Field Lagrangians as systems with infinte degrees of freedom?

[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?

 

[peteratcam]

(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?

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.

 

[pellman]

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

promoting the number of parameters in the way you describe in (1) is playing fast and loose with the structure of classical mechanics.

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

 

[peteratcam]

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

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.