EM Lagrangian

Here’s something I’ve been working on and I found a derivation in a book called Quantum Field Theory Demystified by McMahon. Hardly a canonical text, however it has a lot of worked out examples and yes its a bit dumbed down but its good to fall back against when you stumble.

I first came across the idea of a Lagrangian density for EM field theory in a book by Altland and Simons called Condensed Matter Field Theory. I was salivating as they described the criteria they were going to use to construct this Lagrangian.

(1) Lorentz invariance
(2) Gauge invariance
(3) Simplicity

Symmetry and simplicity…oh goodness…and the surprising thing is that it works. I’ll have more to say about these assumptions later in this post or a later one but it has to do with gauge invariance.

Let’s recall that the 4-potential is A^{\mu}=(\phi, \vec{A}) where we’re setting c=1. We can define the electromagnetic field tensor in terms of this:


Also recall the four current J^{\mu}=(\rho, \vec{J}). Then we can consider the following Lagrangian density:


So we’ve include derivatives of the potential (up to second order) but only terms of first order in it. Both of these contracts are Lorentz invariants and so our density is Lorentz invariant. The constants K_1,K_2 will be determined after the fact. A note, if we had included a term of order two in the potential then we would loose gauge invariance which I’ll show below. Gauge invariance is pretty important as I’m starting to realize.

Its also worth noting that our generalized coordinates here are the components of the 4-potential. At each point in space we want to know how that field acts, so at each point in space we construct a Lagrangian. For a scalar field we would have only that one value to vary (to find the ‘shape’ of the field minimizing the action) but for a 4-vector field we have 4 ‘generalized’ coordinates at each point in space. So when we vary our action we need to know what we’re varying.

Given that there exists a field A^{\mu} minimizing the action, we’ll see what requirement we can place on it. When we vary the action we’re going to vary the ‘coordinates’ A_{\mu}, you could do the contravariant components as well but I’m going to go along with the derivation.

The action is,

S=\int K_1F_{\mu\nu}F^{\mu\nu}+K_2J^{\mu}A_{\mu}  d^4x

Where d^4x is the Lorentz invariant 4-volume element. Now we vary the action (this is still something I’m working on as its presented in many ways and the way I usually do it is far from efficient…)

\delta S=\int K_1(\delta F_{\mu\nu})F^{\mu\nu}+F_{\mu\nu}(\delta F^{\mu\nu})+K_2 J^{\mu}\delta A_{\mu} d^4x

Notice that the \delta passed right through the 4-current, but then the 4-current doesn’t depend on the components of the 4-potential. Let’s look at these terms one at a time.

(\delta F_{\mu\nu})F^{\mu\nu}=\delta(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})F^{\mu\nu}

The \delta and the \partial commute so we can interchange them and we can integrate each term by parts to swap derivatives, pick up a minus sign and some boundary terms. We will expect that the variation of the potential goes to zero at the boundary so those terms will fall out. This leaves us with,

-\delta A_{\nu}\partial_{\mu}F^{\mu\nu}+\partial_{\nu}F^{\mu\nu}\delta A_{\mu}

Notice that in each term all the indices are being summed over. So they are dummy indices and we can switch them around as is convenient to us. Let’s take the first term and make a switch of \mu\to \nu and vice versa,

-\delta A_{\mu}\partial_{\nu}F^{\nu\mu}+\partial_{\nu}F^{\mu\nu}\delta A_{\mu}

Now use the antisymmetric nature of the field tensor F^{\mu\nu}=-F^{\mu\nu},

\delta A_{\mu}\partial_{\nu}F^{\mu\nu}+\partial_{\nu}F^{\mu\nu}\delta A_{\mu}

So we can sum these two terms resulting in,

2\delta A_{\mu}\partial_{\nu}F^{\mu\nu}

You can do this to the other term as well, though you have to use the metric tensor to lower indices on the A^{\mu} since we’re not varying those components, not exactly. Then you’ll get something that almost looks like the above but the partials have raised indices as well, just lower them by again using the metric tensor and you’ll have a pair of metric tensors and also lower indices on the field tensor. So group those together and voila, you’ll have another copy of the above. Putting that all together you get,

\delta S =\int (4K_1\partial_{\nu}F^{\mu\nu}+K_2J^{\mu})\delta A_{\mu}d^4x

We require that this variation vanish for arbitrary variations of A_{\mu} and so we result in the following condition on the components of the 4-potential,


I’m going to pause here (the typesetting is killing my enthusiasm). Next time I might unpack these, or you can if you want (just look up the other definition of the field tensor in terms of the E and B fields). But there’s another detail I want to talk about in the future. Why we didn’t include A^{\mu}A_{\mu}. If we do we won’t have gauge invariance (you can check this though I will next time) and gauge invariance and charge conservation go hand in hand it seems. So if we give up gauge invariance we loose charge conservation. It also turns out that if we were to put an unknown constant in front of that term it would be related to the mass of a photon. There have been some experiments to put upper bounds on such a mass (if its not zero) and so far its in the ball park of 21 orders of magnitude below that of an electron. But this is pretty huge, as I see it, the interplay between elegance, symmetry, and experimental consequences comes together here. With any luck I’ll write on that soon.


About because0fbeauty

Let's become better.
This entry was posted in Uncategorized. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s