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AstronomyHow do black holes overcome the Pauli Exclusion principle? (self.askscience)
submitted by QuantumCastration
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When stars undergo gravitational collapse, those with remnants over the Tolman–Oppenheimer–Volkoff limit will collapse to black holes regardless off the degeneracy pressure from the Pauli Exclusion principle.
How is the exclusion principle broken in this case? Isn't it a fundamental law of nature?
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[–]Zelrak 346 points347 points  (70 children)
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No one is answering the actual question of how gravitational collapse can compete with the Pauli exclusion principle. The answer is that it competes with an effective pressure coming from the exclusion principle not the principle itself.
As you try to compress a bunch of fermions, some of them need to move into higher energy levels as you squeeze them into less spatial states. This costs energy and so creates a pressure opposing the compression. If the star is massive enough, there is sufficient gravitational pressure to overcome this and squeeze the fermions into the higher energy levels until the entire star is contained inside the Schwarzschild radius. After that we'll need a theory of quantum gravity to know what happens.
See http://en.wikipedia.org/wiki/Electron_degeneracy_pressure for an example of this effective pressure in the case of electrons.

[–]djimbobHigh Energy Experimental Physics 150 points151 points  (16 children)
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Yup this is the answer. Ordinary normal stars are a fight between gravity and gas/radiation pressure from the hot nuclear core undergoing fusion.
As the stars run out of elements for fusion, the gas pressure can't fight against gravity and the star collapses inward. If its in a certain mass range, it will become a white dwarf. This happens when the electron degeneracy pressure cannot be overcome by gravity. Electron degeneracy pressure means that the exclusion principle prevents electrons from getting closer to each other unless the electrons get excited into a higher energy state. This degeneracy pressure can get overcome once the force of gravity becomes powerful enough to force them to go into a higher energy state so they can be closer together.
If the collapsing star started in a higher mass range, the electron degeneracy pressure would be overcome and it further becomes a neutron star. Then the neutron degeneracy pressure fights against gravity. Basically as the star further collapses, all the electrons get absorbed by protons (inverse beta decay) into neutrons, and these neutrons prevent against further self collapse by rising into higher energy states.
Black holes happen with the largest mass initial stars and are just the last step where neutron degeneracy pressure is overcome and the stars collapse within the event horizon and we really don't know what goes on inside. If all degeneracy pressures are overcome (and presumably you could get a quark degeneracy pressure inside the BH), then you get a singularity which we really don't know much about.

[–]vriggy 8 points9 points  (0 children)
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Excellent information! Thanks, well put too!

[–]forever_stalone 5 points6 points  (14 children)
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So could there be a string degeneracy pressure if the quark degeneracy preassure is overcome?

[–]djimbobHigh Energy Experimental Physics 11 points12 points  (5 children)
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If the strings are fermions, then conceivably yes. Granted, you should recognize that string theory, while an interesting idea doesn't have much acceptance as fact in the physics community as currently it has made no tested experimental predictions. This is to contrast say quarks which are a fundamental part of the standard model and have lots of experimental evidence on top of theory.

[–]Cosmic_DongAstrophysics | Dynamical Astronomy 2 points3 points  (4 children)
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Not only that but string theory assumes SUSY and LHC has pretty much shown that SUSY doesn't hold up.

[–]djimbobHigh Energy Experimental Physics 4 points5 points  (3 children)
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The LHC has found no evidence of SUSY and has largely ruled out some of the simpler SUSY models (e.g., CMSSM), but I wouldn't call SUSY dead just yet. Here's a good review of the state of SUSY from Sept 2013 (remember LHC shutoff for two years from Feb 2013) .

[–]Cosmic_DongAstrophysics | Dynamical Astronomy 2 points3 points  (2 children)
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I'm don't have the time to read a 50 page review and I'm not a particle physicist. But when I talk to my friends across the street they all pretty much agree that SUSY is dead in the manner that any version of SUSY that still hasn't been invalidated is a very contrived version and therefore unlikely to be correct. Is that wrong?

[–]oss1xParticle Physics Detectors 5 points6 points  (1 child)
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I'm neither a theorist nor a SUSY expert, so please take with a grain of salt, but here is my understanding:
SUSY has something like 127 possible parameters. Most models ruled out by now (like CMSSM, M-Sugra etc.) will assume most of these parameters as their "trivial" value (think usually 1 or 0, depending how the parameter acts in the theory), only leaving very few (I don't know exact numbers but I believe usually it is <10) parameters "active". This makes for very well behaving, easily understood models with nice features, like easily solving the finetuning problem - which sadly are not realised in nature in this way.
By "turning the knobs" on all of the 120-something parameters you can generate a vast amount of arbitrarily complex manifestations of SUSY. Those models might not have all of the easily accessible features that the simpler models offer, but who says nature is realised in the most simple way? Who says we do not find a better theory that includes all previous particle physics theory that looks very simple an beautiful (like string theory) some time after an "ugly" SUSY model has been confirmed?
The parameter space searchable (and in some part already ruled out) by LHC is incredibly tiny compared to the full SUSY parameter space. And even the parameter space accessible by LHC has not nearly been exhausted yet. In the LHC design studies, discovery of SUSY was not even predicted for run1 of LHC (and I believe this was a prediction for a flawless run1 @14TeV, they actually ran at 7/8TeV).
So to conclude: SUSY is FAR from dead at all. SUSY accessible at LHC is also not dead yet, but it already took some massive hits, though in the end that does not matter. My old supervisor used to tell me that he believes nature is necessarily supersymmetric, as SUSY is the only theory that gives you a direct link between fermions and bosons. If SUSY only works on scales far higher than what we can probe right now, so what? That just means SUSY does not solve some of our current problems in particle physics.

[–]Cosmic_DongAstrophysics | Dynamical Astronomy 2 points3 points  (0 children)
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Ah ok, that makes sense. Thanks.

[–]gilescorey10 2 points3 points  (7 children)
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This is what I was thinking too. Could there be another degeneracy pressure that would make a black hole not a mathematical singularity, but appear to be one from our perspective because of the inability of light to escape?

[–]djimbobHigh Energy Experimental Physics 6 points7 points  (5 children)
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A Schwarzchild (non-rotating non-charged) BH is basically a singularity in the center of a spherical event horizon. The event horizon marks the distance where you'd need to go faster than the speed of light to reach escape velocity from the black hole.
So without going inside a BH we can't tell what a singularity looks like. Note, for normal (star-sized) BHs, we'd be killed by tidal forces (it's called spaghettification) if we tried going in a BH. With a supermassive BH (the type at the centers of galaxies), we could cross the event horizon but we still eventually be spaghettified when we got close enough to the singularity).

[–]Cosmic_DongAstrophysics | Dynamical Astronomy 4 points5 points  (2 children)
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Saying that there is a singularity at the center of a black hole is incorrect. A fundamental assumption of GR is local flatness, a singularity gives you infinite curvature, i.e. local non-flatness. So, the theory that predicts black holes does cannot contain any singularities. Therefore saying "we can't tell what a singularity looks like" is an incorrect statement, the correct statement is "we have no idea what happens at the center of a black hole because GR is not valid there", there could be a unicorn there for all we know.

[–]djimbobHigh Energy Experimental Physics 1 point2 points  (1 child)
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We agree gravitational singularities pop up in general relativity at the center of BH. This is currently our most accepted theory of gravitation; though we know its incomplete as its not compatible with QM. The places where we need a good theory of quantum gravity the most are at the big bang, singularities in BHs, where quantum and GR effects should both be significant.
Now whether these singularities that pop up in GR are actually singularities (with all the mass condensed to one hyper dense point smaller than a Planck length) or some weird new form of matter or what not or new big bangs or just deus ex machina for 2014 science fiction movies, I agree we do not know. If you go to the end of my first comment, I admit that we don't know much about singularities.
If all degeneracy pressures are overcome ... then you get a singularity which we really don't know much about.
I'm still calling them singularities (as I've always heard it called; per our sloppy language), as that's what our incomplete theories predict -- even though I understand that we have no idea what a singularity will be like.

[–]Cosmic_DongAstrophysics | Dynamical Astronomy 1 point2 points  (0 children)
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We agree gravitational singularities pop up in general relativity at the center of BH.
I have to disagree here. The center of a black hole is a point in which GR is not valid therefore GR cannot make any predictions as to what occurs there. That point in space is outside of GR, just as "the time before BB" is outside of GR.
I'm still calling them singularities (as I've always heard it called; per our sloppy language), as that's what our incomplete theories predict -- even though I understand that we have no idea what a singularity will be like.
That's the problem, ain't it? A lot of bad nomenclature has been grandfathered into astrophysics which causes a lot of confusion, I'm trying to avoid keep doing this because... it's bad.

[–]phungus420 2 points3 points  (1 child)
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A semantic correction (but it makes a difference). You can't go faster than the speed of light, it just doesn't make any sense. The Schwarzchild radius marks the distance where space is so warped that all trajectories in the fabric of spacetime lead toward the singularity. If by some magic you could travel faster than light, and you were inside the Schwarzchild radius, it wouldn't matter, you still couldn't escape, you can only get closer to the singularity - unless you want to say going faster than the speed of light allows you to go back in time (which arguably is what going faster than the speed of light means), but ignoring time travel there simply is no escape from a black hole.

[–]djimbobHigh Energy Experimental Physics 4 points5 points  (0 children)
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I agree with your first part; going faster than c doesn't make sense for massive or massless particles in relativity. When you think of relativity geometrically, the notion of going faster than c is pretty meaningless.
For tachyons (particles with imaginary mass), they will always go faster than c, have an imaginary proper time, would violate causality (e.g., could communicate backwards in time), aren't believed to exist (should create observable Cerenkov radiation if it interacts with other particles), and would lose energy by speeding up towards infinite velocity -- these would be able to escape the event horizon. Granted we don't believe these exist, so shouldn't talk about them.
But I wasn't claiming you can escape by going faster than c, just that it would be required.
There's a nice Newtonian model of a BH, where you just equate kinetic and gravitational energy (to find escape velocity): 1/2 mv2 = GM m/R, set escape velocity to c, and then find R = 2 G M/c2 as the event horizon radius (which actually agrees with the real GR derivation). Obviously we shouldn't put too much faith in this model (e.g., the Newtonian model for deflection of light by gravity is off by a factor of 2), but its useful pedagogically (like teaching the Bohr model of the atom before electron wavefunctions).
The point is that the EH is not a singularity or something physically noticeable to someone who travels to an EH. It's just the "point of no return" and being the furthest you can look in from the outside (as things approaching the EH will seem to move slower as an outside observer). The EH isn't anything physical and is not a singularity (even though it appears as infinite in some metrics, e.g., the Schwarzchild metric but that just means you should a more appropriate metric there).

[–]KeyBorgCowboy 1 point2 points  (0 children)
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This is the only solution that makes sense to me, that there is some stable form of matter more dense than a neutron star that puts the black hole mass radius at some value below the event horizon.

[–]Peace_Panda 11 points12 points  (26 children)
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Eli5 please

[–]Zelrak 85 points86 points  (12 children)
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Picture a bunch of marbles, if you put them on the ground they will all spread out so that none are on top of each other. If you put a rope around the marbles with some sort of slipknot and tighten the rope, the marbles will come together. When you squeeze enough they will bump against each other (this is the exclusion principle) and you won't be able to easily squeeze any more. Except that if you squeeze enough some of the marbles will pile on top of each other (move into higher energy levels) and if you were to let go they would fall back down pushing your rope away -- this is the degeneracy pressure. If you squeeze enough you could get all the marble into a very small but tall pile.
How's that?

[–]Peace_Panda 20 points21 points  (1 child)
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thank you, that was much easier. I'm trying to learn more about physics and quantum mechanics but i barely made it through algebra so some of the big words confuse me.

[–]ManikMiner 7 points8 points  (0 children)
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We're all here to learn! Thanks for the question and the answer guys !

[–]Myperson54 6 points7 points  (6 children)
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Is this why a star must be so-and-so massive to become a black hole? Because if they were less massive, there wouldn't be enough energy to overcome the PEP?

[–]Dont____Panic 4 points5 points  (3 children)
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In theory, any sized object (down to 22 micrograms, the minimum possible size where the equations are solvable) can make a black hole.
It's only that it can't collapse into that state due to gravity alone until the mass reaches a certain quantity. So, yes, the extra mass gives more gravitational force, which pushes it over the degeneracy pressure.

[–]tree-ent 2 points3 points  (1 child)
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What else is required for smaller objects? Compression of some sort or added energy or something like that?

[–]Dont____Panic 1 point2 points  (0 children)
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Yep. Colliding particles at high energy can theoretically make a micro-black hole briefly.
Massive input of energy in the form of, perhaps, kinetic or magnetic form might do it.
I'm not really sure what else could... Theoretically it's possible though.


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    [–]djimbobHigh Energy Experimental Physics 11 points12 points  (12 children)
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    It's not like you are five, but here goes.
    There are two classes of particles -- bosons (particles with integer spin -- prominent examples include photons, gluons, the Higgs) and fermions (particles with half-integer spin -- prominent examples include electrons, protons, neutrons, quarks).
    The Pauli exclusion principle applies to all fermions (and doesn't apply to bosons) and states that two fermions can't be in the same quantum state (meaning have exactly identical quantum numbers) in the same location (that is within one de Broglie wavelength of each other). You typically encounter quantum numbers in chemistry class when you build up the periodic table -- an electron in an atom has four quantum numbers - principal (n), azimuthal (ℓ), magnetic quantum number (m_ℓ -- I'm using the underscore to indicate subscript; this should be an m with a subscript ℓ), and spin quantum number (m_s). Generally, electrons will want to be at the lowest energy levels allowed, so you fill up the lower energy levels first.
    There are rules for quantum numbers; e.g., you have the principal quantum number (n) which determines the gross energy. For example in the hydrogen atom, the energy levels go as E = -13.6 eV/n2 (neglecting fine structure; energy slightly depends on values of other quantum numbers). So for hydrogen n=1 is -13.6 eV, n=2 is -3.4 eV (which is higher than -13.6 eV), n=3 is -1.5eV (even higher), etc. For a given n, the allowed values of ℓ range from 0, 1, 2 ..., n - 1. E.g., if n = 2, then ℓ can be 0 or 1. If n = 5, ℓ can be any of {0,1,2,3,4}. For a given value of ℓ, then m_ℓ ranges from -ℓ ... -1, 0, 1, ..., ℓ. E.g., for ℓ=1, the allowed values of m_ℓ are -1,0,1. And m_s is always either +1/2 or -1/2. All these seemingly arbitrary rules actually come out quite nicely if you solve the Schrodinger equation for the hydrogen atom, but to do that you need to really take a full quantum course and understand solving partial differential equations. Anyhow the point is in the n=1 energy level, you are allowed only two electrons with (n, ℓ, m_l, m_s) being (1,0,0,1/2) and (1,0,0,-1/2). This is why the first row of your periodic table has two elements. In the n=2 energy level, you are allowed eight elements -- two with ℓ=0: (2,0,0,1/2), (2,0,0,-1/2) and six with ℓ=1: (2,1,1, +/- 1/2), (2,1,0, +/- 1/2) and (2,1,-1, +/-1/2). This is why in the next row of elements you have Li, Be (the elements with n=2, ℓ=0 being the last shell filled) and then six more elements (the six with n=2, ℓ=1). Note physicists go by quantum numbers; chemists go by shells - where ℓ=0 corresponds to s-shell, ℓ=1 is the p-shell, ℓ=2 is the d shell, ℓ=3 is the f-shell. So the 3d shell for example, corresponds to n=3, ℓ=2 and allows up to 10 unique sets of quantum numbers in it (as m_l has five values 2,1,0,-1,-2 and m_s has two values for each m_l). See this periodic table where in each region they show the shell of the last electron in each energy level. Ok, enough digression into quantum numbers and building the periodic table.
    Now for a star that runs out of fuel for fusion, it will start to cool. It can no longer give a strong enough force to counteract the inward force of gravity, so it starts self-collapsing and occupying a smaller and smaller volume. Eventually, the only thing that prevents further collapse is this Pauli exclusion principle on the electrons in the star. The electrons want to go into the same location as other electrons, but to do that they have to get excited to another higher energy level. This is called degeneracy pressure. (Degenerate is a fancy word for two particles having the same quantum numbers -- and degeneracy pressure is the effective outward force that comes from particles in the same state not being able to be squeezed more tightly together). You can overcome degeneracy pressure once you have enough energy to raise the energy levels of the electrons so they can be in the different quantum states and thus be in the same location.
    (Warning: The following is a gross simplification as the electrons in a white dwarf should be treated as a free electron gas and will not have energy levels similar to an isolated hydrogen atom). If you imagine the quantum numbers and energy levels for electrons in the white dwarf behaved just like electrons in a hydrogen atom with E=-13.6 eV/n2 , you could see once gravity is powerful enough (from starting with a high enough initial mass) to give each electron up to 13.6eV of energy, then you can completely overcome the electron degeneracy pressure as electrons can be in a ton of different quantum states. E.g., some electron could go to the n=10 energy requiring about 13.5eV (because -13.6eV/102 ~ 0.1 eV), going to the n=100 energy level would require 13.6 eV (-13.6eV/1002 ~ 0.001 eV) and going to the n=1000000 energy level wouldn't require much more energy either. (Really to properly analyze this should deal with other concepts like Fermi energy, but just giving a rough idea for intuition.)

    [–]Greifenhorst 0 points1 point  (9 children)
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    Does the de Broglie wavelength of an electron change as it reaches a higher energy level/quantum state? I'm having trouble understanding the concept of electrons in a standing wave configuration taking up 'less space' by moving further away from the nucleus/into a higher orbital. How are the atoms able to be compressed further if the kinetic energy of the electron is higher?

    [–]djimbobHigh Energy Experimental Physics 0 points1 point  (8 children)
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    The Pauli exclusion principle only excludes particles from overlapping when they are in the same quantum state (all the quantum numbers are the same - the particles are degenerate) and the same place. Otherwise, the particles are more than happy to share the exact same space with a different particle as long as they are in a different quantum state.
    At the quantum level don't think of particles as physical balls or anything. They are just fields that interact by these often strange rules of quantum mechanics.

    [–]Greifenhorst 0 points1 point  (7 children)
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    Interesting. This raises many questions for me (physics undergrad) so any information you can impart would be greatly appreciated.
    How does the PEP affect the nuclei of two neighbouring/compressed atoms? Are the nucleons able to occupy the same space and achieve a different quantum state in order to accommodate the PEP? I imagine that there can be no change to their principal quantum number - do their azimuthal/magnetic/spin QN change, and if so, in a general sense, how does this affect the behaviour of the nucleons with respect to each other and the atom as a whole?
    At what point do two nuclei occupying the same space with differing quantum numbers transition into nuclear fusion? When they become degenerate? Is the energy released in this reaction proportional to the input energy required to align the nuclear quantum numbers?

    [–]djimbobHigh Energy Experimental Physics 0 points1 point  (6 children)
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    In the core of the sun ( ~105 kg/m3), electron/neutron degeneracy is insignificant. In white dwarfs (~109 kg/m3), nuclear degeneracy is insignificant and there's not the right fuel/temperature for fusion. Neutron stars where neutron degeneracy matters are at densities of ~1017 kg/m3.
    Remember electrons are about 2000 less massive than nucleons, so electrons have a much larger de Broglie wavelength (lambda ~ h/p) so degenerate effects happen first.
    I was being a little deceptive when I talked about the quantum numbers for the hydrogen atom. Really (n, ℓ, m_ℓ, m_s) are meaningless quantum numbers in a white dwarf; but they are the only connection people have to quantum numbers and Pauli exclusion principle. In a white dwarf electrons aren't bound to individual atoms, so being in a particular orbital or energy level from a nucleon doesn't make sense. There are quantum numbers though (different ones) and the same PEP applies preventing two from being in the same quantum state.
    Fusion is a completely different beast than degeneracy pressure. You just have a hot nuclear core at densities much less than degeneracy and the nucelons undergo say a proton-proton chain reaction. You just need a temperature to get protons to overcome the electrostatic repulsion and then you have to get lucky to go down a stable chain to get to stable alpha particles.

    [–]Greifenhorst 0 points1 point  (5 children)
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    Hmm. So the electron degeneracy pressure of a white dwarf is due to non-localised/non-uniform electron distribution? What happens to the electrons? Do they move outward toward the surface?
    How does neutron degeneracy manifest as pressure? If you compress them, what change do they undergo in order to accommodate the PEP? Since there is no electrostatic repulsion, what is the agent of degeneracy pressure?
    Thanks you for your responses, by the way.

    [–]djimbobHigh Energy Experimental Physics 0 points1 point  (4 children)
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    Since there is no electrostatic repulsion, what is the agent of degeneracy pressure?
    Sigh. I feel like you misunderstood all my posts. Degeneracy pressure has to do only with the Pauli exclusion principle which prevents identical fermions (same particle in same quantum state) from being in the same spot. It's a strictly quantum mechanical phenomenon arising as spin-half particles have anti-symmetric wavefunctions (and half-integer spin being anti-symmetric arises naturally in QFT). Nothing to do with electromagnetism.

    [–]Greifenhorst 0 points1 point  (3 children)
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    I understand the difference between electrostatic repulsion and degeneracy pressure, I was merely referring to the repulsive force that prevents protons from fusing, and asking how neutrons are prevented from doing the same given their neutral charge. I know that neutron degeneracy pressure is the answer, but what is the actual change that occurs to provide this pressure? In electrons it is the input energy required to increase their energy level - what happens with neutrons? Do they, too, increase in energy level, and if so, what does that mean being that they, unlike electrons, cannot move further away from the 'nucleus'?

    [–]Gelsamel 0 points1 point  (1 child)
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    Wait, if you should treat it as a free electron gas then is there even an issue with degeneracy then? Aren't free particles not quantized?

    [–]djimbobHigh Energy Experimental Physics 2 points3 points  (0 children)
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    Maybe calling it free isn't the best term as we aren't neglecting the Pauli Exclusion principle and they are still bound gravitationally. More non-interacting electron gas. You basically deal with the Fermi energy (the minimum energy of the highest energy particles) of N particles confined to a volume V. One way to do it is essentially solving the particle in a box for Schrodinger and then seeing how high the energy levels have to go to fit N particles in the box all at different energy levels.
    See: http://farside.ph.utexas.edu/teaching/qmech/Quantum/node66.html
    If you solve it you'll end up with a Fermi energy proportional to N2/3/R2.
    But to see overcoming degeneracy pressure you really have to be in the relativistic regime. Following Landau's argument: you have N particles in a star of radius R, that the number density is n ~ N/R3. Volume per fermion is 1/n, or in one dimension delta-x ~ n-1/3 so by the uncertainty principle, momentum is of order p ~ hbar/(delta x) ~ hbar n1/3.
    The fermi energy for relativistic particles is E_f ~ p c ~ hbar n1/3 c = hbar c N1/3/R. Gravitational energy per particle goes as E_g = - G M mp / R = - G N mp2/R, (where mp is mass of a particle; note M = mp N).
    So total energy goes as:
    E = hbar c N1/3/R - G N mp2/R
    Note that N goes in both terms (positive term as N1/3 , negative term as N) and both terms have the same 1/R dependence. If N is very large, then this simplifies to E ~ (negative)/R ; if N is smaller, then it simplifies to E ~ (positive)/R
    Physically, we reach equilibrium when we minimize energy. If you try minimizing a system with E ~ (positive)/R , you'll lower energy by increasing R. Eventually you'll get out of the relativistic regime, where the assumption E ~ (delta-p) c was valid. Then the Fermi energy depends on 1/R2 so you can actually find an equilibrium radius. If you try minimizing a system with E ~ (negative)/R, you'll lower energy by decreasing R. This will keep the problem in the relativistic regime, and keep raising the Fermi energy (meaning degeneracy pressure is being overcome by gravitational energy as electrons keep getting pushed into higher and higher quantum states).

    [–]StarkRG 4 points5 points  (1 child)
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    If I understand it correctly, basically once everything's inside the Schwarzschild radius we really have no clue what goes on. On the one hand physics suggests matter collapses into a zero-width point. But on the other hand physics also tells us it CAN'T collapse into a zero-width point (ie matter takes up space). Physics, as it currently stands, is simply unable to resolve this conflict. For all we know all that matter is converted, through some unknown mechanism, into energy which can very happily exist in a point. However, from the perspective of someone outside, whatever happens inside has absolutely no effect outside, as long as gravity remains the same.

    [–]keeead 1 point2 points  (1 child)
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    Yeah, when I wondered about this, I imagined neutron stars and 'pasta matter.' The uncertainty principle, Pauli Exclusion and the Dirac Sea imposing a repulsive force that keeps the star puffed up.
    What fundamentally changes that overwhelms that repulsive force as matter collapses further than a Neutron Star into a Black Hole?

    [–]Zelrak 6 points7 points  (0 children)
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    Nothing fundamentally changes as far as we know. It's just that the gravitational pressure gets stronger faster than the degeneracy pressure does when you increase the mass of the star. At some mass the gravity wins, but bellow that mass the degeneracy pressure is enough to stabilise the system.
    Think of a weight at the end of an elastic. As you increase the weight the equilibrium point will get lower as the elastic force balances gravity. Eventually if you increase it enough the elastic will break when the elastic force can no longer compensate. (Don't take this analogy too seriously it's just the first thing that came to mind.)

    [–]Froztwolf 0 points1 point  (14 children)
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    until the entire star is contained inside the Schwarzschild radius.
    Is it possible to have a stable situation S where only a part of a star is inside the radius?

    [–]lookmeat 2 points3 points  (4 children)
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    Stable? Probably not.
    Have you ever asked yourself how strong gravity would be at the center of the sun?
    Assuming a perfect sphere the answer is 0.
    A way to think about it is that, when you are inside a sphere there's gravitational force of whatever is between you and the center of the sphere, but also what is between you and the surface of the sphere! An easy way to visualize this is imagine that the sphere is actually two things: a small sphere whose surface is where you are and a shell, which goes from the surface until where you are; both with the same center. The gravitational pull of a spherical shell when you are inside is 0, everything cancels itself out, no matter where you are.
    Lets assume an equal density sphere. The mass of a sphere is linearly proportional to the volume, which itself is proportional to its radius r3. Since your mass and the gravitational constant remain the same, we can claim that the part of the gravitational pull that changes is V/r2, where V is the volume of the sphere, since the sphere only has its radius changing, we get something like: r3/r2 which becomes r. This means that, as long as we are within as sphere, the gravitational pull decreases linearly the closer we are at the center, becoming 0 at the center.
    So our star has less of a gravitational pull at the center than at the surface!
    Ah but stars don't have uniform density! This is true, but the point remains, at the surface you receive the gravitational effect of the center plus the surface. Unless our star is filled with anti-gravity (I separate this from negative mass which is a quantum thing and therefore does not consider how it works with gravity) it will always have a stronger pull at the surface than at the center.
    It makes sense, that's why things don't have blackholes at their center: once you get to the Schwarzchild radius for the thing, the amount of gravitational pull has decreased, and so has the radius!
    So as our star's density increases, it's full radius decreases too. Until it becomes too much and it collapses. The part that actually collapses first is not the center it's the surface! The surface is constantly pushing down with huge force on the core.
    Because stars are not of uniform density, the highest density material goes to the center. As the pressure increases, the highest density remains on the center. When the density is enough to surpass the Exclusion Principle, the core (which is where it is) collapses. The surface immediately follows. The thing is that the core quickly stabilizes itself into a new, more resistant form (generally made of neutrons in a supernova) which leads to the surface "bouncing" off the star. What remains is generally a neutron star, or if it's small enough a black hole. Sometimes a neutron star remains, but as the matter expelled falls back into the neutron star it collapses into a black hole.
    So it's impossible, because for the surface to be stable that would imply that the surface was feeling no force, but the core was. In reality the surface is the one with the strongest force of gravity (just not the highest pressure).

    [–]Froztwolf 0 points1 point  (3 children)
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    Your answer is very informative. Thanks a lot for taking the time.
    My question seems to stem from me misunderstanding what the Schwarzchild radius is. Didn't realize that it essentially just describes the size at which a massive body becomes a black hole.
    until the entire star is contained inside the Schwarzschild radius.
    This made me think that maybe there were situations where only a part of the star was contained inside the radius. I was probably conflating it with the Event Horizon, getting confused and not realizing that every item that's not a black hole is partially contained within it's Schwarzchild radius.
    While there would be no gravity at the center of the sun, surely there would be immense pressure from the outer layers, wouldn't there? That is, from the parts of the sun that ARE subject to gravity and are being pushed towards the center.
    once you get to the Schwarzchild radius for the thing, the amount of gravitational pull has decreased, and so has the radius!
    Not sure I understand. Why would the gravitational pull have decreased when the thing has reduced below its Schwarzchild radius?
    Edit: Foamratting

    [–]lookmeat 1 point2 points  (2 children)
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    Not sure I understand. Why would the gravitational pull have decreased when the thing has reduced below its Schwarzchild radius?
    So, to simplify our thing again. The Schwarzchild radius is the distance for a mass were the escape velocity is equal to c. In a black hole the event horizon would be on the Schwarzschild radius.
    Now lets imagine our even density star. Lets say that it has a mass equal to the sun, so it has a schwarzschild radius of 3km. Now here's a fun thing, the sun actually has a density lower than Earths. Lets assume our star also has the density of the sun (only it's evenly spread). So we travel down the star until we are only 3km away from the center. Notice that the outside of the star is not pulling us toward the center anymore, only the 3km radius star, which means the mass that is pulling us is now much less than the who star's, only 58336269741kg. In order for the escape velocity of that mass to be equal to c, the radius would have to be about2.35149895×10-16 m! So we get closer, but as we get closer the amount of star that is pulling us to the center keeps getting smaller and smaller.
    So as we keep moving closer to a star to get to it's schwarzchild radius, we move into the star, and the part that is pulling us in becomes lighter and lighter. We can never reach a true point were we cannot escape unless the star was smaller than it's schwarzschild radius to begin with.

    [–]Froztwolf 0 points1 point  (1 child)
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    So, I understand that if we were to move inside a star, the further in we get, the less the gravitational pull would be. No problems there.
    But if the star itself shrinks, surely the gravitational pull at a point on the new, smaller surface would not be smaller than the pull at the old, larger surface? Since the same mass is pulling at the point.
    Seeing as how gravity drops off with distance, surely it would be greater, no?

    [–]lookmeat 0 points1 point  (0 children)
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    If the star increases it's density, then yes. If the density is high enough that all the mass of the star falls within the radius, it becomes a black hole.

    [–]cricketwisperer 1 point2 points  (2 children)
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    You mean during gravitational collapse? Black holes form when a sufficiently large star runs out of fuel and all the matter falls inward because there's no force strong enough to oppose gravity. There's no chance of this stable situation where some of the star matter somehow stays outside of event horizon (Schwarzchild radius).
    Or do you mean a different situation? It's hard to tell from your question. A black hole from the perspective of gravity is no different than another massive body with the same mass. If you replaced our sun with a black hole of the same mass, the Earth and other planets would continue the same orbits. Black holes are not some cosmic vacuum cleaners.

    [–]Froztwolf 0 points1 point  (1 child)
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    My question seems to stem from me misunderstanding what the Schwarzchild radius is. Didn't realize that it's essentially just describes the size at which a massive body becomes a black hole.
    Are there similar ways to find the radius at which stars become neutron stars?

    [–]cricketwisperer 1 point2 points  (0 children)
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    For a star to collapse into a neutron star, you need a remnant between 1.4 to 3 times the mass of the Sun. (I can't remember the exact size, but I believe this means the star needs to be about 8-10 times the mass of the Sun before it dies, cos lots of the stuff gets blown off during the supernova explosion).
    The Schwarzchild radius is mostly theoretical for bodies smaller than large stars. Practically, the only way we've seen matter being compacted to that degree is through gravity, meaning you need a ton of mass. So, even though the Earth (or a human body, a cell phone, etc.) has a Schwarzchild radius, there's no means to actually compact the earth into that small space.

    [–]PM_ME_UR_REDDIT_GOLD 0 points1 point  (5 children)
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    sure, A part of every star is inside the Schwarzchild radius all the time

    [–]TheCat5001Computational Material Science | Planetology 2 points3 points  (4 children)
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    No, it's not. Because not all mass is at the center.

    [–]astrawnomore 0 points1 point  (0 children)
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    Thanks! I've read about this before in my course work, but this made it really click. :)

    [–]thiosk 0 points1 point  (0 children)
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    Hi,
    How MUCH energy are we talking here to over come the this pressure? Ramming fermions together like this-- are we talking megajoules/mol? petajoules/mol?


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      [–][deleted] 251 points252 points  (121 children)
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      We'd have to get a better idea of the internal physics of black holes before this question could be given the answer it deserves, I think. It's a good question though, and one of the big ones that needs to be explained.
      However... given what we suspect about black holes, e.g. that their mass may be condensed into an infinitely small point (a gravitational singularity), it's quite likely that their mass no longer takes the form of matter as we know it. We already know that the exclusion principle fails in degenerate objects that cross a particular threshold in the 2-3 Solar mass range, even if we don't yet know the exact TOV limit. At such high gravities, it's possible that the fermion vs. boson distinction itself might just fall apart, making exclusion irrelevant - so many other aspects of established physics (e.g. Newtonian physics in general) fall apart at the event horizon, so maybe those do as well?
      TL;DR: We're not sure yet, but we can speculate!

      [–]whooosh32 65 points66 points  (100 children)
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      Could our universe be inside a giant blackhole?

      [–]o0shad0o 138 points139 points  (85 children)
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      When you plug the estimated mass and size of the universe into the equation for calculating the Schwartzchild radius, they more or less match; so it's entirely possible!

      [–]TheOneTrueTrench 217 points218 points  (66 children)
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      Universal Gravitational Constant: G (6.673×10−11 N*m2 / kg2 )
      Speed of Light in a vacuum: c
      Approximate mass of the universe: m (0.92*1053 kg)
      Formula for the Schwarzschild radius: 2*G*m / c2
      Which comes out to: 1.366×1026 meters, or 14.44 billion lightyears.
      The approximate age of the universe is 13.8 billion years, so they are pretty fucking close.
      Wolfram Alpha Calculation
      So, yeah, the age of the universe in years is pretty close to the Schwarzschild radius of the mass of the universe in light years, pretty cool huh?
      >.>
      <.<
      Except that the actual radius of the universe is 46.5 billion lightyears due to expansion, so none of this actually works out, and the observable universe is bigger than the Schwarzschild radius of its mass.
      I love finding things that make people happy and ruining them with math...
      EDIT: I love it when people ruin my math with facts.
      According to Wolfram Alpha, wtfisthat it's 533 billion ly
      Now, if the baryonic mass of 0.921053 kg is accurate, dark matter and matter together are about 6.461053 kg. I don't believe that dark energy contributes to the gravitational effect across the universe. It would approximately match the size of the observable universe.
      -BUT- the estimates of the mass of the universe range over 10 orders of magnitude. So, hell, I don't know.

      [–]macutchi 30 points31 points  (6 children)
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      Is that the mass of the baryonic matter alone? Isn't there dark matter and dark energy variables to be considered?
      Serious question.

      [–]Bausse 45 points46 points  (2 children)
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      The mass of 0.92 x 1053 kg includes baryonic matter and dark matter, but excludes dark energy.
      Let's add dark energy, shall we?
      Sir Fred Hoyle calculated the mass of an observable steady-state universe using the formula:
       (4/3) * π * ρ * (c / H)^3
      This can also be stated as
       c^3 / 2*G*H
      H = Hubble Constant, ρ = Hoyle's value for density, G = Gravitational Constant, and c = speed of light.
      This calculation yields approximately 0.92×1053 kg; however, this represents all energy/matter and is based on the Hubble volume (the volume of a sphere with radius equal to the Hubble length of about 13.8 billion light years). The Hoyle equation mass/energy result must be adjusted for increased volume due to the comoving distance radius of 46.6 billion light years.
      The comoving distance radius gives a volume about 39 times greater (46.7 cubed divided by 13.8 cubed). However, as volume increases, ordinary matter and dark matter would not increase; only dark energy increases with volume. Thus, assuming ordinary matter, neutrinos, and dark matter are 31.7% of the total mass/energy, and dark energy is 68.3%, the amount of total mass/energy for the steady-state calculation would be:
      1. mass of ordinary matter and dark matter (31.7% times 0.92×1053 kg <--- /u/TheOneTrueTrench 's value of mass)
      2. plus the mass of dark energy (68.3% times 0.92×1053 kg)
      3. times increased volume (39).
      This equals: 2.48×1054 kg. Ordinary matter is 4.8% of all energy/matter. If the Hoyle result is multiplied by this percent, the result for ordinary matter is 1.20×1053 kg.
      So basically it's only a difference of one order of magnitude. You will notice that a lot of sources, ranging from NASA to the Planck Institute to Wikipedia, give different constants for a lot of the arguable variables (percentages of matter types, cosmological constants, masses of the universe, radius of the universe, and density of the universe) that vary within a few orders of magnitude. This is extremely normal for astrophysical calculations, so don't be alarmed!
      edit: formatting and additional links, clarity.

      [–]ombx 0 points1 point  (0 children)
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      The mass-energy ratio of atleast the Observable Universe:
      4% - Visible matter (approx)
      26% - Dark matter (approx)
      70% - Dark Energy (approx)
      These are all approximate but close percentages, and the actual percentages may vary (but not by much).

      [–]drpeterfosterGenetics | Cell biology | Bioengineering 12 points13 points  (13 children)
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      Well, aren't you only considering the mass of the observable universe? The CMV is the "edge" of what we can see, but we have no reason to think it is the edge of the universe. Rather, that is as "far back in time/distance" that we can observe. Because of the expansion of spacetime, there are untold numbers of galaxies that have already been pushed beyond the threshold of observation.
      As I understand it, this means we have NO IDEA how big the WHOLE universe is; we can only estimate the mass of the chunk of spacetime in which light is still able to reach us. Overtime, expansion will slowly push more and more galaxies out of our limit of observation and the observable universe will appear to lose mass.
      Right?

      [–]ceilte 6 points7 points  (5 children)
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      As spacetime is warped in the presence of gravity, is it possible that there's more space on the inside of a black hole or our universe than the outside (the Schwarzchild radius)?

      [–]Shattered_Sanity 2 points3 points  (3 children)
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      I wonder if it's like Gabriel's Horn, except in an extra dimension. Gabriel's horn is easy to visualize: graph y = 1/x then cut out everything before x = 1, and spin what's left around the x-axis. The volume of the tapering cone is π, the area is infinite. If the universe was 2-D and all of the mass in a black hole really is concentrated in an infinitely small point, it might look similar. In 2-D, of course, space is measured by area, not volume.

      [–]ceilte 0 points1 point  (2 children)
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      I've poked around at the concept of Gabriel's Horn before, studying the apparent paradox of a finite surface area but infinite volume, but I know my limits. ;)
      Fortunately, the concepts and math of real spacetime being TARDIS-like in nature have already been approached with, fortunately for mankind, far greater minds than mine.
      The only significant difference in this case is that I'm asking whether or not these TARDIS-like spaces exist within an event horizon: A common example given to beginning physics student is that space is like a "rubber sheet", where it deforms around massive object (read: objects with mass), but I never got a straight answer to whether that deformation actually included stretching of spacetime also.
      Given that gravity and spacetime don't seem to make much distinction between miniscule massive objects (electrons, for instance) versus such gargantuan objects as a galactic core black hole, it would seem to me that a human-scale experiment could be devised to determine if the phenomenon actually exists.

      [–]Shattered_Sanity 0 points1 point  (1 child)
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      I never got a straight answer to whether that deformation actually included stretching of spacetime also.
      To the best of my understanding (not a physicist), the rubber sheet is spacetime, thought in 2-D. Think about this: light always travels in a straight line. Photons have no mass, so gravity shouldn't have any effect on them. But massive objects bend the paths of light rays by bending the spacetime they're in. See Gravitational Lensing.
      I'm not totally sure whether gravity stretches spacetime, but you may find this interesting. Stretching itself is very well substantiated in cosmology via redshift. Imagine a photon as a wave. The longer it travels through the expanding universe, the more it's stretched out, and the redder the light appears. By measuring positions of characteristic spectral lines, we can get a good idea of how long the photons have been traveling and thus how far away the source is.
      As for electrons v. black holes, you're venturing into uncharted territory. Every experiment run to test physics on atomic scales has substantiated quantum mechanics, while everything that falls into the realm of general relativity (planetary orbits, cosmological observations, etc.) has substantiated relativity. Headway has been made in uniting special relativity with quantum mechanics, but GR and QM are fundamentally incompatible. A century later, we're still looking for some way to reconcile the two or a better theory that does away with both, much like Einstein disproved Newton. That's about as far as my understanding goes.
      it would seem to me that a human-scale experiment could be devised to determine if the phenomenon actually exists.
      We're still working on LIGO.

      [–]ceilte 0 points1 point  (0 children)
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      Really, the only part of the question I'm concerned with (and I apologize, I tried to make it clearer before, but did a poor job of that) is whether the presence of mass stretches spacetime in addition to distorting it. I don't think we will find any firm boundaries to quantum gravity once we have a unified theory, and expect that future experimentation will prove that even the smallest massive objects will distort spacetime, albeit to an incomprehensibly small degree.
      With the sheer amount of stuff in the universe (~1e53kg), the amount of fabric stretching, if it exists, should be observable.
      Maybe /u/djimbob could offer his opinion on this?

      [–]StarkRG 1 point2 points  (0 children)
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      That's an interesting thought. The issue would be how to maintain the strength of gravity. As long as there was a mechanism to maintain the same amount of gravity at the event horizon you can do whatever you want with the inside of a black hole. Nothing inside an event horizon can have any impact on the outside (that's why it's called an event horizon, it's a horizon (something you can't see beyond) of events).

      [–]wtfisthat 2 points3 points  (9 children)
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      Wolfram Alpha also provides a mass estimate for the universe that is larger than the one you're using (it returns 3.4x1054 kg). Are you including dark matter and dark energy?
      http://www.wolframalpha.com/input/?i=mass+of+observable+universe&lk=4&num=1
      Plugging that mass in gives a Swartzchild radius of 533 billion ly...
      http://www.wolframalpha.com/input/?i=%282+G+3.4*10%5E54+kg%29%2Fc%5E2

      [–]Ch3mee 0 points1 point  (8 children)
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      Then the universe is denser than a black hole? As far as we can tell?

      [–]randomguy186 1 point2 points  (0 children)
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      the observable universe is bigger than the Schwarzschild radius of its mass.
      This is the key fact.
      Question - does the approximate mass include dark matter? Ninja edit: Wikipedia says your figure excludes dark matter.

      [–]RagingOrangutan 1 point2 points  (1 child)
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      I mean... That's only off by a factor of 4. By cosmic standards that's negligible.

      [–]o0shad0o 0 points1 point  (0 children)
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      And the diameter numbers given by u/TheOneTrueTrench are also off by a factor of 4. And according to the equation the numbers are related linearly.

      [–]Thimoteus 4 points5 points  (11 children)
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      I've never seen that 46.5 billion lightyears number, I'd be interested in a reference if you have one.

      [–]jonahedjones 11 points12 points  (0 children)
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      http://www.astro.ucla.edu/~wright/cosmology_faq.html#DN
      Basically the universe is 13.6 Billion years old but and the most distant objects we can observe emitted the light we see today 13.6 Billion years ago. But in that time they have moved away due to the expansion of the universe, which expands as t2/3 .

      [–]ombx 4 points5 points  (2 children)
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      The radius of the Observable Universe is approximately 46–47 billion light-years, and the diameter is approximately 93 billion light years.
      This is due to metric expansion of space.

      [–]mjedm6 3 points4 points  (0 children)
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      http://en.wikipedia.org/wiki/Observable_universe
      Fourth paragraph.
      The diameter is more oft quoted versus the radius as ~90 billion light years

      [–]Smithium 0 points1 point  (2 children)
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      Is the Schwarzschild radius measured from outside the black hole (half it's apparent diameter) or the distance from the event horizon to the singularity at the center (potentially infinite distance?).

      [–]Joff_Mengum 0 points1 point  (1 child)
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      How would it be infinite? The Event horizon will have a finite apparent radius so the distance from it to its center will be finite as well.

      [–]Smithium 0 points1 point  (0 children)
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      Spacetime stretches under the extreme gravitational effects. The distance through is greater than the apparent distance from outside.

      [–]demandingsmudge 0 points1 point  (1 child)
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      where does one learn this what education do you have lol last time i learned planets was 6th grade

      [–]JulitoCG 0 points1 point  (1 child)
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      Ok, so I had heard once that, within a black hole, dimensions kind of change. I believe it was an article in Scientific America that mentioned that the usual time dimension becomes "space-like," while the spacial dimension representing movement along the radius (up-down, I figure) becomes "time-like." Does this relate whatsoever to what we're discussing here?

      [–]TheOneTrueTrench 0 points1 point  (0 children)
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      Inside a Schwarzschild radius, all paths must lead into the singularity. And since the edge of the universe recedes from us at the speed of light, no path can lead out of the universe, and the edge of what we actually see as we look out is the beginning of the universe.
      So, it sort of doesn't.
      Time like, in this case sort of means that progress on that axis must be towards the center of the black hole.
      So, the center of the observable universe is always where you are, so it sort of does apply?

      [–]avabit 7 points8 points  (0 children)
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      The Schwartzchild radius reasoning is about static solution of GE equations and does not apply to rapidly expanding matter. See this and this for more detailed arguments against the "Universe is a black hole" thesis.

      [–]pananana1 1 point2 points  (1 child)
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      Isn't the size of the universe just the size of the observable universe? Just because c x (age of the universe) happens to be close to the Schwartzhild radius, why is this viewed as evidence at all, considering c x (age of the universe) has nothing to do with the actual size of the universe?

      [–]o0shad0o 0 points1 point  (0 children)
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      I never got into those details, I'm afraid... When I discovered the formula for calculating the Schwartzchild radius of a black hole was linearly related to mass, I looked up estimated sizes and masses other people had proposed, and when plugged into the equation they seemed to fit together.


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        [–]Jaccep 4 points5 points  (5 children)
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        There's a theory that all universes are inside black holes, and universes with better conditions for black hole creation are able to create more "children" universes in a weird sort of natural selection.

        [–]kalimashookdeday 1 point2 points  (3 children)
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        I saw a doc on that too - was so crazy to think about how evolutionary properties could be observed on cosmic levels too (possibly).

        [–]kareesmoon 0 points1 point  (2 children)
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        You remember the name of that doc?

        [–]kalimashookdeday 0 points1 point  (1 child)
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        I wish I could - I have a terrible memory with these things. I watch only educational stuff so everything ends up blending together.
        Here's an article about the same topic via NatGeo - I want to say the documentary might have been about - specifically new information on black holes (look in 2013/2014 for release date) and in particular talked about the MW theories. I want to say it was on NatGeo or could have been the revisted "The Universe" series that had some episodes come in the past year or two. Basically artists have depicted the concept like this.

        [–]benkuykendall 4 points5 points  (5 children)
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        What would this imply about our universe that is different from our current understanding?

        [–]yeast_problem 7 points8 points  (3 children)
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        Our universe could be gaining mass from an external accretion disk that we could never observe?

        [–]Inane_newt 3 points4 points  (0 children)
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        The source of dark energy. From outside our black hole Universe, we are a singularity, so any mass added to our singularity is uniformly added to our interior Universe, which of course matches the nature of dark energy which is constant metric expansion of our space.
        If we ever got really good at measuring dark energy, we might even detect that it isn't constant, but varies depending on how much material is falling into our black hole.
        Right now, we are deriving the constant by averaging it over billions of years, which doesn't allow for much granularity.

        [–]ronin1066 2 points3 points  (0 children)
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        Couldn't we say it is a black hole in that light can't escape? Even though it may be only for the reason that it's expanding faster than light?

        [–]Hosko817 5 points6 points  (5 children)
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        given what we suspect about black holes, e.g. that their mass may be condensed into an infinitely small point (a gravitational singularity), it's quite likely that their mass no longer takes the form of matter as we know it.
        This might be a dumb question but, if their mass can be (theoretically) condensed into an infinitely small point, how and why do the get so large?

        [–]LTailsL 12 points13 points  (2 children)
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        Their area of influence on the surrounding space increases while still being a singularity. It's not the black part of a black hole is made of something, it's that nothing within that area can escape.

        [–]wabalaba1 2 points3 points  (0 children)
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        The mass, however large is confined to an infinitely small point but the influence of that mass is proportional. The event horizon, that perfect black sphere you see in pictures, isn't a physical surface or anything. In a sense, there's nothing there. It's sort of like a shadow, metaphorically speaking.

        [–]Shekt 0 points1 point  (0 children)
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        The singularity is small, bit the eventos horizon isn't. The giant black hole probably have the same singularity size of all others, but bigger masses imply in larger event horizons.

        [–]astrawnomore 4 points5 points  (0 children)
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        Not a bad answer, but I have to nitpick at this statement:
        given what we suspect about black holes, e.g. that their mass may be condensed into an infinitely small point (a gravitational singularity)
        This is currently not the consensus. It is painfully understood that General Relativity is an incomplete theory, especially with respect to black holes, so true spacetime singularities are not thought to be physically realistic. There have been several attempts to describe black holes semi-classically, and many of the formulations lead to spacetime singularities being impossible. One of these formulations was a motivator in the recent paper that suggested that there has been infinite time since the Big Bang.

        [–]Nougat 1 point2 points  (1 child)
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        Wouldn't a singularity be smaller than Planck Space, and would that mean "anything goes?"

        [–]Smithium 0 points1 point  (0 children)
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        We need observations before we can speculate on that. Be a dear and go check for, us, won't you?

        [–]pbhj 4 points5 points  (3 children)
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        Perhaps instead of compressions of matter blackholes are expansions in to other dimensions than the 4 we normally experience directly?
        Is that possible, instead of staying in one of these dimensions then matter gets bumped in to a dimension that is normally tightly wrapped (for us). You could maintain fermionic exclusion whilst occupying less real space than would ordinarily allow it.
        Someone do the maths and tell me that's crazy.

        [–]andpassword 26 points27 points  (0 children)
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        Are you putting forth a scientific argument for Hermione's Undetectable Expansion Charm?

        [–]SenorPuff 0 points1 point  (0 children)
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        String theory is predicated on multiples of dimensions being necessary for gravity to make sense.


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          [–]Rufus_Reddit 35 points36 points  (8 children)
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          The Pauli Exclusion principle only applies to half-integer spin particles, but pairs of fermions can act like bosons. This happens, for example in low temperature superconductors. (http://en.wikipedia.org/wiki/Cooper_pair)
          It's also worth noting that the 'number of states' encompassed in a black hole could be quite large even if they are spatially small.

          [–]TomatoAintAFruit 24 points25 points  (5 children)
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          This is a little misleading actually. Two fermions don't actually form a proper boson in the sense that the Cooper pair obeys the usual Bose-Einstein statistics.
          The fermions are still there. They still obey the exclusion principle. The fermions are in a bound state, but this does not change their statistics. The bound state acts boson-like (but not exactly).
          Put differently, the many-body state of the Cooper pairs can be approximated by a boson-like wavefunction (i.e. symmetric with respect to the interchange of cooper pairs). But this is only approximate. If you write the wavefunction in terms of the electrons, then you recover the asymmetric nature of the wavefunction due to exclusion principle.
          The fermionic nature of the electrons does not vanish when the electrons form a bound state.

          [–]lurkingowl 0 points1 point  (0 children)
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          The electrons are already gone at this point. The fermions we've got (neutrons) are already composite fermions (3x 1/2 spin quarks to make a 1/2 spin neutron.)
          So there's a lot of room for the neutrons to break down into quark Cooper pairs (mesons) and keep Pauli happy.
          Once we're at the point where quarks are starting to overlap, we're well outside of what the standard model can tell us without a theory of quantum gravity.

          [–]Zelrak 0 points1 point  (2 children)
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          How does that jive with the idea that a superconductor is a condensate of these cooper pairs? Isn't the whole reason that the superconductivity appears because the fermionic statistics of the individual electrons is replaced by bosonic statistics of the cooper pairs (at least in some regime where this effective description is accurate).

          [–]TomatoAintAFruit 2 points3 points  (0 children)
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          Great question.
          Isn't the whole reason that the superconductivity appears because the fermionic statistics of the individual electrons is replaced by bosonic statistics of the cooper pairs
          No, the reason that superconductivity arises is because the fermions form a fermionic condensate. This condensate can only arise when the Cooper pairs are formed. To some degree we can think of it as if the Cooper pairs have condensed into a Bose-Einstein condensate. But if we take into account the electrons, then we see that this is not the case.
          Once the fermionic condensate forms the energy spectrum will have an energy gap. We are then in the situation where there is a lowest energy state (the ground state) and a gap to the nearest excited state. This gap is the reason why the system is immune to dissipation.
          This situation is very similar to what happens in a Bose Einstein condensate. There we also have a ground state and an energy gap between the ground state and the first excited state (the gap "protects" the ground state). In this case all bosons sit in the same quantum state.
          But it's not the same as what happens in a superconductor. Specifically, the electrons do not occupy the same quantum state. In reality, there is a Fermi sphere. The ground state does not correspond to a single quantum state, but to a filled sphere.
          This is the main difference between the two systems: the ground state of a bosonic condensate is a single quantum state occupied by many bosons. The ground state of the fermionic condensate consists of as many unique quantum states as there are electrons.

          [–]EtherDaisTransmission Electron Microscopy | Spectroscopic Ellipsometry 0 points1 point  (0 children)
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          I think it's useful to remember that identifying the statistics of a particle's behavior is always measured in a context. We have abstracted the statistics observed from the context to identify certain particles as 'fermions' or 'bosons', but this may be an artifact of speech patterns in that "particles which obey fermi dirac statistics" is more tedious to repeat than 'fermions'. What begins in shorthand jargon can through time and repetition become a separate concept which assists in description but may imply a fundamental property which is unjustified given the abstraction from a context.

          [–]browb3aten 1 point2 points  (1 child)
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          What determines when this pairing takes place? Why don't these pairs form in normal matter and cause everything to spontaneously condense?

          [–]UnclePat79Physical Chemistry 1 point2 points  (0 children)
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          Simply speaking the thermal energy breaks these pairs up. Often the "pairing energy" is very small so that very low temperatures are required to allow for pairing. The Cooper pairs are as already mentioned one example and often exist only up to a few K above absolute zero.
          To your second question: Bosons are most often interacting or tightly coupled with fermions. For example, in the helium atom the individual electrons still behave partly as fermions even though they are spin paired. However, when cooling these atoms down the excited states and thus fermionic contributions become less populated until at one critical temperature the particles can "condense" into one coherent bosonic quantum state. Typical examples are suprafluid helium or a Bose-Einstein condensate.

          [–]Astronom3rAstrophysics | Supermassive Black Holes 12 points13 points  (1 child)
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          I'm going to quote a good explanation of why black holes don't have issue with the Pauli Exclusion principle from stackexchange:
          The Pauli exclusion doesn't quite say that fermions can't be squeezed together in space. It says that two fermions can't share the same quantum state (spin included). A black hole has an enormous amount of entropy (proportional to its area, from the famous Bekenstein-Hawking formula S=A4) and hence, its state count is ∼eA. Now, this might not seem like a big deal since usual matter has entropy proportional to volume. However, volume of such collections is also proportional to the mass. This means that a counting of the number of states goes as eM. For a black hole, it's Schwarzschild radius is proportional to the mass, hence A∼M2. So, the number of states scales as eM2 which is much much more than ordinary matter, especially if the mass is "not small". So there seem to be a lot of quantum states into which one can shove the fermions. So it seems like the fermions should have an easier time in a black hole than in (say) a neutron star.

          [–]Greifenhorst 0 points1 point  (0 children)
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          What happens to the electrons in a main sequence star? Do any of them have sufficient kinetic energy to achieve ionisation? I would infer that those nearest the core would have the highest KE and would be more likely to propagate through the interior of the star toward the surface, or be absorbed via K-capture, resulting in an electron deficiency that is ~inversely proportional to radial excursion. Assuming that they do, and that they enter a plasmatic state, would they not then move to the surface of the star and distribute ~evenly as they would on the surface of a charged sphere?

          [–]adamsolomonTheoretical Cosmology | General Relativity 21 points22 points  (0 children)
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          The Pauli exclusion principle is a consequence of quantum theory, describing subatomic particles. It might not be a firm, always-true law of nature. In particular, quantum theory does not take into account gravitation, which is what's responsible for black holes. Indeed, we don't even have a working theory which combines both quantum mechanics and gravity. Until we know what that theory looks like, we can't confidently answer this question.

          [–]anoneuclidean 2 points3 points  (0 children)
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          The best answer I could find is that when the pressure of a gravitational collapse is greater than the pressure arising from Pauli exclusion of electrons, there is enough energy to drive the collision of protons and electrons to form neutrons and antineutrinos, which can be viewed as a rearrangement of the beta decay process. The 'rearrangement' is afforded by the energy put into the process, which is provided by the gravitational collapse. When this happens, the electrons that would have been constrained by the Pauli exclusion principle are annihilated. This would avoid a violation of the exclusion principle.

          [–]bwainfweeze 2 points3 points  (3 children)
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          I wish I could find it, but I recall last fall reading a fringe theory that maybe singularities aren't a stable state of matter. That the star first collapses, then bounces back out due to the mechanics at the quantum level. But! the time dilation is so severe that it's an unimaginably slow explosion from a human perspective.
          In other words, by their theory the stability of a black hole is subjective, and objectively it's unstable.

          [–]dgm42 1 point2 points  (2 children)
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          I remember that paper. Pauli Exclusion preserves a small core that is in "normal" time. Then when the black hole shrinks, due to Hawkings radiation, until the event horizon gets down to the size of the inner core the core remnant bursts out and the black hole disappears.
          From the point of view of an outside observer this take trillions and trillions of years but from the point of view of matter that falls into the hole this only takes 10-23 seconds so it feels like a collapse and bounce.

          [–]bwainfweeze 1 point2 points  (1 child)
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          Tangentially, when people calculate heat death of the universe do they account for the time it will take all the galactic central black holes to explode, reform smaller black holes and decay again and again until there's nothing left?

          [–]malenkylizards 2 points3 points  (1 child)
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          Follow-up question: Is this discrepancy related to the general discrepancies between GR and QM?

          [–]adamsolomonTheoretical Cosmology | General Relativity 10 points11 points  (0 children)
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          Absolutely. Near the singularity of a black hole, where the OP's question is most relevant, we can't trust either GR or QM. Since Pauli exclusion comes from QM, it might not apply deep inside a black hole.

          [–]KalterTod 1 point2 points  (0 children)
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          The simple fact of the matter is that the Degeneracy Pressure has a finite value, based upon the density of free electrons (or other fermionic particles) in the volume.
          If the force exerted from the inward gravitational pull exceeds this value, it collapses.
          Also important to note that black holes are not the only situation where Pauli's Principle is broken. See also neutron stars/pulsars. They are held together by neutron degeneracy pressure, but not necessarily Electron Degeneracy.

          [–]kanzenryu 1 point2 points  (0 children)
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          The formula for the probability of finding two fermions a distance d apart has a term in it that exponentially decreases as d decreases. So in other words it becomes very unlikely to find these two particles close together. But if gravity is high enough then we can have another term in the equation that overwhelms the earlier exponential term and makes the overall probability higher for finding the particles closer together. Of course gravity has to be extremely strong for this effect to occur.

          [–]kukulaj 1 point2 points  (2 children)
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          My related question: how do Cooper pairs overcome the Pauli exclusion principle? A pair of electrons - a pair of fermions - somehow turns into a boson. Then this great heap of bosons can all condense and voila a superconductor!
          Probably the same thing goes on with mesons. Aren't quarks fermions? But then a pair of quarks becomes a meson, which is a boson.
          This is all surely related with the spin statistics theorem. I always wanted to understand this stuff, but life is just so short.

          [–]Snuggly_Person 1 point2 points  (0 children)
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          The symmetries of the system that treat both electrons as a unit will be bosonic, so the overall quasiparticle is a boson. The symmetry is always there (i.e. swapping any two pairs of electrons is always a bosonic symmetry, regardless of what they're doing) but is only relevant for describing particle condensation/interactions to whatever level of approximation the bound electrons can be treated as one particle in the first place. At scales similar to the size of the bound electron pair the pauli exclusion principle will become more obvious. Large scale bosonic dynamics, but small scale fermionic dynamics as the separation between the electrons in the pair becomes apparent.


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