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)

 

[TysonM8]

This appears to be an application of the reverse product rule. Starting from the second step and working backwards;

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

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

I'm not sure about this last step...

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

Hmm, I understand it's the last step you don't understand. Well it must be valid if the derivation is correct. You'll just have to figure out why;

$∂_{μ}(δ(\phi)) = δ(∂_{μ}\phi)$

Sorry I can't help with that, I don't know anything about classical field theory or the Euler-Lagrange equations.

 

[muzialis]

Let me preface I know very little about Field Theory, so please take my reply with caution.
On the other hand, the step in question seems to me stemming from basic calculus of variations (but care needs to be exterted as there are different definitions of variations around).
If you search under "commutative rule" (i.e. on the property of certain variations to commute, i.e. the variation of the velocity is equal to the time-derivative of the variation, for example in classical mechanics) in Calculus of variations you will find relevant material.