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!

Variation of the action using tensor algebra.

  1. Dec 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Hi, I have a problem calculating the variation of the action using tensor algebra because two derivative indices are causing some problem.

    2. Relevant equations

    Generally you have the action [itex]S = \int L(A^{\mu}, A^{\mu}_{\;,\nu}, x^{\mu})d^4x [/itex]
    where:
    [itex] A ^{\mu}= A^{\mu}(x^{\nu}) [/itex]
    [itex] A ^{\mu}_{\;,\nu} = \frac{\partial A^{\mu}}{\partial x^{\nu}} [/itex]
    [itex] x^{\nu} = (x^0, x^1, x^2 ,x^3) [/itex]
    [itex] d^{4}x = dx^0 dx^1dx^2dx^3 [/itex]

    Would I be correct in stating that the variation of the action is [itex] \delta S = \int ( \frac{\partial L}{\partial A^{\mu} } \delta A^{\mu} + \frac{\partial L}{\partial A ^{\mu}_{\;,\nu} } \delta A ^{\mu}_{\;,\nu} ) d^{4} x [/itex] ?

    3. The attempt at a solution

    Say that our function L looks like this:
    [itex] L = A_{\mu, \nu}-A_{\nu, \mu} [/itex]
    where [itex] A_{\mu} = \eta _{\mu \nu} A^{\nu} [/itex] and [itex] \eta_{\mu \nu} [/itex] is the Minkowski metric tensor.
    How do I make sense of this in context of the variation [itex]\delta S[/itex] above? More specifically what should the derivative index of the variation [itex] \delta A^{\mu}[/itex] be? Because L is the [itex]\nu[/itex] derivative of [itex]A_{\mu}[/itex] minus [itex]A_{\nu}[/itex] derivated with respect to the [itex]\mu[/itex] derivative.
    Specifically what I want to accomplish is to rewrite the right hand side of [itex]\delta S[/itex] as the sum of two parts that looks something like this
    [itex]\frac{\partial L}{\partial A_{\mu , \nu} } \delta A_{\mu, \nu}=\frac{\partial }{\partial x^{\nu}} (\frac{\partial L}{\partial A_{\mu,\nu}} \delta A_{\mu})-\delta A_{\mu} \frac{\partial }{\partial x^{\nu}}( \frac{\partial L}{\partial A_{\mu,\nu}}) [/itex] but for me to be able to do that I need a common derivative index in L which I don't have. I have two separate derivative indexes and I have not idea what to do with them. Thank you for your help.

    I know the equations are physically nonsensical since I have removed all the clutter beside the actual problem, so this is mainly a mathematical question. I could not decide whether it should be in the Physics section or Math section. If a moderator think it should be moved somewhere else please feel free to do it there or tell me and I'll do it later.
     
    Last edited: Dec 9, 2013
  2. jcsd
  3. Dec 9, 2013 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Your Lagrangian doesn't make sense, because it's a 2nd-rank tensor but should be a scalar (density). The action for the free em. field in Heaviside-Lorentz units with [itex]c=1[/itex] reads
    [tex]\mathcal{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu},[/tex]
    where
    [tex]F_{\mu \nu}=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}.[/tex]
    Then you can use the antisymmetry of this tensor to write
    [tex]\delta \mathcal{L}=-\frac{1}{2} F^{\mu \nu} \delta F_{\mu \nu}=-F^{\mu \nu} \partial_{\mu} \delta A_{\nu},[/tex]
    where I've also taken into account that in the Hamilton principle the space-time variables are not varied, i.e., that [itex]\delta (\partial_{\mu} A_{\nu})=\partial_{\mu} \delta A_{\nu}[/itex].
     
  4. Dec 9, 2013 #3
    Thank you! It did not answer what I was looking for but it still helped in another way. Which is good since my question, as you pointed out was badly phrased. I will try and derive the result you posted as I cannot see the connection by just looking at it.

    Regarding my original question. It does not have to be related to the Lagrangian. If you think of L as just some 2nd rank tensor, would it make more sense then or is there some other error which makes the expression illegal even if one just rewrites [itex] \delta S \rightarrow \delta S^{\mu \nu} [/itex]?

    Another question, my teacher did this on the whiteboard some lecture ago without motivation.
    [itex] \frac{\partial F_{\mu \nu}}{\partial A^{\mu}_{\;,\nu}} = \frac{\partial F_{\mu \nu}}{\partial A_{\mu, \nu}} [/itex]

    Is this true generally for any function or only in the case where A is the 4-potential and F is Maxwell field tensor?

    *edit*

    I managed to solve the problem. Thank you for the help! But the question above still stands :)
     
    Last edited: Dec 9, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Variation of the action using tensor algebra.
  1. Tensor algebra (Replies: 7)

  2. Tensor Algebra (Replies: 2)

Loading...