# Schwarzschild radius equals spatial stretch at surface?

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

• You'll have to explain what you mean by the stretching of space. I can't think of any sense in which this is related to the position of the event horizon. – John Rennie Aug 1 '17 at 18:38
• For how and why the curvature causes a gravitational acceleration see How does “curved space” explain gravitational attraction? – John Rennie Aug 1 '17 at 18:41
• I have seen in several places that the warping/curvature of space at the surface of the earth, caused by the mass of the earth, is about 0.88cm. And this number also is the calculated Schwarzschild radius for an object with the mass of the earth. I took the warping/curvature of space as the stretching of space since coordinate distances near a grav. mass are smaller than the proper distances. So it isn't a question about what is going on at the event horizon, but whether it is coincidence or not that the curvature at the surface of a grav. mass exactly matches to Rs calculation. Thanks. – John Nolan Aug 1 '17 at 21:00
• You'll need to provide a link to where you saw it said that the stretching of space at the surface of the Earth is $0.88$cm. On its own the statement is meaningless. – John Rennie Aug 2 '17 at 3:54
• askamathematician.com/2010/12/… – John Nolan Aug 2 '17 at 20:42

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.

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.

• Thanks very much, I appreciate your help. I'm afraid I went down a blind alley. I realize that curvature of space and curvature of time cannot be completely isolated from each other, especially for relativistic settings. But is a low mass system, like the solar system, isn't time curvature almost entirely responsible for gravity? I thought that in that situation the space curvature Rs/r might be distinguishable. How would one use that value of 4E06 to arrive at a distance quantity that might indicate the strength/severity of space curvature at the location? – John Nolan Aug 4 '17 at 20:27
• Last question (thanks for your patience), can it be said that gravity on Earth is due to gravitational time dilation which creates a time gradient radially away from the earth. The difference in the rate of time across an object induces, via conservation of energy, acceleration towards the center of the Earth. Thanks very much. – John Nolan Aug 4 '17 at 20:28
• @JohnNolan: kind of, but it isn't as simple as that. A full answer would be long enough to warrant asking it as a separate question. – John Rennie Aug 5 '17 at 4:34
• OK. I'll try to express it better than this question. Thanks. – John Nolan Aug 5 '17 at 15:19