Einstein and Gravitation — Question 2

Question 2

The equivalence principle has a measurable finite-laboratory limit

Two free particles are initially side by side, separated horizontally by $\ell\ll R$, with zero relative speed near a spherical planet. In the weak-field local free-fall frame, use the transverse tidal equation $\ddot\ell\simeq-(g/R)\ell$. Take $g=10\,\mathrm{m/s^2}$ and $R=6.4\times10^6\,\mathrm m$.

1. Estimate the fractional change in separation over a short duration $t$.

2. Find the largest short duration for the magnitude of that fractional change to stay below $10^{-8}$, and interpret the result.

Question 2 - Solution

The equivalence principle has a measurable finite-laboratory limit

(a) Derivation and calculation

Using the supplied tidal equation and zero initial relative speed, $$\ell(t)\simeq\ell_0+\frac12\ddot\ell_0t^2 =\ell_0\left(1-\frac{gt^2}{2R}\right).$$ Hence $$\boxed{\frac{\Delta\ell}{\ell_0}\simeq-\frac{gt^2}{2R}}.$$ The negative sign describes transverse convergence.

(b) Result and interpretation

Require the magnitude to remain strictly below $10^{-8}$: $$\frac{gt^2}{2R}<10^{-8}\quad\Longrightarrow\quad t<\sqrt{\frac{2R\,10^{-8}}g}.$$ $$\boxed{t<\sqrt{\frac{2(6.4\times10^6)10^{-8}}{10}} =0.113137\ \mathrm s}.$$ The boundary value is the limiting duration, excluded by the strict inequality. A falling frame cannot remove this relative acceleration.

Physical check

At the bound, $gt^2/R=2\times10^{-8}\ll1$, consistent with the short-time expansion.