9.4 Electrodynamics

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4-Tensor fields

A 4-tensor field of type $(r,s)$ consists of a map $T:M{V}^{(r,s)}$ . We can think of this as a 4-tensor assigned at each point of space-time. The components of a 4-tensor field are functions of space-time coordinate

$$T_{\rho \sigma \dots }^{\mu \nu \dots }=T_{\rho \sigma \dots }^{\mu \nu \dots }(x^{\alpha }).$$

Define the gradient of a 4-tensor field $T$ to be the 4-tensor field of type $(r,s+1)$ ), having components

$$T_{\rho \sigma \dots ,\tau }^{\mu \nu \dots }=\frac{\partial }{\partial x^{\tau }}{T}_{\rho \sigma \dots }^{\mu \nu \dots }.$$

This is a 4-tensor field since a Poincaré transformation Eq. (9.7) induces the transformation

$$\begin{array}{rl}& T_{{\rho }^{\prime },\dots ,{\tau }^{\prime }}^{{\mu }^{\prime }\dots }=\frac{\partial }{\partial x^{\prime {\tau }^{\prime }}}(T_{\beta \dots }^{\alpha \dots }{L}_{\alpha }^{{\mu }^{\prime }}\dots L_{{\rho }^{\prime }}^{\prime \beta }\dots )\\ & =\frac{\partial x^{\gamma }}{\partial x^{\prime {\tau }^{\prime }}}\frac{\partial }{\partial x^{\gamma }}(T_{\beta \dots }^{\alpha \dots }{L}_{\alpha }^{{\mu }^{\prime }}\dots L_{{\rho }^{\prime }}^{\prime \beta }\dots )\\ & =T_{\beta ,\dots ,\gamma }^{\alpha \dots }{L}_{{\tau }^{\prime }}^{\prime \gamma }{L}_{\alpha }^{{\mu }^{\prime }}\dots L_{{\rho }^{\prime }}^{\prime \beta }\dots \end{array}$$

For example, if $f:M\to \mathbb{R}$ is a scalar field, its gradient is a 4-covector field,

$$f_{,\mu }=\frac{\partial f(x^{\alpha })}{\partial x^{\mu }}.$$

Example 9.3

A 4-vector field, $J=J^{\mu }(x^{\alpha })e_{\mu },$ , is said to be divergence-free if

$$J_{,\mu }^{\mu }=0.$$

Setting $j_i=J^i(i=1,2,3)$ and $\rho =\frac{1}{c}{J}^4$ , the divergence-free condition reads

$$\frac{\partial \rho }{\partial t}+\nabla \cdot \mathbf{j}=0,\text{\hspace{1em}(9.36)}$$

known both in hydrodynamics and electromagnetism as the equation of continuity. In terpreting $\rho$ as the charge density or charge per unit volume, $j$ is the current density. The charge per unit time crossing unit area normal to the unit vector $n$ is given by $jn$. Equation (9.36) implies conservation ofcharge – the rate of increase of charge in a volume equals the flux of charge entering through the boundary surface :

$$\frac{\mathrm{d}q}{\mathrm{d}t}={\int }_{\mathcal{V}}\frac{\partial \rho }{\partial t}\mathrm{d}V=-{\int }_{\mathcal{V}}\nabla \cdot \mathbf{j}\mathrm{d}V=-{\int }_{\mathcal{S}}\mathbf{j}\cdot \mathrm{d}\mathbf{S}.$$

Electromagnetism

As in Example 9.3, let there be a continuous distribution of electric charge present in Minkowski space-time, having charge density $\rho (\mathbf{r},t)$ and charge flux density or current density $\mathbf{j}=\rho \mathbf{v},$ where $\mathbf{v}(\mathbf{r},t)$ is the velocity field of the fluid. The total charge of a system is a scalar quantity – else an unionized gase would not generally be electrically neutral. Charge density in a local instantaneous rest frame of the fluid at any event $p$ is denoted ${\rho }_0(p)$ and is known as proper charge density. It may be assumed to be a scalar quantity, since it is defined in a specific inertial frame at $p.$ . On the other hand, the charge density $\rho$ is given by

$$\rho =\underset{\Delta V\to 0}{\lim }\frac{\Delta q}{\Delta V}$$

where, by the length–volume contraction effect Eq. (9.18),

$$\Delta V=\frac{1}{\gamma }\Delta V_0.$$

Since charge is a scalar quantity, $\Delta q=\Delta q_0$ , charge density and proper charge density are related by

$$\rho =\underset{\Delta V_0\to 0}{\lim }\frac{\Delta q_0}{V_0/\gamma }=\gamma {\rho }_0.$$

If the charged fluid has a 4-velocity field $V=V^{\mu }(x^{\alpha })e_{\mu }$ , define the 4-current $J$ to be the 4-vector field having components

$$J^{\mu }={\rho }_0{V}^{\mu }.$$

From $Eq.$ . Eq. (9.24) together with the above we have

$$J^{\mu }=(\mathbf{j},\rho c),$$

and by Example 9.3, conservation of charge is equivalent to requiring the 4-current be divergence-free,

$$J_{,\mu }^{\mu }=0⟺\nabla \cdot \mathbf{j}+\frac{\partial \rho }{\partial t}=0.$$

In electrodynamics we are given a 4-current field $J=J^{\mu }{e}_{\mu }$ representing the charge density and current of the electric charges present, also known as the source field, and an antisymmetric 4-tensor field $\mathcal{F}=F_{\mu \nu }(x^{\alpha }){\epsilon }^{\mu }\otimes {\epsilon }^{\nu }$ such that $F_{\mu \nu }=-F_{\nu \mu }$ , known as the electromagnetc field, satisfying the Maxwell equations:

$$F_{\mu \nu ,\rho }+F_{\nu \rho ,\mu }+F_{\rho \mu ,\nu }=0,\text{\hspace{1em}(9.37)}$$

$$F_{,\nu }^{\mu \nu }=\frac{4\pi }{c}{J}^{\mu },\text{\hspace{1em}(9.38)}$$

where $F^{\mu \nu }=g^{\mu \alpha }{g}^{\nu \beta }{F}_{\alpha \beta }$ . Units adopted here are the Gaussian units, which are convenien for the formal presentation of the subject.

The first set Eq. (9.37) is known as the source-free Maxwell equations, while the second set Eq. (9.38) relates electromagnetic field and sources. It is common to give explicit symbols for the components of the electromagnetic field tensor

$$F_{\mu \nu }=\left(\begin{array}{cccc}0& B_3& -B_2& E_1\\ -B_3& 0& B_1& E_2\\ B_2& -B_1& 0& E_3\\ -E_1& -E_2& -E_3& 0\end{array}\right),\text{ i.e. set }{F}_{12}=B_3,\text{ etc. }\text{\hspace{1em}(9.39)}$$

The 3-vector fields $\mathbf{E}=(E_1,E_2,E_3)$ and $\mathbf{B}=(B_1,B_2,B_3)$ are called the electric and magnetic fields, respectively. The source-free Maxwell equations Eq. (9.37) give non-trivia equations only when all three indices $\mu ,\nu$ and $\rho$ are unequal, giving four independen equations

$$(\mu ,\nu ,\rho )=(1,2,3)⟹\nabla .\mathbf{B}=0,\text{\hspace{1em}(9.40)}$$

$$(\mu ,\nu ,\rho )=(2,3,4),\text{ etc. }⟹\nabla \times \mathbf{E}+\frac{1}{c}\frac{\partial \mathbf{B}}{\partial t}=0.\text{\hspace{1em}(9.41)}$$

The second set ofMaxwell equations Eq. (9.38) imply charge conservation for, on commuting partial derivatives and using the antisymmetry of $F^{\mu \nu }$ , we have

$$J_{,\mu }^{\mu }=\frac{c}{4\pi }{F}_{,\nu \mu }^{\mu \nu }=\frac{c}{8\pi }(F^{\mu \nu }-F^{\nu \mu }{)}_{,\mu \nu }=0.$$

Using $F_{i4}=-F_{4i}=E_i$ and $F_{ij}={ϵ}_{ijk}{B}_k$ , Eqs. Eq. (9.38) reduce to the vector form ofMaxwel equations

$$\nabla \cdot \mathbf{E}=4\pi \rho ,\text{\hspace{1em}(9.42)}$$

$$-\frac{1}{c}\frac{\partial \mathbf{E}}{\partial t}+\nabla \times \mathbf{B}=\frac{4\pi }{c}\mathbf{j}.\text{\hspace{1em}(9.43)}$$

Exercise

Show Eqs. Eq. (9.42) and Eq. (9.43).

There are essentially two independent invariants that can be constructed from an elec tromagnetic field,

$$F_{\mu \nu }{F}^{\mu \nu }\text{ and }\ast F_{\mu \nu }{F}^{\mu \nu }$$

where the dual electromagnetic tensor $F_{\mu \nu }$ is given in Example 8.8. Substituting electric and magnetic field components we find

$$F_{\mu \nu }{F}^{\mu \nu }=2({\mathbf{B}}^2-{\mathbf{E}}^2)\mathrm{and}\ast F_{\mu \nu }{F}^{\mu \nu }=-4\mathbf{E}\cdot \mathbf{B}.\text{\hspace{1em}(9.44)}$$

Exercise

Show that the source-free Maxwell equations Eq. (9.37) can be written in the dual form

$$\ast F_{,\nu }^{\mu \nu }=0.$$

The equation of motion of a charged particle, charge $q_{:}$ , is given by the Lorentz force equation

$$\frac{\mathrm{d}}{\mathrm{d}\tau }{P}_{\mu }=\frac{q}{c}{F}_{\mu \nu }{V}^{\nu }=F_{\mu }\text{\hspace{1em}(9.45)}$$

where the 4-momentum $P_{\mu }$ has components $(\mathbf{p},\nabla -\mathcal{E}/c)$ . Energy is written here as $\mathcal{E}$ so that no confusion with the magnitude ofelectric field can arise. Using Eq. (9.34) for components of the 4-force $F_{\mu }$ we find that

$$\mathbf{f}=\frac{\mathrm{d}\mathbf{p}}{\mathrm{d}t}=q\left(\mathbf{E}+\frac{1}{c}\mathbf{v}\times \mathbf{B}\right)\text{\hspace{1em}(9.46)}$$

and taking $v$ of this equation gives rise to the energy equation (see Problem 9.16)

$$\frac{\mathrm{d}\mathcal{E}}{\mathrm{d}t}=\mathbf{f}\cdot \mathbf{v}=q\mathbf{E}\cdot \mathbf{v}.$$

Potentials and gauge transformations

The source-free equations Eq. (9.37) are true if and only if in a neighbourhood of any event there exists a 4-covector field $A_{\mu }(x^{\alpha })$ , called the 4-potential, such that

$$F_{\mu \nu }=A_{\nu ,\mu }-A_{\mu ,\nu }.\text{\hspace{1em}(9.47)}$$

The if part of this statement is simple, for Eq. (9.47) implies, on commuting partial derivatives,

$$F_{\mu \nu ,\rho }+F_{\nu \rho ,\mu }+F_{\rho \mu ,\nu }=A_{\nu ,\mu \rho }-A_{\mu ,\nu \rho }+A_{\rho ,\nu \mu }-A_{\nu ,\rho \mu }+A_{\mu ,\rho \nu }-A_{\rho ,\mu \nu }=0.$$

The converse will be postponed till Chapter 17, Theorem 17.5.

Exercise

Setting $A_{\mu }=(A_1,A_2,A_3,-\varphi )=(\mathbf{A},-\varphi )$ , show that Eq. (9.47) reads

$$\mathbf{B}=\nabla \times \mathbf{A},\mathbf{E}=-\nabla \varphi -\frac{1}{c}\frac{\partial \mathbf{A}}{\partial t}.\text{\hspace{1em}(9.48)}$$

$A$ is known as the vector potential, and $\varphi$ as the scalar potential.

If the 4-vector potential of an electromagnetic field is altered by addition of the gradien of a scalar field $\psi$

$${\tilde{A}}_{\mu }=A_{\mu }+{\psi }_{,\mu }\text{\hspace{1em}(9.49)}$$

then the electromagnetic tensor $F_{\mu \nu }$ remains unchanged

$${\tilde{F}}_{\mu \nu }={\tilde{A}}_{\nu ,\mu }-{\tilde{A}}_{\mu ,\nu }=A_{\nu ,\mu }+{\psi }_{,\nu \mu }-A_{\mu ,\nu }-{\psi }_{,\mu \nu }=F_{\mu \nu }.$$

A transformation Eq. (9.49), which has no effect on the electromagnetic field, is called a gauge transformation.

Exercise

Write the gauge transformation Eq. (9.49) in terms of the vector and scalar potential,

$$\tilde{\mathbf{A}}=\mathbf{A}+\nabla \psi ,\tilde{\varphi }=\varphi -\frac{1}{c}\frac{\partial \psi }{\partial t},$$

and check that $E$ and $B$ given by Eq. (9.48) are left unchanged by these transformations.

Under a gauge transformation, the divergence of $A^{\mu }$ transforms as

$${\tilde{A}}_{,\mu }^{\mu }=A_{,\mu }^{\mu }+\square \psi$$

where

$$\square \psi ={\psi }_{\mu }^{,\mu }=g^{\mu \nu }{\psi }_{,\mu \nu }={\nabla }^2\psi -\frac{1}{c^2}\frac{{\partial }^2\psi }{\partial t^2}.$$

The operator $\square$ is called the wave operator or d’Alembertian. If we choose $\psi$ to be any solution of the inhomogeneous wave equation

$$\square \psi =-A_{,\mu }^{\mu }\text{\hspace{1em}(9.50)}$$

then ${\tilde{\mathbf{A}}}^{\mu }{}_{,\mu }=0$ . Ignoring the tilde over $A$, any choice of 4-potential $A^{\mu }$ that satisfies

$$A_{,\mu }^{\mu }=\nabla \cdot \mathbf{A}+\frac{1}{c}\frac{\partial \varphi }{\partial t}=0\text{\hspace{1em}(9.51)}$$

is called a Lorentz gauge. Since solutions of the inhomogeneous wave equation Eq. (9.50) are always locally available, we may always adopt a Lorentz gauge if we wish. It should, however, be pointed out that the 4-potential $A_{\mu }$ is not uniquely determined by the Lorentz gauge condition Eq. (9.51), for it is still possible to add a further gradient ${\bar{\mathbf{\psi }}}_{,\mu }$ provided $\hat{\psi }$ is a solution of the wave equation, $\begin{array}{r}⨆\bar{\psi }=0\end{array}$ . This is said to be the available gauge freedom in the Lorentz gauge.

In terms of a 4-potential, the source-free part of the Maxwell equations Eq. (9.37) is auto matically satisfied, while the source-related part Eq. (9.38) reads

$$F_{,\nu }^{\mu \nu }=A_{,\nu }^{\nu ,\mu }-A_{,\nu }^{\mu ,\nu }=\frac{4\pi }{c}{J}^{\mu }.$$

If $A^{\mu }$ is in a Lorentz gauge Eq. (9.51), then the first term in the central expression vanishes and the Maxwell equations reduce to inhomogeneous wave equations,

$$\square A^{\mu }=-\frac{4\pi }{c}{J}^{\mu },A_{,\mu }^{\mu }=0,\text{\hspace{1em}(9.52)}$$

or in terms of vector and scalar potentials

$$\square \mathbf{A}=-\frac{4\pi }{c}\mathbf{j},\square \varphi =-4\pi \rho ,\nabla \cdot \mathbf{A}+\frac{1}{c}\frac{\partial \varphi }{\partial t}=0.\text{\hspace{1em}(9.53)}$$

In the case of a vacuum, $\rho =0\mathrm{and}\mathbf{j}=0$ , the Maxwell equations read

$$\square \mathbf{A}=0,\square \varphi =0.$$

Problems

Problem 9.17

Show that with respect to a rotation Eq. (9.8) the electric and magnetic fields $E$ and $B$ transform as 3-vectors,

$$E_i^{\prime }=a_{ij}{E}_j,B_i^{\prime }=a_{ij}{B}_j.$$

Problem 9.18

Under a boost Eq. (9.13) show that the 4-tensor transformation law for $F_{\mu \nu }$ or $F^{\mu \nu }$ gives rise to

$$\begin{array}{l}{E}_1^{\prime }=F_{14}^{\prime }=E_1,E_2^{\prime }=\gamma (E_2-\frac{v}{c}{B}_3),E_3^{\prime }=\gamma (E_2+\frac{v}{c}{B}_2),\\ B_1^{\prime }=F_{23}^{\prime }=B_1,B_2^{\prime }=\gamma (B_2+\frac{v}{c}{E}_3),B_3^{\prime }=\gamma (B_2-\frac{v}{c}{E}_2).\end{array}$$

Decomposing $E$ and $B$ into components parallel and perpendicular to $\mathbf{v}=(v,0,0)$ , show that these transformations can be expressed in vector form:

$${\mathbf{E}}_{\parallel }^{\prime }={\mathbf{E}}_{\parallel },{\mathbf{E}}_{\perp }^{\prime }=\gamma ({\mathbf{E}}_{\perp }+\frac{1}{c}\mathbf{v}\times \mathbf{B}),$$ $${\mathbf{B}}_{\parallel }^{\prime }={\mathbf{B}}_{\parallel },{\mathbf{B}}_{\perp }^{\prime }=\gamma ({\mathbf{B}}_{\perp }-\frac{1}{c}\mathbf{v}\times \mathbf{E}).$$

Problem 9.19

It is possible to use transformation of $E$ and $B$ under boosts to find the field of a uniformly moving charge. Consider a charge $q$ travelling with velocity $v$, which without loss of generality may be taken to be in the $x$-direction. Let $\mathbf{R}=(x-vt,y,z)$ be the vector connecting charge to field point $\mathbf{r}=(x,y,z)$ . In the rest frame of the charge, denoted by primes, suppose the field is the coulomb field

$${\mathbf{E}}^{\prime }=\frac{q{\mathbf{r}}^{\prime }}{r^{\prime 3}},{\mathbf{B}}^{\prime }=0$$

where

$${\mathbf{r}}^{\prime }=(x^{\prime },y^{\prime },z^{\prime })=\left(\frac{x-vt}{\sqrt{1-v^2/c^2}},y,z\right).$$

Apply the transformation law for $E$ and $B$ derived in Problem 9.18 to show that

$$\mathbf{E}=\frac{q\mathbf{R}(1-v^2/c^2)}{R^3(1-(v^2/c^2){\sin }^2\theta {)}^{3/2}}\text{ and }\mathbf{B}=\frac{1}{c}\mathbf{v}\times \mathbf{E},$$

where $\theta$ is the angle between $R$ and $v$. At a given distance $R$ where is most of the electromagnetic field concentrated for highly relativistic velocities $v\approx c2$

Problem 9.20

A particle of rest mass $m$, charge $q$ is in motion in a uniform constant magnetic field $\mathbf{B}=(0,0,B)$ . Show from the Lorentz force equation that the energy of the particle is constant, and its motion is a helix about a line parallel to $B$, with angular frequency

$$\omega =\frac{qcB}{\mathcal{E}}.$$

Problem 9.21

Let $E$ and $B$ be perpendicular constant electric and magnetic fields, $\mathbf{E}\cdot \mathbf{B}=0.$

(a) If $B^2>E^2$ show that a transformation to a frame $K^{\prime }$ having velocity $\mathbf{v}=k\mathbf{E}\times \mathbf{B}$ can be found such that $E$ vanishes.

(b) What is the magnitude of $B$ after this transformation?

(c) If $E^2>B^2$ find a transformation that makes $B$ vanish.

(d) What happens if $E^2=B^2?$

(e) A particle of charge $q$ is in motion in a crossed constant electric and magnetic field $\mathbf{E}\cdot \mathbf{B}=0,B^2>E^2$ . From the solution of Problem 9.20 for a particle in a constant magnetic field, describe its motion.

Problem 9.22

An electromagnetic field $F_{\mu \nu }$ is said to be of ‘electric type’ at an event $p$ if there exists a unit timelike 4-vector $U_{\mu }$ at $p,U_{\alpha }{U}^{\alpha }=-1$ , and a spacelike 4-vector field $E_{\mu }$ orthogonal to $U^{\mu }$ such that

$$F_{\mu \nu }=U_{\mu }{E}_{\nu }-U_{\nu }{E}_{\mu },E_{\alpha }{U}^{\alpha }=0.$$

(a) Show that any purely electric field, i.e. one having $\mathbf{B}=0$ , is of electric type.

(b) If $F_{\mu \nu }$ is of electric type at $p,$ show that there is a velocity $v$ such that

$$\mathbf{B}=\frac{\mathbf{v}}{c}\times \mathbf{E}(|\mathbf{v}|<c).$$

Using Problem 9.18 show that there is a Lorentz transformation that transforms the electromag netic field to one that is purely electric at $p$ .

(c) If $F_{\mu \nu }$ is of electric type everywhere with $U^{\mu }$ a constant vector field, and satisfies the Maxwell equations in vacuo, $J^{\mu }=0,$ , show that the vector field $E^{\mu }$ is divergence-free, $E_{\nu }^{\nu }=0$

Problem 9.23

Use the gauge freedom $\sqcap \psi =0$ in the Lorentz gauge to show that it is possible to set $\varphi =0$ and $\nabla \cdot \mathbf{A}=0$ . This is called a radiation gauge.

(a) What gauge freedoms are still available to maintain the radiation gauge?

(b) Suppose $A$ is independent of coordinates $x$ and $y$ in the radiation gauge. Show that the Maxwell equations have solutions of the form

$$\mathbf{E}=(E_1(u),E_2(u),0),\mathbf{B}=(-E_2(u),E_1(u),0)$$

where $u=ct-z$ and $E_i(u)$ are arbitrary differentiable functions.

(c) Show that these solutions may be interpreted as right-travelling electromagnetic waves.