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Consider a system of three masses (all $m$) connected by four springs (all with spring constant $k$), and the ends are fixed to walls, like

Wall --- (k) --- [m] --- (k) --- [m] --- (k) --- [m] --- (k) --- Wall

The lagrangian is $$L=\frac{1}{2}m(v_1^2+v_2^2+v_3^2)-\frac{k}{2}[(x_3-x_2)^2+(x_2-x_1)^2+x_3^2+x_1^2]$$

In the standard steps, we have $$L=\frac{1}{2}m v^TIv-\frac{k}{2}x^TKx$$

where $v=(v_1,v_2,v_3)^T$ and $K=\left \{ \begin{matrix} 2& -1&0 \\ -1& 2&-1 \\ 0&-1&2 \end{matrix} \right\} $

However, to recover our quadratic lagrangian, we can also have other kind of matrix $K$,e.g. $K=\left \{ \begin{matrix} 2& -2&0 \\ 0&2&-2 \\ 0&0&2 \end{matrix} \right\} $

But if we use this matrix, we can't find right normal modes.

So my question is: Is there physical reasons tell us matrix $K$ should be symmetric? (Why physical reasons? Because we have explicit explaination of normal modes, like phonon, so I think we may not treat normal modes as some mathematical trick only)

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The equations of motion are what matter here, since the standard method for solving the system involves plugging an ansatz of the form $\mathbf{x}(t) = e^{i \omega t} \mathbf{x}$ into the equations of motion. If our Lagrangian is of the form you propose, $$ L = \frac{m}2 \sum_{i} \dot{x}_i^2 - \frac{k}2 \sum_{ij} x_i K_{ij} x_j $$ then it is not hard to show that the resulting Euler-Lagrange equations are $$ m \ddot{x}_i = - k \sum_j \frac12 (K_{ij} + K_{ji}) x_j $$ or, in linear algebra terms, $$ m \ddot{\mathbf{x}} = - \frac{k}{2} (K + K^T) \mathbf{x}. $$ In other words, even if you put a non-symmetric matrix $K_{ij}$ into the potential term of the Lagrangian, it's the "symmetrized" version $\tilde{K} = \frac12(K + K^T)$ of that matrix that actually matters for the equations of motion and the existence of normal modes. We usually just skip a step and assume that $K$ is symmetric to begin with; but if some reason you were to start with a non-symmetric $K$, you would have to symmetrize it before applying the eigenvalue procedure.

Similar considerations apply to cases where the kinetic energy is of the form $$ T = \frac{m}{2} \sum_{i,j} \dot{x}_i M_{ij} \dot{x}_j, $$ and can be used to show that only the "symmetrized" portion of $M$ affects the equations of motion.

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  • $\begingroup$ Is Onsager's reciprocity related to this natural symmetry of the equations of motion? $\endgroup$ Commented Jul 3 at 19:19
  • $\begingroup$ @hyportnex: I'm not familiar enough with Onsager's reciprocal relations to say one way or the other. That said, I don't think they'd give out a Nobel Prize for something this simple. :-) $\endgroup$ Commented Jul 3 at 20:09
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You don't need symmetric matrices, for evaluating modes. Symmetric matrices result from Lagrangian mechanics of mechanical systems, after linearization.

Once you have symmetric mass and stiffness matrixes you can exploit this properties, namely orthogonality w.r.t. massa and stiffness matrices, in the evaluation of modes.

Here some notes I wrote about

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the K matrix mustn't be symmetric.

Example

$$\mathbf K=\left[ \begin {array}{ccc} 1&-2&0\\ 0&2&-3 \\ 0&0&3\end {array} \right] $$ the equation of motion are

$$\mathbf M\,\mathbf{\ddot{q}}+\mathbf K\,\mathbf q=\mathbf 0$$

Notice

If the matrix $\mathbf K$ , is not symmetric $\mathbf K\ne \mathbf K^T$ this means that the force is not conservative , in this case you don't have potential $U=U(\mathbf x)$.

thus only for conservative forces the matrix $\mathbf K$ is symmetric

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The symmetric version is not necessary (to solve for the motion of the system), but it is always available and is easier to work with. The reason the symmetric version is always available is, I suspect, Newton's third law, but I have not checked that point.

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    $\begingroup$ You can write down a symmetric potential term in a Lagrangian that does not obey Newton's Third Law; for example, $U(x_1, x_2) = x_1 x_2$. It's only if the interaction term is a function of $x_1 - x_2$ only that you get conservation of momentum, since then (per Noether's theorem) you have translational invariance. $\endgroup$ Commented Jul 3 at 18:44

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