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Next: The Transformation of Electromagnetic Up: The Lorentz Group Previous: Thomas Precession   Contents

Covariant Formulation of Electrodynamics

We are now ready to get serious about electrodynamics. We have developed a beautiful, geometric system for describing the coordinates in terms of which electrodynamics must be formulated for the speed of light to be an invariant. We have developed a group of coordinate transformations that preserves that invariance. Now we must face up to the fact that our original equations of electrodynamics are not in a ``covariant'' formulation that makes these constraints and transformation properties manifest. For example, we do not yet know how the electric and magnetic fields themselves transform under a LT!

Let us then reformulate our basic equations in 4-tensor form. We will make the equations themselves 4-scalars, 4-vectors, or 4-tensors of higher rank so that we can simply look at them and deduce their transformation properties. In addition, we will simplify the notation when possible.

We begin at the beginning. All we really know about electromagnetic fields is their (defined) action on a charged particle:

\frac{d {\bf p}}{dt} = q \left( {\bf E} + \frac{\bf v}{c} \times {\bf B}
\end{displaymath} (16.135)

(in 3-vector notation). Well, we know that the 3-vector momentum is just part of a 4-vector momentum:
p^\alpha = (p^0,{\bf p}) = m(U^0,{\bf U})
\end{displaymath} (16.136)

(where $p^0 = E/c$). Also, we don't know what ``t'' is (since that depends on the choice of frame) so we need to use ``$\tau$'' instead in our definition.

Thus we can write

\frac{d {\bf p}}{d\tau} = \frac{q}{c} \left( U^0 {\bf E} + {\bf U} \times
{\bf B} \right) .
\end{displaymath} (16.137)

The left hand side tells us the rate of change of the (spatial) momentum, which is just part of a four vector. The time component should tell us how the energy changes with proper time:
\frac{dp^0}{d \tau} = \frac{q}{c} {\bf U} \cdot {\bf E}.
\end{displaymath} (16.138)

Now, if this energy-force 4-vector equation is to be covariant (so its transformed form is still a 4-vector) then the right hand sides must form a 4-vector too. Thus we must be able to express it (as a contraction of co and contra variant tensors) so that this property is ``manifest''. We know (experimentally) that charge is a Lorentz scalar; that is, charge is invariant under LT's. $(U^0,{\bf U})$ forms a contravariant 4-vector.

From this we can deduce the 4-tensor form for the electromagnetic field! Since the space parts ${\bf U} \cdot {\bf E}$ form the time component of a four vector, E must be the time-space part of a tensor of rank two. That is,

{\bf E} \cdot {\bf U} = F^{0 \beta}U_\beta .
\end{displaymath} (16.139)

We could easily find B in a similar fashion and could eventually work out the electromagnetic field strength tensor. However, it is more constructive to keep on making four vectors, etc. out of the rest of the relations at hand.

For example, we already have observed that the continuity equation is a covariant 4-scalar:

\frac{\partial \rho}{\partial t} + \mbox{\boldmath$\nabla$}\cdot {\bf J} = 0 .
\end{displaymath} (16.140)

To make it's covariance manifest, we define a 4-current
J^\alpha = (c \rho, {\bf J})
\end{displaymath} (16.141)

so that
\partial_\alpha J^\alpha = 0
\end{displaymath} (16.142)

is the continuity equation. Note that (as Jackson remarks) this only works because electric charge is a Lorentz invariant and so is a four-dimensional volume element (since $\det A = +1$).

Next, consider the wave equations for the potentials in the Lorentz gauge (note well that Jackson for no obvious reason I can see still uses Gaussian units in this part of chapter 11, which is goiing to make this a pain to convert below - bear with me):

$\displaystyle \frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2} - \nabla^2\phi$ $\textstyle =$ $\displaystyle \frac{\rho}{\epsilon_0} = \frac{J^0}{\epsilon_0 c}$  
  $\textstyle =$ $\displaystyle \mu_0 \frac{J^0}{\mu_0 \epsilon_0 c} = \mu_0 (c J^0)$ (16.143)

so that:
$\displaystyle \frac{1}{c^2}\frac{\partial^2 (\phi/c)}{\partial t^2} - \nabla^2(\phi/c)$ $\textstyle =$ $\displaystyle \mu_0 J^0$ (16.144)
$\displaystyle \frac{1}{c^2}\frac{\partial^2 \mbox{\boldmath$A$}}{\partial t^2} - \nabla^2\mbox{\boldmath$A$}$ $\textstyle =$ $\displaystyle \mu_0\mbox{\boldmath$J$}$ (16.145)

Therefore, if we form the 4-vector potential

A^\alpha = (\frac{\phi}{c},{\bf A})
\end{displaymath} (16.146)

then we can write the various 4-relations:
\partial_\alpha A^\alpha = \frac{1}{c}\frac{\partial A^0}{\partial t} + \mbox{\boldmath$\nabla$}\cdot {\mbox{\boldmath$A$}}
\end{displaymath} (16.147)

(which is the 4-scalar Lorentz gauge condition)
\Box A^\alpha = \partial_\beta \partial^\beta A^\alpha = \mu_0 J^\alpha
\end{displaymath} (16.148)

(the 4-vector inhomogeneous electromagnetic wave equation constructed from the 4-scalar D'Lambertian wave operator - the set of four wave equations for $\phi$ and the components of $\mbox{\boldmath$A$}$ above).

Now we can construct the components of E and B from the covariant 4-vector potential. For example, we know that:

\mbox{\boldmath$E$}= -\mbox{\boldmath$\nabla$}\phi - \frac{\partial \mbox{\boldmath$A$}}{\partial t}
\end{displaymath} (16.149)

where $\phi = cA^0$, so
E_x = - c \frac{\partial A_x}{\partial (c t)} - \frac{\partial c A^0}{\partial x} = -
c (\partial^0 A^1 - \partial^1 A^0)
\end{displaymath} (16.150)

and similarly, since $\mbox{\boldmath$B$}= \mbox{\boldmath$\nabla$}\times \mbox{\boldmath$A$}$:
B_x = \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} = -(\partial^2 A^3 -
\partial^3 A^2)
\end{displaymath} (16.151)


The components of the electric and magnetic fields (all six of them) thus transform like the components of a second rank, antisymmetric, traceless field strength tensor16.7 :

F^{\alpha\beta} = \partial^\alpha A^\beta - \partial^\beta A^\alpha.
\end{displaymath} (16.152)

In explicit component form,
F^{\alpha \beta} =
\left( \begin{array}{cccc}
0 & -E_x/...
...& 0 & -B_x \\
E_z/c & -B_y & B_x & 0
\end{array} \right) .
\end{displaymath} (16.153)

The tensor with two covariant indices (formed by two contractions with $g$) is obtained by replacing $\mbox{\boldmath$E$}$ with $-\mbox{\boldmath$E$}$.
F_{\alpha \beta} =
\left( \begin{array}{cccc}
0 & E_x/c...
... 0 & -B_x \\
-E_z/c & -B_y & B_x & 0
\end{array} \right) .
\end{displaymath} (16.154)

Another important version of this tensor is the dual field strength tensor ${\cal F}^{\alpha\beta}$. In terms of the totally antisymmetric tensor of the fourth rank and the normal field strength tensor it is given by:

{\cal F}^{\alpha\beta} = \frac{1}{2} \epsilon^{\alpha \beta...
... & E_x/c \\
B_z & E_y/c & -E_x/c & 0
\end{array} \right) .
\end{displaymath} (16.155)

This is obtained from the basic contravariant field strength tensor by the substitutions $\mbox{\boldmath$E$}\to \mbox{\boldmath$B$}, \mbox{\boldmath$B$}\to -\mbox{\boldmath$E$}$. Consideration of the section on magnetic monopoles shows that this is indeed a particular duality transformation obtained in free space with the ``rotation'' parameter equal to $\pi/2$ (in J6.151).

Finally, we must write Maxwell's equations in covariant form. The inhomogeneous equations are (recall)

$\displaystyle \mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$E$}$ $\textstyle =$ $\displaystyle \frac{\rho}{\epsilon_0}$ (16.156)
$\displaystyle \mbox{\boldmath$\nabla$}\times \mbox{\boldmath$B$}- \frac{1}{c^2} \frac{\partial \mbox{\boldmath$E$}}{\partial t}$ $\textstyle =$ $\displaystyle \mu_0\mbox{\boldmath$J$}$ (16.157)

The quantity on the right is proportional to the four current. The quantity on the left must therefore contract a 4-derivative with the field strength tensor. You should verify that
\partial_\alpha F^{\alpha \beta} = \mu_0 J^\beta
\end{displaymath} (16.158)

exactly reconstructs the inhomogeneous equation for each component of $J^\beta$.

The homogeneous equations

$\displaystyle \mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$B$}$ $\textstyle =$ $\displaystyle 0$ (16.159)
$\displaystyle \mbox{\boldmath$\nabla$}\times \mbox{\boldmath$E$} + \frac{\partial \mbox{\boldmath$B$}}{\partial t}$ $\textstyle =$ $\displaystyle 0$ (16.160)

also form a four vector (of zero's) and must hence be the contraction of a field strength tensor. But which one? Well, the second homogeneous equation requires that ${\bf B} \to -{\bf E}$ and both require that ${\bf E} \to {\bf
B}$, so it must be the dual:
\partial_\alpha {\cal F}^{\alpha \beta} = 0 .
\end{displaymath} (16.161)

If we feel compelled to write everything in terms of the field strength tensor itself, this can be done. The result is the four equations
\partial^\alpha F^{\beta \gamma} + \partial^\beta F^{\gamma \alpha} +
\partial^\gamma F^{ \alpha \beta} = 0
\end{displaymath} (16.162)

where $\alpha, \beta, \gamma$ are any three of the four indices 0,1,2,3. However, this equation is a third rank tensor on the left, and its reduction by symmetry to a tensor of first rank is not manifest. It is ugly, if you will.

Now that we have written Maxwell's equations (and the consequences of ME) in four dimensional form (remarking all the while that they are unusually beautiful and concise in this notation) we are done. Before we go on to deduce (from these results) how electric and magnetic fields LT, however, we should complete the question with which we began the discussion, namely, how does Newton's law become covariant? The answer is (now that we know what the field strength tensor is)

\frac{dp^\alpha}{d \tau} = m \frac{dU^\alpha}{d \tau} = \frac{q}{c}
F^{\alpha \beta} U_\beta .
\end{displaymath} (16.163)

The time-component equation is just the work-energy theorem, and the space equations are Newton's law.

As a postscript to our discussion (recalling that sometimes the fields propagate in some medium and not free space) we note that in this case the homogeneous equation(s) remain unchanged, but the inhomgeneous equations are modified (by using $\mbox{\boldmath$H$}$ and $\mbox{\boldmath$D$}$ instead of $\mbox{\boldmath$B$}$ and $\mbox{\boldmath$E$}$). The inhomogeneous equation is then

\partial_\alpha G^{\alpha\beta} = \mu J^\beta
\end{displaymath} (16.164)

where hopefully the definition of $G^{\alpha\beta}$ is obvious (that is, substitute $v = \sqrt{1/\epsilon\mu}$ for $c$ throughout in appropriate places, or if you prefer recapitulate the entire derivation using $\mbox{\boldmath$H$}$ and $\mbox{\boldmath$D$}$ from the beginning).

Let us pause for a moment of religious silence and contemplate a great wonder of nature. This is the scientist's version of ``prayer in school''.

next up previous contents
Next: The Transformation of Electromagnetic Up: The Lorentz Group Previous: Thomas Precession   Contents
Robert G. Brown 2007-12-28