1
$\begingroup$

In chapter 8.2.2 of Chaikin&Lubensky [1], they argue why the addition of a term $ -\Omega_e L$ to the Hamiltonian can be used to model a rotating system.

A simplified version of the argument, as I understand it, goes like this. Consider a collection of rods in the rest frame $(x,y)$ that can rotate around their individual centers of mass at fixed positions $\{l\}$. We can assign the coordinate $\theta_l$ to each rotor, which is its angle wrt. the x-axis. The Lagrangian for this system will thus be

$$ \mathcal L = \sum_l \frac{1}{2} I \dot \theta_l^2, $$

where $I$ is its moment of inertia of one rod. The Hamiltonian is found to be

$$ \mathcal H = \sum_l \frac{p_l^2}{2I}, $$

where $p_l = \frac{\partial \mathcal L}{\partial \dot \theta_l} = I \dot \theta_l$. Now, if we introduce a rotating coordinate system $(x',y')$, given by $\theta_l = \theta_l' + \Omega_e t$, we can express the same Lagrangian as follows

$$\mathcal L = \sum_l \frac{1}{2} I (\dot \theta_l' + \Omega_e)^2$$

while the Hamiltonian for these coordinates is given by

$$\mathcal H_T = \sum_l \frac{{p_l'}^2}{2I} - \Omega_e L$$

where $L = \sum_l p_l'$ and $p_l' = \frac{\partial \mathcal L}{\partial \dot \theta_l'} = I(\dot \theta_l' + \Omega_e) = I \dot \theta_l= p_l$. Thus, $L$ is the angular momentum in the rest frame. So we can write

$$ \mathcal H_T = \mathcal H - \Omega_e L. $$

The question is, how can we infer meaning from the Hamiltonian of the $(x',y')$ system expressed in coordinates of the $(x,y)$ system? It seems to me that the authors claim that this shows the energetic contribution from rotating a system, but this is not obvious to me, as the Hamiltonian of the $(x',y')$ system does not describe the evolution of the system in the $(x,y)$-system.

But I have seen this type of argument before, so there is clearly some deeper understanding that I have not properly understood.

[1] Chaikin, P. M., & Lubensky, T. C. (1995). Principles of condensed matter physics. Cambridge University Press.

$\endgroup$
0
$\begingroup$

Depending on the type of coordinate transformation, Hamiltonians in different frames of reference are more or less closely related, so I will have to answer a little differently for each of them:

  • Galilei- or Lorentz-Transformations:

    In this case, it really is as simple as kinetic terms emerging or disappearing due to transformations, because it does not matter which inertial frame of reference is chosen for describing a system. This is commonly used for example to describe the motion of rigid bodies as pure rotations in the center-of-mass reference frame and then, when the solution is found, just add the translation motion of the center of mass.

  • Transformations between frames of reference, one of which is not an inertial frame of reference:

    This case is a little more complicated and your example of a rotating frame of reference belongs to this category. In general, when transforming into a non-inertial frame of reference, fictitious forces need to be taken into account. In the case of rotations with constant angular velocity, these are the centrifugal and the Coriolis force. Those fictitious forces lead to additional energies of masses, so in general it is not sufficient to just add the (rotational) kinetic terms.

    In your example of the rotating rods the centers of mass of which are fixed in place, the energies due to the fictitious forces are ignored, because if the rods' centers of mass are on the rotation axis of the coordinate transformation, there are no such forces acting and even in case that would not be true, the circular paths (in the rotating frame of reference) of said centers of mass are already known from them being fixed in the inertial frame of reference and consequentially need not be determined using Hamiltonian mechanics. In fact, you could, for each rod, first translate your origin to it's center of mass and then start rotating the coordinate system, to remove all fictitious forces for that rod.

Remark:

Technically, since the rods have a non-zero length, even if their centers of mass are on the coordinate rotation axis, centrifugal forces act at their ends (and any point which is not the center of mass). Those forces are compensated by the atomic or molecular forces of the material the rods are made of, but in classical mechanics of rigid bodies this is not of relevance and can be ignored, just like the kinetic terms corresponding to the circular motion of the centers of mass if these are not on the rotation axis of the coordinate system.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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