Talk:General Relativity/Contravariant and Covariant Indices

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Vector Spaces and Basis Vectors - This section is a bit heavy on mathematical jargon. Perhaps someone can rewrite it so that it is a bit more accessible to the mathematically challenged (like me).

How's this? Boud 13:51, 12 January 2006 (UTC)

Someone changed it after you. It's pretty inaccessible right now. 1.28.09

[edit] Contravariant and Covariant Vectors

The covariant and contravariant example makes no sense. With w, if you increase displacement, temperature goes up. With v, if you increase time, displacement goes up. The author is either not being clear, or this is vandalism.

The author references two figures at the end of the Contravariant and Covariant Vectors section, however, the figures are not available.

Rabadi 05:54, 7 April 2006 (UTC)

Can one think of Contravariant Tensors as an example of proportionality relationship while the covariant Tensors are inversly proportional. Like Xa is proportional to something while Xa is inversly proportional to the same thing, personally i don't know because i am trying to learn about the stuff myself.

harryortho 16:49, 22/12/06 (GMT)

On the next section i wondered is it like having matrices v = v and w = v * which is a conjugate matrix, like 1x2.2x1 and when you multiply them together they end up makeing a number with no units, except in Tensors there is another tensor without units like before the inverse times the original equals 1 but in a Tensor kind of way? is there any sense here?

harryortho 17:04, 22/12/06 (GMT)

[edit] Confusion

Last line "Finally, if we define \mathbf{e}_\alpha (\mathbf{\omega}^\beta) = \delta_\alpha^\beta, we see that \mathbf{v} (\mathbf{\omega}) = \mathbf{\omega} (\mathbf{v})."

What is meant by that? Should it be \mathbf{v}(\mathbf{\sigma}) = \mathbf{\sigma} (\mathbf{v})? Also, isn't the multiplication commutative by definition? If it is not too important, can we omit it? --Hirak 99 11:42, 29 March 2007 (UTC)