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Electromagnetic Lagrangian

  1. Jan 17, 2014 #1
    If you don't have any charges or currents, the electromagnetic Lagrangian becomes ##\mathcal{L}=-\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}##. The standard way to derive Maxwell's equations in free space is to replace ##F_{\alpha\beta}## by ##\partial_\alpha A_\beta -\partial_\beta A_\alpha## and apply the Euler-Lagrange equations. But why can't you just apply the Euler-Lagrange equations right away? I see no reason why this shouldn't work, except that it gives the wrong answer:
    [tex]\frac{\partial\mathcal{L}}{\partial F_{\beta\gamma}} -\partial_\alpha\left(\frac{\partial\mathcal{L}}{\partial (\partial_\alpha F_{\beta\gamma})}\right)=-\frac{1}{4}F^{\beta\gamma}=0.[/tex]

    What's wrong with my math?
     
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  3. Jan 17, 2014 #2

    vanhees71

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    You treat the components of [itex]F_{\mu \nu}[/itex] as independent quantities, which they are not!
     
  4. Jan 17, 2014 #3
    Thanks.
     
  5. Jan 18, 2014 #4

    Bill_K

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    True, but irrelevant. By the same reasoning, dEdt could have written L = ½(E2 - B2) in which all six variables, the components of E and B, are independent. This, of course, would still lead to incorrect Euler-Lagrange equations.

    The point is that the Lagrangian is a functional form whose purpose in life is to be varied, not a value. Replacing it with some other expression that happens to have the same value but a different functional form does not in general lead to the same or equivalent Euler-Lagrange equations.
     
  6. Jan 18, 2014 #5
    Obviously what you're saying is correct, but I still don't see why. If two actions with different functional forms have the same value, then shouldn't the trajectory the minimizes one minimize the other as well?
     
  7. Jan 18, 2014 #6

    vanhees71

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    The action written simply as
    [tex]A=\int \mathrm{d}^4x (\vec{E}^2-\vec{B}^2)[/tex]
    Has no kinetic term. So you never get equations of motion. They are NOT independently to vary quantitities in the correct action functional. These are only the potentials (or better said the electromagnetic four-potential), i.e., you should write
    [tex]A[A^{\mu}]=-\frac{1}{4}\int \mathrm{d}^4 x F_{\mu \nu} F^{\mu \nu} \quad \text{with} \quad F_{\mu \nu}=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}.[/tex]
    This gives the correct field equations which read
    [tex]\partial_{\mu} F^{\mu \nu}=0,[/tex]
    because we have the free em. field here.

    The [itex]F_{\mu \nu}[/itex] (or equivalently [itex]\vec{E}[/itex] and [itex]\vec{B}[/itex]) are constraint by the fact that they are derivable from a vector potential in the form given above!
     
  8. Jan 18, 2014 #7

    BruceW

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    The important thing (as vanhees says) is that there are certain constraints we must place on ##F^{\nu \mu}##, while there are no constraints on the four-potential. The two constraints (I think) are:
    [tex]\partial_{[\alpha} F_{\beta \gamma ]} = 0[/tex]
    and the other one is:
    [tex]F^{\alpha \beta} = - F^{\beta \alpha}[/tex]
    So if you're going to use a Lagrangian with ##F^{\nu \mu}## then you need to include these constraints, by using a Lagrange multiplier for each:
    [tex]\mathcal{L}=-\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta} + \lambda_{\nu \mu} (F^{\nu \mu} + F^{\mu \nu}) + \omega^{\delta \sigma \kappa} \partial_{[\delta} F_{\sigma \kappa ]} [/tex]
    Where omega and lambda are two separate Lagrange multipliers, that are constant tensors. Now, I haven't actually tried this out, but it should work. And the reason you don't need to include these constraints in the Lagrangian for the four-potential, is because they are already automatically zero, due to the definition of the four-potential.
     
  9. Jan 18, 2014 #8

    BruceW

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    and similarly, if you wanted to re-write the Lagrangian in terms of the electric and magnetic fields, there are two constraint equations:
    [tex]\nabla \cdot \vec{B} = 0[/tex]
    and the other constraint equation (well, it's 3 really, since it is a vector equation), is:
    [tex]\nabla \wedge \vec{E} + \frac{\partial \vec{B}}{\partial t} = 0 [/tex]
    So, you need to include these in the Lagrangian, along with Lagrange multipliers to get the correct Lagrangian:
    [tex]\mathcal{L}= 1/2(E^2 - B^2) + \lambda ( \nabla \cdot \vec{B} ) + \vec{\omega} \cdot ( \nabla \wedge \vec{E} + \frac{\partial \vec{B}}{\partial t} ) [/tex]
    Where here, lambda is simply a constant scalar Lagrange multiplier, and omega is a vector Lagrange multiplier (i.e. just 3 independent Lagrange multipliers, one in each direction). Anyway, I haven't tried out this Lagrangian either, but I think it should work.
     
  10. Jan 18, 2014 #9

    BruceW

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    wait, maybe I'm being stupid. uh... I think we allow ##\lambda## and ##\vec{\omega}## to be fields, which can depend on time and space (they are not constants). And since the derivatives of ##\lambda## and ##\vec{\omega}## don't appear in the Lagrangian itself, this is what makes them Lagrange multipliers. arg, I should really be able to remember stuff like this by now.
     
  11. Jan 19, 2014 #10

    BruceW

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    yeah, now I've thought about it, I think that's right. we just treat the Lagrange multipliers as fields themselves which can vary. Another way you could write the Lagrangian is just to include ##F_{\alpha \beta} = \partial_\alpha A_\beta -\partial_\beta A_\alpha## as a constraint, which gives this equation:
    [tex]\mathcal{L}=-\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta} + \lambda^{\nu \mu} (F_{\nu \mu} - \partial_\nu A_\mu +\partial_\mu A_\nu ) [/tex]
    where ##\lambda^{\nu \mu}## are the 16 Lagrange multipliers, which are fields which can be varied just like if they were 'real' fields. Again, this Lagrangian should work. But in this case, it is clear from the start that we could just make our lives easier by replacing ##F_{\alpha \beta}## with ##\partial_\alpha A_\beta -\partial_\beta A_\alpha##. But you always have the option of doing the Lagrangian by varying ##F_{\alpha \beta}## if you want to.
     
  12. Jan 19, 2014 #11

    BruceW

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    The point is that from the start we can either:
    1) treat all the ##A_\beta## and ##F_{\nu \mu}## as independent (i.e. forget that we know ##F_{\nu \mu} = \partial_\nu A_\mu -\partial_\mu A_\nu##). And so we must introduce the term ##\lambda^{\nu \mu} (F_{\nu \mu} - \partial_\nu A_\mu +\partial_\mu A_\nu )## into the Lagrangian, which ensures that the constraint is satisfied.
    Or, we can:
    2) from the start, make use of the fact that all the ##F_{\nu \mu}## are not independent, but all the ##A_\beta## are independent, so if we can re-write our Lagrangian in terms of only the ##A_\beta##, then we don't need any constraint.

    p.s. when I say ##A_\beta## are independent, I really mean that there is no restriction on the values that each can take (with respect to the others). And there is no restriction on what values their derivatives can take (with respect to the others). i.e. ##F_{\nu \mu}## are not independent, since there is a restriction on the values. And also from before, ##\vec{B}## are not independent, since ##\nabla \cdot \vec{B}=0## places a restriction on what values the derivatives can take. So, uh, I guess this is not the standard meaning of 'independent'.

    edit: man, I've written a lot. Sorry about that. It's probably because I have been trying to reason it out in my own head, while writing. If I thought about it all before starting to type, it could all have been a lot more concise.
     
  13. Jan 19, 2014 #12

    Bill_K

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    If you really want to do this, you don't need Lagrangian multipliers, just write

    L =(1/2)FμνFμν - Fμν(Aμ,ν - Aν,μ)

    and vary Fμν and Aμ independently. Variation of Fμν yields the constraint Fμν - Aμ,ν + Aν,μ = 0, and variation of Aμ yields the remaining Maxwell equation Fμν,ν = 0.
     
  14. Jan 19, 2014 #13

    BruceW

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    yeah. In this particular case, the Lagrange multipliers are in fact proportional to ##F_{\mu \nu}##. The equation involving Lagrange multipliers was this:
    [tex]\mathcal{L}=-\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta} + \lambda^{\nu \mu} (F_{\nu \mu} - \partial_\nu A_\mu +\partial_\mu A_\nu )[/tex]
    And since ##F_{\mu \nu}## has no derivatives in the Lagrangian, variation of ##F_{\delta \gamma}## gives:
    [tex]-1/2 F^{\delta \gamma} + \lambda^{\delta \gamma} = 0[/tex]
    And this then leads to something similar to your equation. So in a way, the Lagrange multiplier formalism gives a 'reason' for the equation which you wrote.

    edit: although maybe, the equation you wrote has some fundamental reasoning behind it.. like maybe one term can be thought of as potential and the other as kinetic? I don't know, so I would prefer the Lagrangian -(1/4)FμνFμν, along with constraint.
     
  15. Jan 20, 2014 #14

    dextercioby

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    So you're trying to write some sort of Palatini formalism for the e-m field. Definitely the lagrangian in post #13, is 'pushed': you typed in the constraint within, and recovered it afterwards. It makes no sense. A better version would be to use a modified version of what Bill wrote:

    [tex] \mathcal{L}_{\mbox{mod}} = -\left[\frac{1}{2}F_{\mu\nu}F^{\mu\nu} - \frac{1}{4}F_{\mu\nu}\left(T^{\mu\nu}{}{}_{\alpha\beta} \partial^{\alpha}A^{\beta}\right)\right] [/tex]

    where the T is constant (depends only on the metric, epsilon and delta Kronecker) and shares the symmetries of the Riemann tensor ((2,2) Young tableau).
     
    Last edited: Jan 20, 2014
  16. Jan 20, 2014 #15

    BruceW

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    I think the OP'er simply wanted to write the Lagrangian for electromagnetism in Minkowski space. But the crucial part, is that he wanted to write it in terms of ##F_{\mu \nu}## instead of ##A_\alpha## (or at least, he wanted to know why it was not possible by the method in post #1).

    I guess it is possible to use the Palatini action. I did not know what this is, but I just read that it is equivalent to the Einstein-Hilbert action. So this would give a general-relativistic answer, right? But this doesn't really address the OP's question, as far as I can understand.

    The Lagrangian that I hope answers the OP's question would be something like this:
    [tex]\mathcal{L}=-\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta} + \lambda_{\nu \mu} (F^{\nu \mu} + F^{\mu \nu}) + \omega^{\delta \sigma \kappa} \partial_{[\delta} F_{\sigma \kappa ]}[/tex]
    yes, a lot of Lagrange multipliers. And I know it is not nice, but that is the most straightforward way I can think of to make the Lagrangian in terms of just ##F_{\nu \mu}##
     
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