\(\bar{V_{\alpha}}\) 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
\(\bar{V^{\alpha}}\)
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 :)
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...
Thanks Opalg
The second part of the question asks to repeat the same process but for:
\(p=x\)
\(q = x \cos\theta + y \sin\theta\)
Therefore:
\(x=p\)
And:
\(y=\frac{q}{\sin\theta}\ - p \frac{\cos\theta}{\sin\theta}\)
Which can be written as:
\(y=\frac{q}{\sin\theta}\ -p \cot\theta\)
Thank you Opalg, after establishing the metric tensor the question goes on to say:
Suppose the q axis goes through the point (5,12), what is the metric tensor now in the (p,q) system?
Using:
\(r^2 = x^2 + y^2\)
\(r^2 = (5)^2 + (12)^2\)
\(r^2 = 25 +144\)
\(r^2 = 169\)
\(r = 13\)...
Re: Flat two dimensional Plane (please Help)
Hi Opalg, having taken on board what you said and found the differential coefficients. What is the next step? Do I now substitute them into
\(ds^2 = dx^2 + dy^2\)
Thanks again :)
Re: Flat two dimensional Plane (please Help)
Hi Hallsoofivy, I'm sorry about the attachment, I wanted to upload it in two parts but there was insufficient space. As for the drop down window, there is now answer there. That is as far as I have gotten.
I'm not sure how to substitute the...
Hi guys, first time poster here. I have been struggling with the following problem for a few days now and could use some guidance. I believe I have the first part correct but the second is causing me trouble (see attachment). Is what I'm doing correct and how would I go about finding the metric...
Just like to pick up in this old thread, still having trouble with the question.
Using what I have already done:
δrθ=(∂r/∂x . ∂x/∂θ) + (∂r/∂y . ∂y/∂θ) + (∂r/∂z . ∂z/∂θ) (1)
Where:
x=r sin θ cos φ and y= r sin θ sin φ z= r cos θ
Would (1) then become:
δyx= = ((sin...