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I Lagrangian differentiation

  1. Apr 7, 2017 #1
    I'm just in need of some clearing up of how to differentiate the lagrangian with respect to the covariant derivatives when solving the E-L equation:

    Say we have a lagrangian density field
    \begin{equation}
    \mathcal{L}=\frac{1}{2}(\partial_{\mu}\hat{\phi})(\partial^{\mu}\hat{\phi})
    \end{equation}

    When solving for E-L why does:

    \begin{equation}
    \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)}= \partial^{\mu}\hat{\phi}
    \end{equation}

    i.e Why do we lose the factor of 1/2
     
  2. jcsd
  3. Apr 7, 2017 #2

    jedishrfu

    Staff: Mentor

    Could it be due to a power rule like ##y=(1/2) * x^2## to ##dy/dx = (1/2) * 2 * x = x## ?
     
  4. Apr 7, 2017 #3
    I always found it useful to actually write out the expression in summation notation to see what's happening.
     
  5. Apr 8, 2017 #4

    haushofer

    User Avatar
    Science Advisor

    Indeed. The rules for functional differentiation follow in many respects the rules for ordinary differentation.

    Use the product rule and the fact that derivatives of the fields are linearly independent. Then you get two terms which are exactly the same. Hence the factor of 2, which cancels the factor 1/2.
     
  6. Apr 8, 2017 #5

    haushofer

    User Avatar
    Science Advisor

    By the way, why do you say "covariant derivatives"? Those are just partial derivatives. But in the end that doesn't matter: the covariant derivatives contain a partial derivative on phi plus a connection*phi. Phi and the partial derivative of phi are considered to be linearly independent. Just to be clear (I use deltas for functional derivation), say
    $$D_a \phi = \partial_a \phi - A_a \phi $$
    then
    $$\frac{\delta }{\delta (\partial_a)\phi} \Bigl(\partial_b \phi - A_b \phi \Bigr) = \delta_b^a - 0 = \delta_b^a $$

    so

    $$\frac{\delta }{\delta (\partial_c)\phi} \Bigl(\eta^{ab} D_a \phi D_b \phi\Bigr) = \eta^{ab} \Bigl(\delta_a^c D_b \phi + D_a \phi \delta_b^c \Bigr) = 2 D_c \phi $$
     
  7. Apr 8, 2017 #6
    Write it as
    $$\frac{1}{2} (\partial_{\mu} \phi) (\partial^{\mu} \phi) = \frac{1}{2} (\partial_{\mu} \phi) (\partial_{\nu} \phi)g^{\mu \nu}$$
    now differentiate.
     
  8. Apr 8, 2017 #7
    He didn't elaborate on his question but I don't think we're helping. From the question it appears to me that the OP is just learning to deal with the summation notation which can be very confusing to students who have never seen it. Suppose this is just a classical field, no metric, no complex fields. Then the repeated lower and upper index means sum over that index. So, for example, if [tex]\varphi = \varphi ({x_1},{x_2},{x_3})[/tex] then [tex]\frac{1}{2}({\partial _\mu }\varphi )({\partial ^\mu }\varphi ) = \frac{1}{2}\left( {\frac{{\partial \varphi }}{{\partial {x_1}}}} \right)\left( {\frac{{\partial \varphi }}{{\partial {x_1}}}} \right) + \frac{1}{2}\left( {\frac{{\partial \varphi }}{{\partial {x_2}}}} \right)\left( {\frac{{\partial \varphi }}{{\partial {x_2}}}} \right) + \frac{1}{2}\left( {\frac{{\partial \varphi }}{{\partial {x_3}}}} \right)\left( {\frac{{\partial \varphi }}{{\partial {x_3}}}} \right)[/tex] and [tex]\frac{{\partial L}}{{\partial ({\partial _1}\varphi )}} = \frac{{\partial L}}{{\partial \left( {\frac{{\partial \varphi }}{{\partial {x_1}}}} \right)}} = \frac{{\partial \varphi }}{{\partial {x_1}}}[/tex] etc.

    I really think his issue is notation and I suspect that's what the first response was suggesting but I think we scared the guy away.
     
  9. Apr 9, 2017 #8

    No you didn't scare me away. I do understand what the notation represents too. However, I saw in my notes that they had differentiated quickly the klein gordon lagrangian to get the klein gordon equation.

    It is defnitely a notation thing though. I don't really understand the difference between
    \begin{equation}
    \partial_{\mu} \\ \partial^{\mu}
    \end{equation}

    except the factor of -1 in the spacial part.

    However, if we are differentiating with respect to the covariant derivative (subscript mu) why do we differentiate the contravariant derivative (superscript mu). Or am I missing somehting entirely? Is it just that we can switch between the supersript and subscript using the raising operator?
     
    Last edited: Apr 9, 2017
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