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Derivative of lagrangian density

  1. Nov 25, 2015 #1
    i have a mathematical question which is quite similar to one asked before, still a bit different
    https://www.physicsforums.com/threa...agrangian-density-for-real-k-g-theory.781472/


    the first term of KG-Lagrangian is: [itex]\frac{1}{2}(\partial^{\mu} \phi)(\partial_{\mu} \phi)[/itex]
    when i try do find [itex]\frac{\partial L}{\partial(\partial^{\mu}\phi)}[/itex], i have two different options, where of course only one can be right.

    1) [itex]\partial^{\mu} \phi[/itex] and [itex]\partial_{\mu} \phi[/itex] are different things, so one gets: [itex]\frac{\partial L}{\partial(\partial^{\mu}\phi)} = \frac{1}{2}\partial_{\mu} \phi[/itex]

    2) [itex]\partial_{\mu} \phi = \eta_{\mu \nu} \partial^{\nu} \phi \rightarrow \frac{\partial L}{\partial(\partial^{\mu}\phi)} = \frac{\partial}{\partial(\partial^{\mu}\phi)}(\frac{1}{2}\eta_{\mu \nu}(\partial^{\nu} \phi)(\partial^{\mu} \phi)) = \partial_{\mu} \phi[/itex]

    i am confused. several sources tell me that 2) is right, but my understanding of partial derivative tells me to do 1).

    as i think of it, one may not simply put equations into others when doing the partial derivative, because it then changes. like z(x, y(x)) = 2x + y(x). partial derivative with respect to x gives 2, but it changes when i put in y(x).

    help please :)
    thank you
     
  2. jcsd
  3. Nov 25, 2015 #2

    Orodruin

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    Option 2 is correct. The quantities ##\partial_\mu\phi## and ##\partial^\mu\phi## may not be equal, but they do not correspond to different degrees of freedom (they are related by a sign). Compare this to differentiating ##-x## with respect to ##x##, ##-x## is not equal to ##x##, but it is directly dependent.
     
  4. Nov 25, 2015 #3

    stevendaryl

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    No, they are definitely not independent. If you write it with the metric, then this becomes more obvious:

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

    So [itex]\frac{\partial \mathcal{L}}{\partial (\partial^{\sigma} \phi)} = \frac{1}{2}g^{\mu \nu} (\delta^\sigma_\mu \partial_{\nu} \phi + \delta^\sigma_\nu \partial_{\mu} \phi) = \frac{1}{2} (g^{\sigma \nu} \partial_{\nu} \phi + g^{\mu \sigma} \partial_{\mu} \phi) = \frac{1}{2} (\partial^{\sigma} \phi + \partial^{\sigma} \phi) = \partial^{\sigma} \phi [/itex]
     
  5. Nov 25, 2015 #4
    i do not say they are independent. y(x) and x are not indepentent either, still in a partial derivative of z with respect to x, one does not consider the y(x).

    so orodruin, i see that they are related by the metric. y(x) is also related to x by its defining function. so i do still not get the point, where my z-function analogy is failing.
    do y and x correspond to different degrees of freedom in z?

    regards, kawillzocken
     
  6. Nov 25, 2015 #5

    stevendaryl

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    Let's make it a lot simpler, by considering only 1 space dimension and 1 time dimension. Then the lagrangian is:

    [itex]\mathcal{L} = \frac{1}{2}(\frac{\partial \phi}{\partial t} \frac{\partial \phi}{\partial t} - \frac{\partial \phi}{\partial x} \frac{\partial \phi}{\partial x})[/itex]

    I think it's pretty obviousthat [itex]\frac{\partial \mathcal{L}}{\partial (\frac{\partial \phi}{\partial t})} = \frac{\partial \phi}{\partial t}[/itex]. Right?
     
  7. Nov 25, 2015 #6

    stevendaryl

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    In a second-order differential equation for [itex]\phi[/itex] (which is what the lagrangian equations of motion produce), to know [itex]\phi[/itex] everywhere, it is good enough to specify [itex]\phi, \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}, \frac{\partial \phi}{\partial t}[/itex] at a single point [itex](x_0, y_0, z_0, t_0)[/itex]. Those are 5 completely independent choices.
     
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