In David Tong: Lectures on General Relativity in section 6.2.5 there's discussion of extremal black holes.

Metric takes the form:

$$ ds^2 = -\left(1 - \frac{GM}{r}\right)^2 dt^2 + \left(1 - \frac{GM}{r}\right)^{-2} dr^2 + r^2 d \Omega_2^2 $$

After that Tong calculate spatial distance:

After that he concluded:

Horizon of an extremal black hole lies at infinite spatial distance.

How understand this statement? Really, BH horizon in infinity? So BH is everywhere?

How to interpret this result?


enter image description here ($i^+$ is a typo, it should read $i^0$, too difficult to correct the picture :))

The text you quote is a bit sloppy but correct. A Carter-Penorse diagram could help to understand the structure of the spacetime.

A Killing horizon $H^+ \cup H^-$ is a null 3-surface separating the external spacetime from the region containing the singularity (which is timelike in the extremal Reissner-Nordstrom case as in the picture though a timelike singularity also appears for other types of also non-extremal black holes). It is made of the Killing time vector $\partial_t$ which becomes lightlike and tangent exactly to that surface. One is actually interested in the 2-spatial surface obtained as the intersection of the Killing horizon and a spatial slice $\Sigma_t$ orthogonal to the Killing time vector (which becomes the standard Minkowski time far from the horizon). This notion of space is natural, because thereon the spatial metric is stationary with respect to the Killing time (in the external region for standard black holes but in the whole spacetime for extremal ones). Usually this intersection called the Killing bifurcation, exists. For the standard Schwarzshild bh it is a spacelike 2-sphere. However dealing with extremal black holes, the said intersection is infinitely far from any point in the slice: the Killing bifurcation is out of the spacetime, it does not exist (that is one of the reasons why extremal black holes are claimed to have zero Hawking temperature). I indicated by $\infty$ that missed region in the picture.

There is an important remark however. The worldline of a massive particle (purple line in the figure) starting from the external region may cross the Killing horizon in a finite interval of proper time exactly as it happens for all non-extremal black holes. Another interesting fact: as the singularity is timelike a particle can avoid it after crossing the Killing horizon in the case of the extremal RN black hole I considered in the picture. That is impossible, e.g., in the Schwarzschild black hole but it is possible e.g. in the case of a rotating ("Kerr") black hole.

  • $\begingroup$ Beutiful answer! Could you give some references? $\endgroup$ – Nikita Nov 9 '20 at 23:08
  • $\begingroup$ Sorry, actually I don't remember a precise textbook about the Killing horizon of extremal horizons (maybe Wald's book says something but I am not sure). I suppose it is a part of my knowledge of GR I have learned in view of my research activity (part of my past research activity concerned QFT in curved spacetime, especially in bh background) and (also present) teaching duties. $\endgroup$ – Valter Moretti Nov 9 '20 at 23:23
  • 1
    $\begingroup$ Your remarks regarding the timelike nature of the singularity can currently be misinterpreted as suggesting that this is a feature unique to extremal black holes. Of course, this is feature is common to any charged (or rotating) BH. It may be worth clarifying this. $\endgroup$ – mmeent Nov 10 '20 at 10:38
  • 1
    $\begingroup$ It still kind of sounds like you are implying that the singularity of a non-extremal RN black hole is not timelike. $\endgroup$ – mmeent Nov 10 '20 at 10:54
  • 1
    $\begingroup$ I think now it is clear. $\endgroup$ – Valter Moretti Nov 10 '20 at 11:15

It means exactly what it says.

For any point outside the horizon, the distance to the horizon (as measured along a slice of constant $t$) is infinite.

Note that this statement depends on the choice of time coordinate, you can choose other coordinates for such an extremal spacetime such that the distance to the horizon at fixed time is always finite. (If you would follow a null curve, the distance would, of course be zero.)

  • $\begingroup$ So, where is horizon located? In infinity? Or near BH? $\endgroup$ – Nikita Nov 9 '20 at 15:44
  • 1
    $\begingroup$ @Nikita near the BH obviously. $\endgroup$ – mmeent Nov 9 '20 at 16:47
  • $\begingroup$ "near the BH" does not mean anything as the BH is a region of a spacetime... $\endgroup$ – Valter Moretti Nov 10 '20 at 10:23
  • $\begingroup$ @ValterMoretti Technically, the horizon is simply the boundary of the BH region. Hence, for any reasonable concept of "near" the horizon is "near the BH". $\endgroup$ – mmeent Nov 10 '20 at 10:33
  • 1
    $\begingroup$ In fact, it does not add any piece of information. Ah, "obviously" was used in that sense? A joke more or less. OK in that sense it has a meaning. $\endgroup$ – Valter Moretti Nov 10 '20 at 10:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.