Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
It may help to remember that proper time is an arc-length parametrization. – Neal May 18 '12 at 21:36
I'm finding it a bit confusing to see what the canonical definitions are, since a lot of articles (and books) appear to be quite circular! How would you define (a) timelike/spacelike/null geodesics, (b) proper time? Thanks! – Edward Hughes May 18 '12 at 21:41
First fix the signature of your metric. Say, $-1,1,1,1$. With this convention, a timelike vector $v$ has $g(v,v) > 0$, a spacelike vector $v$ has $g(v,v)<0$, and a null vector $v$ has $g(v,v) = 0$. A geodesic $\gamma$ is timelike/spacelike/null provided $\gamma'(t)$ is timelike/spacelike/null (need only for one time, since $\gamma'$ is parallel along $\gamma$). If $c$ is some curve, not necessarily geodesic, then proper time $\tau$ is arc-length along $c$. – Neal May 19 '12 at 19:19

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.

share|cite|improve this answer
Thanks Nick. Just one question: What would you take as the canonical definition for timelike, spacelike and null curves? Would you define proper time as simply the parameterisation which makes $g_{ab}\dot{x}^a\dot{x}^b=0,1,-1$? Many thanks! – Edward Hughes May 18 '12 at 21:33
@hgbreton: The wikipedia article on spacetime gives good elaboration. But it's geometrically clear: You have vectors at, in and outside of the cone at every point and these are light-, time- and space-like respecively. This doesn't even have to do with the lenght of the vector, all the vectors $v=const.\times(1,0.4,-0.1,0)$ are all timelike, independend of $const.$ and any vector, which is timelike can be written as $(v^0,0,0,0)$ in some frame of reference. Therefore, there are only three normal forms, which pointwise characterize spacetime. And proper time is only in the time direction. – NikolajK May 19 '12 at 0:15
So with this in mind is it consistent for me to take my definitions to be the following: $\mathrm{d}s^2=0$ for null particles $\mathrm{d}s^2>0$ for spacelike particles $\mathrm{d}s^2<0$ for timelike particles and define proper time $\tau$ to be the parameter such that $g_{ab}\dot{\mathrm{d}x^a}\dot{\mathrm{d}x^b}=0,1,-1$ for null, spacelike and timelike particles. We may always choose this by your diagonalization argument. How do I now regain the statement that $\mathrm{d}s^2=-\mathrm{d}\tau^2$? Many thanks! – Edward Hughes May 19 '12 at 10:23
@hgbreton: The expression with $\text{d}$ and dots over $x$ is strange, but the idea is right: With the dot's (i.e. without the $\text{d}$'s), $g(\dot{x},\dot{x})$, the equation as written holds. And the other way around (without the dots but with $\text{d}$), $g_{ab}\text{d}x^a\text{d}x^b$, it's just the line element squared $\text{d}s^2$. Once you choose a path $s\in M$ and then a parametrzation $s(t), t\in \mathbb{R}$, the line element becomes a one form w.r.t. one parameter t: $ds=\ ...=f(t)dt$. Proper time parametrization $s(\tau)$, or arc lenght $ds=\pm d\tau$, has trivial $f$. – NikolajK May 19 '12 at 14:03
@hgbreton: As a remark, I think the problem I have with your notation $ds^2=\pm 1$, i.e. "the infinitesimal squared equals some finite number" is just that $ds^2$ really denotes $g_{ab}\text{d}x^a\text{d}x^b$ and this is an object in some bundle, not a single real value. It's just that you can feed it a curve with a parametrization and then you get a number. With this abuse of notation in mind, yes $ds^2<0$ for timelike, $ds^2>0$ for spacelike, and $ds^2=0$ for lightlke curves (or rather along lightlike curves). – NikolajK May 19 '12 at 14:13
up vote 0 down vote accepted

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!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.