So that is my question. If I have a manifold with Lorentz metric, how do I know the dimension of the space of null geodesics. For example, in the general relativity the space of null geodesics is 5... I don't know how they get to this number. SO please, if some could help me that be great.

$\begingroup$ Each point in a spacelike slice (dimension 3) has a celestial sphere (dimension 2). $\endgroup$ – S. Carnahan♦ Dec 1 '13 at 9:35
For a Minkowski space $M^n$ the dimension of the null geodesics is $2n3$. It is naturally a contact manifold (for general Lorentzian space in general relativity you need a condition for the space of geodesics to be a manifold, but that's another story).
This is how you can see it: consider the space of all parametrized geodesics in $M^n$, that is the space of curves $t \mapsto x + tv$, this space is equivalent to $TM$ ($x$ and $v$), dimension $2n$. Now identify two lines $c : t \mapsto x + tv$ and $c' : t \mapsto x' + tv'$ having the same trajectory, that is the same image $\{x+tv \mid t \in {\bf R}\}$, that gives you the space of unparametrized geodesics. But two such lines have the same image iff $c'(t) = c(at+b)$ where $a \in {\bf R}$ and $a \neq 0$ and $b \in {\bf R}$. Thus these unparametrized geodesics are the orbits of the affine group ${\rm Aff}({\bf R})$. That gives you $x'=x+bv$ and $v' = av$. The quotient $TM/{\rm Aff}({\bf R})$ is equivalent to $TS^{n1}$. Now in $TM$ the space of null geodesics is the $2n1$ subspace defined by $v \cdot v = 0$ (hypersurface), the affine group acts on this subspace, and its action is still free, then the dimension of your quotient, the space of null unparametrized geodesics, is $2n3$. And indeed for $n=4$ you get 5, what you were looking for.
P.S. Long time ago, I wrote a small text on this construction that I never published. But I recently put it on line. It was a letter to a colleague of mine. You can find the manuscript here