A General relativity -- Who reports the components of ds^2?

1. Apr 13, 2017

Jonsson

Hello there, suppose we take $M$ to denote the spacetime manifold. Suppose also that $ds^2 = g_{\mu \nu} dx^\mu dx^\nu$. I have some confusions with regards to the metric and the line elements.

My main confusion is at which points in the manifold are $ds^2$ defined? Is it correct that if $p \in M$, then it only makes sense to fix $ds^2$ for each tangent space $T_pM$, or is it fixed for each $v \in T_pM$? More specifically:

(1) If there is some position dependent line element, such as the Schwartzschild line element, then $g_{\mu \nu}= g_{\mu \nu}(t,r, \theta, \phi)$. Are these $(r, \theta, \phi)$ the coordinates of some $(t,r, \theta, \phi) \in M$, or are they the coordinates of some $(t,r, \theta, \phi) \in T_pM$?

(2) Are the differentials $dx^\mu(x)$ evaluated at each $x \in M$ or $x \in T_pM$?

Last edited: Apr 13, 2017
2. Apr 13, 2017

PeroK

I'm not sure I can give you the full answer, but here's part of the answer.

In curved spacetime, you have the notion of distance only by integrating over the line element. In that sense, $ds^2$ is defined everywhere in spacetime, and relates to the continuous changes in distance along a path. In answer to question 1) you can calculate the distance along a path in spacetime by integrating $ds$ without any recourse to tangent spaces. The points and line element are defined therefore on the spacetime manifold.

The problem is how to define vectors. A vector has a length and a direction. But, if spacetime is curved, how do you define a vector of finite length in a given direction? This is the main motivation for tangent spaces. There is a tangent space at every point, and each coordinate system defines a set of basis vectors for the tangent space.

In answer to 2), you only have a single spacetime point in each tangent space, so the line element $ds^2$ doesn't apply in the tangent space.

3. Apr 13, 2017

pervect

Staff Emeritus
I believe ds^2 is usually taken to be a scalar, which means it exists at a point regardless of the tangent space.

If one wanted to write a map from a pair of vectors to a scalar, which would depend on the tangent space, one would probably write $ds \otimes ds$, where the operation $\otimes$ is the tensor product.

There's more potential for confusion without the square, ds could be a map from a vector to a scalar, or it could be just a scalar. I've seen various conventions, one is to write the scalar valued map from a vector (which is a dual vector) with a boldface d, ds, and the scalar in non-boldface, ds.