Blackhole Entropy, LQG and Super String
There is recently a heated debate between the super string camp, Lubos et al, the loop quantum gravity (LQG) camp, Smolin et al, and the Other, Peter Woit et al, triggered by this review paper by Hermann Nicolai, Kasper Peeters, Marija Zamaklar:
http://www.arxiv.org/abs/hep-th/0501114
See:
http://motls.blogspot.com/2005/01/troubled-loop-gravity.html
and:
http://www.math.columbia.edu/~woit/blog/archives/000145.html
Note that both the super string camp and the LQG camp claimed their derivations of the Bekenstein-Hawking black hole entropy as their biggest success of their theories. In my judgement, claiming the derivation of Bekenstein Hawking entropy, such a trivial feat, as their biggest success, is completely "childish" and only shows the lack of "innate" ability on the part of each camp to comprehend what is the REAL physics behind the blackhole entropy!
I am going to show one very trivial derivation of the black hole entropy and how it is proportional to the event horizen surface area divided by Planck area. One that is different from Hawking's but much simpler.
But first, one has to realize two things:
1.Hawking entropy is not an empirical experimental evidence, but merely the result of a gedanken "experiment", e.g., mind exercise.
2. The entropy is a DIMENTIONLESS physical quantity.
Since Hawking entropy is just a mind exercise instead of empirical experimental result. Any claim of deriving the same result as Hawking merely shows that your theory does not have any logical inconsistency or conflict against the line of logic that Hawking's gedanken mind exercise. That's all. You still have not made any connection with the physics reality, unless, of course, that the Hawking formula is confirmed by a REAL experiment, not a gedanken one.
So, if a theory is able to derive Hawking entropy, it's good but really not a big deal. But if it can not, then there is a huge trouble in that it is logically inconsistent with Hawking's reasoning and what Hawing based his reasonings on.
Actually, any consistent theory at all will always leads to the Hawking formula, give or take a trivial numerical factor which is of the order of one, a numerical factor that both LQG and super string had struggled a bit to get right.
Now back to the dimentionless-ness of entropy. Given some basic physics quantities and known physics constants for a black hole, how would you construct an entropy formula that gives a dimentionless quantity? To construct such a formula, all the units should cancel out. That gives you pretty good clue to almost certainly arrive at the only correct answer.
We are given:
1.G, the gravity constant that is involved in any thing related to gravity
2.hbar, anything that envolved entropy needs to count quantum micro states.
3.C, light speed is certainly involved in anything related to spacetime.
4.M, the mass of black hole. We certainly need that.
There is nothing more we need. How would you construct a dimentionless number out of these 4 quantities? An immediate possibility is similar to the electromagnetic coupling constant, the fine structure constant, we can construct a gravity coupling constant here using:
S = G*M^2/(hbar*C) (1)
And that is the Hawking entropy formula, give or take a numerical factor!!! Actual it is only bigger than the Hawking entropy by a factor of 4*PI. Actually that is the only simple way to get a dimentionless number out of the 4 quantities!!!
How come? We know the radius of a black hole is proportional to its mass:
R = 2*G*M/C^2 (2)
So:
M = R*C^2/(2*G) (3)
So the (1) becomes:
S = (1/4)*R^2/(G*hbar/C^3) = (1/4) * R^2/lp^2 = (1/(4*PI)) * (1/4) * A/lp^2 (4)
So it differ from Hawking formula by 1/(4*PI).
Now, let me try to use a totally different but much simpler gedanken experiment to derive an entropy formular similar to Hawkings. It's trivial. Any one can think out a hundred different gedanken experiments, all arrive at the same result, differing by only a numerical factor.
Let's start with a black hole of almost zero mass, and gradually increase its size by throwing in photons in appropriate wavelength.
We do not want to throw in photons whose wavelength is much smaller than the size of the blackhole, since they will lose a great portion of their energy by gravity red-shift, and we do not know how much the mass of blackhole is increase. We do not want to throw in photons of wavelenth much larger than the blackhole size either, since then the photo will diffract around the blackhole completely, without absorbtion.
Let's choose photo wavelength
Lambda = diameter of blackhole = 2*R (5)
Such photons will be absorbed with its energy largely unchanged, increasing the blackhole mass by the equivalent mass of the photon energy. The increase of mass is:
delta M = delta E * C^-2 = 2*PI*hbar * C^-1/lambda = 2*PI*hbar*C^-1/(2*R) (6)
We know that each photon carries an entropy of exactly one, regardless of the photon's energy, so the increase of black hole entropy for each photon is one:
delta S = 1 (7)
Therefore:
dM = PI*hbar*C^-1/R * dS (8)
dS = 1/(PI*hbar) * C * R * dM (9)
Now, since
R = 2*G*M/C^2, which is M = C^2 * R/(2G) (10)
dM = C^2/(2G) * dR (11)
Put it into equ. (9):
dS = 1/(2*PI*G*hbar) * C^3 * R * dR (12)
Integrate equ. (12) from zero:
S = (1/(4*PI)) * R^2/ (hbar*G/C^3) (13)
S = (1/4*PI^2)^2 * (1/4)* A / lp^2 (14)
Again we ontained virtually the same Hawking entropy, except for a small numerical factor. I could easily get the factor correct if I am willing to try a little bit numerology like the LQG and super string camp did in their crackpot theories.
So you see it is really not a big deal at all to have derived the black hole entropy proportional to horizen area divided by Planck area.
Quantoken
posted by Quantoken at 1:55 AM
52 Comments:
Could you save anything you post in Motl's blog; or post it on your blog as well. I managed to read the following, before he erased it: petty prof. Very correct, and scientifically sound. But he obviously does not want to look foolish :)
Lubos the simpleton: You should have better physics instinctions to know this is a mere a media hype and there is nothing useful or practical.
Patents are too many. I myself have several industry patents in my name and used in widely available actual products. Big corporations love to apply for patents even for the most trivial and seemingly not very useful inventions. It costs them basically nothing to apply even if it is useless, and if it turn out to be something useful and they never applied, then it would have costed them dearly for the mistake.
In the HP crossbar invention. I do not see it as anything cheaper or faster than current technology. If anything it would be many times more expensive and a couple thousand times slower, and the device will has a much better chance to malfunction.
Keep in mind the traditional Si technology tries to move electrons and holes, while the HP technology tries to move Na ions, which is tens of thousands times more heavier to move than an electron!!! They need to apply a voltage 10 volts and up, which is a lot more than the state of art of Si technology, which now takes just about 1 volts.
You know they are now struggling very hard to remove every bit of heat dissipation efficiently, to keep the Si based CPU speed up. With the HP technology, the salt crystal (NaCl) will long be melted instantly way below the performance of today's Si technology.
Plus a little bit of moisture will be all it takes to dis-solve the Na ions and damage the device!
I do not see how it could be useful at all.
12:01 PM
Quantoken, perhaps you could put some background info into your profile or even a picture. Like, who you are, what you do, your background etc. I appreciate that you may wish to remain anonymous but most blogs give some background info on the host. What are your patents, what do you actually do, what is your past involvement in physics or engineering, computing or industry? (Even if just in general terms). Not asking for any specific or personal information, but very general background info or profile. While I don't agree with the majority of what you say I am much more willing to engage in dialogue or argument if I know a bit more about the blog host their background, and where he (she?) is coming from and what is motivating them etc.
Regards
2:16 PM
Quantoken, you assume that each time a photon is dropped into the black hole, its entropy increases by 1. Would you also say that if I have a large empty container, and I continually drop new gas molecules into it one-by-one, at the end of this process the entropy of the gas-filled container will simply be the entropy of each gas molecule times the number of molecules? If so, I don't think you understand the basic idea of entropy very well.
9:43 PM
JesseM:
Looks like you are really lacking in basic physics instinctions. It took me less than a second to come up with an answer for you.
You have to know the difference between a Boson and a Fermion!!! Bosons can occupy the same state but Fermions can never. That's a huge difference in terms of evaluating entropies.
So yes, you can sum up the entropy by counting the entropy increase one by one as photons are thrown into a blackhole. And NO, you can not do the same and count the entropy of the whole system by summing up the entropy of individual Fermion gas molecules! The entropy of the Fermion gas system would increase much faster than the increase of number of molecules.
Certainly, Fermions can couple together and form Bosons. That way, their entropy will be greatly reduced, since Bosons can occupy the same state! You should know better! That's called Bose-Einstein condensation. When Bose-Einstein condensation happens it falls into a state of greatly reduced entropy! Reduction of entropy is the whole meaning what condensation means!!!
Quantoken
10:29 PM
On second thought, for ideal gas system, one that there are not too many gas molecules per unit volume. Yes, the entropy will be proportional to number of gas molecules, assuming the size of container and the temperature does not change!!!
But as you pile up more molecules into the limited volume, eventually all the lower energy states will be occupied. Remember it's Fermion, so as you add more molecules, they have to occupy higher and higher energy level states. That's when the entropy begin to increase faster than the increase of number of molecules. Keep in mind increase of entropy times temperature equals to increase of energy, when the temperature is constant.
Certainly that can not be compared with the process of adding photons to blackholes. You can not keep increasing a blackhole's mass while keep its temperature unchanged. Hawking told us blackholes of specific mass always has corresponding specific temperature, never higher and never lower.
Quantoken
10:47 PM
You have to know the difference between a Boson and a Fermion!!! Bosons can occupy the same state but Fermions can never. That's a huge difference in terms of evaluating entropies.
Some atoms are bosons, such as Helium-4. In any case, it is not true for either fermions or bosons that if you put a bunch in a box the entropy is proportional to the number of particles, not even close. Consider this--how many possible ways are there to put a single checker on a checkerboard? How many possible ways are there to put two checkers on a checkerboard? (the answer will be slightly different depending on whether you allow the two pieces to be stacked on the same square, of course, but either way it will not be twice the number of ways there are to put one piece on the board)
8:11 AM
2. The entropy is a DIMENTIONLESS physical quantity.This is also wrong (and you mispelled 'dimensionless')--the expression for entropy S of a given macrostate in statistical mechanics is S=k*ln(N), where N is the number of microstates consistent with that macrostate...ln(N) is a dimensionless number, but k is Boltzmann's constant, which has units of Joules/Kelvin. You can also see that entropy must have these units if you use the expression for the entropy change of a heat bath in classical thermodynamics, dS = dQ/T, where dQ is the heat energy released into the bath, and T is the bath's temperature.
9:42 AM
JesseM:
Entropy is still a dimeniontionless quantity. The boltzman constant, k, is not a physical constant! It is a man made convertion constant, it is just like the constant 2.54 cm/Inch, which is made necessary because the humen happened to use different measuring unit system.
Likewise, the Boltzmann constant k becomes necessary only because we have one set of arbitrary unit to measure temperature in Kelvin, and there is another arbitrary set of unit Joules to measure energy. The two are different and we need a conversion constant to convert between the two.
Remember there is nothing physical that temperature must be measured in Kelvin or Celsius or Farenheit or anything. It can well be measured in Joules, in which case the Boltzmann conversion constant k will be simply one. So k is not really a physical constant, but merely a concersion constant same as 2.54cm = 1 inch.
The entropy is still a dimentionless quantity once you remove the none-physical factors. And to obtain that dimentionless quantity out of the basic quantities of a blackhole. The only feasible way is still:
G*M*M/(hbar*C)
Quantoken
12:41 PM
JesseM said:
"Some atoms are bosons, such as Helium-4. In any case, it is not true for either fermions or bosons that if you put a bunch in a box the entropy is proportional to the number of particles, NOT EVEN CLOSE. Consider this--how many possible ways are there to put a single checker on a checkerboard? How many possible ways are there to put two checkers on a checkerboard? (the answer will be slightly different depending on whether you allow the two pieces to be stacked on the same square, of course, but either way it will not be twice the number of ways there are to put one piece on the board)"
JesseM you really need to study your checkerboard example more carefully. Keep in mind entropy is proportional to the LOG of all possible configurations, not proportional to number of configurations.
Suppose you have w1 different ways to put the first chess. You will have w2 different ways to put the second chess, etc. The totall possibilities will be all the w's multiplied:
W = w1 * w2 * w3 * w4....
The entropy is porportional to the LOG of the number of possibilities:
S = ln(W) = ln (w1*w2*w3...)
= lnw1 + lnw2 + lnw3...
You see how the entropy is obtained by adding the entropy of individual chesses?
Certainly the situation is slightly different when two chesses are not allowed to stack together, and when there are enough of chesses to have significant possibility than an arbitrary way will result in stacked chesses.
Quantoken
12:50 PM
Ah, I didn't think that through, of course you're correct that ln(64^2) = ln(64) + ln(64), so the pseudo-entropy here would just be proportional to the number of pieces, at least if you allow them to stack. But an additional complication with the black hole example is that the BH's volume increases as more particles are thrown in--if the size of the checkerboard increased each time you added a piece, obviously the entropy would not be proportional to the number of pieces. Of course photons in a black hole presumably don't have the same amount of freedom as photons in a hollow sphere of equal volume, but this just shows why your analysis is overly simplistic.
1:13 PM
JesseM:
Each photon has exactly the entropy of ONE, regardless of the photon's energy, and regardless whether you put the photon in a bigger box or a smaller box. (certainly you can not put a photon into a box smaller than its wavelength.)
To see that, remember entropy times temperature equals to energy:
s*(kT) = E
A photon's temperature equals to the blackbody temperature at which it is emitted, which means kT = E. So S equals ONE.
So a photon always carry an entropy of exactly one, no more and no less. That's not surprising since photons are the fundamental element for exchange of quantum information in the universe.
That's also the basis of thinking in my GUITAR (Generalize Universal Information Theory And Relativity) theory. That's why quantum information is a more fundamental physics entity than space, time, energy etc. Please note the fact that under Lorentz transformation, a photon may seem to have a different energy and different wavelength in different reference frame, but its entropy is still always exactly ONE, a truely Lorentz invariant quantity!!!
Also, talking about where a photon may be located is meaningless, least to say within a blackhole there is no way of knowing anything about the insider of blackholes.
1:27 PM
Each photon has exactly the entropy of ONE, regardless of the photon's energy, and regardless whether you put the photon in a bigger box or a smaller box. (certainly you can not put a photon into a box smaller than its wavelength.)
To see that, remember entropy times temperature equals to energy:
s*(kT) = E
A photon's temperature equals to the blackbody temperature at which it is emitted, which means kT = E. So S equals ONE.Well, this argument wouldn't work for a collection of photons with a non-blackbody spectrum. And even if you assume a collection of photons in a box have a blackbody spectrum, if you expand the volume of the box won't the temperature of this collection of photons go down, since temperature is basically a measure of energy per degree of freedom, and expanding the box gives the photons more degrees of freedom? The cosmic microwave background radiation has a blackbody spectrum, but its temperature drops as the universe expands.
6:52 PM
Quantoken said:
" Entropy times temperature equals energy so S*(kT)=E. A photon's temperature equals the black body radiation at which it is emitted which means kT=E so S=1"
...............
This is incorrect. The entropy of a single mode of thermal blackbody radiation depends only on the average number of photons in the mode. The entropy S is correctly given by
S= d(logZ)/dT
where Z is the partition function of a black body photon gas, and the frequency w of the mode is held constant. The partition function Z for a single mode at temperature T and frequency w is
Z=1/[1-exp(-hbar*w/kT)].
The average occupation number of photons in the mode is
n=1/[exp(hbar*w/kT)-1]
so that (n+1)/n=exp(hbar*w/kT) or hbar*w/kT=log((n+1)/n)
The entropy derivative becomes
S= logZ - d/dT [log(1-exp(-hbar*w/kT))]
This is easy but quite messy to work out but comes out to (after 4 or five lines). You do the derivative but you have all the peices you need already to get a dimensionless expression in terms of n only. It is
S= (n+1)log(n+1)-nlog(n)
The average entropy per photon in the mode at temp T is then
s= S/n = [(n+1)/n]log(n+1)- log(n)
which is not one, so deltaS in your equation 7 is not one either. Statistical mechanics is an advanced subject and what I outlined would be a typical simple homework or exam problem for 2nd or 3rd year undergrads.
As regards the entropy-area relation for black holes anyone can piss around with the basic constants and get the black hole entropy, with hindsight and having seen how Hawking worked out black hole thermodynamics. It is practically a trivial exercise. It is the deep underlying quantum gravitational explanation and justification that matters and this is still a hard problem.
Applying ideas of quantum information to the problem is potentially interesting--people have thought about this--but I doubt if you have done that in the theory you claim to have. One would also expect Von neumann entropies to be involved, or Shannon entropies, in ANY theory based on "quantum information".
regards
Steve
7:14 PM
Farthermore, you have
dM=dE c^{2} at your equation 6, so M=Ec^{2}?? Assuming it's a typo and you meant c^{-2} then you get instead
dM=Ec^{-2} = hbarw c^{-2} = hbar c^{-1}/ lambda
using E=hbarw=hbar*c/lambda for a photon so that
dM =hbar/[c lambda],
which is a debroglie wavelength and correct units. The rest of your derivation then makes no sense. Assuming dS=1 (which it doesnt!) then continuing your original argument gives
dM=hbar*dS/[c lambda]so dS=(c lambda)/hbar)dM
lambda=2R= 4GM/c^{2}
so dS=dM* [4GM^{2}/c*hbar], integrating
S=2GM^{2}/c hbar
which is just bullshit.
Also you havent taken into account that fact that you said you started off with a black hole of very small almost zero mass (your words)...say the Planck mass or heavier? Then the temperature of the hole is already
T=hbar c^{3} /(8*pi*G*M)
Temperature of the hole is inversely proportional to mass. If the mass changes by dM then so does its temperature by dT, so you have'nt addressed whether the entropy change in absorbing a photon(s) is adiabatic and so on and so on. For a vacuum hole this small it is also emitting more than it is absorbing and there is a huge backreaction of the gravitational field since there is massive radius of curvature. The blackbody photon gas would have to have a temperature greater than the hole temperature otherwise the small mass hole cant absorb one of your photons. If the temps are the same they come to thermal equilibrium. You cant ignore all this! 1st and second laws and heat flows from hot to cold basically.
Thats also why Hawkings calculation is semiclassical and breaks down--no-one can do the full calculation because of the gravitational backreaction. Thats why quantum gravity poeple argue about it. Of course you can use a big "cold" massive black hole that is almost flat at the horizon but your photon gets violet shifted from the point of view of your coordinate frame outside the hole (from which you are measuring w) as it approaches the hole. Basically, you have to taken into account variations of the total entropy of the hole-photon bath system
dStot= dSphoton + dShole = dSphoton + {1/4}c^{3}A/G\hbar
This stuff has been known for 30 years
and the very simple calculation you are trying to do involving photons, hole mass, entropy, photon wavelengths and horizon radius, has already been done to death many many many times! So have all kinds of gedanken experiments involving lowering a photon gas in a box on a "rope" down to the black hole and releasing it. It is not new at all. It is also an infinitely more complex and subtle problem than you realise.
10:17 PM
Anonymous:
It really does not take much effort for me to repute you. Your derivation is indeed messy and it is also wrong.
You see it is wrong because you obtained an entropy of an individual photon, which is dependent on n, which in turn depends on the blackbody radiation system from which the photon come from. That is wrong. If I give you a photon and tell you it's frequency, that should be all you need to know to figure out its entropy.
Because regardless of where the photon originally came from, it all look the same!!! So why should a photon's entropy depend on n, a parameter which is only meaningful if you study the photon's birth history. That's why you were wrong when you showed me an entropy formula that depends on n or Z.
Where did you make the mistake? You made the mistake because your derivation would have to assume a equilibrium state. The distribution of photons in the system would have to agree with the partition function exactly to be in an equilibrium state. Now, you remove one single photon from the equilibrium system. It's NO LONGER in equilibrium any more so your derivation is no longer valid. Certainly the deviation from equilibrium is very small if we are talking about taking out just one photon from a system of many many millions of photons. But then the entropy change is also very small comparing with the total entropy of the system.
Once again remember, one photon has entropy of exactly ONE, regardless of its energy. I leave it as a homework for you to derive that result.
My theory is based on the principle of conservation of quantum information. It is solid and backed up by all experimental evidences. One amazing result is I have derived the exact ratio between the mass of neutron and electron, to the accuracy of 10 (TEN) decimal places matching the best known experimental value. I will be revealing the details pretty soon on this BLOG.
Quantoken
10:47 PM
I did make a typo in the middle of derivation in my original message. I have corrected it now. But it really does not change the end result. I still get something similar to Hawking formula, differ by a small numerical factor.
I already said I do not care that small numerical factor. You said:
"S=2GM^{2}/c hbar
which is just bullshit."
Why is it bullshit? You get the right result. I assume you mean
S=2GM^{2}/(c*hbar), that is eaxtly right. That's horizen area divided by Planck area, and differ by just a numerical factor.
I guess you don't really understand the power of dimentional analysis. As long as the calculation gives the correct unit for end result, you there is only one reasonable possibility:
S = constant * G*M^2/(hbar*C)
The rest is just trying to figure out that trivial dimentionless numerical constant.
This is NOT hind sight after Hawking!!! No definitely not. Any one know a little bit of dimentional analysis can reach the same thing, without Hawking's gedanken experiment.
Actually I should further say that Hawking had been wrong in saying the entropy is proportional to the hirizental area. That's just a coincident! The Hawking relationship would not work for tiny blackholes that exists for a very brief period of time (a few Planck time). It only works when it can be considered that the expected lasting time of the blackhole is extremely long, as long as the age of the universe.
The correct blackhole entropy formula is:
S = Surface Area of the 4-D spacetime region that the existence of blackhole is within, measured in natural unit. (Natural unit equals to the classical electron radius, and the corresponding time for light to come across that radius)
Note it's surface of 4-D spacetime, not 3-D space-only. At the large blackhoel approximation where the existence of the blackhole extends the whole time axis of the age of the universe, it degenerate into the Hawking blackhole entropy formula we know.
11:18 PM
JesseM said:
"And even if you assume a collection of photons in a box have a blackbody spectrum, if you expand the volume of the box won't the temperature of this collection of photons go down, since temperature is basically a measure of energy per degree of freedom, and expanding the box gives the photons more degrees of freedom? The cosmic microwave background radiation has a blackbody spectrum, but its temperature drops as the universe expands."
Once again each individual photon always has entropy of ONE. Also talk about the collective temperature of a group of photons in a box is meaningless. Photons do not interact with each other and they can not reach thermo equilibrium amoung themselves. The concept of temperature is only valid when you have reached thermal equilibrium.
And don't start to talk about Big Bang explaination of CMB. Big Bang is one of the biggest crackpot in establishment camps. I have shown how my theory leads to the EXACTLY correct CMB temperature, completely within the experimental error of the best observed data (2.725K +- 0.005K. My result is 2.7243K). Big Bang never even come close to estimate what the CB temperature should be.
Big Bangers must be brain dead in not realizing that in BigBang model, hydrogen atoms could not exist today. We all know that given sufficient temperature and density, nuclear fusion happens and smaller nuclears fuse into larger ones, to lower energy. From the origin of BigBang to today's universe, it must have experienced a period when it is cool enough for quarks to form protons and neutrons, but still hot enough for fusion to happen to form heavier atoms. The fact Hydrogen still exist today in largen quatity has fausified the Big Bang.
I shall discuss this further in the future.
Quantoken
11:44 PM
Once again each individual photon always has entropy of ONE.That is what you are trying to prove, but since we are arguing about whether you are correct, you can't assume that's true in your proof.
When you talk about the entropy of a single photon, what is the macrostate that you are counting microstates for? For a collection of particles, you'd ordinarily use continuous macroscopic parameters like temperature and pressure, and look at the number of microstates for a given value of temperature, pressure, etc....but concepts like temperature and pressure aren't intrinsic to a particle (although if you have a container perhaps you could define the temperature and pressure of a system composed of one particle confined in that container, I'm not sure).
Also talk about the collective temperature of a group of photons in a box is meaningless. Photons do not interact with each other and they can not reach thermo equilibrium amoung themselves. The concept of temperature is only valid when you have reached thermal equilibrium.If photons can be absorbed and reemitted by the inner wall of the container, then the container will eventually reach thermal equilibrium. this page on blackbody radiation explains: In contrast to the molecules in a gas, the different wavelengths of radiation do not collide with each other in the middle of the oven. Nevertheless, energy can shift from one mode to the other through the intermediary of the oscillators in the wall - which is to say, just the atoms and molecules the wall is made of. An electromagnetic wave in the oven can set a charge oscillating in the wall, energy can be transferred to other charges in the wall, which will in general oscillate and radiate at different frequencies. Thus the different frequencies of radiation inside the oven will come to thermal equilibrium. And don't start to talk about Big Bang explaination of CMB. Big Bang is one of the biggest crackpot in establishment camps.That's irrelevant to this particular argument, I was only using the CMBR to illustrate the point that physicists seem to agree with my intuition that the temperature of a collection of photons with a blackbody spectrum will decrease as the volume available to them increases, unless I am misunderstanding something.
2:33 AM
Hey, how come my paragraph breaks almost never show up? Am I supposed to use HTML tags or something?
2:39 AM
For breaking paragraphs, you probably just need to leave one extra blank line in. I just started my BLOG recently and have yet to get to speed in configuring it.
You said:"When you talk about the entropy of a single photon, what is the macrostate that you are counting microstates for? For a collection of particles, you'd ordinarily use continuous macroscopic parameters like temperature and pressure, and look at the number of microstates for a given value of temperature, pressure, etc....but concepts like temperature and pressure aren't intrinsic to a particle (although if you have a container perhaps you could define the temperature and pressure of a system composed of one particle confined in that container, I'm not sure)."
I am talking about quantum entropy (or quantum information in another word), not just the thermodynamic entropy. So your discussion of macrostates would not apply. You've got to understand that quantum information is the most fundamental building block of our universe, spacetime, mass-energy, are all secondary physics properties derived from quantum information.
Comparingly, the thermodynamic entropy is only an extremely tiny portion of the total quantum information of the universe. That's why we can have a thermodynamic entropy that only increases and never decrease, while at the same time the total quantum entropy of the universe is conserved and never increase or decrease.
The total quantum entropy of the universe is equal to the Hawking entropy of a black hole the size of our universe. That's 2*PI^3*N^3. N = PI*exp(2/(3*alpha))1.4898253x10^40. So the quantum entropy of the universe is about 2x10^122. That's an enomous amount of entropy comparing with the total mass-energy of the universe, which is only proportional to N^2. The size of the universe is proportional to N^1 (PI*N exactly, measured using the natural length of electron radius).
Seth Lyyod calculated the same 10^122 quantity, see his "The computational Universe":
http://www.edge.org/3rd_culture/lloyd2/lloyd2_p2.html
For now, remember the universal constant N = PI*exp(2/(3*alpha)). It is used in a lot of amazing accurate calculations, including the derivation of CMB temperature to within the error margin of the best observed value, and accurate calculation of neutron to electron mass ratio to 10 decimal places accuracy!!!
Quantoken
7:52 AM
Photon entropies can only be computed properly from the partition function of a blackbody photon gas. It is not one.
I concede that the BH entropy formula is correct but even then all you have done is simply reproduce a basic and well-known heuristic argument that has been known for 30 years, which had me confused since you put dM=dEc^{2}. Dimensionally, all you do is set a thermal wavelength hbarc/kT equal to the horizon radius 2GM/c^{2} to get the hole temperature, which is
T=h*bar*c^{3}/(8 pi G M).
Since the hole has a temperature it must have an entropy and vice versa. The area of the horizon is
A=4 pi(2GM/c^{2})^{2}=4pi R^{2}. From the first law of thermodynamics
dE=TdS
Using E=Mc^{2} and T= (hbar*c^{3}/(8 pi G M) gives
for black holes
d(Mc^{2}) = [c^{6}/32 G^{2}pi M]dA = TdS
so the entropy of the hole is identified as
S= (kc^{3}/4G*hbar)A.
No big deal! Since T varies as 1/M, black holes have a negative specific heat--as they lose energy/mass they get hotter. Hawking explained the deeper physics(correctly) as pair creation in a strong gravitational field using seniclassical quantum gravity. A black hole of temp T in contact with a photon bath at T' will come to equilibrium with the bath.
You have'nt done or said anything thats new or said anything important. And if you had a real theory of "quantum information" (which you don't) you would be able to explain--very rigorously!--why in black hole evaporation and information loss, pure states appear to evolve into mixed states violating unitarity. Incidently the term "quantum information" can only be given any kind of meaning whatsoever in terms of the Von Neumann entropy which you have not discussed.
9:23 AM
Anonymous said: "Photon entropies can only be computed properly from the partition function of a blackbody photon gas. It is not one."
Not true. That's only one of many possible ways of calculating it. Whatever way you do the calculation, if you do it correctly you get the same result of entropy ONE. Your partition function calculating is wrong as I pointed out, The system in equilibrium to start with is no longer in equilibrium when you take just one single photon out. It is off equilibrium just slightly, but that slight inbalance is enough to render your calculation of single photon entropy invalid, since you obtain it by considering the change of the TOTAL entropy of the system, before and after that one photon is taken away.
Your partition function logic would leads to different photon entropy for a photon emitted from the sun or a photon emitted from a stove. But if you look at the individual photons, they really look the same no matter where they came from. So the entropy should also be the same.
Not only entropy of photons of same energy are the same. Entropies of photons of different energy are also the SAME. That's because the same photon's energy may look differently from different reference frames in special relativity. But since the entropy is a dimentionless quantity, it should NOT have a dependence on relative speed of reference frames. Therefore entropy of photons MUST be Lorentz invariant and be a constant, for photons of differen energy and different origin.
Dimentional analysis is a very powerful tool and too bad you could not appreciate what you can do with it. And the methodogy wasn't invented by Hawking. Let me show you how it works:
Assuming we know absolutely nothing of Hawking or any one's derivation of blackhole entropy. But by logic we know the entropy S must depend on the following things:
1.G, Gravity constant
2.C, light speed
3.hbar, the Plank constant.
4.M, mass of black hole.
Then we can write entropy S as certain products of these values:
S = beta * G^a * C^b * hbar^c * M^d
With beta a trivial dimentionless numerical factor. And a,b,c,d are powers of the variable.
The above formula must end up giving the correct unit to the calculation result. That is the constraining condition that allows one to determine a,b,c,d.
The units, in MKS international unit set are:
G = [M^3*Kg^-1*S^-2]
C = [M^1*S^-1]
hbar = [Kg^1*M^2*S^-1]
M = [Kg^1]
Now, the end result, entropy, should be unit-less. So when multiplied by the powers a,b,c,d, the units should all cancel out:
G^a*C^b*hbar^c*M^d
= [M^3*Kg^-1*S^-2]^a * [M^1*S^-1]^b *
[Kg^1*M^2*S^-1]^c * [Kg^1]^d
= [Kg^0 * M^0 * S^0]
Look at the units of Kg, M, S respectively, each should give zero:
-a + c + d = 0 (for kg)
3a + b + 2c = 0 (for M)
-2a - b - c = 0 (for Sec)
All of the possible solutions to a,b,c,d can be expressed as:
(a,b,c,d) = (1,-1,-1,2) * gamma
gamma could be any number. But the most simple and sensible solution would be gamma = 1. so a,b,c,d are
1, -1, -1, 2, respectively. So:
S = beta * G*M^2/(hbar*C)
is the ONLY correct answer.
You see the above analysis is purely based on dimentional analysis and has absolutely nothing to do with Hawking's reasoning.
The point I want to make is, no matter what method you use to calculate the blackhole entropy, and no matter whether you theory is right or wrong. As you as you do not make the elementary school mistakes of getting the unit wrong, you would ALWAY arrive at an entropy proportional to horizenal area divided by planck area.
S ~ GM^2/(hbar*C)
is guaranteed to be correct by analysing the units along!!!!
That's why I say suerstring and LQG camps had been childish in claiming a success of deriving entropy proportional to horizental area of blackholes. It's absolutely no big deal at all and of course they will always get that result. They could not possibly get anything else at all, unless they made a low level primary school mistake by getting the units wrong!!!!
Quantoken
10:35 AM
I am talking about quantum entropy (or quantum information in another word), not just the thermodynamic entropy. So your discussion of macrostates would not apply. You've got to understand that quantum information is the most fundamental building block of our universe, spacetime, mass-energy, are all secondary physics properties derived from quantum information.
Comparingly, the thermodynamic entropy is only an extremely tiny portion of the total quantum information of the universe. That's why we can have a thermodynamic entropy that only increases and never decrease, while at the same time the total quantum entropy of the universe is conserved and never increase or decrease.If when you say "the entropy of a photon is one" you are not using the word "entropy" in the thermodynamics sense, then you have no justification for using this to conclude anything about the thermodynamic entropy of a black hole. And didn't your derivation of the claim that the entropy of a photon is one involve only thermodynamic arguments? It seems then that you use thermodynamic arguments to justify the claim, but that when it is argued that your thermodynamic arguments are incorrect (because, for example, the temperature of a group of photons in a container should depend on the volume of the container), you retreat to the position that it doesn't matter anyway, because your claim is not really about thermodynamics at all. But if that's the case, then how do you justify "the entropy of a photon is one" in the first place? How would you convince Seth Lloyd or some other quantum information theorist that this claim is correct, or even meaningful? I'm pretty sure they would not agree that the entropy of a photon is one and that this is all you need to explain black hole entropy (in fact, I read an article by Lloyd recently where he tried to explain the black hole entropy formula in terms of quantum information theory, and his argument involved creating a network of satellites to map the geometry of spacetime, and noting that if you try to measure the geometry too precisely using closely-spaced satellites, the energy density of the satellites' computations will cause a black hole to form).
11:58 AM
"From Big Bang to todays universe must have experienced a period in which it was hot enough for quarks to form protons and neutrons but still hot enough for fusion to happen. The amount of hydrogen around today falsifies the big bang"
.....................
An rediculous statement. If you are going to refute the BB you should at least understand the theory in great detail (which you clearly don't).
Its basic nuclear physics. Quarks condense into nucleons, yes. Subsequent fusion into heavier nuclei is not possible for a simple reason. During the first second after the BB, any heavier nuclei that formed would have been blasted apart by the very high flux of photons (photo-dissociation). Also at this time protons were constantly being transformed into neutrons and vice versa, due to interactions mediated by the weak nuclear force. After one second, the rate of weak interactions become slower than the cosmic expansion rate, at which point the N/P ratio froze out.
Free neutrons decay back into protons and electrons in about 14 miniutes and 49 seconds. However fusion reactions still proceeded in the heat of the BB, while the level of destruction from photon blasting soon dropped below that which the strong nuclear binding force could withstand since everything is expanding. The strong force was then free to bind protons and neutrons together, drawing neutrons into the sanctuary of atomic nuclei. This process began after the first second and was operating in full by about 10 seconds. This lead to deuterium, then 2 dueterium nuclei fuse making make helium4--this is how most of the dueterium ended up plus traces of tritium. But it is by now no longer hot or dense enough top make heavier nuclei. The expansion is diluting things out. At this stage we have about 25% helium, the rest hydrogen and traces of Lithium and deuterium. This agrees with what is observed, taking into account subsequent element production in the stars.
Also, there are no stable nuclei with atomic weights A=5 or A=8 so it is virtually impossible to build up elements heavier than helium by alpha-alpha, proton-alpha, n-alpha collisions in the early universe. The density of the expanding universe is at about 10^9 K at this time, is too low to allow much helium burning to occur. The production of heavier elements can only bridge this A=5,A=8 barrier by certain process that can only occur in stars...this will come later.
What you think is irrelevant. What has been demonstrated by the most rigorous calculations and observation is.
12:43 PM
quantoken, your "dimensional analysis" argument is silly (and you misspelled dimensional again--look it up, there is no "t" in dimensional), because there was no a priori reason to expect the entropy of a black hole would simply be a function of its mass. For example, it's not true that the entropy of a box filled with gas is simply a function of the mass of the gas, regardless of volume or pressure or temperature. Before Hawking came up with a real argument for why the entropy of a BH should be a function only of its area (or, equivalently, of the square of its mass), why should any physicist have trusted your dimensional argument, when such an argument would clearly fail for well-understood systems like a container of gas?
2:32 PM
JesseM:
Then tell me what other parameters the entropy of a blackhole could possibly depend on? It depends on the exchange ratio of euro to dollars, for example, you say? :-)
There are really not many parameters of blackhole you can talk about. Electric charge, maybe. But we know blackhole still is blackhole and still holds entropy, even with no charge.
You may also say it also depends on the radius of the black hole in some way. But since mass and radius are dependent on each other, you can only pick one of them for your dimentional analysis, but not both. Either one you pick, you come to the same conclusion at the end. You should be able to try out G, C, hbar, R by yourself and see what you get.
As for the BigBang argument, you are simply copying standard textbooks without giving it a thought yourself. What you said is crap:
"Free neutrons decay back into protons and electrons in about 14 miniutes and 49 seconds. However fusion reactions still proceeded in the heat of the BB, while the level of destruction from photon blasting soon dropped below that which the strong nuclear binding force could withstand since everything is expanding. The strong force was then free to bind protons and neutrons together, drawing neutrons into the sanctuary of atomic nuclei. This process began after the first second and was operating in full by about 10 seconds. This lead to deuterium, then 2 dueterium nuclei fuse making make helium4--this is how most of the dueterium ended up plus traces of tritium. But it is BY NOW no longer hot or dense enough top make heavier nuclei."
How soon after the big bang is your "BY NOW"? I assume you are talking about the first 10 or 20 minutes. Let me give you plenty of room, let's say it is one full day later, 24 hours. By that time the universe would be no bigger than the light speed times 24 hours, which is roughly 86400 s * 3x10^8 m/s = 2.6x10^13 m.
The volume of the universe would be 7x10^40 m^3. Since all the protons and netrons in the universe would have been created by now, that's roughtly 10^80 particles. Divide that by the volume of the universe, it is an incredibly high density!!! 10^39 particles per m^3, or
10^9 particles per volume of one single hydrogen atom. That's a density far exceeing the threshold of triggering thermal fusion!
How could you say the universe has diluted enough to prevent fusion from happening? And I am already giving you plenty of room for calculating using 24 hours, instead of just 20 minutes.
Use your own brain, do not copy textbook, OK?
Quantoken
4:23 PM
Let me reiterate for the very LAST time!
Fusion is still happening later at about 3 minutes, but it is not possible to make anything heavier than helium (save some traces of lithium) and there is a barrier to fusing into heavier nuclei since there are NO stable atomic weights of A=5 and A=8, and large Coulomb barriers exist. This is verified FACT of nuclear physics. I will exlain more below. This is also a PRECISE and beautiful series of calculations done by some of the very best minds physics ever produced (Gamow, Peebles, Hoyle, Wagoner, Fowler, Weinberg, Alpher, Bethe...etc etc) involving nuclear rate reactions, statistical mechanics and general relativity, and it agrees incredibly well with what is observed. Over the decades the precision has only gotton better. It is one of the key successes of the big bang. In fact, it is so well known --to anyone with a basic physics/cosmology background--I don't even need to refer to a textbook.
The scale factor R(t) used to estimate the density at time t is a solution of the Einstein equations and R(t) grows either as an exponential(in infltion) or a power law. It is pretty big after 180seconds, which is a massively long length scale in early cosmology. The energy density of radiation scales as R^{-4} and the rest mass scales as R^{-3}. As a result, at t=1 sec the density is about 5 x10^5g cm^{-3} and temp T ~10^10K. At t=4 secs it is about 10^{4}g cm^{-3} and T is about 10^{9}K. At around t ~180secs nucleosysnthesis produces He4 nuclei.
Very little nucleosynthesis occurs before 3 minutes since deuterium, D2, plays a VERY CRUCIAL role in the creation of helium by fusion, but the abundance of D2 is VERY LOW until the temp drops to 10^{9}K, when it is "cool enough" for D to survive and be produced in sufficient amounts via collisions of p and n. (Stuff gets blasted apart again by high energy photons.)
Deuterium has a low binding energy and serves as a block or bottle neck that delays the formation of more complex nuclei UNTIL T drops to near 10^{9}K. Onlt at this temp and below does D2 becomes abundant enough.
FACT!-->Complex nuclei must be built up from 2-body reactions in successive steps such s
1) p + n -->D
2) D + D--> He3 + n <--> H3 + p
3) He3+d <--> He4+n and so on
(if you dont believe these reactions then check out some H-bomb explosion films)
These CANNOT proceed until the abundance of aforementioned Deuterium D is high enough to allow these D-D, D-P, D-n reactions
to actually proceed at an adequate rate. NO nucleosynthesis of elements beyond He4 occurs at this stage of the universe due to large Coloumb barriers and the lack of stable nuclei at atomic weights A=5 and A=8. (Only later in red giants will this be possible where the conditions are different). The neutrons left at the proton-neutron ratio freezout, which did not decay are mostly converted to He4. The COULOMB BARRIERS in the reactions to heavier nuclei like lithium and berylium
4) He4+H3-->Li7
5) He4+He3-->Be7
prevent them from energetically competeting with the reactions
6) p+H3-->He4
7) n+He3-->He4.
The effect of the nuclear reactions (1),2),3) is very rapidly to incorporate ALL remaining available neutrons into He4 nuclei, which have the highest binding energies of nuclei whose atomic weights are less than A=5.
About 25% of the material in the universe is then helium, with the rest mostly hydrogen, plus smaller traces of isotopes of He3 and D2. and Li7. That is EXACTLY what is observed, whether you like it or not.
Thats basic!I will refer to textbooks, journals or original papers if want the technical fine details. Nucleosynthesis is a very mature, experimentally verified and exact science...you can say things with a very great deal of certainty--its not at all like string theory say.
Basically you have zero understanding and appreciation of the these matters. You are a heckler out to disagree with anything and everything and anyone that doesnt fit into your niave views and misunderstandings.
There is also no point in arguing with you over since you have never actually heard of any of this stuff until I told you about it just there! I've already wasted enough valuable time here.
6:53 PM
Then tell me what other parameters the entropy of a blackhole could possibly depend on? It depends on the exchange ratio of euro to dollars, for example, you say? :-)You are assuming that a black hole would have entropy in the first place, and that a black hole's entropy would be a property that could be knowable to us. With the benefit of hindsight it's easy to make these assumptions, but there's no a priori reason why it should be meaningful to assign an entropy to a black hole at all, and there were good reasons for thinking this idea was unworkable--if it has an entropy, it would have to have a temperature in order for the "generalized second law" to hold (this law says that the sum of the entropy of everything outside a BH plus the entropy of the BH itself must always increase), but until the discovery of Hawking radiation this seemed impossible (see the first two paragraphs of this lecture by Stephen Hawking). Likewise, there's no a priori reason why black hole's entropy couldn't depend on hidden aspects of its internal state, and therefore be unknowable to us. It's only because of arguments by physicists like Hawking and Bekenstein that physicists came to accept that black holes do have an entropy, and that any two black holes with identical external properties should have identical entropies.
Also, you keep confusing my posts with those of others, I wasn't the one who wrote the post explaining the errors in your comments about big bang nucleosynthesis. But I did write an earlier post which you haven't addressed, questioning why you derive the "entropy of a photon" from thermodynamic equations but then retreat to the position that you are not talking about thermodynamic entropy when it's pointed out your derivation doesn't make sense in thermodynamic terms, yet still insist that your non-thermodynamic notion of entropy is somehow relevant to understanding the entropy of a black hole.
8:22 PM
You also assumed that the universe expands linearly as
R(t)=ct in your totally wrong estimation of the density. In the period we have discussed the universe is very hot and RADIATION DOMINATED not matter dominated. That means that the energy density scales as an R^{4} law and not an R^{3} law. That is
rho * R^{4}=const and NOT rho R^{3}=constant
From the Einstein equations, the cosmic scale factor for a radiation dominated universe actually grows like
R(t)=[(4K)^{1/4}] t^{1/2}
where K=8pi*rho r^{4}/3
and NOT linearly. (In inflation models R(t) can grow exponentially before going over to a power law and outrun R=ct, without violating relativity). At temperature T, massive particles (like protons and neutrons) whose mass-energy is MUCH LESS than kT effectively act as zero mass particles and are included in the radiation density as just another form of radiation.
For such thermal radiation the density rho is given by the following basic expression from quantum statistical mechanics (FACT!!)
rho = sum_{i=1}^{n}a_{i}g_{i} (pi^{2}/30hbar^{3}c^{5}(kT^{4})
where n is the species number of particles, g the spin degeneracy (eg g=1 for photons) and a=1 for bosons and 7/8 for fermions. These equations together imply that
T ~ rho^{1/4} ~ R^{-1}
so the radiation is redshifted as R increases. The universe becomes matter dominated when R reaches about 1/1000 of its present value and comes out of thermal equilibrium and cools off. (Only then can you can talk about there being 10^{80} nucleons or whatever.)
Nuclear fusion can't occur until about 3 minutes at which T~10^{9}K for reasons already discussed regarding deuterium and the 2-body collision processes required to make He4. Again, this is the first time you have heard of any of this stuff.
9:32 PM
Anonymous:
This is definitely not the first time I heard about this crap. When I learned those stuff, and actually believed it, you probably hasn't gone to high school yet. But as I am beginning to think with my own brain, I start to reject these craps from textbooks.
Whatever happens to conservation of energy, the most fundamental physics law, eh? When you say:
rho * R^{4}=const and NOT rho R^{3}=constant
The total CMB energy would equal to rho times volume of universe, which is R^3. So rho*R^3 would have to be constant for energy conservation.
The way BigBang describe the CMB, it considered that the total number of CMB photons remain largely unchanged since it "decouple from matter". Then since the average energy of each photon is reducing (CMB temperature is cooling down), where does the energy go?
Are you going to give up on energy conservation?
Back to backholes. Certainly without Hawking's work, it is still natural to realize that blackholes have entropy, and an enormous amount of it. The entropy is not zero because the blackhole at least carries some information which is measurable: Its mass and size. Once you realize blackholes have a none-zero entropy, the rest of dimentional analysis is easy.
As for the currently known Generalized Second Law. It's wrong, I tell you. Entropy certainly could never be reduced. But the total quantum entropy of the universe could never be increased either!!! The correct statement would be the total quantum entropy of the universe is a conserved quantity.
Putting the GSL and Einstein's Universe Equation together. You will see they conflict with each other. Einstein allows a possible solution, with proper cosmological constant, that the universe could go from a Big Bang and eveutually end up in a Big Crunch. In another word it's entropy go from almost zero, to a huge value, then it crunch back to zero. Does it not violate the Generalize Second Law? You can't believe in both. Actually both are wrong.
Quantoken
11:49 PM
Energy conservation is a fair enough question. But if you had studied this stuff as you say then you should know about the differences of energy densities of relativistic versus ordinary matter. The cosmological energy conservation equation is simply
R^{3}dP/dt=d/dt[R^{3}(P+rho)]
where P is the pressure of the matter and rho its density. This is the same as saying
d/dR[rho R^{3}] = - 3P R^{2}
You then need an equation of state relating P and rho so that P=P(rho). You then determine rho as a function of R. If the energy density of the universe is dominated by nonrelativistic matter with negligible pressure then
rho ~ R^{3} for p<< rho
as you would expect. BUT if the energy density is dominated by relativistic particles like photons (and remember massive particles can be treated as radiation when mc^{2} < < kT in the early universe)then P=rho/3 for a relativistic gas of radiation.
This immediately gives
rho ~ R^{-4} for P=rho/3
Energy conservation is fine.
6:22 AM
Anynymous:
Again you are quoting textbooks that I already know a long time ago. It's wrong.
Basically you are arguing using the pressure. You are saying that there is a pressure, pushing the edge of the universe to expand at light speed. And that pressure and pushing would do some work, costing some energy, and hence lowering the remaining total energy. If you sum up the work done by the pressure and the remaining energy, then the total is still conserved.
That argument is wrong!!! There is nothing to "push" at the edge of the universe, since there is no "edge" of the universe. So there is NO work to be done there, whether the universe expands or not. Any work done within the universe by any pressure or pushing, remains part of the energy within the universe and does NOT change the total mass-energy of the universe!
BigBang breaks that fundamental physics law. If you compare the total mass-energy of the universe at different stages of BigBang, they are not conserved. This is impossible unless the universe has energy exchange or other interactions with the "outside" of the universe, which is un-physical and does not exist!
6:54 AM
I want you to sum up the TOTAL mass-energy of everything, not just radiation, and see if the mass-energy is conserved or not during different stages of the BigBang. It's NOT conserved! That's a big problem.
6:58 AM
This is a web site of Big Bangers ADMIT that they indeed violated energy conservation:
http://www.astronomycafe.net/qadir/ask/a11609.html
How could some one with basic physics training, would accept the violation of energy conservation so lightly and easily, in order to accept the validity of Big Bang theory?
Which one would you choose if you can only choose either energy conservation, or Big Bang, but not both? I would say I keep energy conservation.
Quantoken
7:02 AM
Fair enough. Energy conservation is indeed a thorny issue in general relativity and no-one is saying it is the complete story or that big bang cosmology is complete. It is not a case of giving up energy conservation just in order to believe in the big bang. Certainly, there is no theory of quantum gravity that can explain the very earliest times, particle production or resolve the initial singularity problems. One also has to account for present day observed fluctuations in the CMB, galaxy formation, presence of phase transitions and topological defects, cold dark matter and dark energy/cosmological constant, cosmic acceleration, inflation baryon asymmetry etc. The amount of cosmological data accumulated is large! Thinking about the very early universe takes the human mind to its limit after all. We can only push our theories and physical concepts back so far before they break down. No-one said it should be easy! Why should it be? The correct nucleo-synthesis prediction of elements, redshift and cosmic acceleration and CMB remain a success, however in regimes where we can actually calculate with some confidence (t=3 minutes) and get an agreement with experiment and observation. Thats what matters. There are simply no serious competing theories here that have any credence.
Incidently, I don't just believe anything that happens to be in textbooks. In reading a textbook or paper you must try and work through all the details and hard calculations yourself and try the problems. You have to do the work! You seem to dismiss virtually everything in textbooks and papers without having done this hard ground work in detail and thus you continiously make fundamental errors and wrong statements in attempting to debunk well-estblished theories, when you would know better if you had done the work and actually studied the theories in considerable depth properly. If you had actually studied this stuff and seen it before, like you claim, you would not have made the statements you made earlier in this thread about nuclear fusion in cosmology and the "presence of hydrogen falsifying the BB" since you would already know about the very crucial and subtle role of deuterium in fusion reactions, and its temperature dependence, and the lack of stable nuclei with atomic weights 5 and 8. Nuclear physics we can do here on Earth and understand very well. These reactions are some of the miracles of nature and there is some really beautiful physics here. You would also have already known that rhoR^{4}=const for a photon/radiation gas in Friedmann cosmology and that P=rho/3 for such a relativistic gas, even if you dont believe it. If you dont believe rhoR^{4}=const then thats fine but you had never heard of that until I told you. (Incidently NO other theory can explain why there is 25% helium4 in the universe and the rest hydrogen to and isotopes to this level of accuracy.)
Textbooks are not bibles of course but to do something new or extend current knowledge or question current knowledge and its claims, you have to really know and understand in VERY considerable technical detail what has already been done and what its claims are, even if you don't actually agree with any of it. Peter woit does'nt believe in string theory but he has studied it and its claims carefully in enough technical depth to be in a position to make his comments about it. I have worked through all this cosmological stuff very many times to convince myself, publishing too, and it holds up well. Any reasonable person will find that it does although no-one is seriously expecting them to believe it is the full or complete picture...something we may never find. As much as I like discussing this stuff...even to people who dont buy it...I cant really spare any more time on it.
Regards
steve
8:17 AM
Anonymous:
See my new post how I calculated the CMB temperature to be 2.724K, completely agree with experimental value? I presumed CMB is star radiation and obtained that accurate value. No Big Banger has ever come close to estimate the value of CMB temperature!!!
You may dispute the notion that CMB is star radiation and discard my result as pure coincidence. But you have to admit that stars DO radiate energy and that energy do gets accounted for correctly using CMB energy. If CMB is not star radiation, then you have to exexplain away WHERE does the energy radiated by stars go??? The should not disappear and should still be in the sky. However the bulk of radiation energy is CMB and there is virtually nothing else!!!
Is it not a fair request that you need to account for start radiation energy by your observation data? If you counted CMB for Big Bang remains already, you do not have anything left to account for the amount of energy radiated by stars!!!
Quantoken
8:39 AM
Back to backholes. Certainly without Hawking's work, it is still natural to realize that blackholes have entropy, and an enormous amount of it.
But again, it's not natural to assume that the entropy depends only on observable properties of the BH, it's more natural to guess it would depend on unknowable details about its internal state. The reason this is not natural is because if you assume it is only dependent on observable properties, then as you say it's fairly simple to come up with the guess that it would most likely depend on the area, and yet if the black hole has no temperature this would mean you can come up with a situation where entropy of outside world + entropy of BH actually [i]de[/i]creases, in violation of the second law of thermo.
As for the currently known Generalized Second Law. It's wrong, I tell you. Entropy certainly could never be reduced. But the total quantum entropy of the universe could never be increased either!!! And yet you are apparently unwilling or unable to give a precise definition of what you mean by "quantum entropy", why you say photons have a quantum entropy of 1, and what the connection is between quantum entropy and thermodynamic entropy (even if quantum entropy can never increase, why does this prove thermodynamic entropy can't increase?)
Putting the GSL and Einstein's Universe Equation together. You will see they conflict with each other. Einstein allows a possible solution, with proper cosmological constant, that the universe could go from a Big Bang and eveutually end up in a Big Crunch. In another word it's entropy go from almost zero, to a huge value, then it crunch back to zero. Does it not violate the Generalize Second Law?No, because there's no reason to believe that there would be this sort of symmetry between the expanding phase and the contracting phase--most physicists believe that even if there was a big crunch, entropy would continue to increase during the contraction.
12:41 AM
JesseM you need to exam your logic in your latest messages.
You keep bringing up the issue "if the black hole has no TEMPERATURE". You have got to understand, temperature is a secondary statistics property, defined in terms of other properties. There is ALWAYS a temperature, although I guess what you actually mean is temperature equals zero. Temperature is defined as the derivative of change of energy divided by change of entropy. If entropy is constantly zero, then temperature of a blackhole is infinite, not zero, based on thermodynamics definition.
The trouble is then blackholes VIOLATE the thermodynamics laws by being a heat sink of infinite temperature where it sucks energy and heat from its surrounding of lower temperature. Thermodynamics tells us heat should never flow from low temperature to high temperature. Since blackholes are pretty good at sucking in heats and energies, it's temperature must be very cool and so it must have entropy, and an enormous amount comparing with its energy.
We know blackholes are clearly capable of absorbing photons. Common senses tell us that things always happen symmetrically in nature. If one thing is capable of absorbing photons, it must also be capable of emitting photons. Though how it could have happened remained to be answered by a detail proposed by Hawking as we knew it.
I think physics is the science of observables in nature. None observables, like God or spirit or pure imaginary things are not in the domain of physics. You could say within one photon there is one whole universe inside and we could not see anything of that universe. And there will be an astronomical amount of entropy associated with the unknown details of that universe. But that will be none-physical and physics is really not interested in none-observables. So talking about entropies associated with un-seeable details is meaningless.
You say "And yet you are apparently unwilling or unable to give a precise definition of what you mean by "quantum entropy", why you say photons have a quantum entropy of 1, and what the connection is between quantum entropy and thermodynamic entropy (even if quantum entropy can never increase, why does this prove thermodynamic entropy can't increase?)"
My definition of quantum entropy is the same conventional definition of quantum entropy, but only when you have reaches the smallest possible scales and there is no further microscopic structure or details to be discovered.
To explain what it means, let's look at the conventional definition of Von Neumann quantum entropy. First, you look at molecular scale, you define Von Neumann entropy to be the trace of the matrix obtained by multiplying the logarism of each states's distrubution probability times the states' weighing factor, which is also the same probability. Here you talk about quantum states of individual atoms.
Then you further go down to atomic scale, you can define another Von Neumann entropy by considering the electrons' states around the nuclear, which is now considered a point particle.
And then you could find more states within the nuclear. You find it's composed of protons and quarks and there are states related to that. So you have yet another new definition of Von Meumann entropy. You can further go down this path until you reach the Quark and whatyouknow composes quark, etc etc...
So Von Meumann entropy is really ambiguious until you have decided at what scale you want to calculate it. My definition of quantum information or quantum entropy would be simply the Von Meumann entropy at the smallest possible scale, where no further sub structure could be found. Such a definition of quantum information is valid since we know the smallest structure must exist, you can't push it smaller infinitely.
Such a quantum information encompasses all the thermal entropy and quantum mechanics entropy we encounter at different scales. It is the sum of every things. And everything is constructed from this basic element of quantum information, which is discrete. Spacetime is constructed and described by the quantum information. And mass-energy is further constructed and described by the spacetime geometry, by the curvature of spacetime.
That's the whole idea of GUITAR. It is still a theory in development but I have already obtained many amazingly precise numerical results, including the calculation of G, CMB temperature, baryon density of the universe, solar constant, the precise value of proton and neutron mass, etc. etc.
Among those results the most accurate result is the neutron to electron mass ratio, accurate to 10 decimal places! The least accurate one is G, I am off by 2%. But in both case the limiting factor is really the experimental accuracy of the measurements. G is the least accurately known. Different groups reported experimental results differe by up to 0.7%. I think the known value is really off by 2%, if we had a chance to do a more accurate measurement, at a location far away from the gravity force of the galaxy.
You also said "most physicists believe that even if there was a big crunch, entropy would continue to increase during the contraction."
The I really do not know they define the entropy, which is astronomical amount, confined within a very narrow spacetime upon big crunch. And what is the ruler they would use in nature to measure spacetime and tell us the size of the universe upon cruch? I guess they never thought about that when they mermered that nonsense.
Quantoken
5:58 AM
You keep bringing up the issue "if the black hole has no TEMPERATURE". You have got to understand, temperature is a secondary statistics property, defined in terms of other properties. There is ALWAYS a temperature, although I guess what you actually mean is temperature equals zero. Temperature is defined as the derivative of change of energy divided by change of entropy. If entropy is constantly zero, then temperature of a blackhole is infinite, not zero, based on thermodynamics definition.Uh, I never said the entropy of a black hole was zero, so I don't know what you're talking about there. The point is that if you speculate that the entropy is simply dependent on the BH's area (and if entropy depends only on observable properties, as you say it is pretty trivial to show it is likely to be proportional to mass squared, or area), and yet the BH does not have the ability to supply heat to the external environment (which is what I meant by 'has no temperature'), then it's possible to come up with a situation where the entropy of the external environment plus the entropy of the BH will decrease. As Hawking says in this lecture:
black hole had a physical entropy, it would also have a physical temperature. If a black hole was in contact with thermal radiation, it would absorb some of the radiation, but it would not give off any radiation, since by definition, a black hole was a region from which nothing could escape. If the thermal radiation was at a lower temperature than the black hole, the loss of entropy down the black hole, would be greater than the increase of horizon area.
This would be a violation of the generalized Second Law, that Bekenstein proposed. With hind sight, this should have suggested that black holes radiate. But no one, including Bekenstein and myself, thought anything could get out of a non rotating black hole. So you see, when you proclaim that it's a very simple result that the BH's entropy is proportional to its area, you're missing the point. It is a trivial result if you already assume the BH has an entropy that is dependent only on observable properties, but this is itself a highly nontrivial assumption which most physicists would not have been willing to make, because without the BH being able to heat up its external environment this would violate the 2nd law of thermodynamics.
Common senses tell us that things always happen symmetrically in nature. If one thing is capable of absorbing photons, it must also be capable of emitting photons. Though how it could have happened remained to be answered by a detail proposed by Hawking as we knew it.That isn't common sense at all--you might as well say that since we know nebulas can be pulled together into stars by gravity, we should believe that nebulas can be formed by the atoms in stars slowly drifting apart in a reversed version of the way they form (as opposed to a nova, which is quite different than a reversed version of star formation).
I think physics is the science of observables in nature. None observables, like God or spirit or pure imaginary things are not in the domain of physics. You could say within one photon there is one whole universe inside and we could not see anything of that universe. And there will be an astronomical amount of entropy associated with the unknown details of that universe. But that will be none-physical and physics is really not interested in none-observables. So talking about entropies associated with un-seeable details is meaningless.Even if the entropy of a BH depended on facts which were unknowable to external observers, you could see it in principle, if you are willing to sacrifice your life by diving into the BH yourself. And unlike groundless speculations about the insides of fundamental particles (which would be problematic anyway, since the exclusion principle for fermions is only supposed to apply to states which are precisely identical), we have very good reason for thinking there must be something going on inside a black hole, since we can watch things dropping in and calculate that they will continue to exist after they cross the event horizon, from their own point of view.
You say "And yet you are apparently unwilling or unable to give a precise definition of what you mean by "quantum entropy", why you say photons have a quantum entropy of 1, and what the connection is between quantum entropy and thermodynamic entropy (even if quantum entropy can never increase, why does this prove thermodynamic entropy can't increase?)"
My definition of quantum entropy is the same conventional definition of quantum entropy, but only when you have reaches the smallest possible scales and there is no further microscopic structure or details to be discovered.
To explain what it means, let's look at the conventional definition of Von Neumann quantum entropy. First, you look at molecular scale, you define Von Neumann entropy to be the trace of the matrix obtained by multiplying the logarism of each states's distrubution probability times the states' weighing factor, which is also the same probability. Here you talk about quantum states of individual atoms.
Then you further go down to atomic scale, you can define another Von Neumann entropy by considering the electrons' states around the nuclear, which is now considered a point particle.
And then you could find more states within the nuclear. You find it's composed of protons and quarks and there are states related to that. So you have yet another new definition of Von Meumann entropy. You can further go down this path until you reach the Quark and whatyouknow composes quark, etc etc...
So Von Meumann entropy is really ambiguious until you have decided at what scale you want to calculate it. My definition of quantum information or quantum entropy would be simply the Von Meumann entropy at the smallest possible scale, where no further sub structure could be found. Such a definition of quantum information is valid since we know the smallest structure must exist, you can't push it smaller infinitely.But the conventional notion of entropy in QM is calculated for systems which can be in multiple possible states, just like in classical thermodynamics. You could find the entropy of a system consisting of just one particle confined to a small volume, but you can't assume that the entropy of a large volume containing multiple particles is simply the sum of the entropy for each particle if it was confined to a smaller volume. Like I said before, I'm fairly certain the Von Neumann entropy of a cavity filled with blackbody radiation would change if you increased the size of the cavity, for the same reason that the classical entropy of a gas-filled container would change if the volume of the container changed. So your claim that the entropy of a collection of photons is simply proportional to the number of photons is unlikely to be true if you are talking about Von Neumann entropy.
You also said "most physicists believe that even if there was a big crunch, entropy would continue to increase during the contraction."
The I really do not know they define the entropy, which is astronomical amount, confined within a very narrow spacetime upon big crunch. And what is the ruler they would use in nature to measure spacetime and tell us the size of the universe upon cruch? I guess they never thought about that when they mermered that nonsense.
I doubt you can define entropy at the moment of the Big Crunch or Big Bang, they are singularities after all. But entropy can be defined throughout the contracting phase in exactly the same way it's defined throughout the expansion phase.
4:27 PM
JesseM said:
"Even if the entropy of a BH depended on facts which were unknowable to external observers, you could see it in principle, if you are willing to sacrifice your life by diving into the BH yourself. And unlike groundless speculations about the insides of fundamental particles (which would be problematic anyway, since the exclusion principle for fermions is only supposed to apply to states which are precisely identical), we have very good reason for thinking there must be something going on inside a black hole, since we can watch things dropping in and calculate that they will continue to exist after they cross the event horizon, from their own point of view."
Once again it's unphysical to talk about none-observables. Talking about the insider of blackholes is as meanless and groundless as talking about a presumed universe insider a particle.
You may risk your life to jump into a blackhole and observe what's insider, but it's meaningless unless you can live to come back and tell us the stuff you see. It's the same as asking you to experience death, and then come back alive to tell us what the hell looks like. It's just unthinkable and it's not physics to even think about such unthinkable things.
Quantoken
2:59 PM
Once again it's unphysical to talk about none-observables. Talking about the insider of blackholes is as meanless and groundless as talking about a presumed universe insider a particle.
Well, then according to your definition of "unphysical", it should also be unphysical to talk about the entropy of a BH unless you can show that it will depend only on observable qualities of the BH, and until the realization that BHs can emit radiation, there was very good reason to believe it wouldn't.
Are you going to address my comment about how the Von Neumann entropy of a box filled with blackbody radiation would almost certainly inrease as the size of the box increases?
4:17 PM
JesseM said:
"Are you going to address my comment about how the Von Neumann entropy of a box filled with blackbody radiation would almost certainly inrease as the size of the box increases?"
I think I have made it clear that the entropy of a single photon is a constant regardless where you place that photon, or where that photon came from.
In the case of a box filled with blackbody radiation, and box size increases. You've got to understand that the box must have maintained an energy and entropy EXCHANGE with the outside, to maintain a constant temperature, also the number of photons will also increase. So you have to carefully count all these in in all calculations.
If you do that and make sure everything is counted, you still arrive at the conclusion that the total entropy is exactly porportional to the number of photons in the box, and the entropy of each photon is still a constant ONE.
5:08 PM
In the case of a box filled with blackbody radiation, and box size increases. You've got to understand that the box must have maintained an energy and entropy EXCHANGE with the outside, to maintain a constant temperature
No, I'm talking about a case where the contents of the box are completely isolated from the outside, and you increase the size without messing with the contents in any way. In this case the temperature will not stay constant, it will drop.
also the number of photons will also increase. So you have to carefully count all these in in all calculations.
If you do that and make sure everything is counted, you still arrive at the conclusion that the total entropy is exactly porportional to the number of photons in the box, and the entropy of each photon is still a constant ONE.I don't believe that you have done the calculations that would be needed to prove this--figuring out the number of photons would presumably require quantum electrodynamics, and I doubt you are well-versed in quantum field theory, judging from your comments.
7:47 PM
JesseM said:
"No, I'm talking about a case where the contents of the box are completely isolated from the outside, and you increase the size without messing with the contents in any way. In this case the temperature will not stay constant, it will drop."
If you are talking about a box which has NO interaction with the photons in the box, then, once you increase the box size, the photons in the box no longer exhibit an equilibrium blackbody spectrum. Since the photons do not interact with each other, they could not reach thermal balance amoung themselves. The same number and energy distribution of the photons remain, but their density per volume dilutes due to the increase box size.
If the photons do interact with the wall of the box, then there is entropy and energy change. The wall will help split higher energy photons into more photons of lower energy. If you carefully calculate everything, each photon's entropy are eaxctly the same.
You also said:
"I don't believe that you have done the calculations that would be needed to prove this--figuring out the number of photons would presumably require quantum electrodynamics, and I doubt you are well-versed in quantum field theory, judging from your comments."
Wrong and wrong again. You've got to know that the guy who first derived blackbody radiation and other thermodynamics stuffs did not know anything about QED at all, since QED wasn't even invented at that time yet. There is no need to involve QED in our discussion of entropy here. Clearly, since a photon's entropy does not depend on alpha, the characteristic number describing QED, so it really has nothing to do with QED. A neutrino, assuming it is indeed massless and travel at light speed, also has an entropy of ONE, even though it is not described by QED.
Quantoken
10:49 PM
If you are talking about a box which has NO interaction with the photons in the boxI said the contents of the box have no interaction with the outside world. The inside walls of the box are not part of the "outside world". Remember, I was responding to your comment "You've got to understand that the box must have maintained an energy and entropy EXCHANGE with the outside, to maintain a constant temperature". An interaction between the photons and the inside walls of the box does not constitute "an energy and entropy exchange with the outside".
If the photons do interact with the wall of the box, then there is entropy and energy change. The wall will help split higher energy photons into more photons of lower energy. If you carefully calculate everything, each photon's entropy are eaxctly the same.I don't believe you have "carefully calculated everything" here.
Wrong and wrong again. You've got to know that the guy who first derived blackbody radiation and other thermodynamics stuffs did not know anything about QED at all, since QED wasn't even invented at that time yet. Yes, and photons weren't invented at that time either. To calculate the number of electromagnetic quanta (photons) in a box filled with blackbody radiation, you need a quantum theory of elecromagnetism, I can't see how there could be any other valid way to calculate this.
11:33 PM
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12:00 AM
OK, I did some research and found out that what I said above about needing QED is wrong, all you really need is the fact that E=hf for an individual photon, so if you know the total energy at each cavity mode (given here), you can divide by the energy of an individual photon at that frequency to get the number of photons at that frequency, and sum over all the different modes. This page gives the average number of photons per mode as a function of temperature (and also a formula which can be rearranged to tell you the temperature as a function of total energy and volume), so if you just sum over the allowable modes you should be able to get the total number of photons.
12:24 AM
Jesse, the descriptions in the links you give are absolutely correct as is the entropy formula for blackbody photons. Its just basic stuff really, something that would be covered in a couple of lectures in any undergraduate course on thermodynamics or statistical mechanics. Predictably, quantoken will argue (erronously) that it is all wrong.
10:35 AM
Anonymous said: "Predictably, quantoken will argue (erronously) that it is all wrong."
Your prediction is wrong. I carefully examined the link JesseM provided the first time I saw JesseM posted it, and I did NOT find anything wrong in it. So I did not say anything about it.
http://www.sparknotes.com/physics/thermodynamics/stats/section2.rhtml
It is correct. You would be able to derive a constant entropy for an individual photon, out of that relationship. The two are completely compatible.
11:11 AM
One thing I wondered about after looking at that page is that the formula for the total energy U contained in a given volume as a function of the volume V and the temperature T is U = (pi^2 * T^4 * V) / (15 * hbar^3 * c^3). So if you assume that the entropy of a box filled with blackbody radiation is simply the total energy in that box divided by the temperature, you'd get S = (pi^2 * T^3 * V) / (15 * hbar^3 * c^3). But the formula for entropy they give is S = (4 * pi^2 * V / 45) * (T / hbar * c)^3, which is almost exactly what I got by taking U/T except for an extra factor of 4/3...where does that factor come from?
Anyway, they also give the number of photons in a given mode as 1/(e^(hbar*omega/T) - 1), or 1/(e^(hbar*2pi*f/T) - 1). If you have a box with sides of lenght L, will all available modes have wavelengths of L/n (where n is some positive integer), or f = nc/L? If so, to find the total number of photons in the box you'd have to do a sum over each possible n of 1/(e^(hbar*2pi*n*c/L*T) - 1)...they said that "the sum over positive n in three dimensions becomes 1/8 * (integral from 0 to infinity of 4*pi*n^2 dn)", I'm not quite sure how that would work, do you just substitute 4*pi*n^2 in the place of n in the above equation and then integrate, or something else? Meanwhile, you can also use the Stefan-Boltzmann equation to get T as a function of energy U and volume V=L^3, or T=squareroot(15*hbar^3*c^3*U/pi^2*L^3), and plug that into the function for the number of photons. So, this should give the total number of photons in a box as a function of energy U and side length L, so you can figure out what happens to that number as you keep U constant and vary L...can anyone help me out with how you'd do the integral over n to get this function?
12:35 PM
JesseM:
I am not your physics teacher and do not intend to turn this BLOG into a free tutoring station. But let me give you a small hint.
It is not T*S = U, The formula is:
T*dS = dU
If you different S and different U on that two equation given at the URL, you see they do match.
I assure you the stuff that URL talks is correct. You need to do a little bit exercise and calculate number of photons and how much entropy each photon has, etc. Once you get your calculation correct you get the conclusion that each photon always carry an entropy of ONE, regardless of its energy or the size, temperature of the box it came from. Ask your physics teacher if you need help.
Quantoken
12:47 PM
Quantoken, I know that dS = dQ/T, but according to this page you can also say that S = Q/T, where S is the entropy and Q is the total heat content of the system. But I did a little more research and figured out that "heat content" is not the same thing as the total energy U of the system--the heat content is another word for the enthalpy, which is defined as U + PV, where P is pressure and V is volume.
As for my other question, if you have already done the integral to find the total number of photons in the box, it should be no problem for you to reproduce the mathematics here, I'm not convinced that the answer will in fact be proportional to the entropy as you vary L. If you're unwilling to do this, maybe some other reader of this blog can do so.
1:42 PM
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