Recent content by andrey21

\(\bar{V_{\alpha}}\) 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: \(r^2 = x^2 + y^2\) \(r^2 = (5)^2 + (12)^2\) \(r^2 = 25 +144\) \(r^2 = 169\) \(r = 13\)...

(Rofl) I saw my mistake just as you replied. Given that is the final answer, what part would I use for the metric tensor? Thanks for the LaTex tip

So using:\(dx = \frac{\partial x}{\partial p}dp + \frac{\partial x}{\partial q}dq\) Where: \(\frac{\partial x}{\partial p}=1\) \(\frac{\partial x}{\partial q}=cos\theta\) \(\frac{\partial y}{\partial p}=0\) \(\frac{\partial y}{\partial q} = sin\theta\) Substituting into: \(ds^2 = dx^2 +...

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...

So I have the correct answer, that's great. Thanks for all your help :smile:

I understand so you end up with: W2= -(J10P3+J30P1+J31P0) Correct?

So using that information and that: ε1234=1 and ε1243=-1 W2= -1/2 (-J10P3+ J30P1+J10P3 +J31P0-J30P1-J31P0) =-1/2 (0)

:smile: Thats what I meant... So M31= - M13 for example...