# General relativity change of observer

I have a problem in calculating the module of the velocity of a particle measured by a static observer in a specific metric. This metric is

$$ds^2=(r^2-R^2)dt^2-\frac{dr^2}{r^2-R^2}-r^2d\varphi^2$$

Because both $$\partial_t$$ and $$\partial_\varphi$$ are Killing vectors, we have two conserved quantities

$$E=-g_{\mu\nu}U^\mu\delta^{\nu t}=(r^2-R^2)\frac{\partial t}{\partial \tau}$$ $$L=g_{\mu\nu}U^\mu\delta^{\nu \varphi}=r^2\frac{\partial\varphi}{\partial\tau}$$

where $$U^\mu$$ is the four velocity. The equations of motion for a massive particle come from $$U^2=1$$ and are

$$E^2=\left(\frac{\partial r}{\partial\tau}\right)^2+(r^2-R^2)(1+\frac{L^2}{r^2}).$$

My problem is that i dont understand how to calculate the velocity of that a static observer in $$r_1>r_0$$ (where $$r_0$$ is the initial point of the particle).

Any help is appreciatted.

• From the two conserved quantities and the fact that $d\tau^2=(r^2-R^2)dt^2-\frac{dr^2}{r^2-R^2}-r^2d\varphi^2$, you can calculate $dr/d\tau$ as a function of $r$, $E$, and $L$. Dec 3, 2023 at 14:36
• Do you mean the local 3-velocity v relative to a static observer at the same coordinate as the test particle so that a photon has v=c? Then the covariant momentum pᵤ=vᵤ√|gᵤᵤ|/√(1-μ²|v|²) with |μ|=1 for particles and 0 for photons. The coordinate velocity dx/dt may be smaller and the proper velocity dx/dτ higher than the local v. Dec 3, 2023 at 18:28