I think I may have some fundamental misunderstanding about what $dt, dx$ are in general relativity.

As I understand it, in special relativity, $ds^2=dt^2-dx^2$, we call this the length because it is a quantity that is invariant under Lorentz boosts. If a ball is moving in space and I want to calculate the $ds$ for the ball to travel from point A to point B, then $d\tau=ds$ (where $\tau$ is proper time) because according to the ball, $dx=0$, since point A and point B are at the same place in its own frame.

Assuming this is correct, this is all fine with me. Now fast forward to general relativity. $ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$...so if a ball is travelling in spacetime under whatever metric, then to me it would seem like, in the ball's frame, we should set $dx^i=0$, and then we get $ds^2=g_{00}d\tau^2$...however, from what I am told and from what I have read, $ds^2=d\tau^2$. When I look at the Schwarzschild metric, for example, $g_{00}$ doesn't appear to be 1 in a travelling ball's frame.

What have I misunderstood about the interval $ds$?. Is it that $ds$ only purely has to do with geometry of spacetime, and doesn't quite represent the distance between events?

  • $\begingroup$ I'm not sure what "only purely has to do with geometry of spacetime, and doesn't quite represent the distance between events?" is supposed to mean, but quite apart from that: the time component of the metric is generally written $\mathrm{d}s^2 = g_{00} \mathrm{d}t^2+\cdots$, reserving the symbol $\tau$ for the proper time, which is given by $\mathrm{d}s^2=\mathrm{d}\tau^2$ (in your metric sign convention). In a frame comoving with an object the time coordinate is the proper time and you will have $g_{00}=1$. $\endgroup$
    – Michael
    Commented Feb 20, 2013 at 23:05
  • $\begingroup$ If you have coordinates such that $\mathrm{d}s^2 = g_{00} \mathrm{d}t^2 + \cdots$ then the time experienced by an observer whose $x,y,z$ are constant is $\tau = \int \sqrt{g_{00}} \mathrm{d}t$. $\endgroup$
    – Michael
    Commented Feb 20, 2013 at 23:08
  • 1
    $\begingroup$ @MichaelBrown that could be the basis of an answer ;-) $\endgroup$
    – David Z
    Commented Feb 20, 2013 at 23:17

2 Answers 2


The problem arose when you wrote $ds^2 = g_{00} d\tau^2$. Generally one of your coordinates $x^\mu$ will be timelike, and the others spacelike, but the timelike one is not in general the proper time of someone whose spatial coordinates are not changing. That is, $t \neq \tau$. Using your sign convention,1 $d\tau^2 = ds^2$, so the (arbitrarily large) lapse in proper time between two events A and B is $$ \Delta\tau = \int\limits_\text{path} \sqrt{ds^2} = \int_{\sigma_\mathrm{A}}^{\sigma_\mathrm{B}} \sqrt{g_{\mu\nu} \frac{\mathrm{d}x^\mu}{\mathrm{d}\sigma} \frac{\mathrm{d}x^\nu}{\mathrm{d}\sigma}} \ \mathrm{d}\sigma. $$ Here $\sigma$ is any parameter that parametrizes the specific path (required to be timelike) taken from A to B, and your location along the path at parameter $\sigma$ has coordinates $x^\mu(\sigma)$.

Suppose $x^i \equiv 0$ along the path. Also call $x^0$ by the name $t$. Then most of the terms in the sum vanish and we have $$ \Delta\tau = \int_{t_\mathrm{A}}^{t_\mathrm{B}} \sqrt{g_{00}} \ \mathrm{d}t. $$ In fact we could have gotten this just by examining $$ d\tau^2 = g_{\mu\nu} \mathrm{d}x^\mu \mathrm{d}x^\nu. $$

Now often in GR we do call that timelike coordinate $t$, as I have done. Usually when this is done, as in the Schwarzschild metric $$ ds^2 = \left(1 - \frac{2M}{r}\right) \mathrm{d}t^2 - \left(1 - \frac{2M}{r}\right)^{-1} \mathrm{d}r^2 - r^2 \left(\mathrm{d}\theta^2 + \sin^2(\theta) \ \mathrm{d}\phi^2\right), $$ that $t$ becomes arbitrarily close to the $\tau$ of a local observer "at rest" in these coordinates as you move into the flat regime. In this case, as you move away from the mass at small $r$, $t \to \tau$. In other words, you can only neglect the $g_{00}$ term in some asymptotic cases (which you'll note correspond to $g_{00} \to 1$). In your SR example, you happened to use coordinates that matched those of the observer whose proper time you cared to measure, but you could have boosted to a new frame, in which case the $\mathrm{d}t^2$ part of the metric would have had some prefactor involving $\gamma$.

1 I will point out that most often in GR proper, $ds^2$ is negative for timelike separations, with $d\tau^2 = -ds^2$. The convention you adopted is more common in particle physics.

  • $\begingroup$ What's going on with the index mismatch in the equation $\Delta\tau = \int\sqrt{g_{\mu\nu}} dt$? $\endgroup$ Commented Feb 21, 2013 at 0:32
  • $\begingroup$ @joshphysics oops $\endgroup$
    – user10851
    Commented Feb 21, 2013 at 1:20
  • $\begingroup$ I think I get it. The $dt, dr, d\theta, d\phi$ in the Schwarzschild metric refer to $(t,r,\theta,\phi)$, which are the coordinate functions of your chart. These coordinates don't cover the whole spacetime, but at every point in spacetime there is a neighbourhood homeomorphic to $\mathbb{R}^4$, and in each neighbourhood we can use those coordinates so that the metric takes that form. But in general $(t,x,y,z)$, don't have to represent time and space, so unless you happen to be using a coordinate system where t is the time of an observer, you can't set dx=0 to find $d\tau$. Is that right? $\endgroup$
    – JLA
    Commented Feb 21, 2013 at 6:49
  • $\begingroup$ @JLA Right. And $t$ in these coordinates is the proper time only for an "observer at infinity" ($r\to\infty$), at which point $r$ also behaves more like proper distance to the center. $\endgroup$
    – user10851
    Commented Feb 21, 2013 at 6:55
  • $\begingroup$ One last thing...will an observer necessarily be at rest in his own frame, if he uses coordinates (to describe the spacetime around him) that make the metric take that form? By at rest, I mean his spacelike coordinates will be zero. $\endgroup$
    – JLA
    Commented Feb 21, 2013 at 16:47

Expanded on Michael Brown's comments, you seem to have confused $dt$ with $d\tau$. They are not the same. It is true that $ds^2 = dt^2$, but also true that $ds^2 = g_{tt} dt^2 + \ldots$ (other terms omitted).

What this means physically is that, even for two events that take place at the same location but different coordinate times (according to a given frame), there is a difference between elapsed coordinate time and actual (or proper) elapsed time.

This can happen in special relativity, too. Choose a time coordinate $\bar t$ such that $dt = \alpha \, d\bar t$, and the metric would look like $ds^2 = \alpha^2 \, d\bar t^2 = d\tau^2$. What this should emphasize is that one's choice of time coordinate can end up being quite poor, not corresponding well (if at all!) to the physically meaningful quantity, proper time.

While it may be confusing for the time coordinate to not correspond exactly to proper time, it nevertheless gives an enormous amount of freedom to recast the metric in a way that is convenient--you do not have to have proper time be your time coordinate if you don't want it to be, and it will always have a well-defined expression regardless of your choice of coordinates, as all physical quantities must have.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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