# Lagrangian density for the EM field in a dielectric medium

1. Jan 8, 2007

Why does the Lagrangian density for the EM field in a dielectric medium take the form $$d^3 \bf x \left[ \epsilon \bf E^2 - \bf B^2 \right]$$? I can see that the expression for Lagrangian density has units of energy per unit volume as you would expect but that's about it. Much appreciated.

2. Jan 9, 2007

### dextercioby

It comes from the relativistic lagrangian describing an em field in a dielectric medium. I think (not 2 sure, though) there's a chapter on this in Griffiths text on electrodynamics.

Daniel.

3. Jan 9, 2007

hmm would the following reasoning be sufficient?

1. The lagrangian density for an EM field in a dielectric medium has to have dimesions of energy per unit volume.

2. Lagrangians and such involve a number of terms.

3. Each term must be a contribution from each field involved and the terms must reflect (only dimensionally) the physics of the system. In this case the properties/aspects/entities of the system are: the dielectric property of the medium ($$\epsilon$$), the magnetic field (B), and the electric field (E). In cgs units energy density for B field is given here: http://scienceworld.wolfram.com/physics/MagneticFieldEnergyDensity.html
and energy density for E field is given here: http://en.wikipedia.org/wiki/Energy_density
Using dimensional analysis we find that the terms of the lagrangian density are as given in the OP

4. Considering that all lagrangians involve atleast one negative and one positive term and that it's just a matter of convention which term is which we arrive at the given lagrangian density.

It's all quite handwavey but all based on sound physical principles applied in a sound manner. Since Langrangians and such are rather fundamental principles that cannot be derived I'm pretty sure I'm justified in using this logic to answer my question in the OP. Comments would be much appreciated. Thanks :)

Last edited: Jan 9, 2007
4. Jan 9, 2007

### masudr

If you look at it from the point of view of special relativity, the Lagrangian has to be a scalar quantity. Given that the field tensor is a 2-form, that seriously limits the possible combinations of the field tensor you can think of. Perhaps the simplest is:

$$L = k F^{\alpha\beta}F_{\alpha\beta}$$

where k is a constant. This does indeed yield Maxwell's equations.

5. Jan 10, 2007

### dextercioby

There's actually more to it, Masud. One should build a lagrangian that incorporates the gauge invariance of the theory. Yours does. However, there's one more restriction: it should have at most 2 derivatives. That is the eqna of motion must be at most second order in derivatives. If that hadn't been an issue, then more general lagrangian (densities) could be built.

For example: $\mathcal{L} =k\left(F^{\mu\nu}F_{\mu\nu}\right)^{2}$.

Daniel.

6. Jan 10, 2007

### vanesch

Staff Emeritus
Yes, but we are working in a dielectric, which has a preferred (rest) frame to it, so we are not obliged anymore to respect Lorentz invariance, no ?