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Calculating derivatives of a Lagrangian density

  1. Dec 11, 2013 #1
    Hey everyone,

    I wasn't really sure where to post this, since it's kind of classical, kind of relativistic and kind of quantum field theoretical, but essentially mathematical. I'm reading and watching the lectures on Quantum Field Theory by Cambridge's David Tong (which you can find here), and I'm kind of stuck in calculating derivatives of the Langrangian density with respect to the derivatives of the field. For instance, the following Lagrangian (Maxwell in a vacuum):
    $$ \mathcal{L} = -\frac{1}{2}(\partial_\mu A_\nu)(\partial^\mu A^\nu) + \frac{1}{2}(\partial_\mu A^\mu)^2 = -\frac{1}{2}(\partial_\mu A_\nu)(\eta^{\mu \alpha} \eta^{\nu \beta} \partial_\alpha A_\beta) + \frac{1}{2}(\eta^{\mu \nu}\partial_\mu A_\nu)^2 $$
    I'm supposed to calculate [itex]\partial \mathcal{L}/\partial(\partial_\mu A^\nu)[/itex]. Now, I have two ways to think about this, one of them is to think of each term as a homogeneous function of degree 2 of the variable [itex]\partial_\mu A_\nu[/itex], and using Euler's theorem on homogeneous functions, multiply by 2 and take the derivatives of each term with respect to [itex]\partial_\mu A_\nu[/itex] as if it wasn't being summed over, giving:
    $$ \frac{\partial \mathcal{L}}{\partial(\partial_\mu A^\nu)} = -(\eta^{\mu \alpha} \eta^{\nu \beta} \partial_\alpha A_\beta) + (\eta^{\alpha \beta}\partial_\alpha A_\beta)(\eta^{\mu \nu}) = -\partial^\mu A^\nu + \eta^{\mu \nu} \partial_\alpha A^\alpha $$
    This is a bit shady in my head, but apparently it works. The other way I think about it is significantly longer in terms of algebra, but basically involves actually differentiating with respect to a different set of indices, let's say [itex]\partial \mathcal{L}/\partial(\partial_\alpha A_\beta)[/itex], use the product rule and the fact that [itex]\partial(\partial_\mu A_\nu)/\partial(\partial_\alpha A_\beta) = \delta^\alpha_\mu \delta^\beta_\nu[/itex], which gives the same result, but seems unnecessarily long.

    So, how do you think about it? What's the quickest but still somewhat rigorously pleasing way to think about it that gives the right answer?

  2. jcsd
  3. Dec 11, 2013 #2
    I use the second method and do all the calculations in my had to keep it from feeling unnecessarily long.
  4. Dec 12, 2013 #3
    Ok, thank you for the feedback. BTW, the following Lagrangian gives rise to the same equations of the motion as the above one: $$ \mathcal{L} = -\frac{1}{2}(\partial_\mu A_\nu)(\partial^\mu A^\nu) + \frac{1}{2}(\partial_\nu A_\mu)(\partial^\mu A^\nu) $$
    So they should differ only by a total derivative. I can test if something is a total derivative by plugging it into the Euler-Lagrange equations, and see if I get something that can be reduced to [itex]0=0[/itex]. In this case, there should be some vector [itex]B^\mu[/itex] such that the difference between the two Lagrangians, is equal to [itex]\partial_\mu B^\mu[/itex], right? My question is, how can you find [itex]B^\mu[/itex] (up to the addition of a divergence-less term)?
  5. Dec 15, 2013 #4
    Should I reask this question in the Relativity sub-forum? I don't seem to be having much success here...
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