In my lecturer's notes on Lagrangian mechanics in the chapter on normal modes they state the following:

Instead of rigid constraints, let us now consider a situation where the constraints are flexible so that the particles can move around their equilibrium positions. We assume that the system is described by $N$ generalised coordinates $q_i$. We also assume that it is natural, which means that the kinetic energy is a quadratic homogeneous function of the generalised velocities. We can write this as $$T = \frac{1}{2}\sum_{ij}a_{}(q_1, \ldots,q_N)\dot{q_i}\dot{q_j},$$ where the coefficients a_{ij} can depend on the coordinates $q_i$ but not on velocities $\dot{q_i}$. They can be chosen to be symmetric $(a_{ji}=a_{ij})$ without any loss of generality.

I am curious about two things here:

1) How can we just assume this form of the kinetic energy, what is the basis of this assumption? If I were to look at this subject myself with no notes, I would hardly just come up with this form.

2) $a_{ij}$ can be "chosen" to be symmetric, is this not necessary true from the commutative property of multiplication?


Non-canonical Lagrangians (Lagrangians with non-quadratic kinetic term) arise in some applications such as plasma physics. Quadratics usually imply small perturbations from equilibrium; i.e., something like Hooke's law. Here is an example of non-symmetric kinetic energy: https://aip.scitation.org/doi/abs/10.1063/1.4773440.

If A and B are symmetric then AB is symmetric iff A and B commute.

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