Jackson’s Laplacian in spherical Coordinates [Proof]

If you took a look at one of the previous posts on how to remember the Laplacian in different forms by using a metric,  you will notice that the form of  the Laplacian that we get is:

\nabla^2 \psi = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial \psi}{\partial r} \right) + \frac{1}{r^2 \sin(\theta)} \frac{\partial}{\partial \theta} \left( sin(\theta)  \frac{\partial \psi}{\partial \theta} \right)   + \frac{1}{r^2 \sin^2(\theta)} \frac{\partial^2 \psi}{\partial \phi^2}   

But in Jackson’s Classical Electrodynamics, III edition he notes the following:


This is an interesting form of the Laplacian that perhaps not everyone has encountered. This can obtained from the known form by making the substitution u = r \psi and simplifying. The steps to which have been outlined below:





Prof.Ghrist at his best!

Screenshot from 2017-07-28 10:55:27

To understand why this is true, we must start with the Fundamental Theorem of Vector calculus. If F  is a conservative field ( i.e F = \nabla \phi  ), then

\int\limits_{A}^{B} F.dr = \int\limits_{A}^{B} \nabla\phi .dr = \phi_{A} - \phi_{B}

What this means is that the value is dependent only on the initial and final positions. The path that you take to get from A to B is not important.

Screenshot from 2017-07-28 11:29:46

Now if the path of integration is a closed loop, then points A and B are the same, and therefore:

\int\limits_{A}^{A} F.dr = \int\limits_{A}^{A} \nabla\phi .dr = \phi_{1} - \phi_{1} = 0

Now that we are clear about this, according to Stokes theorem the same integral for a closed region can be represented in another form:

\int_{C} F.dr = \int\int_{A} (\nabla X F) .\vec{n} dA  = 0

From this we get that Curl = \nabla X F = 0 for a conservative field (i.e F = \nabla \phi ). Therefore when a conservative field is operated on by a curl operator (\nabla X ), it yields 0.

Bravo Prof.Ghrist! Beautifully said 😀


Solving the Laplacian in Spherical Coordinates (#1)

In this post, let’s derive a general solution for the Laplacian in Spherical Coordinates. In future posts, we shall look at the application of this equation in the context of Fluids and Quantum Mechanics.


x = rsin\theta cos\phi
y = rsin\theta cos\phi
z = rcos\theta


0 \leq r < \infty
0 \leq \theta \leq \pi
0 \leq \phi < 2\pi

The Laplacian in Spherical coordinates in its ultimate glory is written as follows:

\nabla ^{2}f ={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial f}{\partial r}}\right)+{\frac {1}{r^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial f}{\partial \theta }}\right)+{\frac {1}{r^{2}\sin ^{2}\theta }}{\frac {\partial ^{2}f}{\partial \phi ^{2}}} = 0

To solve it we use the method of separation of variables.

f = R(r)\Theta(\theta)\Phi(\phi)

Plugging in the value of f into the Laplacian, we get that :

\frac{\Theta \Phi}{r^2} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right) + \frac{R \Phi}{r^2 sin \theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{d\theta} \right) + \frac{\Theta R}{r^2 sin^2 \theta} \frac{d^2 \Phi}{d\phi^2} = 0

Dividing throughout by R\Theta\Phi and multiplying throughout by r^2, further simplifies into:

\underbrace{ \frac{1}{R} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right)}_{h(r)} + \underbrace{\frac{1}{\Theta sin \theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{d\theta} \right) + \frac{1}{\Phi sin^2 \theta} \frac{d^2 \Phi}{d\phi^2}}_{g(\theta,\phi)} = 0

It can be observed that the first expression in the differential equation is merely a function of r and the remaining a function of \theta and \phi only. Therefore, we equate the first expression to be \lambda = l(l+1) and the second to be -\lambda = -l(l+1). The reason for choosing the peculiar value of l(l+1) is explained in another post.

\underbrace{ \frac{1}{R} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right)}_{l(l+1)} + \underbrace{\frac{1}{\Theta sin \theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{d\theta} \right) + \frac{1}{\Phi sin^2 \theta} \frac{d^2 \Phi}{d\phi^2}}_{-l(l+1)} = 0 (1)


The first expression in (1) the Euler-Cauchy equation in r.

\frac{d}{dr} \left( r^2\frac{dR}{dr} \right) = l(l+1)R

The general solution of this has been in discussed in a previous post and it can be written as:

R(r) = C_1 r^l + \frac{C_2}{r^{l+1}}


The second expression in (1) takes the form as follows:

\frac{sin \theta}{\Theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ l(l+1)sin^2 \theta + \frac{1}{\Phi} \frac{d^2 \Phi}{d\phi^2} = 0

The following observation can be made similar to the previous analysis

\underbrace{\frac{sin \theta}{\Theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ l(l+1)sin^2 \theta }_{m^2} + \underbrace{\frac{1}{\Phi} \frac{d^2 \Phi}{d\phi^2}}_{-m^2} = 0 (2)


The first expression in the above equation (2) is the Associated Legendre Differential equation.

\frac{sin \theta}{\Theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ l(l+1)sin^2 \theta = m^2

sin \theta \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ \Theta \left( l(l+1)sin^2 \theta - m^2 \right) = 0

The general solution to this differential equation can be given as:
\Theta(\theta) = C_3 P_l^m(cos\theta) + C_4 Q_l^m(cos\theta)


The solution to the second term in the equation (2) is a trivial one:

\frac{d^2 \Phi}{d\phi^2} = m^2 \Phi
\Phi(\phi) = C_5 e^{im\phi} + C_6 e^{-im\phi}


Therefore the general solution to the Laplacian in Spherical coordinates is given by:

R\Theta\Phi = \left(C_1 r^l + \frac{C_2}{r^{l+1}} \right) \left(C_3 P_l^m(cos\theta) + C_4 Q_l^m(cos\theta \right) \left(C_5 e^{im\phi} + C_6 e^{-im\phi}\right)