Page:BumsteadContraction.djvu/7

Rh transverse mass greater than the longitudinal, whereas the opposite is the case with the apparent mass due to electrical charges. A closer consideration however shows that this is an error arising from the application of the ideas of rigid dynamics to a body which is changing its shape. The path of any particle of the bar, if measured by a scale carried along with the earth, will appear to be the circle $$\scriptstyle{A B^\prime C}$$; if measured with reference to a scale at rest however it will be the ellipse $$\scriptstyle{ABC}$$ in which $$\scriptstyle{OB=\sqrt{1-\beta^{2}}\,OA}$$. For brevity we shall refer to these as the "apparent" and the "true" paths. In case (1), let $$\scriptstyle{P_1}$$ be the true position of the particle, $$\scriptstyle{P_1^\prime}$$; its apparent position; let $$\scriptstyle{OM_1=x}$$; $$\scriptstyle{M_1 P_1=y}$$; $$\scriptstyle{\measuredangle AOP_1=\theta_1}$$; $$\scriptstyle{\measuredangle AOP_1^\prime=\theta_1^\prime}$$. In case (2), let $$\scriptstyle{OM_2=x_2}$$; $$\scriptstyle{M_2 P_2=y_2}$$; $$\scriptstyle{\measuredangle BOP_2=\theta_2}$$; $$\scriptstyle{\measuredangle BOP_2^\prime=\theta_2}$$. The potential energy of the twisted wire in either case depends on the apparent angle $$\scriptstyle{\theta_1^\prime}$$ or $$\scriptstyle{\theta_2^\prime}$$. This is seen if we consider two pointers attached to the wire, one along $$\scriptstyle{OA}$$ when the wire is untwisted and the other along $$\scriptstyle{OB}$$; if the wire is now given any twist the two apparent angles $$\scriptstyle{\theta_1^\prime}$$ and $$\scriptstyle{\theta_2^\prime}$$ will be the same, but the real angles $$\scriptstyle{\theta_1}$$ and $$\scriptstyle{\theta_2}$$ will be different as well as the two elliptical arcs traced out by the ends of the pointers. As the apparent motion is isochronous we may put the potential energy equal to $$\scriptstyle{\frac{1}{2}k\theta^{\prime 2}}$$. In position (1) we have For small oscillations, $$\scriptstyle{x_1=a}$$; $$\scriptstyle{\tan{\theta^\prime_1}=\theta^\prime_1}$$, and  Thus the potential energy is $$\scriptstyle{\frac{1}{2}\frac{k}{a^{2}(1-\beta^{2})}y_{1}^{2}}$$; the equation of motion of the particle becomes  and the period of oscillation In case (2) $$\scriptstyle{x_2=a\sin\theta^\prime_2=a\theta^\prime_2}$$ for small oscillations. The potential energy is thus $$\scriptstyle{\frac{1}{2}\frac{k}{a^{2}}x_{2}^{2}}$$, and the period