# Proper time difference in Schwarzschild geometry: stationary observer vs. observer on a circular orbit

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 dont 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 $$2$$ doesnt 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 $$2\pi$$.

• 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. – Javier Jan 7 at 19:03
• Okay thank you....But my problem is that I dont know how to find the relation between $\tau_{1}$ and $t$.... – Udalricus.S. Jan 7 at 19:04
• Use the angular momentum to find $d\varphi/dt$, that will give you a relation between $t$ and $r$. – Javier Jan 7 at 19:06
• 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... – Udalricus.S. Jan 7 at 19:14

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