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).

  • $\begingroup$ Should this be on Mathematics SE? $\endgroup$ – Aaron Stevens Aug 2 '19 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.

  • $\begingroup$ Thank you for your answer. Isn't there $\delta \phi$ missing in $\int d^4x(c/2)$? Should it rather be $\int d^4x[(c/2)\delta\phi]=c/2$? $\endgroup$ – Kathi Aug 3 '19 at 13:00
  • $\begingroup$ I fixed the typo, thanks. $\endgroup$ – Valter Moretti Aug 3 '19 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 $\frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu} \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)} = 0$.

  • $\begingroup$ Thank you for your answer. How can I prove that there exists a $\delta \phi$ for which the integral doesn't vanish? $\endgroup$ – Kathi Aug 2 '19 at 18:40
  • 2
    $\begingroup$ @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. $\endgroup$ – Javier Aug 2 '19 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.