# I Radius in Schwarzschild metric

#### kelly0303

Hello! I am a bit confused about the definition of the radius in Schwarzschild metric. In the Schutz book on GR (pg. 264, General rules for integrating the equations) he says: "A tiny sphere of radius $r = \epsilon$ has circumference $2\pi\epsilon$, and proper radius $|g_{rr}|^{1/2}\epsilon$ (from the line element). Thus a small circle about $r = 0$ has ratio of circumference to radius of $2\pi|g_{rr}|^{−1/2}$". I am a bit confused. Is the proper radius the one measured by a local observed (walking from the center to that radius) and the actual radius, the same distance measured by an observer at infinity? And the same for the circle circumference? Thank you!

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#### PeterDonis

Mentor
I am a bit confused about the definition of the radius in Schwarzschild metric.
It depends on the choice of coordinates; "radius" here is a coordinate, not a physical distance.

In the Schutz book on GR (pg. 264, General rules for integrating the equations) he says: "A tiny sphere of radius $r = \epsilon$ has circumference $2\pi\epsilon$, and proper radius $|g_{rr}|^{1/2}\epsilon$ (from the line element).
This means that Schutz is using the coordinate choice usually called "Schwarzschild coordinates", where the radial coordinate $r$ (what you are calling the "radius" although that is really a misnomer, see above) is defined so that the circumference is $2 \pi r$. And this definition requires that the proper radius ("proper distance from the center" would be a better term) is what Schutz says.

Is the proper radius the one measured by a local observed (walking from the center to that radius) and the actual radius, the same distance measured by an observer at infinity? And the same for the circle circumference?
An observer at infinity can't directly measure the distances. The proper radius (see above for a better term) and the circle circumference are the actual distances measured by a local observer.

#### kelly0303

It depends on the choice of coordinates; "radius" here is a coordinate, not a physical distance.

This means that Schutz is using the coordinate choice usually called "Schwarzschild coordinates", where the radial coordinate $r$ (what you are calling the "radius" although that is really a misnomer, see above) is defined so that the circumference is $2 \pi r$. And this definition requires that the proper radius ("proper distance from the center" would be a better term) is what Schutz says.

An observer at infinity can't directly measure the distances. The proper radius (see above for a better term) and the circle circumference are the actual distances measured by a local observer.
Thank you for your reply! I am a bit confused by " the radial coordinate $r$ (what you are calling the "radius" although that is really a misnomer, see above) is defined so that the circumference is $2 \pi r$". The circumference of what?

#### PeterDonis

Mentor
The circumference of what?
The circumference of the circle at radial coordinate $r$.

#### Ibix

$r$ is the circumference, $C$, of a circle divided by $2\pi$ and, roughly speaking, the proper radius (or proper distance to the centre) is the distance you have to walk to get to the centre.

On a flat piece of paper these are identical. But think of the rim of a bowl - the distance to the middle of the bowl along the surface is greater than $C/2\pi$. And think of the rim of the noisy end of a trumpet - the distance to the middle along the surface isn't defined.

In spacetime there are analogous issues. If you measure the circumference of a solid non-rotating sphere, then measures the distance through a narrow hole drilled through it, that distance will be slightly greater than $C/\pi$. And if you repeat the exercise with a spherical shell around a black hole, you cannot even define the "distance through the middle". If you can draw a sphere, though, you can always define $C/2\pi$, which is why it is used as a radial coordinate. For an infinitesimal sphere (and with caveats about the choice of coordinates, as Peter says) the proper distance to the middle is $\sqrt{|g_{rr}|}\epsilon$.

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#### pervect

Staff Emeritus
Hello! I am a bit confused about the definition of the radius in Schwarzschild metric. In the Schutz book on GR (pg. 264, General rules for integrating the equations) he says: "A tiny sphere of radius $r = \epsilon$ has circumference $2\pi\epsilon$, and proper radius $|g_{rr}|^{1/2}\epsilon$ (from the line element). Thus a small circle about $r = 0$ has ratio of circumference to radius of $2\pi|g_{rr}|^{−1/2}$". I am a bit confused. Is the proper radius the one measured by a local observed (walking from the center to that radius) and the actual radius, the same distance measured by an observer at infinity? And the same for the circle circumference? Thank you!
I don't have Schhutz, but that passage is certainly confusing. I'm not even sure what context it is in, since I don't own the book.

Consider a set of points of constant Scharzshilc coordinate "r" and, to be very specific, some constant Schwarzschild time coordinate "t". This set of points is two dimensional surface, and has exactly the same geometry as some sphere of radius r in 3 dimensional space in normal, flat, space-time. A point on the sphere is located by the $\theta$ and $\phi$ coordinates.

The surface area of this sphere is $4\pi r^2$ and the circumference of the sphere at the "equator" is $2 \pi r$.

However, the radial distance between such a sphere of radius r and another such sphere of radius r+dr, measured along the radial direction is given by the metric ds.

$$ds^2 = \frac{1}{1- \frac{r}{r_s}} dr^2$$

Thus we can write

$$ds = \sqrt{ \frac{1}{1- \frac{r}{r_s}} } dr$$

Here ds is the radial distance between the two spheres, and $r_s$ is the Schwarzschild radius, $r_s = 2GM / c^2$, M being the mass of the black hole.

Because the geometry of space-time is distorted, while it is perfectly fine to imagine that the points of some constant schwarzschild r coordinate have the same basic geometry as the points that form a sphere in a normal flat Minkowskii space-time, the formula for finding the radial distance between two such spheres is not the same as it is in flat space-time, because of the distortions introduced by gravity and one's particular choice of coordinates.

In particular, it is very confusing to talk about the "radius", especially since the above expression for ds becomes singular at $r=r_s$ and imaginary below it. I would therefore suggest not even thinking about the "radius" of the sphere, just it's circumference and surface area.

#### kelly0303

$r$ is the circumference, $C$, of a circle divided by $2\pi$ and, roughly speaking, the proper radius (or proper distance to the centre) is the distance you have to walk to get to the centre.

On a flat piece of paper these are identical. But think of the rim of a bowl - the distance to the middle of the bowl along the surface is greater than $C/2\pi$. And think of the rim of the noisy end of a trumpet - the distance to the middle along the surface isn't defined.

In spacetime there are analogous issues. If you measure the circumference of a solid non-rotating sphere, then measures the distance through a narrow hole drilled through it, that distance will be slightly greater than $C/\pi$. And if you repeat the exercise with a spherical shell around a black hole, you cannot even define the "distance through the middle". If you can draw a sphere, though, you can always define $C/2\pi$, which is why it is used as a radial coordinate. For an infinitesimal sphere (and with caveats about the choice of coordinates, as Peter says) the proper distance to the middle is $\sqrt{|g_{rr}|}\epsilon$.
Thank you for this. So in the bowl example, $r$ is the radius of the top circle (i.e. if you put an elastic membrane on the top, $r$ is the radius of the created disk). But the proper distance is the distance you have to walk to get to the center of the bowl (which would be the bottom part of it) and the distance, which is obviously greater, is given by $\sqrt{|g_{rr}|}r$. Is this right?

#### Ibix

Thank you for this. So in the bowl example, $r$ is the radius of the top circle (i.e. if you put an elastic membrane on the top, $r$ is the radius of the created disk).
Yes. The bowl is a 2d surface embedded in a 3d space, so you can interpret $C/2\pi$ as the radius of a circle in this manner. Spacetime, however, is not embedded in a higher dimensional space as far as we are aware. So $r$ doesn't have this physical interpretation in general relativity - it's simply the circumference of the circle divided by $2\pi$.
But the proper distance is the distance you have to walk to get to the center of the bowl (which would be the bottom part of it) and the distance, which is obviously greater, is given by $\sqrt{|g_{rr}|}r$. Is this right?
Actually, it's given by $\int\sqrt{|g_{rr}|}dr$ for the bowl and other such macroscopic entities. Schutz is ducking the integral by working with an infinitesimal-sized sphere.

And, again, this depends on a fairly specific choice of coordinates. Since Schutz is looking at stellar interiors (justifying $m(r)|_{r=0}=0$ as a boundary condition, in fact) and using the obvious Schwarzschild-esque coordinate system, it's fine here. Just be aware that the general expression is more complex.

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