Expanding delta in Field Theory Derivation of Euler-Lagrange Equations

[chuchi]

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:

\frac{∂L}{∂(∂_{μ}\phi)} δ(∂_{μ}\phi)

= -∂_{μ}( \frac{∂L}{∂(∂_{μ}\phi)}) δ(\phi) + ∂_{μ} (\frac{∂L}{∂(∂_{μ}\phi)} δ(\phi))

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.)

 

[Avodyne]

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 \frac{∂L}{∂(∂_{μ}\phi)}, so I will just call it V^\mu.

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

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

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

The rest is trivial calculus. By the product rule for derivatives, \partial_\mu(V^\mu\delta\phi)=(\partial_\mu V^\mu)\delta\phi+V^\mu(\partial_\mu\delta\phi). Rearrange that to get your equation.

 

[dextercioby]

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