The proper distance from $R$ to $R+\Delta R$ in Schwarzschild metric is given by

$$\displaystyle l = \int_R^{R+\Delta R} \frac{1}{\sqrt{1-\frac{r_s}{r}}} \; dr.$$

If static observer whose radial coordinate is $R$ send light in a radial direction (for example using fiber cable) then the coordinate time interval would be

$$\displaystyle \Delta t = \int_R^{R+\Delta R} \frac{1}{{1-\frac{r_s}{r}}} \; dr.$$

The proper time interval that this observer would measure should be
$$\displaystyle \Delta \tau =\sqrt{1-\frac{r_s}{R}}\Delta t= \sqrt{1-\frac{r_s}{R}}\int_R^{R+\Delta R} \frac{1}{{1-\frac{r_s}{r}}} \; dr.$$

So the spatial distance that this observer would measure should be $$\displaystyle d=c\Delta \tau = c\sqrt{1-\frac{r_s}{R}}\int_R^{R+\Delta R} \frac{1}{{1-\frac{r_s}{r}}} \; dr.$$

witch is different from the proper distance $l$.

So how we actually measure proper distance?


In Schwarzschild coordinates $(t, r, \theta, \phi)$ the proper radial distance is defined along $r$ with $t, \theta, \phi$ constant. The first equation is correct. The second equation for the coordinate time interval is correct as well.

However the third equation expressing the proper time interval vs. the coordinate time interval is not. Reason being that the proper time interval vs. the coordinate time interval requires the other coordinates $r, \theta, \phi$ to be constant. Instead in the integral $r$ is varying from $R$ to $R + \Delta R$.

If you assume $\Delta R$ to be infinitesimal, i.e. $dr$, you recover the correct expression for the proper time interval, that you can integrate and get again the first equation for the proper distance.

Note: In the formulas you assume natural units, that is $c = G =1$. To be consistent you should have $c = 1$ also in the last statement.

  • $\begingroup$ In the third expression the integral is referring to the photon radial coordinate not to the observer one which is constant $\endgroup$ – amilton moreira Oct 7 '18 at 18:33
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    $\begingroup$ The proper time at the stationary observer is not the proper time of other stationary observers along the photon path. That is why you should approach the description in a differential way. $\endgroup$ – Michele Grosso Oct 8 '18 at 9:44

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