1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Euler-Lagrange Equations and Derivatives

  1. Apr 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Hi. I am attempting to get the Euler-Lagrange equations of motion for the following Lagrangian:

    L(ψ[itex]^{μ}[/itex]) = -[itex]\frac{1}{2}[/itex] ∂[itex]_{μ}[/itex] ψ[itex]^{\nu}[/itex] ∂[itex]^{μ}[/itex] ψ[itex]_{\nu}[/itex] + [itex]\frac{1}{2}[/itex] ∂[itex]_{μ}[/itex] ψ[itex]^{\mu}[/itex] ∂[itex]_{\nu}[/itex] ψ[itex]^{\nu}[/itex] + [itex]\frac{m^{2}}{2}[/itex] ψ[itex]_{\nu}[/itex] ψ[itex]^{\nu}[/itex]

    2. Relevant equations

    So, I want to get [itex]\frac{∂}{∂(∂_{\mu}ψ)}[/itex] (L). My issue is that I'm not sure how this interacts with the [itex]∂^{\mu}[/itex] term.

    3. The attempt at a solution

    I think that it's probably one of these things.

    Either [itex]∂^{\mu} ψ_{\nu}[/itex] is treated as independent to [itex]∂_{\mu} ψ^{\nu}[/itex] , i.e. [itex]\frac{∂}{∂(∂_{\mu}ψ)}[/itex] ([itex]∂^{\mu} ψ_{\nu} a[/itex]) = 0, or it is derived as -1 times the derivative of [itex]∂_{\mu} ψ^{\nu}[/itex], i.e. [itex]\frac{∂}{∂(∂_{\mu}ψ)}[/itex] ([itex]∂^{\mu} ψ_{\nu} a[/itex]) = -a.

    Any help on how to get this Euler-Lagrange would be really appreciated.

    Last edited: Apr 9, 2012
  2. jcsd
  3. Apr 9, 2012 #2
    Here's how I think about this--it's a bit ugly, though, so somebody else can chime in if there's a better way.

    Consider the simpler case [itex]\mathcal{L} = \frac{1}{2}\partial_\mu\phi\partial^\mu\phi[/itex]. In this case, [itex]\frac{1}{2}\partial_\mu\phi\partial^\mu\phi = \frac{1}{2}g^{\mu\nu}\partial_\mu\phi\partial_\nu \phi[/itex]. Now, since the metric is diagonal, the only terms you really care about are the ones where [itex]\mu=\nu[/itex], so you can write this in a somewhat kooky way as [itex]\frac{1}{2}g^{\mu\mu}\partial_\mu\phi\partial_\mu \phi = \frac{1}{2}g^{\mu\mu}(\partial_\mu\phi)^2[/itex]. This makes sense intuitively, because if you break out the original equation, it's the sum of only four terms, not the full 16, and they have the same signs as the ones implied by this equation.

    Now the derivative is easy to perform: [itex]\frac{\partial}{\partial(\partial_\mu\phi)} \mathcal{L} = \frac{\partial}{\partial(\partial_\mu\phi)}\frac{1}{2}g^{\mu\mu}(\partial_\mu\phi)^2 = g^{\mu\mu}\partial_\mu\phi=\partial^\mu\phi[/itex], which is the right answer, because as part of the E-L equation you're going to take the derivative of that, which is [itex]\partial_\mu\partial^\mu\phi = \Box^2\phi[/itex] like you would expect.

    I don't know if there's a cleaner way to do it than that, but it seems to work. You should be able to do something similar for your case where the field itself is a vector.
    Last edited: Apr 9, 2012
  4. Apr 10, 2012 #3
    That's great! Thanks for the help!

    I changed my Lagragian to be:

    L([itex]\phi^{\mu}) = - \frac{1}{2} \partial_{\mu} g^{\mu \nu} g^{\mu \mu} \phi^{\mu} g^{\mu \mu} \partial_{\mu} g^{\mu \nu} \phi^{\mu} + \frac{1}{2} \partial_{\mu} \phi^{\mu} g^{\mu \nu} g^{\mu \mu} \partial_{\mu} g^{\mu \nu} g^{\mu \mu} \phi^{\mu} + \frac{m^{2}}{2} g^{\mu \mu} \phi^{\mu} \phi^{\mu}[/itex]

    =[itex] - \frac{1}{2} g^{\mu \mu} g^{\mu \nu} g^{\mu \mu} g^{\mu \nu} \partial_{\mu} \phi^{\mu} \partial_{\mu} \phi^{\mu} + \frac{1}{2} g^{\mu \mu} g^{\mu \nu} g^{\mu \mu} g^{\mu \nu} \partial_{\mu} \phi^{\mu} \partial_{\mu} \phi^{\mu} + \frac{m^{2}}{2} g^{\mu \mu} \phi^{\mu} \phi^{\mu}[/itex]

    =[itex] \frac{m^{2}}{2} g^{\mu \mu} (\phi^{\mu})^{2}[/itex]

    I'm supposed to then show that this means that [itex]∂_{\mu} \phi^{\mu}[/itex] is 0, then find the canonical momenta. But the E-L equations of my Lagrangian above just gives me that [itex]m^{2} \phi_{\mu} = 0[/itex]. While this does give me my desired quality, it seems to trivialize the following questions, and this makes me think I have gone wrong somewhere.

    As always, I will be incredibly grateful for any help.
    Last edited: Apr 10, 2012
  5. Apr 10, 2012 #4
    Hmm, I guess that doesn't generalize to the vector case as easily as I thought it did. Oh well, when all else fails, you can always fall back on the underlying definitions of the tensor operations to figure this out. For your first term, you can expand it out as follows (for brevity's sake I'm going to pretend we're in 1+1 dimensional space instead of a 1+3 dimensional space, but the generalization is immediate):

    [tex]\frac{1}{2}\partial_\mu\Psi^\nu\partial^\mu \Psi_\nu = \frac{1}{2}\partial_0\Psi^0\partial^0\Psi_0 + \frac{1}{2}\partial_0\Psi^1\partial^0\Psi_1 + \frac{1}{2}\partial_1\Psi^0\partial^1\Psi_0 + \frac{1}{2}\partial_1\Psi^1\partial^1\Psi_1\\
    = \frac{1}{2}\partial_0\Psi^0\partial_0\Psi^0 - \frac{1}{2}\partial_0\Psi^1\partial_0\Psi^1 -\frac{1}{2}\partial_1\Psi^0\partial_1\Psi^0 + \frac{1}{2}\partial_1\Psi^1\partial_1\Psi^1\\
    = \frac{1}{2}(\partial_0\Psi^0)^2 - \frac{1}{2}(\partial_0\Psi^1)^2 - \frac{1}{2}(\partial_1\Psi^0)^2 + \frac{1}{2}(\partial_1\Psi^1)^2

    Now, break out the derivatives:
    \frac{\partial\mathcal{L}}{\partial(\partial_0 \Psi^0)} = \partial_0\Psi^0 = \partial^0\Psi_0\\
    \frac{\partial\mathcal{L}}{\partial(\partial_0 \Psi^1)} = -\partial_0\Psi^1 = \partial^0\Psi_1\\
    \frac{\partial\mathcal{L}}{\partial(\partial_1 \Psi^0)} = -\partial_1\Psi^0 = \partial^1\Psi_0\\
    \frac{\partial\mathcal{L}}{\partial(\partial_1 \Psi^1)} = \partial_1\Psi^1 = \partial^1\Psi_1

    So we can conclude that:
    \frac{\partial}{\partial(\partial_\mu \Psi^\nu)}\frac{1}{2}\partial_\mu\Psi^\nu\partial^\mu \Psi_\nu = \partial^\mu\Psi_\nu

    You should be able to do the same thing for the other term in the Lagrangian, although you'll need to use integration by parts to flip the derivatives around to the other fields first (hopefully you've seen this trick used by now, if not I can show you how it's done.)

    In general, it seems to be the case that if you are taking the derivative of something in a Lagrangian, you can treat the other half of the term as an independent variable and get the right answer, as long as you remember that you're taking the derivatives of quadratic forms, so a factor of 2 will come down. This seems sloppy and counterintuitive, since the other term isn't really independent, but it seems to give the right answer. So:

    \frac{\partial}{\partial(\partial_\mu \Psi^\nu)}\frac{1}{2}\partial_\mu\Psi^\nu\partial^\mu \Psi_\nu = \partial^\mu\Psi_\nu\\
    \frac{\partial}{\partial(\partial_\mu \Psi^\nu)}\frac{1}{2}\partial_\mu\Psi^\nu\partial_\nu \Psi^\mu = \partial_\nu\Psi^\mu\\
    \frac{\partial}{\partial\Psi^\nu}\frac{1}{2}m^2 \Psi^\nu\Psi_\nu = m^2 \Psi_\nu

    You can use a similar bit of ad-hockery to minimize a complex field, by assuming that [itex]\Psi[/itex] and [itex]\Psi^*[/itex] are independent fields. I'm not sure how to prove rigorously that this insane-sounding thing actually works reliably, but in all of the cases I've seen it used so far, it appears to do the job.
    Last edited: Apr 10, 2012
  6. Apr 12, 2012 #5
    I see.

    So I would get the E-L equation to be:

    [itex]-(∂_{\mu} ∂^{\mu} + m^{2}) ψ_{\nu} + ∂_{\mu} ∂_{\nu} ψ^{\nu} = 0[/itex]

    The first bit is just the Klein Gordon equation. I'm supposed to be able to show that my E-L equation implies that [itex]∂_{\nu} ψ^{\nu} = 0[/itex].

    Presumably I can't just claim that the Klein Gordon equation works here, implying [itex]∂_{\mu} ∂_{\nu} ψ^{\nu} = 0[/itex].

    Thanks again, you've been a great help!
  7. Apr 12, 2012 #6
    Almost. Check your tensor indices in the second term.

    Now that you mention it, the arguments I've heard for this do basically boil down to "well, we want [itex]\psi[/itex] to be a solution to the Klein-Gordon equation, so for that to work we need [itex]\partial_\mu \psi^\mu=0[/itex]." I'm not sure if there's some more rigorous argument that can be made to show that there are no possible solution for the equation in the presence of that term, but it's certainly the case that there are no solutions which are also solutions to the free Klein-Gordon equation unless the four-divergence vanishes, so perhaps that's rigorous enough.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook