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Lagrangian density for continuous distribution of matter

  1. Jan 15, 2014 #1
    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.
     
    Last edited: Jan 15, 2014
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  3. Jan 15, 2014 #2

    pervect

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    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}##
     
  4. Jan 15, 2014 #3
    But the density isn't a scalar, so how can it be equal to the Lagrangian density?
     
  5. Jan 15, 2014 #4

    pervect

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    I should have said ##\rho_0##, the density in the rest frame of the matter.
     
  6. Jan 15, 2014 #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.
     
  7. Jan 16, 2014 #6

    samalkhaiat

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    See this View attachment Jackiw fluid.pdf
     
  8. Jan 16, 2014 #7

    pervect

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    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.
     
  9. Jan 20, 2014 #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?
     
  10. Jan 20, 2014 #9

    vanhees71

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    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).
     
  11. Jan 20, 2014 #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?
     
  12. Jan 20, 2014 #11

    dextercioby

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    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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