Talk:General Relativity/Metric tensor

if dx is contravariant, then surely g must be covariant, or vice versa? To be honest I can never remember which is which, but I'm sure they must be different.

dx is contravariant, g is covariant and symmetric. The way to remember is that dx'a=dx'a/dxb·dxb   by the chain rule This is the change-of-coordinate equation of a contravariant tensor, and thus dx is contravariant. Hope this helps. - Itzamna 2 Feb 2008

I also don't understand the last example. How could that work, since ds is a scalar, 2 is a scalar, and dx and dy are contravariant vectors?

ANSWER: Remember Pythagoras in 2D. Say you have a point (3,4) in the Cartesian system. The squared length of a line connecting this point with the origin (0,0) is: 3^2 + 4^2 = 25; so the length of the connecting line is 5.

The point (3,4) can also be represented by a positionvector: P = 3.i + 4.j, with 3 and 4 being the components of the vector P, and i and j being basisvectors of the Cartesians system with length 1!!! In fact, the positionvector P is the connecting line I just mentioned. So the length of this vector must have the same length as the length of the connecting line.

The squared length of a positionvector is found by taking the scalar-product: P. P = (3.i x 3.i) + (4.j x 4.j) = (9 x i^2) + (16 x j^2) = (9 x 1) + (16 x 1) = 25. So the length of the positionvector is indeed 5. This is the scalar you were talking about! It is produced by squaring 3 (= dx) and 4 (= dy) [they are components of a vector; not vectors themselves!!!], and squaring the basisvectors i and j [they make up the metric tensor, with only 1's on its diagonal, in the case of the Cartesian system]. The metric tensor is a matrix. So, multiplying the ELEMENTS of a matrix with the COMPONENTS of a vector gives a scalar!

The metric tensor is always the product of two basisvectors!

This page will be edited, possibly even re-written.

-blathem