2
$\begingroup$

I'm a layperson who just read Brief Answers to Big Question and in the book, Hawking proposes that:

A typical black hole is a star so massive that it has collapsed in on itself. It’s so massive that not even light can escape its gravity, which is why it’s almost perfectly black. Its gravitational pull is so powerful, it warps and distorts not only light but also time. To see how, imagine a clock is being sucked into it. As the clock gets closer and closer to the black hole, it begins to get slower and slower. Time itself begins to slow down. Now imagine the clock as it enters the black hole—well, assuming of course that it could withstand the extreme gravitational forces—it would actually stop. It stops not because it is broken, but because inside the black hole time itself doesn’t exist. And that’s exactly what happened at the start of the universe.

I think I understand the notion of time dilation as the clock approaches the black hole but I'm having trouble wrapping my head around the idea that time literally stops inside the black hole. Hawking then uses this idea to propose that there was no time before the Big Bang:

As we travel back in time towards the moment of the Big Bang, the universe gets smaller and smaller and smaller, until it finally comes to a point where the whole universe is a space so small that it is in effect a single infinitesimally small, infinitesimally dense black hole. And just as with modern-day black holes, floating around in space, the laws of nature dictate something quite extraordinary. They tell us that here too time itself must come to a stop. You can’t get to a time before the Big Bang because there was no time before the Big Bang. We have finally found something that doesn’t have a cause, because there was no time for a cause to exist in. For me this means that there is no possibility of a creator, because there is no time for a creator to have existed in.

I've done a little bit of googling on the topic and it seems that time doesn't stop in a black hole. Nothing in high school physics or introductory physics in college prepared me for this idea.

Does time stop at the center of a black hole? If not, why does Hawking present this idea? How else might can we describe the idea of "there was no time before the Big Bang"?

$\endgroup$
3
$\begingroup$

Hawking is using nontechnical language to talk about things that are hard to describe correctly in nontechnical language. He's also not being particularly careful about making fine distinctions.

You need to distinguish between (1) being inside a black hole's event horizon and (2) hitting the black hole's singularity.

Taking the two quotes together, it's clear that Hawking is talking about singularities. When you hit the singularity of a black hole, time stops for you simply because you're annihilated. This is similar to what happens at the big bang singularity: any observer would have been annihilated by the conditions of the early universe in which the temperature and density diverged to infinity.

Time does not stop for you just because you fall through a black hole's event horizon. For a black hole of typical size, you have something like a few milliseconds of free-fall during which you continue to experience time before you hit the singularity. However, you can't send signals to an outside observer once you cross the horizon, and as an outside observer monitors your signals, your signals appear to become infinitely slowed down as you approach the horizon. So to them, it's very much like seeing your time slow to a stop.

$\endgroup$
  • $\begingroup$ Isn't it better to say that we can't continue geodesics beyond the singularity (by definition--but the important point is that such singularities exist) rather than saying that time stops for an observer because the observer would be annihilated at the singularity? In particular, in GR (which of course is known to be flawed), there literally doesn't exist any continuation of the geodesic beyond the singularity and thus, at the singularity, time literally does stop. Please correct me if I am misunderstanding stuff. $\endgroup$ – Feynmans Out for Grumpy Cat Mar 16 at 17:56
  • $\begingroup$ @DvijMankad: In GR, "observer" is often used as not much more than a synonym for "timelike world-line." I don't think there is really much of a distinction to be made between time ending and time no longer existing for an observer because their world-line is future incomplete. The operational definition of time is really just that it's what an observer sees when they look at their watch. $\endgroup$ – Ben Crowell Mar 16 at 18:55
-1
$\begingroup$

"Time is stopping" is figure of speech. Actual time of observer, sitting on falling into BH object, will not stop. In fact, every object, beginning from some distance, will fall into BH within limited time period.

"Time is slowing" also for rays of light, and only when their direction is outside of BH. "To slow the time" for light is obtaining red shift. Speed of light is the same, but at the same moment, curvature is high. To move against that curvature is to obtain elongated wavelength.

But, it is true that time does not exist inside BH as the concept. The concept of time is to be able to synchronize each and everyones clock following "synchronize procedure". You cannot synchronize any two clocks laying on the event horizon. But, instead, you can only care about single one clock, associated with falling object. Falling object will not freeze in time, only light emitted by that object will froze, and you will observe that light. For outside observer, object will forever lay on the surface, redshifted and distorted by gravitational lensing. You can also see multiple copies of the same object, because event horizon is surface of infinite light rotation. Light can have few rotations and only after get to you.

Because event horizon is surface of infinite light rotation, we do not have the logically correct synchronize procedure for the surface of BH. But we can "syncronize hardly", it is not guaranteed to work, since event horizon has zero thickness. Slightly off-course emitted light will nor circle the BH, but swept inside. Because curvature flow is almost fully directed inside BH, rather then outside or alongside event horizon.

PS. Inside Kerr BH, where inside shell exist, time could become real number again at some point inside BH. But how to measure it? As you approach the second horizon, you will again expirience strong curvature.

Also, speaking of classical BH. The falling observer could live up until he meets the singularity point. Nothing really interesting will happen for him when he crosses the event horizon, except that he will not see anyting from below, only the light falling inside from the outer layers. Not being able to syncronize his clocks will not being noticable anyhow. Because anyway, he is already at near-speed-of-light velocity and by that point started to emit X-rays.

$\endgroup$
  • $\begingroup$ You cannot synchronize any two clocks laying on the event horizon. This is false. Synchronizing clocks is always locally possible. The equivalence principle says that space is always locally Minkowski. $\endgroup$ – Ben Crowell Mar 16 at 17:43
  • $\begingroup$ @BenCrowell an outside observer wouldn't be able to tell if the equivalence principle failed at the event horizon ... and an inside observer wouldn't be able to share their findings with the world, even if they survived all the way to the horizon. $\endgroup$ – John Dvorak Mar 16 at 17:47
  • $\begingroup$ @BenCrowell Time is imaginary number inside BH. How to deal with it? And light could not have the path. Procedure exists, but there is no light path for it to work. $\endgroup$ – sanaris Mar 16 at 17:47
  • $\begingroup$ Light doesn't rotate at the horizon, because the speed of light at the horizon is zero in any reference frame. $\endgroup$ – safesphere Apr 7 at 8:10

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.