In the book The science of interstellar by Kip thorne can be found the following:

There is a maximum spin rate that any black hole can have. If it spins faster than that maximum, its horizon disappears, leaving the singularity inside it wide open for all the universe to see; that is, making it naked.

Can someone explain to me why and how a black hole spinning that fast would have its horizon disappear?


2 Answers 2


The spacetime geometry around a rotating uncharged black hole is described by the Kerr metric. I'll give this below, and it will look terrifying, but bear with me because there's only one small bit of the equation we need to see why the horizon disappears. Anyhow, the Kerr metric is:

$$\begin{align} ds^2 &= -(1 - \frac{r_s r}{\rho^2})dt^2 \\ &+ \frac{\rho^2}{\Delta}dr^2 \\ &+ \rho^2d\theta^2 \\ &+ (r^2 + \alpha^2 + \frac{r_s r\alpha^2}{\rho^2}\sin^2\theta)\sin^2\theta d\phi^2 \\ &+ \frac{2r_sr\alpha\sin^2\theta}{\rho^2}dt d\phi \end{align}$$


$$\begin{align} r_s &= 2M \\ \alpha &= \frac{J}{M} \\ \rho^2 &= r^2 + \alpha^2\cos^2\theta \\ \Delta &= r^2 - r_sr + \alpha^2 \end{align}$$

In the equation $J$ is the angular momentum of the black hole, $r$ is the distance from the centre of the black hole, $\theta$ is the latitude, $\phi$ is the longitude and $t$ is time. The parameter being calculated $ds$ is the total distance moved if you move by a distance $dr$ and angles $d\theta$ and $d\phi$ in a time $dt$.

Now, suppose we stay at some fixed angle relative to the black hole so $d\theta = d\phi = 0$, and we measure the distance along the radius to the centre of the black hole. We'll choose some fixed time $t$ for the measurement, so $dt = 0$. With all these restrictions the metric simplifies drastically to:

$$ ds^2 = \frac{\rho^2}{\Delta}dr^2 $$

and this is what we need to understand the behaviour of the horizon, because the event horizon radius is the value of $r$ for which the value of $ds^2$ goes to infinity. This happens when $\Delta = 0$, because then we get a division by zero. So to find the event horizon radius we just have to solve the equation:

$$ \Delta = r^2 - r_sr + \alpha^2 = 0 $$

and this is just a quadratic in $r$, like we all learned to solved at school. Using the quadratic formula the solution is (given by the larger root):

$$ r = \frac{r_s + \sqrt{r_s^2 - 4\alpha^2}}{2} \tag{1} $$

And the variation of the event horizon radius $r$ with $\alpha/r$ looks like:

Event horizon radius

Note that the line stops at $r/r_s = 0.5$ and $\alpha/r_s = 0.5$. The line stops here because beyond this point the equation (1) for $r$ has no real roots, and this means there is no event horizon. But remember that $\alpha$ is linked to the angular momentum $J$ by:

$$ \alpha = \frac{J}{M} $$

So for any value of the angular momentum $J > M$ there is no event horizon, and this is why the event horizon disappears when you spin the black hole too fast.

However there are good reasons to suppose that a black hole can never spin this fast, and the disappearance of the event horizon isn't real, but rather a sign that we've tried to apply the Kerr metric to a system that cannot physically exist. There is an article here (160KB PDF) analysing the physics, and the conclusions are that it is physically impossible to spin a black hole that fast.

  • 1
    $\begingroup$ It's also worth pointing out that if you take the other root in that quadratic formula, it has a physical meaning too, as the radius of the "inner horizon", a separate event horizon that would within the outer horizon of an ideal rotating black hole. The inner horizon is a Cauchy horizon, meaning that closed timelike curves (worldlines of objects traveling in a time loop) can pass through any point inside the inner horizon, but not points outside. $\endgroup$
    – Hypnosifl
    Nov 16, 2014 at 16:20
  • $\begingroup$ Also, the singularity inside the inner horizon is a "spacelike" one--unlike the "timelike" singularity in a non-rotating black hole, it isn't part of the inevitable future of anyone inside. Theoretically they could even fly around it and look at it--from their point of view it would be a "naked" singularity. So one way of thinking of about the math is that the radial distance between the outer and inner horizons shrinks as angular momentum increases, and for an extremal black hole it goes to zero, so there's no longer any "veil" between the naked singularity and the outside. $\endgroup$
    – Hypnosifl
    Nov 16, 2014 at 16:25
  • 1
    $\begingroup$ The Chappell paper isn't dated, but from the references it looks like it's from about 1998, which means it's extremely out of date. There has been a lot more work on this kind of thing recently by people like Hubeny and Poisson. Some more recent references show that the situation is far from settled: arxiv.org/abs/1211.3889 , arxiv.org/abs/1309.2027 $\endgroup$
    – user4552
    Nov 16, 2014 at 18:45

It all arises from the Kerr metric which describes rotating black holes. On the equation when the value of delta is zero we find the radius $$r_\pm=M\pm\sqrt{M^2-a^2}$$ where $a$ is the angular momentum. When $a$ is less than $M$, it results into a black hole with two horizons. When $a$ equals $M$, it results into an extreme Kerr black hole with one event horizon. When $a$ is greater than $M$, the radius of the equation will be an imaginary number and so we have no horizon hence a naked singularity. So this is what I think he means by the horizon disappearing.

  • $\begingroup$ Welcome to Physics! We do have MathJax here that makes your equations look better. You can see the notation page in help center for details, if you aren't familiar with it. $\endgroup$
    – Kyle Kanos
    Apr 11, 2019 at 17:11

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.