Let us say we have a one dimensional system of masses connected by springs. All the masses are equal and the springs alternate with spring constants of $k$ and $6k$. The appropriate way to analyze this system is to take the coordinates of two adjacent masses as $u(n,t)$ and $v(n,t)$($n$ is the index for the $n$th particle)and then writing down Newton's laws for each, we obtain the two modes of oscillation from a quadratic in $\omega$.

If we somehow make the masses move in such a way that two adjacent masses have a symmetric coordinate $u(n,t)$ and $u(n+1,t)$ and solve a single force equation, we can see that the resulting angular frequency $\omega$ is complex. Physically seeing this, the exponent $e^{iwt}$ would now turn into a phase factor and an exponentially damping term.

My query is whether our imposition of symmetry on an inherently asymmetrical dynamic system has led to the ideal mass spring system to suddenly start losing energy exponentially?

EDIT: One argument which might be put forward to account for the energy lost would go like "since the hand or whatever that moves the two masses in a symmetric fashion, the hands do work to keep them moving in the same way and considering the hand and the mass-spring system into a larger system, the net system loses energy". I am OK with this, but my question is, do we need to always expend energy to make an asymmetric system completely symmetric?

EDIT 2: I agree with the fact that the system is not completely asymmetric. But any two adjacent masses will not move synchronously, i.e., they wont have the same amplitude structure. I want to make them move with the same structure, hence, I end up expending energy. Does this always happen with an asymmetric system? For instance, an asymmetric see-saw can never stand on its fulcrum in equilibrium. But, we could pull one end up and keep it symmetric and in equilibrium, but at the expense of our energy.

  • $\begingroup$ I think that " a feature of losing energy to the surroundings" causes the system to lose energy exponentially. I do not see where this system is not symmetric, it has a discrete translation symmetry - which should hold even after the addition of damping. $\endgroup$
    – Neuneck
    Feb 26, 2015 at 7:17
  • $\begingroup$ What do you mean exactly by symmetric coordinates (do you mean symmetry by a translation of two springs)? $\endgroup$
    – kristjan
    Feb 27, 2015 at 10:45

1 Answer 1


Energy conservation of the system in your example restricts solutions which will make energy dependent on time. Conservation laws are conclusions of symmetries of equations of motion, and for system of springs it seems that energy conservation is a valid law. It is not possible to write down trajectory that would have any sort of damping and solve equations of motion (unless this damping doesn't affect energy, but in our case it isn't; it is easy to check). So I doubt that you can write down solution with dissipative behavior.

It is worth noticing that we can only talk about symmetries for a whole system, which is defined by describing a forces on the right hand side of Newton's equation. That means that system with some external forces (e.g. your hand keeping unbalanced see-saw standing on its fulcrum in equilibrium) is different physical system compared to isolated see-saw. That implies different equations, space of solutions can contain solutions with symmetries broken in ``original'' system, usual conservation laws are invalid now (the force you apply means that you translate your body muscle energy to see-saw, so only total energy conserves).

  • $\begingroup$ I think I agree with you. But I am confused. Do you say that because the system I chose was ideal, any two adjacent springs can never move like $u(s)$ and $u(s+1)$ but must always move like $u(s)$ and $v(s)$? Here u,v are coordinates and s represents the s'th particle's index. $\endgroup$ Feb 28, 2015 at 0:19
  • $\begingroup$ @TorstenHĕrculĕCärlemän such configuration, if once present, will not persist. It doesn't restrict such configuration in some point of time, but it will be momentary (like horizontal position for see-saw with unequal masses on it) $\endgroup$
    – Vladimir
    Feb 28, 2015 at 22:05
  • $\begingroup$ But won't the system lose energy between the time it is in the symmetric state and the point from which such a motion "would not persist"? $\endgroup$ Mar 2, 2015 at 6:49

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