Is it really, really hot in a black hole? Or really, really cold? Or does temperature no longer exist?
The best we can do is define the temperature of the entire black hole, as a single entity. Depending on the size of the black hole, the temperature can be anywhere from vanishingly small to hotter than the core of the sun:
T = 1/(8piM)
where T is temperature and M is the mass of the black hole. For your supermassive black holes, this is negligible. For the hypothesized micro black holes, it would be very high indeed. In practice, you can’t get it arbitrarily high, since there will come a point when runaway Hawking radiation causes complete evaporation of the hole in an explosion.
This temperature is called the Hawking Temperature, and it was conceptually realized when Bekenstein and Hawking studied the thermodynamics of black holes. The wiki article on black hole thermodynamics is about as straightforward as this stuff gets.
Spatial Rift, while that’s all correct, I’m not sure it’s what the OP was asking. That’s what outside observers would observe the temperature of the hole to be, but it looks to me like he was asking about an inside observer. Of course, we can’t really answer that question, since there’s no way to do the experiment (or at least, no way to publish the results). But we can talk about what an observer a little bit outside the hole would observe. If the observer is just falling straight in, he won’t observe anything out of the ordinary, and in fact, in the infalling reference frame, the Hawking radiation doesn’t even exist. If, however, the observer is just outside and prevented from falling in (an extremely powerful rocket, say), then any light that reaches him from distant points (stars, say) will be extremely blueshifted, with an effect the same as if the sources of the light were much hotter.
The Hawking temperature is basically the ‘surface temperature’ of the event horizon due to gravitational effects. For black holes formed from stars, thsi is generally so small as to be negligble due to their size. It is at least conceivable that their may exist primordial black holes of much smaller sizes and these black holes would have a much,much higher Hawking temperature.
Whilst the Hawking temperature is of a black hole is independent of the observer, as Chronos notes the actual details are little more observer-dependent. As he notes a free-falling observer would not locally observe any Hawking radiation being emitted, even as they crossed the event horizon.
Genrally speaking around a black hole there will be an accretion disk of in-falling matter. Due to friction this gets very hot indeed and will generally be the dominant factor in determing the actual temperarture ‘felt’ by an observer near the horizon.
I think an observer in the event horizon trying to measure the local temperature would find it certainly hotter than normal empty space due to the presecence of infalling matter.
Well, if the singularity of a black hole is infinitely dense then I suppose it is infinitely hot. Whatever it is that super dense thing is going to be blazingly hot.
Except, heat cannot escape the singularity so, other nasty effects aside, if you were standing next to the singularity I guess you would measure no temperature. If you entered the singularity it was be stupendously hot bit honestly you have other problems at that point anyway.
As for the space between the event horizon and the singularity I suppose it depends if matter is infalling or not. If the black hole is feeding them presumably there will be a lot of stuff traveling to the singularity and that stuff will probably be hot (for an average black hole) to not hot at all (for a supermassive black hole…you can pass its event horizon without norticing).
The close it gets to the singularity the hotter it will get I would think.
At least, for the ones we know about. It’s really hard to detect a black hole without an accretion disk, so we don’t know how many of those there are, and it might plausibly be more than the number with.
The temperature outside of a Black Hole may be a mess to describe. It requires statistics and maybe string or loop theory – if in fact not more.
What is inside of a Black Hole however is reasonably simple.
I hypothesize that the only particle in the universe that (actually) has rest mass, is the neutrino. And furthermore that this mass “m” diminishes in size as the scale of the visible universe grows. When the visible universe only consisted of a Planck particle, the Planck mass was simply “m”.
Such as the above was derived from the Chandrasckhar Limit as:
M(universe) = [m(Planck)]^3/m^2
Mu = [mp]^3/m^2 For any Black Hole embedded within the Visible Universe:
M = [mp]^3/(m/L)^2, where L = [1 - (v/c)^2]^0.5
The idea is quite simple. For any star-like mass M there would exist an (m/L) massed particle able to prevent a mathematical singularity from existing. Infinite anything is most likely a fiction. If (m/L) = the mass of a neutron, then M is about a couple of solar masses. For example.
By current theory, the smallest possible massive particle in the universe is the neutrino. I merely postulate that it is quite hardy enough to exist within the Black Hole, as a sort of gas, or as a super dense substance. A gas shrinking to a small ball may or may not have Planck Density within a star sized Black Hole. But probably would have that density at Hubble Time = zero.
Mc^2 = (compaction energy) + N(m/L)*c^2, The great unknown is “N”. With assumptions it might be numerically computable. But for the moment we don’t need it.
(m/L)^2 = [mp]^3/M
(L/m)^2 = M/[mp]^3
L/m = (M/[mp]^3)^0.5
L = m*(M/[mp]^3)^0.5
E = kT = [mc^2)/L
T = [mc^2]/[kL] = [mc^2]/{km*(M/[mp]^3)}
T = c^2/{k*(M/[mp]^3)^0.5} = (c^2/k)*([mp]^3/M)^0.5
The lower limit for M is M = mp
mp = 2.1765E-8 kg, the Planck Mass
k = 1.3806E-23 J/K, the Boltzmann constant
c = 2.99792E8 m/s, the speed of light
For the lower limit of M, T = 1.41687E32 K or simply the Planck Temperature
Let M = Mu = 8.74E52 kg
Then T = 70.7 K The expected temperature of the Visible Universe if the Light Dominated Phase of expansion did not transition the the Matter Dominated Phase of expansion.
All of the above is of course a hypothesis. But a testable one. It is only necessary to actually measure the rest mass of a neutrino, perhaps with a carefully conducted heavy hydrogen experiment and compare it with:
m = ([mp]^3/Mu)^0.5 = 1.09E-38 kg
1 kg = 5.610E35 eV
m = (1.09E-38 kg)*(5.610E35 eV/kg) = 0.00611 eV
I believe that the best current estimates of the neutrino mass, with a lot of assumptions about just how many relic neutrinos that the universe has is 0.06 eV. Or maybe that was an error estimate.
With or without an error in estimating “m”, the equation for “T” of at least a gas stage version of a Black Hole seems quite valid.
Whaddya know? My “temperature inside a black hole” sez 98.8 degrees.
What, please, is the basis of any of these hypotheses? How might they be tested?
If the fundamental laws of physics break down inside a black hole, then it is a moot point. It’s not just unobservable, it is completely indescribable. Properties like temperature, mass, speed, time etc. no longer apply…
[Things might not be that bad. As I see it, we inhabit (in a random gas distributed state) the volume of a very large Black Hole. It is a place that we are comfortable in, and call home.]
QUOTE=Chronos;18559245]What, please, is the basis of any of these hypotheses? How might they be tested?
[/QUOTE]
[Dear Chronos, essentially everything that I am writing about in these posts came from merely assuming that one mere equation that an astronomer Prof. Chandrasckhar came up with to describe the unstable mass build up of a growing neutron star, was not just true for his famous Chandrasckhar Limit – but also true of any star composed of “any” fundamental particle. In the simplest form, for a star composed of primarily neutrons, much of the complicating Quantum Mechanical considerations of Chandrasckhar’s Limit simply go away. We are left with:
M = (constant)*[mp]^3/m^2, where [mp] is the Planck Mass, and in this simplified case at least, “m” is no more than the mass of a neutron. The (constant) term, has a number of things within it, but as can be seen is free from having any units, and its worked out magnitude is a number not far removed from unity. For simplicity I merely let it be equal to “1”. You can see why I did that. If the test particle “m” were allowed to take on the value of [mp] then the formula says that “M” would then be equal to [mp]. In the Planck state, to a Planck Particle, the “visible” universe that it inhabits is merely the volume of its own existence. This, despite there obviously being untold numbers of Planck Volumes nearby to it. The same is true for us, just on a larger scale. We should keep in mind that a Planck Volume, has a ghost-like nature to it. If Captain Kirk says to the crew of the Enterprise that they are to journey with their faster than light “warp speed” device to a planet “X” on the very edge of the “visible” universe. Then when they get there and look back, a lot of old stars are gone, and a lot of new stars are visible. In our new position, just as on the popular Geek Comedy show “The Big Bang Theory”, the universe is still the famous 13.7 billion years old – even though we can not see things that we KNOW to exist (thanks to Captain Kirk’s warp-speed order). Einstein said that from his GR Theory, that worm holes (or warp drives) are at least mathematically possible. But actually making one is another matter.
Anyway Chandrasckhar was a rare super genus that in one of his minor works derived from the formidable equations of General Relativity his limit statement about stars (compact stars at any rate). He faced a certain amount of Hell in doing so. Scientists at the time were relatively cool with the idea that stellar objects could have such a strong gravity field that even light could not escape from them. But what Chandrasckhar found (seemingly at the time) was that totally scary things like “singularities” could (possibly) exist within the interiors of the newly named star-like objects. None of them could warm up to the idea that mathematical infinities could be a part of reality. And they could also find no errors in his derivations and calculations. They thus hated the poor man. And some of them even on a racial level. The rest of his life was warped by the effects of that hate – even when he reached the very top recognition and status of his profession.
I can not solve these GR equations. Or at least I have not put in the time (thus far) to make that sort of special effort. That said. I’d like to point out, that just possibly Chandrasckhar, if he knew of particles more fundamental than neutrons (like neutrinos) could have closed the same scary door with “singularities”, that he had opened in the first place. His fellow astronomers would have seen the scary door close up and lionized him for it. But at the time no one knew that neutrinos had to have mass, nor had clue one as to how to even estimate that mass.
My minor contribution was three little observations: (1) that the Chandrasckhar Limit was subject to simple algebra to solve for “m” in terms of “M” rather than the other way around; (2) That if a majority of the mass of even galaxies (and Quasars) was tied up in star-like Black Hole objects that a really good estimate for the mass of a fundamental particle might be made; (3) That the more massive the star-like Black Hole is, the better the upper bound estimate of the fundamental particle would be expected to be – until the mass of the “visible” universe itself was employed.
Worrisome was the result that that as the universe grew older, and its “visible” scale and mass increased, correspondingly the mass of the neutrino would shrink. A little bit at a time, rather than in a quantum effect, mass was being converted to energy. And old relic neutrinos didn’t have much mass or energy to begin with.
Well observe that for a long long time we have accepted this energy loss from photons. Super energetic photons (or even Cosmic rays) make peel off particles as they travel through space. This generally argued because space isn’t truly a total vacuum and this high energy stuff reacts with this non-vacuum. But even low energy photons loose energy with the expanding universe, so that on average those photons would have current back-ground radiation energy of
E= k*T, where it is observed that currently T = 2.71 K
Without the matter making phase of the universe, which produced a cloud of unbounded electrons, which much interacted with light (for a time anyway), the expected background temperature of the universe would otherwise be given by:
T = (c^2/k)*([mp]^3/M)^(1/2)
With the (current) observed “visible” mass of the universe being (about) M = 8.74E52 kg, the corresponding temperature of the background photons of the universe should be:
T = 70.7 K.
The kinetic energy of any neutrinos from the early universe, since they never significantly have reactions with electrons should be (presently) associated with that specified temperature.
Kinetic Energy = kT = (1.3806E-23 J/K)(70.7 K) = 9.76E-22 Joules/neutrino
(9.76E-22 J)*(6.242E18 eV/J) = 0.00609 eV
Please note that this is nothing other than the exact same mass prediction given by
m = {[mp]^3/M}^(1/2)
m = {[2.1765E-8 kg]^3/8.74E52 kg}^(1/2) = 1.08E-38 kg
(1.08E-38 kg)*(5.610E35 eV/kg) = 0.00606 eV
Chandrasckhar and Einstein paid their considerable dues with differential equations extracted and solved from the General Relativity Field Equations. They did all the heavy lifting. I merely did simple algebra based upon what they solidly established. My little contributing thought:
That the kinetic energy imparted from the Big Bang to relic neutrinos and to newly created neutrinos (of today) is exactly the same as the rest mass energy of ALL neutrinos.
Yes, this does create a seeming paradox with Special Relativity. A particle’s rest mass (energy) and kinetic (energy) should not be exactly the same. The answer would have to come from Special Relativity somehow being embedded in General Relativity. Or at least that is my starting point guess.
As to how you may know if this set of conjectures is valid or not. The answer seems quite simply. Actually find an experiment to as directly as possible, measure the rest mass of one or more types of neutrinos. Along those lines, the most careful current research seems to suggest three observed versions of the Higgs Particle. Just as there are three versions of neutrinos. And we “know” that neutrinos must have mass since one type can transition to another type. Well neutrinos simply can’t do that trick if they are always going at exactly the speed of light. And yet the difference of their speed from the speed of light, given their incredibly slight mass, would require an experimental race between the two particles, on the scale of a galaxy to comfortably measure.
Here is an alternative proposal. If there really are three different types of Higgs Particles, then just maybe nature gives them three different types of jobs to do. Maybe each type of Higgs Particle is responsible for its own separate Higgs Field. And maybe the mass like or kinetic energy effect of each of the three types of neutrinos is attached to one and only one Higgs Field at a time. If the speed of some effect in a Higgs Field is the speed of light (or one of several speeds of light) then maybe as long as the transition from one type of neutrino to another is brief enough (Planck time brief perhaps) – that neutrinos have kinetic energy but not mass. All of that is a fuzzy suggestion of course. But what else to do with three types of Higgs Particles? That seems to be too much of a good thing to have discovered. And to try desperately to un-discover it, as seems to be the case. The dog ate my experimental data. Forget it, never mind the seeming discovery.]
*********************** [The above added later than what follows.]
I actually composed several pages of involved calculations to flesh out and back up my hypothesis. Unfortunately the dog ate my homework, when the web page timed out. I tried, but failed to cut and paste all those wonderful pages into a new quoted document. Don’t assume that I didn’t have something with meticulous detail.
A source, I won’t hunt it up again reported an estimate that there are about 330 million relic neutrinos per cubic meter of space. I computed that there were more like 8E11 neutrinos per cubic meter of space. One of the potential sources of my neutrinos was the freeze out into existence of the Higgs Particle when the entire “visible” universe was only about 7.86E-19 meters in radius.
If we use the formula M = [mp]^3/m^2 to approximate the mass of the “visible” universe when the first neutrons froze out of the cosmic fireball, with [mp] = 2.1765E-8 kg = the Planck Mass, and
m = 1.67E-27 kg = the neutron’s mass, then
M = 3.70E30 kg = the mass of the “visible” universe at the freeze out point. From Chandrasckhar’s Limit, not an unexpected result. The sun has approximate mass of 2E30 kg.
M = (critical density)(volume) = [3/(8piGt^2)](4/3)pi(ct)^3
M = [c^3/(2*G)]*t
t = M*[(2G)/c^3] = (3.70E30 kg)[(26.6738E-11 m^3kg^(-1))/2.99792E8^3] = 1.83E-5 sec.
The radius of the “visible” universe when neutrons froze out was c*t = 5486 meters
Volume of the “visible” universe was V = (4/3)pi(5486 m)^3 = 6.916E11 m^3
Density = M/V = (3.70E30 kg)/(6.916E11 m^3) = 5.35E18 kg/m^3
This density is about 6 times what one might compute for a neutron, but is just about the limit of density in the interior of neutron stars – before it might gravitationally collapse. So it has merit.
One might easily do similar calculations for the Higgs Particle to predict its size and density. Something quite unlikely to find in a lab, since Higgs Particles are so super rare, and last so very briefly 7.86E-19 m = c*t, t = (7.86E-19 m)/(2.998E8 m/s) = 2.62E-27 seconds.
The current age of the universe is about:
(13.7E9 yr)(365 da)(24 hr/da)*(3600 s/hr) = t = 4.32E17 seconds
The (critical) density of the universe = d = (3/(8piG*t^2))
d = (3/(8pi6.67E-11*4.32E17^2)) = 9.59E-27 kg/m^3
The current rest mass of a neutrino is m = [mp]^(3/2)/M^(1/2) where M(universe) = 8.74E52 kg
m = (2.18E-8 kg)^(3/2)/(8.74E52 kg)^(1/2) = 1.09E-38 kg/neutrino
The number of relic neutrinos per cubic meter is N = d/m = 8.80E11 neutrinos/m^3
8.80E11/330E6 = 2667 I predict a lot more relic neutrinos than they do, but I am quite comfortable in doing so. Understand that they made their prediction from the unstable conditions near to the freeze out point for neutrons (and protons and electrons of course). That instability according to their arguments generated several million times more neutrinos than protons and neutrons. Thus they argue for 0.5 atoms/m^3 while arguing for 330 million neutrinos/m^3.
I don’t argue differently than they do. Not really. They were not [aware] of the very heavy Higgs Particle that froze out at a much earlier and much more dense era of the early universe. Unstable neutrons and the like, are not even nearly as unstable as Higgs Particles. The measure of the instability of the Higgs Particle is approximated by the Heisenberg Uncertainty Principle, one form of which is:
dtdE >= (h/(2pi))/2 Or dt <= [(h/(2*pi))/2]/dE
The mass of a Higgs Particle is m = 125 GeV = 125E3 MeV*(1.783E-30 kg/MeV)
m = 2.23E-25 kg Thus dE = m*c^2 = 2.01E-8 Joules
h = 6.6262E-34 J*s
dt <= [(h/(2*pi))/2]/dE = 2.62E-27 seconds
dx = dt*c = 7.86E-19 m
Any way, I think it utterly obvious that these early researchers should redo their statistical calculations about how many relic neutrinos resulted from unstable neutrons and protons as well as the much more unstable Higgs Particles. Particles that they NOW know exists. When a Higgs Particle decays it would need neutrinos to balance out energy and momentum equations every bit as much as protons, neutrons and electrons. It would be poor science to assume otherwise.
So with thousands of times as many relic neutrinos as previously thought. And with the inarguable observations of various neutrinos fluxes from the sun, all evolving from the neutrino having mass. Only one conclusion is possible, that neutrinos have thousands of times less mass than current Cosmological deductions suggest.
As the kinetic energy of a relic neutrino is of the order E = kT = (1.38E-23 J/K)(2.71 K)
E = (3.74E-23 J)*(6.242E12 MeV/J) = 2.33E-10 MeV = 2.33E-4 eV
By current methods, neutrinos with less than 0.511 MeV are impossible to detect, since they can not be used to create an electron (or anti-electron). And even if they could be detected, the flux of relic neutrinos would be completely dwarfed by solar and radioactive mineral neutrinos. An impossible combination.
It is commonly conjectured that really large Black Holes and Quasars did not come into existence by consuming space dust and stars. It is only logical to conclude that they gained their mass at a much earlier era than the freezing out of neutrons. Moreover, it is being taught that something like half of the mass of most galaxies is tied up in giant Black Holes. No great answers seem in the offering for Dark Matter in the universe. I simply assign the likelihood that the majority of the unseen mass in the universe is in the form of neutrinos. And by Chandrasckhar’s Limit what other known particle could prevent Black Hole “stars” from geometrically folding up and exiting the universe without so much as a trace that they were even there?
Here is a minor contribution to Black Hole lore. Let Mu = mass of the universe = 8.74E52 kg. Then there exists some mass “m” that solves:
Mu = [mp]^3/m^2, where [mp] = 2.1765E-8 kg = the Planck Mass.
I simply assume that “m” is the mass of the smallest and lightest known particle, the neutrino. A seemingly very safe assumption. But isn’t the beginning particle for the universe [mp]? Sure it is. But when that was true, Mu = [mp] = m. A “visible” universe of only one particle.
For M < Mu, there exists (m/L) such that:
M = [mp]^3/(m/L)^2, where alternatively L might be defined as L = (1 - (v/c))^0.5.
(v/c) is not of much interest, except as a scaling device trick. “L” is of much more interest.
(m/L)^2 = [mp]^3/M, “M” perhaps a star-like object.
(L/m)^2 = M/[mp]^3
L/m = {M/[mp]^3}^0.5
L = m*{M/[mp]^3}^0.5
E = kT = (mc^2)/L
T = (mc^2)/(kL) = (mc^2)/[km*(M/[mp]^3)^0.5]
T = (c^2/k)*([mp]^3/M)^0.5
The limit for M, is M = [mp]
T(limit) = (c^2/k)*[mp] = 1.41687E32 K = the Planck Temperature
This time assume that M = Mu = 8.74E52 kg then
T = 70.7 K, The average temperature of an expanded universe with using only the Light Dominated Phase. That is, that no atoms froze out to absorb expansion energy.
The above equations have the universe as sort of a balloon with an average temperature and no core to the Black Hole. No ball of “singularium”.
If a neutrino of mass “m” is allowed to fall from infinity (several times the Schwarzchild radius) into the Black Hole, then if the Black Hole is very small we might write:
kT = GMm/r where “r” satisfies M = (Planck Density)(4/3)pir^3
This is an inexact procedure for estimating “T” the temperature of the core of a mature Black Hole. When “M” is about 15 million solar masses the problem glaringly shows itself. In that event “T” is of the order of the Planck Temperature.
It is better to numerically integrate in many steps of size dr from “R” of several Schwarzchild radii down to “r”. From step to step the energy expressed as velocity picks up. To prevent it from building up too quickly use the following little increment:
dE = FdrL = FdrL, where L starts out = to one, and then shrinks.
E(new) = E(old) + dE
T = (1/k)*E(new)
L = (1 - (T/Tp)^2)^0.5, where Tp = the Planck Temperature.
E(old) = E(new), unless the looping steps are completed.
Then print E(new).
The guess is that as T approaches Tp that L will mirror the apparent shrinking of the length of dr in a roughly Relativistic manner.