How to Convert Maxwell's Equations into Differential Form

Gauss' Law

Begin with Gauss' law in integral form. ∮ S E ⋅ d S = Q ϵ 0 {\displaystyle \oint _{S}\mathbf {E} \cdot \mathrm {d} \mathbf {S} ={\frac {Q}{\epsilon _{0}}}} \oint _{{S}}{\mathbf {E}}\cdot {\mathrm {d}}{\mathbf {S}}={\frac {Q}{\epsilon _{0}}}

Rewrite the right side in terms of a volume integral. ∮ S E ⋅ d S = ∫ V ρ ϵ 0 d V {\displaystyle \oint _{S}\mathbf {E} \cdot \mathrm {d} \mathbf {S} =\int _{V}{\frac {\rho }{\epsilon _{0}}}\mathrm {d} V} \oint _{{S}}{\mathbf {E}}\cdot {\mathrm {d}}{\mathbf {S}}=\int _{{V}}{\frac {\rho }{\epsilon _{0}}}{\mathrm {d}}V

Recall the divergence theorem. The divergence theorem says that the flux penetrating a closed surface S {\displaystyle S} S that bounds a volume V {\displaystyle V} V is equal to the divergence of the field F {\displaystyle \mathbf {F} } {\mathbf {F}} inside the volume. ∮ S F ⋅ d S = ∫ V ( ∇ ⋅ F ) d V {\displaystyle \oint _{S}\mathbf {F} \cdot \mathrm {d} \mathbf {S} =\int _{V}(\nabla \cdot \mathbf {F} )\mathrm {d} V} \oint _{{S}}{\mathbf {F}}\cdot {\mathrm {d}}{\mathbf {S}}=\int _{{V}}(\nabla \cdot {\mathbf {F}}){\mathrm {d}}V

Use the divergence theorem to rewrite the left side as a volume integral. ∫ V ( ∇ ⋅ E ) d V = ∫ V ρ ϵ 0 d V {\displaystyle \int _{V}(\nabla \cdot \mathbf {E} )\mathrm {d} V=\int _{V}{\frac {\rho }{\epsilon _{0}}}\mathrm {d} V} \int _{{V}}(\nabla \cdot {\mathbf {E}}){\mathrm {d}}V=\int _{{V}}{\frac {\rho }{\epsilon _{0}}}{\mathrm {d}}V

Set the equation to 0. ∫ V ( ∇ ⋅ E ) d V − ∫ V ρ ϵ 0 d V = 0 ∫ V ( ∇ ⋅ E − ρ ϵ 0 ) d V = 0 {\displaystyle {\begin{aligned}\int _{V}(\nabla \cdot \mathbf {E} )\mathrm {d} V-\int _{V}{\frac {\rho }{\epsilon _{0}}}\mathrm {d} V&=0\\\int _{V}\left(\nabla \cdot \mathbf {E} -{\frac {\rho }{\epsilon _{0}}}\right)\mathrm {d} V&=0\end{aligned}}} {\begin{aligned}\int _{{V}}(\nabla \cdot {\mathbf {E}}){\mathrm {d}}V-\int _{{V}}{\frac {\rho }{\epsilon _{0}}}{\mathrm {d}}V&=0\\\int _{{V}}\left(\nabla \cdot {\mathbf {E}}-{\frac {\rho }{\epsilon _{0}}}\right){\mathrm {d}}V&=0\end{aligned}}

Convert the equation to differential form. The above equation says that the integral of a quantity is 0. Because the only quantity for which the integral is 0, is 0 itself, the expression in the integrand can be set to 0. ∇ ⋅ E − ρ ϵ 0 = 0 {\displaystyle \nabla \cdot \mathbf {E} -{\frac {\rho }{\epsilon _{0}}}=0} \nabla \cdot {\mathbf {E}}-{\frac {\rho }{\epsilon _{0}}}=0 This leads to Gauss' law in differential form. ∇ ⋅ E = ρ ϵ 0 {\displaystyle \nabla \cdot \mathbf {E} ={\frac {\rho }{\epsilon _{0}}}} \nabla \cdot {\mathbf {E}}={\frac {\rho }{\epsilon _{0}}}

Gauss' Law for Magnetism

Begin with Gauss' law for magnetism in integral form. ∮ S B ⋅ d S = 0 {\displaystyle \oint _{S}\mathbf {B} \cdot \mathrm {d} \mathbf {S} =0} \oint _{{S}}{\mathbf {B}}\cdot {\mathrm {d}}{\mathbf {S}}=0

Invoke the divergence theorem. ∫ V ( ∇ ⋅ B ) d V = 0 {\displaystyle \int _{V}(\nabla \cdot \mathbf {B} )\mathrm {d} V=0} \int _{{V}}(\nabla \cdot {\mathbf {B}}){\mathrm {d}}V=0

Write the equation in differential form. As with Gauss' Law, the same argument used above yields our answer. ∇ ⋅ B = 0 {\displaystyle \nabla \cdot \mathbf {B} =0} \nabla \cdot {\mathbf {B}}=0

Faraday's Law

Begin with Faraday's law in integral form. ∮ C E ⋅ d l = − d d t ∫ S B ⋅ d S {\displaystyle \oint _{C}\mathbf {E} \cdot \mathrm {d} \mathbf {l} =-{\frac {\mathrm {d} }{\mathrm {d} t}}\int _{S}\mathbf {B} \cdot \mathrm {d} \mathbf {S} } \oint _{{C}}{\mathbf {E}}\cdot {\mathrm {d}}{\mathbf {l}}=-{\frac {{\mathrm {d}}}{{\mathrm {d}}t}}\int _{{S}}{\mathbf {B}}\cdot {\mathrm {d}}{\mathbf {S}}

Recall Stokes' theorem. Stokes' theorem says that the circulation of a field F {\displaystyle \mathbf {F} } {\mathbf {F}} around the loop C {\displaystyle C} C that bounds a surface S {\displaystyle S} S is equal to the flux of curl ⁡ F {\displaystyle \operatorname {curl} \mathbf {F} } \operatorname {curl}{\mathbf {F}} over S . {\displaystyle S.} S. ∮ C F ⋅ d l = ∫ S ( ∇ × F ) ⋅ d S {\displaystyle \oint _{C}\mathbf {F} \cdot \mathrm {d} \mathbf {l} =\int _{S}(\nabla \times \mathbf {F} )\cdot \mathrm {d} \mathbf {S} } \oint _{{C}}{\mathbf {F}}\cdot {\mathrm {d}}{\mathbf {l}}=\int _{{S}}(\nabla \times {\mathbf {F}})\cdot {\mathrm {d}}{\mathbf {S}}

Use Stokes' theorem to rewrite the left side as a surface integral. ∫ S ( ∇ × E ) ⋅ d S = − d d t ∫ S B ⋅ d S {\displaystyle \int _{S}(\nabla \times \mathbf {E} )\cdot \mathrm {d} \mathbf {S} =-{\frac {\mathrm {d} }{\mathrm {d} t}}\int _{S}\mathbf {B} \cdot \mathrm {d} \mathbf {S} } \int _{{S}}(\nabla \times {\mathbf {E}})\cdot {\mathrm {d}}{\mathbf {S}}=-{\frac {{\mathrm {d}}}{{\mathrm {d}}t}}\int _{{S}}{\mathbf {B}}\cdot {\mathrm {d}}{\mathbf {S}}

Set the equation to 0. ∫ S ( ∇ × E ) ⋅ d S + d d t ∫ S B ⋅ d S = 0 ∫ S ( ∇ × E + ∂ B ∂ t ) ⋅ d S = 0 {\displaystyle {\begin{aligned}\int _{S}(\nabla \times \mathbf {E} )\cdot \mathrm {d} \mathbf {S} +{\frac {\mathrm {d} }{\mathrm {d} t}}\int _{S}\mathbf {B} \cdot \mathrm {d} \mathbf {S} &=0\\\int _{S}\left(\nabla \times \mathbf {E} +{\frac {\partial \mathbf {B} }{\partial t}}\right)\cdot \mathrm {d} \mathbf {S} &=0\end{aligned}}} {\begin{aligned}\int _{{S}}(\nabla \times {\mathbf {E}})\cdot {\mathrm {d}}{\mathbf {S}}+{\frac {{\mathrm {d}}}{{\mathrm {d}}t}}\int _{{S}}{\mathbf {B}}\cdot {\mathrm {d}}{\mathbf {S}}&=0\\\int _{{S}}\left(\nabla \times {\mathbf {E}}+{\frac {\partial {\mathbf {B}}}{\partial t}}\right)\cdot {\mathrm {d}}{\mathbf {S}}&=0\end{aligned}}

Convert the equation to differential form. ∇ × E + ∂ B ∂ t = 0 ∇ × E = − ∂ B ∂ t {\displaystyle {\begin{aligned}\nabla &\times \mathbf {E} +{\frac {\partial \mathbf {B} }{\partial t}}=0\\\nabla &\times \mathbf {E} =-{\frac {\partial \mathbf {B} }{\partial t}}\end{aligned}}} {\begin{aligned}\nabla &\times {\mathbf {E}}+{\frac {\partial {\mathbf {B}}}{\partial t}}=0\\\nabla &\times {\mathbf {E}}=-{\frac {\partial {\mathbf {B}}}{\partial t}}\end{aligned}}

The Ampere-Maxwell Law

Begin with the Ampere-Maxwell law in integral form. ∮ C B ⋅ d l = μ 0 ∫ S J ⋅ d S + μ 0 ϵ 0 d d t ∫ S E ⋅ d S {\displaystyle \oint _{C}\mathbf {B} \cdot \mathrm {d} \mathbf {l} =\mu _{0}\int _{S}\mathbf {J} \cdot \mathrm {d} \mathbf {S} +\mu _{0}\epsilon _{0}{\frac {\mathrm {d} }{\mathrm {d} t}}\int _{S}\mathbf {E} \cdot \mathrm {d} \mathbf {S} } \oint _{{C}}{\mathbf {B}}\cdot {\mathrm {d}}{\mathbf {l}}=\mu _{0}\int _{{S}}{\mathbf {J}}\cdot {\mathrm {d}}{\mathbf {S}}+\mu _{0}\epsilon _{0}{\frac {{\mathrm {d}}}{{\mathrm {d}}t}}\int _{{S}}{\mathbf {E}}\cdot {\mathrm {d}}{\mathbf {S}}

Invoke Stokes' theorem. ∫ S ( ∇ × B ) ⋅ d S = μ 0 ∫ S J ⋅ d S + μ 0 ϵ 0 d d t ∫ S E ⋅ d S {\displaystyle \int _{S}(\nabla \times \mathbf {B} )\cdot \mathrm {d} \mathbf {S} =\mu _{0}\int _{S}\mathbf {J} \cdot \mathrm {d} \mathbf {S} +\mu _{0}\epsilon _{0}{\frac {\mathrm {d} }{\mathrm {d} t}}\int _{S}\mathbf {E} \cdot \mathrm {d} \mathbf {S} } \int _{{S}}(\nabla \times {\mathbf {B}})\cdot {\mathrm {d}}{\mathbf {S}}=\mu _{0}\int _{{S}}{\mathbf {J}}\cdot {\mathrm {d}}{\mathbf {S}}+\mu _{0}\epsilon _{0}{\frac {{\mathrm {d}}}{{\mathrm {d}}t}}\int _{{S}}{\mathbf {E}}\cdot {\mathrm {d}}{\mathbf {S}}

Set the equation to 0. ∫ S ( ∇ × B ) ⋅ d S − μ 0 ∫ S J ⋅ d S − μ 0 ϵ 0 d d t ∫ S E ⋅ d S = 0 ∫ S ( ∇ × B − μ 0 J − μ 0 ϵ 0 ∂ E ∂ t ) ⋅ d S = 0 {\displaystyle {\begin{aligned}\int _{S}(\nabla \times \mathbf {B} )\cdot \mathrm {d} \mathbf {S} -\mu _{0}\int _{S}\mathbf {J} \cdot \mathrm {d} \mathbf {S} -\mu _{0}\epsilon _{0}{\frac {\mathrm {d} }{\mathrm {d} t}}\int _{S}\mathbf {E} \cdot \mathrm {d} \mathbf {S} &=0\\\int _{S}\left(\nabla \times \mathbf {B} -\mu _{0}\mathbf {J} -\mu _{0}\epsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}\right)\cdot \mathrm {d} \mathbf {S} &=0\end{aligned}}} {\begin{aligned}\int _{{S}}(\nabla \times {\mathbf {B}})\cdot {\mathrm {d}}{\mathbf {S}}-\mu _{0}\int _{{S}}{\mathbf {J}}\cdot {\mathrm {d}}{\mathbf {S}}-\mu _{0}\epsilon _{0}{\frac {{\mathrm {d}}}{{\mathrm {d}}t}}\int _{{S}}{\mathbf {E}}\cdot {\mathrm {d}}{\mathbf {S}}&=0\\\int _{{S}}\left(\nabla \times {\mathbf {B}}-\mu _{0}{\mathbf {J}}-\mu _{0}\epsilon _{0}{\frac {\partial {\mathbf {E}}}{\partial t}}\right)\cdot {\mathrm {d}}{\mathbf {S}}&=0\end{aligned}}

Convert the equation to differential form. ∇ × B − μ 0 J − μ 0 ϵ 0 ∂ E ∂ t = 0 ∇ × B = μ 0 J + μ 0 ϵ 0 ∂ E ∂ t {\displaystyle {\begin{aligned}\nabla &\times \mathbf {B} -\mu _{0}\mathbf {J} -\mu _{0}\epsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}=0\\\nabla &\times \mathbf {B} =\mu _{0}\mathbf {J} +\mu _{0}\epsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}\end{aligned}}} {\begin{aligned}\nabla &\times {\mathbf {B}}-\mu _{0}{\mathbf {J}}-\mu _{0}\epsilon _{0}{\frac {\partial {\mathbf {E}}}{\partial t}}=0\\\nabla &\times {\mathbf {B}}=\mu _{0}{\mathbf {J}}+\mu _{0}\epsilon _{0}{\frac {\partial {\mathbf {E}}}{\partial t}}\end{aligned}}