Einstein and Gravitation — Question 3

Question 3

A height difference accumulates into a clock offset

Two stationary clocks are separated vertically by $H=1000\,\mathrm m$ in a weak nearly uniform field $g=10\,\mathrm{m/s^2}$. Use the supplied fractional rate difference $(\text{upper}-\text{lower})/\text{lower}\simeq\Delta\Phi/c^2=gH/c^2$, with $c=3\times10^8\,\mathrm{m/s}$. Ignore rotation.

1. Find the rate difference and the accumulated clock difference over $86400\,\mathrm s$.

2. Explain why adding a common constant to both Newtonian potentials does not change this prediction.

Question 3 - Solution

A height difference accumulates into a clock offset

(a) Derivation and calculation

The supplied weak-field rate difference is $$\delta=\frac{gH}{c^2} =\frac{(10)(1000)}{(3\times10^8)^2} =\boxed{1.11111\times10^{-13}}.$$ Over the stated interval, $$\Delta\tau\simeq\delta t=(1.11111\times10^{-13})(86400) =\boxed{9.6\times10^{-9}\ \mathrm s=9.6\ \mathrm{ns}}.$$ The upper clock gains.

(b) Result and interpretation

Under a common change of potential reference, $$\Phi'_u=\Phi_u+C,\quad\Phi'_l=\Phi_l+C,$$ $$\Delta\Phi'=\Phi'_u-\Phi'_l=\Phi_u-\Phi_l=\Delta\Phi.$$ Thus the predicted relative rate $\delta\simeq\Delta\Phi/c^2$ and accumulated offset are unchanged.

Physical check

$gH$ has units of squared speed, making $gH/c^2$ dimensionless and very small.