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Calculating the energy-momentum tensor for Maxwell Lagrangian

  1. Dec 3, 2011 #1
    Hi guys, can you help me with this?

    I'm supposed to calculate the energy momentum for the classic Maxwell Lagrangian, [itex]\mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}[/itex] , where [itex]F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu[/itex]
    with the well known formula:
    [tex]T^{\sigma\rho}=\frac{\delta\mathcal{L}}{\delta \partial_{\sigma} A_\gamma}\partial^\rho A_\gamma-\mathcal{L}g^{\sigma\rho}[/tex]

    The point is that i'm not sure on how should I calculate the [tex]\frac{\delta\mathcal{L}}{\delta\partial_\sigma A_\gamma}\partial^\rho A_\gamma=-\frac{1}{4}\frac{\delta\left[(\partial^\mu A^\nu-\partial^\nu A^\mu)(\partial_\mu A_\nu -\partial_\nu A_\mu)\right]}{\delta\partial_\sigma A_\gamma}\partial^\rho A_\gamma[/tex] term; i cannot figure out on which component should i derive.


    Can you help me?
     
    Last edited: Dec 3, 2011
  2. jcsd
  3. Dec 3, 2011 #2

    Fredrik

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    I don't think that should be a functional derivative, but to be honest, I have never bothered to study functional derivatives, so maybe I'm wrong. Anyway, if you can interpret it as a partial derivative, then it's easy. You just have to understand what function you're talking partial derivatives of. It's just a polynomial in several variables, and you know how to take partial derivatives of polynomial. The expression [tex]\frac{\partial\mathcal L}{\partial(\partial_\sigma A_\gamma)}[/tex] looks scary, but it's just the nth partial derivative (for some integer n) of the polynomial [itex]\mathcal L[/itex].

    Consider this simpler example first:
    The denominator of [tex]\frac{\partial L}{\partial\dot q}[/tex] just tells you which partial derivative you're dealing with.
    [tex]\frac{\partial L}{\partial\dot q}=\frac{\partial}{\partial b}\bigg|_{a=q(t),\ b=\dot q(t),\ c=t}L(a,b,c)=(mb)_{a=q(t),\ b=\dot q(t),\ c=t}=m\dot q(t)[/tex]
    But maybe you knew all this already. I just explained the part that confuses everyone at first. I realize now that this is a bit tricky even if you know this.
     
  4. Dec 3, 2011 #3

    dextercioby

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    It is a partial derivative there, don't use the delta, that symbol should stand for the functional (Ga^teaux derivative). So the partial derivative obeys the product law. Start differentiating from the left.
     
  5. Dec 3, 2011 #4
    Well, that is a functional derivative, being [itex]\partial_\mu \phi[/itex] a function!
     
  6. Dec 3, 2011 #5

    samalkhaiat

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    Lagrangians are local functions not functionals. So, you are dealing with partial derivatives. Cast the em Lagrangian in the form
    [tex]
    \mathcal{L}= - \frac{1}{2} \eta^{\rho \sigma} \eta^{\mu \nu}( \partial_{\rho}A_{\mu} \partial_{\nu}A_{\sigma} - \partial_{\rho}A_{\mu}\partial_{\sigma}A_{\nu}).
    [/tex]
    Now use
    [tex]
    \frac{\partial}{\partial ( \partial_{\lambda}A_{\tau})} \left( \partial_{\alpha}A_{\beta}\right) = \delta^{\lambda}_{\alpha}\delta^{\tau}_{\beta}.
    [/tex]

    Sam
     
  7. Dec 4, 2011 #6

    Fredrik

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    You need to look at what I said about how Lagrangians are polynomials. In the simple example I posted, [itex]\dot q[/itex] is a function but [itex]\partial/\partial \dot q[/itex] is just the partial derivative operator with respect to the second variable, and the function L that it's supposed to be applied to, is just a polynomial in 3 variables. You might as well write it as [itex]D_2[/itex], or [itex]D_2|_{(q(t),\ \dot q(t),\ t)}[/itex] to indicate at what point the partial derivative of the polynomial is to be evaluated.
     
    Last edited: Dec 4, 2011
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