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Why do we always observe spontaneous symmetry breaking in nature and not restoration? Does there exist some argument with the 2nd law of thermodynamics and the entropy of the universe increasing? If yes, it will be great if someone can refer me a mathematical proof of the above.

Is it the same reason as a hot body always spontaneously cools down, because we expect greater symmetries in higher temperature.

Consider a particle physics symmetry breaking, like the simple $\mathrm{SU}(2)_\mathrm L \times \mathrm U(1)_Y \to \mathrm U(1)_\mathrm{em}$. In this context by what argument I can say that the entropy of the universe increases?

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who says that we don't? If you have a critical temperature f.i. then you can restore the symmetry by heating the system above Tc – Noldig Apr 20 at 9:21
So long as breaking the symmetry increases the entropy, then the 2nd law means that we cannot get spontaneous restoration. – Ihle Apr 20 at 9:36
Of course it is spontaneous. The spontaneity of some thermodynamical transformation is defined wrt some parameter as the temperature and pressure of the system. Above $Tc$ you don't need to use a catalyst to induce the transition, the system spontaneously transforms to its symmetrical phase. The fact that you have to apply energy only means that you don't have the right temperature for the transition to happen spontaneously. If the temperature is high enough for it to happen (e.g. in the first stages of the evolution of the universe) it will indeed happen on its own. – Giorgio Comitini Apr 20 at 9:48
The answer to your question is: because the temperature/pressure is too low. It is not nature itself, it's its current thermodynamical state. It doesn't happen for the same reason that you won't find ice in the desert. – Giorgio Comitini Apr 20 at 9:55
@GiorgioComitini: That seems to be an answer, not a comment! – ACuriousMind Apr 20 at 10:34

We do observe spontaneous symmetry restorations in nature. This is called an emergent symmetry. See e.g. this post. A system posses an emergent symmetry if it appears symmetric at large (coarse-grained) scales although the apparent symmetry is explicitly broken by the microscopic description (typically the Hamiltonian or Lagrangian).

I can give two examples from the top of my head:

  • There are many discrete deposition models that belong to the KPZ universality class. These are defined in terms of particles falling on a discrete lattice (the 1d case looks like a game of tetris). The large scale features of many of these models are identical to the what you would get by simulating the continuous KPZ differential equation, $$ \partial_t h = \frac{\lambda}{2}\left(\nabla h\right)^2+ \nabla^2h + \eta \, ,$$ with a stochastic noise, $\eta$. In particular although they are defined on a discrete space lattice (no translation invariance), KPZ equation is invariant under space translations. Note that standard KPZ equation (with white noise) posses many exact symmetries. In particular the statistical tilt (or Galilee) symmetry as well as the time reversal (for 1d systems) symmetry can be broken by noise correlations (the delta functions correspond to white noise) $$\langle \eta(t,x) \eta(t',x')\rangle = F(t-t',x-x') \neq \delta(t-t') \delta(x-x') \, .$$ It turns out that these are restored at large scale. The KPZ fixed point is reached under iteration of the Renormalisation Group (RG) flow for a wide class of functions $F(t-t',x-x')$.

  • Another example can be found in the field of driven-dissipative Bose gases such as exciton-polariton condensates. See these two papers. In this case, the fluctuation-dissipation relation is expressed as a consequence of a 'Thermal equilibrium symmetry'. See this. Once more, even though the microscopic action may not exhibit a fluctuation-dissipation relation, a temperature emerges at large scales because the thermal equilibrium RG fixed point is attractive enough. The non-equilibrium components die out as the RG flow is iterated.

I give here two examples from non-equilibrium statistical physics and I am aware that thermodynamic arguments will not apply to them. Sorry, that is my field of research and I do not dare to write about things that I do not know well. I am however confident that you can find many more examples online.

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