# Expanding delta in Field Theory Derivation of Euler-Lagrange Equations

• chuchi
In summary, the conversation discusses a difficulty in understanding the derivation of the Euler-Lagrange Equations in classical field theory on page 15 of Peskin & Schroeder. The equation involves a delta term and the speaker is unsure how to extract anything from it. The expert summarizes the process of replacing the field \phi with \phi+\delta\phi and the use of the product rule for derivatives in solving the equation. The conversation also clarifies that \mathcal{L} can represent either the Lorentz group or the Lagrangian density in this context.
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.)

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.

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

## 1. What is the importance of the expanding delta in field theory derivation of Euler-Lagrange equations?

The expanding delta in field theory is a mathematical concept that represents a variation in a field. It is essential in the derivation of the Euler-Lagrange equations because it allows us to account for infinitesimal changes in the field and calculate the corresponding changes in the action functional.

## 2. How is the expanding delta defined in field theory?

The expanding delta, denoted as Δ, is defined as a function that takes in a field φ and a variation δφ and produces a new field φ' that represents a small change in the original field.

## 3. What role does the expanding delta play in the Euler-Lagrange equations?

The expanding delta is crucial in the Euler-Lagrange equations because it allows us to calculate the functional derivative of the action, which is necessary for finding the extremum of the action. It also helps us to account for variations in the field that satisfy the boundary conditions.

## 4. How is the expanding delta used in the derivation of the Euler-Lagrange equations?

In the derivation of the Euler-Lagrange equations, the expanding delta is used to vary the field and calculate the corresponding changes in the action functional. It is then substituted into the principle of least action, and the resulting equation is set to zero to find the extremum of the action, which gives us the Euler-Lagrange equations.

## 5. Are there any limitations to using the expanding delta in field theory derivation of Euler-Lagrange equations?

The expanding delta is a useful mathematical tool, but it has its limitations. It can only be applied to fields that satisfy certain smoothness conditions, and it may not be applicable in all cases, such as when dealing with discontinuous or non-differentiable fields.

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