Penrose's Cosmic Censorship Hypothesis doesn't hold for naked singularities which means that at least light can escape the singularity. But, if we calculate escape velocity with the given mass only, light shouldn't escape.

How to calculate real escape velocity of naked singularities? Also, tell me if there's a physical sense to that other than a mathematical solution of General Relativity.

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    $\begingroup$ What makes you think that a naked singularity should have a well-defined escape velocity? By the simplest definition of a naked singularity, the big bang singularity actually qualifies. (Some extra work is required in order to define a naked singularity that forms by gravitational collapse from nonsingular initial conditions.) Even in Newtonian gravity, escape velocity depends on where you start -- how far from the gravitating object. $\endgroup$
    – user4552
    Nov 16, 2014 at 16:07
  • $\begingroup$ @Ben 1. Big Bang isn't a naked singularity by definition of naked part. There was nothing outside which could really see it. 2. Does Escape Velocity really need a definition? 3. When start point isn't given, it's understood that you are escaping from the surface. In case of Singularity, escape from the point. $\endgroup$ Nov 16, 2014 at 16:14
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    $\begingroup$ For the definition of a naked singularity, see Penrose, "Gravitational Collapse," 1973, adsabs.harvard.edu/full/1974IAUS...64...82P $\endgroup$
    – user4552
    Nov 16, 2014 at 20:44
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    $\begingroup$ ...probably can't form a naked singularity without exotic matter, so a negative mass wouldn't necessarily be surprising.) $\endgroup$
    – user4552
    Nov 17, 2014 at 21:06
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    $\begingroup$ @BenCrowell: all your comments would make a good answer. Given the vagueness of the question I think an answer that discusses the issues, rather than just give a formula, is entirely reasonable. I'd upvote it :-) $\endgroup$ Nov 18, 2014 at 17:29

3 Answers 3


A quick note in the light of some of the comments: I'm interpreting the question to be asking about the escape velocity from a black hole containing a naked singularity rather than the escape velocity from the singularity itself. The escape velocity at the singularity is undefined as GR cannot describe the geometry at that point.

Anyhow, a convenient way to describe the escape velocity from a black hole is to write the metric using the Gullstrand-Painlevé coordinates. In these coordinates spacetime is flowing inwards towards the black hole, and the escape velocity is simply the velocity of the inflow. This is commonly known as the River Model, because the analogy is with objects being swept along by a flowing river. For the brave, the details are given in the paper The river model of black holes.

The river model can be used to describe rotating black holes like Gargantua, but the maths involved is scary hard. However there is another class of black hole that can have naked singularities, and that's the charged non-rotating black hole described by the Reissner-Nordström metric. This is a great deal simpler, and the escape velocity can be easily calculated.

So, if you're happy for your naked singularity to be changed and non-rotating, instead of uncharged and rotating, here's how to calculate the escape velocity.

In Gullstrand-Painlevé coordinates coordinates the inflow velocity of a Reissner-Nordström black hole as a function of radial distance $r$ is (see the River Model paper for details):

$$ v^2 = \frac{2G}{r} \left (M - \frac{Q^2}{8\pi\epsilon_0 c^2 r} \right) $$

Since we're only interested in the general behaviour I'll convert this to geometrised units, and normalise the Schwarzschild radius to unity. This simplifies the equation to:

$$ v^2 = \frac{1}{r} - \frac{Q^2}{r^2} $$

In these units the extremal black hole has a charge of $Q = 0.5$, so $Q \lt 0.5$ looks from outside to be ordinary black hole with an event horizon and $Q \ge 0.5$ is a naked singularity. If I graph $v^2$ (you'll see why I graph $v^2$ in a moment) against $r$ for a range of different charges the results look like:

Escape velocity

The red line is an uncharged black hole, and as we expect the escape velocity goes to $1$ (i.e. $c$) at $r = 1$ (i.e. $r = r_s$).

The green line is for $Q = 0.4$, and the escape velocity goes to $c$ at $r = 0.8r_s$, so the event horizon has contracted a bit. However if you look at the behaviour inside the horizon the escape velocity rises then falls again and returns to $c$ at $r = 0.2r_s$. This is the location of the inner event horizon.

For the extremal black hole, $Q = 0.5$, the escape velocity rises to $c$ at $r = 0.5r_s$ but then falls again. There is a single horizon at $r = 0.5r_s$.

Finally for the naked singularity $Q = 0.6$, the escape velocity never reaches $c$ so there is no horizon.

However something rather odd happens at small $r$ for all the charged black holes. $v^2$ falls to zero then goes negative. A negative value of $v^2$ means the escape velocity is imaginary. This is generally interpreted as meaning that the Reissner-Nordström metric ceases to be physically meaningful at smaller values of $r$.

So, subject to worries about the behaviour at small $r$, it's straightforward to calculate the escape velocity and it doesn't do anything particularly weird. In principle the same calculation can be done for a rotating black hole, but as I said out the outset it's a big jump in difficulty so I shall leave it to Kip Thorne.

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    $\begingroup$ If I'm understanding the analysis correctly, doesn't this mean that for this type of naked singularity, this analysis fails to determine an escape velocity from the singularity, which I assume is at $r=0$? $\endgroup$
    – user4552
    Nov 18, 2014 at 23:35
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    $\begingroup$ In the non-naked $Q=0.4$ case, is it really meaningful to talk about escape velocities at $r<0.2r_s$? For example, a ray of light from this region can't actually escape to infinity, can it? Wouldn't it would rise outward but fall back in before reaching the inner horizon? $\endgroup$
    – user4552
    Nov 18, 2014 at 23:37
  • $\begingroup$ @BenCrowell: I've been scratching my head trying to work out if the imaginary escape velocity has any physical meaning. I suspect it means nothing can approach the singularity closer than the point where $v^2 = 0$ i.e. beyond that point the gravity from electric field outside is too strong. If you can shine any light on this I'd be very interested. $\endgroup$ Nov 19, 2014 at 6:43
  • $\begingroup$ @BenCrowell and others: I assume Sachin meant escape velocity from black hole with a naked singularity rather than the escape velocity from the singularity itself. The latter is of course undefined. $\endgroup$ Nov 19, 2014 at 6:45
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    $\begingroup$ Imaginary escape velocity simply means that particles placed there will shoot outward to infinity (an another universe, perhaps?). The RN singularity is REPULSIVE. That being said, anything within the inner horizon isn't thought to be RN geometry, as "mass inflation" results. $\endgroup$ Dec 18, 2014 at 1:05

The local escape velocity (relative to a ZAMO) of an angular momentum free test particle in the vicinity of a naked singularity with spin a and charge ℧ can be derived with the Kerr Newman metric by setting the total conserved specific energy of the test particle (in natural units of G=M=c=kc=1)

$${\rm E} = g_{\rm t t} \ \dot {\rm t}+g_{\rm t \phi} \ \rm \dot \phi+\rm q \ A_{t}= $$

$$= \rm \left(1-\frac{2 r-\mho ^2}{\Sigma }\right) \dot{t}+ \frac{a \sin ^2 \theta \left(2 r-\mho ^2\right)}{\Sigma } \ \dot{\phi}+q \ \frac{\mho \ r }{\Sigma} = $$

$$= {\rm \sqrt{\frac{\Delta \ \Sigma}{(1- \mu^2 v^2) \ \chi}} + \omega \ L_z+ q \ \frac{\mho \ r }{\Sigma}}$$

to 1 and the angular momentum Lz to 0, and solve for the the local velocity v. Then we get

$$\rm v_{esc}=v_{r}= |\frac{\sqrt{a^4 \cos ^4 \theta \ \zeta +2 a^2 r \cos ^2 \theta (q \chi \mho + r \ \zeta )+r^2 \left(-q^2 \chi \mho ^2+2 q r \chi \mho +r^2 \ \zeta \right)}}{\sqrt{\chi } \left(a^2 \cos ^2 \theta +r (r-q \mho )\right)}|$$

where the abbreviated terms are

$$\rm \Sigma =a^2 \cos ^2 \theta +r^2, \ \Delta =a^2+r^2-2 r+\mho ^2, \ \chi =\left(a^2+r^2\right)^2-a^2 \Delta \sin ^2 \theta , \ \zeta = \Delta \Sigma -\chi $$

and q is the specific charge of the test particle. That holds for neutral and oppositely charged test particles all the way down an infinitesimal above the singularity (on the singularity itself it is, like John Rennie already mentioned, undefined due to the division by 0, but you can take the limit, which can be anything from 0 to c, depending on the spin and charge), while for test particles with the same charge as the black hole or naked singularity the optimal configuration for the test particle contains angular momentum.

It might be that there are some shorter equations that can do the same, I let Mathematica solve it for me which doesn't always give the shortest possible form. The analytical solution for the required vertical launch angle for charged test particles is too long to post here, but it can be found here. For neutral test particles it would simply be 90° (locally, relative to a ZAMO), while chargend particles need a local velocity along the φ axis in order to have zero axial angular momentum because of the qAφ term in the equation for Lz

$${\rm L_z} = |g_{\phi \phi}| \ \dot \phi - |g_{\rm t \phi}| \ \rm \dot t - q \ A_{\phi}=$$

$$= {\rm \frac{\dot \phi \ \chi \sin ^2 \theta }{\Sigma }-\frac{\dot t \ a \ \sin ^2 \theta \left(2 r-Q^2\right)}{\Sigma }+\frac{a \ r \ \mho \ q \ \sin ^2 \theta }{\Sigma }} = $$

$$= {\frac{{\rm v_{\phi}} \ \sqrt{|g_{\phi \phi}|} }{\rm \sqrt{1-\mu^2 \ v^2}}+\rm \frac{(1-\mu^2 v^2) \ a \ r \ \mho \ q \ \sin ^2 \theta }{\Sigma }}$$

where the velocity components v=√(vr²+vθ²+vφ²) are measured in the local tetrad of a ZAMO. In the case of an uncharged test particle and a nonrotating naked singularity or black hole the equations reduce to those given by John Rennie in the answer above.

Here for example is a plot of the escape velocity of an angular momentum free test particle from a naked singularity with spin a=2 in the plane θ=80°, the x axis is r (the singularity is at r=0, R=a):

naked singularity escae velocity plot

As you can see in this scenario the escape velocity is decreasing again when you get close to the singularity (that is because the singularity can be repulsive, see d²r/dτ² in the numerical display below), here the particle escapes from r=0.001, θ=80° (it doesn't look like it though because the Boyer Lindquist radius is not the same as the cartesian radius) and gets accelerated first before it decelerates again:

naked singularity escae velocity animation

The local velocity components on the bottom of the last row are measured relative to local ZAMOs. The black ring represents the singularity, and the blue and red surfaces are the outer and inner ergospheres (more examples with other configuratons can be found here). To enlarge click on the image.


A naked singularity is created the same way as an ordinary singularity, the only difference is that if the angular momentum $J$ is greater than the mass squared, $M^{2}$, then the event horizons (which would be at \begin{equation} r = M(1 \pm \sqrt{(1 - (J/M^{2}})^{2} \end{equation} cannot exist. Also, There's no such thing as escape velocity for a black hole since the definition of a black hole singularity is that any line going to the singularity ends there.

Lines not going to the singularity (however close they may get) don't have to end.

The following paper seems to be very interesting, though I can't tell for sure as I'm not a complete expert on Naked Singularity theory, or indeed Loop Quantum Cosmology.

http://arxiv.org/abs/gr-qc/0506129 Quantum evaporation of a naked singularity Rituparno Goswami, Pankaj S. Joshi, Parampreet Singh

Abstract "We investigate here gravitational collapse of a scalar field model which classically leads to a naked singularity. We show that non-perturbative semi-classical modifications near the singularity, based on loop quantum gravity, give rise to a strong outward flux of energy. This leads to the dissolution of the collapsing cloud before a naked singularity can form. Quantum gravitational effects can thus censor naked singularities by avoiding their formation. Further, quantum gravity induced mass flux has a distinct feature which can lead to a novel observable signature in astrophysical bursts."

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    $\begingroup$ A naked singularity is created... This is not right. You're describing one hypothetical type of naked singularity, not a naked singularity in general. It may be possible to create a naked singularity by overspinning a black hole, but that's still being studied and there is not yet a conclusive answer. Overcharging may also be possible. $\endgroup$
    – user4552
    Nov 17, 2014 at 16:10
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    $\begingroup$ Also, There's no such thing as escape velocity for a black hole since the definition of a black hole singularity is that any line going to the singularity ends there. It's true that geodesic incompleteness is the definition of a singularity. However, a timelike singularity can cause geodesics to be past-incomplete, which means that stuff can emerge from them. The default expectation is given in a famous description by Earman: "all sorts of nasty things - TV sets showing Nixon's 'Checkers' speech, green slime, Japanese horror movie monsters, etc. - emerge helter-skelter from the singularity." $\endgroup$
    – user4552
    Nov 17, 2014 at 16:13
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    $\begingroup$ In addition to what @BenCrowell, it is also the case that timelike singularties can exert repulsive forces, so that a geosdesic coming in to the singularity will bounce and emerge from a future horizon. Extended solutions for non-trivial black holes are weird. $\endgroup$ Nov 18, 2014 at 17:02

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