Excuse me if the question is naïve. I am not a physicist.

Let's imagine formation of a black hole from a homogeneous collapsing star. At certain moment when enough of matter is inside of a small enough spherical area the event horizon starts to exist. When does the singularity start to exist and why?

The matter below the event horizon still continues to collapse towards the centre (e.g. in the reference frame of the falling matter). What are the necessary conditions for the singularity to appear?

I have found two similar questions but the answers did not really explain this problem to me:

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    $\begingroup$ The singularity is an unphysical artifact of the theory. One shouldn't take it too seriously. The singularity already disappears in the physical case of a charged rotating black hole, which is a good indicator that it can't be expected to be a stable physical phenomenon. What is "inside" a black hole (and if there is even an "inside") is an open questions, which probably won't find an answer until we can make telescopes that can image the event horizon of a black hole in detail. $\endgroup$
    – CuriousOne
    Commented Sep 28, 2014 at 5:31
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    $\begingroup$ OK, but within the GR theoretical construct how long does it take? I guess it depends where the observer is too. A good question. $\endgroup$
    – ProfRob
    Commented Sep 28, 2014 at 8:53
  • $\begingroup$ @CuriousOne I understand that general relativity fails to describe the interior of a black hole but I think it is still interesting to see which "weirdnesses" show from the failing theory. --- Are there any promising results clarifying more of the problem from the theories trying to unify GR and QFT like the string theory? $\endgroup$ Commented Sep 28, 2014 at 8:58
  • $\begingroup$ @pabouk: I can't tell you much about theory, I am an experimentalist. As an experimentalist I can tell you that so far 0 experiments have been performed on black holes. All we have is basically unverified models and a few observations of matter falling into black holes. The signatures of that fall are, of course, generated mostly far away from the event horizon. I think there is one sign that what we are looking at are really black holes, and that's the spectral properties of the x-rays and gamma rays generated by the falling matter, but that's nowhere close enough to probe EH physics. $\endgroup$
    – CuriousOne
    Commented Sep 28, 2014 at 19:08
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    $\begingroup$ I understand that general relativity fails to describe the interior of a black hole Not true. It only fails to describe the singularity. $\endgroup$
    – user4552
    Commented Sep 28, 2014 at 22:17

2 Answers 2


If you're not up to speed with general relativity this is going to be hard to explain, but I'll give it a go. The more determined reader may want to look at this PDF (just under 1MB in size) that describes the collapse in a rigorous way.

A couple of points to make before we start: you're being vague about the distinction between the singularity and the event horizon. The singularity is the point at the centre of the black hole where the curvature becomes infinite. The event horizon is the spherical surface that marks the radial distance below which light cannot escape. As you'll see, these form at different times.

The other point is that to make the calculation possible at all we have to use a simplified model. Specifically we assume the collapsing body is homogeneous (actually I see you anticipated that in your answer) and is made up of dust. In general relativity the term dust has a specific meaning - it means matter that is non-interacting (apart from gravity) and has zero pressure. This is obviously very different from the plasma found in real stars.

With the above simplifications the collapse is described by the Oppenheimer-Snyder model, and it turns out that the size of the collapsing object is described by the same equation that describes the collapse of a closed universe. This equation is called the FLRW metric, and it gives a function called the scale factor, $a(t)$, that describes the size of the ball of dust. For a closed universe the scale factor looks something like:

Scale factor

(image from this PDF)

A closed universe starts with a Big Bang, expands to a maximum size then recollapses in a Big Crunch. It's the recollapse, i.e. the right hand side of the graph above, that describes the collapse of the ball of dust.

The radius of the ball is proportional to $a(t)$, so the radius falls in the same way as $a(t)$ does, and the singularity forms when $a(t) = 0$ i.e. when all the matter in the ball has collapsed to zero size.

As always in GR, we need to be very careful to define what we mean by time. In the graph above the time plotted on the horizontal axis is comoving or proper time. This is the time measured by an observer inside the ball and stationary with respect to the grains of dust around them. It is not the same as the time measured by an observer outside the ball, as we'll see in a bit.

Finally, we should note that the singularity forms at the same time for every comoving observer inside the ball of dust. This is because the ball shrinks in a homogeneous way so the density is the same everywhere inside the ball. The singularity forms when the density rises to infinity (i.e. the ball radius goes to zero), and this happens everywhere inside the ball at the same time.

OK, that describes the formation of the singularity, but what about the event horizon. To find the event horizon we look at the behaviour of outgoing light rays as a function of distance from the centre of the ball. The details are somewhat technical, but when we find a radius inside which the light can't escape that's the position of the event horizon. The details are described in the paper by Luciano Rezzolla that I linked above, and glossing over the gory details the result is:

Horizon formation

This shows time on the vertical axis (Once again this is comoving/proper time as discussed above) and the radius of the ball of dust on the horizontal axis. So as time passes we move upwards on the graph and the radius decreases.

It's obviously harder for light to escape from the centre of the ball than from the surface, so the event horizon forms initially at the centre of the ball then it expands outwards and reaches the surface when the radius of the ball has decreased to:

$$ r = \frac{2GM}{c^2} $$

This distance is called the Schwarzschild radius and it's the event horizon radius for a stationary black hole of mass $M$. So at this moment the ball of dust now looks like a black hole and we can no longer see what's inside it.

However note that when the event horizon reaches the Schwarzschild radius the collapse hasn't finished and the singularity hasn't formed. It takes a bit longer for the ball to finish contracting and the singularity to form. The singularity only forms when the red line meets the vertical axis.

Finally, one last note on time.

Throughtout all the above the time I've used is proper time, $\tau$, but you and I watching the collapse from outside measure Schwarzschild coordinate time, $t$, and the two are not the same. In particular our time $t$ goes to infinity as the ball radius approaches the Schwarzschild radius $r = 2GM/c^2$. For us the part of the diagram above this point simply doesn't exist because it lies at times greater than infinity. So we never actually see the event horizon form. I won't go into this any further here because it's been discussed to death in previous questions on this site. However you might be interested to note this is one of the reasons for Stephen Hawking's claim that event horizons never form.

  • $\begingroup$ Thank you very much for nice reply and the detailed explanation of the terms. In the question I probably did not express enough that I expect that the event horizon and the singularity do not form simultaneously. ------ I did not expect that the matter can really simply continue to contract to a space of zero size. Does the theory describe the zero size because it is caused just by the extremely curved space (the matter in different coordinate system will not be at a single spot)? Or could the contraction just be possibly stopped by some unknown repulsive forces stronger than gravity? $\endgroup$ Commented Sep 30, 2014 at 18:36
  • $\begingroup$ I think that there is an illogicality or something unexplained in your description: I think that the singularity should form before the red line meets the vertical line (zero size). --- Why should the singularity form only after the matter from the stellar surface collapses to the central point? Before that moment in the central point there should already be the matter which was below the stellar surface. -- Or let's see it differently: When the star starts to collapse with enough of velocity we can start to remove some matter from the stellar surface and a smaller black hole will form anyway. $\endgroup$ Commented Sep 30, 2014 at 18:50
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    $\begingroup$ @pabouk: we are using a simplified model that assumes the collapsing object is homogeneous and made of non-interacting matter i.e. dust. In this case the object remains homogeneous at all stages of the collapse, so the singularity forms at a single point in (proper) time everywhere inside the collapsing body. A real star would have some density profile i.e. it would b denser at the centre than at the surface. In this case it's likely the singularity would form while the outer layers were still at non-zero $r$. $\endgroup$ Commented Sep 30, 2014 at 19:48
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    $\begingroup$ hey wait .. "It's obviously harder for light to escape from the centre of the ball than from the surface, so the event horizon forms initially at the centre of the ball then it expands outwards and reaches the surface" - but wait, surely inside that smaller inner ball, there is considerably less mass, so why would that small inner ball form an event horizon? $\endgroup$
    – Fattie
    Commented May 14, 2016 at 17:18
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    $\begingroup$ Hmm, in that example the door is at a certain radius. It's always there and it closes at a certain time. the event horizon forms initially at the centre of the ball then it expands outwards and reaches the surface that would be like the door sliding along from -10m to the final position of 0m. the event horizon forms initially at the centre if the event horizon at some point is say radius 10m, then, there must be a black hole inside that radius. It's obviously harder for light to escape from the centre .. than from the surface ...why? at center of the Earth weight==zero, not more $\endgroup$
    – Fattie
    Commented May 16, 2016 at 16:44

Each collapsing star will have its own time table depending on mass. This video simulation by NASA of two subcritical mass neutron stars collapsing towards each other gives a time line:

"At 13 milliseconds, the more massive star has accumulated too much mass to support it against gravity and collapses, and a new black hole is born," NASA officials said in a statement. "The black hole's event horizon — its point of no return — is shown by the gray sphere. While most of the matter from both neutron stars will fall into the black hole, some of the less-dense, faster-moving matter manages to orbit around it, quickly forming a large and rapidly rotating torus."

For these distances and these masses,

It starts off with two neutron stars — the city-size, dense remnants of a violent supernova explosion — separated by about 11 miles (18 kilometers), NASA officials said. One object contains about 1.7 times the mass of our sun, while the other weighs in at 1.4 solar masses.

the order of time for an event horizon to form is 13 milliseconds. What happens within is not accessible to mathematical simulation. The formation of the event horizon during the collapse of a star implies a black hole "singularity", as in the video, but singularities could exist without an event horizon.

In general relativity, a naked singularity is a gravitational singularity without an event horizon. In a black hole, the singularity is completely enclosed by a boundary known as the event horizon, inside which the gravitational force of the singularity is strong enough so that light cannot escape. Hence, objects inside the event horizon—including the singularity itself—cannot be directly observed. A naked singularity, by contrast, is observable from the outside.

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    $\begingroup$ It is the existence of the even horizon that defines the black hole "singularity". No, the event horizon and the singularity are two different things. $\endgroup$
    – user4552
    Commented Sep 28, 2014 at 22:19
  • $\begingroup$ Thank you for the interesting article and video but I agree with Ben Crowell that event horizon is not enough for the singularity to exist. $\endgroup$ Commented Sep 30, 2014 at 18:17

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