Deriving the Helmholtz Theorem
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
- bring the derivative operator with respect to r into the integral
- switch the derivative from r to r’ with a cost of a minus sign
- integrate by parts
- 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.