Lagrangian density for continuous distribution of matter

In summary: But then you are no longer dealing with point particles only, but with extended objects, and the equations of motion become more formidable.
  • #1
dEdt
288
2
The Lagrangian for a point particle is just [tex]L=-m\sqrt{1-v^2}.[/tex] 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.
 
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  • #2
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
But the density isn't a scalar, so how can it be equal to the Lagrangian density?
 
  • #4
I should have said ##\rho_0##, the density in the rest frame of the matter.
 
  • #5
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
dEdt said:
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
 
  • #7
dEdt said:
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.
 
  • #8
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
I don't know what [itex]\rho_0[/itex] might be.

The Lagrangian for the electromagnetic field (minimally) coupled to the (necessarily conserved) electromagnetic current reads (using Heaviside-Lorentz units with [itex]c=1[/itex] and the west-coast convention for the metric, [itex]\eta_{\mu \nu}=\text{diag}(1,-1,-1,-1)[/itex]
[tex]\mathcal{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu}+J_{\mu} A^{\mu}.[/tex]
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
[tex]L=-m \sqrt{1-\dot{\vec{y}}^2},[/tex]
and the current density is given by
[tex]j^{\mu}(x)=q \int_{\mathbb{R}} \mathrm{d} t \; \dot{\vec{y}} \delta^{(4)}(x-y).[/tex]
Here, [itex]\vec{y}(t)[/itex] is the trajectory of the particle as a function of the coordinate time in the inertial reference frame with [itex]y^0=t[/itex].

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
[tex]S_{\text{f}0}[A]=\int \mathrm{d}^4 x \left (-\frac{1}{4} F_{\mu \nu} F^{\mu \nu} \right ),[/tex]
the free action of the particle
[tex]S_{\text{p0}}[\vec{y}]=-m \int \mathrm{d} t \sqrt{1-\dot{\vec{y}}^2},[/tex]
and the interaction term
[tex]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].[/tex]
Taking the variation with respect to [itex]A^{\mu}[/itex] you get Maxwell's equations with the current given by the single-particle current and variation with respect to [itex]\vec{y}[/itex] 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 [itex]\vec{y}(t)[/itex], 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
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
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.
 
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1. What is Lagrangian density for continuous distribution of matter?

Lagrangian density for continuous distribution of matter is a mathematical concept used in theoretical physics to describe the dynamics of a continuous distribution of matter. It is a function that describes the energy of a system at each point in space and time.

2. How is Lagrangian density different from Lagrangian mechanics?

Lagrangian density is an extension of Lagrangian mechanics which is used to describe the dynamics of discrete particles. While Lagrangian mechanics deals with the motion of a finite number of particles, Lagrangian density deals with the motion of an infinite number of particles in a continuous medium.

3. What is the importance of Lagrangian density in physics?

Lagrangian density is important in physics as it allows for the formulation of equations of motion for a continuous distribution of matter. This is particularly useful in fields such as fluid mechanics, where the motion of a fluid is described by a continuous distribution of matter.

4. How is Lagrangian density used in the study of quantum field theory?

In quantum field theory, Lagrangian density is used to describe the interactions between particles and fields. It allows for the development of the mathematical framework for understanding the behavior of subatomic particles and their interactions.

5. Can Lagrangian density be extended to include other physical phenomena?

Yes, Lagrangian density can be extended to include other physical phenomena such as electromagnetic fields and gravity. This allows for a more comprehensive understanding of the dynamics of various systems in physics.

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