# Using metric to raise differential operator

Given $\mathbf{L} = -i \mathbf {r} \times \nabla$ we can express this in polar coordinates as $$\mathbf {L} = i \mathbf{e_\theta} \frac{1}{\sin\theta}\frac{\partial}{\partial\phi} - i \mathbf{e_\phi} \frac{\partial}{\partial\theta}.$$ If follows that $$\mathbf{ L}^2 = -\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}(\sin\theta\frac{\partial}{\partial \theta}) - \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\tag{1}$$ where one must be careful taking derivatives with respect to the basis vectors themselves.

Alternatively, in index notation, I can express $\bf L$ as a 3 vector $L_i$ in the spherical space defined by metric $g_{ij}$. When I then go to take the inner product I get $$\mathbf{ L}^2 = L_iL^i = g^{ij}L_iL_j = \frac{1}{r^2}L_2L_2 + \frac{1}{r^2\sin^2\theta}L_3L_3\tag{2}.$$

Clearly (2) does not agree with (1). How does one handle the metric when dealing with operators, and how can one arrive at (2) using the metric in index notation?

Related: The placement of the metric in polar coordinates now seems to matter as $L_i(g^{ij}L_j) \ne g^{ij}L_iL_j$.

First the "related" question: in this case it does matter because, in general, $g_{ij}=g_{ij}(x^i)$ and because $L_i$ are differential operators, it truly is different to do $LgL$ or $gLL$ (schematically).

Now, I will try to solve a similar problem and leave the one you are working on as an exercise. Consider the Laplace operator in spherical coordinates. How do we get to it through index notation? Remember that the Laplacian is nothing more than the divergence of a gradient. Thus, for a general orthogonal coordinate system,

$$\nabla^2\equiv\frac{1}{\sqrt{g}}\frac{\partial}{\partial x^i}\left(\frac{\sqrt{g}}{g_{ii}}\frac{\partial}{\partial x^i}\right)\ .$$

The above equation follows from the fact that the gradient can be written as $\nabla^i=g^{ij}\frac{\partial}{\partial x^j}=\frac{1}{g_{ii}}\frac{\partial}{\partial x^i}$ and the (covariant) divegence of a vector $\vec{F}$ can be written as $\nabla^i F_i=\frac{1}{\sqrt{g}}\frac{\partial}{\partial x^i}\left(\sqrt{g}\, F^i\right)$. Then pick $\vec{F}$ to be the gradient and voilà.

For the particular case of spherical symmetry, we have $g_{ij}=\text{diag}(1,r^2,r^2\,\sin^2\theta)$ and $x^i=(r,\theta,\varphi)$. $g$ denotes the determinant of $g_{ij}$ which is, in this case, $r^4\sin^2\theta$. Then we have

\begin{align*} \nabla^2 &=\frac{1}{r^2\sin\theta}\left[\frac{\partial}{\partial r}\left(r^2\sin\theta\frac{\partial}{\partial r}\right)+\frac{\partial}{\partial \theta}\left(\frac{r^2\sin\theta}{r^2}\frac{\partial}{\partial \theta}\right)+\frac{\partial}{\partial \varphi}\left(\frac{r^2\sin\theta}{r^2\sin^2\theta}\frac{\partial}{\partial \varphi}\right)\right]\\ &=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\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\varphi^2}\ . \end{align*}

It shouldn't be too difficult to carry on the calculation you want now.

• Thanks for this example. Can you explain how the covariant derivative in polar coordinates acting on a scalar function $\nabla_i \phi$ is supposed to behave? Naively I want to say $\nabla_i \phi = \frac{\partial}{\partial x^i}\phi$ - this is not correct but I do not see the gap in logic. I can get the correct form for the gradient if I do a coordinate transformation from cartesian to polar, but I don't think this should be a necessary step. – Lone Wolf Oct 17 '17 at 14:58
• In a cartesian coordinate system, $\nabla_i\phi=\frac{\partial}{\partial x^i}\phi$. In a different coordinate system, we must include the Jacobian of the transformation $\tilde{\nabla}_i\phi=\frac{\partial \tilde{x}^j}{\partial x^i}\tilde{\partial}_j\phi$ (the tildes indicate, for example, polar coordinates). I do not see a way around this at the moment. – L. Werneck Oct 17 '17 at 16:10