$\delta S=0$ only for $\frac{\partial\mathcal{L}}{\partial\phi}-\partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}=0$?

Question

Condition for the variation of action is:

$$0=\delta S$$

$$=\int d^4 x [\frac{\partial \mathcal{L}}{\partial \phi}\delta\phi-\partial_\mu(\frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)})\delta \phi+\partial_\mu (\frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)}\delta \phi)].$$

It is clear that the last summand is zero because $\delta \phi=0$ at the beginning (B) and ending (E):

$$\int_{x_B}^{x_E} d x (\frac{\partial \mathcal{L}}{\partial \dot\phi}\delta\phi)|_{t_B}^{t_E}+\int_{t_B}^{t_E} d t (\frac{\partial \mathcal{L}}{\partial \phi'}\delta\phi)|_{x_B}^{x_E}=0.$$

In every source I find they say only that $$\int d^4 x [(\frac{\partial \mathcal{L}}{\partial \phi}-\partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)})\delta \phi]$$ is zero if $$\frac{\partial \mathcal{L}}{\partial \phi}-\partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)}=0.$$ That is obvious. But is it provable that $$\int d^4 x [(\frac{\partial \mathcal{L}}{\partial \phi}-\partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)})\delta \phi]\neq 0$$ for $$\frac{\partial \mathcal{L}}{\partial \phi}-\partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)}\neq0~? $$

I tried to integrate this, for example with partial integration, but I don't get far with this and I don't have a clue how to proove this (if there is any proof).

Answer 1

Let us assume that the Lagrangian is $C^2$ and the fields are $C^2$ as well. As a consequence the function $$x \mapsto \left( \frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu} \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)}\right)|_x $$ is atleast continuous.

Suppose now that there is $x_0$ where $$\left( \frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu} \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)}\right)|_{x_0} = c\neq 0\:.$$ Let us assume that $c>0$ since for $c<0$ the argument is identical.

Since the function is continuous, for every $\epsilon>0$ there must be an open set $A_\epsilon$ including $x_0$ where $$c+ \epsilon > \left( \frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu} \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)}\right) > c -\epsilon\:.$$ We fix in particular $\epsilon=c/2$ so that $$\left( \frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu} \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)}\right) > c/2 >0$$ in $A_\epsilon$.

It is possible to construct a smooth function $\delta \phi$ such that it vanishes outside $A_\epsilon$ and $$\int \delta \phi d^4x =\int_{A_\epsilon} \delta \phi d^4x =1\:.$$ (Start from a Gaussian centered on $x_0$, next make it smoothly vanishing outside $A_\epsilon$, finally re-normalize it to obtain a total integral $1$.) As a consequence $$\int d^4x \left[ \left( \frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu} \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)}\right) \delta\phi \right] = \int_{A_\epsilon} d^4x \left[ \left( \frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu} \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)}\right) \delta\phi \right] \geq \int d^4x (c/2) \delta \phi= c/2 >0\:.$$ This is impossible because we have supposed that $$\int d^4x \left[ \left( \frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu} \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)}\right) \delta\phi \right] =0$$ for every choice of $\delta \phi$. As a consequence $x_0$ does not exist so that $$\left( \frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu} \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)}\right)=0$$ everywhere.

Answer 2

The point here is that $\int d^4x \left[ \left( \frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu} \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)}\right) \delta\phi \right] = 0$ for all $\delta \phi$. It means that for a specific $\delta \phi$, the integral can be $0$ without satisfying the variational condition, but if so, you can prove that there exists a $\delta \phi$ for which the integral won't vanish anymore. The only way for the integral to be identically $0$ for any $\delta \phi$ is if $\frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu} \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)} = 0$.