Talk:Spontaneous symmetry breaking
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- 1 Quantum Vacuum Collapse
- 2 vacuum states
- 3 langrangians
- 4 Mexican hat potential
- 5 ferromagnetic materials
- 6 Many examples are not spontaneous
- 7 Why does it matter?
- 8 Transcribed from Comments page
- 9 Isn't symmetry added also?
- 10 Does symmetry breaking involve release of energy?
- 11 Ball example
- 12 Merge proposal
- 13 Symmetry breaking vs. Wavefunction collapse
Quantum Vacuum Collapse
QVC redirects to this page. A more explicit explanation of why would be nice.
This article uses the term vacuum state that is inconsistent with the definition at Vacuum state: a vacuum state is supposed to not contain any particles. Wouldn't it be more appropriate to speak of the energy minimum or the ground state? --18.104.22.168 21:47, 13 November 2006 (UTC)
A Lagrangian isn't even necessary for the process of spontaneous symmetry breaking. Neither is a vacuum. What's necessary is simply a solution to symmetric dynamical equations which isn't invariant under the same symmetries. Phys 19:04, 20 Feb 2004 (UTC)
- I have, as may be obvious, only an amateur interest in physics, so I'm sorry for my somewhat loose explanation. I put in the Lagrangian bit so that I could provide a more concrete example of spontaneous symmetry breaking, and also because this is how I learned about it. I have reworded the article ("one way of seeing" instead of "is accomplished by") in order to be more accurate. If you have further suggestions for restating it, I definitely welcome them. Thanks for the input. Jcobb 05:12, Feb 21, 2004 (UTC)
Is it supposed to be the lagrangian or the lagrangian density? The lagrangian is generally not lorentz invariant. JeffBobFrank 23:43, 19 Mar 2004 (UTC)
- Technically you are correct that it is more accurately described as the Lagrangian density. However, as can be seen in the Lagrangian article itself, many different entities end up being called just the plain Lagrangian (for another example the n-form L dx^0 wedge dx^1 wedge ... wedge dx^n is also commonly described as the Lagrangian even though it is something slightly different). In the context of this article, I don't think it is necessary to make the distinction between Lagrangian and Lagrangian density. Jcobb 07:53, Mar 21, 2004 (UTC)
What's wrong with my example on the Earth's gravitational field? It's not a scalar field, but it's still an example, one everyone is familiar with. Phys 00:20, 15 Aug 2004 (UTC)
- What's the example with Earth's field? Wilgamesh 21:14, 17 Sep 2004 (UTC)
Well, you know, you've been around WP long enough to catch on to the requirements of the articles. It may have had a point to make, but it certainly didn't make it clearly. Please don't add chat to articles when it really belongs to the talk page. Charles Matthews 08:33, 15 Aug 2004 (UTC)
Hi! I added another example I think is worthwhile because it's easily imaginable/testable, and yet still non-intuitive (like the ruler example). I may have made a mistake, but I think it's right, I remember seeing it in an undergraduate mechanics class with the pot cover, drill and marble, and it was fantastic! Jesse 12:58 2 Apr 2004 (Paris Time?)
the article says: In this state the Lagrangian has a U(1) symmetry. However, once it falls into a specific stable vacuum state (corresponding to a choice of θ) this symmetry will be lost or spontaneously broken.
but is this correct ? the lagrangian is U(1) symmetric independent of the state what is probably meant is the symmetry of the state itself and not of the lagrangian
so I would suggest: In this state the system has a U(1) symmetry. However, once it falls into a specific vacuum state (corresponding to a choice of θ) this symmetry will be lost or spontaneously broken.
Mexican hat potential
It sounds clear enough to me. Penrose's picture looks more like a hat though. David R. Ingham 01:14, 26 August 2006 (UTC)
Well, it was not clear enough to me. Sure, it was clear why it was called "Mexican hat" (it does _sort_ of look like a Mexican hat), but what is not clear is the axes. Please label the axes, and/or which angle is Phi, which Theta, what coordinates _are_ you using, anyway? —Preceding unsigned comment added by 22.214.171.124 (talk) 19:04, 13 February 2008 (UTC)
It is not mentioned that the lowest energy configuration must also minimize the magnetic field energy. I am not sure that that is needed in such a brief mention of this example.
Perhaps because he started as a mathematician, Roger Penrose got that wrong in The Road to Reality. Soft iron can be easily magnetized but spontaneously loses most of its magnetization when taken out of the magnetic field. For an isolated ferromagnetic material, the stable configurations have the magnetization of the domains pointed around in small closed loops, so that the field energy is nowhere great. Small iron particles are "super-paramagnetic" with the whole particle having the same direction, but they are not good examples of topologically stable grain boundaries either. He should have used a different type of crystal with only short range interactions. David R. Ingham 01:14, 26 August 2006 (UTC)
Many examples are not spontaneous
Many of the "other examples" are not really examples of _spontaneous_ symmetry breaking (other than in a very convoluted sense which does not help to explain the concept). The examples of an observer on earth are not about spontaneous breaking, here the question should be why there are planets at all, and not just evenly distributed matter. The heated fluid has its symmetry broken from outside, and so on. --126.96.36.199 (talk) 15:53, 25 August 2010 (UTC)
I am in agreement completely with the comment above. Specifically, the example of the thin fluid covering an infinitely large Euclidean plane is one that is easily, and accurately modeled by incorporating the curl of the resultant vector field (del x F) or in the case of this two dimensional example the component called vorticity. Because the heating is continuous (temperature is a continuous function whether you like it or not) the curl exists everywhere on the vector field AND it is not an emergent phenomenon that arises once enough heat has been added. It is simply a consequence of the small vorticity patterns becoming apparent to the eye. Additionally, since this example used an infinitely large Euclidean plane, the observed covection cells are only semi-stable and will evolve as heat is added. In any case, this is absolutely NOT an example of symmetry breaking. And so I have deleted it. Do not forget that each individual point has its own vorticity value that in this non-discontinuous system behaves nicely. Other energy addition materia may not behave so well. — Preceding unsigned comment added by 188.8.131.52 (talk) 21:56, 14 April 2013 (UTC)
- I am in agreement with Lugia2453's reversion of your peremptory deletions. In particular, I see nothing in the above arguing for your wholesale deletion of the best intuitive example of symmetry breaking in simple mechanical systems, to my taste (which the Weak interactions book of Commins and Bucksbaum bothers to highlight: "Take a thin cylindrical plastic rod and push both ends together. Before buckling, the system is symmetric under rotation, and so visibly cylindrically symmetric. But after buckling, it looks different, and asymmetric. Nevertheless, features of the cylindrical symmetry are still there: ignoring friction, it would take no force to freely spin the rod around, displacing the ground state in time, and amounting to an oscillation of vanishing frequency, unlike the radial oscillations in the direction of the buckle. This spinning mode is effectively the requisite Nambu–Goldstone boson." If you may improve this, or if you have a better non-technical example in simple engineering which exemplifies a Goldstone mode and which does not require QFT or even QM, go ahead. But as you notice from the above pleas for nontechnical examples, non-physicists often come here for insight.Cuzkatzimhut (talk) 22:22, 14 April 2013 (UTC)
Why does it matter?
Symmetry is a fairly intuitive concept. What has always puzzled me (as a non-physicist) is why symmetry breaking is considered so significant. Of course a ball rolling off the top of a symmetric mound is no longer in a position to roll in any direction. And of course a pencil, once balanced on its tip but now fallen over, is no longer in a position to fall in any direction. That's common sense. Why is that so important to physicists? And perhaps more to the point, what mathematical or theoretical leverage does symmetry breaking provide? RussAbbott (talk) 04:23, 12 June 2008 (UTC)
- Even though inexpert, your question is by far the best question on this page at this point in time. I tried to address it by some tweaks. What the article did not emphasize was that the "breaking" is only apparent, and the symmetry is still present, albeit "hidden" and acting in mathematically powerful ways, in the "Nambu-Goldstone mode", fleshed out in Goldstone boson. The article further had a couple of unsound confusions between the Higgs mechanism (requiring SSB), and the Higgs boson (an optional feature of implementation, at least theoretically): I removed those. It also had too many illustrations of the breaking of discrete symmetries, which do not require Goldstone bosons and cannot thus be utilized in the Higgs mechanism.
- In condensed matter physics, SSB underlies the characterization of phases, and massless excitations. In particle physics, it is the foundation of the Theory of Electroweak Interactions and further describes the lowest energy sector of the Strong interactions, namely chiral symmetry breaking, for which Nambu was awarded the Nobel prize discussed. Cuzkatzimhut (talk) 16:15, 4 February 2011 (UTC)
Transcribed from Comments page
I'm surprised that this article is rated A. I came to this page because I wanted to understand symmetery in physics and symmetry breaking. This article didn't help. The overview and the first section are not useful to someone who doesn't already understand the concept. The examples are useful, but they are incomplete. So I looked further and found a couple of other interesting articles: The Future of Language: Symmetry or Broken Symmetry? and The Symmetry-Breaking Paradigm by Jim Coplien. These are where I came to understand that symmetry is invariance under a transformation. RussAbbott (talk) 05:42, 12 June 2008 (UTC)
Isn't symmetry added also?
This may seem like a foolish question, so I apologize in advance.
When water vapor condenses to liquid or when (liquid) water freezes to ice, symmetry is broken. (Right?) There are fewer possibilities after the state transition than before. It's like the pencil falling down.
But symmetry is added also, isn't it? In the case of ice there is a new symmetry that characterizes how the atoms in the ice crystal relate to each other. They are always in a fixed relationship, no matter how the ice is moved around in space. So isn't that an additional symmetry that is added when the symmetry going from water to ice is lost?
Or is that better expressed as a new "conservation law?" The new conservation law states that the atoms in the ice crystal will retain the same mutual relationship no matter how the ice crystal itself moves. If symmetry is invariance under a transformation, isn't this "conservation law" an example of symmetry?
Does symmetry breaking involve release of energy?
Please forgive me if this question is ignorant because I am uninformed on this subject, but do symmetry breaking phase transitions always involve release of energy? Examples of symmetry breaking I have heard of that involve the release of energy are: freezing (crystallizing) of water, superconductivity, ferromagnetism, magnetic refrigeration, and (I think) the phase transition to the Guth cosmic inflation period in the early universe. Is this always a feature of symmetry breaking? If so, should it be added to the article? --ChetvornoTALK 22:10, 24 November 2008 (UTC)
- The symmetry-breaking concept is one of reducing (trading) potential energy for energy in another form, kinetic, for example. If there's no change in the potential, then no energy will be transacted. Phase change would essentially be a different energy configuration. Since some phase changes can occur continuously, I don't see why there'd be a necessary discontinuous phase change. That is to say, I can't see why you'd release any more energy than just what would be released by reducing the potential independent of a change in phase. 184.108.40.206 (talk) 05:30, 11 August 2010 (UTC)
The "ball sitting on top of a hill" example is good for symmetry. It is not, however, a good example for spontaneous symmetry breaking because it does not show spontaneity or independence from external causes.Lestrade (talk) 19:52, 30 November 2008 (UTC)Lestrade
Also, a dome is the upper half of a sphere-like surface, so all domes are upward unless stated otherwise. Therefore, saying "upward dome" obscures the discourse. My two cents. — Preceding unsigned comment added by 220.127.116.11 (talk) 10:47, 6 July 2012 (UTC)
I have suggested that the article Broken symmetry be merged into this one because it covers much the same stuff, focusing on translational and rotational symmetry breaking. Reyk YO! 00:22, 7 December 2008 (UTC)
- Agree. I'd like to see info on translational and rotational breaking in this article. 18.104.22.168 (talk) 08:28, 9 August 2010 (UTC)
Symmetry breaking vs. Wavefunction collapse
Any similarity? I found reading these 2 subjects separately confusing.
That is a profound and deep question. When we make a measurement the probabilistic state of possibilities of the system collapses to distinct values. In symmetry breaking possible system values are realized which no longer express the invariant state of the system. The former is not reversible, yet the latter, is.
The whole article is a hodge podge of tinkered together, ontologically different, concepts. Should be rewritten from scratch. — Preceding unsigned comment added by Ahhaha (talk • contribs) 14:58, 16 November 2013 (UTC)