0
$\begingroup$

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: $$ L\left(q_{i}, \dot{q}_{i}, t\right)=L_{0}(q, t)+\dot{q}_{i} a_{i}(q, t)+\dot{q}_{i}^{2} T_{i}(q, t)\tag{8.22}$$ 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$$L(q, \dot{q}, t)=L_{0}(q, t)+\tilde{\dot{\mathbf{q}}} \mathbf{a}+\frac{1}{2} \tilde {\dot{\mathbf{q}}} \mathbf{T} \dot{\mathbf{q}},\tag{8.23}$$

It's then said that $\mathbf{T}$ is symmetric. Why is it symmetric? Please give me a hint or an answer.

$\endgroup$
7
  • $\begingroup$ You didn't define $\mathbf{T}$, just some numbers $T_i$ (not components of a matrix $T_{ij}$). What's the definition of $\mathbf{T}$? $\endgroup$
    – ACuriousMind
    Commented Dec 22, 2021 at 13:39
  • $\begingroup$ I've copied the whole passage. That's all what the author said. $\endgroup$
    – Kashmiri
    Commented Dec 22, 2021 at 13:41
  • 3
    $\begingroup$ It is without loss of generality symmetric. $\endgroup$ Commented Dec 22, 2021 at 13:59
  • $\begingroup$ Even if you include the antisymmetric part, $\boldsymbol{T} = \boldsymbol{T}_S + \boldsymbol{T}_A$, it never contributes ($\bar{\boldsymbol{a}} \boldsymbol{T}_A \boldsymbol{a} = 0$ for any vector $\boldsymbol{a}$). $\endgroup$
    – tueda
    Commented Dec 22, 2021 at 14:18
  • $\begingroup$ Related: physics.stackexchange.com/q/607437/2451 $\endgroup$
    – Qmechanic
    Commented Dec 22, 2021 at 14:25

2 Answers 2

2
$\begingroup$

Let us be general and suppose that you have a real column vector $x$ and a real matrix $M$. Define $$Q_M(x)=x^T Mx\tag{1}.$$

Now as is well-known you can split $M = S+A$ where $S$ is symmetric, $S^T=S$ and $A$ is anti-symmetric $A^T=-A$. One just has to define $$S = \dfrac{M+M^T}{2},\quad A=\dfrac{M-M^T}{2}\tag{2}.$$

Now observe that $$Q_M(x)=x^T Sx+x^T Ax\tag{3}.$$

It is easy, though, to see that the second term is zero. Indeed $x^T A x$ is a number, and hence is its own transpose. Therefore $$x^T Ax=(x^T Ax)^T= x^T A^T (x^T)^T=x^T(-A)x=-x^T A x\tag{4}$$

Equation (4) means that $x^TAx=0$ as claimed. Therefore $Q_M(x)=x^T S x$.

In other words: $Q_M(x)$ depends just on the symmetric part of $M$. Since the anti-symmetric part doesn't matter, we can without loss of generality assume that $M$ is symmetric, meaning that we have discarded the non-contributing anti-symmetric part.

$\endgroup$
1
  • $\begingroup$ +1. If anyone's worried about the "transpose" of a number, note that while $x\cdot Mx$ is a number $x^TMx$ is a $1\times1$ matrix. See also here. $\endgroup$
    – J.G.
    Commented Dec 22, 2021 at 14:59
2
$\begingroup$

You're supposed to define $T$ as the diagonal matrix with entries $T_i(q,t)$ along the diagonal. So in this fixed basis for the $\dot{q}$ we have $\dot{q}Tq = \dot{q}_i^2 T_i$, and $T$ is obviously symmetric. In some other basis, you get a general $T$ as the transform of the diagonal matrix. In particular, $T$ is therefore in general a diagonalizable real matrix with real eigenvalues $T_i$ and orthonormal eigenspaces (spanned by the initial choice of $\dot{q}_i$). Now, a real matrix is symmetric if and only if its eigenspaces are orthogonal and span the full vector space (see e.g. this math.SE question and its linked questions), therefore $T$ is symmetric.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.