# Field Lagrangians as systems with infinte degrees of freedom?

1. Jun 4, 2010

### pellman

There is nothing particular quantum about this question but I'm posting it here because I think the quantum folks are likely more familiar with the topic. Hope that's ok.

There are two ways of looking at field Lagrangian densities in relation to particle Lagrangians.

(1) A particle (one coordinate) action looks like

$$S=\int L\left(q(t),\dot{q}(t)\right)dt$$

t is a continuous parameter. and

$$\dot{q}=\frac{dq}{dt}$$

We can easily generalize this to 4 continuous parameters and write

$$S=\int L\left(q(x),\partial_{\mu} q(x)\right)d^4 x$$

where now $$x=(x^0, x^1, x^2, x^3)$$ and $$\partial_{\mu} q$$ stands for all four partial derivatives. We then call q a field and L a Lagrangian density. But from this viewpoint the names suggest more distinction than is warranted. It is just a matter of how many continuous parameters we consider. The first is a system with one coordinate and one parameter, the second one coordinate and four parameters.

(2) In the second approach we first expand to systems of N coordinates and write

$$S=\int L\left(q_1,...,q_N,\dot{q_1},...,\dot{q_N}\right)dt$$

Then we suppose that N goes to (continuum) infinity. We replace the discrete $$q_j(t)$$ with $$q(\vec{x},t)$$, where x is just a "label" to idenfity one of the infinitely many degrees of freedom. The Lagrangian becomes

$$L=\int \mathcal{L}\left(q(\vec{x},t),\dot{q}(\vec{x},t),\frac{\partial q}{\partial x^j}\right)d^3 x$$

When we put this L into the action integral we get something like what we got in (1) above.

This second approach is common in the physics literature and is suggestive of physical signficance: fields are systems with infinitely many degrees of freedom. However, I find this approach troubling in two ways. First, L is represented as an integral, which is a limit of a sum. Yet there is no summation in the definition of L in the N degrees of freedom case. So how is one the limiting case of the other?

Secondly, how do the partial derivatives with respect to x come in? They are not analogous to anything in the discrete case.

Is this approach inherently misleading? (yet historically useful in an accidental sort of way) Or am I missing something?

Last edited: Jun 4, 2010
2. Jun 9, 2010

### peteratcam

Yes, you're missing the benefit of a specific example. Take the harmonic chain. The discrete degrees of freedom are masses joined by springs.
The Lagrangian (proper) is
$$L = \sum_i \left[\frac{m}{2}\dot q_i^2 - \frac{k}{2}(q_{i+1}-q_i)^2\right]$$
Loosely taking a continuum limit gives
$$L = \int dx \left[\frac{m}{2}\dot q(x)^2 - \frac{ka^2}{2}(\frac{dq}{dx})^2\right]$$
where a is some lattice spacing.
The thing in square brackets is the Lagrangian density $$\cal L$$, since if you integrate it over space you get the Lagrangian.

I find this version much more appealing than the verision (1) you described - promoting the number of parameters in the way you describe in (1) is playing fast and loose with the structure of classical mechanics.

3. Jun 9, 2010

### pellman

thanks! I had forgotten how differences between different q in the discrete case can approach derivatives in the continuum limit.

I'm curious about what you mean. Please elaborate if you have time.

4. Jun 9, 2010

### peteratcam

If I have more time, I will expand more, but my initial reaction is this:
The Hamiltonian is constructed as the Legendre transform of L with respect to the variables $$\{\dot q_i\}$$, or in the continuum case, $$\dot q(x)$$. Thinking that it's ok to just increase the number of parameters from 1 to 4 doesn't respect the fact that the 1 time parameter is rather special in the Hamiltonian formalism.