# Line element to polar coordinates [closed]

I'm calculating the effective metric for a vortex in polar coordinates. The velocity and the potential is:

$$$$\mathbf{v}=\frac{A}{r} \hat{r} + \frac{B}{r}\hat{\theta}$$$$

So:

$$$$\mathbf{v}=\boldsymbol{\nabla} \psi \longrightarrow \psi= A ~log r + B~\theta$$$$

And I have the line element in cartesian coordinates $$(t,x^1,x^2,x^3)=(t,x,y,z)$$:

$$$$ds^2 = \dfrac{\rho_0}{c_s} \left[ - \left( c_s^2-v_0^2\right) dt^2 - v_0^i dt dx^i - v_0^j dt dx^j + \delta_{ij} dx^i dx^j \right]$$$$

I need to obtain the following line element, effective metric acoustic $$(t,r,\theta)$$:

$$$$ds^2 = - \left( c_s^2-\frac{A^2+B^2}{r^2}\right) dt^2 +dr^2 - 2\frac{A}{r}dtdr + r^2d\theta-2Bdtd\theta$$$$

Without $$z$$ because vortex is axially symmetric. I don't know how can I do it. I would appreciate some help to get started, what do I do with the terms with $$i$$.

## closed as off-topic by Pieter, Kyle Kanos, Gert, ZeroTheHero, AccidentalFourierTransformDec 9 '18 at 2:29

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• Something for the mathematics SE. – Pieter Dec 2 '18 at 18:43
• When you say "the terms with $i$", do you mean things like $v_0^i\, dt\, dx^i$? If so, you're supposed to assume an implicit sum, so that really mean $v_0^x\, dt\, dx + v_0^y\, dt\, dy + v_0^z\, dt\, dz$. Similarly, the $\delta{ij}$ term involves a sum over both $i$ and $j$. – Mike Dec 2 '18 at 19:18

We can start by writing everything in Cartesian coordinates. Given your expression for $$\mathbf{v}$$, we have $$$$\mathbf{v} \, = \, \frac{1}{x^2 \, + \, y^2} \, \left[ \left(A \, x \, - \, B \,y\right) \, \mathbf{\hat{i}} \, + \, \left(A \, y \, + \, B \, x \right) \, \mathbf{\hat{j}} \right] \, .$$$$ We should assume then that $$$$v_1 \, = \, \frac{A \, x \, - \, B \, y}{x^2 \, + \, y^2} \quad , \quad v_2 \, = \, \frac{A \, y \, + \, B \, x}{x^2 \, + \, y^2} \quad , \quad v_3 \, = \, 0 \quad \mathrm{and} \quad \mathbf{v}^2 \, = \, \frac{A^2 \, + \, B^2}{x^2 \, + \, y^2} \, .$$$$
Now we can consider the transformation from Cartesian coordinates to Cylindrical coordinates. This is $$$$g_{\mu' \nu'} \, = \, \frac{\partial \xi^{\mu}}{\partial \chi^{\mu'}} \, \frac{\partial \xi^{\nu}}{\partial \chi^{\nu'}} \, g_{\mu \nu} \, ,$$$$ with $$\begin{eqnarray} \xi^{\mu} &=& \left( t,\, x,\, y,\, z \right) \, = \, \left( t ,\, r \, \cos \theta ,\, r \, \sin \theta ,\, z \right) \, , \\ \chi^{\mu'} &=& \left( t ,\, r ,\, \theta ,\, z \right) \, , \end{eqnarray}$$ and $$$$g_{\mu \nu} \, = \, \frac{\rho_0}{c_s} \, \begin{bmatrix} - \, c^2_s \, + \, \mathbf{v}^2 & - \, v_1 & - \, v_2 & - \, v_3 \\ - \, v_1 & 1 & 0 & 0 \\ - \, v_2 & 0 & 1 & 0 \\ - \, v_3 & 0 & 0 & 1 \end{bmatrix} \, .$$$$
Following the transformation rule you find $$$$g_{\mu' \nu'} \, = \, \frac{\rho_0}{c_s} \, \begin{bmatrix} - \, c^2_s \, + \, \frac{A^2 \, + \, B^2}{r^2} & - \, \frac{A}{r} & - \, B & 0 \\ - \, \frac{A}{r} & 1 & 0 & 0 \\ - \, B & 0 & r^2 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \, ,$$$$ which, up to the factor of $$\frac{\rho_0}{c_s}$$, is the result you suggest.
• $-B\mapsto -B/r$ ? – Eli Dec 3 '18 at 15:55
• @Eli You mean in the final result for $g_{\mu' \nu'}$? When I did the multiplication yesterday it seemed to produce what I wrote (I am lazy and did it in Mathematica). The dimensional analysis seems ok to me as well... – secavara Dec 3 '18 at 17:59
• @secavare, my mistake, i got it with this transformation matrix J=$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos(\varphi) & -r\,\sin(\varphi) & 0 \\ 0 & \sin(\varphi) & r\,\cos(\varphi) & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$ so $G'=J^T G J$ – Eli Dec 3 '18 at 19:55