"John Larkin" wrote in message news: snipped-for-privacy@4ax.com...
Well, if one wanted to be perfectly obtuse...
Quantum spin is in units of hbar, which is J.s = kg.m^2 / s, angular momentum. Inertia is kg.m^2, so a quantity of 1/s (exactly, angular frequency) is required. This would ordinarily be the rotation frequency of the object, but a photon isn't exactly a spherical chicken, with a well-defined shape and [angular] velocity. It's questionable whether one could use the frequency itself here (i.e., c / lambda = 600THz); a freely propagating photon tends to be a linearly oscillating phenomenon, but it also does a fine job oscillating in place if confined to a resonator, so it might not be too horrible.
Moments of inertia of geometric shapes have the form I = a * MR^2, where /a/ depends on the mass distribution. If we use the wavelength as the radius and this inertia, we get a = 1, which shouldn't be surprising as:
I = a * MR^2 and I = k * lambda / c M = k / (lambda * c) R = lambda
k * lambda a * k * lambda^2
------------ = ------------------ c lambda * c
a = 1, basically what we started with.
At best, this implies that all the photon's angular momentum is carried on the periphery (a thin ring or hoop spinning on axis).
Tim