# Is any continuous transformation a symmetry of action?

Consider a continuous transformation $$\phi \rightarrow \phi+ \delta\phi$$, where $$\phi$$ is a field operator and $$\delta \phi$$ is a infinitesmal change. If such continuous transformation is applied to a system with Lagrangian density $$L(\phi,\partial_\mu \phi)$$, the deviation of Lagrangian density is

$$\delta L = \frac{\partial L}{\partial \phi} \delta \phi + \frac{\partial L}{ \partial (\partial_\mu \phi)} \partial_\mu \delta \phi \\ = \left(\frac{\partial L}{\partial \phi} - \partial_\mu\frac{\partial L}{ \partial (\partial_\mu \phi)}\right) + \partial_\mu \left( \frac{\partial L}{ \partial (\partial_\mu \phi)} \delta\phi\right).$$ When equation of motion is satisfied, the first term of last line is zero, we have

$$\delta L = \partial_\mu \left( \frac{\partial L}{ \partial (\partial_\mu \phi)} \delta\phi\right).$$

$$\delta L$$ is a form of total derivative. Change of action is $$\delta S = \int d^4 x \ \delta L$$. By using Stokes' theorem, the bulk integral can be transformed into surface integral, which will not affect the form of equation of motion. Thus given any continuous transformation, we can prove that it is a symmetry.

I think this strange and should be wrong, but I do not know which step above is mistaken. Please give me some hint.

• What you showed is trivial since along the equations of motion $\delta S=0$ always. – Davide Morgante Feb 14 at 9:05

• Oh, I see. For time-space translational symmetry, $\delta L = \epsilon^\nu \partial_\nu L$. It is a total derivative, thus action will not change without satisfying EOM. – lsdragon Feb 14 at 8:44