How to Convert Maxwell's Equations into Differential Form

Step-by-Step Guide

1. 1

   Step 1: Begin with Gauss' law in integral form.

   ∮SE⋅dS=Qϵ0{\displaystyle \oint _{S}{\mathbf {E} }\cdot {\mathrm {d} }{\mathbf {S} }={\frac {Q}{\epsilon _{0}}}}
2. 2

   Step 2: Rewrite the right side in terms of a volume integral.

   ∮SE⋅dS=∫Vρϵ0dV{\displaystyle \oint _{S}{\mathbf {E} }\cdot {\mathrm {d} }{\mathbf {S} }=\int _{V}{\frac {\rho }{\epsilon _{0}}}{\mathrm {d} }V} , The divergence theorem says that the flux penetrating a closed surface S{\displaystyle S} that bounds a volume V{\displaystyle V} is equal to the divergence of the field F{\displaystyle {\mathbf {F} }} inside the volume. ∮SF⋅dS=∫V(∇⋅F)dV{\displaystyle \oint _{S}{\mathbf {F} }\cdot {\mathrm {d} }{\mathbf {S} }=\int _{V}(\nabla \cdot {\mathbf {F} }){\mathrm {d} }V} , ∫V(∇⋅E)dV=∫Vρϵ0dV{\displaystyle \int _{V}(\nabla \cdot {\mathbf {E} }){\mathrm {d} }V=\int _{V}{\frac {\rho }{\epsilon _{0}}}{\mathrm {d} }V} , ∫V(∇⋅E)dV−∫Vρϵ0dV=0∫V(∇⋅E−ρϵ0)dV=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}}} , 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−ρϵ0=0{\displaystyle \nabla \cdot {\mathbf {E} }-{\frac {\rho }{\epsilon _{0}}}=0} This leads to Gauss' law in differential form. ∇⋅E=ρϵ0{\displaystyle \nabla \cdot {\mathbf {E} }={\frac {\rho }{\epsilon _{0}}}} , ∮SB⋅dS=0{\displaystyle \oint _{S}{\mathbf {B} }\cdot {\mathrm {d} }{\mathbf {S} }=0} , ∫V(∇⋅B)dV=0{\displaystyle \int _{V}(\nabla \cdot {\mathbf {B} }){\mathrm {d} }V=0} , As with Gauss' Law, the same argument used above yields our answer. ∇⋅B=0{\displaystyle \nabla \cdot {\mathbf {B} }=0} , ∮CE⋅dl=−ddt∫SB⋅dS{\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} }} , Stokes' theorem says that the circulation of a field F{\displaystyle {\mathbf {F} }} around the loop C{\displaystyle C} that bounds a surface S{\displaystyle S} is equal to the flux of curlF{\displaystyle \operatorname {curl} {\mathbf {F} }} over S.{\displaystyle S.} ∮CF⋅dl=∫S(∇×F)⋅dS{\displaystyle \oint _{C}{\mathbf {F} }\cdot {\mathrm {d} }{\mathbf {l} }=\int _{S}(\nabla \times {\mathbf {F} })\cdot {\mathrm {d} }{\mathbf {S} }} , ∫S(∇×E)⋅dS=−ddt∫SB⋅dS{\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} }} , ∫S(∇×E)⋅dS+ddt∫SB⋅dS=0∫S(∇×E+∂B∂t)⋅dS=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}}} , ∇×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}}} , ∮CB⋅dl=μ0∫SJ⋅dS+μ0ϵ0ddt∫SE⋅dS{\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} }} , ∫S(∇×B)⋅dS=μ0∫SJ⋅dS+μ0ϵ0ddt∫SE⋅dS{\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} }} , ∫S(∇×B)⋅dS−μ0∫SJ⋅dS−μ0ϵ0ddt∫SE⋅dS=0∫S(∇×B−μ0J−μ0ϵ0∂E∂t)⋅dS=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}}} , ∇×B−μ0J−μ0ϵ0∂E∂t=0∇×B=μ0J+μ0ϵ0∂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}}}
3. 3

   Step 3: Recall the divergence theorem.

4. 4

   Step 4: Use the divergence theorem to rewrite the left side as a volume integral.

5. 5

   Step 5: Set the equation to 0.

6. 6

   Step 6: Convert the equation to differential form.

7. 7

   Step 7: Begin with Gauss' law for magnetism in integral form.

8. 8

   Step 8: Invoke the divergence theorem.

9. 9

   Step 9: Write the equation in differential form.

10. 10

    Step 10: Begin with Faraday's law in integral form.

11. 11

    Step 11: Recall Stokes' theorem.

12. 12

    Step 12: Use Stokes' theorem to rewrite the left side as a surface integral.

13. 13

    Step 13: Set the equation to 0.

14. 14

    Step 14: Convert the equation to differential form.

15. 15

    Step 15: Begin with the Ampere-Maxwell law in integral form.

16. 16

    Step 16: Invoke Stokes' theorem.

17. 17

    Step 17: Set the equation to 0.

18. 18

    Step 18: Convert the equation to differential form.

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