# Geodesic vs null geodesic

I'm just wondering about the specific example in the textbook...
Sorry to be a bit scattered, I have a number of relativity texts and have been looking at them, just don't have the time today to go through them to find a close example. I thought the problem and the question made sense in general, as martinbn said:
Yes, but the question was in principle. If the point and the vector at the point are given, there is a geodesic through that point with the given vector as a tangent vector, for any metric.
This is what I thought martinbn meant in his initial post#2. My issue is with the null vector. I understand the meaning of "initial point"; what is the meaning of initial "null vector"? I would assume this is an initial "zero tangent". Is this a correct assumption? The expression for the zero tangent is
##g_{ij}\frac{dx^i}{d\lambda}\frac{dx^j}{d\lambda}=0## (the vector has zero length)
which is confusing, because in it I see derivatives of the coordinates w/respect to the parameter ##\lambda##, while the initial conditions given in that example were time derivatives ##\dot{r}=\dot{z}=0##.
I guess then we can still write
##g_{ij}\frac{dx^i}{d\lambda}\frac{dx^j}{d\lambda}=g_{ij}\frac{dx^i}{dt}\frac{dx^j}{dt}(\frac{dt}{d\lambda})^2=0##

But the vector is null, which is one more condition.
martinbn, could you explain the condition for the "null vector"...

martinbn