It’s said that from an outside observer’s reference frame, a black hole (as well as the event horizon) takes forever to form because of runaway time dilation, which causes falling matters to slow down exponentially as they approach the Schwarzschild radius. However, according to Hawking’s theory, black holes emit radiation and hence have finite lifespans. Besides only objects with event horizons can emit Hawking radiation (which excludes neutron stars). So here comes a problem. If the event horizon takes forever to form, a collapsing star will never start to emit Hawking radiation and will have an infinite lifespan, which contradicts with Hawking’s prediction. So there must be a moment in the finite future when a collapsing star starts to emit Hawking radiation. Can we say at this moment the collapsing star is a de facto black hole?

  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/5031/44126 $\endgroup$
    – rob
    Commented Jun 21, 2022 at 1:19
  • $\begingroup$ Thanks. There are many detailed answers there, but I found little discussions on Hawking radiation. $\endgroup$
    – 哲煜黄
    Commented Jun 21, 2022 at 1:30
  • 1
    $\begingroup$ The horizon needed for the Hawking radiation is not the same type of the horizon that never forms. There are at least a dozen different types of horizons in GR. $\endgroup$
    – safesphere
    Commented Jun 22, 2022 at 5:18

2 Answers 2


The temperature of Hawking radiation goes like $T\sim 1/M$: more massive black holes have lower temperatures. For a given spherically-symmetric mass distribution, the pressure at the origin diverges when the radius $R$ falls below some multiple of the Schwartzchild radius $r_S \sim M$. But the density goes like $M/R^3$, so more-massive black holes can form at lower densities than less-massive black holes.

Whether Hawking radiation “wins” over infalling radiation is a thermodynamical question: the net energy flow is from hot to cold. (Note that all astrophysical black holes are colder than the cosmic microwave background, and many astrophysical black holes are surrounded by hot accretion disks, so probably Hawking radiation is not currently shrinking any black holes in our visible universe.) Its possible to imagine density fluctuations in the collapsing fluid temporarily creating small black holes which are hot enough to evaporate again. But mountain-sized black holes are cooler than stellar interiors, and so density fluctuations at that scale will form black holes which will be net absorbers of both matter and radiation. The size (and temperature) of the initial horizon is a computational problem which depends on the equation of state for the collapsing matter.

For a discussion of how hot infalling matter disappears into the cold Hawking radiation in finite time, see this question by me.


I'm not sure it's particularly helpful to think from the outside observer's reference frame because what's observed is the effect and not the cause of the black hole. If a dead star is massive enough, an event horizon will form immediately. Outgoing light rays that start at the horizon remain there indefinitely. There is no reason to assume a finite lifespan for these objects.


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