Graduate Why is this matrix symmetric here?

[Kashmir]

Goldstein 3rd Ed, pg 339

"In large classes of problems, it happens that $L_{2}$ is a quadratic function of the generalized velocities and $L_{1}$ is a linear function of the same variables with the following specific functional dependencies:
$L\left(q_{i}, \dot{q}_{i}, t\right)=L_{0}(q, t)+\dot{q}_{i} a_{i}(q, t)+\dot{q}_{i}^{2} T_{i}(q, t)$

where the $a_{i}^{\prime} s$and the $T_{i}$'s are functions of the $q$ 's and $t$.

Under the given assumptions the Lagrangian can be written as$L(q, \dot{q}, t)=L_{0}(q, t)+\tilde{\dot{q}} \mathbf{a}+\frac{1}{2} \tilde{\mathbf{q}} \mathbf{T} \dot{\mathbf{q}}$"

It's then said that $\mathbf{T}$ is symmetric.
Why is it symmetric? Is it because it's always diagonal? The author didn't say that,instead says it's just a n by n square matrix.

Please give me a hint or an answer. Thank you

 

[ergospherical]

With moving (i.e. time-dependent) constraints $\mathbf{r}_a = \mathbf{r}_a(\mathbf{q},t)$, the kinetic energy is indeed generally of the form $T = \dot{\mathbf{q}}^T \mathbf{A} \dot{\mathbf{q}} + \mathbf{B}^T \dot{\mathbf{q}} + C$, where $\mathbf{A}$, $\mathbf{B}$ and C are all functions of $\mathbf{q}$ and $t$.

This follows from the expression $\dot{\mathbf{r}}_a = \sum_i \dfrac{\partial \mathbf{r}_a}{\partial q^i} \dot{q}^i + \dfrac{\partial \mathbf{r}_a}{\partial t}$ for the particle velocities. Using this to write out the kinetic energy $T = \frac{1}{2} \sum_a m_a \dot{\mathbf{r}}_a \cdot \dot{\mathbf{r}}_a$ of the system, do you see why $A_{ij} = A_{ji}$ (i.e. $\mathbf{A}$ is a symmetric matrix and the quadratic form $\dot{\mathbf{q}}^T \mathbf{A} \dot{\mathbf{q}}$ is homogeneous?).

 

[Kashmir]

With moving (i.e. time-dependent) constraints $\mathbf{r}_a = \mathbf{r}_a(\mathbf{q},t)$, the kinetic energy is indeed generally of the form $T = \dot{\mathbf{q}}^T \mathbf{A} \dot{\mathbf{q}} + \mathbf{B}^T \dot{\mathbf{q}} + C$, where $\mathbf{A}$, $\mathbf{B}$ and C are all functions of $\mathbf{q}$ and $t$.

This follows from the expression $\dot{\mathbf{r}}_a = \sum_i \dfrac{\partial \mathbf{r}_a}{\partial q^i} \dot{q}^i + \dfrac{\partial \mathbf{r}_a}{\partial t}$ for the particle velocities. Using this to write out the kinetic energy $T = \frac{1}{2} \sum_a m_a \dot{\mathbf{r}}_a \cdot \dot{\mathbf{r}}_a$ of the system, do you see why $A_{ij} = A_{ji}$ (i.e. $\mathbf{A}$ is a symmetric matrix and the quadratic form $\dot{\mathbf{q}}^T \mathbf{A} \dot{\mathbf{q}}$ is homogeneous?).

Thankyou sir.

 

[robphy]

From an old thread
https://www.physicsforums.com/threads/quadratic-forms-and-kinetic-energy.826884/
I wrote

https://www.google.com/search?tbm=bks&hl=en&q=riemann+metric+-robot+"kinetic+energy"

The search is inspired by this statement:
"Kinetic energy is a Riemann Metric on Configuration Space"
- Geometrical Mechanics, Lectures by Saunders MacLane, Dept. of Math, U. Chicago, 1968.
http://www.worldcat.org/title/geome...-of-chicago-winter-quarter-1968/oclc/63616841