Problem Calculate the metric in elliptical coordinates
\begin{align*} x & =\frac{a}{2}\cosh \mu \cos \theta \\ y & =\frac{a}{2}\sinh \mu \sin \theta \end{align*}
Solution
The coordinates in the Cartesian system are \(\zeta ^{1}=x,\zeta ^{2}=y\) and the coordinates in the other system (Elliptic) are \(x^{1}=\mu ,x^{2}=\theta \). The relation between these must be known and invertible also, meaning \(\zeta \equiv \zeta \left ( x\right ) \) and \(x\equiv x\left ( \zeta \right ) \). This relation is given to use above as \begin{align*} \zeta ^{1} & =\frac{a}{2}\cosh \mu \cos \theta \\ \zeta ^{2} & =\frac{a}{2}\sinh \mu \sin \theta \end{align*}
The first step is to determine the metric tensor \(g_{ij}\) for the Polar coordinates. This is given by\[ g_{kl}=\delta _{ij}\frac{\partial \zeta ^{i}}{\partial x^{k}}\frac{\partial \zeta ^{j}}{\partial x^{l}}\] The above using Einstein summation notation.
\begin{align*} g_{11} & =\frac{\partial \zeta ^{1}}{\partial x^{1}}\frac{\partial \zeta ^{1}}{\partial x^{1}}+\frac{\partial \zeta ^{2}}{\partial x^{1}}\frac{\partial \zeta ^{2}}{\partial x^{1}}\\ & =\frac{\partial \zeta ^{1}}{\partial \mu }\frac{\partial \zeta ^{1}}{\partial \mu }+\frac{\partial \zeta ^{2}}{\partial \mu }\frac{\partial \zeta ^{2}}{\partial \mu }\\ & =\left ( \frac{\partial \zeta ^{1}}{\partial \mu }\right ) ^{2}+\left ( \frac{\partial \zeta ^{2}}{\partial \mu }\right ) ^{2}\\ & =\left ( \frac{a}{2}\sinh \mu \cos \theta \right ) ^{2}+\left ( \frac{a}{2}\cosh \mu \sin \theta \right ) ^{2}\\ & =\frac{a^{2}}{4}\left ( \sinh ^{2}\mu \cos ^{2}\theta +\cosh ^{2}\mu \sin ^{2}\theta \right ) \\ & =\frac{a^{2}}{4}\left ( \left ( \cosh ^{2}\mu -1\right ) \cos ^{2}\theta +\cosh ^{2}\mu \left ( 1-\cos ^{2}\theta \right ) \right ) \\ & =\frac{a^{2}}{4}\left ( \cosh ^{2}\mu \cos ^{2}\theta -\cos ^{2}\theta +\cosh ^{2}\mu -\cosh ^{2}\mu \cos ^{2}\theta \right ) \\ & =\frac{a^{2}}{4}\left ( \cosh ^{2}\mu -\cos ^{2}\theta \right ) \end{align*}
And\begin{align*} g_{12} & =\frac{\partial \zeta ^{1}}{\partial x^{1}}\frac{\partial \zeta ^{1}}{\partial x^{2}}+\frac{\partial \zeta ^{2}}{\partial x^{1}}\frac{\partial \zeta ^{2}}{\partial x^{2}}\\ & =\frac{\partial \zeta ^{1}}{\partial \mu }\frac{\partial \zeta ^{1}}{\partial \theta }+\frac{\partial \zeta ^{2}}{\partial \mu }\frac{\partial \zeta ^{2}}{\partial \theta }\\ & =\left ( \frac{a}{2}\sinh \mu \cos \theta \right ) \left ( -\frac{a}{2}\cosh \mu \sin \theta \right ) +\left ( \frac{a}{2}\cosh \mu \sin \theta \right ) \left ( \frac{a}{2}\sinh \mu \cos \theta \right ) \\ & =0 \end{align*}
The above is as expected since the coordinate system is orthogonal. And\begin{align*} g_{21} & =\frac{\partial \zeta ^{1}}{\partial x^{2}}\frac{\partial \zeta ^{1}}{\partial x^{1}}+\frac{\partial \zeta ^{2}}{\partial x^{2}}\frac{\partial \zeta ^{2}}{\partial x^{1}}\\ & =\frac{\partial \zeta ^{1}}{\partial \theta }\frac{\partial \zeta ^{1}}{\partial \mu }+\frac{\partial \zeta ^{2}}{\partial \theta }\frac{\partial \zeta ^{2}}{\partial \mu }\\ & =\left ( -\frac{a}{2}\cosh \mu \sin \theta \right ) \left ( \frac{a}{2}\sinh \mu \cos \theta \right ) +\left ( \frac{a}{2}\sinh \mu \cos \theta \right ) \left ( \frac{a}{2}\cosh \mu \sin \theta \right ) \\ & =0 \end{align*}
The above is as expected since the coordinate system is orthogonal. It is also because \(g_{ij}\) is symmetric and we already found that \(g_{12}=0\). And finally\begin{align*} g_{22} & =\frac{\partial \zeta ^{1}}{\partial x^{2}}\frac{\partial \zeta ^{1}}{\partial x^{2}}+\frac{\partial \zeta ^{2}}{\partial x^{2}}\frac{\partial \zeta ^{2}}{\partial x^{2}}\\ & =\frac{\partial \zeta ^{1}}{\partial \theta }\frac{\partial \zeta ^{1}}{\partial \theta }+\frac{\partial \zeta ^{2}}{\partial \theta }\frac{\partial \zeta ^{2}}{\partial \theta }\\ & =\left ( \frac{\partial \zeta ^{1}}{\partial \theta }\right ) ^{2}+\left ( \frac{\partial \zeta ^{2}}{\partial \theta }\right ) ^{2}\\ & =\left ( -\frac{a}{2}\cosh \mu \sin \theta \right ) ^{2}+\left ( \frac{a}{2}\sinh \mu \cos \theta \right ) ^{2}\\ & =\frac{a^{2}}{4}\left ( \cosh ^{2}\mu \sin ^{2}\theta +\sinh ^{2}\mu \cos ^{2}\theta \right ) \\ & =\frac{a^{2}}{4}\left ( \cosh ^{2}\mu \left ( 1-\cos ^{2}\theta \right ) +\left ( \cosh ^{2}\mu -1\right ) \cos ^{2}\theta \right ) \\ & =\frac{a^{2}}{4}\left ( \cosh ^{2}\mu -\cosh ^{2}\mu \cos ^{2}\theta +\cosh ^{2}\mu \cos ^{2}\theta -\cos ^{2}\theta \right ) \\ & =\frac{a^{2}}{4}\left ( \cosh ^{2}\mu -\cos ^{2}\theta \right ) \end{align*}
From the above we see that\begin{align*} g_{ij} & =\begin{pmatrix} g_{11} & g_{12}\\ g_{21} & g_{22}\end{pmatrix} \\ & =\frac{a^{2}}{4}\begin{pmatrix} \cosh ^{2}\mu -\cos ^{2}\theta & 0\\ 0 & \cosh ^{2}\mu -\cos ^{2}\theta \end{pmatrix} \end{align*}
That there are different ways to write the above, and they are all the same. For example, we can write\begin{align*} g_{ij} & =\frac{a^{2}}{4}\begin{pmatrix} \left ( 1+\sinh ^{2}\mu \right ) -\left ( 1-\sin ^{2}\theta \right ) & 0\\ 0 & \left ( 1+\sin ^{2}\mu \right ) -\left ( 1-\sin ^{2}\theta \right ) \end{pmatrix} \\ & =\frac{a^{2}}{4}\begin{pmatrix} \sinh ^{2}\mu +\sin ^{2}\theta & 0\\ 0 & \sinh ^{2}\mu +\sin ^{2}\theta \end{pmatrix} \end{align*}
Or we could use the double angle relations \(\cos ^{2}\theta =\frac{1}{2}\left ( 1+\cos \left ( 2\theta \right ) \right ) \) and \(\cosh ^{2}\mu =\frac{1}{2}\left ( 1+\cosh \left ( 2\theta \right ) \right ) \) to obtain\begin{align*} g_{ij} & =\frac{a^{2}}{4}\begin{pmatrix} \frac{1}{2}\left ( 1+\cosh \left ( 2\theta \right ) \right ) -\frac{1}{2}\left ( 1+\cos \left ( 2\theta \right ) \right ) & 0\\ 0 & \frac{1}{2}\left ( 1+\cosh \left ( 2\theta \right ) \right ) -\frac{1}{2}\left ( 1+\cos \left ( 2\theta \right ) \right ) \end{pmatrix} \\ & =\frac{a^{2}}{8}\begin{pmatrix} \cosh \left ( 2\theta \right ) -\cos \left ( 2\theta \right ) & 0\\ 0 & \cosh \left ( 2\theta \right ) -\cos \left ( 2\theta \right ) \end{pmatrix} \end{align*}
Problem Show that in a general coordinates system \(\epsilon _{i_{1}\cdots i_{N}}=g\epsilon ^{i_{1}\cdots i_{N}}\) where the covariant form is obtained by lowering the indices on the contravariant form.
Solution
In tensor analysis, contravariant components of a tensor uses upper indices and covariant components uses lower indices. Given a tensor in contravariant form \(\epsilon ^{i}\) then the covariant form \(\epsilon _{i}\) is obtained using\[ \epsilon _{i}=g_{ij}\epsilon ^{j}\] Where on the right side the sum is taken over \(j\) since it is the repeated index. This operation is called index contracting.
Therefore extending the above to all indices in \(\epsilon _{i_{1}\cdots i_{N}}\) results in\begin{equation} \epsilon _{i_{1}i_{2}\cdots i_{N}}=g_{i_{1}j_{1}}g_{i_{2}j_{2}}\cdots g_{i_{N}j_{N}}\epsilon ^{j_{1}j_{2}\cdots j_{N}} \tag{1} \end{equation} But we know that, from page 123 in the Matrices notes, that the determinant of the metric can be written using Levi-Civita tensor as\begin{equation} g=\sum _{i_{1}i_{2}\cdots i_{N}}g_{1i_{1}}g_{2i_{2}}\cdots g_{Ni_{N}}\epsilon ^{i_{1}i_{2}\cdots i_{N}} \tag{2} \end{equation} Comparing (1) and (2) shows that\begin{align*} \epsilon _{123\cdots N} & =g_{1i_{1}}g_{2i_{2}}\cdots g_{Ni_{N}}\epsilon ^{i_{1}i_{2}\cdots i_{N}}\\ & =k\epsilon ^{i_{1}i_{2}\cdots i_{N}} \end{align*}
Where \(k\) is constant, which in the case of \(\epsilon _{123\cdots N}\), this constant is \(g\). Now need to show that the constant is \(g\) for all cases of indices in \(\epsilon _{i_{1}i_{2}\cdots i_{N}}\) and not for the case \(\epsilon _{123\cdots N}\).
Looking at the case of \(N=2\), and let us see what happens if we change the order of the indices.\[ \epsilon _{i_{1}i_{2}}=g_{i_{1}j_{1}}g_{i_{2}j_{2}}\epsilon ^{j_{1}j_{2}}\] And\[ \epsilon _{i_{2}i_{1}}=g_{i_{2}j_{2}}g_{i_{1}j_{1}}\epsilon ^{j_{2}j_{1}}\] But \(g_{i_{1}j_{1}}g_{i_{2}j_{2}}\) is the same as \(g_{i_{2}j_{2}}g_{i_{1}j_{1}}\). So the ordering of indices does not change the constant \(k\). And since we found that this constant is \(g\) from above, therefore we conclude that\begin{equation} \epsilon _{i_{1}i_{2}\cdots i_{N}}=g\epsilon ^{j_{1}j_{2}\cdots j_{N}} \tag{3} \end{equation}
Problem Compute all components of the affine connection in polar coordinates.
Solution
In polar coordinates \(x^{1}=r,x^{2}=\theta \), the relation to the Cartesian coordinates is\begin{align*} x & =r\cos \theta \\ y & =r\sin \theta \end{align*}
Using\begin{equation} \Gamma _{jk}^{i}=\frac{1}{2}g^{li}\left ( \frac{\partial g_{kl}}{\partial x^{i}}+\frac{\partial g_{jl}}{\partial x^{k}}-\frac{\partial g_{jk}}{\partial x^{l}}\right ) \tag{1} \end{equation} We know that in polar coordinates the metric tensor is \(g_{11}=g_{rr}=1,\) and \(g_{12}=g_{r\theta }=0,\) and \(g_{21}=g_{\theta r}=0,\) and \(g_{22}=g_{\theta \theta }=0\) or in matrix form\[ g_{ij}=\begin{pmatrix} 1 & 0\\ 0 & r^{2}\end{pmatrix} \] Hence \(g^{ij}\) is its inverse\[ g^{ij}=\begin{pmatrix} 1 & 0\\ 0 & \frac{1}{r^{2}}\end{pmatrix} \] Using (1), let \(i=r,j=r,k=r\) then\[ \Gamma _{rr}^{r}=\frac{1}{2}g^{lr}\left ( \frac{\partial g_{rl}}{\partial r}+\frac{\partial g_{rl}}{\partial r}-\frac{\partial g_{rr}}{\partial x^{l}}\right ) \] The sum is now over \(l\), which goes from \(r,\theta \) since these are the only coordinates. Hence the above becomes\begin{align} \Gamma _{rr}^{r} & =\frac{1}{2}g^{rr}\left ( \frac{\partial g_{rr}}{\partial r}+\frac{\partial g_{rr}}{\partial r}-\frac{\partial g_{rr}}{\partial r}\right ) +\frac{1}{2}g^{\theta r}\left ( \frac{\partial g_{rr}}{\partial r}+\frac{\partial g_{rr}}{\partial r}-\frac{\partial g_{rr}}{\partial \theta }\right ) \nonumber \\ & =\frac{1}{2}\left ( 1\right ) \left ( 0+0-0\right ) +\frac{1}{2}\left ( 0\right ) \left ( \frac{\partial g_{rr}}{\partial r}+\frac{\partial g_{rr}}{\partial r}-\frac{\partial g_{rr}}{\partial \theta }\right ) \nonumber \\ & =0\tag{2} \end{align}
Using (1), let \(i=r,j=\theta ,k=r\) then\[ \Gamma _{\theta r}^{r}=\frac{1}{2}g^{lr}\left ( \frac{\partial g_{rl}}{\partial r}+\frac{\partial g_{\theta l}}{\partial r}-\frac{\partial g_{\theta r}}{\partial x^{l}}\right ) \] The sum is now over \(l\), which goes from \(r,\theta \) since these are the only coordinates. Hence the above becomes\begin{align} \Gamma _{\theta r}^{r} & =\frac{1}{2}g^{rr}\left ( \frac{\partial g_{rr}}{\partial r}+\frac{\partial g_{\theta r}}{\partial r}-\frac{\partial g_{\theta r}}{\partial r}\right ) +\frac{1}{2}g^{\theta r}\left ( \frac{\partial g_{r\theta }}{\partial r}+\frac{\partial g_{\theta \theta }}{\partial r}-\frac{\partial g_{\theta r}}{\partial \theta }\right ) \nonumber \\ & =\frac{1}{2}\left ( 1\right ) \left ( 0+0-0\right ) +\frac{1}{2}\left ( 0\right ) \left ( \frac{\partial g_{r\theta }}{\partial r}+\frac{\partial g_{\theta \theta }}{\partial r}-\frac{\partial g_{\theta r}}{\partial \theta }\right ) \nonumber \\ & =0\tag{3} \end{align}
Using (1), now let \(i=r,j=\theta ,k=\theta \) then\[ \Gamma _{\theta \theta }^{r}=\frac{1}{2}g^{lr}\left ( \frac{\partial g_{\theta l}}{\partial \theta }+\frac{\partial g_{\theta l}}{\partial \theta }-\frac{\partial g_{\theta \theta }}{\partial x^{l}}\right ) \] The sum is now over \(l\), which goes from \(r,\theta \) since these are the only coordinates. Hence the above becomes\begin{align} \Gamma _{\theta \theta }^{r} & =\frac{1}{2}g^{rr}\left ( \frac{\partial g_{\theta r}}{\partial \theta }+\frac{\partial g_{\theta r}}{\partial \theta }-\frac{\partial g_{\theta \theta }}{\partial r}\right ) +\frac{1}{2}g^{\theta r}\left ( \frac{\partial g_{\theta \theta }}{\partial \theta }+\frac{\partial g_{\theta \theta }}{\partial \theta }-\frac{\partial g_{\theta \theta }}{\partial \theta }\right ) \nonumber \\ & =\frac{1}{2}\left ( 1\right ) \left ( \left ( 0\right ) +\left ( 0\right ) -\frac{\partial r^{2}}{\partial r}\right ) +\frac{1}{2}\left ( 0\right ) \left ( \frac{\partial g_{r\theta }}{\partial r}+\frac{\partial g_{\theta \theta }}{\partial r}-\frac{\partial g_{\theta r}}{\partial \theta }\right ) \nonumber \\ & =\frac{1}{2}\left ( -2r\right ) \nonumber \\ & =-r\tag{4} \end{align}
Using (1), now let \(i=r,j=r,k=\theta \). Hence we need to find \(\Gamma _{r\theta }^{r}\). But due to symmetry in lower indices, then \(\Gamma _{r\theta }^{r}=\Gamma _{\theta r}^{r}\) which we found in (3) to be zero. Hence\begin{equation} \Gamma _{r\theta }^{r}=0\tag{5} \end{equation} Using (1), now let \(i=\theta ,j=r,k=r\) then\[ \Gamma _{rr}^{\theta }=\frac{1}{2}g^{l\theta }\left ( \frac{\partial g_{rl}}{\partial \theta }+\frac{\partial g_{rl}}{\partial r}-\frac{\partial g_{rr}}{\partial x^{l}}\right ) \] The sum is now over \(l\), which goes from \(r,\theta \) since these are the only coordinates. Hence the above becomes\begin{align} \Gamma _{rr}^{\theta } & =\frac{1}{2}g^{r\theta }\left ( \frac{\partial g_{rr}}{\partial \theta }+\frac{\partial g_{rr}}{\partial r}-\frac{\partial g_{rr}}{\partial r}\right ) +\frac{1}{2}g^{\theta \theta }\left ( \frac{\partial g_{r\theta }}{\partial \theta }+\frac{\partial g_{r\theta }}{\partial r}-\frac{\partial g_{rr}}{\partial \theta }\right ) \nonumber \\ & =\frac{1}{2}\left ( 0\right ) \left ( \frac{\partial g_{rr}}{\partial \theta }+\frac{\partial g_{rr}}{\partial r}-\frac{\partial g_{rr}}{\partial r}\right ) +\frac{1}{2}\left ( \frac{1}{r^{2}}\right ) \left ( 0+0-0\right ) \nonumber \\ & =0\tag{6} \end{align}
Using (1), now let \(i=\theta ,j=\theta ,k=r\) then\[ \Gamma _{\theta r}^{\theta }=\frac{1}{2}g^{l\theta }\left ( \frac{\partial g_{rl}}{\partial \theta }+\frac{\partial g_{\theta l}}{\partial r}-\frac{\partial g_{\theta r}}{\partial x^{l}}\right ) \] The sum is now over \(l\), which goes from \(r,\theta \) since these are the only coordinates. Hence the above becomes\begin{align} \Gamma _{\theta r}^{\theta } & =\frac{1}{2}g^{r\theta }\left ( \frac{\partial g_{rr}}{\partial \theta }+\frac{\partial g_{\theta r}}{\partial r}-\frac{\partial g_{\theta r}}{\partial r}\right ) +\frac{1}{2}g^{\theta \theta }\left ( \frac{\partial g_{r\theta }}{\partial \theta }+\frac{\partial g_{\theta \theta }}{\partial r}-\frac{\partial g_{\theta r}}{\partial \theta }\right ) \nonumber \\ & =\frac{1}{2}\left ( 0\right ) \left ( \frac{\partial g_{rr}}{\partial \theta }+\frac{\partial g_{\theta r}}{\partial r}-\frac{\partial g_{\theta r}}{\partial r}\right ) +\frac{1}{2}\frac{1}{r^{2}}\left ( 0+\frac{\partial r^{2}}{\partial r}-0\right ) \nonumber \\ & =\frac{1}{2}\frac{1}{r^{2}}\left ( 2r\right ) \nonumber \\ & =\frac{1}{r}\tag{7} \end{align}
Using (1), now let \(i=\theta ,j=r,k=\theta \) which finds \(\Gamma _{r\theta }^{\theta }\) but due to symmetry this is the same as \(\Gamma _{\theta r}^{\theta }\) which is found above. Hence \begin{equation} \Gamma _{r\theta }^{\theta }=\frac{1}{r}\tag{8} \end{equation} Using (1), now let \(i=\theta ,j=\theta ,k=\theta \) then\[ \Gamma _{\theta \theta }^{\theta }=\frac{1}{2}g^{l\theta }\left ( \frac{\partial g_{\theta l}}{\partial \theta }+\frac{\partial g_{\theta l}}{\partial \theta }-\frac{\partial g_{\theta \theta }}{\partial x^{l}}\right ) \] The sum is now over \(l\), which goes from \(r,\theta \) since these are the only coordinates. Hence the above becomes\begin{align} \Gamma _{\theta \theta }^{\theta } & =\frac{1}{2}g^{r\theta }\left ( \frac{\partial g_{\theta r}}{\partial \theta }+\frac{\partial g_{\theta r}}{\partial \theta }-\frac{\partial g_{\theta \theta }}{\partial r}\right ) +\frac{1}{2}g^{\theta \theta }\left ( \frac{\partial g_{\theta \theta }}{\partial \theta }+\frac{\partial g_{\theta \theta }}{\partial \theta }-\frac{\partial g_{\theta \theta }}{\partial \theta }\right ) \nonumber \\ & =\frac{1}{2}\left ( 0\right ) \left ( \frac{\partial g_{\theta r}}{\partial \theta }+\frac{\partial g_{\theta r}}{\partial \theta }-\frac{\partial g_{\theta \theta }}{\partial r}\right ) +\frac{1}{2}\frac{1}{r^{2}}\left ( 0+0-0\right ) \nonumber \\ & =0\tag{9} \end{align}
This completes the computation. In summary\begin{align*} \Gamma _{rr}^{r} & =0\\ \Gamma _{\theta r}^{r} & =0\\ \Gamma _{\theta \theta }^{r} & =-r\\ \Gamma _{r\theta }^{r} & =0\\ \Gamma _{rr}^{\theta } & =0\\ \Gamma _{\theta r}^{\theta } & =\frac{1}{r}\\ \Gamma _{r\theta }^{\theta } & =\frac{1}{r}\\ \Gamma _{\theta \theta }^{\theta } & =0 \end{align*}
Problem Calculate the gradient curl and divergence and Laplacian in spherical coordinates using tensor analysis.
Solution
The following coordinates system convention is used
The coordinates in the Cartesian system are \(\zeta ^{1}=x,\zeta ^{2}=y,\zeta ^{3}=z\). And the coordinates in the Spherical system are \(x^{1}=\phi ,x^{2}=r,x^{3}=\theta \). The relation between these is known as (Note that the following depends on convention used for which is \(\theta \) and which is \(\phi \). Physics convention as shown in the diagram above is used here).\begin{align*} \zeta ^{1} & =r\sin \theta \cos \phi \\ \zeta ^{2} & =r\sin \theta \sin \phi \\ \zeta ^{3} & =r\cos \theta \end{align*}
The first step is to determine the metric tensor \(g\) for the Spherical coordinates. This is given by\[ g_{kl}=\delta _{ij}\frac{\partial \zeta ^{i}}{\partial x^{k}}\frac{\partial \zeta ^{j}}{\partial x^{l}}\] Since the coordinate system are orthogonal, \(g_{kl}\) will be diagonal. Hence only \(g_{11},g_{22},g_{33}\) are non zero.\begin{align*} g_{11} & =g_{\phi \phi }\\ & =\frac{\partial \zeta ^{1}}{\partial x^{1}}\frac{\partial \zeta ^{1}}{\partial x^{1}}+\frac{\partial \zeta ^{2}}{\partial x^{1}}\frac{\partial \zeta ^{2}}{\partial x^{1}}+\frac{\partial \zeta ^{3}}{\partial x^{1}}\frac{\partial \zeta ^{3}}{\partial x^{1}}\\ & =\frac{\partial \zeta ^{1}}{\partial \phi }\frac{\partial \zeta ^{1}}{\partial \phi }+\frac{\partial \zeta ^{2}}{\partial \phi }\frac{\partial \zeta ^{2}}{\partial \phi }+\frac{\partial \zeta ^{3}}{\partial \phi }\frac{\partial \zeta ^{3}}{\partial \phi }\\ & =\left ( \frac{\partial \zeta ^{1}}{\partial \phi }\right ) ^{2}+\left ( \frac{\partial \zeta ^{2}}{\partial \phi }\right ) ^{2}+\left ( \frac{\partial \zeta ^{3}}{\partial \phi }\right ) ^{2}\\ & =\left ( -r\sin \theta \sin \phi \right ) ^{2}+\left ( r\sin \theta \cos \phi \right ) ^{2}+\left ( 0\right ) ^{2}\\ & =r^{2}\sin ^{2}\theta \sin ^{2}\phi +r^{2}\sin ^{2}\theta \cos ^{2}\phi \\ & =r^{2}\sin ^{2}\theta \left ( \sin ^{2}\phi +\cos ^{2}\phi \right ) \\ & =r^{2}\sin ^{2}\theta \end{align*}
And\begin{align*} g_{22} & =g_{rr}\\ & =\frac{\partial \zeta ^{1}}{\partial x^{2}}\frac{\partial \zeta ^{1}}{\partial x^{2}}+\frac{\partial \zeta ^{2}}{\partial x^{2}}\frac{\partial \zeta ^{2}}{\partial x^{2}}+\frac{\partial \zeta ^{3}}{\partial x^{2}}\frac{\partial \zeta ^{3}}{\partial x^{2}}\\ & =\frac{\partial \zeta ^{1}}{\partial r}\frac{\partial \zeta ^{1}}{\partial r}+\frac{\partial \zeta ^{2}}{\partial r}\frac{\partial \zeta ^{2}}{\partial r}+\frac{\partial \zeta ^{3}}{\partial r}\frac{\partial \zeta ^{3}}{\partial r}\\ & =\left ( \frac{\partial \zeta ^{1}}{\partial r}\right ) ^{2}+\left ( \frac{\partial \zeta ^{2}}{\partial r}\right ) ^{2}+\left ( \frac{\partial \zeta ^{3}}{\partial r}\right ) ^{2}\\ & =\left ( \sin \theta \cos \phi \right ) ^{2}+\left ( \sin \theta \sin \phi \right ) ^{2}+\left ( \cos \theta \right ) ^{2}\\ & =\sin ^{2}\theta \cos ^{2}\phi +\sin ^{2}\theta \sin ^{2}\phi +\cos ^{2}\theta \\ & =\sin ^{2}\theta \left ( \cos ^{2}\phi +\sin ^{2}\phi \right ) +\cos ^{2}\theta \\ & =\sin ^{2}\theta +\cos ^{2}\theta \\ & =1 \end{align*}
And\begin{align*} g_{33} & =g_{\theta \theta }\\ & =\frac{\partial \zeta ^{1}}{\partial x^{3}}\frac{\partial \zeta ^{1}}{\partial x^{3}}+\frac{\partial \zeta ^{2}}{\partial x^{3}}\frac{\partial \zeta ^{2}}{\partial x^{3}}+\frac{\partial \zeta ^{3}}{\partial x^{3}}\frac{\partial \zeta ^{3}}{\partial x^{3}}\\ & =\frac{\partial \zeta ^{1}}{\partial \theta }\frac{\partial \zeta ^{1}}{\partial \theta }+\frac{\partial \zeta ^{2}}{\partial \theta }\frac{\partial \zeta ^{2}}{\partial \theta }+\frac{\partial \zeta ^{3}}{\partial \theta }\frac{\partial \zeta ^{3}}{\partial \theta }\\ & =\left ( \frac{\partial \zeta ^{1}}{\partial \theta }\right ) ^{2}+\left ( \frac{\partial \zeta ^{2}}{\partial \theta }\right ) ^{2}+\left ( \frac{\partial \zeta ^{3}}{\partial \theta }\right ) ^{2}\\ & =\left ( r\cos \theta \cos \phi \right ) ^{2}+\left ( r\cos \theta \sin \phi \right ) ^{2}+\left ( -r\sin \theta \right ) ^{2}\\ & =r^{2}\cos ^{2}\theta \left ( \cos ^{2}\phi +\sin ^{2}\phi \right ) +r^{2}\sin ^{2}\theta \\ & =r^{2}\cos ^{2}\theta +r^{2}\sin ^{2}\theta \\ & =r^{2} \end{align*}
Hence \(ds^{2}\) in Spherical coordinates is\begin{align*} ds^{2} & =g_{kl}dx^{k}dx^{l}\\ & =g_{11}\left ( dx^{1}\right ) ^{2}+g_{22}\left ( dx^{2}\right ) ^{2}+g_{33}\left ( dx^{3}\right ) ^{2}\\ & =g_{11}\left ( d\phi \right ) ^{2}+g_{22}\left ( dr\right ) ^{2}+g_{33}\left ( d\theta \right ) ^{2}\\ & =r^{2}\sin ^{2}\theta \left ( d\phi \right ) ^{2}+\left ( dr\right ) ^{2}+r^{2}\left ( d\theta \right ) ^{2} \end{align*}
From the above we see that, using the order \(\phi ,r,\theta \) for the rows and columns\begin{align*} g_{ij} & =\begin{pmatrix} g_{11} & g_{12} & g_{13}\\ g_{21} & g_{22} & g_{23}\\ g_{31} & g_{32} & g_{33}\end{pmatrix} \\ & =\begin{pmatrix} r^{2}\sin ^{2}\theta & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & r^{2}\end{pmatrix} \end{align*}
Therefore the determinant is \(g=r^{4}\sin ^{2}\theta \) and \(h_{i}\) are given by the square root of the diagonal elements of \(g_{ij}\)\begin{align} h_{1} & =r\sin \theta \tag{A}\\ h_{2} & =1\nonumber \\ h_{3} & =r\nonumber \end{align}
\[ \nabla =\left ( \frac{1}{h_{1}}\frac{\partial }{\partial x^{1}},\frac{1}{h_{2}}\frac{\partial }{\partial x^{2}},\frac{1}{h_{3}}\frac{\partial }{\partial x^{3}}\right ) \] Where \(h_{i}\) are given in (A) and \(x^{1}=\phi ,x^{2}=r,x^{3}=\theta \). Therefore\[ \nabla =\left ( \frac{1}{r\sin \theta }\frac{\partial }{\partial \phi },\frac{\partial }{\partial r},\frac{1}{r}\frac{\partial }{\partial \theta }\right ) \] Hence given a function scalar \(f\left ( \phi ,r,\theta \right ) \) then\[ \nabla f=\frac{1}{r\sin \theta }\frac{\partial f}{\partial \phi }\hat{e}_{\phi }+\frac{\partial f}{\partial r}\hat{e}_{r}+\frac{1}{r}\frac{\partial f}{\partial \theta }\hat{e}_{\theta }\]
Using \(h_{i}\) in (A) and \(x^{1}=\phi ,x^{2}=r,x^{3}=\theta \) then\begin{align*} \left ( \vec{\nabla }\times \vec{V}\right ) _{1} & =\frac{1}{h_{2}h_{3}}\left ( \frac{\partial }{\partial x^{2}}\left ( h_{3}V_{3}\right ) -\frac{\partial }{\partial x^{3}}\left ( h_{2}V_{2}\right ) \right ) \\ \left ( \vec{\nabla }\times \vec{v}\right ) _{\phi } & =\frac{1}{r}\left ( \frac{\partial \left ( rV_{\theta }\right ) }{\partial r}-\frac{\partial V_{r}}{\partial \theta }\right ) \end{align*}
And\begin{align*} \left ( \vec{\nabla }\times \vec{V}\right ) _{2} & =\frac{1}{h_{3}h_{1}}\left ( \frac{\partial }{\partial x^{3}}\left ( h_{1}V_{1}\right ) -\frac{\partial }{\partial x^{1}}\left ( h_{3}V_{3}\right ) \right ) \\ \left ( \vec{\nabla }\times \vec{V}\right ) _{r} & =\frac{1}{r^{2}\sin \theta }\left ( \frac{\partial }{\partial \theta }\left ( r\sin \theta V_{\phi }\right ) -\frac{\partial }{\partial \phi }\left ( rV_{\theta }\right ) \right ) \\ & =\frac{1}{r\sin \theta }\left ( \frac{\partial \left ( \sin \theta V_{\phi }\right ) }{\partial \theta }-\frac{\partial V_{\theta }}{\partial \phi }\right ) \end{align*}
And\begin{align*} \left ( \vec{\nabla }\times \vec{V}\right ) _{3} & =\frac{1}{h_{1}h_{2}}\left ( \frac{\partial }{\partial x^{1}}\left ( h_{2}V_{2}\right ) -\frac{\partial }{\partial x^{2}}\left ( h_{1}V_{1}\right ) \right ) \\ \left ( \vec{\nabla }\times \vec{V}\right ) _{\theta } & =\frac{1}{r\sin \theta }\left ( \frac{\partial }{\partial \phi }\left ( V_{r}\right ) -\frac{\partial }{\partial r}\left ( r\sin \theta V_{\phi }\right ) \right ) \\ & =\frac{1}{r}\left ( \frac{1}{\sin \theta }\frac{\partial V_{r}}{\partial \phi }-\frac{\partial \left ( rV_{\phi }\right ) }{\partial r}\right ) \end{align*}
Therefore given a vector \(\vec{V}\), its curl is\[ \vec{\nabla }\times \vec{V}=\frac{1}{r}\left ( \frac{\partial \left ( rV_{\theta }\right ) }{\partial r}-\frac{\partial V_{r}}{\partial \theta }\right ) \hat{e}_{\phi }+\frac{1}{r\sin \theta }\left ( \frac{\partial \left ( \sin \theta V_{\phi }\right ) }{\partial \theta }-\frac{\partial V_{\theta }}{\partial \phi }\right ) \hat{e}_{r}+\frac{1}{r}\left ( \frac{1}{\sin \theta }\frac{\partial V_{r}}{\partial \phi }-\frac{\partial \left ( rV_{\phi }\right ) }{\partial r}\right ) \hat{e}_{\theta }\]
\begin{equation} \nabla \cdot V=\nabla _{i}V^{i}=\frac{\partial }{\partial x^{i}}V^{i}+\Gamma _{ij}^{i}V^{j} \tag{1} \end{equation} Where \(\Gamma _{ij}^{i}=\frac{1}{2}g^{li}\left ( \frac{\partial g_{jl}}{\partial x^{i}}+\frac{\partial g_{il}}{\partial x^{j}}-\frac{\partial g_{ij}}{\partial x^{l}}\right ) =\frac{1}{2}g^{li}\left ( \frac{\partial g_{il}}{\partial x^{j}}\right ) \) which simplifies to as shown in class notes page 143 to hence above becomes\[ \Gamma _{ij}^{i}=\frac{1}{\sqrt{g}}\frac{\partial }{x^{j}}\left ( \sqrt{g}\right ) \] Hence (1) becomes\begin{align} \nabla \cdot V & =\frac{\partial }{\partial x^{i}}V^{i}+\frac{1}{\sqrt{g}}\frac{\partial }{x^{j}}\left ( \sqrt{g}\right ) V^{j}\nonumber \\ & =\frac{1}{\sqrt{g}}\frac{\partial }{x^{i}}\left ( \sqrt{g}V^{i}\right ) \nonumber \end{align}
Using the covariant form the above becomes\[ \nabla \cdot V=\frac{1}{\sqrt{g}}\frac{\partial }{x^{i}}\left ( \frac{\sqrt{g}}{\sqrt{g_{ii}}}V_{i}\right ) \] Where in class notes \(h_{i}\) is used in place of \(\sqrt{g_{ii}}\), but it is it the same.
The sum is over \(i\). From above, the spherical coordinates are \(x^{1}=\phi ,x^{2}=r,x^{3}=\theta \). And \(g=r^{4}\sin ^{2}\theta \). Hence the above becomes after expanding\begin{align*} \nabla .V & =\frac{1}{\sqrt{r^{4}\sin ^{2}\theta }}\left ( \frac{\partial }{\partial \phi }\left ( \frac{\sqrt{r^{4}\sin ^{2}\theta }}{\sqrt{g_{\phi \phi }}}V_{\phi }\right ) +\frac{\partial }{\partial r}\left ( \frac{\sqrt{r^{4}\sin ^{2}\theta }}{\sqrt{g_{rr}}}V_{r}\right ) +\frac{\partial }{\partial \theta }\left ( \frac{\sqrt{r^{4}\sin ^{2}\theta }}{\sqrt{g_{\theta \theta }}}V_{\theta }\right ) \right ) \\ & =\frac{1}{r^{2}\sin \theta }\left ( \frac{\partial }{\partial \phi }\left ( \frac{r^{2}\sin \theta }{r\sin \theta }V_{\phi }\right ) +\frac{\partial }{\partial r}\left ( \frac{r^{2}\sin \theta }{1}V_{r}\right ) +\frac{\partial }{\partial \theta }\left ( \frac{r^{2}\sin \theta }{r}V_{\theta }\right ) \right ) \\ & =\frac{1}{r^{2}\sin \theta }\left ( \frac{\partial }{\partial \phi }\left ( rV_{\phi }\right ) +\frac{\partial }{\partial r}\left ( r^{2}\sin \theta V_{r}\right ) +\frac{\partial }{\partial \theta }\left ( \sin \theta V_{\theta }\right ) \right ) \\ & =\frac{\partial }{\partial \phi }\left ( \frac{1}{r\sin \theta }V_{\phi }\right ) +\frac{1}{r^{2}}\frac{\partial }{\partial r}\left ( r^{2}V_{r}\right ) +\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left ( \sin \theta V_{\theta }\right ) \end{align*}
The Laplacian is given by \[ \nabla ^{2}=\frac{1}{\sqrt{\det \left ( g\right ) }}\frac{\partial }{\partial x_{i}}\left ( \frac{\sqrt{\det \left ( g\right ) }}{g_{ii}}\frac{\partial }{\partial x^{i}}\right ) \] Hence\begin{align*} \nabla ^{2} & =\frac{1}{\sqrt{r^{4}\sin ^{2}\theta }}\frac{\partial }{\partial x_{1}}\left ( \frac{\sqrt{r^{4}\sin ^{2}\theta }}{g_{11}}\frac{\partial }{\partial x^{1}}\right ) +\frac{1}{\sqrt{r^{4}\sin ^{2}\theta }}\frac{\partial }{\partial x_{2}}\left ( \frac{\sqrt{r^{4}\sin ^{2}\theta }}{g_{22}}\frac{\partial }{\partial x^{2}}\right ) +\frac{1}{\sqrt{r^{4}\sin ^{2}\theta }}\frac{\partial }{\partial x_{3}}\left ( \frac{\sqrt{r^{4}\sin ^{2}\theta }}{g_{33}}\frac{\partial }{\partial x^{3}}\right ) \\ & =\frac{1}{r^{2}\sin \theta }\frac{\partial }{\partial \phi }\left ( \frac{r^{2}\sin \theta }{r^{2}\sin ^{2}\theta }\frac{\partial }{\partial \phi }\right ) +\frac{1}{r^{2}\sin \theta }\frac{\partial }{\partial r}\left ( \frac{r^{2}\sin \theta }{1}\frac{\partial }{\partial r}\right ) +\frac{1}{r^{2}\sin \theta }\frac{\partial }{\partial \theta }\left ( \frac{r^{2}\sin \theta }{r^{2}}\frac{\partial }{\partial \theta }\right ) \\ & =\frac{1}{r^{2}\sin \theta }\frac{\partial }{\partial \phi }\left ( \frac{1}{\sin \theta }\frac{\partial }{\partial \phi }\right ) +\frac{1}{r^{2}\sin \theta }\frac{\partial }{\partial r}\left ( r^{2}\sin \theta \frac{\partial }{\partial r}\right ) +\frac{1}{r^{2}\sin \theta }\frac{\partial }{\partial \theta }\left ( \sin \theta \frac{\partial }{\partial \theta }\right ) \\ & =\frac{1}{r^{2}\sin ^{2}\theta }\frac{\partial ^{2}}{\partial \phi ^{2}}+\frac{1}{r^{2}}\left ( 2r\frac{\partial }{\partial r}+r^{2}\frac{\partial ^{2}}{\partial r^{2}}\right ) +\frac{1}{r^{2}\sin \theta }\left ( \cos \theta \frac{\partial }{\partial \theta }+\sin \theta \frac{\partial ^{2}}{\partial \theta ^{2}}\right ) \\ & =\frac{1}{r^{2}\sin ^{2}\theta }\frac{\partial ^{2}}{\partial \phi ^{2}}+\frac{2}{r}\frac{\partial }{\partial r}+\frac{\partial ^{2}}{\partial r^{2}}+\frac{\cos \theta }{r^{2}\sin \theta }\frac{\partial }{\partial \theta }+\frac{1}{r^{2}}\frac{\partial ^{2}}{\partial \theta ^{2}}\\ & =\frac{\partial ^{2}}{\partial r^{2}}+\frac{2}{r}\frac{\partial }{\partial r}+\frac{1}{r^{2}}\left ( \frac{\cos \theta }{\sin \theta }\frac{\partial }{\partial \theta }+\frac{\partial ^{2}}{\partial \theta ^{2}}\right ) +\frac{1}{r^{2}\sin ^{2}\theta }\frac{\partial ^{2}}{\partial \phi ^{2}} \end{align*}
Therefore\begin{align*} \nabla ^{2}u & =\frac{\partial ^{2}u}{\partial r^{2}}+\frac{2}{r}\frac{\partial u}{\partial r}+\frac{1}{r^{2}}\left ( \frac{\cos \theta }{\sin \theta }\frac{\partial u}{\partial \theta }+\frac{\partial ^{2}u}{\partial \theta ^{2}}\right ) +\frac{1}{r^{2}\sin ^{2}\theta }\frac{\partial ^{2}u}{\partial \phi ^{2}}\\ & =u_{rr}+\frac{2}{r}u_{r}+\frac{1}{r^{2}}\left ( \frac{\cos \theta }{\sin \theta }u_{\theta }+u_{\theta \theta }\right ) +\frac{1}{r^{2}\sin ^{2}\theta }u_{\phi \phi } \end{align*}