Topic is that last mentioned in the so far underwhelmingly responded to entry here Referencing mainly to fig.9 and accompanying in the article http://www.ate.uni-duisburg-essen.de/data/postgraduate_lecture/AJP_2009_Griffiths.pdf , it will be helpful to work from a Cartesian coordinate system(adsbygoogle = window.adsbygoogle || []).push({}); xyz, with unit vectorxhorizontal to the right,yvertically up,zout of the page. let's consider what happens when the rectangularish loop current shown in fig.9 there (height w, lengthl) is made to precess at some constant angular velocityωabout an axis alongyand coinciding with the center of loop left vertical leg. We have from (16) there the general resultp_{hid}= 1/c^{2}m×Efor a magnetic momentmimmersed in a uniformEfield, it being understood the moving charges comprising loop currentmmust feel the appliedE. That rules out an ordinary conducting loop current. At the instantp_{hid}acts along axisy(fig.9) there must be exerted on the loop a forceF_{hid}equal to minus that rate-of-change-of-p_{hid}, given by

F_{hid}= -dp_{hid}/dt = -ω×p_{hid}= -ω×(1/c^{2}m×E) =z1/c^{2}ωwlIE - (1), and thus a torqueζ_{hid}acting abouty

ζ_{hid}=r×F_{hid}= (1/2lx)×(z1/c^{2}ωwlIE) = -y1/2w(l/c)^{2}ωIE - (2)

That factor of 1/2 owes to thatF_{hid}acts through the center of the loop^{[1]}. Naturally one expects conservation of angular momentum holds, requiring an equal and opposite torque to be exerted by some other internal agency. There is indeed an opposing internally sourced torque, seen by moving into the instantaneous rest frame S' of the right vertical leg. In S', moving at velocityv=lω×y= -zlω relative to rest frame S,Eseen in S is transformed into anE',B' given by

E' = γE,B' = -1/c^{2}v×E' = -x1/c^{2}lωE' - (3)

ThisB' acts onI' = -yγI in that leg to give a magnetic Lorentz force

F_{em}= (I'w)×B' = I'w(-y)×(-x1/c^{2}lωE') = -zI'w1/c^{2}lωE' = -γ^{2}F_{hid}- (4)

AsF_{em}acts along the direction of relative motion, it transforms back into S unchanged. The resulting torqueζ_{em}exerted on the loop owes toF_{em}acting over the full moment arm=lxl, and by inspection is therefore

ζ_{em}=r×F_{em}= (xl)×(-yγ^{2}F_{hid}) = -2γ^{2}ζ_{hid}- (5)

Net torque acting is thus

ζ=ζ_{hid}+ζ_{em}=ζ_{hid}(1-2γ^{2}) ~ -ζ_{hid}(v<<c) - (6)

At this point it must be asked whether there is an overall conservation of angular momentum. The one 'reasonable' place left to look^{[2]}is the source ofE, which we can take to be two very large uniformly charged circular capacitor plates spaced equally above and below the precessing loop. That any equal and opposite back-reaction torque exists on the sum of these charged plates can be quickly ruled out. That's because the loop sweeps out the form of a toroidal solenoid each revolution. There can be no requisite circular E field generated by such a toroid - rotating or not, unless one wishes to challenge validity of ∇×E= -∂B/∂t. Hence there is zero net torque exerted on the source charges giving rise to the appliedE=yE. Sure looks like conservation of angular momentum is in serious trouble!

Returning to the loop itself, a non-zeroζmeans a mechanical angular power generation

W =ζ.ω= 1/(2c^{2})ω^{2}wl^{2}IE(2γ^{2}-1) - (7)

The sign of W can be changed at will given we are free to alter independentlyω,E, and sense of I. Can we at least rescue conservation of energy here - there surely must be a power-drain in the circulating I exactly off-setting the mechanical W in (7)? The one seeming possibility is that E fields acting in the loop's two vertical legs are effectively different, and to the right degree. This is not evident in lab frame S, whereEis uniform, and what's more the charge flow rates must be equal. The Lorentz force lawF= q(E+v×B) explicitly forbids anyy-acting 'boost' in one leg owing to it's relative velocity in S. (it does not however prevent existence of a transversex-acting reaction force in supporting tube - necessary in order to explainF_{em}in (4)) Hence in S we have no net power drain in circulating current I owing toE. In S'E' is a factor gamma greater, but the resulting forces qE' transform back in S to give just qE. Nope, looks a lot like conservation of energy, and thus in any other frame, energy-momentum, is in deep pooh too. Goodness me, can someone point out my obvious blundering here? A simple sign error maybe, or worse, just wrong-headed thinking? Well judging by the neg-response in that other thread, this might be a long wait. Maybe the topic is just too boring for the erudite readership here at PF. Or something.

[1]: Assuming E generates all the pressure differential between upper and lower horizontal legs,F_{hid}is the sum of uniform precessional force densities that are opposed and weighted in favor of the upper leg. The center ofF_{hid}thus acts half-way between the two vertical legs. We ignore anyp_{hid}induced in the right vertical leg owing to centripetal acceleration from rotation at ω aboutyas it makes no contribution to thex-actingp_{hid}of interest here.

[2]: There are perhaps two other conceivable places to look:

a) - Precession of SFM (stored field momentum) of density ε_{0}E×B, which as shown in eq'n (57c) in above cited article, is numerically equal and opposite top_{hid}in the situations covered there. But what can one say here? Where is this precessing SFM able to physically manifest? We have applied the usual expressions involving Lorentz force and SR transformations, and there is no room at all for any 'cancellation' ofF_{hid}by precessing SFM. Hence an unphysical quantity; at least in this situation.

b) - Self-interaction between charges flowing in the loop. Any such interactions must go as the square of charge density for a given mean flow speed, hence as the square of I. Whereas the expressions for force and torque above found are directly proportional to I. So that is ruled out on functional dependence grounds. And ruled out completely if our model was based on an 'incompressible' charged fluid for which any EM self-interactions would be identical to that with no imposedE, which is an essential forp_{hid}etc.

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# Not so hidden momentum

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