ジェフィメンコ方程式
d ρ ( s , t − ) = ∇ ′ ρ ⋅ d s + ∂ ρ ∂ t − d t − = ∇ ′ ρ ⋅ d s + ∂ ρ ∂ t − ( ∂ t − ∂ t d t + ∂ t − ∂ R d R ) = ∇ ′ ρ ⋅ d s + ∂ ρ ∂ t − ( d t − 1 c d R ) = ∇ ′ ρ ⋅ d s + ∂ ρ ∂ t − [ d t − 1 c ( ∇ R ⋅ d r + ∇ ′ R ⋅ d s ) ] {\displaystyle {\begin{aligned}\mathrm {d} \rho ({\boldsymbol {s}},\,{t}_{-})&=\nabla '\rho \cdot \mathrm {d} {\boldsymbol {s}}+{\frac {\partial \rho }{\partial {t}_{-}}}\mathrm {d} {t}_{-}\\&=\nabla '\rho \cdot \mathrm {d} {\boldsymbol {s}}+{\frac {\partial \rho }{\partial {t}_{-}}}\left({\frac {\partial {t}_{-}}{\partial t}}\mathrm {d} t+{\frac {\partial {t}_{-}}{\partial {\mathfrak {R}}}}\mathrm {d} {\mathfrak {R}}\right)\\&=\nabla '\rho \cdot \mathrm {d} {\boldsymbol {s}}+{\frac {\partial \rho }{\partial {t}_{-}}}\left(\mathrm {d} t-{\frac {1}{c}}\mathrm {d} {\mathfrak {R}}\right)\\&=\nabla '\rho \cdot \mathrm {d} {\boldsymbol {s}}+{\frac {\partial \rho }{\partial {t}_{-}}}\left[\mathrm {d} t-{\frac {1}{c}}(\nabla {\mathfrak {R}}\cdot \mathrm {d} {\boldsymbol {r}}+\nabla '{\mathfrak {R}}\cdot \mathrm {d} {\boldsymbol {s}})\right]\\\end{aligned}}\,\!}
で与えられる。ここで、 ∇ ′ {\displaystyle \nabla '} は、 s {\displaystyle {\boldsymbol {s}}} に関する勾配微分作用素を意味する。今、
となる。ここで、
ρ ̇ ( s , t − ) = ∂ ρ ( s , t ) ∂ t − {\displaystyle {\dot {\rho }}({\boldsymbol {s}},\,{t}_{-})={\frac {\partial \rho ({\boldsymbol {s}},\,t)}{\partial {t}_{-}}}\,\!}
を意味する。また、
∇ R ≡ R ^ {\displaystyle \nabla {\mathfrak {R}}\equiv {\hat {\boldsymbol {\mathfrak {R}}}}}
とする。
以上から、 φ {\displaystyle \phi } の勾配ベクトル場は、
∇ φ ( r , t ) = 1 4 π ε 0 ∫ V [ − ρ ̇ ( s , t − ) c R ^ R − ρ ( s , t − ) ( R ^ R 2 ) ] d 3 s {\displaystyle \nabla \phi ({\boldsymbol {r}},\,t)={\frac {1}{4\pi \epsilon _{0}}}\int _{\boldsymbol {V}}\left[-{\frac {{\dot {\rho }}({\boldsymbol {s}},\,{t}_{-})}{c}}{\frac {\hat {\boldsymbol {\mathfrak {R}}}}{\mathfrak {R}}}-\rho ({\boldsymbol {s}},\,{t}_{-})\left({\frac {\hat {\boldsymbol {\mathfrak {R}}}}{{\mathfrak {R}}^{2}}}\right)\right]\,\mathrm {d} ^{3}{\boldsymbol {s}}\,\!}