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When considering a spring-like object, it seems almost intuitive that it would be "springy", i.e., harmonically oscillate. However, these oscillations occur at a macroscopic scale, unlike the normal phonons found in every solid (the spring still "dings" when hit, but only microscopic degrees of freedom oscilate).

While trying to explain this behaviour, I've come to think about the helical-periodical shape of the spring as a macro-scale lattice, breaking the continuous symmetries of one dimensional translation and rotation (via the "phase" of the helix).

This continuous symmetry breaking then implies the existence of two Goldstone modes, that match the longtitudal and transversal modes of the spring.

Is this a viable and physically accurate way to describe the system?

If so, can it offer some added values? For instance, can the spring coefficient K or any other interesting phenomenon be derived from it?

Edit (Clarification after Kian Maleki's feedback): The continuous translational symmetry is replaced by a discrete translational symmetry, as the system is now invariant to a translation by one coil, or to rotation by one coil (360°). Of course, a continuous symmetry still exists (translate+rotate), but the "effective" breaking of the symmetries may be treated like the translation symmetry breaking in a lattice (as a spring is a macro-scale 1d lattice of sort). Micro-scale lattices (solids) exhibit phonons as goldstone modes to translational symmetry, and I thought the same might be true for spring oscillations.

A. Ok
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    Description in terms of Goldstone modes would be appropriate only for scales much larger then the separation between coils – nwolijin Dec 23 '20 at 23:32
  • Will that description hold for a long enough spring with many coils (say, a slinky)? – A. Ok Dec 23 '20 at 23:41
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    Recall Goldstone modes require zero frequency of oscillation, so infinite period of macroscopic oscillations. Without touching upon your "model", there is a well-known model of macroscopic SSB. Namely, pushing down on a vertical flexible rod, so it buckles, SBreaking axial symmetry; the radial oscillation mode thereof is the "massive" mode, and the free rotation is the Goldstone mode with infinite period. – Cosmas Zachos Dec 24 '20 at 17:14
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    The "helical-periodical shape of the spring" is in reality a continuously-infinite number of degrees of freedom that are treated as (a 'rigid body' with) one degree of freedom, if what you're trying to do made sense it would also apply to a pendulum with small oscillations, nothing to do with a "helical-periodical shape". SSB occurs for a classical oscillator when you go to $x^4$ level e.g. here or in a falling pencil. – bolbteppa Dec 24 '20 at 17:37
  • I completely understand and agree with the examples of macro-SSB given here, however I would love an explanation as to what is the difference to the case of phonons in solid as discussed here - as far as I can think, in the same way solid lattices break translational symmetry, thus creating goldstone bosons in the form of phonons, a spring with periodic coiling breaks the symmetry in the same manner. Consider an infinite spring for simplicity, why is it not a macro-1d-lattice? And why doesn't it create macro-phonons, that may be considered goldstone? – A. Ok Dec 24 '20 at 18:48

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Nope , it does not resemble the Goldstone modes. Because of you pick a phase on the helix you will still have the transnational symmetry (at least unto the separation between round) OR i did not understand your example correctly.

I think Goldstone modes are a bit different. The potential should be symmetric to begin with, only then by choosing a phase you break the symmetry.

This is how I would relate the spring the Goldstone modes. slightly compress a spring between your index finger and thumb. The system now has azimuthal symmetry. Then, poke the middle of the spring without moving your fingers. The sprig will form a curve between your fingers. This resembles a broken symmetry. If the spring is very thin and long, it would be hard to compress it with two fingers without breaking the azimuthal symmetry. So when the spring is compressed and it is straight, it resembles an unstable point and slight force can let it curve and break the symmetry.

Kian Maleki
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  • I believe the continuous translational symmetry is replaced by a discrete translational symmetry, as the system is now invariant to a translation by one coil, or to rotation by one coil (360°). Of course, a continuous symmetry still exists (translate+rotate), but the "effective" breaking of the symmetries may be treated like the translation symmetry breaking in a lattice (as a spring is a macro-scale 1d lattice of sort). Micro-scale lattices (solids) exhibit phonons as goldstone modes to translational symmetry, and I thought the same might be true for spring oscillations. – A. Ok Dec 24 '20 at 06:01