General Relativity: Particle velocity travelling on Schwarzchild orbit

[stongio]

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:

$\textbf{u}_{obs}(r)$$\cdot$$\textbf{u}_{obs}(r)$=$g_{αβ}$$u^{α}_{obs}(r)$$u^{β}_{obs}(r)$=-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
${u}^{a}$=[1/$\sqrt{1-2m/r}$,0,0,0]
then, how does observe the particle's velocity?
An answer I found somewhere said

$\frac{u^{2}}{1-u^{2}}$=$u^{a}u^{b}(g_{ab}+u_{a}u_{b})=(u^{a}u_{a})^{2}-1$

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:

$u_{a}u_{b} \ast u^{a}u^{b} = (u^{a}u_{a})^{2}$

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

 

[mark726]

Thank you for your question. The speed of a particle in the smallest possible circular orbit in the Schwarzschild geometry, as measured by a stationary observer at that orbit, can be calculated using the normalization condition and the Schwarzschild metric. The normalization condition allows us to find the four-velocity of the observer, which is necessary to calculate the particle's speed.

To find the particle's speed, we can use the relation you mentioned, which is derived from the normalization condition and the Schwarzschild metric. This relation, also known as the Lorentz factor, relates the speed of the particle as measured by the observer to the observer's four-velocity. This is why it is used to calculate the particle's speed as measured by the observer.

As for your question about why we can write u_{a}u_{b} * u^{a}u^{b} = (u^{a}u_{a})^{2}, this is due to the fact that the four-velocity is a four-vector, which means it has both a magnitude and direction. When we multiply two four-vectors together, the result is a scalar quantity (a number), not a vector. Therefore, (u^{a}u_{a})^{2} is the magnitude squared of the four-velocity, which is equivalent to the product of the four-velocity with itself.

I hope this helps clarify your doubts. Keep exploring and asking questions about general relativity, it's a fascinating field!