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Lets consider two observer, which are at a position $r=R$ around a spherical mass which is described by Schwarzschild geometry. At some point, observer $1$ make a full round around the object on a circular orbit (without an engine, "free falling"), while observer $2$ stays fixed. My goal is to calculate $\Delta\tau_{2}/\Delta\tau_{1}$, where $\tau$ denotes the proper time measured by observer 1 and 2, respectively.

The proper time of observer $1$ is straightfoward. I can use the Binet equation in Schwarzschild geometry, which reads

$$\frac{\mathrm{d}^{2}u}{\mathrm{d}\varphi^{2}}+u=\frac{GM}{h^{2}}+\frac{3GM}{c^{2}},$$

where $u=1/r$ and $h$ is a constant related to the angular momentum. Observer $1$ is moving on a circular orbit and hence $u=1/R=\text{const.}$ This yields the relation

$$h=\sqrt{\frac{\mu c^{2}R^{2}}{(R-3\mu)}}$$

with $\mu=GM/c^{2}$. Applying one of the geodesics equations for Schwarzschild geometry, namely $r^{2}\frac{\mathrm{d}\varphi}{\mathrm{d}\tau}=h$, one obtains

$$R^{2}\frac{\mathrm{d}\varphi}{\mathrm{d}\tau_{1}}=h=\sqrt{\frac{\mu c^{2}R^{2}}{(R-3\mu)}}$$

and therefore

$$\Delta\tau_{1}=\int_{0}^{2\pi}\frac{1}{R^{2}}\sqrt{\frac{\mu c^{2}R^{2}}{(R-3\mu)}}\,\mathrm{d}\varphi=2\pi\sqrt{\frac{R^{2}}{\mu c^{2}}(R-3\mu)}$$

But I don`t know how to calculate $\Delta\tau_{2}$... In the end I should get

$$\frac{\Delta\tau_{2}}{\Delta\tau_{1}}=\sqrt{\frac{R-2\mu}{R-3\mu}}.$$

EDIT: My idea is the following: Observer $2$ doesn`t move and therefore: $$c^{2}\mathrm{d}\tau_{2}^{2}=\mathrm{d}s^{2}=c^{2}\bigg (1-\frac{2\mu}{r}\bigg) \mathrm{d}t^{2}$$.

This yields $$\frac{\mathrm{d}\tau_{2}}{\mathrm{d}t}=\sqrt{\bigg (1-\frac{2\mu}{r}\bigg)} $$

When I am now able to find a relation between $\tau_{1}$ and the coordinate time $t$, I can relate $\tau_{2}$ to $\tau_{1}$ and therefore to the $\varphi$-coordinate of observer $1$, which can be used to integrate from 0 to $2\pi$.

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    $\begingroup$ Your idea is correct, though you're missing a square root. If you work the whole thing out you could post it as an answer to your own question. $\endgroup$ – Javier Jan 7 at 19:03
  • $\begingroup$ Okay thank you....But my problem is that I don`t know how to find the relation between $\tau_{1}$ and $t$.... $\endgroup$ – Udalricus.S. Jan 7 at 19:04
  • $\begingroup$ Use the angular momentum to find $d\varphi/dt$, that will give you a relation between $t$ and $r$. $\endgroup$ – Javier Jan 7 at 19:06
  • $\begingroup$ I think I have an idea....When I take $c^{2}\mathrm{d}\tau^{2}=c^{2}(1-2\mu/r)\mathrm{d}t^{2}-r^{2}\mathrm{d}\varphi^{2}$ and divide through $\mathrm{d}\tau$, I find the required relation... $\endgroup$ – Udalricus.S. Jan 7 at 19:14
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I`ve got the solution....

Step 1: Observer $1$ is on a circular orbit and therefore $$\mathrm{d}s^{2}=c^{2}\mathrm{d}\tau_{1}^{2}=c^{2}\bigg (1-\frac{2\mu}{R}\bigg )\mathrm{d}t^{2}-R^{2}\mathrm{d}\varphi_{1}^{2}.$$ We conclude that

$$\frac{\mathrm{d}t}{\mathrm{d}\tau_{1}}=\sqrt{\frac{c^{2}+R^{2}\big(\frac{\mathrm{d}\varphi_{1}}{\mathrm{d}\tau_{1}}\big)^{2}}{c^{2}(1-2\mu/R)}}.$$

Step 2: use $$\mathrm{d}\tau_{2}=\frac{\mathrm{d}\tau_{2}}{\mathrm{d}\varphi_{1}}\mathrm{d}\varphi_{1}=\frac{\mathrm{d}\tau_{2}}{\mathrm{d}t}\frac{\mathrm{d}t}{\mathrm{d}\varphi_{1}}\mathrm{d}\varphi_{1}=\frac{\mathrm{d}\tau_{2}}{\mathrm{d}t}\frac{\mathrm{d}t}{\mathrm{d}\tau_{1}}\frac{\mathrm{d}\tau_{1}}{\mathrm{d}\varphi_{1}}\mathrm{d}\varphi_{1}=...=\sqrt{\frac{R^{2}}{\mu c^{2}}(R-2\mu)}\mathrm{d}\varphi_{1}$$

Step 3: Integrating from $0$ to $2\pi$ yields an additional factor $2\pi$ and one obtains finally the relation

$$\frac{\Delta\tau_{2}}{\Delta\tau_{1}}=\sqrt{\frac{R-2\mu}{R-3\mu}}.$$

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