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Building a Relativistic Field Theory

We have not quite finished the job of building a proper relativistic field theory of electromagnetism. That is because we would like to be able to obtain all of the equations of motion (that is, physics) describing the system from a covariant action principle. We have done that for the particles in the fields, but what about the fields themselves? In fact, since the particles produce (and hence modify) the fields, we do not even have the correct solutions for the particles alone, yet. Let us see if we can develop a suitable Lagrangian for the fields that leads, ideally, to Maxwell's equations.

The Rules for building a field theory Lagrangian are of interest in and of themselves, since they are quite general. The rules are:

  1. Take the position and velocity coordinates for continuous space time and replace them with field variables.
  2. Label the field variables with discrete (coordinate direction) labels and with continuous (position) variables.
  3. Replace the ``velocity'' with the 4-gradient.
  4. Require the action to be stationary w.r.t. variations in the field variables themselves and their gradients.
That is,
$\displaystyle i$ $\textstyle \to$ $\displaystyle x^\alpha,k$ (17.57)
$\displaystyle q_i$ $\textstyle \to$ $\displaystyle \phi_k(x)$ (17.58)
$\displaystyle \dot{q}_i$ $\textstyle \to$ $\displaystyle \partial^\alpha \phi_k(x)$ (17.59)
$\displaystyle L = \sum_i L_i(q_i,\dot{q}_i)$ $\textstyle \to$ $\displaystyle \int {\cal L}(\phi_k,\partial^\alpha
\phi_k) d^3x$ (17.60)
$\displaystyle \frac{d }{dt}\left( \frac{\partial L}{\partial \dot{q}_i} \right) =
\frac{\partial L}{\partial q_i}$ $\textstyle \to$ $\displaystyle \partial^\beta \frac{\partial \cal L}{\partial \partial^\beta
\phi_k} = \frac{\partial \cal L}{\partial \phi_k} .$ (17.61)

When we make an action integral, we integrate ${\cal L}$ over time, making the total integral four dimensional. We therefore call ${\cal L}$ the Lagrangian density in four dimensions. Note that the action will be covariant provided the Lagrangian density is a 4-scalar. This is what I have meant whenever I have inadvertantly called the ``Lagrangian'' a scalar. Good, clean, relativistic theories with or without particles are made out of scalar Lagrangian densities, not Lagrangians per se:

A = \int \int {\cal L} d^3x dt = \int {\cal L} d^4 x .
\end{displaymath} (17.62)

We now do the usual dance. We know that ${\cal L}$ for the fields must be a scalar. We also know that Maxwell's equations relate the fields to the currents that produce them, and also link the electric and magnetic fields. We thus need to build a theory out of $F^{\alpha \beta}, A^\alpha, J^\alpha$. Various ways we can do this include

F_{\alpha \beta} F^{\alpha \beta}

J_\alpha A^\alpha

F_{\alpha \beta} {\cal F}^{\alpha \beta}

and still messier pieces like

F_{\alpha \beta} J^\alpha A^\beta .

The first two terms are invariant under the transformations of the full Lorentz group. The third is not a scalar under inversion, but a pseudoscalar (odd under inversion). We reject it. The last is a mess. We reject it. We want a term quadratic in the 4-gradients of the fields. This is the first term. We want a source term to couple the fields and the particles. The second term does that.

So, we try a Lagrangian density with just these two terms, with unknown constants $Q$ and $R$ that have to be determined so that they correctly reconstruct Maxwell's equations in whatever system of units we like:

{\cal L} = - Q F_{\alpha \beta} F^{\alpha\beta} - R J_\alpha A^\alpha .
\end{displaymath} (17.63)

We need to take derivatives of ${\cal L}$ with respect to $\partial^\beta A^\alpha$, so it is useful to write this Lagrangian in the form:

{\cal L} = - Q g_{\lambda \mu} g_{\nu \sigma} (
...bda A^\nu -
\partial^\nu A^\lambda) - R J_\alpha A^\alpha .
\end{displaymath} (17.64)

When we form $\partial {\cal L}/\partial (\partial^\beta A^\alpha)$ we get delta functions whenever $\alpha$ and $\beta$ are equal to a pair of the indices above. We therefore get four terms:
\frac{\partial \cal L}{\partial (\partial^\beta A^\alpha)} ...
\delta_\alpha^\lambda F^{\mu \sigma} \right\}
\end{displaymath} (17.65)

where the first two terms come from delta functions formed from the first term and the second two terms come from delta functions formed from the second term.

$g_{\alpha \beta}$ is symmetric (in fact, diagonal). The $F^{\alpha\beta}$ is antisymmetric. When we do the sums against the $\delta$-functions, the four terms make identical contributions:

\frac{\partial \cal L}{\partial (\partial^\beta A^\alpha)} = - 4Q
F_{\beta \alpha} = 4 Q F_{\alpha \beta} .
\end{displaymath} (17.66)

The other part of the E-L equation (which corresponds in position space to the ``potential'', or ``force'' term) is
\frac{\partial \cal L}{\partial A^\alpha} = - R J_\alpha.
\end{displaymath} (17.67)

Therefore the equations of motion for the electromagnetic field can be written
4Q \partial^\beta F_{\beta \alpha} = R J_\alpha .
\end{displaymath} (17.68)

If one checks back in one's notes, one sees that this is indeed the covariant form of the inhomogeneous Maxwell's equations if $Q = 1/4$ and $R = \mu_0$:

\partial^\beta F_{\beta \alpha} = \mu_0 J_\alpha
\end{displaymath} (17.69)

follows from the Lagrangian density:
{\cal L} = - \frac{1}{4} F_{\alpha \beta} F^{\alpha\beta}
- \mu_0 J_\alpha A^\alpha .
\end{displaymath} (17.70)

Therefore the Lagrangian we have constructed yields the inhomogeneous Maxwell equations, but not the homogeneous ones. That is okay, though, because we have constructed the $F^{\alpha\beta}$ in terms of the $A^\alpha$ in such a way that the homogeneous ones are satisfied automatically! To observe that this miracle is true, we recall the covariant form of the homogeneous equations:

\partial_\alpha {\cal F}^{\alpha \beta} = 0 .
\end{displaymath} (17.71)

{\cal F}^{\alpha \beta} = \frac{1}{2} \epsilon^{\alpha \beta \gamma
\delta} F_{\gamma \delta} .
\end{displaymath} (17.72)

$\displaystyle \partial_\alpha {\cal F}^{\alpha \beta}$ $\textstyle =$ $\displaystyle \frac{1}{2} \partial_\alpha
\epsilon^{\alpha \beta \gamma \delta} F_{\gamma \delta}$  
  $\textstyle =$ $\displaystyle \partial_\alpha \epsilon^{\alpha \beta \gamma \delta}
\partial_\gamma A_\delta$  
  $\textstyle =$ $\displaystyle \epsilon^{\alpha \beta \gamma \delta} \partial_\alpha
\partial_\gamma A_\delta$ (17.73)

is the first term. But $\partial_\alpha \partial_\gamma$ is symmetric, while $\epsilon^{\alpha \beta \gamma \delta}$ is antisymmetric in the same two indices, so the contraction on the two indices vanishes (work it out term by term if you doubt it).

Thus the homogeneous equations are satisfied by our definition of ${\cal F}^{\alpha\beta}$ quite independent of any dynamics. In four dimensions, all of the inhomogeneous source terms must appear in equations with the form of the inhomogeneous equation above, and only one of these equations can result from the action principle. The similarity transformation to the fields we observe is thus the ``natural'' form of the ME's, and in four dimensions the homogeneous equations are really not independent as far as the action principle is concerned. Note that this is fundamentally because our field strength tensor derives from the definition of the magnetic field as the curl of the vector field $\vec{A}$ (which is divergenceless) which is built into the definition.

As a final quixotic note, observe that if we take the 4-divergence of both sides of the inhomogeneous Maxwell equations:

\partial^\alpha \partial^\beta F_{\beta \alpha} =
\mu_0 \partial^\alpha J_\alpha
\end{displaymath} (17.74)

the left hand side vanishes because again, a symmetric differential operator is contracted with a completely antisymmetric field strength tensor. Thus
\partial^\alpha J_\alpha = 0 ,
\end{displaymath} (17.75)

which, by some strange coincidence, is the charge-current conservation equation in four dimensions. Do you get the feeling that something very deep is going on? This is what I love about physics. Beautiful things are really beautiful!

We will now blow off the ``proca'' Lagrangian, which would be appropriate if the photon had a mass. It doesn't, but if it did you would need to read this chapter. It might, of course, so you should probably read the chapter anyway, but it currently (bad pun) doesn't so I'm going to make light of it (worse pun) and continue.

If we had one more month, we would now study the covariant forms of the stress tensor. It is important, but it is also quite difficult, and necessitates a broader discussion than we can now afford. To treat the subject properly, we would need to treat parts of chapter 17 simultaneously, and we would need to do a lot of algebra. This would mean that we would miss (in all probability) being able to learn the Liénard-Wiechart potential, which is far more important. We will therefore content ourselves with merely defining the stress tensor, remarking on some of its properties without proof, and moving on. You are responsible for working your way through this chapter, according to your needs, inclinations, and abilities, on your own.

next up previous contents
Next: The Symmetric Stress Tensor Up: Relativistic Dynamics Previous: Motion of a Point   Contents
Robert G. Brown 2007-12-28