Laplacian in Spherical Coordinates — MTH 496

These are notes on how to express the Laplacian in spherical coordinates.

The first step is to define the coordinates ${(r,\theta,\phi)}$ via the following identities:

$\displaystyle \mathbf{x}=\begin{pmatrix}x\\ y\\ z \end{pmatrix}=\begin{pmatrix}r\sin\theta\cos\phi,\\ r\sin\theta\sin\phi,\\ r\cos\theta. \end{pmatrix}. \ \ \ \ \ (1)$

Here ${r}$ is the radius from the origin, ${\theta\in[0,\pi]}$ is the angle from the ${z}$-axis and ${\phi\in[0,2\pi]}$ is the azimuthal angle from the ${x}$-axis. The map ${(r,\theta,\phi)\mapsto\mathbf{x}}$ is a one-to-one invertible map from ${(0,\infty)\times(0,\pi)\times(0,2\pi)}$ onto ${\mathbb{R}^{3}}$ minus the ${z}$-axis. The mapping is singular on the ${z}$-axis since the value of ${\phi}$ is undetermined there.

If we use (1) to express ${x,y,z}$ in terms of ${r,\theta,\phi}$, then the chain rule leads to the identities:

$\displaystyle \begin{array}{rcl} \frac{\partial}{\partial r} & = & \sin\theta\cos\phi\frac{\partial}{\partial x}+\sin\theta\sin\phi\frac{\partial}{\partial y}+\cos\theta\frac{\partial}{\partial z}=\frac{x}{r}\frac{\partial}{\partial x}+\frac{y}{r}\frac{\partial}{\partial y}+\frac{z}{r}\frac{\partial}{\partial z},\\ \frac{\partial}{\partial\theta} & = & r\cos\theta\cos\phi\frac{\partial}{\partial x}+r\cos\theta\sin\phi\frac{\partial}{\partial y}-\sin\theta\frac{\partial}{\partial z}=\frac{zx}{\sqrt{r^{2}-z^{2}}}\frac{\partial}{\partial x}+\frac{zy}{\sqrt{r^{2}-z^{2}}}\frac{\partial}{\partial y}-\frac{\sqrt{r^{2}-z^{2}}}{r}\frac{\partial}{\partial z}\\ \frac{\partial}{\partial\phi} & = & -r\sin\theta\sin\phi\frac{\partial}{\partial x}+r\sin\theta\cos\phi\frac{\partial}{\partial y}=-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}. \end{array},$

In principle we could use these identities to express the Laplacian ${\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}}}$ in terms of ${r,\theta,\phi}$ by first inverting the formulae to obtain ${\frac{\partial}{\partial x}}$, ${\frac{\partial}{\partial y}}$, ${\frac{\partial}{\partial z}}$, in terms of ${\frac{\partial}{\partial r}}$, ${\frac{\partial}{\partial\theta}}$, ${\frac{\partial}{\partial\phi}}$ etc. and then squaring. However, this computation is boring and not very enlightening. There is a more geometric approach.

To obtain this approach it is useful to think first about the gradient ${ }$operator ${\nabla.}$ We are used to thinking of the gradient ${\nabla f}$ of a function as the vector field

$\displaystyle \nabla f=\frac{\partial f}{\partial x}\widehat{\mathbf{e}}_{x}+\frac{\partial f}{\partial y}\widehat{\mathbf{e}}_{y}+\frac{\partial f}{\partial z}\widehat{\mathbf{e}}_{z}$

where ${\widehat{\mathbf{e}}_{x}}$, ${\widehat{\mathbf{e}}_{y}}$, ${\widehat{\mathbf{e}}_{z}}$ are unit vectors in the directions of the positive ${x}$, ${y}$ and ${z}$ axes respectively. However, it is also useful to think of ${\nabla f}$ somewhat differently. Specifically given ${\mathbf{x}}$ and ${\mathbf{v}}$ in ${\mathbb{R}^{3}}$, ${\nabla f\left(\mathbf{x}\right)\cdot\mathbf{v}}$ gives the directional derivative of ${f}$ at ${\mathbf{x}}$ in direction ${\mathbf{v}}$, i.e.,

$\displaystyle \nabla f\left(\mathbf{x}\right)\cdot\mathbf{v}=\left.\frac{df(\mathbf{x}+t\mathbf{v})}{dt}\right|_{t=0}. \ \ \ \ \ (2)$

We will now express the gradient operation using a moving frame that is more suited to the spherical coordinates. Namely

$\displaystyle \begin{array}{rcl} \widehat{\mathbf{r}} & =\frac{\partial\mathbf{x}}{\partial r} & =\sin\theta\cos\phi\widehat{\mathbf{e}}_{x}+\sin\theta\sin\phi\widehat{\mathbf{e}}_{y}+\cos\theta\widehat{\mathbf{e}}_{z},\\ \widehat{\mathbf{\theta}} & =\frac{1}{r}\frac{\partial\mathbf{x}}{\partial\theta} & =\cos\theta\cos\phi\widehat{\mathbf{e}}_{x}+\cos\theta\sin\phi\widehat{\mathbf{e}}_{y}-\sin\theta\widehat{\mathbf{e}}_{z},\\ \widehat{\mathbf{\phi}} & =\frac{1}{r\sin\theta}\frac{\partial\mathbf{x}}{\partial\phi} & =-\sin\phi\widehat{\mathbf{e}}_{x}+\cos\phi\widehat{\mathbf{e}}_{y}. \end{array} \ \ \ \ \ (3)$

Note that

$\displaystyle \left|\widehat{\mathbf{r}}\right|=\left|\widehat{\mathbf{\theta}}\right|=\left|\widehat{\mathbf{\phi}}\right|=1,$

while

$\displaystyle \widehat{\mathbf{r}}\cdot\widehat{\mathbf{\theta}}=\widehat{\mathbf{r}}\cdot\widehat{\mathbf{\phi}}=\widehat{\theta}\cdot\widehat{\mathbf{\phi}}=0.$

That is ${\widehat{\mathbf{r}}}$, ${\widehat{\mathbf{\theta}}}$, ${\widehat{\mathbf{\phi}}}$ is an orthonormal frame. It is a moving frame because the choice of frame depends on our position in configuration space.

Combining eqs. (2) and (3) we see that

$\displaystyle \begin{array}{rcl} \nabla f\left(\mathbf{x}\right)\cdot\widehat{\mathbf{r}} & = & \frac{\partial f\left(\mathbf{x}\right)}{\partial r},\\ \nabla f\left(\mathbf{x}\right)\cdot\widehat{\mathbf{\theta}} & = & \frac{1}{r}\frac{\partial f\left(\mathbf{x}\right)}{\partial\theta},\\ \nabla f\left(\mathbf{x}\right)\cdot\widehat{\mathbf{\phi}} & = & \frac{1}{r\sin\theta}\frac{\partial f\left(\mathbf{x}\right)}{\partial\phi}. \end{array}$

Since ${\widehat{\mathbf{r}}}$, ${\widehat{\mathbf{\theta}}}$, ${\widehat{\mathbf{\phi}}}$ is an orthonormal frame, it follows that

$\displaystyle \nabla f\left(\mathbf{x}\right)=\widehat{\mathbf{r}}\frac{\partial f\left(\mathbf{x}\right)}{\partial r}+\widehat{\mathbf{\theta}}\frac{1}{r}\frac{\partial f\left(\mathbf{x}\right)}{\partial\theta}+\widehat{\mathbf{\phi}}\frac{1}{r\sin\theta}\frac{\partial f\left(\mathbf{x}\right)}{\partial\phi}.$

In other words, we can express the gradient operator as follows

$\displaystyle \nabla=\widehat{\mathbf{r}}\frac{\partial}{\partial r}+\widehat{\mathbf{\theta}}\frac{1}{r}\frac{\partial}{\partial\theta}+\widehat{\mathbf{\phi}}\frac{1}{r\sin\theta}\frac{\partial}{\partial\phi}.$

Now to find the Laplacian we need only express ${\nabla\cdot\nabla}$ as

$\displaystyle \nabla\cdot\nabla=\left(\widehat{\mathbf{r}}\frac{\partial}{\partial r}+\widehat{\mathbf{\theta}}\frac{1}{r}\frac{\partial}{\partial\theta}+\widehat{\mathbf{\phi}}\frac{1}{r\sin\theta}\frac{\partial}{\partial\phi}\right)\cdot\left(\widehat{\mathbf{r}}\frac{\partial}{\partial r}+\widehat{\mathbf{\theta}}\frac{1}{r}\frac{\partial}{\partial\theta}+\widehat{\mathbf{\phi}}\frac{1}{r\sin\theta}\frac{\partial}{\partial\phi}\right). \ \ \ \ \ (4)$

However in expanding this expression we need to be mindful of the fact that ${\widehat{\mathbf{r}}}$, ${\widehat{\mathbf{\theta}}}$, ${\widehat{\mathbf{\phi}}}$ is a moving frame and thus certain derivatives of these vectors are non-zero:

$\displaystyle \begin{array}{ccc} \frac{\partial\widehat{\mathbf{r}}}{\partial r}=0, & \frac{\partial\widehat{\theta}}{\partial r}=0, & \frac{\partial\widehat{\mathbf{\phi}}}{\partial r}=0,\\ \frac{\partial\widehat{\mathbf{r}}}{\partial\theta}=\widehat{\mathbf{\theta}}, & \frac{\partial\widehat{\mathbf{\theta}}}{\partial\theta}=-\widehat{\mathbf{r}}, & \frac{\partial\widehat{\mathbf{\phi}}}{\partial\theta}=0,\\ \frac{\partial\widehat{\mathbf{r}}}{\partial\phi}=\sin\theta\widehat{\mathbf{\phi}}, & \frac{\partial\widehat{\mathbf{\theta}}}{\partial\phi}=\cos\theta\widehat{\mathbf{\phi}}, & \frac{\partial\widehat{\mathbf{\theta}}}{\partial\phi}=\cos\theta\widehat{\mathbf{\phi}}. \end{array}$

Consider, for example, the term ${\widehat{\mathbf{\theta}}\frac{1}{r}\frac{\partial}{\partial\theta}\cdot\widehat{\mathbf{r}}\frac{\partial}{\partial r}.}$ When this acts on a function ${f\left(\mathbf{x}\right)}$ we have, by the product rule,

$\displaystyle \widehat{\mathbf{\theta}}\frac{1}{r}\frac{\partial}{\partial\theta}\cdot\widehat{\mathbf{r}}\frac{\partial}{\partial r}f\left(\mathbf{x}\right)=\widehat{\theta}\cdot\widehat{\mathbf{r}}\frac{1}{r}\frac{\partial}{\partial\theta}\frac{\partial}{\partial r}f\left(\mathbf{x}\right)+\widehat{\theta}\cdot\frac{\partial\widehat{\mathbf{r}}}{\partial\theta}\frac{1}{r}\frac{\partial}{\partial r}f\left(\mathbf{x}\right)=\frac{1}{r}\frac{\partial}{\partial r}f\left(\mathbf{x}\right)$

since ${\widehat{\theta}\cdot\widehat{\mathbf{r}}=0}$ and ${\frac{\partial\widehat{\mathbf{r}}}{\partial\theta}=\widehat{\mathbf{\theta}}}$ . Thus

$\displaystyle \widehat{\mathbf{\theta}}\frac{1}{r}\frac{\partial}{\partial\theta}\cdot\widehat{\mathbf{r}}\frac{\partial}{\partial r}=\frac{1}{r}\frac{\partial}{\partial r}.$

When expanded fully, the right hand side to (4) has nine terms:

$\displaystyle \begin{array}{rcl} \widehat{\mathbf{r}}\frac{\partial}{\partial r}\cdot\widehat{\mathbf{r}}\frac{\partial}{\partial r} & = & \frac{\partial^{2}}{\partial r^{2}},\\ \widehat{\mathbf{r}}\frac{\partial}{\partial r}\cdot\widehat{\mathbf{\theta}}\frac{1}{r}\frac{\partial}{\partial\theta} & = & 0,\\ \widehat{\mathbf{r}}\frac{\partial}{\partial r}\cdot\widehat{\mathbf{\phi}}\frac{1}{r\sin\theta}\frac{\partial}{\partial\phi} & = & 0,\\ \widehat{\mathbf{\theta}}\frac{1}{r}\frac{\partial}{\partial\theta}\cdot\widehat{\mathbf{r}}\frac{\partial}{\partial r} & = & \frac{1}{r}\frac{\partial}{\partial r},\\ \widehat{\mathbf{\theta}}\frac{1}{r}\frac{\partial}{\partial\theta}\cdot\widehat{\mathbf{\theta}}\frac{1}{r}\frac{\partial}{\partial\theta} & = & \frac{1}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}},\\ \widehat{\mathbf{\theta}}\frac{1}{r}\frac{\partial}{\partial\theta}\cdot\widehat{\mathbf{\phi}}\frac{1}{r\sin\theta}\frac{\partial}{\partial\phi} & = & 0,\\ \widehat{\mathbf{\phi}}\frac{1}{r\sin\theta}\frac{\partial}{\partial\phi}\cdot\widehat{\mathbf{r}}\frac{\partial}{\partial r} & = & \frac{1}{r}\frac{\partial}{\partial r},\\ \widehat{\mathbf{\phi}}\frac{1}{r\sin\theta}\frac{\partial}{\partial\phi}\cdot\widehat{\mathbf{\theta}}\frac{1}{r}\frac{\partial}{\partial\theta} & = & \frac{\cos\theta}{r^{2}\sin\theta}\frac{\partial}{\partial\theta},\\ \widehat{\mathbf{\phi}}\frac{1}{r\sin\theta}\frac{\partial}{\partial\phi}\cdot\widehat{\mathbf{\phi}}\frac{1}{r\sin\theta}\frac{\partial}{\partial\phi} & = & \frac{1}{r^{2}\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}. \end{array}$

$\displaystyle \Delta=\nabla\cdot\nabla=\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}}+\frac{\cos\theta}{r^{2}\sin\theta}\frac{\partial}{\partial\theta}+\frac{1}{r^{2}\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}.$
Observing that ${\frac{\partial r^{2}}{\partial r}=2r}$ and ${\frac{\partial\sin\theta}{\partial\theta}=\cos\theta}$ this can be re-expressed as
$\displaystyle \Delta=\frac{1}{r^{2}}\frac{\partial}{\partial r}r^{2}\frac{\partial}{\partial r}+\frac{1}{r^{2}\sin\theta}\frac{\partial}{\partial\theta}\sin\theta\frac{\partial}{\partial\theta}+\frac{1}{r^{2}\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}.$