Finding Ratio Between Reduced Circumference & Radius of a Black Hole

• CJames
In summary, the concept of distance in general relativity is complex and depends on the observer's frame of reference and the method used to measure it. There are various methods, such as using rulers or light time, to measure the radial distance between two points. However, due to the warping of space, these methods may give different results and there is no one "actual radius". The ratio between the reduced circumference and the actual radius also depends on the frame of reference and the method used. Ultimately, the concept of distance in general relativity is not straightforward and can be challenging to define precisely.
CJames
Much of general relativity is expressed in terms of the r-coordinate, or the reduced circumference. That is, since space is warped, the actual radius will be different than the reduced circumference, which is found using C = 2 pi r. Does anybody know how to find the ratio between the reduced circumference and the actual radius?

CJames said:
Much of general relativity is expressed in terms of the r-coordinate, or the reduced circumference. That is, since space is warped, the actual radius will be different than the reduced circumference, which is found using C = 2 pi r. Does anybody know how to find the ratio between the reduced circumference and the actual radius?

There's a problem here with the concept of "actual radius". It's like asking for the distance on a flat map between two points on the Earth. It all depends on which map projection you use.

There are various possible methods of measuring radial distances, including for example rulers, light time, and other methods which relate to displacement within a background coordinate system (the "map"). However, none of these currently qualifies to be called the "actual radius".

CJames said:
Much of general relativity is expressed in terms of the r-coordinate, or the reduced circumference. That is, since space is warped, the actual radius will be different than the reduced circumference, which is found using C = 2 pi r. Does anybody know how to find the ratio between the reduced circumference and the actual radius?

As Jonathon Scott has said, the concept of spatial distance is somewhat problematic in general relativity. In this case, the spacetime inside the event horizon is not stationary, $r$ is a timelike coordinate inside the event horizon, (the singularity at) $r=0$ is a line "on" a spacetime diagram, not a point at the centre of a sphere, etc.

Here are the details for a couple of the methods that Jonathon mentioned.

Consider two observers outside the horizon that hover at constant $r=r_1$ and $r=r_2$, with $r_2 > r_1$, and with both observers having the same constant values of $\theta$ and $\phi$. The observer at $r_2$ drops a calibrated plumb line (a tape measure!) down to the observer at $r_1$ such that the line ends up stationary with respect to both observers. How much tape does the observer at $r_2$ unwind?

$$\left[ r\sqrt{1-\frac{2M}{r}}+M\ln \left( r-M+r\sqrt{1-\frac{2M}{r}}\right) \right] _{r_{1}}^{r_{2}}$$

This is found by integrating the Schwarzschild metric with $dt=d\theta =d\phi =0$,

$$ds = \left( 1-\frac{2M}{r}\right) ^{-\frac{1}{2}}dr,$$

from $r_1$ to $r_2$.

Another method is for one hovering observer to bounce a light signal off a mirror held by the other observer, and to define distance between the observers as half the round trip light travel time (with $c=1$).

I gave the details for this https://www.physicsforums.com/showpost.php?p=928277&postcount=19".

Not that these two methods give different answers!

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George Jones said:
$$\left[ r\sqrt{1-\frac{2M}{r}}+M\ln \left( r-M+r\sqrt{1-\frac{2M}{r}}\right) \right] _{r_{1}}^{r_{2}}$$

This is found by integrating the Schwarzschild metric with $dt=d\theta =d\phi =0$,

$$ds = \left( 1-\frac{2M}{r}\right) ^{-\frac{1}{2}}dr,$$

from $r_1$ to $r_2$.

May I ask why c and G are left out in the calculations? I'm trying to solve a similar problem where I have to express actual length with terms of coordinate distance, so it would help me to understand!

May I ask why c and G are left out in the calculations?
In the above M=Gm/c^2, where m is the mass.

Mentz114 said:
In the above M=Gm/c^2, where m is the mass.

Okay thanks! How would you correctly derive the Newtonian approximation for the distance with this integral? I would only guess that the mass either would have to be relatively small or that the observers r1 and r2 stay far away from the mass.

How would you correctly derive the Newtonian approximation for the distance with this integral? I would only guess that the mass either would have to be relatively small or that the observers r1 and r2 stay far away from the mass.
If you go far enough away so M/r << 1 you're be in flat space and the distance is r1- r2. But I doubt if that solves your problem, which is - you haven't defined 'actual length'. There is 'proper length' which is defined operationally as George Jones has clearly explained.

CJames said:
Much of general relativity is expressed in terms of the r-coordinate, or the reduced circumference. That is, since space is warped, the actual radius will be different than the reduced circumference, which is found using C = 2 pi r. Does anybody know how to find the ratio between the reduced circumference and the actual radius?

George Jones has probably already answered the question but here's a few equations that you might find of interest-

In all cases a=J/mc and M=Gm/c2 Horizon to crunch distance for an object falling radially from rest at infinity (rain frame) for a static black hole-

$$\tau_{rain}(2M \rightarrow 0)=\frac{4}{3}M$$

divide by c to get the horizon to crunch time for an object falling radially from rest at infinity.

Horizon to crunch distance for an object falling radially from rest at the event horizon (drip frame) for a static black hole-

$$\tau_{max}(2M \rightarrow 0)=\pi M$$

divide by c to get the horizon to crunch time for an object falling radially from rest at the event horizon. On a slightly different note, for an observer hovering at a small Schwarzschild distance $\Delta r$ above the horizon of a black hole, the radial distance $\Delta r'$ to the event horizon with respect to the observer's local coordinates would be-

$$\Delta r' =\frac{\Delta r}{\sqrt{1-\frac{2M}{2M+\Delta r}}}$$

This shows that coordinate-wise an observer could 'hover' within 100 mm of the event horizon of a 3.7e+6 sol static black hole but from the observers perspective, the distance would appear to be ~33 km to the EH.

While Dr’ doesn’t give you a specific radial distance from a static observer to the singularity, it does at least give the proper distance from a static observer to the EH. This distance is relative to the velocity of the approaching observer and will reduce exponentially(?) the more the observer picks up speed, eventually reducing to radial distance=coordinate distance (i.e. dr’=dr) when the speed of the infalling observer matches that of the rain frame. This of course changes once the event horizon has been crossed.As stated, the reduced circumference for a static black hole is $r=C/2\pi$. In the respect of a rotating black hole-

Coordinate radius at event horizon (r)-

$$r_+=M+\sqrt{M^2-a^2}$$

But the reduced circumference (R) at a specific coordinate radius (r) is-

$$R^2=\frac{\Sigma^2}{\rho^2}sin^2\theta$$

Where

$$\] \Sigma^2=(r^2+a^2)^2-a^2\Delta sin^2\theta\\ \\ \Delta= r^{2}+a^{2}-2Mr\\ \\ \rho^2=r^2+a^2 cos^2\theta\\ \[$$

Which is related to the frame-dragging effect (you can see the equation for the reduced circumference expressed clearly in some forms of Kerr metric). The reduced circumference (in the azimuth plane at least) at the event horizon of a rotating black hole $(R_+)$ is equal to the event horizon radius for a static black hole $(R_s)$ of the same mass which means the Schwarzschild radius is still related to the event horizon in a rotating black hole even though the EH appears to reduce coordinate wise $(r_+)$. This applies every time regardless of mass and spin.

Incidentally, the same applies to the (inner) Cauchy horizon $\left(r_-=M-\sqrt{M^2-a^2}\right)[/tex], regardless of how small the coordinate radius is (i.e. for a spin of a/M=0.1, the inner Cauchy horizon coordinate radius would be fairly insignificant compared to the outer event horizon) the reduced circumference of the Cauchy horizon [itex](R_-)$ also equals the Schwarzschild radius (in the azimuth plane). This may be just a geometric curiosity.

When $a$ reduces to zero, the equation for R reduces to $r=C/2\pi$.

It’s worth noting that the reduced circumference for a rotating black hole is not the proper distance from the observer to the singularity. In some hypotheses, mass inflation at the Cauchy horizon causes the curvature scalar to diverge at the IH so technically r=0 (in respect of proper distance) might be at the IH.

In respect of the azimuth plane only, the equation for R can be reduced to-

$$R^2=r^2+a^2+\frac{2\ Ma^2}{r}$$

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1. What is the formula for finding the ratio between reduced circumference and radius of a black hole?

The formula for finding the ratio between reduced circumference and radius of a black hole is C/R = 2π/GM, where C represents the reduced circumference, R represents the radius of the black hole, G is the gravitational constant, and M is the mass of the black hole.

2. How is the reduced circumference of a black hole calculated?

The reduced circumference of a black hole is calculated by dividing the actual circumference of the black hole by the square root of 1-((2GM)/(c^2R)), where G is the gravitational constant, M is the mass of the black hole, and c is the speed of light.

3. What does the ratio between reduced circumference and radius tell us about a black hole?

The ratio between reduced circumference and radius can tell us about the size and mass of a black hole. A larger ratio indicates a larger mass and a smaller ratio indicates a smaller mass. This can also give us an idea of the strength of the black hole's gravitational pull.

4. How does the ratio between reduced circumference and radius change as a black hole grows?

As a black hole grows, the ratio between reduced circumference and radius decreases. This is because the mass of the black hole increases, causing a larger gravitational pull and a smaller circumference. The radius also increases, but not as much as the mass, resulting in a smaller ratio.

5. What other factors can affect the ratio between reduced circumference and radius of a black hole?

The ratio between reduced circumference and radius can also be affected by the spin of a black hole. A rotating black hole has a different ratio than a non-rotating black hole. Additionally, the presence of other objects near the black hole can also affect the ratio as their gravitational effects can distort the space-time around the black hole.

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