Einstein and Gravitation — Question 5

Question 5

Gravitational and motion-induced clock shifts can cancel

Use the supplied weak-field slow-motion rate $d\tau/dt\simeq1+\Phi/c^2-v^2/(2c^2)$, where $t$ is time referenced to rest at infinity. A nonrotating spherical planet has radius $R$ and $\mu=GM$. Compare a stationary surface clock with a clock on a circular orbit of radius $r$, where $\Phi=-\mu/r$ and $v^2=\mu/r$.

1. Find their first-order fractional rate difference, orbital minus surface.

2. Find the orbit radius where the two rates match and determine the sign on either side.

Question 5 - Solution

Gravitational and motion-induced clock shifts can cancel

(a) Derivation and calculation

Insert circular speed and potential into the supplied rate law: $$\left(\frac{d\tau}{dt}\right)_{\mathrm{orb}} \simeq1-\frac\mu{rc^2}-\frac{\mu/r}{2c^2} =1-\frac{3\mu}{2rc^2}.$$ The stationary surface rate is $1-\mu/(Rc^2)$, so to first order $$\boxed{\delta_{\mathrm{orb-surf}}\simeq\frac\mu{c^2} \left(\frac1R-\frac3{2r}\right)}.$$

(b) Result and interpretation

Matching rates requires $$\frac1R=\frac3{2r}\quad\Longrightarrow\quad\boxed{r=\frac32R}.$$ Thus $$\delta\begin{cases}<0,&R<r<3R/2,\\=0,&r=3R/2,\\>0,&r>3R/2.\end{cases}$$ The gravitational gain and kinematic loss must both be included. Normalizing the rate difference by the surface rate changes only higher-order terms.

Physical check

The derivation assumes $\mu/(Rc^2)\ll1$ and nonrelativistic orbital speed, as required by the supplied approximation.