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Homework Help: Differentiating Lagrangian in Tensor Notation

  1. May 29, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi all, I'm trying to learn how to manipulate tensors and in particular to differentiate expressions. I was looking at a Lagrangian density and trying to apply the Euler-Lagrange equations to it.

    2. Relevant equations
    Lagrangian density:
    [tex]\mathcal{L} = -\frac{1}{2} \partial_{\alpha} \phi^{\beta} \partial^{\alpha} \phi_{\beta}
    + \frac{1}{2} \partial_{\alpha} \phi^{\alpha} \partial_{\beta} \phi^{\beta}
    + \frac{1}{2}\mu^2 \phi^{\alpha} \phi_{\alpha}[/tex]

    [tex]\frac{\partial \mathcal{L}}{\partial \phi^i} = \partial^k \frac{\partial \mathcal{L}}{\partial \phi^{i,k}}[/tex]

    3. The attempt at a solution
    I have attempted to differentiation the expression several times; I can compute [tex]\frac{\partial \mathcal{L}}{\partial \phi^i}[/tex] with no problems and can compute [tex]\frac{\partial \mathcal{L}}{\partial \phi^{i,k}}[/tex] for the first and third terms.

    However, I just cannot figure out how to differentiate the middle term. My attempt:

    [tex]\mathcal{L}_2 = \frac{1}{2} \partial_{\alpha} \phi^{\alpha} \partial_{\beta} \phi^{\beta}
    = \frac{1}{2} g_{\alpha \lambda} g_{\beta \sigma} \partial^{\lambda} \phi^{\alpha} \partial^{\sigma} \phi^{\beta}[/tex]
    [tex] \frac{\partial \mathcal{L}_2}{\partial \phi^{i,k}} =
    \frac{1}{2} g_{\alpha \lambda} g_{\beta \sigma} \left(
    \delta_k^{\lambda} \delta_i^{\alpha} \partial^{\sigma} \phi^{\beta} +
    \delta_k^{\sigma} \delta_i^{\beta} \partial^{\lambda} \phi^{\alpha}
    = \frac{1}{2} \left(
    g_{i k} \partial_{\beta} \phi^{\beta} +
    g_{i k} \partial_{\alpha} \phi^{\alpha}
    = g_{i k} \phi_i \phi^i

    So as you can see, I have somehow picked up this additional factor of the metric. I'm not sure what to do with it, or where I have gone wrong!

    Best wishes,
  2. jcsd
  3. May 29, 2015 #2


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    The final step should be ##g_{ik}\partial_j\phi^j##. But other than the ##\partial## that got swapped for a ##\phi## (typo?) and the choice to call the summation index i (the same as one of the free indices), I don't see anything wrong with what you did. Are you sure that the answer isn't supposed to be ##g_{ik}\partial_j\phi^j##?

    I think it's pretty natural for the metric to show up there. If the metric is diagonal, the expression you're differentiating with respect to ##\partial^i\phi^k## contains no factors of the form ##\partial^l\phi^m## with ##l\neq m##, so the derivative should contain a factor that's zero when ##i\neq k##.
  4. May 29, 2015 #3
    Oh, yes, that is a typo. It is meant to be a ##\partial## not a ##\phi##. My decision to call the sum index ##i## was just because the derivative was taken with respect to ##\phi^{i,j}## but this might be bad practice. I'm not familiar with the conventions people use when using tensors because I've really only been doing it a few days. I actually couldn't find anything online through google about doing this kind of thing, which is what led me to ask here!

    As for the answer being "wrong", I suppose it might not be – I just didn't expect to have a metric in the result, since the next step is to calculate ##\partial^k## of the answer as per the E-L equation. With the metric term, this would give

    [tex]\partial^k g_{ik}\partial_j \phi^j = \partial_i \partial_j \phi^j[/tex]. Is this a legitimate expression?

    Taken with the other terms, the final E-L would then be:
    [tex]g_{i j} \mu^2 \phi^j = \partial^k (-\partial_k \phi_i + g_{ik} \partial_j \phi^j) = - \square \phi_i + \partial_i \partial_j \phi^j = - g_{i j} \square \phi^j + \partial_i \partial_j \phi^j[/tex]
    or, rearranging
    [tex][g_{i j}(\mu^2 + \square) - \partial_i \partial_j]\phi^j = 0[/tex]

    I think this might be right actually, but I'm still uneasy about the double covariant derivative product; what does that actually represent physically?
  5. May 30, 2015 #4


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    I found a simpler way to get the result you got. We have
    $$\mathcal L_2=\frac 1 2 \partial_\alpha\phi^\alpha \partial_\beta\phi^\beta =\frac 1 2(\partial_\alpha\phi^\alpha)^2.$$ So by the chain rule, we have
    $$\frac{\partial\mathcal L_2}{\partial(\partial_\mu\phi^\nu)}=\partial_\alpha\phi^\alpha \frac{\partial}{\partial(\partial_\mu\phi^\nu)}\partial_\beta\phi^\beta =\partial_\alpha\phi^\alpha\delta^\mu_\nu.$$ My left-hand side is however different from yours. If I just change your Latin indices to Greek, your left-hand side is ##\frac{\partial\mathcal L_2}{\partial(\partial^\mu\phi^\nu)}##. We get
    $$\frac{\partial\mathcal L_2}{\partial(\partial^\mu\phi^\nu)}=g_{\mu\rho} \frac{\partial\mathcal L_2}{\partial(\partial_\rho\phi^\nu)}=g_{\mu\rho} \partial_\alpha\phi^\alpha\delta^\rho_\nu =g_{\mu\nu}\partial_\alpha\phi^\alpha.$$ I would write the Euler-Lagrange equation for ##\mathcal L_2## as
    $$0=\frac{\partial L_2}{\partial \phi^\nu}-\partial_\mu\left(\frac{\partial\mathcal L_2}{\partial(\partial_\mu\phi^\nu)}\right) =-\partial_\mu\left(\partial_\alpha\phi^\alpha \delta^\mu_\nu\right) =-\delta^\mu_\nu\partial_\mu \partial_\alpha\phi^\alpha =-\partial_\nu\partial_\alpha\phi^\alpha.$$

    Yes, that's actually the reason why you can't use i or j as the summation index, but any other symbol will do.

    You should also start following the convention to have Latin indices run from 1 to 3 and Greek indices from 0 to 3. For example ##A^i_i## and ##A^\mu_\mu## don't mean the same thing. The former means ##A^1_1+A^2_2+A^3_3## and the latter means ##A^0_0+A^1_1+A^2_2+A^3_3##.

    Yes. But the indices should be Greek throughout the calculation. Otherwise your sums will be interpreted as having three terms rather than four, and your result will be interpreted as three equations rather than four.
    Last edited: May 30, 2015
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