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There are several points over which I stumble when studying Goldstein, 3rd ed., chap 8.1, concerning the matrix representation of the hamiltonian formalism. In (8.22) he assumes the lagrangian to be (with Einstein's convention)

$ L(q_i,\dot{q}_i,t)=L_0(q,t)+ \dot{q}_i a_i (q,t)+ \dot{q}^2_i T_i(q,t). \qquad \qquad $ (1)

From this he derives the corresponding matrix representation

$ L(q_i,\dot{q}_i,t)=L_0(q,t)+ \mathbf{\tilde{\dot{q}}} \mathbf{a} + \frac{1}{2} \mathbf{\tilde{\dot{q}}} \mathbf{T} \mathbf{\dot{q}} \qquad \qquad $ (2)

with the row vectors being denoted by a tilde and $ \mathbf{T} $ being an $ n \times n $ matrix. I understand the term $ \dot{q}^2_i T_i(q,t) $ to be a quadratic form representing the kinetic energy of the system. In matrix representation this can be written as $ \mathbf{\tilde{\dot{q}}} \mathbf{T} \mathbf{\dot{q}} $.

Q1: But where does the factor $ \frac{1}{2} $ in (2) come from?

Q2: Goldstein says that the symmetrical matrix $ \mathbf{T} $ is most often a diagonal matrix, but not always. But according to the term $ \dot{q}^2_i T_i(q,t) $, it must be a diagonal matrix, doesn't it? So what does Goldstein mean here? In which cases is it not diagonal?

From (2) Goldstein derives the hamiltonian to be

$ H = \mathbf{\tilde{\dot{q}}} (\mathbf{p}-\mathbf{a}) - \frac{1}{2} \mathbf{\tilde{\dot{q}}} \mathbf{T} \mathbf{\dot{q}} -L_0. \qquad \qquad $ (3)

Q3: But it remains totally unclear to me how from this he gets

$ \mathbf{p} = \mathbf{T} \mathbf{\dot{q}} + \mathbf{a} $.

Goldstein's wording here is obscure to me, too.

Finally, Goldstein says that $ \mathbf{T} $ is invertible because of the "positive definite property of the kinetic energy". I assume he means that with respect to the quadratic form $ \dot{q}^2_i T_i(q,t) $, and I assume that this means, the corresponding matrix $ \mathbf{T} $ is positive definite, too (although I cannot find this equivalence in wikipedia).

Q4: Do we see the positive definite property directly from this term $ \dot{q}^2_i T_i(q,t) $, and if yes, how? Or are we making additional assumptions here about the kinetic energy? I also cannot find in wikipedia the relationship between a positive definite matrix and its invertability.

Can anyone help me back on the track?

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  • $\begingroup$ $p=\dfrac{\partial L}{\partial \dot q}=a +T\, \dot q$ $\endgroup$
    – Eli
    Commented Jan 13, 2021 at 18:20
  • $\begingroup$ It seems OP's questions mainly concern various typos in the 3rd edition of Goldstein. $\endgroup$
    – Qmechanic
    Commented Jan 13, 2021 at 18:36
  • $\begingroup$ @Eli, this is what Goldstein says. But I don't see how the second equal sign follows from the vector equations (2) or (3). $\endgroup$ Commented Jan 13, 2021 at 19:03
  • $\begingroup$ @Qmechanic, I've checked the errata for the 3rd ed. on Safko's site. My questions are not related to any of them. $\endgroup$ Commented Jan 13, 2021 at 19:31
  • $\begingroup$ T matrix is the metric, and the metric is positive definite snd always invariable $\endgroup$
    – Eli
    Commented Jan 13, 2021 at 19:41

2 Answers 2

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  1. Qmechanic is correct, eq. (1) has a typo and should have a factor of $1/2$ in front of the $T$. If you don't find this satisfactory, then you might also argue that the $T$ in (1) has not been properly defined yet, so that multiplicative factors are irrelevent. In eq. (2), we then choose $\mathbf{T}$ to have a coefficient of $1/2$. This is handy for calculations such as $$ \frac{\partial}{\partial\dot{\mathbf{q}}}\left(\dot{\mathbf{q}}^T\mathbf{T}\dot{\mathbf{q}}\right) = 2\mathbf{T}\dot{\mathbf{q}} $$ as the factors of $2$ will cancel. I promise there is nothing physical about the erroneous factor!

  2. $\mathbf{T}$ is positive definite and symmetric and so it is always possible to find a basis $\mathbf{q}$ in which it is diagonal. In the diagonal basis we can write $\mathbf{T} = \text{diag}(T_{11}, \ldots, T_{nn})$ and then the quadratic form is $$ \dot{\mathbf{q}}^T\mathbf{T}\dot{\mathbf{q}} = \dot{q}_i T_{ij} \dot{q}_j = \sum_i \dot{q}_i T_{ii} \dot{q}_i \sim \dot{q}_i^2 T_{i}(q, t) $$ where the last part is simply a shorthand for a diagonal quadratic form.

In summary, the answer to your questions 1 and 2 is that Goldstein's equation (8.22) is really a shorthand for the more technically correct $$ L(q_i, \dot{q}_i, t) = L_0(q, t) + \dot{q}_i a_i(q, t) + \frac{1}{2}\dot{q}_i T_{ij}(q, t) \dot{q}_j. $$

  1. This follows directly from the definition $$ \mathbf{p} = \frac{\partial L}{\partial \dot{\mathbf{q}}} $$ and equation (1). In case you aren't too comfortable with index manipulations, the calculation is \begin{align} p_k &= \frac{\partial L}{\partial \dot{q}_k} = \frac{\partial}{\partial \dot{q}_k} \left(L_0(q, t) + \dot{q}_i a_i(q, t) + \frac{1}{2}\dot{q}_i T_{ij}(q, t) \dot{q}_j \right) \\ &= \delta_{ik}a_i(q, t) + \frac{1}{2}\left(\delta_{ik}T_{ij}q_{j} + q_i T_{ij}\delta{jk}\right) \\ &= a_k(q, t) + T_{kj}q_j. \end{align}

  2. Physically, the kinetic energy must be positive or bad things will happen e.g. infinitely accelerated motion. This means that the quadratic form $\dot{q}_i T_{ij}(q, t) \dot{q}_j$ must be positive definite, which is equivalent to all the eigenvalues $\lambda_i$ of $T$ being positive. Since $T$ is symmetric, it has a basis of eigenvectors. Thus $\det \mathbf{T} = \prod_i \lambda_i >0$ which is non-zero so $\mathbf{T}$ is invertible.

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    $\begingroup$ Thank you very much! You gave exactly the explanations which I needed. I will look into all that in detail tomorrow morning, it is late here. $\endgroup$ Commented Jan 13, 2021 at 23:00
  • $\begingroup$ I'm still having problems with understanding the representation of the kinetic energy. 1.) If the kinetic energy is positive, why does the quadratic form resp. the matrix $ \mathbf{T} $ have to be positive definite? 2.) If it is positive definite, it is symmetric and thus diagonalizable. But what do you mean by the resp. "diagonal" basis being $ \mathbf{q} $? $ \mathbf{q} $ s the vector of the generalized coordinates, not n vectors forming a basis of the configuration space. Do you mean the canonical basis for these coordinates? (BTW, the missing factor 1/2 in G.'s (8.22) comes from (1.71).) $\endgroup$ Commented Jan 15, 2021 at 10:05
  • $\begingroup$ 1. Suppose that $\mathbf{T}$ has a negative eigenvalue $\lambda_0<0$, and then suppose that the system is excited in the eigenmode corresponding to this eigenvalue, and all other modes are not at all excited. Then the kinetic energy will be of the form $-|\lambda_0|\dot{q}_0^2$ which is negative. $\endgroup$
    – xzd209
    Commented Jan 15, 2021 at 15:52
  • $\begingroup$ 2. Yep that's what I meant :) $\endgroup$
    – xzd209
    Commented Jan 15, 2021 at 15:56
  • $\begingroup$ Thanks a lot again for your kind help! $\endgroup$ Commented Jan 15, 2021 at 17:12
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The Hamiltonian is:

$$H=\boldsymbol p^T\,\dot{\boldsymbol{q}}-L(\boldsymbol q~,\boldsymbol p)$$

with

$$L=L_0(\boldsymbol q)+\dot{\boldsymbol{q}}^T\,\boldsymbol a+\frac 12 \dot{\boldsymbol{q}}^T\,\boldsymbol T\,\dot{\boldsymbol{q}}$$

Step I

$$\boldsymbol p=\frac{\partial L}{\partial \dot{\boldsymbol{q}}}=\boldsymbol T\,\dot{\boldsymbol{q}}+\boldsymbol a$$ $\Rightarrow$ $$\dot{\boldsymbol{q}}=\boldsymbol T^{-1}(\boldsymbol p-\boldsymbol a)$$

Step II

$$L(\boldsymbol p~,\boldsymbol q)= L_0(\boldsymbol q)+\boldsymbol a^T\,\boldsymbol T^{-1}(\boldsymbol p-\boldsymbol a)\,+\frac 12 \dot{\boldsymbol{q}}^T\,\boldsymbol T\,\dot{\boldsymbol{q}}$$

Thus:

$$H=\underbrace{(\boldsymbol T\,\dot{\boldsymbol{q}}+\boldsymbol a)^T\,(\boldsymbol T^{-1}(\boldsymbol p-\boldsymbol a))}_{Z}\, -\left[L_0(\boldsymbol q)+\boldsymbol a^T\,\boldsymbol T^{-1}(\boldsymbol p-\boldsymbol a)\,+\frac 12 \dot{\boldsymbol{q}}^T\,\boldsymbol T\,\dot{\boldsymbol{q}}\right]$$

$$Z=(\dot{\boldsymbol{q}}^T\,\boldsymbol T^T+\boldsymbol a^T)\,\boldsymbol T^{-1}(\boldsymbol p-\boldsymbol a)=\dot{\boldsymbol{q}}^T\,(\boldsymbol p-\boldsymbol a) +\boldsymbol a^T\,\boldsymbol T^{-1}(\boldsymbol p-\boldsymbol a)$$

with $~\boldsymbol T^T=\boldsymbol T~,\boldsymbol T~$ is symetric.

thus:

$$H=\dot{\boldsymbol{q}}^T\,(\boldsymbol p-\boldsymbol a)-\left[L_0(\boldsymbol q)+ \frac 12 \dot{\boldsymbol{q}}^T\,\boldsymbol T\,\dot{\boldsymbol{q}}\right]~\surd$$

remarks:

the kinetic for particle is:

$$T_K=m\,\frac 12\boldsymbol v^T\,\boldsymbol v$$

where $~v$ is the velocity

$$\boldsymbol v=\underbrace{\frac{\partial \boldsymbol R(\boldsymbol{q})}{\partial \boldsymbol q}}_{\boldsymbol J}\,\dot{\boldsymbol{q}}$$

where $\boldsymbol R$ is the position vector of the particle

thus

$$T_K=m\,\frac 12\dot{\boldsymbol{q}}^T\,\boldsymbol J^T\,\boldsymbol J\,\dot{\boldsymbol{q}}$$

so $~\boldsymbol T=\,\boldsymbol J^T\,\boldsymbol J~$ symmetric and positive definite

I use this notation $~(..)^T~$ T stay for transpose , for a column vector you get row vector, for a matrix changing rows with column.

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