Expanding delta variations in Euler-Lagrange field theory derivation

[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.