Post by: Zer0 on 31/10/2022 18:02:03
P.S. - Can We ever know the Answer for Certain?
Post by: Kryptid on 31/10/2022 22:24:58
Post by: Eternal Student on 01/11/2022 01:25:06
An alternative way of stating things:
"The observable universe" is something we can see. We're all fairly sure that it's a real thing.
"The Universe" is just a word with various definitions, no-one has seen it. It's generally understood to mean "everything, absolutely everything". This concept on its own is not a simple one. Can we say it's a set of things, a really big set that has everything in it? In Mathematics, most set theories are incompatible with the existence of such a set (see Wikipedia's entry if you're interested https://en.wikipedia.org/wiki/Universal_set ). To say this more strongly, the ZFC system of Mathematics - which is what you would have studied at school whether you knew it by the name ZFC or not - this does not allow the existence of such a set.
Best Wishes.
Post by: Halc on 01/11/2022 02:38:56
Hi.Hi ES! Glad to see you up and around and posting now and then again.
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The observable universe" is something we can see.More than that I think.
I can see the sun of 9 minutes ago, but I can't see the sun now. I just presume from seeing the sun in the past that it is probably there still now, and thus part of the observable universe.
So what we see is for the most part confined to the edge of our past light cone, which at no point gets further away from 'here' than about 6 billion light years away. That's the limit of what we can see. But the visible universe is like the sun now, presumed to be there because we can see parts of its past. That's the usual definition of the visible universe, and it is about 48 billion light years in radius, about 8 times the maximum proper distance from Earth any light falling on it has had.
Similarly, I can't see a live T-rex, but they're part of the visible universe. Similarly, there are plenty of nearby burnt out stars that we can't see because their light has already completely passed us by. Those stars, like the T-rex, are part of the observable universe despite not being in any way observable.
Can We ever know the Answer for Certain?
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The Universe" is just a word with various definitionsExactly. Given that there are various definitions (and interpretations of empirical physics for that matter), there is no one correct answer to such a question. 'The universe' is just what you want it to be. It's just a word, or two of them actually.
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It's generally understood to mean "everything, absolutely everything"Most often not I think, since then it would not make any sense to talk about a different universe, or the multiverse, or whatever, both of which are 'more of something', limiting 'the universe' to a subset of all that. I would say it generally is meant to refer classically to our particular chunk of 4D spacetime, but not other structures with the same or different physics/properties.
Post by: Zer0 on 01/11/2022 07:18:31
Appreciate your Inputs Eternal
Pfft @ Hal
Fascinating to know that We cannot see the Whole Universe...
But a bit saddening too.
P.S. - So there Really might be a Flying Spaghetti Monster hidden from sight somewhere far far away receding at FTL speeds...hmm!
Maybe it's caught a Cold & keeps Sneezing, that would explain Dark Energy.
And maybe it's Spaghetti Tentacles are warped around in higher dimensions across Galaxies, that explains Dark Matter.
It Surely must have Farted long ago, hence Inflation...
YaaY!
(i finally know it all now)
Post by: alancalverd on 01/11/2022 17:36:21
It is entirely possible, indeed it would be vainly anthropocentric and ludicrous to think otherwise, that there can be a whole lot of stuff outside, that we will never know about and will never affect us in any way.
Post by: Zer0 on 03/11/2022 13:40:16
Post by: Kryptid on 03/11/2022 16:53:02
Gravity perhaps cannot work FTL.
Gravitational fields don't have a speed. Changes in a gravitational field, on the other hand, do travel at the speed of light.
Post by: Eternal Student on 03/11/2022 17:03:53
There's all sorts of stuff out there that has at least a gravitational influence on everything else, and as far as we know the observable universe therefore extends to the Schwarzchild radius of everything from here outwards.
1. Could you check that you got the right words in this sentence?
The observable universe has a reasonably precise definition. You can't increase the size of the observable universe by inferring the existence of something else, either light has been able to reach us from that place or it hasn't. The universe is a different thing, you can try to make inferences about that.
2. Inferring the existence of something in our universe based on gravitational effects is still limited. An eternal black hole would be a good example here. There's a gravitational pull from it. In all respects it looks like there should be some mass at the centre of it. However the interior of such a black hole is not in our observable universe (by definition, since light could never reach us from there). For an eternal black hole, the interior never has been, is not now and never will be in our observable universe. Applying what we believe about eternal black holes, the interior isn't even something you can describe with just space co-ordinates. It's a place and a time and just a time right at the central singularity. Anyway, hopefully you get my point. Inferring that it is a place in our universe (the wider universe, not just our observable universe) just by the gravitational effects from it could lead you astray. It is some gravitational effect with "cause unknown" or "no physical thing that can be located with just spatial co-ordinates".
3. The only flexibility we have in defining the "observable universe" is whether we consider it to describe just some space (just pure space) or a region of spacetime (space and time co-ordinates). If it describes some space then we must understand it to mean the space that exists now. If it describes some points (x,t) in spacetime then we have less of a problem. Non eternal black holes are one example of something that causes these problems, i.e. black holes that weren't eternal but did have a formation time (so typically they form from the collapse of massive stars). When a black hole forms, a region of space is simply lost. To see it more easily, we can't get light from that region any longer, so you could certainly say it isn't in our observable universe now. If we allow the observable universe to be a set of spacetime coordinates (rather than just space co-ordinates), then (x= location of the black hole, t) is in the observable for universe for t< time of formation but (x, T) for T > time of formation is not. Meanwhile, if we regard "the observable universe" as a set of space co-ordinates only then we must consider it to be a subset of the space that exists now (e.g. take a time-slice through spacetime) and we see that x=location of the black hole is not any sort of location that makes sense with t= now. Any object with a worldine that passed through (x, time of formation) simply terminated there and cannot be located at (x, t=now).
Anyway, I'm not sure what "Schwarzschild radius" you were talking about. There isn't just one boundary for the observable universe, there are multiple boundaries. Every time we hit a black hole, we have a boundary. So the observable universe is like Swiss cheese. It's not even all that spherical at what you might describe as the particle horizon. The universe is expanding at slightly different rates in different places, so the observable universe is a hole-y and mis-shapen blob of space. It is usually growing in size as time passes but not always, when a new black hole forms then some of it is lost and we have to consider the possibility that there could be other GR effects which could reduce the size of our observable universe sometimes.
Best Wishes.
Post by: Zer0 on 04/11/2022 10:45:01
Gravity perhaps cannot work FTL.
Gravitational fields don't have a speed. Changes in a gravitational field, on the other hand, do travel at the speed of light.
But Gravitational Fields might be having a Limit.
Post by: Eternal Student on 04/11/2022 15:32:12
Gravity perhaps cannot work FTL.I don't know. Nothing much works FTL (Faster than light). All things should travel at light speed or less.
However there is some theory about the Alcubierre drive. See https://en.wikipedia.org/wiki/Alcubierre_drive
from which this quote is taken:
The Alcubierre drive ([alˈkubie:re]) is a speculative warp drive idea according to which a spacecraft could achieve apparent faster-than-light travel by contracting space in front of it and expanding space behind it, ....
It requires something like negative mass or a configurable energy field with less energy than the vaccum, i.e. something we haven't identified yet.
It is very much based on the theory of gravity (General Relativity). So you'd have some grounds for arguing that GR does hold FTL since it is precisely what makes this FTL system possible.
I'm not going to say anymore about the Alcubierre drive since it remains speculative (plus - I just don't know much about it). This is the main section of the science forum and not the "new theories" or "just chat" section, after all.
Best Wishes.
Post by: paul cotter on 04/11/2022 17:05:48
Post by: Eternal Student on 04/11/2022 17:29:40
if travelling at greater than c as observed by static observers what becomes of the Lorentz factor?I've got to keep this short:
1. It's a thread started by someone else, we can't hijack it.
2. It's not mainstream science.
3. I'm not an expert on the Alcubierre drive.
The Lorentz factor and things like Lorentz contraction describe what happens in flat space. In curved space, an object may appear contracted (or enlarged) even when it is has zero velocity but just by virue of being in a different location to the observer and thus where the metric differs. As such the short answer to your question is just, Lorentz contraction, time dilation and all the effects we're familiar with from special relativity don't have such simple formulas in General Relativity. The Lorentz factor γ ≠ 1 / √(1 - v2 / c2 ) in curved space.
Extract from Wikipedia:
.... With regard to certain specific effects of special relativity, such as Lorentz contraction and time dilation, the Alcubierre metric has some apparently peculiar aspects.....
Taken from the section under the sub-title "Physics" in the Wikipedia entry. https://en.wikipedia.org/wiki/Alcubierre_drive#Physics
Best Wishes.
Post by: paul cotter on 04/11/2022 17:59:43
Post by: Eternal Student on 04/11/2022 23:34:39
I'm not sure that the Wiki article says all that much about Lorentz contraction. I may have already picked out the most salient sentences. However, it does have references and links of its own, if time is something you have plenty of.
I'm also sorry to hear about your illness. Rest up and get well soon.
Best Wishes.
Post by: alancalverd on 05/11/2022 01:02:54
However the interior of such a black hole is not in our observable universe (by definition, since light could never reach us from there).I smell philosophy! Red rag!
The inside of a rock is not observable (obviously) but that doesn't exclude the rock from being inside our observable universe, and since the interior is contained within the exterior, the interior is also inside the OU.
Set theory trumps philosophy. And a lot of engineering is based on "black box" models: we can know and exploit the transfer function of a gizmo (like an automatic gearbox, for instance) without knowing what's inside it.
The Schwarzchild radius doesn't have to be isotropic or even continuous. You can define the SR of Swiss cheese as the distance from where you are to the boundary between cheese and air.
Post by: Zer0 on 05/11/2022 03:44:09
Yes indeed, i'm guilty of diverting this thread, apologies to Zero. Thanks very much for your reply, Eternal Student, I will follow up the link you provided.
No Apologies warranted & None accepted.
Please don't view the OP as an Aeroplane, & do Not paint yourselves as Terrorists.
Rather, Please consider the OP as a plot of land demarcated for Gardening.
Sow your own Seeds & let them bud & Blossom.
🌹
Also, i have a Confession to make...
I'm a Trekkie..
& I Simply Adore Warp Drives!
🥰
P.S. - Hope & Wish for Mr & Mrs Cotter to Get Well Soon!
🖖
Post by: Eternal Student on 05/11/2022 16:49:02
Hi.
I smell philosophy! Red rag!It's not philisophy. It's just mathematics.
The OU is understood to be limited by distance. Specifically, the OU from here on earth is all those places in space where light could have travelled the distance from there to here since the big bang (or since an early time like recombination when space first became transparent enough for light to travel).
So my kitchen is in the OU even when I close the curtains and turn the lights off. We don't care if the region actually did emit some light, only that light could have travelled the distance if it had been emitted. Similarly, we don't care if something could have absorbed the light in some ordinary way along that journey (the curtains in my example).
Thus, the interior of your rock is in the OU because the distance from me to the interior of the rock is quite small and we don't care if some other bits of rock could have absorbed the light along the way or if the interior of the rock did emit any light to begin with.
However, a black hole is something completely different. The distance from the interior of a black hole to me is too large.
Here's the Schwarzschild metric:
dS2 = (terms in dt) + (1-rs/r)-1 dr2 + (terms in dθ and dφ)
Where we are using the Schwarzschild co-ordinates (t, r, θ, φ) = (time, radial co-ord, angle θ off the x-axis, angle φ off the z axis).
The bit in red is what's important just because when r ≈ rs = the Scwhrazschild radius that term → ∞.
I'm not going to bore everyone with the integration but it'll be enough to note what the metric is telling us.
LATE EDITING: I SHOULD have done the integration and not just take a short cut to thinking. See https://physics.stackexchange.com/questions/85975/how-much-extra-distance-to-an-event-horizon
It says, that when r ≈ rs then a small change in the radial co-ordinate dr is worth an infinite amount of distance, ds. To say that another way, even if we take a small step across the event horizon (change r from rs-dr to rs+dr ) that is an unbounded amount of physical distance.
Anyway, that's too far (obviously). Anything in the interior of a black hole is too far away from us, it is outside the distance limits permitted by the definition of the OU. Specifically light has not had enough time to travel from there to me here. Since that distance between me and anything in the interior of a black hole is always infinite, then light will never have had enough time to reach me from there, so it will never be in my Observable Universe.
To say this more succinctly, we can exclude the interior of every black hole from being in our OU just on the grounds of physical distance.
Best Wishes.
LATE EDITING: Changed some phrases with "the Observable Universe" to "my Observable Universe". The observable universe is different depending on where you are. If I've left "the OU" in somewhere and there is some doubt about where we were located, then please assume I meant the OU from planet earth.
Post by: Kryptid on 05/11/2022 16:52:38
The inside of a rock is not observable (obviously) but that doesn't exclude the rock from being inside our observable universe
Technically, you could observe the inside of the rock by breaking it open, or via a less destructive method like sound or gamma rays. That's not really an option with black holes.
Post by: alancalverd on 06/11/2022 00:04:25
It is also probably true to say that we are not observable from inside a black hole, but there is no reason why a person inside a black hole couldn't hypothesise the existence of stuff (including people) outside his OU.
As I recall it, the term "Big Bang" was originally coined as an insult by Fred Hoyle, who propounded an alternative "Continuous Creation" theory for which there was, alas, no actual evidence. Now if we were inside a BH, we would be subject to an incoming rain of photons from spaghettified stuff crossing the event horizon. Pretty much like Hoyle's CC source: incoming energy with no information about its origin. And if a few bits coalesced into point masses, we would find stars and black holes within our SR - just as we do. Which rather suggests that the existence of anything can be considered consistent with there being stuff outside your OU, wherever you happen to be, in addition to unobservable entities wirthin the OU.
Post by: Halc on 06/11/2022 00:25:28
I see a semantic problem. We know that there is a black hole around 1600 lightyears away from us. That is well within the radius of the universe about which we can make meaningful statements (48,000,000,000 ly or thereabouts) so it is within the observable universe, even though its contents are not observable.This statement is fairly meaningless since it mixes coordinate systems. In cosmic coordinates, the observable universe relative to Earth currently has a radius of around 48 BLY. In various different coordinate systems, that size might be a lot more or a lot less. Point is, if you're using that coordinate system, black holes don't exist. All events within black holes have no coordinates in that coordinate system. They're all in some kind of causal future of the events which comprise the current state relative to those cosmic coordinates. No black hole event can have a causal effect on an event outside, but the outside events can have a causal effect on the events inside. That puts them all in the future, and thus the problem becomes no different than being able to see Earth, 2024.
So you're right, there are events nearby (e.g. Earth 2024) which are not part of the current observable universe, but one doesn't need to invoke black holes to demonstrate this. Only events before the current time are part of it. By definition, not just events in our past light cone. The universe would be a lot smaller if restricted to only those.
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It is also probably true to say that we are not observable from inside a black holeThat's quite wrong. One can observe the outside just fine from in there. You can still measure the CMB for instance, but it will be even more redshifted than what we see here.
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Now if we were inside a BH, we would be subject to an incoming rain of photons from spaghettified stuff crossing the event horizon.It's tidal forces that spaghettify you. Jump into a large enough black hole and you'd not notice the crossing of the event horizon at all. Nothing of particular note happens to you there, even if you're looking out of the window. Yes, the tidal forces would eventually kill you (inside or outside the BH) before you reach the end of time singularity.
Concerning observation of the inside of a rock, the word 'observe' in 'observable universe' uses the physics definition of observation, not the street definition. The physics definition has nothing to do with humans, eyes, or specifically light.
Post by: Eternal Student on 06/11/2022 01:21:08
You've got to start feeling a little concerned for @alancalverd. It should be made clear that a lot of what they said was right, spelt correctly and quite well written. All of it was valued.
Best Wishes.
Post by: alancalverd on 06/11/2022 09:40:01
In cosmic coordinates, the observable universe relative to Earth currently has a radius of around 48 BLY. In various different coordinate systems, that size might be a lot more or a lot less. Point is, if you're using that coordinate system, black holes don't exist.But we know where some of them are, in our cosmic coordinate system. Knowing the position of something that you know doesn't exist is, to say the least, inconsistent with most definitions of sanity.
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No black hole event can have a causal effect on an event outside, but the outside events can have a causal effect on the events inside.True for events inside the black hole, but haven't we detected the collision of two black holes?
Post by: Eternal Student on 06/11/2022 16:11:20
First of all, I think I've made a mistake. I can only apologise.
In post #17 I said something like this:
I'm not going to bore everyone with the integration but it'll be enough to note what the metric is telling us. It says, that when r ≈ rs then a small change in the radial co-ordinate dr is worth an infinite amount of distance, ds. To say that another way, even if we take a small step across the event horizon (change r from rs-dr to rs+dr ) that is an unbounded amount of physical distance.I really should have bored everyone with the integration, then we'd have got it right. Sorry, it was too late at night when I wrote that post. If you do it carefully, then although we do have ds = f(r) dr and f(r) → ∞ when r → rs it doesn't race off to infinity too fast, recall that for integration we have dr → 0 and overall this integral
remains finite.
Anyone who's interested can find the details here:
https://physics.stackexchange.com/questions/85975/how-much-extra-distance-to-an-event-horizon
Fortunately it doesn't change too much of the spirit of what was said. That integrand becomes imaginary when r<rs which makes the distance from something inside the interior of a black hole to something outside of the event horizon an imaginary number, i.e. it's not an ordinary physical distance (it's a time separation). So we still have the assertion that nothing in the interior of a black hole is included in the Observable Universe (from plant earth).
I have edited the earlier post to show I was an idiot.
Best Wishes.
Post by: Halc on 06/11/2022 16:26:09
I have edited the earlier post to show I was an idiot.We all make mistakes. You're not an idiot for having done one, else I'd be well onto the moron end of the spectrum.
Perhaps the rest of us are the idiots for not spotting the error first (or ever had you given us time).
Post by: evan_au on 06/11/2022 20:08:35
Quote from: Halc
No black hole event can have a causal effect on an event outside
Quote from: alancalverd
haven't we detected the collision of two black holes?The gravitational field of a black hole extends well outside the event horizon.
When two black holes are in close orbit, the rapidly changing gravitational fields shake the fabric of spacetime, radiating away angular momentum in the form of gravitational waves.
Once the two black holes have merged, and the event horizon is a single ellipsoid, the gravitational radiation abruptly ceases.
So I don't think that these two statements are contradictory.
Quote from: Halc
Inside a black hole...You can still measure the CMB for instance, but it will be even more redshifted than what we see here.Scenario 1
I think that if you were able to maintain a constant distance just outside the event horizon, the infalling CMB photons from all directions would be blue-shifted? (or equivalently, your clock would tick slower than a distant clock=Einstein shift?).
- This same effect would occur if you were just inside the event horizon (only "maintaining position" becomes impossible)
Scenario 2
If you are free-falling into the black hole from infinity, you would reach a significant fraction of the speed of light by the time you reached the event horizon, and CMB from behind you would be strongly red-shifted (=Doppler Shift). But CMB from the sides would not be red-shifted by the same amount in classical physics.
- But the time dilation would still occur, so photons from the side would be blue-shifted?
This is a horrible mish-mash of classical and relativistic physics - can you clarify these two scenarios, please, Halc?
Post by: Halc on 06/11/2022 20:53:36
Quote from: Halc
No black hole event can have a causal effect on an event outside
Quote from: alancalverd
haven't we detected the collision of two black holes?Yes, but what we detect is radiation originating outside them. None originate from an event inside.
Keep in mind I'm not an expert, but you asked for my input.
The gravitational field of a black hole extends well outside the event horizon.Well yes, but the gravitational field of an apple extends just as far. A field is by definition 'everywhere', but the influence on that field of said BH extends no further than that of the apple. It arguably stops at the edge of the visible universe, but it also arguably doesn't.
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Once the two black holes have merged, and the event horizon is a single ellipsoid, the gravitational radiation abruptly ceases.Fairly abrupt in terms of our ability to measure it, but technically the radiation continues 'forever' just like the radiation of a light dropped into a black hole. So it fades to immeasurability very quickly.
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First of all, from either of these two observation points, all of the CMB would be visible regardless of the direction from which it comes, in contrast to a telescope on Earth where Earth blocks half of it. But to the observer in question, all the light from anywhere would appear to come from a finite angle of view. If you're hovering like that, it would appear to come from the direction of acceleration, and yes, it would be blueshifted in both cases due to your near infinite acceleration rate. It would be redshifted relative to any inertial observer. Somewhere in between I suppose there's a rate of moderate acceleration where the CMB would appear the same frequency as we see here on Earth.Quote from: HalcInside a black hole...You can still measure the CMB for instance, but it will be even more redshifted than what we see here.Scenario 1
I think that if you were able to maintain a constant distance just outside the event horizon, the infalling CMB photons from all directions would be blue-shifted? (or equivalently, your clock would tick slower than a distant clock=Einstein shift?).
- This same effect would occur if you were just inside the event horizon (only "maintaining position" becomes impossible)
Yes, a clock hovering just outside the black hole runs much slower than a distant clock, just like a clock at the rear of an accelerating rocket runs slower than another clock attached to the front of it.
As for the falling clock with no proper acceleration, which one runs faster is a frame dependent thing.
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Scenario 2Depends how you measure it. Your coordinate speed falls to zero as the EH is approached, which is a lot less than the coordinate speed of light, but still slower than the light falling in with you. So I guess one must ask 'the speed of which light?'.
If you are free-falling into the black hole from infinity, you would reach a significant fraction of the speed of light by the time you reached the event horizon
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and CMB from behind you would be strongly red-shifted (=Doppler Shift). But CMB from the sides would not be red-shifted by the same amount in classical physics.The CMB is visible only from a limited angle of view as the EH is approached. I think (not sure) that CMB light coming from the side (or even from the direction you're falling) would appear just as redshifted as the stuff in the middle.
The angle of view of the outside CMB gets narrower the deeper you fall, and only shrinks to zero degrees at the singularity. They have several really nice simulations of this on youtube.
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But the time dilation would still occur, so photons from the side would be blue-shifted?Nothing blueshifted if it's freefalling.
Post by: yor_on on 30/11/2022 21:50:53
Hi.
An alternative way of stating things:
"The observable universe" is something we can see. We're all fairly sure that it's a real thing.
"The Universe" is just a word with various definitions, no-one has seen it. It's generally understood to mean "everything, absolutely everything". This concept on its own is not a simple one. Can we say it's a set of things, a really big set that has everything in it? In Mathematics, most set theories are incompatible with the existence of such a set (see Wikipedia's entry if you're interested https://en.wikipedia.org/wiki/Universal_set ). To say this more strongly, the ZFC system of Mathematics - which is what you would have studied at school whether you knew it by the name ZFC or not - this does not allow the existence of such a set.
Best Wishes.
Interesting as well as funny ES, as it made me think of https://plato.stanford.edu/entries/goedel-incompleteness/