(Finally got around to it again… Yay, slowpoke :P )
Thus, quite clearly, someone who falls in falls in in a finite time on their own watch, but an infinite time in the coordinate system’s clocks.
That’s exactly my point; if a “finite” amount of subjective time takes an infinite amount of objective time – if it takes a literally endless amount of universal time to move from the last local Planck time to the next – the “finite” amount of subjective time in question can never actually reach its end.
Hence my earlier remark:
Assuming it’s possible to actually cross the event horizon, of course (AFAIK, time should – according to the maths, anyway – stand still at the event horizon, which would seem to make crossing impossible), or to cross it within the finite amount of time the black hole exists. ;)
Motion (in this case, crossing the event horizon) requires time; if time stands still, so does everything else.
Some definitions: dτ is an increment in “proper time,” which is basically what an ideal wristwatch of the person who’s flying around would read. To you, your wristwatch always ticks at 1 second every second. To a distant, stationary observer, it may tick in slow motion. One tick of the clock for a distant, stationary observer is dt.
Note that as r -> 2M, dt -> 0, meaning time stops as recorded by a distant observer. If the time in-between peaks in an emitted wave is P, then what you observe as a remote observer will be Pdτ/dt - the time in-between peaks gets longer as they move in “slow motion,” resulting in “gravitational redshift.” Additionally, the time in-between the emission of photons gets longer as well, meaning they dim. The combined effect is that in reality, it’d looks exactly like things fall in - they fall, then blink out in less than a milisecond. In reality, though, they just started dimming so intensely that you have no hope of ever seeing them again, but if you could somehow measure the extremely small signal from them, you’d find they’re still there, approaching the horizon asymptotically.
“dr” is the change in reduced circumference - because of the warped nature of space, we define “r” as follows: make a full circle around the black hole, then divide the distance (circumference) you traveled by 2pi, and that’s your “r” value. Because space is warped, this is not your actual distance to the singularity. Notice again, that as r -> 2M, dr -> infinity. This means that a stationary observer would measure an infinite distance to the horizon (r = 2M). Interestingly enough, this doesn’t mean you can’t reach it - we’ll get to that, though.
Using geometric units, ie,
M = GM kg/c 2, dτ = c dτ seconds, and dt = c dt seconds
We’ll refer to these as conversions to “geometric units.” Plug them back into the Schwarzchild solution above to get it in units of kg, m, and seconds. Currently we’re using units where c = 1, and everything is described in of meters, hence the name “geometric units” - as the conversions above will show if you punch the units through. Theoretical physicists like to use weird, nonsensical units that make the equations a lot more streamlined :p
Through a rather long and tedius process you can derive that a constant of geodesic (ie, free-fall) motion is
that thing about an infinite dr at the event horizon? Well, special relativistic length contraction will come into play as you fall towards the event horizon, in such a way that you, as reckoned by your wristwatch, will reach the horizon in a finite time. If you’re familiar with Special Relativity, you’ll recognize that (1 - (v/c) 2 ) is the same factor for length contraction, hence the length contraction will cancel the (1 - 2M/r) effect that increases dr.
More rigorously, if you take a summation of the dr/dτ equation, you can find that the wristwatch time dτ from any finite distance to the event horizon is a finite value (This is free-falling straight in, ! So that doesn’t at all say that everything in the universe is doomed to fall in, though that sentence could be misinterpreted that way! :p ).
E here is a general relativistic term for energy - it is a sum of all kinds - rest energy, kinetic, - anything that produces gravity. Thus, for a particle at rest at an infinite distance, the energy is simply the rest energy (m in our units where c = 1), so E/m = 1, thus we can re-write it as:
dτ = (1-2M/r) dt
Plug it back into the Schwarzchild solution, set dϕ = 0 for a radial, plunging direction (straight down, no change in angle in the spherical coordinate system), and you can solve for radial motion as an equation of radial distance:
dr/dt = v coordinate = (1-2M/r) (2M/r) 1/2
Using the equation of geodesic motion from before to switch from the “coordinate clock” to a “wristwatch clock,” from dr/dt to dr/dτ, we get:
dt/dτ = Sqrt(2M/r)
Thus, quite clearly, someone who falls in falls in in a finite time on their own watch, but an infinite time in the coordinate system’s clocks.
But from the frame of reference of the person falling in, it’s a finite time.
A picosecond stretching out for an eternity, figuratively speaking.
A bit less figuratively; as far as the rest of the universe is concerned, you’re very slowly grinding to a halt at the very edge of the event horizon, no? Practically frozen in time until either the black hole or the universe ceases to exist.
@Exhumed Legume
From a coordinate frame of reference, yes - That’s the frame of reference of someone who’s very far away, and at rest with respect to the black hole. But from the frame of reference of the person falling in, it’s a finite time. I’ll show this from the Schwarzchild solution when I can set my laptop up again. At an airport, now. It’ll be a fun exercise :p
For a supermassive black hole like the one at the center of our galaxy (sagatarrius A), you could actually cross the horizon quite nicely without tidal forces being an issue.
Assuming it’s possible to actually cross the event horizon, of course (AFAIK, time should – according to the maths, anyway – stand still at the event horizon, which would seem to make crossing impossible), or to cross it within the finite amount of time the black hole exists. ;)
@Exhumed Legume
Eheheh… I’m on my phone now, but for the sake of fun and science I’ll give this a big, proper reply tonight. This is kind of my specialty :p
For now, though: spaghettification actually depends on the mass of the black hole. Iirc, the gravitational gradient is proportional to the inverse cube, while the radius of the event horizon is proportional to the inverse of the radius to the first power, so the larger the black hole, the less the tidal forces at the horizon.
For typical stellar mass black holes, this is far before the horizon where it’d kill you. For a supermassive black hole like the one at the center of our galaxy (sagatarrius A), you could actually cross the horizon quite nicely without tidal forces being an issue.
Heh… This is comically off topic from the picture, but no matter - Moondancer would appreciate this sort of discussion, anyways, even if she wouldn’t appreciate the edited picture xp
A tad nit-picky perhaps, but I’m just too much of a pedant to not address this bit:
But to someone falling into a black hole, they see it very differently; they cross the horizon without much significant at all happening, in just a few seconds on their watch.
Aside from the phenomenon named – in a fit of morbid humor, I assume – “spaghettification” by some astrophysicist; the gravity gradient pulling the falling person (and their watch) apart into ever smaller pieces until all that’s left is a stream of loose atoms falling downwards.
Which happens well before the event horizon, by the way; the reason it’s called an event horizon is that it’s literally impossible (at least for now – but if that ever changes, it would by definition no longer be an event horizon) to know what happens within that volume of space-time (or indeed if the concept of space-time as we understand it is even applicable within).
You’re correct in that it would happen relatively quickly (and therefore, presumably, with comparatively little suffering) from the perspective of the hapless atomized-addition-to-accretion-disc-to-be, but would seem to take (figuratively) forever from the outside, due to time dilation (and the light taking an increasing amount of time to climb out of the gravity well and into the eyes/cameras/sensors of observers).
Or at least that’s what the mathematics tells us; it’s kind of difficult to test the hypothesis in practice – not to mention extremely unethical.
(Disclaimer: I’m going by memory and my limited understanding, so feel free to not take my word for it.)
Also, what happens to eyesight depends too much on specifically what happens to say. Sometimes it’s more a choking affair, other times it cuts of blood flow - you probably know the whole story :p (those two are assuming there’s no neck-breaking, which, given the lack of any slack, probably won’t happen, here)
@Cirrus Light
Considering its mostly just a recolor, that detail wasn’t taken into . And when hanged, there is no “good eyesight.” Everything goes bright and staticky from the petechiae in the eyes… Still, I agree on at least letting her have her effects…
@Jackalodeath
For what? I was just thinking of the neat kind of twist how solipsism can allow her to not lose a sense of control of her existence, in a sort of mental desperate grasp at feeling in control when obviously you’re not, but also allowing her to get off some last words that would unsettle everyone a little bit. Or a lot.
Really no good reason for her to be executed, no matter what, though. I don’t even think Starlight deserved it and she was a major villain. I think most reasonable people would agree with that… Though that’s a conversation for another day :q
Also, what the heck!? I mean, yeah, she’s going to be strangled to death on the end of a rope - but they took her glasses!? That’s just cruel and unusual. At least let her die with good eyesight XP
(XP face is terribly appropriate for this situation…)
@Cirrus Light
Then to punctuate her defiant last words, she turns to the executioner - “Now free me from the hell you consign yourselves to! Pull the lever!”
“You can hang me, but you’re still nothing more than a character in a fully interactive movie! I remain the supreme God of my own perspective of reality! You hollow fools, you are ‘philosophical zombies’ - the whole lot of you! You don’t even exist! And when I ‘die,’ you do nothing more than free me from my mortal coils, and should I never see you again - end your own existence as I transcend this plane of reality!” XP
@Cirrus Light
An intelligent-ish conversation between two Bronies fueled by semi-barbaric execution without the word “faget” being thrown about all willy-nilly?😂 But I see your point, it did help me understand why I have my “quirks,” and it would be beneficial from a solipsistic POV. Also, I wonder how many people will be Googling solipsism after this. I love it when people are accidentally forced to learn! Kudos on the vocab by the way, I forgot that existed… Enjoy, and keep up the good work!
Or waving a baseball bat in a threatening manner at the Queen of England
Well, she did telekinetically sling Twilight out of the way. Imagine walking through the street and shoving the Queen of England out of the way.
(Finally got around to it again… Yay, slowpoke :P )
Hence my earlier remark:
Huh. Might be a site glitch.
I’ve seen that happen before, but not lately.
Stuff I made up in my head on the fly :q
Also, I think your first link is broken, since it links to LightningBolt’s comment, not mine :p
@Cirrus Light
That is clearly a reference. My question is, what to?
Sure, why not?
Being the target of OP’s edits? :q
Correct, but from your own view, you fall in in a finite (usually a very short) time.
The Schwarzchild solution for a non-spinning, chargeless black hole, reduced to a 2-dimensional plane (which makes life a lot easier) is:
(dτ) 2 = (1-2M/r)(dt) 2 - (1-2M/r) -1 (dr) 2 - r 2 dϕ 2
Some definitions: dτ is an increment in “proper time,” which is basically what an ideal wristwatch of the person who’s flying around would read. To you, your wristwatch always ticks at 1 second every second. To a distant, stationary observer, it may tick in slow motion. One tick of the clock for a distant, stationary observer is dt.
Note that as r -> 2M, dt -> 0, meaning time stops as recorded by a distant observer. If the time in-between peaks in an emitted wave is P, then what you observe as a remote observer will be Pdτ/dt - the time in-between peaks gets longer as they move in “slow motion,” resulting in “gravitational redshift.” Additionally, the time in-between the emission of photons gets longer as well, meaning they dim. The combined effect is that in reality, it’d looks exactly like things fall in - they fall, then blink out in less than a milisecond. In reality, though, they just started dimming so intensely that you have no hope of ever seeing them again, but if you could somehow measure the extremely small signal from them, you’d find they’re still there, approaching the horizon asymptotically.
“dr” is the change in reduced circumference - because of the warped nature of space, we define “r” as follows: make a full circle around the black hole, then divide the distance (circumference) you traveled by 2pi, and that’s your “r” value. Because space is warped, this is not your actual distance to the singularity. Notice again, that as r -> 2M, dr -> infinity. This means that a stationary observer would measure an infinite distance to the horizon (r = 2M). Interestingly enough, this doesn’t mean you can’t reach it - we’ll get to that, though.
Using geometric units, ie,
M = GM kg/c 2,
dτ = c dτ seconds, and
dt = c dt seconds
We’ll refer to these as conversions to “geometric units.” Plug them back into the Schwarzchild solution above to get it in units of kg, m, and seconds. Currently we’re using units where c = 1, and everything is described in of meters, hence the name “geometric units” - as the conversions above will show if you punch the units through. Theoretical physicists like to use weird, nonsensical units that make the equations a lot more streamlined :p
Through a rather long and tedius process you can derive that a constant of geodesic (ie, free-fall) motion is
E/m = (1-2M/r) (dt/dτ) = (1-(v/c) 2 )
*[citation. Note: This article is even beyond _my_ understanding, but the last equation in the linked section is the same as the equation above - except they use the term ‘‘r(s)’’ for what we call “2M” here]^*
that thing about an infinite dr at the event horizon? Well, special relativistic length contraction will come into play as you fall towards the event horizon, in such a way that you, as reckoned by your wristwatch, will reach the horizon in a finite time. If you’re familiar with Special Relativity, you’ll recognize that (1 - (v/c) 2 ) is the same factor for length contraction, hence the length contraction will cancel the (1 - 2M/r) effect that increases dr.
More rigorously, if you take a summation of the dr/dτ equation, you can find that the wristwatch time dτ from any finite distance to the event horizon is a finite value (This is free-falling straight in, ! So that doesn’t at all say that everything in the universe is doomed to fall in, though that sentence could be misinterpreted that way! :p ).
E here is a general relativistic term for energy - it is a sum of all kinds - rest energy, kinetic, - anything that produces gravity. Thus, for a particle at rest at an infinite distance, the energy is simply the rest energy (m in our units where c = 1), so E/m = 1, thus we can re-write it as:
dτ = (1-2M/r) dt
Plug it back into the Schwarzchild solution, set dϕ = 0 for a radial, plunging direction (straight down, no change in angle in the spherical coordinate system), and you can solve for radial motion as an equation of radial distance:
dr/dt = v coordinate = (1-2M/r) (2M/r) 1/2
Using the equation of geodesic motion from before to switch from the “coordinate clock” to a “wristwatch clock,” from dr/dt to dr/dτ, we get:
dt/dτ = Sqrt(2M/r)
Thus, quite clearly, someone who falls in falls in in a finite time on their own watch, but an infinite time in the coordinate system’s clocks.
Here’s a nice illustration of radial velocity (y-axis, where c = 1) plotted against radial distance (in meters, where M is simply set to 1. An example: our sun has a mass M = GM kg / c ^2^ = ~1,500 meters)
So you can set M = 1500 and multiply the two v’s by (299,792,458 m/s) to get the charts to read in meters and meters/second.
A bit less figuratively; as far as the rest of the universe is concerned, you’re very slowly grinding to a halt at the very edge of the event horizon, no? Practically frozen in time until either the black hole or the universe ceases to exist.
From a coordinate frame of reference, yes - That’s the frame of reference of someone who’s very far away, and at rest with respect to the black hole. But from the frame of reference of the person falling in, it’s a finite time. I’ll show this from the Schwarzchild solution when I can set my laptop up again. At an airport, now. It’ll be a fun exercise :p
Eheheh… I’m on my phone now, but for the sake of fun and science I’ll give this a big, proper reply tonight. This is kind of my specialty :p
For now, though: spaghettification actually depends on the mass of the black hole. Iirc, the gravitational gradient is proportional to the inverse cube, while the radius of the event horizon is proportional to the inverse of the radius to the first power, so the larger the black hole, the less the tidal forces at the horizon.
For typical stellar mass black holes, this is far before the horizon where it’d kill you. For a supermassive black hole like the one at the center of our galaxy (sagatarrius A), you could actually cross the horizon quite nicely without tidal forces being an issue.
Heh… This is comically off topic from the picture, but no matter - Moondancer would appreciate this sort of discussion, anyways, even if she wouldn’t appreciate the edited picture xp
A tad nit-picky perhaps, but I’m just too much of a pedant to not address this bit:
Which happens well before the event horizon, by the way; the reason it’s called an event horizon is that it’s literally impossible (at least for now – but if that ever changes, it would by definition no longer be an event horizon) to know what happens within that volume of space-time (or indeed if the concept of space-time as we understand it is even applicable within).
You’re correct in that it would happen relatively quickly (and therefore, presumably, with comparatively little suffering) from the perspective of the hapless atomized-addition-to-accretion-disc-to-be, but would seem to take (figuratively) forever from the outside, due to time dilation (and the light taking an increasing amount of time to climb out of the gravity well and into the eyes/cameras/sensors of observers).
Or at least that’s what the mathematics tells us; it’s kind of difficult to test the hypothesis in practice – not to mention extremely unethical.
(Disclaimer: I’m going by memory and my limited understanding, so feel free to not take my word for it.)
I know, it ‘twas a joke :q
Also, what happens to eyesight depends too much on specifically what happens to say. Sometimes it’s more a choking affair, other times it cuts of blood flow - you probably know the whole story :p (those two are assuming there’s no neck-breaking, which, given the lack of any slack, probably won’t happen, here)
Considering its mostly just a recolor, that detail wasn’t taken into . And when hanged, there is no “good eyesight.” Everything goes bright and staticky from the petechiae in the eyes… Still, I agree on at least letting her have her effects…
For what? I was just thinking of the neat kind of twist how solipsism can allow her to not lose a sense of control of her existence, in a sort of mental desperate grasp at feeling in control when obviously you’re not, but also allowing her to get off some last words that would unsettle everyone a little bit. Or a lot.
Really no good reason for her to be executed, no matter what, though. I don’t even think Starlight deserved it and she was a major villain. I think most reasonable people would agree with that… Though that’s a conversation for another day :q
Also, what the heck!? I mean, yeah, she’s going to be strangled to death on the end of a rope - but they took her glasses!? That’s just cruel and unusual. At least let her die with good eyesight XP
(XP face is terribly appropriate for this situation…)
And now I’m waiting… And watching…
Then to punctuate her defiant last words, she turns to the executioner - “Now free me from the hell you consign yourselves to! Pull the lever!”
I just thought of a rather epic scene >.<
Thanks!
Solipsism is oddly empowering…
your current filter.“You can hang me, but you’re still nothing more than a character in a fully interactive movie! I remain the supreme God of my own perspective of reality! You hollow fools, you are ‘philosophical zombies’ - the whole lot of you! You don’t even exist! And when I ‘die,’ you do nothing more than free me from my mortal coils, and should I never see you again - end your own existence as I transcend this plane of reality!” XP
An intelligent-ish conversation between two Bronies fueled by semi-barbaric execution without the word “faget” being thrown about all willy-nilly?😂 But I see your point, it did help me understand why I have my “quirks,” and it would be beneficial from a solipsistic POV. Also, I wonder how many people will be Googling solipsism after this. I love it when people are accidentally forced to learn! Kudos on the vocab by the way, I forgot that existed… Enjoy, and keep up the good work!