# Flat two dimensional Plane

#### AA23

##### New member
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

#### HallsofIvy

##### Well-known member
MHB Math Helper

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

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

MHB Math Helper

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

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

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

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
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
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
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
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
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
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
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:

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