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I am currently reading the Landua-Lifshitz' field theory book when I came across a concept I don't quite fully understand. In Section 100, pg $323$ (The Centrally Symmetric Gravitational Field) after introducing the Schwarzchild metric they go on to explain the geometric meaning of the spatial metric from obtained from the space time metric, i.e. \begin{equation} \label{eqn:100.15} d l^{2}=\frac{d r^{2}}{1-\frac{r_{g}}{r}}+r^{2}\left(\sin ^{2} \theta d \phi^{2}+d \theta^{2}\right) \end{equation} which I understand where they go this from, however, they then state:

$1)$"The geometrical meaning of the coordinate $r$ is determined by the fact that in the spatial metric (the equation given above) the circumference of a circle with its center at the center of the field is $2\pi r$.

$2)$ But the distance between two points $r_{1}$ and $r_{2}$ along the same radius is given by the integral \begin{equation} \int_{n}^{r_{2}} \frac{d r}{\sqrt{1-\frac{r_{g}}{r}}}>r_{2}-r_{1} \end{equation}

What doe they mean by $1)$, if I recall correctly, we couldn't find a radius vector $r$ which in non-Euclidean space (curvilinear coordinates) exemplified all the properties of the radius vector in Euclidean space, i.e. it is both the distance to the center of a circle and is the length of the circumference which is equal to $C/2\pi$, where C is the circumference. But, I guess this new $r$ satisfies the latter condition?

And for $2)$ did they get that integral from the $ds^{2}$ expression? Which I believe so. And what is the reason that the distance between two points in a gravitational field along the same radius is no longer just the difference between the two points, but now an integral taking to account what seems to be the path and the mass of the object within the field? Can anybody recommend some literature that may point towards solution of these two problems, as I would really appreciate it.

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  1. Consider the two-dimensional surface with $r=$ constant and $t=$ constant, which has the line element $$dl^2=r^2d\theta^2+r^2\sin^2\theta d\phi^2$$ where the metric coefficients are $$g_{ij}=\begin{pmatrix} r^2 & 0 \\ 0 & r^2\sin^2\theta \end{pmatrix}; \quad g=\det(g_{ij})$$ The proper area of this 2-sphere is $$\mathcal{A}=\iint\sqrt{g}\,d\theta d\phi=r^2\iint\sin\theta \,d\theta d\phi=4\pi r^2$$ This shows that $\theta$ and $\phi$ are the standard angular coordinates on a 2-sphere. Now we can obtain the radial coordinate $r$ as $$r=\left(\frac{\mathcal{A}}{4\pi}\right)^{1/2}$$ Thus, the coordinates $(r,\theta,\phi)$ have the natural interpretation in the Euclidean sense. Note that this is not always true for other black hole geometry, for example in the case of Kerr geometry the coordinate $r$ have a completely different interpretation.
  2. Now we again consider the metric, but with $\theta=$ constant and $\phi=$ constant. So the metric is $$dl^2=\dfrac{dr^2}{1-\dfrac{r_g}{r}}$$ from which we can easily obtain $$(l)_{\theta ,\phi\, =\text{constant}}=\int_{r_1}^{r_2}\frac{dr}{\sqrt{1-\dfrac{r_g}{r}}}$$ This is evidently different from what we obtain in the Euclidean case i.e., $(r_2-r_1)$. The reason for this is that in curved spacetime vectors transform differently at different locations in spacetime which you could easily observe while transforming vector differentials from one coordinate system to another, where you get the Christoffel connection terms. This is the same reason that the concept of parallel transport is introduced in general relativity. So the distance between $r_1$ and $r_2$ doesn't depend merely on the spatial interval between them, but also on the geometry of the spacetime.
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  • $\begingroup$ Thank you so much this was extremely helpful! $\endgroup$ – lastgunslinger Jul 18 at 8:16
  • $\begingroup$ Quick question however, why is it that you took the $\sqrt{g}$ and not just $g$ itself in the integration? Is it because taking it that way we get the desired solution of $4\pi r^{2}$ which is the total surface area of the sphere of radius $r$? $\endgroup$ – lastgunslinger Jul 18 at 8:23
  • $\begingroup$ The $\sqrt{g}$ is chosen because $\sqrt{g}d^nx$ is the invariant volume element in n-dimensional space in curved manifold and we need to integrate over this volume element. This is true for any integration in general relativity. In Euclidean space $\sqrt{g}=1$. However this has nothing to do with the desired solution, rather it is the most general convention. In some texts this factor might be taken as $\sqrt{-g}$ depending on the signature of the metric. $\endgroup$ – Richard Jul 18 at 10:19

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