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:

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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:

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Prof.Ghrist at his best!

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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 😀