Null geodesic eqaution / Timelike geodesic equation I'm taking my first course in GR. We were told in lecture that to find null geodesics, we can solve:
$$g_{ab} \dot{x}^{a} \dot{x}^{b} = 0,\tag{1}$$ 
and to find timelike geodesics, we can solve:
$$g_{ab} \dot{x}^{a} \dot{x}^{b} = -1 \tag{2}$$
[if the Minkowski sign convention is (-,+,+,+)].
I had 2 questions about this. 


*

*Is there an equivalent expression for finding spacelike geodesics?

*I am unable to see how these arise from the geodesic equation: $$\ddot{x}^{a} + \Gamma^{a}_{bc} \dot{x}^{b} \dot{x}^{c}= 0.\tag{3}$$ I thought all geodesics should be derivable from this equation. 

 A: So the condition you have been given is necessary but not sufficient.
If you think about Minkowski space, which is a valid GR manifold as well, one can imagine the simple helix$$[ct, R\cos(\omega t),R\sin(\omega t),0]$$ describing a uniform circular motion. This is not a geodesic of that space as that space has no gravity to sustain circular orbits. But it satisfies your first equation just fine. So your first equation is not sufficient for finding geodesics.
Your first equation may still be helpful for finding geodesics, because it may be necessary. This comes from the fact that the free particle Lagrangian is $\frac12 g_{\mu\nu} \dot x^\mu \dot x^\nu$ which leads directly to the geodesic equation as derived in any standard text or many places online. But if you think of it like a completely classical Lagrangian, it has no potential energy term, only the kinetic. A standard classical result is that the Lagrangian $T-U$ conserves the Hamiltonian $T+U,$ and probably you cannot use that classical derivation directly because it makes too many assumptions about the relationship of classical time and space... but there should be a relativistic analogue. The point is that 
$\frac12 g_{\mu\nu} \dot x^\mu \dot x^\nu=\text{constant}$ needs to be viewed as a conservation law satisfied by the geodesic, but not the geodesic itself. Then from the fact that your derivatives are with respect to the arclength do you get the $\pm 1$ normalization for this constant.
A: The expression $g_{ab} \dot{x}^a \dot{x}^b$ can be thought of as being equivalent to the Lagrangian $\mathcal{L}$. The geodesic equation can be be obtained by substituting $\mathcal{L}$ in the Euler-Lagrange equations.
$$g_{ab} \dot{x}^a \dot{x}^b  = 
    \begin{cases}
            -1, &         \text{if timelike },\\
            0, &         \text{if nulllike}, \\
            1, &          \text{if spacelike}.
    \end{cases}$$
Also, maybe this question will help.
