General Relativity - Proper Time Suppose we are working with the signature (- + + +). Then $\mathrm{d}s^2=-1,1,0$ for timelike, spacelike and null curves respectively. We define proper time by $\mathrm{d}\tau^2=-\mathrm{d}s^2$. Suppose we have path $x^a$ parameterized by proper time. The book I'm reading states that $g_{ab}\dot{x}^a\dot{x}^b=-1,1,0$ for $x^a$ timelike, spacelike and null geodesics respectively. I can't immediately see how to derive this from the definitions. Could someone give me a hint? Thanks!
 A: The tensor $g$ is the bilinear form in (pseudo-)Riemannian geometriy and so $g(v,v)$ is the "norm squared" of $v$. That $g_{ab}\dot{x}^a\dot{x}^b=0$ for lightlike curves is then clear by definition of lightlike curves. And notice that in every point you can diagonalize the metric and then in the timelike and spacelike parts of the manifold you can rotate the vector by a Lorentz transformation $L\in \pm SO(1,3)$, such that $x^a$ takes the form $(x^0,0,0,0)$ or $(0,x^1,0,0)$ respectively. 
The second equation of this wikipedia page equals what you put in with what you ask for. The notation $\text{d}s^2=-1,1,0$ is not so nice imho. It also already implies the right parametrization. The norm to $1$ comes from the parametrization by proper time, i.e. you parametrize $\tau$ not by another parameter $t$ like $\tau(t),\ \text{d}(\tau(t))=\tau'(t)\text{d}t$, but you use $\tau$ iself. So the coefficient is simply $1$.
To put the two paragraphs together, the physical trajectories have always constant absolute norm, this is what characterizes them as things, which experience physical time. And $g(\dot{x},\dot{x})=-1$ holds in all frames if it holds in one. I emphasise this fact by using $g(X,Y)$ as often as possible, i.e. not writing $g_{ab}\dot{X}^a\dot{Y}^b$, even if you maybe mean abstract index notation anyway. In computations, the relation $g(\dot{x},\dot{x})=-1$ also becomes important for relations as $g(p,p)=-m^2$ for particles etc., where $p$ is its momentum and $m$ is the mass. The spacelike parametrization to $1$ is less usefull in applications, I'd think. The question has hardly anything to do with general relativity btw., apart from diagonalizing the metric, this is basically a special relativity question.
A: For timelike particles we define the proper time $d\tau^2=-ds^2=-g_{ab}dx^adx^b$. For spacelike particles we define the proper length by $dl^2=ds^2=g_{ab}dx^adx^b$. 
Now from theory we have $ds^2<0$ along timelike curves, $ds^2>0$ along spacelike curves and $ds^2=0$ along null curves. From standard properties of the Lagrangian we deduce that $g_{ab}\dot{x}^a\dot{x}^b=k$ constant. 
For null geodesics it's now easy to see $ds^2=0\Rightarrow g_{ab}\dot{x}^a\dot{x}^b=0$ independent of choice of parameterization. 
For timelike geodesics, parameterize by proper time, then $g_{ab}\dot{x}^a\dot{x}^b=\frac{ds^2}{d\tau^2}=-1$ by definition of proper time above. 
For spacelike geodesics, parameterize by proper length, and $g_{ab}\dot{x}^a\dot{x}^b=\frac{ds^2}{d\tau^2}=1$. 
In summary my confusion was simply a result of my trying to use proper time to parameterize an arbitrary geodesic, which is clearly nonsense! Hope this'll help someone else to understand the issue!
