Read clocks and frequency transfer

Static observers read proper time and compare frequencies through the same interval law. Exact lapse factors and explicit remainder estimates show what the leading redshift formula retains.

15. Rods, clocks and static frequency transfer

The observational readouts retain their measuring protocol: a specified observer, spatial direction, chart, local standard and endpoint or asymptotic condition. A coordinate-length statement is not promoted to an invariant deformation of a material body. The exterior geometry is vacuum; rods, massive particles and light are probes whose backreaction is neglected.

For a radial unit proper rod at fixed static time, the interval readout gives

$$\mathrm{d}\ell^2=A^{-1}\mathrm{d}r^2, \qquad \boxed{\mathrm{d}r_{\rm unit}=\sqrt A =1-\frac\alpha{2r}+O((\alpha/r)^2),} \tag{15.1}$$

recovering Einstein (71). Using the truncated first-order spatial metric as exact would instead give $(1+\alpha/r)^{-1/2}$; its first coefficient agrees and its second coefficient differs. For a tangent direction orthogonal to $n$, (14.8) has unit eigenvalue, recovering Einstein (71a) in that local coordinate arrangement. It does not remove spatial curvature or make arbitrary global distance comparisons Euclidean.

A static clock obeys

$$\boxed{\frac{\mathrm{d}s}{\mathrm{d}t}=\sqrt A,\qquad \frac{\mathrm{d}t}{\mathrm{d}s}=A^{-1/2} =1+\frac\alpha{2r}+O((\alpha/r)^2).} \tag{15.2}$$

The second expression is the coordinate duration of one unit proper clock period, which is the quantity in Einstein (72); the first is its reciprocal rate. For the general weak potential,

$$\frac{\mathrm{d}t}{\mathrm{d}s}=1-\phi+O(\phi^2) =1+\frac\kappa{8\pi}\int\frac{\rho(\mathbf y)}{|\mathbf x-\mathbf y|}\mathrm{d}^3y +O(\phi^2). \tag{15.3}$$

It agrees with (15.2) under the same mass normalization (14.2).

For $0\le x=\alpha/r\le q_0<1$, Taylor’s theorem supplies explicit same-family bounds:

$$\begin{aligned} |\sqrt{1-x}-(1-x/2)|&\le\frac{x^2}{8(1-q_0)^{3/2}},&&\text{(15.4)}\\ |(1-x)^{-1/2}-(1+x/2)|&\le\frac{3x^2}{8(1-q_0)^{5/2}}. \end{aligned}$$

These control the derived exact exterior, not arbitrary unspecified higher-order completions of a truncated metric.

The original paper states the redshift consequence. Its endpoint operation follows explicitly from the same geometry. Staticity conserves the time component $k_t$ of a light covector along its Hamiltonian trajectory, because $q^{\mu\nu}k_\mu k_\nu/2$ has no explicit time dependence. A static unit observer has $U^t=1/\sqrt{g_{44}}$, so measured frequency is proportional to $k(U)=k_t/\sqrt{g_{44}}$. Therefore

$$\boxed{\frac{f_{\rm obs}}{f_{\rm emit}} =\sqrt{\frac{A(r_{\rm emit})}{A(r_{\rm obs})}},\qquad z=\frac{\lambda_{\rm obs}}{\lambda_{\rm emit}}-1 =\sqrt{\frac{A(r_{\rm obs})}{A(r_{\rm emit})}}-1.} \tag{15.5}$$

At first order, $z=\alpha(1/r_{\rm emit}-1/r_{\rm obs})/2+O(\alpha^2)$. Ratios compose through an intermediate static observer because its lapse cancels. Stationary source and receiver, positive lapse and the same normalization are essential; additional Doppler motion and time-dependent backgrounds are not included.