In chapter §84 of Classical Theory of Fields, they introduce the so called spatial metric to measure the distance $dl$ between two infinitesimally close points in space. They send a beam of light from one point, it goes to the second point and reflects back. They write the interval between the two events as:

$ds^2 = g_{ab}dx^adx^b + 2g_{0a}dx^adx^0 + g_{00}(dx^0)^2$

Here the indices $a$ and $b$ range from $1$ to $3$. Since we have a light beam $ds^2=0$. They solve the equation to get:

$dx^0 = \frac{1}{2g_{00}} \left( -2g_{0a}dx^a \pm \sqrt{(2g_{0a}dx^a)^2 - 4g_{00}g_{ab}dx^adx^b} \right)$

The negative root is the time from sending the beam to its reflection from the second point. And the positive root is the time from reflection to going back to the initial point. The time of travel from one point to another is then, half of the difference between the two roots:

$dt=\frac{1}{g_{00}} \left( \sqrt{(g_{0a}g_{0b} - g_{00}g_{ab})dx^adx^b} \right)$

The proper time in a reference frame standing still at the point where the light beam was sent from is $d\tau = \frac{\sqrt{g_{00}}}{c}dt$

Thus the distance $dl=cd\tau$ is

$dl^2 = \left( \frac{g_{0a}g_{0b}}{g_{00}} - g_{ab} \right)dx^adx^b = \gamma_{ab} dx^a dx^b$

The coefficients $\gamma_{ab}$ they call the spatial metric.

But then, they proceed to say it makes no sense to integrate the $dl$'s between two points, because the metric generally depends on $x^0$. But don’t you integrate along constant $x^0$? Thus making the dependence of the metric on $x^0$ irrelevant.

  • $\begingroup$ It's trickier than this. The spatial metric they use is not simply the spatial part of the ordinary metric. Look it up. $\endgroup$ Commented Jun 19, 2017 at 12:04
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    $\begingroup$ It's currently unclear what exactly this question is asking. Please add further information about the context so that potential answerers will know exactly what the issue here is even without having access to the relevant part of Landau-Lifshitz themselves. E.g. include what exactly this "so-called spatial metric" is defined as, and what "two infinitesimally close space-time points" means. $\endgroup$
    – ACuriousMind
    Commented Jun 19, 2017 at 12:09
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    $\begingroup$ Please give the equation numbers you're having problem with. $\endgroup$
    – SRS
    Commented Jun 19, 2017 at 12:18
  • $\begingroup$ I edited the question to make it accessible without the relevant text. $\endgroup$ Commented Jun 19, 2017 at 12:34

1 Answer 1


The relevant quotes from LL are:

We now determine the element $dl$ of spatial distance. In the special theory of relativity we can define $dl$ as the interval between two infinitesimally separated events occurring at one and the same time. In the general theory of relativity, it is usually impossible to do this, i.e. it is impossible to determine dl by simply setting $dx^0 = 0$ in $ds$. This is related to the fact that in a gravitational field the proper time at different points in space has a different dependence on the coordinate $x^0$.


However, we must remember that the $g_{ik}$ generally depend on $x^0$, so that the space metric $\gamma_{ab}$ also changes with time. For this reason, it is meaningless to integrate dl; such an integral would depend on the world line chosen between the two given space points. Thus, generally speaking, in the general theory of relativity the concept of a definite distance between bodies loses its meaning, remaining valid only for infinitesimal distances. The only case where the distance can be defined also over a finite domain is that in which the $g_{ik}$ do not depend on the time, so that the integral $\int dl$ along a space curve has a definite meaning.

The key point in the derivation is where you say that

$$dt = \frac{1}{g_{00}}\sqrt{(g_{0a}g_{0b}-g_{00}g_{ab})dx^a dx^b}$$ and then that $$d\tau = \frac{\sqrt{g_{00}}}{c} dt$$ $$dl = c d\tau$$

That makes sense for infinitesimally nearby events. However, to find the proper distance between two macroscopically separated events, we need to choose a spacetime curve along which to integrate $dl$. If the metric components are time-dependent, then different spacetime curves corresponding to the same space curve (when projected onto 3D space) give different values for the distance between the two points in question.

In other words, the distance between the two points would depend on more than just their coordinates - in which case the whole idea of a definite spatial distance loses its meaning.

Even if we did make a definite choice (such as to only integrate along worldlines of constant $x^0$), this still results in a "distance" between two specific points in space which changes in time - which again, is antithetical to the idea of a definite spatial distance.

  • $\begingroup$ In my opinion it makes absolutely no sense to integrate $dl$ along any other curve than the $t=\text{const}$ one. And also, even when you get a distance that is changing with time, I do not see how this is nonsensical. It is as nonsensical as the distance between two points on a rubber band changing when it is stretched. My question is: mathematically, are you allowed to integrate $dl$? $\endgroup$ Commented Jun 19, 2017 at 14:57
  • $\begingroup$ @RuskoRuskov As to your second point, two points on a rubber band physically move to different spatial coordinates when the rubber band is stretched. This is different - we're talking about static spatial coordinates, like (1,1,0) and (-2,1,5), which stay the same even while the distance between them changes. This means that if I give you the spatial coordinates of two points, you can't tell me the distance between them unless I provide you with more information. $\endgroup$
    – J. Murray
    Commented Jun 19, 2017 at 15:13
  • $\begingroup$ A bit off topic but... Imagine the rubber band has ruler markings, and this is your notion of coordinates. Anyhow, thank you! I think I got the gist of the problem. I may find the interpretation incorrect, but your answer was helpful. $\endgroup$ Commented Jun 19, 2017 at 15:18
  • $\begingroup$ @RuskoRuskov As to your first point, why should I restrict myself to the t=const curve? Imagine watching two people walk from point A to point B along exactly the same path. The first person runs the first half of the way and walks the rest, while the second person walks the first half of the way and runs the rest, so that they leave and arrive at precisely the same time. If the metric is time-dependent, then I would observe the two people having traveled different "distances" in general. This is nonsensical insofar as we're trying to define a meaningful idea of purely spatial distance. $\endgroup$
    – J. Murray
    Commented Jun 19, 2017 at 15:21
  • $\begingroup$ This is correct. A definition of distance, which is dependent on the worldline makes no sense to me too. But, again if you choose the curve $t=\text{const}$, then you get a notion of distance that is: 1 - unambiguous about the worldline taken ; and 2 - it carries a sense of the "distance" now. It may change over time (again the rubber band analogy), which would be against our weak euclidean intuition, but I think it is the best you can get. $\endgroup$ Commented Jun 19, 2017 at 15:30

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