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Natural Science Forum / Physics / Research / September 2007equipartition and rotational instability
This question was posed by John Zwart in the American Journal of Physics, Vol. 70, No. 2, p. 105, February 2002 but has not been answered since: " According to the classical kinetic theory of gases, a rigid nonlinear poly-atomic molecule has six degrees of freedom; three translational and three rotational. The equipartition theorem states that "each degree of freedom has associated with it-on average-an energy of (1/2)kT per molecule." For a free rotation of a rigid body with three nonequal moments of inertia about its principal axes, solving the Euler equations shows that while rotations about the axes with the largest and smallest principal moments are stable, a rotation about the intermediate moment axis is not. Any perturbation in a rotation about this intermediate moment axis grows quickly, leading to a rotation about one of the other axes. If we apply this result to a rigid poly-atomic molecule, it appears that one degree of freedom could not, on average, have as much energy as the other two rotational degrees. Why not?" Can somebody here offer an answer? Thanks. > This question was posed by John Zwart in the American Journal of > Physics, Vol. 70, No. 2, p. 105, February 2002 but has not been [quoted text clipped - 17 lines] > > Can somebody here offer an answer? I don't see why there are three rotational degrees of freedom. The molecule's configuration is specified by its centre-of-mass position (3 degrees of freedom, eg, x,y,z), and its orientation (2 degrees of freedom, eg, theta, phi), giving five rather six degrees of freedom in total. It is worth noting that the equipartition theorem is confined to systems with quadratic Hamiltonians, with a kT/2 contribution per quadratic term. >I don't see why there are three rotational degrees of freedom. The >molecule's configuration is specified by its centre-of-mass position >(3 degrees of freedom, eg, x,y,z), and its orientation (2 degrees of >freedom, eg, theta, phi), giving five rather six degrees of freedom in >total. To specify the orientation of a three-dimensional body requires three angles, not two. Look up "Euler angles" in a mechanics textbook (or on the web, no doubt) for details. The big idea is that you need two angles (theta,phi) to specify the orientation of any given axis through the body, but then you have an extra degree of freedom associated with rotations about that axis. For instance, a mad scientist wanting to reorient the Earth could turn the Earth so that the North Pole pointed in any desired direction (two angles), and then rotate the Earth about the North Pole by any arbitrary third angle. -Ted  Signature[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.] >>I don't see why there are three rotational degrees of freedom. The >>molecule's configuration is specified by its centre-of-mass position [quoted text clipped - 13 lines] > angles), and then rotate the Earth about the North Pole by any > arbitrary third angle. Moreover. The said mad scientist cannot reorient the Earth spin axis because Earth is spinning like a gyroscope in space. Hence, the spinning Earth becomes stiff against any force that might trend to move the axis of spin, relative to its center of mass, creating a torque. The Law of angular momentum conservation requires that to balance any external torque N, planet Earth has to precess. Only with the three Euler angles one can predict Earth motion and understand why it's motion is stable. If the Earth doesn't spin stiff like a gyroscope, the Year's seasons, like Summer and Winter, could change from Year to year, or even have several Summers and Winters every Year. > If we apply this result to a rigid poly-atomic molecule, > it appears that one degree of freedom could not, on average, have as > much energy as the other two rotational degrees. Why not? the collisions between the molecules will "repopulate" this degree of freedom, even if it is mechanically instable. -- | ~~~~~~~~ Martin Ouwehand ~ Swiss Federal Institute of Technology ~ Lausanne __|___________ Email/PGP: http://personnes.epfl.ch/martin.ouwehand ____________ The droning voice of the professor continued to wind itself slowly round and round the coils it spoke of, doubling, trebling, quadrupling its somnolent energy as the coil multiplied its ohms of resistance [James Joyce] > This question was posed by John Zwart in the American Journal of > Physics, Vol. 70, No. 2, p. 105, February 2002 but has not been [quoted text clipped - 19 lines] > > Thanks. If the external torque is zero, then from the point of view of the rotating object, the angular velocity vector makes a loop, eventually coming back to where it started. If the angular velocity vector is near the principal axis with the smallest or largest moments of inertia, it's a tiny loop about the principal axis. If the angular velocity vector is near the intermediate principal axis, the loop is much larger, taking the angular velocity far away from its original value, but it eventually comes back to where it started. Consider an ensemble of rotating molecules. It will include molecules at all stages of this loop. Thus as the angular velocities of some molecules move away from the intermediate principal axis, the angular velocities of other molecules will move toward it. It's unlikely that the angular velocity of a randomly spinning molecule will precess close to the intermediate principal axis, because if the angular velocity isn't on exactly the right path, it will be swept to one side or the other. But this is made up for by the fact that if the angular velocity does get close to that axis, it will stay there for a relatively long period of time. By Liouville's theorem, if the initial probability distribution is uniform in phase space, it has to stay that way. When the molecule collides with another, the external torque will no longer be zero, but I see no reason why the other molecule would be more likely to knock the first molecule's rotation away from the intermediate axis than towards it. SignatureJim E. Black |
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