Kinetic energy in Lagragian mechanics

Question

In my classical mechanics class they asked why the kinetic energy for an holonomic mechanical system has the homogeneous quadratic form. Of course for a autonomous standard system( system that is holonomic and the constraint do no virtual work) you can just calculate that $$ T = \frac { 1 } { 2 } \sum _ { i = 1 } ^ { M } m _ { i } \left( v _ { i } \cdot v _ { i } \right) = \sum _ { j = 1 } ^ { n } \sum _ { k = 1 } ^ { n } a _ { j k } ( q ) \dot { q } _ { j } \dot { q } _ { k } $$ with $$ a _ { j k } ( \boldsymbol { q } ) = \frac { 1 } { 2 } \sum _ { i = 1 } ^ { N } m _ { i } \left( \frac { \partial \boldsymbol { r } _ { i } } { \partial q _ { j } } \cdot \frac { \partial r _ { i } } { \partial q _ { k } } \right) $$ So you can write $$ T = \dot { q } ^ { T } \cdot A \cdot \dot { q } $$ But what is the physical reason?

Answer 1

That the kinetic term $T(q,\dot{q},t)$ is a second-order polynomial in the generalized velocities $\dot{q}$ can be viewed as:

1. a consequence of

   • the holonomic constraint that the positions ${\bf r}_i(q,t)$ do not depend on the generalized velocities $\dot{q}$,

   • and that the non-relativistic kinetic energy is $T=\sum_{i=1}^N\frac{m_i}{2}\dot{\bf r}_i^2$,

   as OP already suggested.

2. merely an ansatz or approximation, partially justified by the fact that the kinetic term should be bounded from below. (This ansatz is famously violated for relativistic point particles.)

References:

1. H. Goldstein, Classical Mechanics, Section 1.6.