Comment by somat
2 days ago
from which perspective? I have yet to wrap my head around it(this usually means I am wrong about something), but there may be no singularity because it takes matter an infinite amount of time to reach the center due to time dilation effects.
https://modern-physics.org/time-dilation-near-massive-bodies...
This is the origin of my favorite science fiction theory. (little to no actual science but you could write a fun space romp around it) If you get a large enough black hole where the tidal forces will not rip you to shreds instantly, you could just scoot across the event horizon right, now what happens? you can still move around, everything feels normal, but really you have lost half a dimension, everything "out" from the center is completely gone from the universe. Now the theory, back to our universe, What happened to time? why does time only go one way? we can accelerate and decelerate along the time axis, but can't reverse it. Where has our missing half of a time dimension gone?
> but there may be no singularity because it takes matter an infinite amount of time to reach the center due to time dilation effects.
The outside observer’s view doesn’t stop physics inside. For a massive black hole, you absolutely do reach the singularity in finite time by your own clock.. likely minutes to hours for the largest ones we've known about so far.
>To an external observer, an object falling into a black hole appears to slow down and never actually crosses the event horizon, seemingly freezing in time.
It takes infinite time to reach event horizon, not the center.
Yeah, that is the tricky part. The problem is that black holes are eldritch interstellar cryptids, and for the most part physics gives up and goes to cry in the corner the minute you start asking about "what's in a black hole?"
But in this specific case, you get one odd conclusion. if it takes forever to enter a black hole. is it impossible for anything to pass the event horizon? It sounds like this is observation dependent. but from an external point of view you are unable to observe anything entering the black hole. and from an internal point of view, the universe will instantly age and die when you try and enter the hole.(and if hawking radiation actually exists you will see the black hole shrink and pop the instant you try and enter it) either way nothing is getting in.
Is most of the mass of the star that formed the black hole actually stuck in a time dilated shell just outside the event horizon? Or perhaps all the mass is eternally stuck collapsing. and never actually reaches the density required to pass the event horizon. is that another way to define the event horizon? the point where time stops.
Time dilation makes my head hurt.
It never actually reaches the density required to form the event horizon.
> It takes infinite time to reach event horizon, not the center.
Even that is only true to a distant observer, not the one crossing the horizon.
> but there may be no singularity because it takes matter an infinite amount of time to reach the center due to time dilation effects.
Wouldn't that just mean that the singularity is located infinitely far into the future?
Isn't it another way of saying that the singularity is never going to exist?
It's not quite true that everything feels normal. If I am standing with my feet toward the singularity, my hand cannot move above my head, the best it can do is fall toward the singularity slower than my head does. Especially at very slow speeds this has some very weird physical effects, not the least of which is the immediate impossibility of all systems that make you 'you' continuing to function.
That doesn't sound right. If you're on the event horizon you're not going at very slow speeds in that sense, the space around you is already falling into the black hole faster than light.
If you're "travelling at 1m/s so you can only raise your hand above your head at 1m/s by expending infinite energy" then you're already travelling at c-1m/s away from the black hole through local space just to 'stay still' at 1m/s 'velocity'. No wonder you need infinite energy to accelerate your arm 1m/s further and things get weird - you're travelling at relativistic velocities.
Is this true?
My understanding is that for extremely large black holes the tidal forces are negligible near the event horizon. So things should function pretty much the same other than you can't move in reverse and get out.
If two rockets fall past the horizon at the same time, one accelerating forward towards the singularity, and the other accelerating backwards away from the singularity, then shouldn't the distance between the rockets increase, even though they are both moving inexorably forward?
If the tidal forces are low, I'd assume that my muscles are still strong enough to "slow down my hand enough" to move it above my head.
The relevant quantities are the curvature scalars near the horizon, and for a sizable black hole they are small there. As an example, consider the Kretschmann scalar (KS). The KS is the sum of the squares of all components of a tensor. In Schwarzschild spacetime KS looks like R_{\mu\nu\lambda\rho}R^{\mu\nu\lambda\rho} = (48 G^2M^2)/(c^4r^6), where R is the Riemann curvature tensor, and we can safely set G=1 and c=1 so (48 M^2)/r^6. In this setting, KS is proportional to the spacetime curvature. At r = 2M, the Schwarzschild radius, the number becomes very small as we increase M, the black hole's mass. However, for any M at r = 0, the Kretschmann scalar diverges.
For a large-M black hole, there is "no drama" for a free-faller crossing the event horizon, as the KS gradient is tiny.
Since the crosser is in "no drama" free-fall he can raise his hands, toss a ball between his hands, throw things upwards above his head, and so forth. The important thing though is that all these motions are most easily thought of in his own local self-centred freely-falling frame of reference, and not against the global Schwarzschild coordinates. His local frame of coordinates is inexorably falling inwards. Objects moving outwards in his local frame are still moving inwards against the Schwarzschild coordinates.
You might compare with a non-freely-falling frame of reference. Your local East-North-Up (ENU) coordinates let you throw things upwards or eastwards, but in less-local coordinates your ENU frame of reference is on a spinning planet in free-fall through the solar system (and the solar system is in free-fall through the Milky Way, and the galaxy is in free-fall through the local group). That your local ENU is not a freely-falling set of coordinates does not change that the planet is in free-fall, and your local patch of coordinates is along for the ride.
A comparison here would be a long-running rocket engine imparting a ~ 10 m s^-1 acceleration to a plate you stand on. In space far from the black hole, you and the rocket engine would tend to move away from the black hole, but you'd be able to do things like juggle or jump up and down, and it'd feel like doing it on Earth's surface. This is a manifestation of the equivalence principle. Inside the horizon the rocket would still be accelerating the plate and you at ~ 10 m s^-1, but you, the plate, and the rocket would all be falling inwards.
Tidal forces are not the constraining factor - the transformation of space into a timeline property is. There is no out, no away direction. All paths lead to singularity. No particle can travel away from singularity .
Two rockets can diverge in distance, because one is slowing itself along the timeline space dimension toward singularity. If you are moving 1 m/s toward singularity, the fastest your hand can raise above your head is 1 m/s with infinite energy expenditure. The same goes for blood pumping to your head, electrical impulses to your brain, etc.
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nor can you swivel your head to look backwards, as all the particles in your head are tidal locked in a falling trajectory towards the center