Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Radius of a black hole

  1. Feb 14, 2008 #1
    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?
     
  2. jcsd
  3. Feb 16, 2008 #2

    Jonathan Scott

    User Avatar
    Gold Member

    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".
     
  4. Feb 17, 2008 #3

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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, [itex]r[/itex] is a timelike coordinate inside the event horizon, (the singularity at) [itex]r=0[/itex] 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 [itex]r=r_1[/itex] and [itex]r=r_2[/itex], with [itex]r_2 > r_1[/itex], and with both observers having the same constant values of [itex]\theta[/itex] and [itex]\phi[/itex]. The observer at [itex]r_2[/itex] drops a calibrated plumb line (a tape measure!) down to the observer at [itex]r_1[/itex] such that the line ends up stationary with respect to both observers. How much tape does the observer at [itex]r_2[/itex] unwind?

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

    This is found by integrating the Schwarzschild metric with [itex]dt=d\theta =d\phi =0[/itex],

    [tex]
    ds = \left( 1-\frac{2M}{r}\right) ^{-\frac{1}{2}}dr,
    [/tex]

    from [itex]r_1[/itex] to [itex]r_2[/itex].

    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 [itex]c=1[/itex]).

    I gave the details for this here.

    Not that these two methods give different answers!
     
    Last edited: Feb 17, 2008
  5. Feb 14, 2009 #4

    Ove

    User Avatar

    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!
     
  6. Feb 14, 2009 #5

    Mentz114

    User Avatar
    Gold Member

    In the above M=Gm/c^2, where m is the mass.
     
  7. Feb 15, 2009 #6

    Ove

    User Avatar

    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.
     
  8. Feb 15, 2009 #7

    Mentz114

    User Avatar
    Gold Member

    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.
     
  9. Feb 16, 2009 #8
    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-

    [tex]\tau_{rain}(2M \rightarrow 0)=\frac{4}{3}M[/tex]

    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-

    [tex]\tau_{max}(2M \rightarrow 0)=\pi M[/tex]

    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 [itex]\Delta r[/itex] above the horizon of a black hole, the radial distance [itex]\Delta r'[/itex] to the event horizon with respect to the observer's local coordinates would be-

    [tex]\Delta r' =\frac{\Delta r}{\sqrt{1-\frac{2M}{2M+\Delta r}}}[/tex]

    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 [itex]r=C/2\pi[/itex]. In the respect of a rotating black hole-

    Coordinate radius at event horizon (r)-

    [tex]r_+=M+\sqrt{M^2-a^2}[/tex]

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

    [tex]R^2=\frac{\Sigma^2}{\rho^2}sin^2\theta[/tex]

    Where

    [tex]
    \]
    \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\\
    \[
    [/tex]

    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 [itex](R_+)[/itex] is equal to the event horizon radius for a static black hole [itex](R_s)[/itex] 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 [itex](r_+)[/itex]. This applies every time regardless of mass and spin.

    Incidentally, the same applies to the (inner) Cauchy horizon [itex]\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_-)[/itex] also equals the Schwarzschild radius (in the azimuth plane). This may be just a geometric curiosity.

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

    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-

    [tex]R^2=r^2+a^2+\frac{2\ Ma^2}{r}[/tex]
     
    Last edited: Feb 16, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?