Classical Electrodynamics · Problem 1.2

Boundary-Value Problems in Electrostatics: I — John David Jackson

Physics — E&M · dirac-delta-function-orthogonal-coordinates

Problem

The Dirac delta function in three dimensions can be taken as the improper limit as α→0\alpha \to 0 of the Gaussian function D(α;x,y,z)=(2π)−3/2α−3exp⁡[−12α2(x2+y2+z2)]D(\alpha; x, y, z) = (2\pi)^{-3/2}\alpha^{-3} \exp\left[-\frac{1}{2\alpha^2}(x^2 + y^2 + z^2)\right] Consider a general orthogonal coordinate system specified by the surfaces u=u = constant, v=v = constant, w=w = constant, with length elements du/Udu/U, dv/Vdv/V, dw/Wdw/W in the three perpendicular directions. Show that δ(x−x′)=δ(u−u′) δ(v−v′) δ(w−w′)⋅UVW\delta(\mathbf{x} - \mathbf{x}') = \delta(u - u')\, \delta(v - v')\, \delta(w - w') \cdot UVW by considering the limit of the Gaussian above. Note that as α→0\alpha \to 0 only the infinitesimal length element need be used for the distance between the points in the exponent.

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