# Geodesic conjugate points

1. Aug 23, 2014

### cr7einstein

Dear all,
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 thhis equation come, though I suspect it just the Frobenius theorem.

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.

2. Aug 23, 2014

### WannabeNewton

Presumably by RNP you mean the Raychaudhuri equation, although the notations you have used are very non-standard and you seem to have redefined the expansion scalar to be its negation. Anyways, there isn't much detail involved. Just use the strong energy condition and the positive norm of the shear to get an differential inequality for the scalar $\rho$ and integrate to get the result. There should be a factor of $1/3$ that you seem to be missing, it follows from the Raychaudhuri equation. By the way this is a caustic not a pair of conjugate points.

Last edited: Aug 23, 2014
3. Aug 24, 2014

### cr7einstein

I have typed the exact same lines as that of the book.I know that the notation is different from the standard Raychauduri Equation, but this is because thiis the unified Raychaudhari-Newman -penrose(RNP) equation...as hawking puts it. the $$\frac{1}{3}$$ you are talking about is $$n=3$$ fot timelike curves. Here's a link to the pdf of the book : http://www.benpadiah.com/otherstuff/elib/HawkingNatureOfSpaceTime.pdf And can you please help me get the desired result from the strong energy condition??? I don't get the head or tail of how it can be done to get the positive norm of shear......

4. Aug 24, 2014

### WannabeNewton

For null geodesic fields you only need the weak energy condition, you don't need the strong energy condition.
The weak energy condition clearly implies that $R_{\alpha\beta}l^{\alpha}l^{\beta} \geq 0$ for any null vector $l^a$ (see p.8 of the pdf you linked).

Furthermore the shear has positive norm $\sigma_{\alpha\beta}\sigma^{\alpha\beta} \geq 0$ because it is space-like in both indices.

Therefore $\frac{d\rho}{dv} \geq \rho^2 \Rightarrow \frac{d\rho}{\rho^2} \geq dv \Rightarrow \rho^{-1}|_{\rho}^{\rho_0} \geq v - v_0 \Rightarrow \rho \geq \frac{1}{\rho_0^{-1} - v + v_0}$.