information:classification, information:extraction
Derivation from the Fourier transform

Note that in the theorem stated here, we have imposed the condition that if F {\displaystyle \mathbf {F} } is not defined on a bounded domain, then F {\displaystyle \mathbf {F} } shall decay faster than 1 / r {\displaystyle 1/r}. Thus, the Fourier transform of F {\displaystyle \mathbf {F} }, denoted as G {\displaystyle \mathbf {G} }, is guaranteed to exist. We apply the convention F (r) = ∭ G (k) e i k ⋅ r d V k {\displaystyle \mathbf {F} (\mathbf {r})=\iiint \mathbf {G} (\mathbf {k})e^{i\mathbf {k} \cdot \mathbf {r} }dV_{k}}

The Fourier transform of a scalar field is a scalar field, and the Fourier transform of a vector field is a vector field of same dimension.

Now consider the following scalar and vector fields: G Φ (k) = i k ⋅ G (k) ‖ k ‖ 2 G A (k) = i k × G (k) ‖ k ‖ 2 Φ (r) = ∭ G Φ (k) e i k ⋅ r d V k A (r) = ∭ G A (k) e i k ⋅ r d V k {\displaystyle {\begin{aligned}G_{\Phi }(\mathbf {k})&=i{\frac {\mathbf {k} \cdot \mathbf {G} (\mathbf {k})}{\|\mathbf {k} \|^{2}}}\\\mathbf {G} _{\mathbf {A} }(\mathbf {k})&=i{\frac {\mathbf {k} \times \mathbf {G} (\mathbf {k})}{\|\mathbf {k} \|^{2}}}\\\Phi (\mathbf {r})&=\iiint G_{\Phi }(\mathbf {k})e^{i\mathbf {k} \cdot \mathbf {r} }dV_{k}\\\mathbf {A} (\mathbf {r})&=\iiint \mathbf {G} _{\mathbf {A} }(\mathbf {k})e^{i\mathbf {k} \cdot \mathbf {r} }dV_{k}\end{aligned}}}

Hence

G (k) = − i k G Φ (k) + i k × G A (k) F (r) = − ∭ i k G Φ (k) e i k ⋅ r d V k + ∭ i k × G A (k) e i k ⋅ r d V k = − ∇ Φ (r) + ∇ × A (r) {\displaystyle {\begin{aligned}\mathbf {G} (\mathbf {k})&=-i\mathbf {k} G_{\Phi }(\mathbf {k})+i\mathbf {k} \times \mathbf {G} _{\mathbf {A} }(\mathbf {k})\\\mathbf {F} (\mathbf {r})&=-\iiint i\mathbf {k} G_{\Phi }(\mathbf {k})e^{i\mathbf {k} \cdot \mathbf {r} }dV_{k}+\iiint i\mathbf {k} \times \mathbf {G} _{\mathbf {A} }(\mathbf {k})e^{i\mathbf {k} \cdot \mathbf {r} }dV_{k}\\&=-\nabla \Phi (\mathbf {r})+\nabla \times \mathbf {A} (\mathbf {r})\end{aligned}}}