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Help with basic tensor algebra

  1. Oct 7, 2005 #1
    Hi, I need some help understanding basic tensor algebra, especially differentiation. The subject I'm studying is quantum field theory, so I'll use examples from there.

    First let's start with a real scalar field. This has a Lagrangian density given by

    [tex]\mathcal{L} = \frac{1}{2}\partial_{\mu}\phi \partial^{\mu}\phi - \frac{1}{2}m^2\phi^2-\frac{\lambda}{4!}\phi^4[/tex]

    where [tex]\lambda[/tex] is just a (coupling) constant. We must then have that the Euler-Lagrange equation

    [tex]\frac{\partial \mathcal{L}}{\partial \phi}-\partial_{\mu}\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)} = 0[/tex]

    coincides with the dynamic equation

    [tex](\square + m^2)\phi = -\frac{\lambda}{6}\phi^3[/tex]

    The first part of the Euler-Lagrange equation is rather easy, differentiate with respect to [tex]\phi[/tex], and this gives

    [tex]\frac{\partial \mathcal{L}}{\partial \phi} = -m^2\phi - \frac{\lambda}{3!}\phi^3[/tex]

    Then, the second part, which I get to be

    [tex]\partial_{\mu}\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)} = \frac{1}{2}\partial_{\mu}\partial^{\mu}\phi[/tex]

    Combining these gives

    [tex]\left( \frac{1}{2}\partial_{\mu}\partial^{\mu}+m^2 \right) \phi = -\frac{\lambda}{6}\phi^3[/tex]

    This is an example from the textbook (Elementary Particles and Their Interactions by Quand Ho-Kim and Pham Xuan Yem), although the calculation has not been carried out explicitly. In this book the operator [tex]\square[/tex] is also defined as [tex]\square = \partial^{\mu}\partial_{\mu}[/tex], which is not quite what I got. Or is it the same after all? If not, what did I do wrong in the calculation? Or is it perhaps a typo in the book? It would be nice if someone could enlighten me on this. I have more examples (with a vector field), but I'll post that after I've, hopefully, gotten some respons to this problem.
  2. jcsd
  3. Oct 7, 2005 #2

    George Jones

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    Saw your similar thread on the Homework: College Level forum. If no one else replies, I will help you either later today or tomorrow. I'm a bit pressed for time right now.

  4. Oct 8, 2005 #3


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    You are mistaken:smile:

    [tex]\partial_{\mu}\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)} = \partial_{\mu}\partial^{\mu}\phi[/tex]
  5. Oct 8, 2005 #4

    George Jones

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    Here are some tips.

    1) Make sure that the indices in the symbol with respect to which you differentiate don't match *any* of the indices in the terms that get differentiated. If the indices match, bad things happen - either an idex that was originally free becomes repeated and thus summed over, or you end up with a term with 3 repeated indices, which is meaningless.

    2) Using 1) will introduce expressions like

    [tex]\frac{\partial a_{\mu}}{\partial a_{\alpha}} = \delta_{\mu}^{\alpha}[/tex],

    and like

    [tex]\delta_{\mu}^{\alpha} b_{\alpha} = b_{\mu}[/tex].

    3) Use the metric to make sure that all indices are in appropriate upstairs or downstairs locations. For example, if you want to differentiate an expression that contains

    [tex]\partial^{\mu} \phi[/tex]

    with respect to

    [tex]\partial_{\alpha} \phi[/tex],

    then the substitution

    [tex]\partial^{\mu} \phi = g^{\mu \beta} \partial_{\beta} \phi[/tex]

    should be made.

    Last edited: Oct 9, 2005
  6. Oct 9, 2005 #5
    Hi, thank you so much for your insight Mr. Jones. I can imagine it wasn't very easy to help when I have such general questions, but I must say, your tips were a very good start.

    Now I'll need to take a look at the examples and exercises again with this in mind, but have no fear, I'll probably be back with more questions, and hopefull they'll be more specific.

    And also thanks to gvk for pointing out that it was indeed I who was mistaking, and not the book. And now that we've located where the problem is... :smile:
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