Suppose you have semigroup instead of typical group construction in Noether theorem. Is this interesting? In fact there is no time-reversal symmetry in the nature, right? At least not in the same meaning as with other symmetries (rotation, translation etc). So why we construct energy as invariant of such kind of group symmetry, not using semigroup instead? Is is even possible? There is a plenty of references about Lie semigroups - it looks like it is active field of mathematics. Is there any kind of Noether theory constructed within such theories?
$\begingroup$
$\endgroup$
4
-
1$\begingroup$ For starters, I think you want a Lie monoid rather than a Lie semigroup, so that there is a unit element. $\endgroup$– Qmechanic ♦Commented Mar 31, 2011 at 20:59
-
$\begingroup$ @Qmechanic - possibly - yes. Semigroup is more general - but when You say - that we have to perform some kind of perturbation around unit element - I probably agree. It has to start with s' = f(s) where f(s) is semigroup action on s and it should be possible to describe f(s) as s itself and small change. Notice - it probably is not simple Taylor expansion, nor any kind of linear transformation ( but maybe it is linear but not reversible?) Yes, I am curious how do they do for Lie monoids ;-) $\endgroup$– kakazCommented Mar 31, 2011 at 21:10
-
$\begingroup$ I have done crosspost mathoverflow.net/q/60238 - it looks like mathematical - that physical question. $\endgroup$– kakazCommented Mar 31, 2011 at 21:48
-
1$\begingroup$ I have found this paper: arxiv.org/abs/math/0604561v1 "Genuine Lie semigroups and semi-symmetries of PDEs" Is it representative for the level of knowledge about this area? If so: it means that such semigroup action on PDE equation solutions are called semi-symmetries. Probably there is a plenty to discover in this area, I suppose;-) $\endgroup$– kakazCommented Apr 2, 2011 at 20:45
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
5
As lurscher says, semigroups are not too interesting in physics. They're not really "symmetries". An important physics example of a semigroup in physics is the (misleadingly called) "Renormalization Group" that allows us to derive effective laws for long distances from the short-distance ones, but this "integrating out" or "flowing" is irreversible - which is exactly why the RG operations deserve to constitute a "semigroup" rather than a "group".
However, more importantly, you are totally confused about the case of Noether's observable called "energy".
Energy is associated with the time-translational-invariance and it is a standard symmetry, expressed by a group and not "just a semigroup". This symmetry says that the laws of physics don't change if the events occur $t$ seconds later. Also, they didn't change if the processes occurred $t$ minutes earlier. The time translations go in both directions, they can be both positive and negative, and they're always symmetries.
This has nothing to do with the irreversibility of the processes or the asymmetries associated with the arrow of time.
However, you are actually wrong even about the time reversal. Even the time reversal symmetry, which is a symmetry of the microscopic laws of physics (if CP holds; or at least CPT is such a symmetry universally) is a symmetry, and it is again expressed by a $Z_2$ group. There is no non-group semigroup anywhere in those spacetime-related operations!
Back to why "semigroups" are not symmetries
Quite generally, objects such as the action cannot be symmetric under semigroups that are not groups. The example of the Renormalization Group clearly shows what happens if the would-be symmetry transformations were irreversible. Irreversibility always means that some information is getting lost, so the contributions to the action from the degrees of freedom whose information is getting lost are inevitably being suppressed, so that action cannot stay constant. If one "flows" to long distances via the Renormalization Group flows, he actually "integrates out" the shortest-distance degrees of freedom which means that they cannot be excited in the new effective theory. So the operation cannot provide us with a one-to-one map for all states which, as explained in the previous sentences, is necessary for a symmetry, and symmetries therefore have to groups and not just semigroups.
answered Mar 31, 2011 at 18:41
-
$\begingroup$ Do You know the notion of active and passive transformation? Is the theory of Noether tied to passive one? Why? And please tell me - instead of particular interpretation - do You have any arguments that world at the time t and the time t' where t'>t are the same? There is a plenty of phenomena for which time reversibility is not true, but they allow description similar to Hamiltonian one - for example by means of Rayleigh function. I do not agree that "this has nothing with irreversibility..." because You simply do not know that. Do You know any example of such description? Just for fun? $\endgroup$– kakazCommented Mar 31, 2011 at 18:50
-
$\begingroup$ Classical Dynamics has nothing to do with CPT theorem. Sorry - it is not important here at all. Please treat is as simply mathematical physics question for references. $\endgroup$– kakazCommented Mar 31, 2011 at 18:51
-
$\begingroup$ š - Yes - You are right - during renormalization flow information is lost.This is exactly what I am looking for, in context of classical dynamics and Noether theorem. Renormalization group is kind of action which leads to certain kind of relation like en.wikipedia.org/wiki/Callan%E2%80%93Symanzik_equation . Of course there are others for beta function and so on. This kind of relation are generalization of invariant of motion equations ( trivial ones, You may also notice that if renormalization action became a group this relations are trivial - couplings are invariant). $\endgroup$– kakazCommented Mar 31, 2011 at 19:12
-
$\begingroup$ Dear kakaz, whether one uses an active or passive transformation is a matter of personal choice. In each situation, he can choose it according to his taste. They differ by being inverse transformations of one another. It's meaningless to ask "what is the right convention for Noether's theorem." Again, Noether's theorem has nothing to do with irreversibility of the macroscopic physics, and it's not true that CPT has "nothing to do with classical dynamics". We can surely discuss CPT transformations in any physical theory. RG is not a "symmetry of the action" in physics' definition of symmetry. $\endgroup$ Commented Apr 1, 2011 at 5:44
-
$\begingroup$ of course You are right - RG is not kind of symmetry of physical system - it is rather kind of symmetry of ts description. Of course You are right- I also believe that QFT is probably the deepest theory we have, so CPT theorem is very fundamental. I am not asking about that. I am asking only about Noether theorem generalization in context of semigrioup, and relate it to some loose idea about time shift. I fact this is question about mathematical physics, and is not related to any kind of physical reality. $\endgroup$– kakazCommented Apr 1, 2011 at 6:02
$\begingroup$
$\endgroup$
4
Here is an argument why "a Noether Theorem with Lie monoid symmetry" essentially wouldn't produce any new conservation laws. Noether's (first) Theorem is really not about Lie groups but only about Lie algebras, i.e., one just needs $n$ infinitesimal symmetries to deduce $n$ conservation laws. If one is only interested in getting the $n$ conservation laws one by one (and not so much interested in the fact that the $n$ conservation laws together form a representation of the Lie algebra), then one may focus on a $1$-dimensional Abelian subgroup of symmetry. The corresponding Lie subalgebra then becomes just $u(1)\cong\mathbb{R}$. Now returning to the question, one may, of course, artificially truncate a Lie group into a Lie monoid, say, if $q$ is a cyclic variable for a Lagrangian $L$, then artificially declare that the symmetry monoid is $q \to q + a$ for only non-negative translations $a\geq 0$, while artificially denying all negative $a<0$. On the other hand, one needs at least access "from one side" because Noether's Theorem is about continuous symmetry. But in practice, one can then always extend, at least infinitesimally, to "the other side" as well, and then one is back to a standard $u(1)$ Lie algebra and a standard Noether Theorem.
answered Mar 31, 2011 at 22:22
-
$\begingroup$ It is a good comment! I do not know how Lie semigroups are described - are You sure that q->q+a a>0 is right there? Could You provide any reference for that? It looks like q->q+a is not very well suited with semigroup approach because I do not know if semigroup allows such representation - in fact relation q->q+a for a>0 looks very artificially. For me, probably better idea is q->a+At where A is not invertible matrix and t is small parameter. Please notice also that on math.stackexchange there is no obvious answer until now - I presume it is not so simple... But of course You may be right! $\endgroup$– kakazCommented Apr 1, 2011 at 6:08
-
$\begingroup$ The Lie semigroup $(]0,\infty[,+)$ and the Lie monoid $([0,\infty[,+)$ are well-defined $1$-dimensional examples. I have no references handy. $\endgroup$– Qmechanic ♦Commented Apr 1, 2011 at 7:48
-
$\begingroup$ Yer they are well defined but what about definition of tangent space of it? And infinitesimal generators? And expansion of small element? Probably You are right that this is well known example (from ergodic theory, for example), but what about such structure in context of Lie algebra etc? $\endgroup$– kakazCommented Apr 1, 2011 at 9:20
-
$\begingroup$ One more notice - I do not expect any new invariant of motion - rather kind of flow remaining Renormalization Group equations. $\endgroup$– kakazCommented Apr 1, 2011 at 12:00
$\begingroup$
$\endgroup$
3
The main problem with semi-groups is that they are not in general invertible, so it doesn't apply well to the notion of a symmetry, which is something that behaves the same after performing an operation that transforms the states into each other
-
1$\begingroup$ You mean not invertible? When You start with Noether theorem You assume You have lagrangian L(q,p) = L(q',p') where s' = s + ðs (and s = p or q). So ðs is infinitesimal symmetry change. There is no requirement apriori that it has to be invertible. Although it is not simple at all to "represent inrreversible" infinitensimal elements because operation "+" is not good here. But here en.wikipedia.org/wiki/… You may look at it in very general description. For me it looks like it could be possible to use semigroup here. $\endgroup$– kakazCommented Mar 31, 2011 at 18:40
-
1$\begingroup$ on the other hand, i wouldn't doubt that a description of thermodynamic (irreversible) processes would make great sense in the semi-group language $\endgroup$– lurscherCommented Mar 31, 2011 at 21:41
-
$\begingroup$ Yes - that is good example. More - dissipative or Hamiltonian dynamics has semigroup as a flow describing motion in phase space. but this is different area - You have phase space volume v and want to know how it flows during dynamics. Then it leads to Liouville equations en.wikipedia.org/wiki/Liouville%27s_theorem_%28Hamiltonian%29 . But here I am asking about something different - just how change of group into semigroup changes Noether theorem, famous torus phase space structure etc. $\endgroup$– kakazCommented Mar 31, 2011 at 21:56