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Three-Dimensional Dirac Delta Function

The three-dimensional Dirac delta function, $ \delta({\bf r}-{\bf r}')$ , has the property

$\displaystyle \delta({\bf r}-{\bf r}') =0$   $\displaystyle \mbox{\hspace{0.5cm}for\hspace{0.5cm}${\bf r}\neq {\bf r}'$}$$\displaystyle .$ (21)

In addition, however, the function is singular at $ {\bf r}={\bf r}'$ in such a manner that

$\displaystyle \int_V \delta({\bf r}-{\bf r}') \,dV = 1.$ (22)

Here, $ V$ is any volume that contains the point $ {\bf r}={\bf r}'$ . (Also, $ dV$ is an element of $ V$ expressed in terms of the components of $ {\bf r}$ , but independent of the components of $ {\bf r}'$ .) It follows that

$\displaystyle \int_V f({\bf r})\,\delta({\bf r}-{\bf r}')\,dV = f({\bf r}'),$ (23)

where $ f{\bf (r})$ is an arbitrary function that is well behaved at $ {\bf r}={\bf r}'$ . It is also easy to see that

$\displaystyle \delta({\bf r}'-{\bf r}) = \delta({\bf r}-{\bf r}').$ (24)

We can show that

$\displaystyle \nabla^{\,2}\left(\frac{1}{\vert{\bf r}-{\bf r}'\vert}\right)=-4\pi\,\delta({\bf r}-{\bf r}').$ (25)

(Here, $ \nabla^{\,2}$ is a Laplacian operator expressed in terms of the components of $ {\bf r}$ , but independent of the components of $ {\bf r}'$ .) We must first prove that

$\displaystyle \nabla^{\,2}\left(\frac{1}{\vert{\bf r}-{\bf r}'\vert}\right)=0$$\displaystyle \mbox{\hspace{0.5cm}for\hspace{0.5cm}${\bf r}\neq {\bf r}'$}$$\displaystyle ,$ (26)

in accordance with Equation (21). If $ R=\vert{\bf r}-{\bf r}'\vert$ then this is equivalent to showing that

$\displaystyle \frac{1}{R^{\,2}}\,\frac{d}{dR}\!\left[R^{\,2}\,\frac{d}{dR}\!\left(\frac{1}{R}\right)\right]=0$ (27)

for $ R>0$ , which is indeed the case. (Here, $ R$ is treated as a radial spherical coordinate.) Next, we must show that

$\displaystyle \int_V \nabla^{\,2}\left(\frac{1}{\vert{\bf r}-{\bf r}'\vert}\right)dV =- 4\pi,$ (28)

in accordance with Equations (22) and (25). Suppose that $ S$ is a spherical surface, of radius $ R$ , centered on $ {\bf r}={\bf r}'$ . Making use of the definition $ \nabla^{\,2}\phi \equiv \nabla\cdot\nabla\phi$ , as well as the divergence theorem, we can write

$\displaystyle \int_V \nabla^{\,2}\left(\frac{1}{\vert{\bf r}-{\bf r}'\vert}\right)dV$ $\displaystyle = \int_V \nabla\cdot\nabla\left(\frac{1}{\vert{\bf r}-{\bf r}'\ve...
... = \int_S \nabla\left(\frac{1}{\vert{\bf r}-{\bf r}'\vert}\right)\cdot d{\bf S}$    
  $\displaystyle = 4\pi\,R^{\,2}\,\frac{d}{dR}\!\left(\frac{1}{R}\right) = -4\pi.$ (29)

(Here, $ \nabla$ is a gradient operator expressed in terms of the components of $ {\bf r}$ , but independent of the components of $ {\bf r}'$ . Likewise, $ d{\bf S}$ is a surface element involving the components of $ {\bf r}$ , but independent of the components of $ {\bf r}'$ .) Finally, if $ S$ is deformed into a general surface (without crossing the point $ {\bf r}={\bf r}'$ ) then the value of the volume integral is unchanged, as a consequence of Equation (26). Hence, we have demonstrated the validity of Equation (25).


next up previous
Next: Solution of Inhomogeneous Wave Up: Maxwell's Equations Previous: Dirac Delta Function
Richard Fitzpatrick 2014-06-27