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Consider the Hamiltonian $$ H=H(x,y,p_x,p_y) $$ which generates the dynamical system $$ \dot{x}=+\frac{\partial H}{\partial p_x} $$ $$ \dot{y}=+\frac{\partial H}{\partial p_y} $$ $$ \dot{p_x}=-\frac{\partial H}{\partial x} $$ $$ \dot{p_y}=-\frac{\partial H}{\partial y} $$ I then discover that this system admits a certain fixed point $$ \vec{z}_0:=(x_0,\,y_0,\,p_{x,0},\,p_{y,0}) $$ To determine the stability properties of $\vec{z}_0$, I compute the Jacobian matrix $J$ associated to the dynamical system, I evaluate it at fixed point $\vec{z}_0$, and I compute the eigenvalues, which are of the type: $$ \lambda_1=+i\omega_a $$ $$ \lambda_2=-i\omega_a $$ $$ \lambda_3=+i\omega_b $$ $$ \lambda_4=-i\omega_b $$ Therefore, I can recognize the presence of two characteristic frequencies, namely $\omega_a$ and $\omega_b$.

So far, so good.

At this point, I try to do the same computation with a different set of dynamical variables, for example making use of polar coordinates instead of cartesian coordinates. So, I start from Hamiltonian $$ H^\prime=H^\prime(r,\theta,p_r,p_\theta), $$ I find the same fixed point found before, but now written in polar coordinates, i.e. $\vec{z}_0^\prime=(r_0,\theta_0,p_{r,0},p_{\theta,0})$. I then compute the Jacobian matrix $J^\prime$ associated to $H^\prime$, evaluate it at fixed point $\vec{z}_0^\prime$ and compute its eigenvalues. The latter have the following structure: $$ \lambda_1^\prime=+i\omega_a^\prime $$ $$ \lambda_2^\prime=-i\omega_a^\prime $$ $$ \lambda_3^\prime=+i\omega_b^\prime $$ $$ \lambda_4^\prime=-i\omega_b^\prime $$ So, i can recognize two characteristic frequencies: $\omega_a^\prime$ and $\omega_b^\prime$.

What I find really counterintuitive is that $$ \omega_a\neq\omega_a^\prime $$ $$ \omega_b\neq\omega_b^\prime $$ that means that, the two eigenfrequencies are different in the two schemes (the cartesian one and the polar one) that I developed. I would have expected them to be the same, i.e. that $\omega_a=\omega_a^\prime$ and $\omega_b=\omega_b^\prime$.

So, here my question comes: is it possible that different set of dynamical variables lead to different characteristic frequencies for a certain fixed point?

To be noted: actually the two sets of eigenfrequencies, i.e. $\{\omega_a, \omega_b\}$ and $\{\omega_a^\prime, \omega_b^\prime\}$ include similar terms and one can obtain the frequencies in the first set by properly combining the ones in the second set, and viceversa. But, as I said, I expected the two sets to be really identical.

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    $\begingroup$ Why do you think they should be identical? $\endgroup$ – QuantumBrick Mar 23 at 13:08
  • $\begingroup$ @QuantumBrick well, I think that these eigenfrequencies correspond to the frequencies of oscillation of the system when it is perturbed from the fixed point. So, I would expect these frequencies not to depend on the chosen set of dynamical variables. Maybe I erroneously connect my problem to the eigenmodes or “natural modes” of a classical or of a quantum system. Indeed, In case I’m wrong, please correct me and my possibly wrong ansatz. $\endgroup$ – AndreaPaco Mar 23 at 13:36
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    $\begingroup$ Think about the frequencies just as eigenvalues (forget the "physics"). Suppose you change coordinates in a hamiltonian: how does its jacobian respond? Will it have the same eigenvalues as before? In answering this, you'll arrive at the true meaningful question: what is the set of transformations which keep the eigenmodes invariant? Not all, quite certainly! $\endgroup$ – QuantumBrick Mar 23 at 13:41
  • $\begingroup$ Thank you for your comment. Can you please deepen it? In particular: is the set of transformations which keep the eigenmodes invariant the set of canonical transformations? Moreover: if you switch from cartesian-like to polar-like dynamical variables, can you expect the eigenfrequencies of a certain fixed point to change? $\endgroup$ – AndreaPaco Mar 23 at 13:55
  • $\begingroup$ Think about it! Take a matrix formed by derivatives and use the chain rule. You'll find out most transformations, including the canonical ones, change the eigenvalues. This is linear algebra. $\endgroup$ – QuantumBrick Mar 23 at 21:19
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If you change the generalized coordinate the eigenvalues of your new equation of motion must be the same otherwise the dynamic of your new system is changed. to obtain the eigenvalues you must linearize your equation of motions.

Example:

$$\underbrace{\begin{bmatrix} \ddot{x} \\ \ddot{y} \\ \end{bmatrix}}_{\vec{\ddot{q}}}+\underbrace{\left[ \begin {array}{cc} 2\,c&-c\\ -c&2\,c \end {array} \right]}_{C}\,\underbrace{\left[ \begin {array}{c} x\\ y\end {array} \right]}_{\vec{q}} =0\tag 1 $$

where C is the stiffness matrix. to obtain the eigenvalues we transform equation (1) to first order differential equation.

$$\vec{\dot{Y}}=\underbrace{\begin{bmatrix} 0_{2\times 2} & -C \\ I_{2\times 2} & 0_{2\times 2}\\ \end{bmatrix}}_{A}\,\vec{Y}$$

the eigenvalues of the matrix A are:

$$\vec{\lambda}=\left[ \begin {array}{c} i\sqrt {c}\\ -i\sqrt {c} \\ i\sqrt {3}\sqrt {c}\\-i\sqrt { 3}\sqrt {c}\end {array} \right] \tag 2$$

if you choose new generalized coordinates ($r\,,\varphi$) ,polar coordinate , for example

$$ \begin{bmatrix} x \\ y \\ \end{bmatrix}= \left[ \begin {array}{c} r\cos \left( \varphi \right) \\ r\sin \left( \varphi \right) \end {array} \right] $$ equation (1) $\mapsto$:

$$J^T\,J\,\underbrace{\begin{bmatrix} \ddot{r} \\ \ddot{\varphi} \\ \end{bmatrix}}_{\vec{\ddot{w}}}+J^T\,C\,J\,\underbrace{\begin{bmatrix} r \\ \varphi \\ \end{bmatrix}}_{\vec{w}}=0\tag 3$$

where J is the Jacobean :

$$J=\left[ \begin {array}{cc} \cos \left( \varphi \right) &-r\sin \left( \varphi \right) \\ \sin \left( \varphi \right) &r\cos \left( \varphi \right) \end {array} \right] $$

we can solve equation (3) for $\vec{\ddot{w}}$ and get:

$$\vec{\ddot{w}}=-\underbrace{J^{-1}\,C\,J}_{C_w}\vec{w}\tag 4$$

to calculate the eigenvalues we linearize the matrix $C_w\bigg|_{\varphi=0}:=C_L$

$ \Rightarrow$ $$C_{L}= \left[ \begin {array}{cc} 2\,c&-rc\\ -{\frac {c}{r} }&2\,c\end {array} \right] $$

equation (4) $\mapsto$:

$$\vec{\ddot{w}}_L=-C_{L}\vec{w}_L\tag 5$$ again transformed to first order differential equation and obtain the eigenvalues give you the same eigenvalues equation (2). (determinate of matrix C is equal to the determinate of matrix $C_L$)

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  • $\begingroup$ Thank you very much for your reply. Please have a look to @QuantumBrick comme, who says that, in general, a change of coordinates does modify the eigenvalues of the Jacobian. As I said, this is counterintuitive to me because the dynamics of a certain Hamiltonian dynamical cannot depend on the choice of the dynamical variables. The invariance of these eigenvalues upon a change of coordinates is what you seem to claim by means of your example. Are you sure that what you’re claiming has a general validity and that it is not restricted to the example and to the change of coordinates you have made? $\endgroup$ – AndreaPaco Mar 25 at 21:50
  • $\begingroup$ Choose new set of coordinates, cannot describe new system, so if you linearize your system at the „same“ point you must get equal eigenvalues, so I think @QuantumBrick is wrong $\endgroup$ – Eli Mar 26 at 6:45
  • $\begingroup$ Ok, but, are you sure that your reasoning has a general validity? I think your stifness matrix $C$ is not general but very particular. As I said in my question, my Hamiltonian $H$ is general and does not necessary produce motion equations of the type $\ddot{x}+2cx-cy=0$, $\ddot{y}-cx+2cy=0$. Which is the reason behind this assumption of your? Or, am I missing something? $\endgroup$ – AndreaPaco Mar 26 at 10:48
  • $\begingroup$ I just noticed the question asked about changing between generalized coordinates (I think, right?). My comments refer to general transformations imposed on a coordinate set... This answer appears to be correct. $\endgroup$ – QuantumBrick Mar 26 at 13:07
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    $\begingroup$ @AndreaPaco this result are valid for any Stiffness Matrix C, I chose this matrix where all the spring constant are the same to make the documentation short. I will try to answer your quotation for a general case , but perhaps you can put your equation also in mathematics Stack Exchange. you get the Hamiltonian H with $L=T-U$ so you have also EOM's ? $\endgroup$ – Eli Mar 26 at 14:10

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