# Lagrangian density for continuous distribution of matter

1. Jan 15, 2014

### dEdt

The Lagrangian for a point particle is just $$L=-m\sqrt{1-v^2}.$$ If instead we had a continuous distribution of matter, what would its Lagrangian density be? I feel that this should be very easy to figure out, but I can't get a scalar Lagrangian density that reduces to the particle Lagrangian in the appropriate limit.

Last edited: Jan 15, 2014
2. Jan 15, 2014

### pervect

Staff Emeritus
IIRC it's just $L = -\rho$, $\rho$ being the density. If you imagine a stationary box of particles with v=0, you just integrate $-\rho$ over the volume to get m, which is also L since v=0.

If you then perform a Lorentz boost, your box Lorentz contracts along the direction of the boost, making the integral of $\rho$ equal to $m \sqrt{1-v^2}$

3. Jan 15, 2014

### dEdt

But the density isn't a scalar, so how can it be equal to the Lagrangian density?

4. Jan 15, 2014

### pervect

Staff Emeritus
I should have said $\rho_0$, the density in the rest frame of the matter.

5. Jan 15, 2014

### dEdt

But then where's the dependency on velocity? If the Lagrangian density really is just the rest frame density, then the Euler-Lagrange equations don't yield any useful information.

6. Jan 16, 2014

### samalkhaiat

See this View attachment Jackiw fluid.pdf

7. Jan 16, 2014

### pervect

Staff Emeritus
To get the total lagrangian L, you integrate the rest density multiplied by the volume element. When you change frames, your volume element shrinks by a factor of gamma because of Lorentz contraction, and you get $L -m / \gamma = -m \sqrt{1-v^2}$. Because it is a scalar, the _rest_ density $\rho_0$ doesn't change.

8. Jan 20, 2014

### dEdt

Pervect, I was eventually able to convince myself that your answer is correct. I have one final question:

The full Lagrangian for electrodynamics is $\mathcal{L}=\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}-A_\alpha J^\alpha-\rho_0$. Varying by $A_\alpha$ gives Maxwell's equations. But I don't know how to recover the Lorentz force law $f^\alpha=F^{\alpha\beta}J_{\beta}$ ($f^\alpha$ being the 4-force density). What should I vary the action by?

9. Jan 20, 2014

### vanhees71

I don't know what $\rho_0$ might be.

The Lagrangian for the electromagnetic field (minimally) coupled to the (necessarily conserved) electromagnetic current reads (using Heaviside-Lorentz units with $c=1$ and the west-coast convention for the metric, $\eta_{\mu \nu}=\text{diag}(1,-1,-1,-1)$
$$\mathcal{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu}+J_{\mu} A^{\mu}.$$
Now you can go one step further and write down the Lagrangian for a charged particle and the electromagnetic current coupled to the electromagnetic field.

$$L=-m \sqrt{1-\dot{\vec{y}}^2},$$
and the current density is given by
$$j^{\mu}(x)=q \int_{\mathbb{R}} \mathrm{d} t \; \dot{\vec{y}} \delta^{(4)}(x-y).$$
Here, $\vec{y}(t)$ is the trajectory of the particle as a function of the coordinate time in the inertial reference frame with $y^0=t$.

Note that the total action is a Lorentz scalar and thus the equations of motion are relativistically consistent although not written in manifest covariant way. The action now consists of three parts:

The free action of the em. field
$$S_{\text{f}0}[A]=\int \mathrm{d}^4 x \left (-\frac{1}{4} F_{\mu \nu} F^{\mu \nu} \right ),$$
the free action of the particle
$$S_{\text{p0}}[\vec{y}]=-m \int \mathrm{d} t \sqrt{1-\dot{\vec{y}}^2},$$
and the interaction term
$$S_{\text{int}}=\int \mathrm{d}^4 x A_{\mu} j^{\mu}=\int \mathrm{d} t A_{\mu}[t,\vec{y}(t)] \frac{\mathrm{d} x^{\mu}}{\mathrm{d} t}=\int \mathrm{d} t \left [q A^0(t,\vec{y}(t))-\frac{\mathrm{d} \vec{y}}{\mathrm{d} t} \cdot \vec{A}(t,\vec{y}(t)) \right].$$
Taking the variation with respect to $A^{\mu}$ you get Maxwell's equations with the current given by the single-particle current and variation with respect to $\vec{y}$ gives you the equation of motion for a particle in this electromagnetic field.

The only trouble with this fully self-consistent equation is that it is plagued with the well-known problems of the self-consistent description of such a classical point-particle-field system. It's the so-called radiation-reaction problem. A very detailed analysis of the problem can be found in the book

F. Rohrlich, Classical Charged Particles, World Scientific (2007)

A good review is also given in Jackson's Classical Electrodynamics and in vol. 2 of Landau/Lifshitz.

Of course you can read off the usual approximations that can be solved in the sense of perturbation theory, i.e., you consider the motion of the particle in a given electromagnetic field or the electromagnetic field of a particle moving along a given trajectory $\vec{y}(t)$, leading to the retarded potentials (Lienard-Wiechert potentials).

Another approach is to use kinetic theory or hydrodynamics (look, e.g., for "magneto hydrodynamics" in the usual textbooks).

10. Jan 20, 2014

### dEdt

Vanhees,

Thank you for your detailed answer, but I already know how to derive the equations of motion for a point particle from the Lagrangian. What I'm trying to figure out is how to derive the equations of motion for a continuous distribution of matter. The $\rho_0$ in the above equation is the mass density of a continuous distribution of matter in its rest frame, which is supposedly the Lagrangian for a continuous distribution of matter.
Do you have any idea how to derive the EOM for a charge distribution?

11. Jan 20, 2014

### dextercioby

See sections 25->30 of Dirac's 1975 brochure on General Relativity. 'Continuous distribution of electrically charged classical matter' can stand for a relativistic ideal electrically charged fluid.