- #1

LarryL

- 1

- 0

derivative is picked out which preserves inner products of vectors which are parallel transported.

What I don't understand is the interpretation of the fact that, using this definition of the derivative,

the covariant derivative of the metric tensor = 0.

In my mind, the covariant derivative applied to any tensor is telling us something about

how the tensor changes from point to point.

Isn't saying that the covariant derivative of the metric tensor = 0 (everywhere, correct?) the same thing

as saying the metric tensor is "constant" (in some sense) throughout spacetime?

But, of course, it does not have to be constant. And wouldn't that imply constant curvature

(or zero curvature) everywhere? This, of course, must be a wrong interpretation (and Riemann and Ricci

tensors are clearly nonzero, and they more directly relate to curvature).

So what is wrong with my interpretation of a derivative as a measure of change?

thanks.