Symmetry and Finite Coupled Oscillators

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SUMMARY

The discussion centers on the behavior of finite coupled oscillators with identical mass and spring constant k, specifically examining the matrix equation of motion represented as \(\ddot{X}=M^{-1}KX\). It is confirmed that for a large but finite number of coupled oscillators, the matrix \(M^{-1}K\) commutes with the translation operator due to the identical nature of the oscillators. Additionally, the eigenvectors can be derived without direct diagonalization, leveraging the properties of the translation operator.

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  • Understanding of matrix equations in physics
  • Familiarity with eigenvalues and eigenvectors
  • Knowledge of coupled oscillators dynamics
  • Concept of translation operators in linear algebra
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Ibraheem
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For an infinite system of coupled oscillators of identical mass and spring constant k. The matrix equation of motion is \ddot{X}=M^{-1}KX

The eigenvectors of the solutions are those of the translation operator (since the translation operator and M^{-1}K commute). My question is, for the case of a large BUT finite number of coupled oscillators, does M^{-1}K still commute with the translation operator? and if not, is there a way to find the eigenvectors of the solutions, besides directly finding them by diagonalizing?
 
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To your first question, "Yes, of course" since you specify "identical" with respect to masses and spring constants, thus there's no positional dependency. Does that also resolve your second question?
 

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