# Spontaneus symetry breaking

1. Jul 18, 2006

### skywolf

whats the deal with spontaneus symetry breaking?
i mean

the analogy they use is where a pencil is standing on its tip, and therefore symetrical, and then it falls in a random direction.

now, if it falls in a random direction, wouldnt it mean it wasnt symetrical to begin with?

i mean, as far as i am aware an object at rest tends to stay at rest.

so confused,
sw

2. Jul 18, 2006

### Danger

I would expect that it depends upon what is meant by 'spontaneous'. Something has to act upon it to upset the balance. In the case of a pencil on it's point, I figure that just the random thermal motion of its own molecules would be enough to tip it, even without air currents or surface vibrations. I'm not sure about that, though.

3. Jul 18, 2006

### quasar987

The internal forces of a system of particle (e.g. a pen) cannot change the position of its center of mass. So the moment a pen in unstable equilibrium falls is not a function of its temperature. :)

4. Jul 18, 2006

### Born2Perform

if for casuality, 49,999999% of particles would be in one side of the pencil, and 50,000001% in the other side, i don't see reasons why it should not fall without help of external forces.

Last edited: Jul 18, 2006
5. Jul 18, 2006

### quasar987

The condition of equilibrium is not a matter of how many particles are on each side, but of the position of the center of mass (CM). The particles can take any position they want, as long as the line joining the CM and the tip of the pencil remain perpendicular to the surface.

To put the pencil in equilibrium, it was necessary that the CM be so located. Then we let go of the pencil and if all external forces and external torques are nul, the CM will remain there. For the proof of that theorem, grab any classical mechanics textbook.

In lights of this theorem, it is evident that the situation you suggest willl never happen, because the line joining the CM of this system with the tip of the pencil would not be perpendicular to the surface. Hence, the CM would have move since the moment the pencil was in equilibrium. This is impossible according to the theorem.

6. Jul 18, 2006

### Office_Shredder

Staff Emeritus
Of course, the thermal energy of the air may knock the pencil over.... but not the thermal energy of the pencil itself

7. Jul 19, 2006

### ZapperZ

Staff Emeritus
I think there's a misunderstanding here on the meaning of "spontaneous symmetry breaking (SSB)", and in what systems are these things manifested. I'm probably one of the last people to cite a "philosophy" encyclopedia to describe what SSB is, but this is actually a good description of it.

http://plato.stanford.edu/entries/symmetry-breaking/#4.2

And note that this is another example of something that came out of condensed matter physics and is now a fundamental part of all of physics.

Zz.

8. Jul 20, 2006

### gato_

the matter of the pencil is an example, but the central point in symmetry breaking is that the state which has all the symmetry of the system in consideration is unstable, and therefore hardly observable, and will relax to something more stable which has but part of the original symmetry. In the pencil example, the standing over the type state is always unstable, but let us take ,for example, a recipient of water. if you ignore the boundaries, the surface is homogeneous, if you take any point in the surface, everything looks the same in any direction, provided you are far from the borders. it is rotational and translational invariant. Now, if you heat that below, there's a point where that uniform state becomes unstable, and convection starts to make patterns in the surface of the fluid, thus BREAKING the original symmetry. In a well controlled experiment, you could see that first patterns to form are hexagonal, wich have as symmetry a subset of the original one

9. Jul 20, 2006