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The Schwarzschild metric states \begin{equation}\tag{I}ds^2=-(1-R_sr^{-1})(c dt)^2 + (dr)^2(1-R_sr^{-1})^{-1} + r^2(d\theta^2 + sin\theta d\phi^2)\end{equation}

For the purpose of this question, we will ignore $\theta$ and $\phi$ and assume \begin{equation}d\theta =d\phi = 0 \end{equation} We now have \begin{equation}\tag{II}ds^2=-(1-R_sr^{-1})(c dt)^2 + (dr)^2(1-R_sr^{-1})^{-1}\end{equation}where we may define

$ds$ is the invariant spacetime interval.

$dt$ is coordinate time, as measured by observer B, on their clock, at "infinity" (in flat space, out of range of the gravity of this object).

$c$ is the speed of light.

There is a circle of circumference $C >> R_s$ (the Schwarzschild radius), concentric with the spherical mass inside it. \begin{equation}r = C/(2\pi) \end{equation} There is another circle, with the same center, whose circumference is $C + dC$ ($dC>0$) \begin{equation}dr = dC/(2\pi)\end{equation}

The left-hand side of (II) can be expressed in terms of observer A, at the circle with circumference $C$, locally measuring the radial distance between the circles, and expanded to: \begin{equation}\tag{III}-(c dt')^2 + (dr')^2 =ds^2\end{equation} where t' and r' are measured by observer A. Hence, \begin{equation}\tag{IV}-(c dt')^2 + (dr')^2 = -(1-R_sr^{-1})(c dt)^2 + (dr)^2(1-R_sr^{-1})^{-1}\end{equation}

I have seen that one implication of the Schwarzschild metric is that when observer A measures the dr' radial distance between the circles, the result is the same as the flat-space measurement would be, except scaled by $\sqrt{(1-R_sr^{-1})^{-1}}$. That is: \begin{equation}\tag{V} (dr')^2 = (dr)^2(1-R_sr^{-1})^{-1}\end{equation} From (IV), that would imply $dt' = dt = 0$ for the measurement. I assume that the radial distance measurement may be thought of as two events, where $dt'=0$ but $dr'>0$.

My question is, how do we know that observer B will also see $dt=0$ for the two events that constitute the radial length measurement? I think that my question assumes that (V) is derived from (I). However, I am new to GR; perhaps (V) is derived first and (I) is derived partially from that?

Mauricio
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David
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2 Answers2

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You made the correct assumption that the manifold is locally Minkowskian.

It is similar to approximate a sphere by a tangent plane locally. Suppose someone at a given point of the surface of the Earth set local Cartesian axis, so that the y-axis points to the North and the x-axis point to the East.

For another observer using global coordinates, $dy = 0 \implies d\theta = 0$, where $\theta$ is the latitude. Then, the relation between $dx$ and $d\phi$ (longitude) is that $dx = Rsin(\theta)d\phi$. Note that the relation was a consequence of this particular choice of $x,y$ orientation.

Claudio Saspinski
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My question is, how do we know that observer B will also see dt=0 for the two events that constitute the radial length measurement?

From the description that you gave, this cannot be known. You did not fully specify the coordinate chart used by observer A. Since you did not specify A's chart then it is impossible to know if $dt'=0$ implies $dt=0$ or not.

However, it is certainly possible to define A's coordinate chart such that $dt \propto dt'$, meaning that A would use the same simultaneity convention as B. Coordinates are something that are chosen as a convention and defined according to our preference. If we define A's chart in that manner, then we know that observer B will also see $dt=0$, by definition.

An easier way to do this is to simply use the standard Schwarzschild coordinates, set $dt=d\phi=d\theta=0$ and calculate $ds^2$ as a function of $dr$. This will calculate the radial physical distance compared to the radial coordinate distance using a single chart.

Dale
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