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Flat two dimensional Plane

AA23

New member
Aug 6, 2012
10
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 :)Revision 4.png
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Re: Flat two dimensional Plane (please Help)

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".
 

AA23

New member
Aug 6, 2012
10
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 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.
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Re: Flat two dimensional Plane (please Help)

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.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
Re: Flat two dimensional Plane (please Help)

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
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.
 

AA23

New member
Aug 6, 2012
10
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 :)
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
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

\(ds^2 = dx^2 + dy^2\)
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 $dq$.
 

AA23

New member
Aug 6, 2012
10
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:


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

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

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 :)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
You forgot about the $dp$ and $dq$.

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:

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

\(ds^2 = \color{red}{dp^2} + 2 \cos\theta \color{red}{dpdq} \color{red}{+dq^2}\)
LaTeX tip: put a backslash before sin and cos to improve their appearance.
 

AA23

New member
Aug 6, 2012
10
(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
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
Given that is the final answer, what part would I use for the metric tensor?
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, $$ ds^2 = dp^2 + 2\cos\theta dpdq + dq^2 = \begin{bmatrix}dp & dq \end{bmatrix} \begin{bmatrix}1 & \cos\theta \\ \cos\theta & 1 \end{bmatrix} \begin{bmatrix}dp \\ dq \end{bmatrix}.$$

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

AA23

New member
Aug 6, 2012
10
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:

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

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

\(r^2 = 25 +144\)

\(r^2 = 169\)

\(r = 13\)

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??
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
Therefore the new metric tensor is:


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

Is this correct??
Looks good to me. :)
 

AA23

New member
Aug 6, 2012
10
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:

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