- #1
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There is a formula for the potential ##\varphi## outside of a homogenous ellipsoid of density ##\mu## in Landau\begin{align*}
\varphi = -\pi \mu abck \int_{\xi}^{\infty} \left(1- \dfrac{x^2}{a^2 + s} + \dfrac{y^2}{b^2 + s} + \dfrac{z^2}{c^2+s} \right) \frac{ds}{R_s} \ \ \ (1)
\end{align*}where ##R_s = \sqrt{(a^2+s)(b^2+s)(c^2+s)}## and ##\xi## satisfies ## \dfrac{x^2}{a^2 + \xi} + \dfrac{y^2}{b^2 + \xi} + \dfrac{z^2}{c^2+\xi} = 1##. How is this formula obtained?
So far I can only find an expression in the limit of ##r \gg a,b,c##. The mass quadrupole tensor is ##Q_{\alpha \beta} = \displaystyle{\int_{\mathcal{V}}} \mu(3x_{\alpha} x_{\beta} - r^2 \delta_{\alpha \beta}) dV## therefore the potential at a point ##\mathbf{r}## has a multipole expansion ##\varphi(\mathbf{r}) = - \dfrac{km}{r} + \dfrac{1}{6} Q_{\alpha \beta} \partial^2_{\alpha \beta} \dfrac{1}{r} + \mathrm{etc}##. Due to the axial symmetry, the matrix ##Q## can be brought to diagonal form by aligning the coordinate system with the principal axes and the non-zero components are\begin{align*}
Q_{xx} &= \mu \int_{\mathcal{V}} d^3 x (2x^2 - y^2 - z^2)\\
&= \mu abc \int_0^{2\pi} d\phi \int_0^{\pi} d\theta \int_0^1 dr\, r^4 ([2a^2 \cos^2{\phi} - b^2 \sin^2{\phi}]\sin^3{\theta} - c^2 \cos^2{\theta} \sin{\theta} ) \\
&= \dfrac{\mu abc}{5} \int_0^{2\pi} d\phi \int_0^{\pi} d\theta \, ([2a^2 \cos^2{\phi} - b^2 \sin^2{\phi}]\sin^3{\theta} - c^2 \cos^2{\theta} \sin{\theta} ) \\
&= \dfrac{2\mu abc}{15} \int_0^{2\pi} d\phi \, (4a^2 \cos^2{\phi} - 2b^2 \sin^2{\phi} - c^2 ) \\
&= \dfrac{4\pi \mu abc}{15} (2a^2 - b^2 - c^2 )
\end{align*}where the "spherical-like" coordinate transformations ##x = ar\sin{\theta} \cos{\phi}##, etc. map the ellipsoid into the unit sphere. Similarly ##Q_{yy} = \dfrac{4\pi \mu abc}{15} (2b^2 - a^2 - c^2 )## and ##Q_{zz} = \dfrac{4\pi \mu abc}{15} (2c^2 - a^2 - b^2 )##. Now\begin{align*}
\partial_{\beta} \dfrac{1}{r} &= - \dfrac{1}{r^2} \cdot \dfrac{1}{2r} \partial_{\beta} r^2 = -\dfrac{1}{2r^3} \partial_{\beta} (x_{\gamma} x_{\gamma}) = -\dfrac{x_{\beta}}{r^3} \\ \\
\implies \partial^2_{\alpha \beta} \dfrac{1}{r} &= -\partial_{\alpha} \dfrac{x_{\beta}}{r^3} = -\frac{1}{r^3} \delta_{\alpha \beta} + \frac{3x_{\alpha} x_{\beta}}{r^5} = \frac{3x_{\alpha} x_{\beta} - r^2 \delta_{\alpha \beta}}{r^5}
\end{align*}therefore ##\partial^2_{xx} \dfrac{1}{r} = \dfrac{2x^2 - y^2 -z^2}{r^5}## and etc. therefore \begin{align*}
\varphi(\mathbf{r}) &= -\dfrac{km}{r} + \dfrac{2\pi \mu abc}{45 r^5} \left\{ (2a^2 - b^2 - c^2 )(2x^2 - y^2 -z^2) + \mathrm{y \ \ first} + \mathrm{z \ \ first} \right\} \\
&= -\dfrac{km}{r} + \dfrac{2\pi \mu abc}{45 r^5} \left\{ 6(a^2 x^2 + y^2b^2 + c^2 z^2) -3((b^2+c^2)x^2 + (a^2+c^2)y^2 + (a^2+b^2)z^2) \right\}
\end{align*}I haven't checked yet if this is consistent with ##(1)## as ##r## gets very big, but I'm more interested to know how Landau derived the exact expression?
\varphi = -\pi \mu abck \int_{\xi}^{\infty} \left(1- \dfrac{x^2}{a^2 + s} + \dfrac{y^2}{b^2 + s} + \dfrac{z^2}{c^2+s} \right) \frac{ds}{R_s} \ \ \ (1)
\end{align*}where ##R_s = \sqrt{(a^2+s)(b^2+s)(c^2+s)}## and ##\xi## satisfies ## \dfrac{x^2}{a^2 + \xi} + \dfrac{y^2}{b^2 + \xi} + \dfrac{z^2}{c^2+\xi} = 1##. How is this formula obtained?
So far I can only find an expression in the limit of ##r \gg a,b,c##. The mass quadrupole tensor is ##Q_{\alpha \beta} = \displaystyle{\int_{\mathcal{V}}} \mu(3x_{\alpha} x_{\beta} - r^2 \delta_{\alpha \beta}) dV## therefore the potential at a point ##\mathbf{r}## has a multipole expansion ##\varphi(\mathbf{r}) = - \dfrac{km}{r} + \dfrac{1}{6} Q_{\alpha \beta} \partial^2_{\alpha \beta} \dfrac{1}{r} + \mathrm{etc}##. Due to the axial symmetry, the matrix ##Q## can be brought to diagonal form by aligning the coordinate system with the principal axes and the non-zero components are\begin{align*}
Q_{xx} &= \mu \int_{\mathcal{V}} d^3 x (2x^2 - y^2 - z^2)\\
&= \mu abc \int_0^{2\pi} d\phi \int_0^{\pi} d\theta \int_0^1 dr\, r^4 ([2a^2 \cos^2{\phi} - b^2 \sin^2{\phi}]\sin^3{\theta} - c^2 \cos^2{\theta} \sin{\theta} ) \\
&= \dfrac{\mu abc}{5} \int_0^{2\pi} d\phi \int_0^{\pi} d\theta \, ([2a^2 \cos^2{\phi} - b^2 \sin^2{\phi}]\sin^3{\theta} - c^2 \cos^2{\theta} \sin{\theta} ) \\
&= \dfrac{2\mu abc}{15} \int_0^{2\pi} d\phi \, (4a^2 \cos^2{\phi} - 2b^2 \sin^2{\phi} - c^2 ) \\
&= \dfrac{4\pi \mu abc}{15} (2a^2 - b^2 - c^2 )
\end{align*}where the "spherical-like" coordinate transformations ##x = ar\sin{\theta} \cos{\phi}##, etc. map the ellipsoid into the unit sphere. Similarly ##Q_{yy} = \dfrac{4\pi \mu abc}{15} (2b^2 - a^2 - c^2 )## and ##Q_{zz} = \dfrac{4\pi \mu abc}{15} (2c^2 - a^2 - b^2 )##. Now\begin{align*}
\partial_{\beta} \dfrac{1}{r} &= - \dfrac{1}{r^2} \cdot \dfrac{1}{2r} \partial_{\beta} r^2 = -\dfrac{1}{2r^3} \partial_{\beta} (x_{\gamma} x_{\gamma}) = -\dfrac{x_{\beta}}{r^3} \\ \\
\implies \partial^2_{\alpha \beta} \dfrac{1}{r} &= -\partial_{\alpha} \dfrac{x_{\beta}}{r^3} = -\frac{1}{r^3} \delta_{\alpha \beta} + \frac{3x_{\alpha} x_{\beta}}{r^5} = \frac{3x_{\alpha} x_{\beta} - r^2 \delta_{\alpha \beta}}{r^5}
\end{align*}therefore ##\partial^2_{xx} \dfrac{1}{r} = \dfrac{2x^2 - y^2 -z^2}{r^5}## and etc. therefore \begin{align*}
\varphi(\mathbf{r}) &= -\dfrac{km}{r} + \dfrac{2\pi \mu abc}{45 r^5} \left\{ (2a^2 - b^2 - c^2 )(2x^2 - y^2 -z^2) + \mathrm{y \ \ first} + \mathrm{z \ \ first} \right\} \\
&= -\dfrac{km}{r} + \dfrac{2\pi \mu abc}{45 r^5} \left\{ 6(a^2 x^2 + y^2b^2 + c^2 z^2) -3((b^2+c^2)x^2 + (a^2+c^2)y^2 + (a^2+b^2)z^2) \right\}
\end{align*}I haven't checked yet if this is consistent with ##(1)## as ##r## gets very big, but I'm more interested to know how Landau derived the exact expression?