Actions, symmetries, and gauge theories

coalquay404

Suppose that we have some theory which is invariant under the action of a gauge group, G. Since this theory is a gauge theory, it can be derived from a singular Lagrangian or Lagrangian density, L. Now suppose that the action for this theory,

S = \int L(q,\dot{q})

is invariant under the action of some group H. What is the relationship between the groups G and H?

I guess the reason I'm interested in asking this is to find out whether or not the symmetries of an action functional must necessarily be the same as the symmetries of the resultant equations of motion for the theory. As an extension, I wonder if the symmetries of an action could generate a different symmetry group in the equations of motion. It strikes me that this is an obvious question, but I can't seem to find any (rigorous) information on it.

Igor Khavkine

On 2007-02-06, coalquay404 <coalquay404.2lkya2@physicsforums.com> wrote:
>
> Suppose that we have some theory which is invariant under the action
> of
> a gauge group, G. Since this theory is a gauge theory, it can be
> derived from a singular Lagrangian or Lagrangian density, L. Now
> suppose that the action for this theory,
>
> S = \int L(q,\dot{q})
>
> is invariant under the action of some group H. What is the
> relationship
> between the groups G and H?

H may be larger than G. For instance, translation invariance is usually
not considered a gauge symmetry. But an action functional may well be
translation invariant and invariant under local gauge transformations.
This is in fact the case for electrodynamics.

> I guess the reason I'm interested in asking this is to find out
> whether
> or not the symmetries of an action functional must necessarily be the
> same as the symmetries of the resultant equations of motion for the
> theory. As an extension, I wonder if the symmetries of an action could
> generate a different symmetry group in the equations of motion. It
> strikes me that this is an obvious question, but I can't seem to find
> any (rigorous) information on it.

A symmetry of the action functional implies a symmetry of the equations
of motion (Noether's theorem). But, a priori, the reverse need not be
true. However, I'm not sure I can provide an example off the top of my
head. Would such an example answer your question, though?

Hope this helps.

Igor

torre@cc.usu.edu

> I guess the reason I'm interested in asking this is to find out whether
> or not the symmetries of an action functional must necessarily be the
> same as the symmetries of the resultant equations of motion for the
> theory. As an extension, I wonder if the symmetries of an action could
> generate a different symmetry group in the equations of motion.

The short answer is that the symmetries of a Lagrangian are always
symmetries
of the Euler-Lagrange equations. But there can be more symmetries of
the equations
than of the Lagrangian.

See for example, P. Olver, "Applications of Lie Groups to Differential
Equations".

charlie torre

torre@cc.usu.edu

> A symmetry of the action functional implies a symmetry of the equations
> of motion (Noether's theorem). But, a priori, the reverse need not be
> true. However, I'm not sure I can provide an example off the top of my
> head.

A standard example is to consider a Lagrangian which yields linear
Euler-Lagrange equations. The equations, being linear, admit a
scaling symmetry. The Lagrangian typically will not admit that
symmetry.

charlie