Categories: Electromagnetism, Physics.

Maxwell’s equations

In physics, Maxwell’s equations govern all macroscopic electromagnetism, and notably lead to the electromagnetic wave equation, which describes the existence of light.

Gauss’ law

Gauss’ law states that the electric flux ΦE\Phi_E through a closed surface S(V)S(V) is equal to the total charge QQ contained in the enclosed volume VV, divided by the vacuum permittivity ε0\varepsilon_0:

ΦE=∮S(V)E⋅dA=1ε0∫VρdV=Qε0\begin{aligned} \Phi_E = \oint_{S(V)} \vb{E} \cdot \dd{\vb{A}} = \frac{1}{\varepsilon_0} \int_{V} \rho \dd{V} = \frac{Q}{\varepsilon_0} \end{aligned}

Where E\vb{E} is the electric field, and ρ\rho is the charge density in VV. Gauss’ law is usually more useful when written in its vector form, which can be found by applying the divergence theorem to the surface integral above. It states that the divergence of E\vb{E} is proportional to ρ\rho:

∇⋅E=ρε0\begin{aligned} \boxed{ \nabla \cdot \vb{E} = \frac{\rho}{\varepsilon_0} } \end{aligned}

This law can just as well be expressed for the displacement field D\vb{D} and polarization density P\vb{P}. We insert E=(D−P)/ε0\vb{E} = (\vb{D} - \vb{P}) / \varepsilon_0 into Gauss’ law for E\vb{E}, multiplied by ε0\varepsilon_0:

ρ=∇⋅(D−P)=∇⋅D−∇⋅P\begin{aligned} \rho = \nabla \cdot \big( \vb{D} - \vb{P} \big) = \nabla \cdot \vb{D} - \nabla \cdot \vb{P} \end{aligned}

To proceed, we split the net charge density ρ\rho into a “free” part ρfree\rho_\mathrm{free} and a “bound” part ρbound\rho_\mathrm{bound}, respectively corresponding to D\vb{D} and P\vb{P}, such that ρ=ρfree+ρbound\rho = \rho_\mathrm{free} + \rho_\mathrm{bound}. This yields:

∇⋅D=ρfree∇⋅P=−ρbound\begin{aligned} \boxed{ \nabla \cdot \vb{D} = \rho_{\mathrm{free}} } \qquad \quad \boxed{ \nabla \cdot \vb{P} = - \rho_{\mathrm{bound}} } \end{aligned}

By integrating over an arbitrary volume VV we can get integral forms of these equations:

ΦD=∮S(V)D⋅dA=∫VρfreedV=QfreeΦP=∮S(V)P⋅dA=−∫VρbounddV=−Qbound\begin{aligned} \Phi_D &= \oint_{S(V)} \vb{D} \cdot \dd{\vb{A}} = \int_{V} \rho_{\mathrm{free}} \dd{V} = Q_{\mathrm{free}} \\ \Phi_P &= \oint_{S(V)} \vb{P} \cdot \dd{\vb{A}} = - \int_{V} \rho_{\mathrm{bound}} \dd{V} = - Q_{\mathrm{bound}} \end{aligned}

Gauss’ law for magnetism

Gauss’ law for magnetism states that magnetic flux ΦB\Phi_B through a closed surface S(V)S(V) is zero. In other words, all magnetic field lines entering the volume VV must leave it too:

ΦB=∮S(V)B⋅dA=0\begin{aligned} \Phi_B = \oint_{S(V)} \vb{B} \cdot \dd{\vb{A}} = 0 \end{aligned}

Where B\vb{B} is the magnetic field. Thanks to the divergence theorem, this can equivalently be stated in vector form as follows:

∇⋅B=0\begin{aligned} \boxed{ \nabla \cdot \vb{B} = 0 } \end{aligned}

A consequence of this law is the fact that magnetic monopoles cannot exist, i.e. there is no such thing as “magnetic charge”, in contrast to electric charge.

Faraday’s law of induction

Faraday’s law of induction states that a magnetic field B\vb{B} that changes with time will induce an electric field EE. Specifically, the change in magnetic flux through a non-closed surface SS creates an electromotive force around the contour C(S)C(S). This is written as:

∮C(S)E⋅dl=−ddt∫SB⋅dA\begin{aligned} \oint_{C(S)} \vb{E} \cdot \dd{\vb{l}} = - \dv{}{t}\int_{S} \vb{B} \cdot \dd{\vb{A}} \end{aligned}

By using Stokes’ theorem on the contour integral, the vector form of this law is found to be:

∇×E=−∂B∂t\begin{aligned} \boxed{ \nabla \times \vb{E} = - \pdv{\vb{B}}{t} } \end{aligned}

Ampère’s circuital law

Ampère’s circuital law, with Maxwell’s correction, states that a magnetic field B\vb{B} can be induced along a contour C(S)C(S) by two things: a current density J\vb{J} through the enclosed surface SS, and a change of the electric field flux ΦE\Phi_E through SS:

∮C(S)B⋅dl=μ0(∫SJ⋅dA+ε0ddt∫SE⋅dA)\begin{aligned} \oint_{C(S)} \vb{B} \cdot d\vb{l} = \mu_0 \Big( \int_S \vb{J} \cdot d\vb{A} + \varepsilon_0 \dv{}{t}\int_S \vb{E} \cdot d\vb{A} \Big) \end{aligned} ∇×B=μ0(J+ε0∂E∂t)\begin{aligned} \boxed{ \nabla \times \vb{B} = \mu_0 \Big( \vb{J} + \varepsilon_0 \pdv{\vb{E}}{t} \Big) } \end{aligned}

Where μ0\mu_0 is the vacuum permeability. This relation also exists for the “bound” fields H\vb{H} and D\vb{D}, and for M\vb{M} and P\vb{P}. We insert B=μ0(H+M)\vb{B} = \mu_0 (\vb{H} + \vb{M}) and E=(D−P)/ε0\vb{E} = (\vb{D} - \vb{P})/\varepsilon_0 into Ampère’s law, after dividing it by μ0\mu_0 for simplicity:

∇×(H+M)=J+∂∂t(D−P)\begin{aligned} \nabla \cross \big( \vb{H} + \vb{M} \big) &= \vb{J} + \pdv{}{t}\big( \vb{D} - \vb{P} \big) \end{aligned}

To proceed, we split the net current density J\vb{J} into a “free” part Jfree\vb{J}_\mathrm{free} and a “bound” part Jbound\vb{J}_\mathrm{bound}, such that J=Jfree+Jbound\vb{J} = \vb{J}_\mathrm{free} + \vb{J}_\mathrm{bound}. This leads us to:

∇×H=Jfree+∂D∂t∇×M=Jbound−∂P∂t\begin{aligned} \boxed{ \nabla \times \vb{H} = \vb{J}_{\mathrm{free}} + \pdv{\vb{D}}{t} } \qquad \quad \boxed{ \nabla \times \vb{M} = \vb{J}_{\mathrm{bound}} - \pdv{\vb{P}}{t} } \end{aligned}

By integrating over an arbitrary surface SS we can get integral forms of these equations:

∮C(S)H⋅dl=∫SJfree⋅dA+ddt∫SD⋅dA∮C(S)M⋅dl=∫SJbound⋅dA−ddt∫SP⋅dA\begin{aligned} \oint_{C(S)} \vb{H} \cdot d\vb{l} &= \int_S \vb{J}_{\mathrm{free}} \cdot \dd{\vb{A}} + \dv{}{t}\int_S \vb{D} \cdot \dd{\vb{A}} \\ \oint_{C(S)} \vb{M} \cdot d\vb{l} &= \int_S \vb{J}_{\mathrm{bound}} \cdot \dd{\vb{A}} - \dv{}{t}\int_S \vb{P} \cdot \dd{\vb{A}} \end{aligned}

Note that Jbound\vb{J}_\mathrm{bound} can be split into the magnetization current density JM=∇×M\vb{J}_M = \nabla \cross \vb{M} and the polarization current density JP=∂P/∂t\vb{J}_P = \ipdv{\vb{P}}{t}:

Jbound=JM+JP=∇×M+∂P∂t\begin{aligned} \vb{J}_\mathrm{bound} = \vb{J}_M + \vb{J}_P = \nabla \cross \vb{M} + \pdv{\vb{P}}{t} \end{aligned}

Redundancy of Gauss’ laws

In fact, both of Gauss’ laws are redundant, because they are already implied by Faraday’s and Ampère’s laws. Suppose we take the divergence of Faraday’s law:

0=∇⋅∇×E=−∇⋅∂B∂t=−∂∂t(∇⋅B)\begin{aligned} 0 = \nabla \cdot \nabla \cross \vb{E} = - \nabla \cdot \pdv{\vb{B}}{t} = - \pdv{}{t}(\nabla \cdot \vb{B}) \end{aligned}

Since the divergence of a curl is always zero, the right-hand side must vanish. We know that B\vb{B} can vary in time, so our only option to satisfy this is to demand that ∇⋅B=0\nabla \cdot \vb{B} = 0. We thus arrive arrive at Gauss’ law for magnetism from Faraday’s law.

The same technique works for Ampère’s law. Taking its divergence gives us:

0=1μ0∇⋅∇×B=∇⋅J+ε0∂∂t(∇⋅E)\begin{aligned} 0 = \frac{1}{\mu_0} \nabla \cdot \nabla \cross \vb{B} = \nabla \cdot \vb{J} + \varepsilon_0 \pdv{}{t}(\nabla \cdot \vb{E}) \end{aligned}

We integrate this over an arbitrary volume VV, and apply the divergence theorem:

0=∫V∇⋅JdV+∂∂t∫Vε0∇⋅EdV=∮SJ⋅dS+∂∂t∫Vε0∇⋅EdV\begin{aligned} 0 &= \int_V \nabla \cdot \vb{J} \dd{V} + \pdv{}{t}\int_V \varepsilon_0 \nabla \cdot \vb{E} \dd{V} \\ &= \oint_S \vb{J} \cdot \dd{S} + \pdv{}{t}\int_V \varepsilon_0 \nabla \cdot \vb{E} \dd{V} \end{aligned}

The first integral represents the current (charge flux) through the surface of VV. Electric charge is not created or destroyed, so the second integral must be the total charge in VV:

Q=∫Vε0∇⋅EdV  ⟹  ∇⋅E=ρε0\begin{aligned} Q = \int_V \varepsilon_0 \nabla \cdot \vb{E} \dd{V} \quad \implies \quad \nabla \cdot \vb{E} = \frac{\rho}{\varepsilon_0} \end{aligned}

And we thus arrive at Gauss’ law from Ampère’s law and charge conservation.