1
$\begingroup$

Consider a Lagrangian theory of fields $\phi^a(x)$. Sometime such a theory posseses a symmetry (let's talk about internal symmetries for simplicity), which means that the Lagrangian is invariant under replacement $\phi^a\to \phi'^a=\phi'^a(\phi,\epsilon)$. Here $\epsilon$ are some continuous transformation parameters. Usually one encounters symmetries that are linear in $\phi$, for example $\phi'=e^{i\epsilon}\phi$ for a single complex scalar or $\phi'^a=\epsilon^{ab}\phi^b$ with orthogonal matrix $\epsilon^{ab}$ for Lagrangian $\mathcal{L}=\frac12\left(\partial_\mu\phi^a\right)^2$.

My question is whether there are examples of non-linear symmetry transformations appearing in physical models? At the time I am mostly concerned with the classical fields but comments on the quantum extensions are surely welcome.


Clarification.

I appreciate references to various actual models. However I would like first to see an example as simple and explicit as possible, where the essence is not obstructed by technicalities. If there are some principle difficulties to construct a really simple example, there must be a reason for that?

Let me also narrow what I mean by a non-linear internal transformation. Assume that replacement $\phi^a(x)\to \phi^a_\epsilon(x)=f(\phi^b(x),\epsilon)$ with some function $f$ leaves Lagrangian invariant $L(\phi,\partial\phi)=L(\phi_\epsilon,\partial\phi_\epsilon)$. Parameter $\epsilon$ could be a vector. Then call such a transformation linear if $\frac{\partial \phi^a_\epsilon}{\partial\phi^b}$ is independent of co-ordinates, $\partial_\mu\frac{\partial \phi^a_\epsilon}{\partial\phi^b}=0$. In this sense, the shift transformation proposed by Andrew is also linear.

$\endgroup$
3
  • 1
    $\begingroup$ This question (v2) seems like a list question. Here is one example of a non-linear gauge symmetry. $\endgroup$
    – Qmechanic
    Commented Oct 22, 2015 at 8:39
  • $\begingroup$ What does this mean? $\endgroup$ Commented Oct 22, 2015 at 8:43
  • $\begingroup$ @Qmechanic the example seems too technical at the first glance. I believe that there must be either a simpler instance, or an argument for why it is not possible. Moreover, I would like a proper symmetry and not a gauge one to be addressed. $\endgroup$ Commented Oct 22, 2015 at 9:26

3 Answers 3

3
$\begingroup$

The simplest example I know of is a shift symmetry, $\phi\rightarrow \phi + c$ for constant $c$. It is not linear in the sense that is not of the form $\phi\rightarrow U\phi$ for some matrix $U$.

And example lagrangian with this symmetry is just a free massless scalar field, $\mathcal{L}=-1/2(\partial_\mu \phi)^2$.

A consequence of the fact that the symmetry is non-linearly realized (to use the jargon) is that we can think of this shift symmetry as a coming from spontaneously breaking another symmetry. Indeed, you can think of $\phi$ as being the Goldstone boson associated with this breaking.

Another related fact is that correlation functions are not invariant under this transformation. For example, the vacuum expectation value $\langle 0 | \phi | 0 \rangle$ is not invariant under $\phi\rightarrow\phi + c$.

$\endgroup$
5
  • $\begingroup$ An interesting example, however not the one I had I mind. Depends on the conventions of course, but I would call this one linear inhomogeneous or something like that. Does not really suit for an application I've been thinking on. $\endgroup$ Commented Oct 22, 2015 at 13:08
  • $\begingroup$ What application do you have in mind, out of curiosity? $\endgroup$
    – Andrew
    Commented Oct 22, 2015 at 13:18
  • $\begingroup$ It seems to me, that invariance of Lagrangian is equivalent to invariance of EOM only if this condition ('linearity' as stated in the appendix to the OP) is satisfied. I wish to check this at some particular examples. Maybe I'll make up a separate question on this. $\endgroup$ Commented Oct 22, 2015 at 13:59
  • 1
    $\begingroup$ @WeatherReport Could you check this with my example? I expect that KPZ equation is invariant under the non-linear symmetry, but I never checked. $\endgroup$ Commented Oct 22, 2015 at 14:48
  • $\begingroup$ @WeatherReport Hm I'm not sure I agree with that, if the actions are the same before and after the transformation then the EOMs derived from those actions will be the same too. A subtlety that may or may not be relevant is that when you have linearly realized symmetries involving multiple fields (such as an internal $SO(2)$ symmetry), the EOMs are technically covariant, not invariant, under the transformation. $\endgroup$
    – Andrew
    Commented Oct 22, 2015 at 15:38
1
$\begingroup$

Consider any non-linear sigma-model with a (pseudo-)Riemannian target space(-time) $(X,g)$. For instance the relativistic particle propagating on a spacetime is an example. And any $p$-brane sigma model is an example. Also the scalar field sector in compactications of higher dimensional (super-)gravity theories are examples.

For these sigma-model field theories, the fields are smooth functions $\Sigma_{p+1} \longrightarrow X$ and the isometries of $(X,g)$ are (induce) non-gauge symmetries of the sigma-model. Any isometry whose underlying diffeomorphism $X \to X$ is, in any set of coordinates, not simply a linear function (and generically it won't) gives an example of a non-linear symmetry.

$\endgroup$
1
  • $\begingroup$ Could you elaborate this explicitly at a simple enough example, say of a single particle? $\endgroup$ Commented Oct 22, 2015 at 11:49
1
$\begingroup$

The stochastic Kardar-Parisi-Zhang (KPZ) equation, $\partial_t h + \frac{\lambda}{2} \left[\vec{\nabla}h\right]^2 = \nu \nabla^2 h + \eta \, ,$ with $\langle \eta \rangle = 0$ and $\langle \eta(t,x) \eta(t',x') \rangle = D \, \delta(t-t') \, \delta(x-x') \, ,$ can be described as a field theory with the action $$ S = \int_{t,x} \tilde{h}\left(\partial_t h + \frac{\lambda}{2} \left[\vec{\nabla}h\right]^2 - \nu \nabla^2 h\right) - D \, \tilde{h}^2 \, . $$ See e.g. this great book for details and further references. $h(t,x)$ is the height field as before and $\tilde{h}(t,x)$ is an auxiliary field that can be used to compute response functions.

This problem posses a non-linear symmetry. Indeed, in it's cole-hopf transformed version, $$h = \frac{2\nu}{\lambda} \, \log\left|w\right| \, , \qquad \tilde{h} = w \, \tilde{w} \, ,$$ the action is $$ S = \int_{t,x} \tilde{w}\left(\partial_t w - \nabla^2 w\right) - \frac{1}{4} \frac{\lambda^2 D}{\nu^3} \left(w \tilde{w}\right)^2 \, .$$ The transformed problem is invariant under the transformation $w(t,x) \rightarrow \tilde{w}(-t,x)$, $\tilde{w}(t,x)\rightarrow w(-t,x)$. When translated back into the original variables, this becomes $$h'(t,x) = -h(-t,x) + \frac{2\nu}{\lambda} \log \left|\tilde{h}(-t,x)\right| \, , \qquad \tilde{h}'(t,x) = \tilde{h}(-t,x) \, , $$ which is non-linear.

See e.g. this paper for details on the stochastic KPZ equation and further references.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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