1
$\begingroup$

If you plot a space-time diagram of an object falling through the event horizon of a black hole, and draw the past and future "light cones" of the object at every point, wouldn't the point infinitely to the event horizon have a light cone which allows light being radiated by the object to reach an observer outside the event horizon at time = infinity? (at the point when the object touches the event horizon, a radiated photon will never reach the observer outside the event horizon) If so, then why don't we, outside the event horizon of black holes being formed by the collapse of stars, observe light from the collapsing star (AKA the light from a supernova forming a black hole) forever? Please let me know if I'm wrong.

$\endgroup$
1
  • 2
    $\begingroup$ What may help you on your quest is looking into Penrose diagrams $\endgroup$
    – Jim
    Aug 14, 2020 at 18:27

2 Answers 2

3
$\begingroup$

As material falls to a horizon, emitting light as it goes, there are three effects to consider: the worldline of the emitter, the redshift of the light, and the intensity of the light (headlight effect).

As the emitter sends out light signals, they get more and more red-shifted and more and more dim and more and more infrequent at a receiver somewhere outside the horizon. For signals setting out from locations close to the horizon, the frequency and the intensity fall exponentially with time at the receiver. Faced with such an exponential decay, you could say it never quite reaches zero, but we don't normally say that for other cases of exponential decay such as atoms decaying to their ground state. We just say the atom decays. So by the same logic we should say the received light from a collapsing star falls to zero intensity, and it is not necessary to wait an infinitely long time for this to be so. Therefore the black hole is indeed black, and in practice the timescale for these decays is short (some tens of microseconds for a one solar mass black hole).

And yet, according to one very natural definition of simultaneity, the falling material does indeed not quite cross the horizon in any finite amount of time registered on the distant clock, so the matter moving to form the black hole never finishes its collapse towards its own horizon. This sounds like a very odd conclusion, but it is owing to the relativity of simultaneity and a time dilation which tends to infinity. There are plenty of other reference frames, and thus definitions of simultaneity, in which the black hole does form in a finite time. And the predictions for what emitted signals do when they arrive elsewhere is independent of such details. The signals die away. The hole is black.

$\endgroup$
8
  • $\begingroup$ So black holes are constantly radiating a faint glow apart from hawking radiation? $\endgroup$
    – user250486
    Aug 16, 2020 at 3:08
  • $\begingroup$ @user250486 "faint glow" is not the right phrase for something like exponential decay. Try calculating $\exp(-1 year/10 microseconds)$ to get an impression; it is vanishing small compared even to the very dim Hawking radiation. $\endgroup$ Aug 16, 2020 at 6:22
  • 1
    $\begingroup$ @user250486 The Hawking radiation has not been observed. Black holes don't glow. $\endgroup$
    – safesphere
    Aug 16, 2020 at 15:35
  • $\begingroup$ "There are plenty of other reference frames, [...], in which the black hole does form in a finite time." - As discussed before, this is a widespread misconception based on ignoring mathematical rigor. No frames exist at the lightlike horizon similarly to no frames existing at the speed of light. Even aside from this point, the event horizon is eternal and cannot form in a star collapse, as explained in this question (not the answers): Can separate manifold regions have the same coordinates? $\endgroup$
    – safesphere
    Aug 16, 2020 at 15:47
  • 1
    $\begingroup$ @AndrewSteane, it is easy to be distracted by artificial definitions of synchroneity based on coordinate time, but they are artificial and do not correspond to what we actually mean by synchronous in any normal context. If you are to talk of the end of the universe then the only reasonable definition uses cosmic time, and it is quite clear that (whether or not the universe is finite) no event horizons form in finite cosmic time. $\endgroup$ Aug 16, 2020 at 17:12
0
$\begingroup$

You are not wrong. If we could live forever, and if we could observe indefinitely small light energies, we would observe the light from material falling into to a forming black hole forever. Because time appears to stop at the Schwarzschild radius, an issue is raised as to whether a singularity can actually form. In 1939 Julius Robert Oppenheimer and one his students, Hartland Snyder, published the seminal paper on gravitational collapse to a black hole (Oppenheimer J. R., Snyder H., 1939, On Continued Gravitational Attraction, Phys. Rev. 56, 455). They concluded that, from the point of view of an exterior observer, “it is impossible for a singularity to form in a finite time.”

The things usually called black holes in modern terminology are stable state solutions of Einstein's equation assuming boundary conditions which never physically appear in practice. As described by Oppenheimer and Snyder, these solutions never occur in our universe.

$\endgroup$
2
  • 1
    $\begingroup$ +1 to cancel another malicious downvote. This site has been hijacked by self appointed "experts" with deep misconceptions downvoting correct answers. $\endgroup$
    – safesphere
    Aug 16, 2020 at 16:10
  • $\begingroup$ What are the black holes that have been discovered to exist according to popular science? $\endgroup$
    – Tavin
    Nov 15, 2022 at 2:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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