MHB Covariant and Contravariant Vector

andrey21
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I have been given the following problem:

The covariant vector field is:

\(v_{i}\) = \begin{matrix} x+y\\ x-y\end{matrix}What are the components for this vector field at (4,1)?

\(v_{i}\) = \begin{matrix} 5\\ 3\end{matrix}

Now I can use this information to solve the following:\(\bar{V_\alpha}\)But am unsure for \(\bar{V^\alpha}\).

I imagine it would be a similar approach with a few changes. Any help would be brilliant thank you :o
 
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If you mean to write a matrix, you could have used
Code:
\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}
to output
$$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$

Let us fix some notation here: are you using lower indexes to indicate covariant tensors and upper to indicate contravariant, or the other way around? Also, what are $\overline{V_\alpha}$ and $\overline{V}^{\alpha}$? I'm a little confused, if you could clarify perhaps we could arrive at an answer together. :D

At a first glance though, your calculations look correct, although I can't see where they headed because I don't understand what the symbols are meant to represent.
 
My calculations are quite long and would take me a very long time to write them out in LaTeX. I would like to include an attachment but am unsure how to delete old ones to make room.

Thanks :)
 
Hey AA23, you still haven't answered my last questions: what do these $\overline{V_\alpha}$, $\overline{V}^{\alpha}$ mean? Also, it seemed you had

$$v_i = \begin{bmatrix} x+y \\ x-y \end{bmatrix},$$

to which you applied at the point $(4,1)$, getting

$$v_i (4,1) = \begin{bmatrix} 5+1 \\ 5-1 \end{bmatrix}.$$

Are there other calculations? Also, we let column matrices denote vectors and we use line matrices to denote covetors, so perhaps that would be

$$v^i = \begin{bmatrix} x+y & x-y \end{bmatrix} .$$
 
\(\bar{V_{\alpha}}\) represents a covariant tensor

\(\bar{V^{\alpha}}\) represents a contravariant tensor Yes I can solve the question to find the components of

\(\bar{V_{\alpha}}\)

But come stuck when finding them for

\(\bar{V^{\alpha}}\)
 
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