1. Jun 22, 2008

### snoopies622

I am not clear on how $$|g_{rr}| \neq 1$$ in the Schwarzschild solution manifests itself. If I approach the event horizon of a black hole while holding a meter stick "vertically", that is, with $$d\theta =0$$ and $$d\phi =0$$ between its endpoints, will it appear to someone far away $$(r \gg r_{s})$$ that its length increases or decreases? Will the center of the black hole actually appear to recede from me as I approach it?

2. Jun 22, 2008

### lbrits

It's not good to compare lengths of things that are separated in space. For instance, when you say "appear", that implies that light from the meter stick has to reach you, and the endpoints travel different paths. Now you are probing a non-local measurement.

Probably the best way to think about it is to put a measuring tape around the black hole, with $$\theta = 0$$. You'll find that the length of the measuring tape required is always $$2\pi r$$, but $$r$$ is not the distance to the center. If you don't like going in the black hole, or event horizon, you could try it with the Earth. Of course, your measuring tape should be pretty accurate. So you could "measure" the distance from the surface to some height, and then "measure" the circumference. There will be a deficit.

3. Jun 22, 2008

### snoopies622

So as measured from far away,

$$\frac {2\pi r}{C}\approx 1$$

but as one approaches the event horizon this ratio increases without bound?

4. Jun 22, 2008

### lbrits

Yes. However, nothing really happens at the event horizon. It's just that $r$ is not a good coordinate to cover the whole space. Sometimes infalling coordinates or tortoise coordinates are used.

5. Jun 22, 2008

### George Jones

Staff Emeritus
Don't you mean "No."?

6. Jun 22, 2008

### snoopies622

Thank you for helping me once again, Ibrits. I will look into those coordinate systems since this is the first I've heard of them.

Edit: George Jones, I didn't mean to ignore your challenge; I simply posted this before I noticed it.

Last edited: Jun 22, 2008
7. Jun 22, 2008

### lbrits

Oops. Thanks George. I was a bit quick to respond. The circumference is always $$2\pi r$$. What I should've said is that the distance between two radial points is
$$R = \int_a^b \frac{1}{1-\frac{2M}{r}} dr$$,
which diverges as you take either point towards the horizon. You don't want the points straddling the horizon either, as you are in different coordinate patches then.

8. Jun 22, 2008

### snoopies622

Now, is that distance R as measured by someone in the immediate neighborhood of the points, or by someone outside the gravitational field altogether?

9. Jun 22, 2008

### lbrits

I think the point of this discussion is that the latter is not well defined. R that I calculated is simply the geometrical or arclength distance at constant $$t$$ between the two points. A person hovering above the horizon would slice spacetime differently than Schwarzschild coordinates. For an infalling observer, the distance that he/she measures is finite.

10. Jun 24, 2008

### snoopies622

I don't mean to be a pest here, but I still don't understand this. If "r" is not the distance between the edge of a circle and its center, what is it, and why call it "r"? Surely the length of a measuring tape running between two such points is well-defined.

11. Jun 24, 2008

### lbrits

It's called $$r$$ because when Schwarzschild wrote the metric, he used spherical coordinates to cover space. But coordinates merely assign numbers to points, and don't talk about distances between them. For instance, I can use polar coordinates on a cone, but circles won't have circumference 2pi r. That's the job of the metric.

The point is that far away from the mass, spherical coordinates "act like" ordinary spherical coordinates, because the metric is very closely equal to the Minkowski metric. But as you approach the mass, curvature becomes significant and Schwarzschild coordinates behave funny.

You are correct that the length of tape is well defined. The problem is that r is not the length of tape, so it needn't be well defined. The reason distance blows up as r approaches the horizon actually has to do with time. The "time" as viewed by a stationary observer at infinity, which is probably what you think of as time, cannot be extended to the other side of the horizon. In fact, that coordinate becomes spacelike inside. So when you take your measuring tape from outside and cross the horizon, you are travelling an infinite distance in time to get there. This sounds a bit mumbo jumbo but that's because it is difficult to understand in this coordinate system.

There are better coordinate systems to use. For instance, Eddington-Finkelstein or Kruskal coordinates or Panleve-Gulstrand coordinates. Most of these are based on the notion of "infalling observers", which have well-defined behaviour crossing the horizon.

12. Jun 24, 2008

### MeJennifer

I find these very instructional.

13. Jun 24, 2008

### snoopies622

Ah. Now I understand. When I asked about changes in $$2\pi r/C$$, I was thinking of 'r' as the distance to the center of the coordinate system. Hence my confusion.

Thanks again.