# Relationship between zero modes and symmetry in a simple system of coupled springs

This Wikipedia page states that "zero modes appear whenever a physical system possesses a certain symmetry," and gives the example of a ring of beads connected by springs having a zero mode associated with rigidly rotating the whole system. It's easy to think of other examples too: most systems of beads connected by springs have zero modes that arise from translating the whole system (i.e. translational symmetry), for example.

Consider a system of four beads, connected by springs in a square at equilibrium. This system has 8 degrees of freedom, so it has 8 modes. The four nonzero modes come from either both the vertical or both the horizontal springs oscillating either in sync or 180 degrees out of sync. Two of the zero modes come from translation of the whole system, and the corresponding symmetry is obvious. One of the zero modes comes from rotating the whole system, and again, the corresponding symmetry is obvious.

However, the final zero mode comes from bringing two of the opposite corners together while pushing the other opposite two corners apart (i.e. squishing the square into a rhombus). So my question is this: what symmetry of the system does this deformation correspond to?

• What do you mean by "zero mode"? Do you mean a mode with zero natural frequency? ie rigid-body motion? Jun 8, 2016 at 21:44
• I mean a normal mode of the system with vanishing eigenvalue. Jun 8, 2016 at 22:08
• So is that the same as a normal mode with zero frequency, or not? Jun 8, 2016 at 22:18
• The eigenvalues of the Hessian == the normal frequencies, so yes, it's the same. Jun 8, 2016 at 22:51
• Although this question is off-topic, take a look in an other related off-topic one here : How to calculate number of modes and oscillation modes of a group of 7 identical masses on a ring connected via springs?. Sep 9, 2018 at 5:47

Evan Rule's answer confuses me because I only think about reflection as a discrete symmetry. We need a continuous symmetry to invoke the Noether/Nambu-Goldstone argument. I will try to do it a little bit better. First of all, the extra symmetry is exactly the transformation which creates the mode. Second, the symmetry should act on every configuration, producing a new configuration which has the same energy.

I will describe the infinitesimal transformation, whose parameter is an infinitesimal angle $\delta \theta$. Let's order the vertices of the quadrilateral 0,1,2,3 (which are not assumed to be in equilibrium position). We consider the 0,1,2 angle $\theta$. There is (almost, see below) always a unique transformation which takes $\theta \mapsto \theta +\delta \theta$ and preserves all lengths, hence preserves the energy but gives us a new configuration not related to the old one by rigid motions because the 0,1,2 angle is now different.

The existence and uniqueness of this transformation has to do with an interesting geometric fact that a generic $n$-gon with fixed edge lengths but freely-pivoting angles has exactly $n-3$ degrees of freedom, plus rigid motions. Only triangles are structurally rigid, and this is why we use them to build stuff.

By the way, the general study of these "tinker toys" is quite interesting, and W. Thurston famously gave a lecture where he proved that every differentiable manifold appears as a component of the moduli space of such things. Check it out.

Actually the moduli space of the rhombus is already quite surprising. See from page 5 here. One might think that as we continue to change the 0,1,2 angle we get a whole circle of configuration space. This is true, but there is certain special points where new degrees of freedom appear. For instance, after flattening a rhombus so that the 0,1,2 angle is zero, there is a new length-preserving motion which changes the 1,2,3 angle while keeping the 0,1,2 angle zero. The total configuration space is a necklace of three kissing circles times group of rigid motions. One circle is the positive area branch of the equilibrium moduli space and two circles are zero are branches. On these two strange branches, a rotation in the 1,2,3 or 2,3,0 angle generates the extra zero mode. The total number of zero modes is always the same except at the three singular points in the equilibrium moduli space, but here one cannot take linear combinations of the new zero modes. So in a way, the total dimension of zero modes remains the same, even though their number does not. It would be interesting to know what the physical consequence of this is!

The final mode corresponds to the reflection symmetry of the system. Using the continuous deformation that you describe you can pinch two of the corners in entirely and invert them, which amounts to a reflection and is inequivalent to any continuous rotation of the system.

• 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 20:58
• 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 21:03
• 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 21:06
• 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 9, 2016 at 16:23
• This doesn't work for trapezoids, though. you can't switch the vertices through its zero mode. Jun 13, 2016 at 15:51