A torsion balance hangs from a vertical fibre

PHYSICS · Multi-part · Fermi University Physics · Scientific Practice & Experimental Methods · Relativity & Gravitation · The equivalence principle and gravity as geometry

Relativity and Gravitation, Ch Experimental foundations: free fall, Lorentz invariance and clock tests

Problem

A torsion balance hangs from a vertical fibre. A rigid beam of negligible mass carries two test bodies of different composition, each of inertial mass m=10.0 gm = 10.0\ \mathrm{g}, at the two ends, each a distance ℓ=3.00 cm\ell = 3.00\ \mathrm{cm} from the fibre. The fibre gives a restoring torque κφ\kappa\varphi for a twist φ\varphi, with κ=2.50×10−8 N m rad−1\kappa = 2.50\times 10^{-8}\ \mathrm{N\,m\,rad^{-1}}. The two bodies fall towards the Sun with laboratory-frame accelerations g1g_1 and g2g_2 whose Eötvös parameter is η\eta; the horizontal component of the Sun's field at the laboratory has magnitude gˉh=5.00×10−3 m s−2\bar g_h = 5.00\times 10^{-3}\ \mathrm{m\,s^{-2}} (the same for both bodies to leading order) and its direction turns through 2π2\pi once per mean solar day, about 86 400 s86\,400\ \mathrm{s}, as Earth turns once relative to the Sun. The beam is horizontal, the fibre's own drift and all other torques on the twist coordinate are ignored, and the balance's free torsional period is much shorter than a day, so the beam follows the slowly varying torque quasi-statically. ((a)) Show that the torque about the vertical fibre axis is Γ=mℓ η gˉhsin⁡θ\Gamma = m\ell\,\eta\,\bar g_h\sin\theta, where θ\theta is the angle between the beam and the horizontal direction to the Sun, and say why the part of the solar field common to the two bodies contributes nothing. Explain in what sense the instrument makes a null measurement. ((b)) Show that the twist is periodic with the solar day and give its amplitude φmax⁡\varphi_{\max} in terms of the given quantities. ((c)) Evaluate φmax⁡\varphi_{\max} for η=1.0×10−13\eta = 1.0\times 10^{-13}. ((d)) If the smallest twist amplitude the instrument can distinguish at the diurnal frequency is 5.0×10−13 rad5.0\times 10^{-13}\ \mathrm{rad}, what upper bound on η\eta does a null result give?

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