemannian spacetime manifold, i.e., a differentiable manifold with (a) a nondegenerate bilinear form (fundamental form, pseudo-metric, for physicists simply metric) of signature (1,3) and (b) the uniquely determined torsion free affine connection that's compatible with the metric in the above discussed sense.

As far as I know, there's no hint from observation that you need a more general affine connection, although it occurs somewhat naturally if you consider GR from the point of view of gauge theories (gauging the Lorentz invariance of Minkowski spacetime), leading to Einstein-Cartan theory.

6. Aug 12, 2017

### PAllen

It depends how you map observables to mathematical objects. Suppose a metric. Then there exists a metric compatible connection, but one may also define a different non compatible connection. Then you have two families of geodesics per the parallel transport definition. If you posit that test bodies follow the non compatible geodesics, you obviously have a very different theory than GR, where free fall does not locally extremize proper time.