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- #1

Thank You in advance for any help

- Thread starter AA23
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- Thread starter
- #1

Thank You in advance for any help

- Jan 29, 2012

- 1,151

This is impossible to read. While I can read most of it after "zooming" to 400%, your answer is in a "drop down" window that does not, here, "drop down".

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- #3

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 differential coefficients into that equation, can you give me some suggestions?

Also is the first part of the answer correct? Finding how the (x,y) coordinates relate to the (p,q) coordinates.

- Feb 29, 2012

- 342

Hey AA23, welcome to the site. You would greatly help us helping you if you could write down the whole question using LaTeX and attach the graph only. These topics will help:

MHB Latex guide PDF

How to use LaTeX on this site.

Cheers.

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- #5

- Feb 7, 2012

- 2,740

Parts (a)(i) and (a)(ii) look correct, but (a)(iii) is not. The angle $\theta$ is supposed to be fixed, so you should not differentiate with respect to it. The variables are changing from $(x,y)$ to $(p,q).$ So you should be looking for the differential coefficients $\frac{\partial x}{\partial p}$, $\frac{\partial x}{\partial q}$, $\frac{\partial y}{\partial p}$ and $\frac{\partial y}{\partial q}$, keeping $\theta$ fixed throughout.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 tensor?

Thank You in advance for any help View attachment 292

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- #7

- Feb 7, 2012

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Use the chain rule in the form $dx = \frac{\partial x}{\partial p}dp + \frac{\partial x}{\partial q}dq$ (and a similar expression for $dy$) to write $dx^2 + dy^2$ in terms of $dp$ and $dq$.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

\(ds^2 = dx^2 + dy^2\)

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- #8

\(dx = \frac{\partial x}{\partial p}dp + \frac{\partial x}{\partial q}dq\)

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:

\(ds^2 = dx^2 + dy^2\)

Gives:

\(ds^2 = (1+cos \theta)^2 + (0+sin\theta)^2\)

\(ds^2 = (1+cos \theta)^2 + (sin\theta)^2\)

Which would give:

\(ds^2 = 1 + 2 cos\theta + cos^2\theta + sin^2\theta \)

\(ds^2 = 2 + 2 cos\theta\)

Is this correct? Thank you in advance

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- #9

- Feb 7, 2012

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LaTeX tip: put a backslash before sin and cos to improve their appearance.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:

\(ds^2 = dx^2 + dy^2\)

Gives:

\(ds^2 = (\color{red}{dp}+\cos \theta \color{red}{dq})^2 + (0+\sin\theta \color{red}{dq})^2\)

Which would give:

\(ds^2 = \color{red}{dp^2} + 2 \cos\theta\color{red}{dpdq} + (\cos^2\theta + \sin^2\theta) \color{red}{dq^2} \)

\(ds^2 = \color{red}{dp^2} + 2 \cos\theta \color{red}{dpdq} \color{red}{+dq^2}\)

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- #10

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- #11

- Feb 7, 2012

- 2,740

I am not very familiar with this topic, but as I understand it the metric tensor in this case should be the $2\times2$ matrix associated with the quadratic form for $ds^2$. In other words, $$ ds^2 = dp^2 + 2\cos\theta dpdq + dq^2 = \begin{bmatrix}dp & dq \end{bmatrix} \begin{bmatrix}1 & \cos\theta \\ \cos\theta & 1 \end{bmatrix} \begin{bmatrix}dp \\ dq \end{bmatrix}.$$Given that is the final answer, what part would I use for the metric tensor?

Thus the metric tensor should presumably be the matrix $ \begin{bmatrix}1 & \cos\theta \\ \cos\theta & 1 \end{bmatrix}.$

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- #12

Suppose the q axis goes through the point (5,12), what is the metric tensor now in the (p,q) system?

Using:

\(r^2 = x^2 + y^2\)

\(r^2 = (5)^2 + (12)^2\)

\(r^2 = 25 +144\)

\(r^2 = 169\)

\(r = 13\)

Substituting this values into:

\(\cos \theta = \frac{x}{r}\)

\(\cos \theta = \frac{5}{13}\)

Therefore the new metric tensor is:

\begin{bmatrix}1 & \frac{5}{13}\\ \frac{5}{13}\ & 1 \end{bmatrix}.

Is this correct??

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- #13

- Feb 7, 2012

- 2,740

Looks good to me.Therefore the new metric tensor is:

\begin{bmatrix}1 & \frac{5}{13}\\ \frac{5}{13}\ & 1 \end{bmatrix}.

Is this correct??

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- #14

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:

\(y=\frac{q}{\sin\theta}\ -p \cot\theta\)