# The integral under the variation $\delta$ sign

In physics books on classical field theory, the authors usually define the action as $$S = \int\mathcal{L(\phi,\partial_\mu\phi)d^4x}$$ where $\mathcal{L}$ is the lagrangian density. Then, they say that the principle of least action states that systems evolve along a path in configuration space for which $S$ is an extremum. So far, so good.

Then, when deriving Euler-Lagrange Equations, they write $$\delta S=0 \\ \iff \int\{\mathcal{\frac{\partial\mathcal{L}}{\partial\phi}\delta\phi+ \frac{\partial L}{\partial(\partial_\mu\phi)}}\delta(\partial_\mu \phi) \} d^4x =0$$

But they do not justify why this is true. Also, I do not have a clear, rigorous notion of what this "variation" $\delta$ is. Also: I have read somewhere that the principle of least action "is valid only in compact supported variations of the fields". Is this true? Although I know what compactly supported means, I do not know how that concept applies to this case.

Note: This argument can be found, for example, in Peskin and Schroeder's book on QFT.

## 1 Answer

Here is the usual ansatz they are referring to. If you are unsatisfied and still interested in learning it very rigorously you should refer to math texts on calculus of variations, but the physics books almost always repeat some variation of the following.

The action $S$ is a functional of your field and $\delta$ is a variation of the path. Specifically by $\delta S=0$ we mean that $S$ has no first order variation (in particular the variation vanishes in all directions).

So for a first order variation in the field (note the field is a function of the coordinates), e.g. $\phi\to\phi+\delta\phi$, you have:

$$\delta S=\int\mathcal{L(\phi+\delta\phi,\partial_\mu\phi+\delta(\partial_\mu\phi))}\,\mathrm{d}^4x-\int\mathcal{L(\phi,\partial_\mu\phi)}\,\mathrm{d}^4x=0$$

and then expanding the first $\mathcal{L}$ in Taylor series and only leaving first order terms, you have your equation. You then use integration by parts on the second term in your equation and drop the boundary terms (since they vanish on the edge of the support). You will later use that the variation vanishes at the end points (in all directions) to eliminate a term and that $\delta\to 0$ to finally get the variant of the Euler-Lagrange equations you're looking for.

• I'd add that the compact support condition is used to drop boundary terms in integration by parts. – J.G. Jul 26 '17 at 23:09
• @J.G. Thanks, added – GPhys Jul 26 '17 at 23:22