# Local curvature of surface just outside Schwarzschild radius

1. Jun 6, 2010

### Jonathan Scott

Suppose one had a solid sphere just slightly larger than its Schwarzschild radius. What would the curvature of the surface look like to a local observer? Would it curve downwards, or appear flat, or curve upwards?

If my brain was working a bit better today, I'd calculate it myself from the Schwarzschild metric or look for it in my old notes, but I'm just hoping someone can save me the trouble. I have a suspicion, without calculating, that it probably approaches being flat, as I think that the curvature of space matches that of the surface. (At the photon sphere, further out, the curvature of space plus that of space-time matches the surface).

2. Jun 6, 2010

### DrGreg

I suspect you may be right. Are you familiar with Flamm's paraboloid? If not, follow the link. As you approach the event horizon the paraboloid surface approaches vertical, which seems to indicate that circles around the sphere become almost geodesic (in space, not spacetime).

3. Jun 6, 2010

### bcrowell

Staff Emeritus
Rays of light can orbit a black hole in (unstable) circular orbits, at R=3M (compared to the Schwarzschild radius R=2M). (These radii are expressed in terms of Schwarzschild coordinates, i.e., the circumference is defined to be $2\pi R$.) So if your sphere had R=3M, I think it would appear spatially flat to an observer on the sphere, in the sense that when he received light rays from points on the sphere, they would arrive from the direction perpendicular to the vertical.

4. Jun 6, 2010

### Jonathan Scott

Thanks, DrGreg, I was fairly familiar with Flamm's Paraboloid but I had forgotten that it actually has the same local shape as the Schwarzschild solution, and I think that confirms my suspicion.
I know what you mean (which is why I mentioned the photon sphere, which is the surface formed by those unstable orbits at R=3M). In an inertial frame, light paths match straight lines. However, that isn't how we normally define straight lines in a frame of reference which feels accelerated. When we are hovering or resting on a surface in a gravitational field, we effectively expect light beams to curve with the same downwards acceleration as bricks (because of the curvature with respect to time), and we do not see the curvature of space in our local rulers.

This means that if we try to construct a local cube with rulers, then relative to a typical external coordinate system the bottom of the cube will be smaller than the top and the sides will diverge a bit, and the lines of the top and bottom will be curved by the spatial curvature. Relative to local rulers, the curvature of the surface will therefore appear less locally than in the external coordinate system, and I specifically suspect that as one gets close to the Schwarzschild radius, the surface of a sphere will appear to become flat locally.

5. Jun 6, 2010

### DrGreg

I guess it depends on what you meant by "look like" in post #1: what you see with your eyes, in which case bcrowell's argument applies, or the geometry of the surface which you measure, in which case mine applies.

EDIT: written before I read post #4!

6. Jun 7, 2010

### bcrowell

Staff Emeritus
[As I try to preview this post, I see that I'm getting the annoying bug in the PF software where my equations are replaced randomly with other equations. If other people viewing this post see them looking incorrect, I can only suggest hitting the Quote button to see my latex source code. After "filling a certain range" below, the symbol should be a theta, but I'm getting it rendered as Gm/c2r.]

OK, to connect this to something we can compute with, consider the analogy with Gaussian curvature for a 2-dimensional space. One way of defining Gaussian curvature is this. From a point P, emit a fan of rays at angles filling a certain range $\theta$ of angles. Let the arc length of this fan at r be L, which may not be equal to its Euclidean value $L_E=r\theta$. Then the Gaussian curvature is $K=-3\frac{d^2}{d r^2} (L/L_E)$.

I believe this definition generalizes in the obvious way to higher dimensions, and what you're referring to is essentially the generalization to three dimensions $K = -d^2/d\rho^2(A/A_E)$, where $\rho$ is the spatial distance measured in the radial direction from some reference point (*not* equal to the Schwarzschild radius r), and $A_E=4\pi \rho^2$. (Because the metric is spherically symmetric, we can just let the fan of rays go out into the full $4\pi$ worth of solid angle. There might be some standard choice of the numerical constant out in front, analogous to the 3 in the 2-dimensional case.)

You could go ahead and calculate this for a Schwarzschild metric. I started, but it was a little messy. The relation between $\rho$ and r is $\rho=\int (1-2m/r)^{-1/2}dr$, which can be integrated but is pretty ugly. You can approach it without using the integration if you use the chain rule, a la $df/d\rho=(df/dr)/(d\rho/dr)$. I doubt that K has a zero at some specific radius, but I could be wrong.

7. Jun 7, 2010

### DrGreg

I got this wrong. Travelling along a geodesic doesn't tell you whether or not you are on a curved 2D surface, not when the 3D space is itself curved.

Actually the answer is very simple.

The metric of a concentric spherical surface in Schwarzschild coordinates is simply

$$ds^2 = r^2 (d\theta^2 + \sin^2 \theta \, d\phi^2)$$​

where r is the constant Schwarzschild r-coordinate of the surface. (Obtained by putting dt=dr=0 in the Schwarzschild metric.)

But that is exactly the same metric as a spherical surface in Euclidean space of Euclidean radius r. That surface certainly doesn't get flatter as r shrinks, the curvature increases.

8. Jun 7, 2010

### DrGreg

This is known bug that has been around for a few months now, which affects previews but not the final post. There is a workaround which isn't as well publicised as it could be -- after you press the "Preview" button, use your browser's command to "refresh" or "reload" the page. You'll be asked whether to resend the data, which you should agree to.

9. Jun 7, 2010

### yuiop

I find in MS Internet Explorer, I have to go into tools/internet options and delete all temporary internet files and cookies and then refresh. It is a bit of a pain.

10. Jun 7, 2010

### yuiop

If you are on the surface of a dense sphere such that your eyes are level with the photon sphere you would see the back of your head smeared horizontally on the horizon and the surface of the sphere would look pretty much like a flat plane. If you stand on a smaller even denser sphere such that your eyes are below r=3m then I suspect that the surface of the sphere will look curved the wrong way, i.e. it will look like you at the base of bowl rather than on a sphere.

Last edited: Jun 7, 2010
11. Jun 7, 2010

### bcrowell

Staff Emeritus
I think we have at least three different ways so far of defining the question:
1. Optically, as in my #3.
2. Variation of surface area with radius, as in Jonathan Scott's #4 and my #6.
3. According to the intrinsic curvature of the 2-surface, as you're doing above, in which case, as you've noted, the answer is trivial.

Thanks for the tip about the math rendering bug!

12. Jun 8, 2010

### Phrak

To all: Below the photosphere, would in not appear as if you were in the bottom of a bowl, so that some but not all of the rays arriving in your upper hemisphere of view (where straight up is radially outward) would intersect the blackhole? This deduction seems to be synonymous with the region between the event horizon and the photosphere.

It seems that at the surface of the photosphere, the event horizon would appear a flat plane, but this might depend upon your state of motion.

13. Jun 9, 2010

### Phrak

No replies usually means either I've said something 1) so inane, it's not worth responding to, or 2) have managed to be completely wrong in such a confusing way that no one can say quite why, or 3) have by blind luck stumbled upon the obvious.

In retrospect, I think that the photosphere appearing as a flat plane should only occur for an observer hovering at the boundary of the photosphere. Is this correct?

14. Jun 9, 2010

### Ich

It's http://en.wikipedia.org/wiki/Photon_sphere" [Broken], not photosphere. In the article, you'll find links that show that what you said is true. Non-stationary observers will see something different due to aberration.

Last edited by a moderator: May 4, 2017
15. Jun 9, 2010

### Jonathan Scott

Yes, "surface area with radius" is loosely the sort of curvature I was looking for.

Specifically, I was looking for the external curvature in the sense that if have two sticks which are each locally vertical a short distance apart, I should be able to measure the ruler distances between the top and bottoms of the sticks and hence determine the angle between them, and combine that with the lengths of the sticks to determine the effective radius of curvature in proper units. I think that Flamm's Paraboloid applies here and shows that the sticks become more and more parallel as the radius decreases towards the Schwarzschild radius.

I'm not interested in the intrinsic curvature, which could for example be measured in terms of how it affects a vector parallel transported around a small area.

16. Jun 10, 2010

### Phrak

Thanks, Ich.

Last edited by a moderator: May 4, 2017
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