Schwarzschild radius equals spatial stretch at surface?

Question

Am I right that the calculated Schwarzschild radius for the moon, Earth and Sun is also the calculated value for the "stretching" of space at the surface of the moon, Earth and Sun (.01cm, .88cm, and 2.96km, respectively)?

If so, I would have thought this would have been an interesting "factoid" to include in an explanation of what GR is, especially for beginners like me. If only to give a sense of how much time curvature contributes to weak-field gravity compared to space curvature.

Time curvature seems to be the whole ballgame for what we earthlings can perceive around us, and yet that also is not explained well in a lot of the introductory literature (an exception, Gravity from the Ground Up). It seems to me that saying that time passes slower at my feet than at my head, and that this gradient is what induces falling when I step off the diving board (I believe by conservation of energy in space-time?), would be accurate (in a weak gravitational field) and quite helpful to the novice.

I realize I may be way off, and so would be grateful to be corrected.

Thanks,

John Nolan

Answer 1

The geometry of spacetime around a spherically symmetric mass is described by an equation called the Schwarzschild metric:

$$ ds^2 = -\left(1-\frac{2GM}{c^2r}\right)c^2dt^2 + \frac{dr^2}{1-\frac{2GM}{c^2r}} + d\Omega^2 \tag{1} $$

And in flat spacetime the geometry is described by the Minkowski metric:

$$ ds^2 = -c^2dt^2 + dr^2 + d\Omega^2 \tag{2} $$

If we rewrite the curved spacetime equation (1) as:

$$ ds^2 = -A(r)c^2dt^2 + \frac{dr^2}{A(r)} + d\Omega^2 \tag{3} $$

where:

$$ A(r) = 1-2GM/c^2r $$

Then if you compare the two equations (3) and (2) you'll see that the difference is that factor of $A(r)$ i.e. if we take the curved space equation (3) and set $A(r) = 1$ we get the flat space equation (2). So you can get a guide as to how curved the spacetime is by evaluating $A(r)$.

A sidenote: you've used the phrase stretching of space. I realise this isn't meant literally, but for the record this factor is not simply the stretching of space because the phrase stretching of space is largely meaningless. The metric allows us to calculate how objects move in a curved spacetime but the metric is a complicated object and can't simply be thought of as how much space has stretched.

Anyhow, back to your question.

In equation (1) $G$ is Newton's constant, $c$ is the speed of light and $M$ is the mass of the object. To simplify the equations we often write:

$$ r_s = \frac{2GM}{c^2} $$

In which case our equation for $A(r)$ simplifies to:

$$ A(r) = 1 - \frac{r_s}{r} $$

And it turns out that $r_s$ is the black hole radius for a black hole with the same mass $M$ as our object. What the paper you cite says is that at the surface of the Sun the factor $r_s/r$ is about $4$ parts per million i.e.

$$ \frac{r_s}{r} \approx 4 \times 10^{-6} $$

What you are doing is taking this number and multiplying it by the radius of the Sun, so you are calculating:

$$\frac{r_s}{r} \times r = r_s $$

And unsurprisingly the result is $r_s$ i.e. the radius of a black hole with the mass of the Sun.

So the answer to your question is that the numbers you are calculating are indeed the black hole radii for black holes with the mass of the Sun, Earth and Moon. But this doesn't mean the stretching of space at the surface of an object is equal to its corresponding black hole radius.