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Expanding delta in Field Theory Derivation of Euler-Lagrange Equations

  1. Feb 23, 2014 #1
    Every time I try to read Peskin & Schroeder I run into a brick wall on page 15 (section 2.2) when they quickly derive the Euler-Lagrange Equations in classical field theory. The relevant step is this:

    [itex]\frac{∂L}{∂(∂_{μ}\phi)} δ(∂_{μ}\phi) [/itex]

    [itex]= -∂_{μ}( \frac{∂L}{∂(∂_{μ}\phi)}) δ(\phi) + ∂_{μ} (\frac{∂L}{∂(∂_{μ}\phi)} δ(\phi)) [/itex]

    How can we even extract anything within that delta? How does that even work? I feel like I'm missing some basic calculus here, but can't find anything in my textbooks or google.

    (those "L"s are supposed to be Lagrangian densities, I just don't know the curly script for L in latex. all of this is inside an integral and there's another term, but neither of those change. I also posted to calculus but no one knew enough.)
     
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  3. Feb 24, 2014 #2

    Avodyne

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    I don't understand what you're asking, but here's what's going on.

    First of all, for the purposes of this equation, we don't need any properties of [itex]\frac{∂L}{∂(∂_{μ}\phi)}[/itex], so I will just call it [itex]V^\mu[/itex].

    Now, the meaning of [itex]\delta X[/itex], where [itex]X[/itex] is anything that depends on the field [itex]\phi[/itex], is this: we take the field [itex]\phi(x)[/itex] and replace it with [itex]\phi(x)+\delta\phi(x)[/itex], where [itex]\delta\phi(x)[/itex] is "small". Then [itex]X[/itex] becomes [itex]X+\delta X[/itex], and we have to work out what [itex]\delta X[/itex] is in terms of [itex]\delta\phi[/itex].

    For your equation, we take [itex]X=\partial_\mu\phi(x)[/itex]. Now this is very simple: [itex]\delta X=\partial_\mu\delta\phi[/itex]. Stare at that until you are sure you understand it.

    The next important fact is that what I am calling [itex]\delta X[/itex] for [itex]X=\partial_\mu\phi[/itex] is what P&S call [itex]\delta(\partial_\mu\phi)[/itex].

    The rest is trivial calculus. By the product rule for derivatives, [itex]\partial_\mu(V^\mu\delta\phi)=(\partial_\mu V^\mu)\delta\phi+V^\mu(\partial_\mu\delta\phi)[/itex]. Rearrange that to get your equation.
     
    Last edited: Feb 24, 2014
  4. Feb 24, 2014 #3

    dextercioby

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    Adding a small thing: it's \mathcal{L} that produces [itex] \mathcal{L}[/itex] to denote either the Lorentz group, or the Lagrangian density.
     
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