# Interpreting the Schwarzschild metric

## Summary:

What a happens to physical distance in the radial direction as one approaches the surface of a Schwarzschild black hole?
As a photon falls radially toward the surface of a Schwarzschild black hole, dr/dt approaches zero. Does this mean that, from the viewpoint of a distant (Schwarzschild) observer, the photon slows down or that the distance covered by successive dr's is getting larger?

Related Special and General Relativity News on Phys.org
Ibix
It means the the coordinate velocity is decreasing. An external observer never sees anything cross the horizon, so the light's ##r## can never reach ##R_S## in finite Schwarzschild time ##t##. That's what ##dr/dt\rightarrow 0## is telling you.

Let me try a variation: Suppose I tried to measure the radial distance to the surface of the black hole by tying a stone to the end of a rope and lowering the stone until it disappeared inside the black hole.

I'm guessing that no matter how much rope I provided (including more rope than my measured circumference around the black hole divided by two pi),

— I would never see the stone disappear, and

— at any point I could start pulling the rope back up and retrieve the stone.

Is this correct?

Ibix
No. You will never see the rock disappear, but if you repeatedly lower and raise and lower a little further there will come a time where you get back a snapped rope. This will take a very long time for a very strong rope, but it's a finite distance to the event horizon (##\int_{R_S}^{R_0}\sqrt{g_{rr}}dr## from ##r=R_0##) so you can get the rock arbitrarily close in finite time.

pervect
Staff Emeritus
Let me try a variation: Suppose I tried to measure the radial distance to the surface of the black hole by tying a stone to the end of a rope and lowering the stone until it disappeared inside the black hole.

I'm guessing that no matter how much rope I provided (including more rope than my measured circumference around the black hole divided by two pi),

— I would never see the stone disappear, and

— at any point I could start pulling the rope back up and retrieve the stone.

Is this correct?
What happens is that the stress in the rope increases without limit as the stone approaches the event horizon. Eventually the rope breaks, as it's impossible for an object with rest mass to be stationary at the event horizon in the Schwarzschild metric. Only "objects" with zero rest mass, such as light, can be stationary at the event horizon.

You can take the mathematical limit as the object is lowered, though, even though the rope eventually breaks. I believe the limit is finite.

As I recall, the length of the rope between r=a and r=b for a black hole with a Schwarzschild radius ##r_s## is given by

$$r = \int_a^b \frac{dr}{\sqrt{1-\frac{r_s}{r}}}$$

While ##\frac{dr}{\sqrt{1-\frac{r_s}{r}}}## approaches infinity, the limit of the integral as r->##r_s## exists.

I

• Ibix
Ah, thank you both!

DrGreg
Gold Member
Eventually the rope breaks, as it's impossible for an object with rest mass to be stationary at the event horizon in the Schwarzschild metric.
Another way to put this is that for an object to hover at a constant height above the event horizon, it must have a proper acceleration upwards which diverges to infinity as the height drops to zero. (Or, in simpler language, its weight approaches infinity.) Of course no rope can provide an infinite force.

Yes, I have no doubt that the rope would eventually break. I was really wondering about the physical distance to the event horizon, or specifically about the relationship between the dr term and measuring distance. The fact that ds represent spacetime rather than space makes it a bit more subtle/confusing for me.

pervect
Staff Emeritus
Yes, I have no doubt that the rope would eventually break. I was really wondering about the physical distance to the event horizon, or specifically about the relationship between the dr term and measuring distance. The fact that ds represent spacetime rather than space makes it a bit more subtle/confusing for me.
For an observer outside the event horizon, it is generally accepted that dr represents the local notion of distance according to a co-located stationary observer at some specific point. A stationary observer in Schwarzschild coordinates is one with constant Schwarzschild coordinates. There are ways to make the notion of a stationary observer less dependent on coordinates, but it would be a digression to get into that, I think.

It's generally accepted that the way to define distance, according to a set of stationary observers, is to create a chain of said observers, each of which measures the distance to the next, and add the distances together. [change]. There is still the element of how to determine the path of course.

The reason we specify observers is that due to things like Lorentz contraction, we need to. Alternatively, we could explore the notion of "proper distance", I suppose. It seems simpler though to specify the observer for now.

Then dr represents "physical distance"for a stationary observer only outside the event horizon.

Stationary observers don't exist at or inside the event horizon.

Getting away from observers, for a moment, in terms of coordinates, and the associated mathematics, dr always represents an interval of one sort or another. The nature of this interval changes, though, at the event horizon.

At the horizon, dr is null. So it mathematically represents something we call a null interval. Two different events on the same light beam in flat spacetime is another example of a null interval.

Inside the event horizon, dr is timelike. So we can regard dr (and r) as measuring time, not space, inside the event horizon.

Thus the interval dr is changing its character depending on the position. Outside the event horizon dr is spacelike, right at the event horizon it's null, and inside the event horizon it's timelike.

Last edited:
Thanks pervect, for some reason I didn't get an email notification of your response. Now working on that definite integral from #4 and #5.

pervect
Staff Emeritus
If you want to get into the math, the transformation
$$r = r_s + \frac{R^2}{4r_s}$$

Eqivalently, we could write

$$R = 2 \sqrt{r_s \left(r-r_s \right) }$$

For events outside the horizon, with r>##r_s##, we can convert the line element to from dt,dr to dt, dR.

It's a somewhat long calculation, but the end result is

$$ds^2 = -\frac{R^2}{R^2 + 4\,r_s^2}\,dt^2 + \left(1 + \frac{R^2}{4\,r_s^2} \right) dR^2 \approx -\frac{1}{4\,r_s^2} \,R^2 \, dt^2 + dR^2$$

where the approximation is valid for R << r_s

The basic idea is that understanding the Schwarzschild metric near the event horizon is closely related to understanding the Rindler metric above the Killing horizon. You can get the result for the behavior through enough series expansions, but it may be simpler to make this transformation.

• snoopies622
PeterDonis
Mentor
2019 Award
As a photon falls radially toward the surface of a Schwarzschild black hole, dr/dt approaches zero. Does this mean that, from the viewpoint of a distant (Schwarzschild) observer, the photon slows down or that the distance covered by successive dr's is getting larger?
Neither. It means that the coordinates you are using are more and more distorted as the horizon is approached. Coordinate speeds in a curved spacetime have no direct physical meaning.

PeterDonis
Mentor
2019 Award
the interval dr is changing its character depending on the position. Outside the event horizon dr is spacelike, right at the event horizon it's null, and inside the event horizon it's timelike.
Note that these statements assume you are using Schwarzschild coordinates; they are not all true in other coordinate charts. For example, there are coordinate charts in which dr is spacelike everywhere.

pervect
Staff Emeritus
Note that these statements assume you are using Schwarzschild coordinates; they are not all true in other coordinate charts. For example, there are coordinate charts in which dr is spacelike everywhere.
Yes, I've acknowledged that I've been talking about the problem in a coordinate dependent way. I felt that trying to add coordinate independence to the discussion would be lengthly, and potentially confusing to the original poster.

But it is worth nothing that the Schwarzschild coordinates themselves are ill-behaved at the horizon. Some confusions can indeed be resolved by using better behaved coordinates.

To be more specific about the ill behavior of the Schwarzschild coordinates, some of the metric coefficients go to zero, an some of the metric coefficients go to infinity at the horizon. However, this problem, and the related problem that some points on the horizon never get coordinate label in the Schwarzschild coordinates while other points get multiple labels, can be resolved by choosing better behaved coordinates that don't have these issues.

2d polar coordinates are another example of a coordinate system that's ill-behaved. In polar coordinates the ill-bheavor occurs only at the origin. If we use coordinates ##r, \theta##, r=0 is the same point no matter what value ##\theta## takes. This makes some of the metric coefficients vanish. Unlike Schwarschild coodinates, polar coordinates don't have the issue of points existing in the manifold that never get assigned coordinate labels, however.

A better behaved set of coordinates for the Schwarzschild geometry would be for instance Kruskal-Szerkes coordinates.

I think the OP is trying to understand the Schwarzschild geometry better, to gain some intuition. But I'm not sure how to help them in this endeavor, though. Some of their confusion is probably related to the coordinates, but I've found in the past that introducing new coordinates rarely seems to help the sorts of confusion the OP is having. I think encouraging them to continue to calculate things mathematically is the best course.

Ibix
Anecdote disagreeing with pervect: I certainly found the Kruskal diagram, wherein lines of constant Schwarzschild coordinates are hyperbolae and their "radii", to be a huge help in understanding Schwarzschild spacetime and how the pathological Schwarzschild coordinates disguise interesting stuff.

PeterDonis
Mentor
2019 Award
I certainly found the Kruskal diagram, wherein lines of constant Schwarzschild coordinates are hyperbolae and their "radii", to be a huge help in understanding Schwarzschild spacetime and how the pathological Schwarzschild coordinates disguise interesting stuff.
That gives me a good excuse to link to my Insights article that discusses this: https://www.physicsforums.com/insights/schwarzschild-geometry-part-3/

This is the 3rd of a 4-part series; all of them might be useful for the OP to read.

DrGreg
Gold Member
I understood Schwarzschild coordinates and black hole event horizons a lot better after I'd studied Rindler coordinates and Rindler horizons.

The relationship between Rindler coordinates and Minkowski coordinates has many similarities to the relationship between Schwarzschild coordinates and Kruskal coordinates.

Good news! An anti-derivative calculator I found online gave me
$$\int (1-a/x)^{(-1/2)}dx = (ax(x/a-1))^{(1/2)}+a (arccosh(\sqrt(x/a))$$
which yields not only a finite radial distance to $r=r_s$ but one that's not even very large. Thanks again everyone, I am happy!

Last edited: