# Lagrangian density, regular Lagrangian, E&M

1. Jul 6, 2014

### HJ Farnsworth

Greetings,

I have two semi-related questions.

1. When making the Lagrangian formalism of electrodynamics, why is it that we use the Lagrangian density $\mathcal{L}$, rather than the plain old regular Lagrangian $L$? Is this something that is necessary, or is it more that it is just very advantageous for simplicity, etc.? To put it another way, somewhere along the way someone must have been trying to set up E&M using Lagrangians and had some realization where they thought, "this would be much better if I use $\mathcal{L}$ instead of $L$." What was this realization?

2. For classical field theories using Lagrangian densities in general, does the difference between a Galilean-invariant theory and a Lorentz-invariant theory come purely from the choice of $\mathcal{L}$, and whether that choice (or the resulting equations of motion) are invariant under Galilean transformations or Lorentz transformations? Or, is there instead an axiom of classical field theory that can be altered to differentiate between Galilean- and Lorent-invariant theories?

Thanks for any help that you can give.

-HJ Farnsworth

2. Jul 6, 2014

### Jano L.

The plain old regular Lagrangian refers to finite number of variables and leads to finite number of ordinary differential equations.

The purpose of the Lagrangian formalism of ordinary electromagnetic theory is to derive all or at least some of the Maxwell equations. These are partial differential equations for fields. Such equations cannot be derived from finite number of ordinary differential equations.

Similarly to continuum mechanics, it was found that they can be derived from a variational principle where integral of certain quantity $\mathcal{L}$ over space and time plays role. In case of EM theory this quantity is a function of finite number of potentials and field strengths. The quantity
$$L=\int\mathcal{L}dV$$
is still called lagrangian, but due to the integral over space it is a functional of the potentials and field strengths so it is not "the plain old regular Lagrangian" in the above sense.

Both $\mathcal{L}$ and $L$ can be used to find the equations for the field.

No. In addition, the transformation properties of the fields have to be assumed. For example, the wave equation for the field $u$
$$\frac{\partial ^2u}{\partial t^2} = c^2\Delta u$$
with $c$ equal to speed of light can be derived from the Lagrangian density
$$\mathcal{L}=(\partial_t u)^2 -c^2(\partial_k u )(\partial_k u)$$
Whether the theory is Galilei or Lorentz invariant depends on how $u$ transforms and this cannot be derived purely from the function $\mathcal{L}$.

3. Jul 7, 2014

### HJ Farnsworth

Hi Jano,

Great answer, thanks for helping me out!