Timeline for Relationship between zero modes and symmetry in a simple system of coupled springs
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jun 13, 2016 at 15:51 | comment | added | Izzhov | This doesn't work for trapezoids, though. you can't switch the vertices through its zero mode. | |
| Jun 9, 2016 at 16:24 | comment | added | Evan Rule | See section 10.6 here www-physics.ucsd.edu/students/courses/fall2010/physics110a/… | |
| Jun 9, 2016 at 16:23 | comment | added | Evan Rule | The symmetry must be continuous and must leave the Lagrangian invariant. The existence of zero modes resulting from symmetry is a consequence of Noether's Theorem, which holds only for continuous symmetries. Even though the triangle has a discrete reflection symmetry, this cannot be promoted to a continuous symmetry in a way that preserves the Lagrangian. The rectangle has this mode and has continuous reflection symmetry; that is, there exists a continuous symmetry of the rectangle which leaves the Lagrangian invariant and reverses the orientation of the rectangle. | |
| Jun 8, 2016 at 21:14 | comment | added | Izzhov | Also, consider that if the system were a rectangle (i.e. you increased the equilibrium length of 2 of the springs), it would still have that zero mode, while losing its reflection symmetry (unless I'm misunderstanding how you're defining reflection symmetry). | |
| Jun 8, 2016 at 21:06 | comment | added | Izzhov | Are you claiming, then, that Wiki's assertion that "zero modes appear whenever a physical system possesses a certain symmetry" is incorrect? | |
| Jun 8, 2016 at 21:03 | comment | added | Evan Rule | In order to continously reflect the triangle, you would have to stretch at least one of its springs and it is no longer a zero mode. The square can do this inversion without any stretching. | |
| Jun 8, 2016 at 20:58 | comment | added | Izzhov | If this is true, then how come an equilateral triangle spring system doesn't have a zero mode corresponding to its own reflective symmetry? | |
| Jun 8, 2016 at 20:48 | history | answered | Evan Rule | CC BY-SA 3.0 |