geodesic conjugate points I was reading "Nature of space and time" by Penrose and Hawking, pg.13, 

If $\rho=\rho_0$ at $\nu=\nu_0$, then the RNP equation
$\frac{d\rho}{d\nu} = \rho^2 + \sigma^{ij}\sigma_{ij} + \frac{1}{n} R_{\mu\nu} l^\mu l^\nu$
  implies that the convergence $\rho$ will become infinite at a point $q$ within an affine parameter distance$\frac{1}{\rho_0}$ if the null geodesic can be extended that far.
if $\rho=\rho_0$ at $\nu=\nu_0$ then $\rho$ is greater than or equal to $\frac{1}{\rho^{-1} + \nu_0-\nu}$. Thus there is a conjugate point before $\nu=\nu_0 + \rho^{-1}$.

I dont understand many terms here. Firstly, what is affine parameter distance? And I am at loss as to how does one get the 2nd relation between $\rho$ and $\frac{1}{\rho^{-1} + \nu_0-\nu}$. How can you derive it? Frankly, I dont understand ANYTHING about how does this equation come, though I suspect it just the Frobenius theorem. Because that is how you get conjugate points in spacetime.
Please give me DETAILED asnwers, as I have mentioned before, I am not too comfortable with it. I dont understand anything in blockquotes other than the RNP equation.
Thanks in advance!!!
 A: This is a statement about a congruence of null geodesics.  We are looking for a conjugate point, which is just a place where the null geodesics cross each other.  The theorem is putting a bound on how far you can advance the affine parameter $\nu$ along the geodesics before the conjugate point occurs (this is what is meant by affine parameter distance).  
To derive the bound, you need to assume the null energy condition, which says that $R_{\mu \nu} l^\mu l^\nu \geq 0$ for all null vectors $l^\mu$.  You are also assuming that the geodesics you are working with are hypersurface orthogonal, which means the twist $\omega_{ij}$ vanishes, and doesn't contribute to the Raychaudhuri equation.  By noting also that the shear $\sigma_{ij}$ is a spatial tensor, so will have a positive definite norm, $\sigma_{ij}\sigma^{ij}\geq 0$, we find that this equation is saying
$$\frac{d\rho}{d\nu}\geq\rho^2.$$
This differential equation is easily solved, and you find that 
$$\rho\geq\frac{1}{\rho_0^{-1}+\nu_0-\nu}.$$
When $\nu-\nu_0=\rho_0^{-1}$, the RHS of this inequality goes to infinity, which means that $\rho$ diverged at some value of $\nu$ before that.  When $\rho$ diverges, it means there was a conjugate point.
