Trouton Noble Revisited
G.R.Dixon, April 16, 2009
Fig. 1 depicts a dielectric, solid rod. It exerts normal, mechanical (i.e., non-electromagnetic) forces on two charges, holding them apart at fixed distance, L. The lines of action of the electrostatic and normal mechanical forces all pass through the rod’s midpoint, and there are no torques.
Figure 1
Rod and Charges in Inertial Frame K
Inertial frame K’ moves at speed v in the negative x-direction of K, and hence the rod and charges move in the positive x’-direction of K’ at speed v. Viewed from K’, each charge experiences a magnetic force whose line of action does not pass through the rod’s midpoint. This magnetic force couplet results in a CCW torque. Yet the rod does not rotate in K’. There must be an opposing torque.
In K’, as in K, the lines of action of the electric forces pass through the rod’s midpoint. Thus the electric forces in K’ do not constitute a couplet. What, then, explains the required CW torque in K’?
A picture of the rod in K’ provides qualitative insight. Fig. 2 illustrates. Note that, owing to length contraction, the rod’s planar ends are not perpendicular to the longitudinal axis as they are in K.
Figure 2
Rod and Mechanical Force Couplet in K’
The mechanical forces (which presumably must again be normal to the planar end surfaces) form a CW torque in K’. And this torque presumably cancels the CCW torque attributable to the magnetic forces.
Of historic note is the fact that this mechanical force couplet and the associated torque were not always taken into account. Indeed it was assumed that the apparatus would rotate in K’, but not in K (when K was the ether rest frame). Trouton and Noble performed an experiment to detect such motion through the hypothetical ether. (They used capacitor plates in lieu of q1 and q2.) No rotation occurred for any orientation, or at different times of the year.
As in the present case, it may not always be obvious, from the Lorentz force transformations alone, that a pair of forces will constitute a torque in K’. But attaching the "tails" of the force vectors to the entities they act upon usually resolves this question.
It is of course tempting to consider the other action-at-a-distance force, gravity. Substituting m1 and m2 for q1 and q2 in Fig. 1, we would again have two attracting particles, held apart in K by a "rectangular" rod. Viewed from K’, the rod would look like Fig. 2, and would exert a CW mechanical torque on the particles. The question in this case would be, "What provides the needed CCW torque, so that the net torque in K’ is zero as it is in K?"