Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Does Hawking Radiation preclude EH formation?

  1. Dec 11, 2012 #1
    Discussions involving Hawking radiation in the study of black holes usually require their preexistence; however, if we apply the Hawking radiation process to the initial stages of a birthing black hole I'm confused about how the theory claims the event horizon would form in the first place.

    In a simple non-rotating non-charged neutron star of, say, 4M the neutron degeneracy pressure would not be exceeded throughout the structure simultaneously; rather, it would occur at the center of the sphere where pressure is greatest. More specifically, it would occur between two neutrons at the center of the sphere, which would then allow further compacting of the surrounding matter, cascading into a traditional black hole. However, the Hawking black hole time-to-dissipate is given by:

    [tex]t_{\operatorname{ev}} = \frac{5120 \pi G^2 M_0^{3}}{\hbar c^4} \;[/tex]

    which is directly related to the mass of the BH as:

    [tex]8.410 \times 10^{-17} \left[\frac{M_0}{\mathrm{kg}}\right]^3 \mathrm{s} \;[/tex]

    Our theoretical "minimalist" black hole would have a mass of two neutrons, or:

    [tex]M_0 = 2*1.6749*10^{-27}{\mathrm{kg}} = 3.3498*10^{-27}{\mathrm{kg}} [/tex]

    Which gives a time-to-dissipate as:

    [tex]8.410 \times 10^{-17} \left[\frac{3.3498*10^{-27}{\mathrm{kg}}}{\mathrm{kg}}\right]^3 \mathrm{s} \; = 3.1612 \times 10^{-96} seconds[/tex]

    In other words, much, much shorter than Planck time! The original formation of the EH would theoretically dissipate more quickly than the cascading compaction could possibly propagate. The result would quickly become a similar structure of reduced mass...with no event horizon :uhh:

    Has this concept been explored?
     
    Last edited: Dec 11, 2012
  2. jcsd
  3. Dec 11, 2012 #2

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    That is interesting. However, if you model formation of horizon from inside out, the horizon itself forms from the inside out. Hawking radiation is a horizon phenomenon. So the radiation from your instantly decaying initial black hole would be emitted in the center of the mass, almost immediately absorbed, and thus play no real role in avoiding mass/density criticality. That is, all mass energy stays put. Obviously, a real answer to what happens at the center of gravitational collapse awaits some solution to QM+GR.
     
  4. Dec 11, 2012 #3

    PeterDonis

    Staff: Mentor

    Interesting idea, but you can't use the standard "dissipation time" result in this scenario because the spacetime inside the neutron star isn't vacuum, and the standard result for the "dissipation time" of a BH due to Hawking radiation depends on the spacetime being vacuum. Even in the case of a fully formed BH, if it is accreting matter faster than energy is being radiated away by Hawking radiation, it won't evaporate; instead it will gain mass. The same would be true for a horizon forming inside a neutron star; it will be "accreting matter" as the neutron star collapses faster than any "evaporation" process can get rid of it.

    At least, that's how I see the standard "semi-classical" treatment of Hawking radiation working in this scenario. A fully quantum mechanical treatment might possibly give a different answer, but we don't have one yet.
     
  5. Dec 11, 2012 #4
    Agreed, but I just wanted to point out that you used the word "even". Since the time-to-dissipate is proportional to the cube of the BH's mass, it isn't surprising that a fully-formed BH would be able to accrete faster than it can dissipate. For two neutrons the timescale makes the process essentially instantaneous. Hawking's equation would have to be SUBSTANTIALLY affected by the local presence of matter for the cascade to be viable.
     
  6. Dec 11, 2012 #5

    PeterDonis

    Staff: Mentor

    It's more than just "substantially altered". The chain of reasoning that leads to Hawking's equation doesn't even go through, as I understand it, unless the quantum field starts out in the vacuum state. So even assuming that there is an analogue of Hawking's equation that holds for the interior of a neutron star is problematic.

    Also, as PAllen pointed out, for the emission of radiation to cause the BH to "evaporate", the radiation has to escape to infinity. (That's another assumption underlying the Hawking calculation, btw; the calculation doesn't work if the radiation doesn't escape to infinity.) Clearly that can't happen at the center of a neutron star. Even if such radiation were to form at the center of a neutron star when the horizon first forms, all it will do is increase the pressure at the center, but if the star starts out larger than the maximum mass, increasing the pressure at the center just makes it collapse faster (because the increased gravity due to the pressure pulls inward more than the increased pressure pushes outward).
     
  7. Dec 11, 2012 #6
    If it only occurred in a pure vacuum state the effect would be meaningless in the real world.
    PeterDonis: I've simply never seen this question explored before, and I have no problem with accepting that we simply don't know yet, but may I ask whether you are arguing from a presumption that BH's exist as traditionally understood (and working backwards from there)?

    Regarding the reabsorption of the radiated energy, I considered that. However, the immediate local reabsorption is not guaranteed; in addition, the pressure must increase without a corresponding increase in temperature (as I understand it) in order to reach neutron degeneracy pressures, and temperature increases with such absorption.
     
  8. Dec 11, 2012 #7

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    In a sense it is guaranteed. Above and beyond normal cross section, the radiation is emitted under conditions where light and matter are nearly trapped. So normal cross sections would be multiplied. However, the main answer is there is no known, valid, way to analyze the center of collapse, at present: these are the exact conditions under which all current theories are presumed to be invalid, to first order. So we're back to: classical GR assuming an equation of state that doesn't violate classical SR (that's what the energy conditions amount to) unambiguously predicts singularity formation by the singularity theorems. Essentially nobody believes this characterizes what happens in the real world. All quantum approaches are approximate, for low energy conditions (compared to Plank energy).

    One thing that won't shed any light on the matter is to cherry pick formulas to apply to a situation tens of orders of magnitude different from their range of applicability.
     
  9. Dec 11, 2012 #8
    I'm not disagreeing but, as I did with PeterDonis, I question whether you are pronouncing that Hawking radiation is not applicable at that scale simply because you disagree with the result.
     
  10. Dec 11, 2012 #9

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    I have no idea what the result is. I've never visited the center of Cygnus X-1. I'm just pointing out that all established quantum methods are presumed by their relevant experts to be invalid under conditions at center of a collapse. This goes for the Standard Model, and all semi-classical approximations. If you think you understand Hawking radiation better than Hawking, go ahead and apply his results where he says they don't apply. As far as what the relevant experts say, all you are possibly left with are the speculative 'beyond SM QG' approaches, none of which claim to provide a clear answer. Some string theorists generally say a horizon with quantum behavior forms (actually, something that is microscopically not a horizon at all, but macroscopically looks like one), and a 'fuzzball' replaces the singularity. But this is not a clear consensus even within this specific QG program, let alone others.
     
    Last edited: Dec 11, 2012
  11. Dec 11, 2012 #10

    Nugatory

    User Avatar

    Staff: Mentor

    There is no such thing as a pure vacuum state in the real world, but enormous swathes of the universe are close enough to a pure vacuum so that calculations assuming a pure vacuum can be used - think planetary motion, for example.

    The question you have to be asking is whether the conditions are close enough to vacuum to trust a vacuum solution to provide accurate enough results, and it is clear that the center of a neutron star on the verge of gravitational collapse is <understatement>not that close to vacuum</understatement>.
     
  12. Dec 11, 2012 #11

    PeterDonis

    Staff: Mentor

    True, but as Nugatory pointed out, there's a big difference between being close enough to vacuum to treat the actual matter present as a small perturbation of the vacuum solution, and being so far away from vacuum that you can't even justify using the vacuum solution as a starting point.

    No; the only thing I'm presuming is that the Einstein Field Equation is valid at the center of a neutron star at the point where a horizon begins to form there. I'm not presuming that a black hole already exists or that one must form; the conclusion that one will form is based on the EFE and what it says about the effects of pressure, specifically that, as I said, increasing the central pressure in a system whose total mass is greater than the maximum mass limit for a neutron star increases the inward pull of gravity more than it increases the outward push of the pressure. Adding "evaporation" at the center doesn't change that; it adds another (speculatively possible) mechanism for generating pressure, but it doesn't change the net effect of increasing pressure.

    It doesn't have to be immediate; all that has to be true is that the mean free path of radiation inside the neutron star has to be much smaller than the star's radius. AFAIK that is true for neutron star matter, but I haven't seen any detailed calculations (not that I'm very familiar with the state of our knowledge about neutron star interiors).

    What I said above does not depend at all on the details of how pressure is generated, or how it relates to temperature (i.e., it doesn't depend on the equation of state). It's a general prediction of the EFE that applies to *any* kind of pressure.
     
  13. Dec 11, 2012 #12
    I'm not claiming I know something you guys don't; to the contrary, I'm not a Physicist while I presume there are some in this thread. That being said, I'm not simple either, and asking questions allows me to separate which parts of Physics are well-probed, proven and accepted vs which parts are more theoretical and speculative. To this point I personally believe black holes are speculative. Do you have reference to Hawking claiming that his calculations are not applicable in the presence of dense matter/energy?
     
  14. Dec 11, 2012 #13

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    I applauded your initial question as the right type of thinking to be doing on your own.

    Hawking radiation is derived via semi-classical methods which are not expected to be valid near Planck energy. The Standard Model is expected to be wrong way before that. It is really common knowledge that no existing theory is expected to be valid in the center of a gravitational collapse. I don't have a convenient set of references handy. Perhaps someone else does.
     
  15. Dec 11, 2012 #14
    Yes you did, PAllen, I appreciate your maturity level and contributions. I wasn't calling you out on anything, and I'm not demanding answers on subjects that currently have none, I just sincerely want to know if Stephen Hawking has made his opinions clear on the domain of applicability of his calculations.
     
  16. Dec 11, 2012 #15

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    Actually, if you look at the (non authoritative) reference you gave in your OP, you see, as an assumption leading the formula you quote:

    "Under the assumption of an otherwise empty universe, so that no matter or cosmic microwave background radiation falls into the black hole, it is possible to calculate how long it would take for the black hole to dissipate:"

    The center of gravitational collapse is hard to accept as an approximation to 'an otherwise empty universe'. It is not 'nearly flat vacuum'. There is no thermal equilibrium.
     
  17. Dec 11, 2012 #16
    Agreed that it is clearly not close to a flat vacuum, but my layman's understanding of Hawking radiation is that it is essentially a thermal balancing, so one could take the power in Watts of the two-neutron BH and compare it to its surroundings (how, exactly??):

    [tex]P = \frac{\hbar c^6}{15360 \pi G^2 M_0^2}\;[/tex]
     
  18. Dec 11, 2012 #17

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    But that power law itself is a further consequence of all the prior assumptions - a black body at Hawking temperature in a classical vacuum.

    Your "how exactly??" gets at the recurring issue: nobody knows; there is no established theory to apply here.
     
  19. Dec 11, 2012 #18

    PeterDonis

    Staff: Mentor

    No, it isn't. Where are you getting that idea? Hawking radiation does mean that a black hole has a finite temperature, which is desirable on thermodynamic grounds since there are good reasons to assign it a finite (very large compared to any other object with the same mass) entropy. But there's no requirement that that temperature "balance" with anything. A black hole can be out of thermal equilibrium with its surroundings just like any other object. I would expect a black hole that is just forming at the center of a neutron star to be far out of thermal equilibrium with its surroundings.
     
  20. Dec 11, 2012 #19
    This is my layman's understanding but it is not completely unfounded.
    Where are you getting your information?
     
  21. Dec 11, 2012 #20

    PeterDonis

    Staff: Mentor

    I didn't say it was *impossible* for a BH to be in thermal equilibrium with its surroundings; I just said it wasn't *necessary*. As the Wikipedia article shows, a BH which happens to have a temperature equal to that of the CMBR will be in thermal equilibrium with the rest of the universe; one with a lower temperature will be, on net, gaining energy, while one with a higher temperature will be, on net, losing energy, i.e., evaporating.

    Of course, all that assumes, once again, that the BH is in vacuum. More precisely, it assumes that the only process happening is thermal radiation, which is only going to be true in vacuum. It certainly doesn't apply to a BH at the center of a neutron star because thermal radiation isn't the only process happening. [Edit: And, of course, we're also assuming that the semi-classical calculation that predicts Hawking radiation is still valid for the interior of a neutron star, which I've already said is problematic.]
     
    Last edited: Dec 11, 2012
  22. Dec 11, 2012 #21

    Nugatory

    User Avatar

    Staff: Mentor

    Well, you do have to be careful about basing your understanding on that Wiki article. The first equation in it is the Schwarzchild metric, and that's a vacuum solution. The calculations and discussions and qualitative stuff further down is all based on that assumption.
     
  23. Dec 11, 2012 #22

    PeterDonis

    Staff: Mentor

    On thinking about the collapsing neutron star scenario some more, I realized that I have been mis-describing how the scenario would actually work; there is a key difference between this scenario and the final evaporation of a BH that hasn't been discussed.

    To see the difference, start with this fact, which I should have brought up earlier: according to our best current understanding, there can't be a BH smaller than the Planck mass. This is because the same general thermodynamic framework which leads to the conclusion that a BH should have a finite temperature, also leads to the conclusion that a BH with the Planck mass has an entropy of zero--i.e., there is only *one* possible internal state for such a BH. Any BH of smaller mass than that can't exist because there is no possible quantum state for such a BH.

    But, you ask, if this is true, how can a BH even start to form? The answer is that the picture rjbeery gave, of a horizon forming at r = 0 at the center of a collapsing neutron star, and a "two neutron mass" BH existing at that point, is not correct. It's true that an event horizon forms at r = 0 at some instant, but there is no singularity at r = 0 at that instant, and there is nothing locally that makes that small patch of spacetime any different from any other small patch with finite density and pressure. If we follow the worldline of a neutron that remains at rest at r = 0, that neutron sees a gradually increasing density and pressure, but at the instant the horizon forms, the density and pressure it sees is still finite; it is not until some time later on that neutron's worldline that a singularity of infinite density and pressure forms (when the surface of the neutron star collapses to r = 0). At that point, it is true, we don't fully understand what happens; we expect quantum effects to become strong, at the latest, when the density at r = 0 becomes comparable to the Planck density (one Planck mass per Planck length cubed). But at that point the entire collapsing neutron star, *all* of its substance, is compressed to a size much smaller than the horizon radius for that mass, so the situation is very different from what we were envisioning before, when a small "kernel" of a BH would form when the event horizon first formed, surrounded by most of the remaining neutron star as normal (well, normal for a neutron star) matter. That doesn't happen.

    The key point here is that the event horizon is a *global* phenomenon: it's the boundary of the region of spacetime (the "black hole" region) that can't send light signals to infinity. So what "a horizon forms at r = 0, and spreads outward" really means is that at some event on the worldline of that neutron that's at rest at r = 0, a signal emitted outward at the speed of light will just barely fail to reach infinity (we consider here an idealized "light signal" that doesn't collide with anything on the way out); instead it will move outward until it reaches the surface of the neutron star, at the exact instant when that surface reaches r = 2M, the horizon radius for the neutron star's total mass. At that point, that light signal is trapped: it can't move outward any further. It stays at r = 2M. But up until that point, it is moving outward; in particular, when it's emitted from r = 0, it is moving outward, locally, at the speed of light, and local observers don't notice anything special about it or about their local region.

    I describe all this in detail to contrast it with what happens when an evaporating BH reaches the Planck mass and (according to the current picture that seems to be the most widely accepted) finally evaporates away. At the instant before it evaporates away, the horizon is still, locally, a trapped surface; outgoing light there can't move outward. Also, there is a singularity of infinite spacetime curvature just inside the horizon. Then the final evaporation occurs, there is a naked singularity for an instant, and then an outgoing flash of light that carries away the final bit of energy from the former BH. So this situation is very different from the situation above, where a newly formed event horizon is not even a locally trapped surface, and the singularity has not yet formed.

    What I said before about the effects of increasing pressure inside a collapsing neutron star that is greater than the maximum mass limit still applies; but based on the above, I don't think there would even be any Hawking radiation generated by the horizon when it first forms, because it isn't yet a trapped surface, and as I understand it the derivation of Hawking radiation assumes that the horizon is a trapped surface (i.e., that outgoing light can't move outward there). So I think the discussion here needs to be re-evaluated in the light of that.
     
  24. Dec 11, 2012 #23

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    Great analsysis! I've made a number of these points in discussing collapsing shells, but blanked on their relevance here.

    While a neutron star is more complex, O-S collapse provides an interesting data point: each 2-sphere of dust reaches and forms the singularity at the same time. This is far away as can be imagined from a singularity that grows in 'mass'.
     
  25. Dec 11, 2012 #24
    I question your explanation on logical grounds. Let's take the 4M neutron star mentioned in the OP, and run it through your scenario in our minds. Fine. Now let's take that 4M star and double it's mass, adding the additional mass to the shell of the first. Are you now suggesting that the *new* surface of the 8M star must reach r=0 before a singularity is created? It doesn't plausible to me. Also, if a Planck mass is the minimum mass for a black hole to form I would be curious to see what kinds of mass are involved in those created by cosmic rays (or theorized to be created at the LHC). A Planck mass is a relatively large amount...perhaps you're right though and energy must be considered. A bit if research is in order

    I'll post more tomorrow, thanks for the feedback :smile:
     
  26. Dec 11, 2012 #25

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    To the extent the O-S collapse is general (I'm not sure which of its features aret typical of realistic collapses), any mass collapsing body would have the feature that all the original mass forms the singularity simultaneously. However, a second shell of matter further away, would reach the singularity later, afterit has formed.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook