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

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).

• Should this be on Mathematics SE? Aug 2, 2019 at 19:23

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.

• Thank you for your answer. Isn't there $\delta \phi$ missing in $\int d^4x(c/2)$? Should it rather be $\int d^4x[(c/2)\delta\phi]=c/2$? Aug 3, 2019 at 13:00
• I fixed the typo, thanks. Aug 3, 2019 at 15:23

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 $$\frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu} \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)} = 0$$.

• Thank you for your answer. How can I prove that there exists a $\delta \phi$ for which the integral doesn't vanish? Aug 2, 2019 at 18:40
• @Kathi if the term in parentheses is nonzero somewhere, by continuity it doesn't change sign in a sufficiently small region. Now consider a $\delta \phi$ which is zero except for a small bump in the region just mentioned. Aug 2, 2019 at 22:13