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I have been reading other questions in this site, but I have not found answers to all my questions about theories without Lagrangians.

What do we mean exactly when we say that they do not have a Lagrangian? Do they still have an action but this action cannot be defined as the integral of a local Lagrangian? Is it then that they do not have a local Lagrangian or they do not even have an action at all?

Qmechanic
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edmateosg
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1 Answers1

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  1. That a set of equations does not have a variational formulation means that there does not exist an action functional, cf. e.g. this Phys.SE post.

  2. The existence of a local functional is typically a separate issue, which is usually required in fundamental physics.

Qmechanic
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  • What can it be said about a system if we know that it cannot have a variational formulation? Could it be that we cannot find such formulation only because we are missing some relevant degrees of freedom in our description? Or are there cases that we couldn't describe even with this additional parameters? Even in these extreme cases, could it be possible to write down some sort of effective Lagrangian that can approximate the solutions in a certain regime even if it fails to describe the system in its entirety? – edmateosg Apr 12 '20 at 23:11
  • As an effective Lagrangian I am thinking of, for example, one written using Lagrange multipliers for the EOM that we know are solutions to the system – edmateosg Apr 12 '20 at 23:24
  • The existence & non-existence of an action formulation is the topic of the linked Phys.SE post. Enforcing dynamical EOMs via Lagrange multipliers has various drawbacks. E.g. it renders the Lagrange multipliers dynamical, and hence changes the dynamical content of the theory. – Qmechanic Apr 12 '20 at 23:41