conversion of laplacian from cartesian to spherical coordinates

Question

In cartesian coordinates, the Laplacian is $$\nabla^2 = \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\qquad(1)$$

If it's converted to spherical coordinates, we get $$\nabla^2=\frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2 \frac{\partial}{\partial r}\right)+\frac{1}{r^2 sin\theta}\frac{\partial}{\partial \theta}\left(sin \theta \frac{\partial}{\partial \theta}\right)+\frac{1}{r^2 sin^2 \theta}\frac{\partial^2}{\partial \phi^2}\qquad(2)$$

I am following the derivation (i.e. the method of conversion from cartesian to spherical) in "Quantum physics of atoms, molecules, solids, nuclei and particles" by Eisberg and Resnick (it's in Appendix M). Their method is to first consider a function of only $r$, $\psi(r)$, then calculate $\frac{\partial^2 \psi}{\partial x^2}$, $\frac{\partial^2 \psi}{\partial y^2}$, $\frac{\partial^2 \psi}{\partial z^2}$. After plugging into (1) and simplifying we end up with $\nabla^2=\frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2 \frac{\partial}{\partial r}\right)$ which is, indeed the first term in (2).

For the second (and third) term they don't show the derivation, they just say "the second and third terms can be obtained by taking $\psi = \psi(\theta)$ and $\psi = \psi(\phi)$.

I am trying to complete the derivation for the other terms, but I don't know how to proceed. Here is what I have so far: $$\frac{\partial \psi}{\partial x} = \frac{\partial \psi}{\partial \theta} \frac{\partial \theta}{\partial x}\qquad (3)$$

$$\frac{\partial\theta}{\partial x} = \frac{xz}{(x^2 + y^2 +z^2)^{3/2}\sqrt{1-\frac{z^2}{x^2+y^2+z^2}}}\qquad (4)$$

putting (4) into (3) and substituting for $r$:

$$\frac{\partial\psi}{\partial x} =\frac{\partial \psi}{\partial\theta} \frac{xz}{r^{3}\sqrt{1-\frac{z^2}{r^2}}}$$

Now, making the conversions $x=rsin\theta cos\phi$ and $z=r cos\theta$ and simplifying I get

$$\frac{\partial\psi}{\partial x}= \frac{\partial\psi}{\partial\theta} \frac{cos \theta cos \phi}{r} $$

My question is how do I proceed?

Answer 1

You're on the right track. Now do what you did for $\partial\psi/\partial y$ and $\partial\psi/\partial z$, then compute the second derivatives and add them up. Note that when computing the second derivatives, you will be able to reuse the results from computing the first derivatives. For example,

$$ \frac{\partial^2\psi}{\partial x^2} = \frac{\partial}{\partial x}\frac{\partial\psi}{\partial x} = \frac{\partial}{\partial x} \Big( \frac{\cos\theta\cos\phi}{r}\,\frac{\partial\psi}{\partial\theta} \Big) $$

To continue from the above, use the chain rule, $$ \frac{\partial^2\psi}{\partial x^2} = \frac{\partial}{\partial x} (\cdot) = \frac{\partial r}{\partial x} \frac{\partial}{\partial r} (\cdot) + \frac{\partial \theta}{\partial x} \frac{\partial}{\partial \theta} (\cdot) + \frac{\partial \phi}{\partial x} \frac{\partial}{\partial \phi} (\cdot) $$

where the $\cdot$ is the term within the parentheses in the first equation above. Note that, in addition to the mixed-coordinate derivatives ($\partial r/\partial x$, etc), you'll need to compute the derivative of a product of functions. For example,

$$ \frac{\partial}{\partial r}\Big( \frac{\cos\theta\cos\phi}{r}\,\frac{\partial\psi}{\partial\theta} \Big) = -\frac{\cos\theta\cos\phi}{r^2}\,\frac{\partial\psi}{\partial\theta} + \frac{\cos\theta\cos\phi}{r}\,\frac{\partial^2\psi}{\partial r\,\partial\theta} $$

Of course, because (in this case) you chose $\psi$ to be a function of only $\theta$, the last term above is zero (since $\psi$ does not depend on $r$).

This problem is not difficult per se but it requires a lot of derivative computations and good organisation. It's a great exercise to improve your computational and organisational skills but you'll learn in the future other methods to find the Laplacian in another coordinate system that are far more efficient and economical.

Oh, and here's a trick to avoid having to deal with that pesky square root in $r$ as a function of $x, y, z$:

$$ \frac{\partial r^2}{\partial x} = 2x $$

but also

$$ \frac{\partial r^2}{\partial x} = 2r\,\frac{\partial r}{\partial x} $$

Thus, combining the two, you get

$$ \frac{\partial r}{\partial x} = \frac{x}{r} $$

Finally, kudos for wanting to do the extra work on your own. Also, the book you mentioned is excellent. If you have your own copy, you might want to keep it. I still have mine (though not with me here), even after 30 years!

Answer 2

Let $F$ be a scalar field on $\mathbb{R}^3$ defined in Cartesian coordinates. Let $M:\mathbb{R}^3\rightarrow \mathbb{R}^3$ be the mapping from spherical coordinates to Cartesian coordinates. Then $S = F\circ M$ is $F$ in spherical coordinates. Using the chain rule for vector fields, $$ S_i = F_j M_i^j. $$ Repeated indices (in this case $j$) are summed over the values 1,2, and 3, subscript $i$ means partial differentiation with respect to variable $i$, and superscript $j$ means coordinate $j$ of the vector field $M.$ This is a system of three linear equations in the three unknowns $F_j$ that could also be written as $DS = (DF) (DM),$ where $DG$ is the total derivative matrix of the vector or scalar field $G.$ The solution is $DF = (DS)(DM)^{-1}.$

Another application of the chain rule yields $$ S_{ik} = F_{jl}M_k^l M_i^j + F_j M_{ik}^j. $$ This is a linear system of nine equations in nine unknowns (the $F_{jl}$). Solving for the $F_{jl}$ using a computer algebra system and computing $F_{11}+F_{22} + F_{33}$ gives the Laplacian as a function of the $S_i$ and $S_{ik}$. Each individual term $F_{ij}$ is extremely complicated, but the sum $F_{11} + F_{22} + F_{33}$ is much simpler.

It is also possible to write the equation for the $F_{jl}$ in matrix form: $$ D^2F = (DM)^{-\text{T}}(D^2S - F_j (D^2M^j))(DM)^{-1} $$ where $D^2G$ is the Hessian matrix of second partial derivatives of the scalar field $G.$