Metric tensor of a non-homogeneous universe

[andrewkirk]

I am a bit perplexed by the consequences of the fact that all covariant first derivatives of the metric tensor are zero. I think I can follow some of the proofs, as presented for example in John Lee "Riemannian Manifolds - An Introduction to Curvature". But intuitively it "seems wrong" to me because it seems to prevent variations in curvature, and I can't understand how it operates in relation to the following scenario of a manifold that is part flat and part curved:

Consider a 2-d manifold M embedded in Euclidean 3-space E3, formed by the revolution of the bump function y=exp(-1/(1-x^2)) about the z axis. This manifold is:
- smooth, because the bump function is smooth
- flat everywhere outside of the unit circle of the x-y plane (x^2+y^2>1)
- curved inside that unit circle

Using the x and y coordinates inherited from E3 as a global coordinate system for M, I would expect the metric tensor to be the identity matrix outside the unit circle but something else inside it. I understand that the matrix representation of the tensor is coordinate dependent, and that it is possible for that to change despite the covariant derivative remaining zero, if the changes in the coordinate representation exactly offset the changes in the basis vectors. Perhaps the answer lies somewhere in that? Also, maybe the curved part of the manifold is not “curved enough” for the metric tensor to be different - after all it is not perfectly spherical.

But what if the manifold within some part of the bump, say the x-y circle of radius 1/2, is the exact shape of part of a sphere? Wikipedia http://en.wikipedia.org/wiki/Bump_function seems to imply that a bump function can be created that does this, while remaining smooth (some clever construction involving convolutions with mollifiers). Isn't the metric tensor within that smaller circle the same as that of a complete sphere, which is fundamentally different from the metric tensor of a flat manifold, irrespective of coordinates?

How does the metric tensor in that case change from the flat space tensor outside the unit circle to the spherical tensor inside of the circle radius 1/2 without its covariant derivative ever being nonzero somewhere in between (ie in the ring between the x-y circles of radius 1/2 and 1)?

Thanks for any help in straightening out my thinking on this.

 

[bcrowell]

The purpose of the covariant derivative is to tell you whether a certain tensor T is really changing, as opposed to just appearing to change because the metric g is changing. If the metric is changing, then T's components are going to change simply because they need to be expressed in a different basis as you slide over to a different point. That wouldn't be a real physical change in T.

When you take the covariant derivative of the metric itself, you're looking for a change in the metric that occurs for some other reason than a change in the metric. That can't happen, so it makes sense that the covariant derivative of the metric itself is zero.

How does the metric tensor in that case change from the flat space tensor outside the unit circle to the spherical tensor inside of the circle radius 1/2 without its covariant derivative ever being nonzero somewhere in between (ie in the ring between the x-y circles of radius 1/2 and 1)?

The metric tensor does change. If you take its plain old partial derivatives, they're nonzero.

 

[andrewkirk]

Thank you very much for your reply bcrowell. If I understand you correctly the covariant derivative of a tensor tells us whether or not the tensor is changing relative to the metric tensor. In other words it tells us the changes in the tensor "net of any changes attributable to the change in the metric tensor". So no matter how many different types of spaces we "sew" together, with all sorts of associated radical changes to the metric between different parts of the manifold, the covariant derivative of the metric will always be zero because it is "changes in the metric tensor net of any changes attributable to changes in the metric tensor", which will be zero by definition.

Have I got that right?