Talk:General Relativity/Contravariant and Covariant Indices


 * 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

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 $$X^{a}$$ is proportional to something while $$X_{a}$$ 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)

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)


 * Yes, it should have been $$\mathbf{v}(\mathbf{\sigma}) = \mathbf{\sigma} (\mathbf{v})$$; I corrected the error.

It can be omitted, I suppose, but for now, I'd like to leave it there. My reasoning is that, for beginners, commutativity may not be obvious or assumed.

-blathem

Some comments
I was getting confused by the lack of bullet points, in one place, so I added them. Roll this back if you don't like it. I am just trying to understand this.

Also does,

''We can now evaluate any functional (covariant vector) applied to any vector (contravariant vector). If $$\mathbf{\sigma} \in V^*$$ and $$\mathbf{v} \in V$$, then by linearity $$\mathbf{\sigma} (\mathbf{v}) = \sigma_\alpha \mathbf{\omega}^\alpha (v^\beta \mathbf{e}_\beta) = \sigma_\alpha v^\beta \mathbf{\omega}^\alpha (\mathbf{e}_\beta) = \sigma_\alpha v^\beta \delta_\beta^\alpha = \sigma_\alpha v^\alpha$$. Finally, if we define $$\mathbf{e}_\alpha (\mathbf{\omega}^\beta) = \delta_\alpha^\beta$$, we see that $$\mathbf{v} (\mathbf{\sigma}) = \mathbf{\sigma} (\mathbf{v})$$.''

mean,

''We can now evaluate any functional (covariant vector) applied to any vector (contravariant vector). If $$\mathbf{\sigma} \in V^*$$ and $$\mathbf{v} \in V$$, then by linearity,''
 * $$\mathbf{\sigma} (\mathbf{v}) = \sigma_\alpha \mathbf{\omega}^\alpha (v^\beta \mathbf{e}_\beta) = \sigma_\alpha v^\beta \mathbf{\omega}^\alpha (\mathbf{e}_\beta) = \sigma_\alpha v^\beta \delta_\beta^\alpha = \sigma_\alpha v^\alpha$$

Also, if we define $$\mathbf{e}_\alpha (\mathbf{\omega}^\beta) = \delta_\alpha^\beta$$ then,
 * $$\mathbf{v} (\mathbf{\sigma}) = v^\alpha \mathbf{e}_\alpha(\sigma_\beta \mathbf{\omega}^\beta) = v^\alpha \sigma_\beta \mathbf{e}_\alpha ( \mathbf{\omega}^\beta) = v^\alpha \sigma_\beta \delta_\alpha^\beta = v^\alpha \sigma_\alpha$$

so,
 * $$\mathbf{v} (\mathbf{\sigma}) = \mathbf{\sigma} (\mathbf{v})$$.

In terms of the beginner the whole linear functional idea hasn't really been explained here, and the reference vector space doesn't help much because it introduces a whole lot of other issues, to confuse the beginner.

I am confused because it is not clear that $$\mathbf{\omega}$$ is a (contra variant) vector.
 * \mathbf{\omega} ... $$\mathbf{\omega}$$ does not look bold.
 * \mathbf{\sigma} ... $$\mathbf{\sigma}$$ does not look bold.
 * \mathbf{e} ... $$\mathbf{e}$$ does look bold.

I would have preferred if you had written $$\mathbf{f}$$ for $$\mathbf{\omega}$$. The look of the thing is that the $$\mathbf{e}$$ are basis vectors, but maybe $$\mathbf{\omega}$$ are something different. I guess they are basis vectors for the dual space that also act as linear functionals.

It seems the important bit is,
 * $$\mathbf{e}_\alpha (\mathbf{\omega}^\beta) = \delta_\alpha^\beta$$
 * $$\mathbf{\omega}_\alpha (\mathbf{e}^\beta) = \delta_\alpha^\beta$$

Can you define these together earlier? Or is one derivable from the other? I am just not sure about your precise argument. Sorry if I am being dense.

Thepigdog (discuss • contribs) 03:59, 3 June 2014 (UTC)