- #1

stongio

- 5

- 0

## Homework Statement

What is the speed of a particle in the smallest possible circular orbit in the Schwarzschild

geometry as measured by a stationary observer at that orbit? Note: The orbit in

question happens to be unstable.

## Homework Equations

Normalization condition:

[itex]\textbf{u}_{obs}(r)[/itex][itex]\cdot[/itex][itex]\textbf{u}_{obs}(r)[/itex]=[itex]g_{αβ}[/itex][itex]u^{α}_{obs}(r)[/itex][itex]u^{β}_{obs}(r)[/itex]=-1

Schwarzschild Metric...

## The Attempt at a Solution

So from the Normalization condition and the Schwarzschild metric it is easy to find the four-velocity of the observer, because the spatial components are zero. My question relates to the observer's measurement of the particle's velocity. If his four-velocity is

[itex]{u}^{a}[/itex]=[1/[itex]\sqrt{1-2m/r}[/itex],0,0,0]

then, how does observe the particle's velocity?

An answer I found somewhere said

[itex]\frac{u^{2}}{1-u^{2}}[/itex]=[itex]u^{a}u^{b}(g_{ab}+u_{a}u_{b})=(u^{a}u_{a})^{2}-1[/itex]

But I have no idea how to arrive at this relation, and why it relates to the observer's measurement of the particle. Is it possibly a derivation of a relation between t and the proper time?

Also, this might seem very rudimentary but I am not sure why we are able to write:

[itex]u_{a}u_{b} \ast u^{a}u^{b} = (u^{a}u_{a})^{2}[/itex]

I'm very new to general relativity, and would really appreciate the help!

Thanks a lot!