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The Symmetric Stress Tensor

Imagine a big blob of jelly. Imagine poking it on a side. The whole thing wiggles and distorts, as the force of your poke acts on the entire blob of jelly. The mathematical mechanism that describes how your poke is distributed is calle the stress tensor of the material. It tells how energy and momentum are connected by the medium itself.

The same concept can be generalized to a four dimensional medium, where the ``jelly'' is space time itself. Let us now study what an electromagnetic stress tensor is, and how it relates to electromagnetic ``pokes''. Recall that

\begin{displaymath}
p_i = \frac{\partial L}{\partial \dot{q}_i}
\end{displaymath} (17.76)

is the canonical momentum corresponding to the variable $q_i$ in an arbitrary Lagrangian. The Hamiltonian is given, in this case, by
\begin{displaymath}
H = \sum_i p_i \dot{q}_i - L
\end{displaymath} (17.77)

as usual. If $\partial L / \partial t = 0$ then one can show that $\partial
H/ \partial t = 0$. For four dimensional fields we should probably have a Lagrangian and Hamiltonian density whose 3-integral are the usual Lagrangian and Hamiltonians. The Hamiltonian is the energy of a particle or system, so it should transform like the zeroth component of a four vector. Thus, since
\begin{displaymath}
H = \int {\cal H} d^3x
\end{displaymath} (17.78)

and $d^4x = d_0x d^3x$, then ${\cal H}$ must transform like the time component of a second rank tensor. If we define the Hamiltonian density ${\cal H}$ in terms of the Lagrangian density ${\cal L}$ of a field, then
\begin{displaymath}
{\cal H} = \sum_k \frac{\partial \cal L}{\partial \left(
\...
...{t}} \right)} \frac{\partial \phi_k}{\partial t} - {\cal L} .
\end{displaymath} (17.79)

Well, great! The first factor in the sum is the conjugate momentum by definition, and the second is the generalized ``velocity''. Since ${\cal H}$ must transform like the time component of a second rank tensor (and the time derivative appears in this equation) it appears that the covariant generalization of the Hamiltonian density is something that puts a covariant derivative there, instead. We try

\begin{displaymath}
T^{\alpha \beta} = \sum_k \frac{\partial \cal L}{\partial (...
...hi_{k})}
\partial^\beta \phi_k - g^{\alpha \beta} {\cal L}.
\end{displaymath} (17.80)

This is called the canonical stress tensor, and is related to the stress tensor defined and studied in Chapter 6. This tensor has the covariant function of telling us how the energy and momentum carried by the electromagnetic field transform.

What is this tensor? It is, in fact, highly non-trivial. The best we can do is note that if we assume that only free fields are present and that the free fields are localized in some finite region of space (neither assumption is particularly physical), then we can show that

\begin{displaymath}
\int T^{00} d^3 x = \frac{1}{2} \int (\epsilon_0{\bf E}^2 +
\frac{1}{\mu_0}{\bf B}^2) d^3 x = E_{\rm field}
\end{displaymath} (17.81)

and
\begin{displaymath}
\int T^{0i} d^3 x = \epsilon_0 c \int ({\bf E} \times {\bf ...
...{c} \int ({\bf E} \times {\bf H})_i d^3 x = c P^i_{\rm field}
\end{displaymath} (17.82)

which are the ``usual'' expressions for the energy and momentum of the free field. At least if I got the change to SI units right...

What, you might ask, is this good for? Well, aside from this correspondance (which is full of holes, by the way), we can write the energy-momentum conservation law

\begin{displaymath}
\partial_\alpha T^{\alpha \beta} = 0.
\end{displaymath} (17.83)

This is proven in Jackson, with a discussion of some of its shortcomings.

One of these is that it is not symmetric. This creates difficulties when we consider the angular momentum carried by the field. Since the angular momentum density is important when we go to create photons (which must have quantized angular momenta), it is worthwhile to construct the symmetric stress tensor

\begin{displaymath}\Theta^{\alpha \beta} = \left( g^{\alpha \mu} F_{\mu
\lambda}...
...1}{4} g^{\alpha \beta}F_{\mu \lambda}F^{\mu
\lambda} \right)
\end{displaymath} (17.84)

in terms of which we can correctly construct a covariant generalization of the energy momentum conservation law
\begin{displaymath}
\partial_\alpha \Theta^{\alpha \beta} = 0
\end{displaymath} (17.85)

and the angular momentum tensor
\begin{displaymath}
M^{\alpha \beta \gamma} = \Theta^{\alpha \beta} x^\gamma - \Theta^{\alpha
\gamma} x^\beta
\end{displaymath} (17.86)

which is therefore conserved. This form of the stress tensor can also be directly coupled to source terms, resulting in the covariant form of the work energy theorem for the combined system of particles and fields.

This is about all we will say about this at this time. I realize that it is unsatisfactory and apologize. If we had one more semester together, we could do it properly, but we don't. Therefore, it is on to


next up previous contents
Next: Covariant Green's Functions Up: Relativistic Dynamics Previous: Building a Relativistic Field   Contents
Robert G. Brown 2013-01-04