# Lagrangian/Hamiltonian formulations in physics

1. Apr 6, 2013

### Logic Cloud

I'm wondering to what extent Lagrangian/Hamiltonian formulations are used outside of mechanics. So far I've found that geometrical optics has such formulations and I can imagine thermodynamics has something like this as well because I have seen the concepts phase space and conjugate variables being used there. I have not, however, been able to find these kind of descriptions for electromagnetism. Is there such a thing as a "Lagrangian/Hamiltonian formulation of electromagnetism"? And if so, could you say that these formulations exist for any subfield of physics?

2. Apr 6, 2013

### kevinferreira

The Lagrangian formulation, and especially the action formulation and the principle of least action, are used in every kind of physics.
Even if in mechanics it is a formulation which is not as fundamental, in the sense that you can work without it, in modern physics it is the fundamental concept, the fundamental key where everything begins. So, even if we don't have a priori a formulation in terms of an action, we always try to have it in order to have an deeper insight. Yes, there is an action formualtion of electromagnetism (quantum electrodynamics), of general relativity, string theory, everything.

3. Apr 6, 2013

### Jano L.

Yes, there are such formulations, especially the Lagrangian formulation is quite developed in electrodynamics.
The Hamiltonian formulation is more problematic.

No, that would not be true. Such formulations occur when some differential equations are to be derived, but not all of physics is differential equations.

Variational formulations have certain advantages, but they are also quite restrictive. Not all subfields of physics can be currently derived from variational principle. Often the basic equations have to be postulated. For example, the Navier-Stokes equations in hydrodynamics, or various models of dissipative dynamics, like Pauli kinetic equation, or time-delayed equations. They are useful equations and "form a subfield of physics", but they do not have apparent variational origins. They may remain on their own, or they may be eventually derived from some other ideas. Although possible, it is not necessary that these subfields will be eventually derived from variational principle.

4. Apr 6, 2013

### HomogenousCow

The Navier-Stokes equation dosen't have a lagrangian? I thought you could find lagrangians for all differential equations.

5. Apr 6, 2013

### WannabeNewton

Well once you start talking about field theories like the classical electromagnetic field there is a difference between the lagrangian for the EM field, which will give you the field equations of EM, and the lagrangian for a charged particle interacting with the field which will give you the equations of motion for that particle.

For example, making no assumptions about non-relatavistic motion, a charged particle interacting with an electromagnetic field will have the usual free lagrangian $L_{free} = \frac{p^{\mu}p_{\mu}}{m\gamma}$ and we can take the interaction term to be $L_{int} = \frac{q}{mc\gamma}A_{\mu}p^{\mu}$ where $A_{\mu}$ is the electromagnetic 4-potential, $p^{\mu}$ is the 4-momentum of the charged particle, and $\gamma$ is the usual gamma factor from special relativity. Thus our total action would be $S = \frac{1}{m}\int(p_{\mu} + \frac{q}{c}A_{\mu}) p^{\mu}\frac{dt}{\gamma} = \int(p^{\mu} + \frac{q}{c}A_{\mu})\mathrm{d} x^{\mu}$. Varying $S$ and using the principle of stationary action will give you the usual equations of motion for the charged particle $\frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau} = \frac{q}{c}(\partial ^{\mu}A_{\nu} - \partial _{\nu}A^{\mu})u^{\nu} = \frac{q}{c}F^{\mu}_{}{}_{\nu}u^{\nu}$ where $F_{\mu\nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$ is the usual electromagnetic field tensor and $u^{\mu}$ is the 4-velocity of the particle.

We can do a similar thing for the electromagnetic field itself in order to get the inhomogenous parts of maxwell's equations (the homogenous parts are immediate from the definition of the electromagnetic field tensor in terms of the 4-potential). Here we first define a lagrangian density $\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + A_{\mu}J^{\mu}$ which is a scalar function of space-time points (here $J^{\mu}$ is the 4-current). The lagrangian is then defined in terms of the lagrangian density as $L = \int \mathcal{L}d^{3}x$ so that the action will be $S = \int Ldt = \int \mathcal{L}d^{4}x$. If you vary the action and apply the principle of stationary action you will in fact get $\partial^{\nu}F_{\mu\nu} = J_{\mu}$ which are in fact the inhomogenous parts of maxwell's equations.

This is but one example of how the lagrangian formulation would be applied outside of mechanics but the example of the electromagnetic field is a very very nice one. You can also, for example, apply the lagrangian formulation to derive Einstein's field equations in general relativity and also apply it to get the equations of motion for a particle in free fall in curved space-time. There variational approach gives a very nice way of exploring the dynamics of the field.

6. Apr 6, 2013

### Jano L.

Why would you think that? Try to find one for the logistic equation:

$$\dot x = x(1-x),$$

or for the (in physics more familiar) equation

$$\ddot x + \gamma \dot x + \omega_0^2 x = 0.$$