Lagrangian density for continuous distribution of matter

[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.

 

[pervect]

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 $m \sqrt{1-v^2}$

 

[dEdt]

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

 

[pervect]

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

 

[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.

 

[samalkhaiat]

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.

See this View attachment Jackiw fluid.pdf

 

[pervect]

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.

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.

 

[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?

 

[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.

The free-particle part reads
$$L=-m \sqrt{1-\dot{\vec{y}}^2},$$
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).

 

[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?

 

[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.