To derive the Helmholtz theorem start first with one representation of the delta-function in 3-dimensions

\[ \nabla^2 \left( \frac{1}{|\vec r – \vec r \, {}’|} \right) = -4 \pi \delta( \vec r – \vec r \, {}’ ) \, .\]

Start with the identity
\[ \vec F(\vec r) = \int_{V} d^3 r’ \delta(\vec r – \vec r \,{}’) F(\vec r \,{}’) \]
for an arbitrary vector field $$\vec F(\vec r)$$ over a given volume $$V$$. Note that time will not be involved in this derivation. Also note that there is a ongoing discussion in the literature about the correct way to extend this theorem for time varying fields. This will be a discussed in a future post.

Using the explicit representation of the delta-function stated above and factoring out the derivatives with respect to the field point $$\vec r$$ yields

\[ \vec F(\vec r) = \frac{ -\nabla^2_{\vec r} }{4 \pi} \int_V d^3 r’ \frac{\vec F(\vec r\,{}’)}{|\vec r – \vec r\,{}’|} \, . \]

Now apply the vector identity $$\nabla^2 = \nabla( \nabla \cdot ) – \nabla \times (\nabla \times)$$. Doing so allows the expression for $$\vec F(\vec r)$$ to take the form
\[ \vec F (\vec r) = \frac{1}{4 \pi} \nabla_{\vec r} \times \vec I_{vector} – \frac{1}{4 \pi} \nabla_{\vec r} I_{scalar} \]
where the integrals
\[ \vec I_{vector} = \nabla_{\vec r} \times \int_v d^3 r ‘ \frac{\vec F (\vec r\,’)}{|\vec r – \vec r\,’|} \]
and
\[ I_{scalar} = \nabla_{\vec r} \cdot \int_v d^3 r ‘ \frac{\vec F (\vec r\,’)}{|\vec r – \vec r\,’|} \; . \]

The strategy for handling these terms is to

  1. bring the derivative operator with respect to r into the integral
  2. switch the derivative from r to r’ with a cost of a minus sign
  3. integrate by parts
  4. apply the appropriate boundary conditions and boundary integral version of the divergence theorem to the total derivative piece

Application of this strategy to the vector (first) integral gives
\[ \vec I _{vector} = \int_V d^3 r’ \frac{ \nabla_{\vec r\,’} \times \vec F (\vec r \, ‘)}{|\vec r – \vec r\,’|} – \int_{\partial V} dS \frac{\hat n \times \vec F(\vec r\,’)}{|\vec r – \vec r\,’|} \; .\]

Likewise, the same strategy applied to the scalar (second) integral gives
\[ I_{scalar} = \int_V d^3 r’ \frac{ \nabla \cdot \vec F ( \vec r \,’)}{|\vec r – \vec r\,’|} – \int_{\partial V} dS \frac{\hat n \cdot \vec F ( \vec r \,’)}{|\vec r – \vec r\,’|} \; . \]

Now the usual case of interest sets the bounding volume to be all space, which requires that the field drop off faster than $$r^{-1}$$. If this condition is met then the surface integrals zero and the original field can be written as
\[ \vec F (\vec r) = -\nabla U(\vec r) + \nabla \times \vec W(\vec r) \]
where
\[ U(\vec r) = \frac{1}{4\pi} \int_V d^3 r’ \frac{ \nabla’ \cdot \vec F ( \vec r \,’)}{|\vec r – \vec r\,’|} \]
and
\[ \vec W(\vec r) = \frac{1}{4\pi} \int_V d^3 r’ \frac{ \nabla’ \times \vec F ( \vec r \,’)}{|\vec r – \vec r\,’|} \; .\]

At this point it is a snap to derive Coulomb’s and Biot-Savart’s laws from the Maxwell equations but that is a post for another time.