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

\begin{equation} \mathbf{v}=\frac{A}{r} \hat{r} + \frac{B}{r}\hat{\theta} \end{equation}

So:

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

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

\begin{equation} 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] \end{equation}

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

\begin{equation} 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 \end{equation}

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 $i$.

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

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  • $\begingroup$ Something for the mathematics SE. $\endgroup$ – Pieter Dec 2 '18 at 18:43
  • $\begingroup$ 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 $j$. $\endgroup$ – Mike Dec 2 '18 at 19:18
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It seems this is mostly about being careful with coordinate transformations and the definitions of the objects you are using.

We can start by writing everything in Cartesian coordinates. Given your expression for $\mathbf{v}$, we have \begin{equation} \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] \, . \end{equation} We should assume then that \begin{equation} 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} \, . \end{equation}

Now we can consider the transformation from Cartesian coordinates to Cylindrical coordinates. This is \begin{equation} g_{\mu' \nu'} \, = \, \frac{\partial \xi^{\mu}}{\partial \chi^{\mu'}} \, \frac{\partial \xi^{\nu}}{\partial \chi^{\nu'}} \, g_{\mu \nu} \, , \end{equation} 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 \begin{equation} 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} \, . \end{equation}

Following the transformation rule you find \begin{equation} 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} \, , \end{equation} which, up to the factor of $\frac{\rho_0}{c_s}$, is the result you suggest.

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  • $\begingroup$ $-B\mapsto -B/r$ ? $\endgroup$ – Eli Dec 3 '18 at 15:55
  • $\begingroup$ @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... $\endgroup$ – secavara Dec 3 '18 at 17:59
  • $\begingroup$ @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 $G'=J^T G J$ $\endgroup$ – Eli Dec 3 '18 at 19:55

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