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# Three-Dimensional Dirac Delta Function

The three-dimensional Dirac delta function, , has the property

 (21)

In addition, however, the function is singular at in such a manner that

 (22)

Here, is any volume that contains the point . (Also, is an element of expressed in terms of the components of , but independent of the components of .) It follows that

 (23)

where is an arbitrary function that is well behaved at . It is also easy to see that

 (24)

We can show that

 (25)

(Here, is a Laplacian operator expressed in terms of the components of , but independent of the components of .) We must first prove that

 (26)

in accordance with Equation (21). If then this is equivalent to showing that

 (27)

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

 (28)

in accordance with Equations (22) and (25). Suppose that is a spherical surface, of radius , centered on . Making use of the definition , as well as the divergence theorem, we can write

 (29)

(Here, is a gradient operator expressed in terms of the components of , but independent of the components of . Likewise, is a surface element involving the components of , but independent of the components of .) Finally, if is deformed into a general surface (without crossing the point ) 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: Solution of Inhomogeneous Wave Up: Maxwell's Equations Previous: Dirac Delta Function
Richard Fitzpatrick 2014-06-27