Helmholtz decomposition
Derivation

Suppose we have a vector function F (r) {\displaystyle \mathbf {F} (\mathbf {r})} of which we know the curl, ∇ × F {\displaystyle \nabla \times \mathbf {F} }, and the divergence, ∇ ⋅ F {\displaystyle \nabla \cdot \mathbf {F} }, in the domain and the fields on the boundary. Writing the function using delta function in the form δ 3 (r − r ′) = − 1 4 π ∇ 2 1 | r − r ′ |, {\displaystyle \delta ^{3}(\mathbf {r} -\mathbf {r} ')=-{\frac {1}{4\pi }}\nabla ^{2}{\frac {1}{|\mathbf {r} -\mathbf {r} '|}}\,,} where ∇ 2:= ∇ ⋅ ∇ {\displaystyle \nabla ^{2}:=\nabla \cdot \nabla } is the Laplace operator, we have

F (r) = ∫ V F (r ′) δ 3 (r − r ′) d V ′ = ∫ V F (r ′) (− 1 4 π ∇ 2 1 | r − r ′ |) d V ′ = − 1 4 π ∇ 2 ∫ V F (r ′) | r − r ′ | d V ′ = − 1 4 π = − 1 4 π = − 1 4 π {\displaystyle {\begin{aligned}\mathbf {F} (\mathbf {r})&=\int _{V}\mathbf {F} \left(\mathbf {r} '\right)\delta ^{3}(\mathbf {r} -\mathbf {r} ')\mathrm {d} V'\\&=\int _{V}\mathbf {F} (\mathbf {r} ')\left(-{\frac {1}{4\pi }}\nabla ^{2}{\frac {1}{\left|\mathbf {r} -\mathbf {r} '\right|}}\right)\mathrm {d} V'\\&=-{\frac {1}{4\pi }}\nabla ^{2}\int _{V}{\frac {\mathbf {F} (\mathbf {r} ')}{\left|\mathbf {r} -\mathbf {r} '\right|}}\mathrm {d} V'\\&=-{\frac {1}{4\pi }}\left\\&=-{\frac {1}{4\pi }}\left\\&=-{\frac {1}{4\pi }}\left\end{aligned}}}

where we have used the definition of the vector Laplacian: ∇ 2 a = ∇ (∇ ⋅ a) − ∇ × (∇ × a) , {\displaystyle \nabla ^{2}\mathbf {a} =\nabla (\nabla \cdot \mathbf {a})-\nabla \times (\nabla \times \mathbf {a})\,}

differentiation/integration with respect to r ′ {\displaystyle \mathbf {r} '} by ∇ ′ / d V ′, {\displaystyle \nabla '/\mathrm {d} V',} and in the last line, linearity of function arguments: ∇ 1 | r − r ′ | = − ∇ ′ 1 | r − r ′ |. {\displaystyle \nabla {\frac {1}{\left|\mathbf {r} -\mathbf {r} '\right|}}=-\nabla '{\frac {1}{\left|\mathbf {r} -\mathbf {r} '\right|}}\.}

Then using the vectorial identities

a ⋅ ∇ ψ = − ψ (∇ ⋅ a) + ∇ ⋅ (ψ a) a × ∇ ψ = ψ (∇ × a) − ∇ × (ψ a) {\displaystyle {\begin{aligned}\mathbf {a} \cdot \nabla \psi &=-\psi (\nabla \cdot \mathbf {a})+\nabla \cdot (\psi \mathbf {a})\\\mathbf {a} \times \nabla \psi &=\psi (\nabla \times \mathbf {a})-\nabla \times (\psi \mathbf {a})\end{aligned}}}

we get F (r) = − 1 4 π. {\displaystyle {\begin{aligned}\mathbf {F} (\mathbf {r})=-{\frac {1}{4\pi }}{\bigg }.\end{aligned}}}

Thanks to the divergence theorem the equation can be rewritten as

F (r) = − 1 4 π = − ∇ + ∇ × {\displaystyle {\begin{aligned}\mathbf {F} (\mathbf {r})&=-{\frac {1}{4\pi }}{\bigg }\\&=-\nabla \left\\&\quad +\nabla \times \left\end{aligned}}}

with outward surface normal n ^ ′ {\displaystyle \mathbf {\hat {n}} '}.

Defining

Φ (r) ≡ 1 4 π ∫ V ∇ ′ ⋅ F (r ′) | r − r ′ | d V ′ − 1 4 π ∮ S n ^ ′ ⋅ F (r ′) | r − r ′ | d S ′ {\displaystyle \Phi (\mathbf {r})\equiv {\frac {1}{4\pi }}\int _{V}{\frac {\nabla '\cdot \mathbf {F} \left(\mathbf {r} '\right)}{\left|\mathbf {r} -\mathbf {r} '\right|}}\mathrm {d} V'-{\frac {1}{4\pi }}\oint _{S}\mathbf {\hat {n}} '\cdot {\frac {\mathbf {F} \left(\mathbf {r} '\right)}{\left|\mathbf {r} -\mathbf {r} '\right|}}\mathrm {d} S'} A (r) ≡ 1 4 π ∫ V ∇ ′ × F (r ′) | r − r ′ | d V ′ − 1 4 π ∮ S n ^ ′ × F (r ′) | r − r ′ | d S ′ {\displaystyle \mathbf {A} (\mathbf {r})\equiv {\frac {1}{4\pi }}\int _{V}{\frac {\nabla '\times \mathbf {F} \left(\mathbf {r} '\right)}{\left|\mathbf {r} -\mathbf {r} '\right|}}\mathrm {d} V'-{\frac {1}{4\pi }}\oint _{S}\mathbf {\hat {n}} '\times {\frac {\mathbf {F} \left(\mathbf {r} '\right)}{\left|\mathbf {r} -\mathbf {r} '\right|}}\mathrm {d} S'}

we finally obtain F = − ∇ Φ + ∇ × A. {\displaystyle \mathbf {F} =-\nabla \Phi +\nabla \times \mathbf {A}.}
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