1
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ What you showed is trivial since along the equations of motion $\delta S=0$ always. $\endgroup$ – Davide Morgante Feb 14 at 9:05
2
$\begingroup$
  1. OP has proven that every infinitesimal transformation of an action is automatically a symmetry on-shell, i.e. the notion of on-shell symmetry is a vacuous notion, and therefore not used in practice.

  2. In contrast, a symmetry of an action is conventionally assumed to hold off-shell. See also this related Phys.SE post.

$\endgroup$
4
  • $\begingroup$ Thank for your answer. Does on-shell means satisfying EOM? If so, a symmetry of action conventionally make a change of Lagrangian a form a total derivative form without need of satisfying EOM? $\endgroup$ – lsdragon Feb 14 at 8:21
  • $\begingroup$ Besides, when I read derivation of energy-momentum tensor, it requires that EOM holds. From your answer, can I say time-space translational symmetry is holds even when EOM does not hold? $\endgroup$ – lsdragon Feb 14 at 8:22
  • $\begingroup$ 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. $\endgroup$ – lsdragon Feb 14 at 8:44
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Feb 14 at 13:55

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.