In the Landau-Lifshitz why do they say it makes no sense to integrate the spatial distance between two points?

Question

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.

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 $x^0$.

and

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.