65  Relativistic Electrodynamics

    Michael Fowler, UVa

Introduction

This lecture is based on relativistic kinematics as presented in lecture 62: you need to be fully familiar with those concepts. To summarize briefly:  positions and intervals Lorentz transform as contravariant vectors, $d{x}^{{\mu }^{\prime }}={\Lambda }^{{\mu }^{\prime }}{}_{\nu }d{x}^{\nu },$ and differential operators transform as covariant vectors, ${\partial }_{{\mu }^{\prime }}={\Lambda }_{{\mu }^{\prime }}{}^{\nu }{\partial }_{\nu }.$ Here ${\partial }_{\mu }=\partial /\partial x^{\mu },$ and ${\Lambda }^{\mu }{}_{\nu }={\left({\Lambda }_{\mu }{}^{\nu }\right)}^{-1}.$ To lower an index, multiply by the metric tensor $g_{\mu \nu },$ to raise one use $g^{\mu \nu },$ both mean (in SR) just multiplying the time-index term by $-1.$  A product of a covariant and a contravariant vector is invariant: $A^{{\mu }^{\prime }}{B}_{{\mu }^{\prime }}={\Lambda }^{{\mu }^{\prime }}{}_{\nu }{\Lambda }_{{\mu }^{\prime }}{}^{\sigma }{A}^{\nu }{B}_{\sigma }={\delta }_{\nu }^{\sigma }{A}^{\nu }{B}_{\sigma }=A^{\sigma }{B}_{\sigma }.$ The two $\Lambda$ 's are inverses of each other, and so ${\delta }_{\nu }^{\sigma }$ is the ordinary Kronecker $\delta ,$ just 1's on the diagonal, zeroes elsewhere. 

Here we’ll formulate Maxwell’s equations in this notation: it leads to new insights.

Four-Current

An important example of a four-vector is the electric current,

$$J^{\mu }={\rho }_0{U}^{\mu }$$

where ${\rho }_0$ is the charge density in its rest frame, and $U^{\mu }$ is the local 4-velocity. The first component ${\rho }_0{U}^0$ gives the local charge density (actually $c\gamma {\rho }_0$) in the (in general) moving frame (note the Lorentz volume contraction), the other three components are the three-dimensional current. (To get the signs right, recall $U^0=d{x}^0/d\tau =cdt/d\tau =c\gamma .$)

The experimentally established conservation of charge is written invariantly (remember ${\partial }_0=\partial /\partial x^0=\left(1/c\right)\partial /\partial t$) :

$$\frac{\partial J^{\mu }}{\partial x^{\mu }}={\partial }_{\mu }{J}^{\mu }=0.$$

This is an "inner product" of the contravariant $J^{\mu }$ and the covariant ${\partial }_{\mu }.$

Wave Equations

Another obvious inner product is:

$${\partial }^{\mu }{\partial }_{\mu }=-\frac{1}{c^2}\frac{{\partial }^2}{\partial t^2}+\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}+\frac{{\partial }^2}{\partial z^2}.$$ This operator will be invariant under a Lorentz transformation.

Recall now the wave equations for the vector and scalar potentials $\overrightarrow{A},\phi$:

$$\begin{array}{c}{\nabla }^2\phi -\frac{1}{c^2}\frac{{\partial }^2\phi }{\partial t^2}=-\frac{\rho }{{\epsilon }_0},\\ {\nabla }^2\overrightarrow{A}-\frac{1}{c^2}\frac{{\partial }^2\overrightarrow{A}}{\partial t^2}=-{\mu }_0\overrightarrow{J}.\end{array}$$ If we divide every term in the first equation by $c,$ the right-hand sides of the two equations together form the four-vector current:

$-{\mu }_0\left(\frac{\rho }{{\epsilon }_0{\mu }_{0}c},\overrightarrow{J}\right)=-{\mu }_0\left(c\rho ,\overrightarrow{J}\right)=-{\mu }_0{J}^{\mu },$

and we can put them together very succinctly:

${\partial }^{\mu }{\partial }_{\mu }\left(\phi /c,\overrightarrow{A}\right)=-{\mu }_0{J}^{\mu }.$

Remember these equations are nothing but a rewriting of Maxwell's equations, so they must be true in all inertial frames.We know the right-hand side is a four-vector (it’s a constant times a four-velocity, so it transforms appropriately) and the differential operator ${\partial }^{\mu }{\partial }_{\mu }$ is invariant, so, from the Lorentz invariance of the whole equation, it follows that

$\left(\phi /c,\overrightarrow{A}\right)=A^{\mu }$

is also a four vector.

There is one loose end here$—$the equations only have this simple "wave equation" form in the Lorenz gauge,

$\overrightarrow{\nabla }\cdot \overrightarrow{A}=-\left(1/c^2\right)\partial \phi /\partial t.$

But this is just

$${\partial }_{\mu }{A}^{\mu }=0,$$

manifestly covariant, and the whole scheme is consistent.

The Field Tensor

The Magnetic Field

Recall now that the magnetic field was generated from the vector potential by $\overrightarrow{B}=\overrightarrow{\nabla }\times \overrightarrow{A},$ or $B_i={\partial }_j{A}_k-{\partial }_k{A}_j$ in an obvious notation. How does this work in four dimensions? To find out, we define the electromagnetic field tensor

$$F^{\mu \nu }={\partial }^{\mu }{A}^{\nu }-{\partial }^{\nu }{A}^{\mu }.$$

(This is the standard notation, with both indices up$—$ so ${\partial }^{\mu }=\partial /\partial x_{\mu },d{x}_{\mu }=g_{\mu \nu }d{x}^{\nu }$ .)

This matrix is constructed from Lorentz vectors, so it will necessarily Lorentz transform as a tensor:

$$F^{{\rho }^{\prime }{\sigma }^{\prime }}={\Lambda }^{{\rho }^{\prime }}{}_{\mu }{\Lambda }^{{\sigma }^{\prime }}{}_{\nu }{F}^{\mu \nu }.$$

Notice that it is an antisymmetric 4 x 4 matrix, so the four diagonal elements are zero, the other elements occur in pairs with opposite signs, and there are only six independent elements.And, we already know three of them:

$\left(F^{23},F^{31},F^{12}\right)=\left(B_x,B_y,B_z\right).$

But what about $F^{01},F^{02},F^{03}$ ?

The first one

$F^{01}={\partial }^0{A}^1-{\partial }^1{A}^0=-\frac{1}{c}\frac{\partial A_x}{\partial t}-\frac{\partial }{\partial x}\frac{\phi }{c},$

remembering ${\partial }^0=-{\partial }_0=-\left(1/c\right)\partial /\partial t.$ This is nothing but the $x$-component of

$-\frac{1}{c}\left(\dot{\overrightarrow{A}}+\overrightarrow{\nabla }\phi \right)=\frac{\overrightarrow{E}}{c},$

so evidently

$$F^{\mu \nu }=\left(\begin{array}{cccc}0& E_x/c& E_y/c& E_z/c\\ -E_x/c& 0& B_z& -B_y\\ -E_y/c& -B_z& 0& B_x\\ -E_z/c& B_y& -B_x& 0\end{array}\right).$$

Exercise: Use $F^{\overline{\rho }\overline{\sigma }}={\Lambda }^{\overline{\rho }}{}_{\mu }{\Lambda }^{\overline{\sigma }}{}_{\nu }{F}^{\mu \nu }$ for the particular case of a boost $\beta =v/c$ along the $x$-axis to find the Lorentz transformation equations for electric and magnetic fields. You should get the equations we derived earlier by considering moving capacitors, etc. (To get you started, ${\overline{F}}^{01}={\Lambda }^0{}_{\mu }{\Lambda }^1{}_{\nu }{F}^{\mu \nu }$ has only two nonzero terms.)

The Levi-Civita Tensor

An interesting higher order tensor (sometimes called the Levi-Civita tensor) is the set of numbers ${\epsilon }^{\mu \nu \rho \sigma }$ defined as follows:

$${\epsilon }^{0123}=1.$$

For other indices, ${\epsilon }^{\mu \nu \rho \sigma }=\pm 1$ if $\left(\mu ,\nu ,\rho ,\sigma \right)$ is $\left(0,1,2,3\right)$ in some order; +1 for an even, -1 for an odd permutation.

But if any of the four indices are equal, then ${\epsilon }^{\mu \nu \rho \sigma }=0.$ Thus, only 24 of the 256 elements are nonzero.

It turns out that this tensor is invariant under Lorentz transformations (although it changes sign under reflection, but that is not our concern here.)

To see how this comes about, look at the transformation equation:

${\epsilon }^{{\alpha }^{\prime }{\beta }^{\prime }{\gamma }^{\prime }{\delta }^{\prime }}={\Lambda }^{{\alpha }^{\prime }}{}_{\mu }{\Lambda }^{{\beta }^{\prime }}{}_{\nu }{\Lambda }^{{\gamma }^{\prime }}{}_{\rho }{\Lambda }^{{\delta }^{\prime }}{}_{\sigma }{\epsilon }^{\mu \nu \rho \sigma }.$

In fact, the expression on the right-hand side is the definition of the determinant $\left|\Lambda \right|$ if we take $\left({\alpha }^{\prime },{\beta }^{\prime },{\gamma }^{\prime },{\delta }^{\prime }\right)$ equal to $\left(0,1,2,3\right),$ and permutation is equivalent to permuting the rows of the matrix, which bring in the appropriate sign change in the determinant. If two are equal, that's the determinant of a matrix with two identical rows, which is zero (in finding a determinant, you can always subtract one row from another).

A Pseudotensor

Actually this $\epsilon$ is sometimes called a pseudotensor, because it changes sign if the Lorentz transformation includes a reflection.

Exercise: how much of this is true in ordinary three-dimensional space of the set ${\epsilon }_{ijk}$ ?

Dual Tensors

The Levi-Civita symbol is important in general relativity, its value here is that we can use it to generate another Lorentz invariant tensor from $F^{\mu \nu }$ :

$$G^{\alpha \beta }=\frac{1}{2}{\epsilon }^{\alpha \beta \gamma \delta }{F}_{\gamma \delta }.$$

This dual tensor is denoted by a script F in Jackson, and some books use $*F$ , the so-called Hodge dual symbol. But the reason we mention this at all is that it illustrates the electric$—$magnetic duality nicely, and it makes some conserved quantities explicit, as we'll see below.

Compare $F^{\mu \nu }$above with

$$G^{\mu \nu }=\left(\begin{array}{cccc}0& B_x& B_y& B_z\\ -B_x& 0& -E_z/c& E_y/c\\ -B_y& E_z/c& 0& -E_x/c\\ -B_z& -E_y/c& E_x/c& 0\end{array}\right).$$

Just as with $F^{\mu \nu },$ the Lorentz transformation of $G^{\mu \nu }$ yields the equations we derived for transformation of electric and magnetic fields between frames.

We've already seen that the magnitude of a vector, say $A^{\mu }{A}_{\mu },$, is invariant under a Lorentz transformation. In the exact same way, we can show that $F^{\mu \nu }{F}_{\mu \nu }$ is invariant. Now

$F^{\mu \nu }{F}_{\mu \nu }=F^{01}{F}_{01}+F^{02}{F}_{02}+F^{03}{F}_{03}+F^{12}{F}_{12}+F^{23}{F}_{23}+F^{31}{F}_{31},$

and recalling that lowering a $0$ suffix brings in a minus sign, we find that

$$-\frac{E^2}{c^2}+B^2$$

is Lorentz invariant.

But what is the significance of this quantity? We'll return to it later: it's the Lagrangian density. (Recall that in classical mechanics the Lagrangian was the difference of kinetic energy and potential energy$—$ something similar is happening here.)

Note this invariance tells us that if in some frame there is only a magnetic field, there is no frame in which there is only an electric field.

The inner product $G^{\alpha \beta }{G}_{\alpha \beta }$ gives the same invariant, by inspection, but $F^{\alpha \beta }{G}_{\alpha \beta }=\overrightarrow{E}\cdot \overrightarrow{B},$ evidently another Lorentz invariant.

Exercise: Check that Maxwell's equations can be written ${\partial }_{\nu }{F}^{\mu \nu }={\mu }_0{J}^{\mu },{\partial }_{\nu }{G}^{\mu \nu }=0$ and that the Lorentz force law is the nonrelativistic limit of $d{p}^{\mu }/d\tau =q{U}_{\nu }{F}^{\mu \nu }.$

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