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Consider a pencil standing in a (ideally) perfectly vertical position.

The gravitational field will the same no matter the (angular) direction it will fall in.

But it will end up falling in a particular direction.

Is this an example of spontaneous symmetry breaking? Or is just that due to quantum fluctuations the pencil will never actually be perfectly vertical?

Qmechanic
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SuperCiocia
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  • Why would the pencil fall in the first place? – zzz Dec 10 '14 at 23:46
  • Touché. So there is not spontaneous symmetry breaking involved after all? – SuperCiocia Dec 10 '14 at 23:49
  • Adding to previous comment, if the reason is something along the lines of uneven weight distribution, then there's no symmetry to start with. If that's not the reason I don't see how else (classically) a perfectly vertical pencil would fall. – zzz Dec 10 '14 at 23:50
  • The reasons I've always seen when authors use this as an example is that there are minor breezes that topple the pencil - meaning that it could fall even if there was initially symmetry. – HDE 226868 Dec 10 '14 at 23:54
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    @bechira: There are a-causal situations even in classical mechanics where no "cause" is required for such a "spontaneous" motion to occur (a famous instance is Norton's dome which would probably be a better model for this question than a pencil). At OP: Spontaneous symmetry breaking is a technical term almost always connected to a phase transition/order parameter, and crucially relies on the notion of "ground state". What would be a "ground state" in this context, and how does it change from invariant to non-invariant? – ACuriousMind Dec 10 '14 at 23:54
  • @ACuriousMind I learned something new today, thanks! – zzz Dec 10 '14 at 23:59
  • Isn't the ground state when the pencil falls on the table? Each angle the pencil falls at is a different ground state, all with the same total energy – SuperCiocia Dec 11 '14 at 00:06

1 Answers1

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It is indeed a an example of symmetry breaking. Basically, the problem (the equactions, and the fields, aven the solution) is symmetric, however because the solution is unstable, in a realistic case where noise is present will break the symmetry into a solution that is not symmetric. The solution "choose" one specific direction to get to a state of lower energy, and because of this choice, the solution is not symmetric anymore. In one of the comments you read

Spontaneous symmetry breaking is a technical term almost always connected to a phase transition/order parameter

which is correct, but not necessarily so. In this case the symmetry breaking is the choice one one particular ground state (the pencil resting horizontal on the table) among equaly probable others of the same energy.

From wikipedia:

Spontaneous symmetry breaking is a mode of realization of symmetry breaking in a physical system, where the underlying laws are invariant under a symmetry transformation, but the system as a whole changes under such transformations, in contrast to explicit symmetry breaking. It is a spontaneous process by which a system in a symmetrical state ends up in an asymmetrical state. It thus describes systems where the equations of motion or the Lagrangian obey certain symmetries, but the lowest-energy solutions do not exhibit that symmetry.

The example asked here staisfies that the equations of motion or the Lagrangian obey certain symmetries, but the lowest-energy solutions do not exhibit that symmetry. As such, it has to be defined as a speontaneous symmetry breaking.

On the other hand, explicit symmetry breaking is the breaking of a symmetry of a theory by terms in its defining equations of motion (most typically, to the Lagrangian or the Hamiltonian) that do not respect the symmetry.

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    If noise causes the symmetry breaking, isn't the situation explict symmetry breaking where we have some interaction term that breaks the symmetry, rather than spontaneous symmetry breaking? – zzz Dec 11 '14 at 00:13
  • I updated my answer. Of course, there might be some branches of physics were noise is noise is not considered a speontaneous process, and then this eaxmple would fall into their definition of explicit symmetry breaking. But I do not believe it is the viewpoint of the majority of physicists. And certainly not the way I learned it. –  Dec 11 '14 at 00:23
  • I'm not too sure about this, when I picture the situation classically, I picture "noise breaking symmetry" as some tiny dust particle knocking over the pencil – zzz Dec 11 '14 at 00:26
  • It doesnt matter for the definition of spentaneous, see here: http://en.wikipedia.org/wiki/Spontaneous_symmetry_breaking, and see updated answer –  Dec 11 '14 at 00:45
  • I would disagree that noise breaking the symmetry is considered a spontaneous symmetry breaking as the noise could have a perfectly well defined cause. I would suggest that if you could completely isolate the pencil from noise, thermal fluctuations, etc etc then the pencil would eventually fall due to quantum fluctuations about its tip and that this would be spontaneous symmetry breaking. – Akoben Dec 11 '14 at 01:12
  • @nmoy then you don not disagree with me, you disagree with the definition. What of the two definitions I gave in my answer do you think cares less about the idea of noise? –  Dec 11 '14 at 01:24
  • To clarify what I meant, if interaction with noise is viewed as some dust speck knocking over the pencil, you can write down the dust speck's dynamics in the lagrangian, and the lagrangian would not have the desired symmetry to start with - we've broken the symmetry explicitly. – zzz Dec 11 '14 at 19:22
  • @bechira as far as I know, the lagrangian for brownian motion has rotational symmetry –  Dec 11 '14 at 19:33
  • @Wolphramjonny As far as I know lagrangian for brownian motion is not well defined? – zzz Dec 11 '14 at 23:08
  • well, you can model radially symmetric noise in the equations of motion or the langragian, right? regerdeless of the term being differentiable or not. –  Dec 11 '14 at 23:20