Thermodynamics   La chaleur est la force vive qui résulte des mouvementsinsensibles des molécules d'un corps. "Mémoire sur la chaleur" (1780) Lavoisier &Laplace
On this site, see also:
Ludwig Boltzmann (1844-1906). Dangerous Knowledge:4 | 5 | 6 | ||||||||||||||||||
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The quantity Cp/ V (which appears as the denominator of thelast term) is the molar heat capacity per molar volume; it's also equalto the mass density multiplied intothe specific heat capacity cp (note lowercase). All told, that term is the inverse of a quantity W homogeneous to a pressure,an elasticity coefficient, or an energy density (more than 10 years ago,I proposed the term thermal wring as a pretext for using the symbol W, which isn'toverloaded in this context):
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One non-hydrostatic example is a battery of electromotive force e (i.e., e is the voltage at the electrodes when no currentflows) andinternal resistance Ras it delivers a charge q in a time t. The longer the time t, the closer we areto the quasistatic conditions which make the transfer of energy approachthe lower limit imposed by the change in G, according to the following inequality:- (2013-01-17)
Adiabatic derivative of Log T with respect to Log 1/V (or Log ). (orth ) to denote this Grüneisen parameter. I beg to differ (to prevent confusion with the adiabatic coefficient).
= V T S = T S 
T V T is the adiabatic ratio of the logarithmic differentials of two quantities: temperature and either density or volume (those two differ by sign only).
The relation to the adiabatic coefficient = Cp / Cv = Ks / KT is simply:
= 1 + T
For condensed states of matter (liquids or solids) the volumetric coefficient of thermal expansion () is quite small and the above adiabatic coefficient remains very close to unity; the Grüneisen parameter is more meaningful (the adiabatic coefficient is traditionally reserved to the study ofgases).
(2006-09-18)
A closed formula for the entropy of a Van der Waals fluid.For one mole of a Van der Waals gas, we have:
( p +a / V2 ) ( V-b ) = RT
( p a / V2 + 2a b / V 3 ) dV + ( V-b ) dp = R dT
Let's combine this with the third relation of Maxwell :
S T = p V = R V T V - b Therefore, S = f (T) + R Log (V-b) for some function f
To be more definite, we resort tocalorimetric considerations,namely:
S V = CV = f ' (T) T T This shows that CV is a function of temperature alone. So, we may as well evaluate it for large molar volumes (very low pressure) and find that:
CV = j/2 R
That relation comes from the fact that, at very low pressure, the energy of interactionbetween molecules is negligible. Therefore, by the theorem of equipartition of energy, the entire energy of a gas is the energy which gets equally distributedamong the j active degrees of freedom of each molecule, including the3 translational degrees of freedom which areused to define temperatureand 0, 2 or 3 rotational degrees of freedom (we assume the temperature is low enough for vibrations modes of the molecules tohave negligible effects; seebelow). All told: j = 3 for a monoatomic gas,j = 5 for a diatomic gas, j = 6 otherwise.
S = S0 + j/2 R Log (T) + R Log (V-b)
There's no way to reconcile this expression with Nernst'sthird lawto make the entropy vanish at zero temperature. That's because the domain of validityof the Van der Waals equation of state does not extend all theway down to zero temperature (there would presumably be a transition to asolid phase at low temperature, which is not accounted for by the model). So, we may as well accept the classical view, which definesentropy only up to an additive constant andchoose the following expression (thestatistical definition of entropy, ultimatelybased onquantum considerations, leaves no such leeway).
Entropy of a Van der Waals fluid ( j = 3, 5 or 6 ) S = R Log [ T j /2 (V-b) ] Thus, the isentropic equation of such a fluidgeneralizes one of the formulations valid for anideal gas, when b = 0 and j/2 = :
T j /2 (V-b) = constant
Unlike CV, Cp = CV is not constant for a Van der Waals fluid,since:
Cp CV = T p V V p T T - (2012-07-03)
The heat capacity per mole is nearly the same (3R) for all crystals. In 1819, Dulong and Petit (respectively the third and the secondholder of the chair of physics at Polytechnique in Paris, France) jointly observed that the heat capacity of metalliccrystals is essentially proportional to their number of atoms. They found it to be nearly 25 J/K/mol for every solid metalthey investigated (this would have failed at very low temperatures).
In 1907,Albert Einstein gave asimplified quantum model,which explained qualitatively why the Dulong-Petit law fails at low temperature. His model also linked the empirical values of the molar heat capacity (at high temperatures) to theideal gas constant R :
3 R = 24.943386(23) J/K/mol
In 1912,Peter Debye devised an evenbetter model (equating the solid's vibrational modes with propagating phonons ) which is also good at low temperatures. Its limited accuracy at intermediate temperatures is entirely due to the simplifying assumptionthat all phonons travel at the same speed. When applied to a gas of photons, that statement is trueand the model then describes blackbody radiation perfectly,explainingPlanck's law !
- wangshu(2010-12-19)
What do the following energy levels contribute to the heatcapacity of carbon dioxide at 400 K? En = (n+1) h where = 20 THz The ratio h / kT = 2.4 indicates that the quanta involved are similar to the average energyof athermal photon (2.7 kT).
- (2005-06-25)
Asentropy varies in a change of phase,someheat must be transferred. The British chemistJosephBlack (1728-1799) is credited with the1754 discovery of fixed air (carbon dioxide) which helpeddisprove the erroneous phlogiston theory of combustion. James Watt (1736-1819)was once his pupil and his assistant. Around 1761, Black observed that a phase transition (e.g., from solid to liquid) must be accompanied by a transfer of heat, which is now called latent heat. In 1764, he first measured thelatent heat of steam.
Thelatent heat L is best described as the difference H in theenthalpy (H=U+pV) of the two phases, which accuratelyrepresents heat transferred under constant pressure (as this voids the second term in dH = TdS + Vdp).
Under constant pressure,phase transistion occurs at constant temperature. So, the free enthalpy (G=H-TS) remains constant (as dG = -SdT + Vdp).
Consider now how this free enthalpy G varies alongthe curve which gives the pressure p as a function of the temperature T when the two phases 1 and 2 coexist. Since G is the same on either side of this curve, we have:
dG = -S1 dT + V1 dp
dG = -S2 dT + V2 dpTherefore, dp/dT is the ratio S/V of the change in entropy to the change in volumeentailed by the phase transition. Since TS = H, we obtain:
The Clausius-Clapeyron Relation : T dp/dT = H / V = L / V That relation is one of the nicest resultsof classical thermodynamics.
- (2020-01-17)
Either diamond or graphite can be metastable in the domain of the other. The boundary between the domains of graphite and diamond is well-known: In the plot of Log(p) as a fubction of T, tt's a straight line between the point at 0 K of temperature and 1.7 GPaof pressure and the diamond-graphite-liquid triple point (5000 K, 12 GPa).
Meeting at that triple-pont on either side or that boundary are two other straight linesbetween which one carbon allotrope is metastable in the domain where the otheris stable. Thus, diamond is ordinarily metastable but wouldmorph into graphite at high temperature and low pressure.
- (2016-08-01)
Entropy doesn't increase when identical substances are mixed. - (2016-08-01)
Partial pressure is proportional to molar fraction. 
(2016-07-28)
How an equilibrium changes when temperature varies.- (2006-09-23)
A flow expansion of areal gasmay cool it enough to liquefy it. Joule Expansion & Inner Pressure
Expanding dS along dT and dV, the expression dU = T dS - p dV becomes:
dU = T S V dT + T S T p dV T V = CV dT + T p V p dV T This gives the following expression (vanishing for a perfect gas) of theso-calledJoule coefficient which tellshow the temperature of a fluid varies when it undergoes a Joule expansion, where the internal energy (U) remains constant. An example of a Joule expansion is the removal of a separation between the gasand an empty chamber.
T U = 1 T p V p V Cv T The above square bracket is often called the internal pressure or inner pressure. It's normally a positive quantity which repays study. Let's see what it amounts to in the case of aVan der Waals fluid;U T = T p T p = RT p = a V T V-b V2 By integration, this yields: U = U0(T)a /V. Thelatent heat of liquefaction (L) is obtained in term of themolar volumes of the gaseous and liquid phases (VG,VL) either asH = U+pV or as TS (using theabove expression for S):
L = p (VG-VL) +a (1/VL-1/VG) = RT Log [ (VG-b) / (VL-b) ]
Joule-Thomson (Joule-Kelvin) Expansion Flow Process
The Joule-Kelvincoefficient () pertains to an isenthalpic expansion. Its value is obtained as above (from an expression of dH instead of dU):
| = | T | H | = | 1 | T | V | p | V | = | V | ( T 1 ) | ||
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| p | Cp | T | Cp | ||||||||||
vanishes forperfect gasesbut allows an expansionflow process which can cool many real gasesenough to liquefy them, if the initial temperature is below theso-called inversion temperature, which makes positive.
More precisely, the inversion temperature is a function of pressure. In the (T,p) diagram, there is a domain where isenthalpic decompression causes cooling. The boundary of that domain is called the inversion curve. In the example of a Van der Waals fluid, the equation of the inversion curve isobtained as follows:
0 = T V p V = R T V T p a / V2 + 2a b/ V3 This gives a relation which we may write next to the equation of state:
R T = p V a / V + 2a b/ V2 R T + p b = p V + a / V a b/ V2 By eliminating V between those two equations, we obtain a single relationwhich is best expressed in units of the critical point (pc = 1,Tc = 1):
T = 15 / 4 + p / 24 9 - p If T is above T (or below T ) then decompression won't cool the gas.
At fairly low pressures, the inversion temperature is approximately:
Ti = 6.75 Tc
The ratio observed for most actual gases is lower than 6.75: Although it's 7.7 for helium, it's only 6.1 for hydrogen, 5.2 for neon, 4.9 for oxygen or nitrogen, and 4.8 for argon.
A Joule-Thomson cryogenic apparatus has no moving parts at low temperature (note that the cold but unliquefied part of the gas is returned in thermal contact withthe high-pressure intake gas, to pre-cool it before the expansion valve).
The effect was described in 1852 by William Thomson (before he becameLord Kelvin). So was the basic design, with the cooling countercurrent. Several such cryogenic devices can be "cascaded" so thatone liquefied gas is used to lower the intake temperature of the next apparatus... Liquid oxygen was obtained this way in 1877,byLouis Paul Cailletet (France)andRaoul Pierre Pictet (Switzerland). Hydrogen was first liquefied in 1898, by Sir James Dewar (1842-1923). Finally, helium was liquefied in 1908, by the Dutch physicist Heike Kamerlingh Onnes (1853-1926; Nobel 1913).
In 1895, the German engineerCarl von Linde (1842-1934)designed anair liquefaction machine based on this throttling process,which is now named after him (Linde's method).
(2019-07-30)
Reversible thermoelectric phenomena.The direct conversion of heat into electricity at the junction of two different metals wasdiscovered in 1794 by Alessandro Volta (1745-1827). This was rediscovered in 1821 by Thomas Johann Seebeck(1770-1830) who observed only the ensuing deflection of a nearby compass needleand termed the effect thermomagnetic. Ørsted (who had discovered only a few monthsearlier that an electric current has magnetic effects) realized thatan electric current was the proper cause and he wisely renamed theeffect thermoelectric, which now stands.
The effect happens to be thermodynamically reversible so that if two junctions of unlike metals are connected in a circuit, then one of the is heated and the other is cooled (according to the direction of the current). The ensuing possibility of direct electric cooling was discovered in 1834by the French physicist Jean-Charles Peltier and the phenomenonis now called Peltier effect in his honor.
- (2005-06-25)
Heat is not the same "form of energy" as Hamiltonian energy. We have introduced relativistic thermodynamicselsewherein the case of apointlike systemwhose rest mass may vary. The fundamental relativistic distinction betweenthe total Hamiltonian energy (E) and the internal energy (U) was also heraldedabove.
Now, it should be clear from itsstatistical definitionthat entropy is a relativistic invariant,since the probability of a well-defined spacetime event cannot depend on thespeed of whoever observes it. Mercifully, all reputable authors agree on this one... They haven't always agreed on the following (correct) formula for the temperature T of a body moving at speed v whose temperature is T0 in its rest frame :
Mosengeil's Formula (1906) T = T0 1 v/c The invariance of the entropy S means that aquantity of heat (Q = T dS) transforms like the temperature T. So do all the thermodynamic potentials,including internal energy (U) free energy (F = U-TS) Helmholtz'enthalpy (H = U+pV) and Gibbs'free enthalpy (G = H-TS)...
Q = Q0 1 v/c Note that theideal gas law (pV = RT) is invariant because the pressure p is invariant, whereas the temperature T transforms like the volume V. (The same remark would hold for any gas obeyingJoule's second law.)
One of several ways to justify the above expression for the temperature of a movingbody is to remark that thefrequency of a typical photon from a movingblackbody is proportional to its temperature. Thus, if it can be defined at all, temperature musttransform like a frequency. This viewpoint was first expounded in 1906 by Kurd von Mosengeil and adopted at once byPlanck,Hasenöhrl andEinstein(who would feel a misguided urge to recant, 45 years later).
In 1911, Ferencz Jüttner (1878-1958) retrieved the same formula for a moving gas,using a relativistic variant of an earlier argument ofHelmholtz. He derived the relativistic speeddistribution function recently confirmed numerically (2008-04-23) by Constantin Rasinariu in the case of a 2-dimensional gas. (In his1964 paperentitled Wave-mechanical approach to relativistic thermodynamics,L. Gold gave a quantum version of Jüttner's argument.) Mosengeil's (correct) formula was also featured in the textbook published by Max von Laue in 1924.
In 1967, under the supervision of Louis de Broglie, Abdelmalek Guessous completed a full-blown attack,using Boltzmann's statistical mechanics (reviewed below). This left no doubt that thermodynamical temperature must indeed transform as stated above (in modern physics, other flavors of temperature are not welcome).
Equating heat with a form of energy was once amajor breakthrough, but thefundamentalrelativistic distinctionbetween heat and Hamiltonian energynoted by most pioneers (including Einstein in his youth) was butchered by others (including Einstein in his old age) before its ultimate vindication...
A few introductions to Relativistic Thermodynamics :
- Richard Chace Tolman (1881-1948)
"Relativity, Thermodynamics, and Cosmology" (1934)
Oxford University Press (Dover 1987 Reprint: ISBN 0-486-65383-8) - Louis de Broglie (1892-1987;Nobel 1929)
"La Thermodynamique de la particule isolée" (1964)
Gauthier-Villars, Paris 1964. - Tom W.B. Kibble, FRS (1933-2016)
Relativistic TransformationLaws for Thermodynamic Variables
Nuovo Cimento 41, 72-78 (1966) [rebutting Ott, Arzeliès & Gamba] - Abdelmalek Guessous [Amazon|Google]305 pages (93 from thesis)
"Thermodynamique relativiste" Gauthier-Villars, Paris 1970. - Wilfried Schröder and Hans-Jürgen Treder
The "Einstein-Laue"discussion. [about a 1992 paper by Chuang Liu] - Covariant Thermodynamicsof an Object with Finite Volume
by Tadas K. Nakamura (arXiv, 2005-05-05). - Inverse Temperature 4-vector inSpecial Relativity
by Zhong Chao Wu (arXiv, 2008-04-25).
In 1968,Pierre-V.Grosjean called the waves of controversies aboutMosengeil's formulathe temperature quarrel... In spite or because of its long history, that dispute is regularly revived by authors who keep discarding one fundamental subtlety of thermodynamics: Heat doesn't transform like a Hamiltonian energy (which is the time-component of an energy-momentum 4-vector) butlike aLagrangian. Many essays are thus going astray, including:
- (1923) by Arthur S. Eddington(1888-1944). Page 34 of the 1954 edition (Cambridge).
- by Danilo Blanusa(1903-1987). Glasnik mat.-fiz i astr.,2 (1947).
- last paper of Heinrich Ott (1894-1962). Zeitschrift für Physik 175, 70-104 (1963).
- by Henri Arzeliès(1913-2003) and Nuovo Cimento 35, 792-804(1965).
- by A. Gamba. Nuovo Cimento 37, 1792-1794 (1965).
- (1967) bythe Danish physicist Christian Møller (1904-1980).
- by Sean A. Hayward (arXiv, 1999-01-22).
- by W.Z. Jiang (arXiv, 2003-05-14).
- by Isak Avramov (2003).
by Bernhard Rothenstein and Ioan Zaharie (Theoretics, 2003)
by Bernhard Rothenstein andGeorge J. Spix (arXiv, 2005-07-27).
by Manfred Requardt (arXiv, 2008-01-17).- by Phil Schewe (AIP, 2007-10-19).
by C.A. Farías, P.S. Moya & V.A. Pinto (arXiv, 2008-03-20).
by Julio Güémez (Physics Recearch International, 2011-04-05).
by Victor A. Pinto & Pablo S. Moya (Nature, 2017-12-15).
Schewe's article prompted a Physics Forums discussion on2007-10-27. Related threads: 2007-06-29,2008-01-11,2008-04-14. Other threads include: 2007-10-24,2007-12-01,2013-05-02,etc.
I stand firmly by the statement that if temperature can be defined at all, it must obeyMosengeil's formula. The following articles argue against the premise of that conditionalproposition, at least for one type of thermometer:
- Doesa Moving Body appear Cool?
by Peter T. Landsberg,1922-2010 (Nature, 1967-05-27). - Temperature andRelativity
by Sandro S. Costa and George E.A. Matsas (1995-05-24). - Laying the ghostof the relativistic temperature transformation
by Peter T. Landsberg & George E.A. Matsas (arXiv, 1996-10-23). - The impossibility of auniversal relativistic temperature transformation
by Peter T. Landsberg and George E.A. Matsas (Physica, 2004). - Entropy of aRelativistic Mono-Atomic Gas
by Edward Bormashenko (Entropy, 2007).
- Richard Chace Tolman (1881-1948)
In 1970, Guessous published an expanded version of the thesis as abook entitled "Thermodynamique relativiste", prefaced by his famous adviser Louis de Broglie (Nobel 1929). That work is still quoted by some scholars,likeSimon Lévesque,fromCégepde Trois-Rivières (2007-01-13) although the subject is no longer fashionable.
The original dissertation of Abdelmalek Guessous appears verbatim asthe first five chapters of the book (93 out of 305 pages). Unfortunately, Guessous avoids contradicting (formally, at least) what was established by his adviser for pointlike systems. This impairs some of the gems found in Chapter VI and beyond,because the author retains his early notations even as he shows them to be dubious. For example, the paramount definition of internal energyas a thermodynamical potential (transforming like a Lagrangian) doesn't appear until Chapter VI where it's dubbed U', since U was used throughout the thesis forthe Hamiltonian energy E (not clearly identified as such). More importantly, Guessous runs into the correct definitionof the inertial mass (namely, the momentum to velocity ratio) but keeps calling it "renormalized mass" (denoted M' ) while awkwardly retaining the symbol M as a mere name for E/c (denoted U/c by him) which downplays the importance of the aforementioned true inertial mass m = M'. So, Guessous missed (albeit barely so) the revolutionary expression of theinertia of energy for N interacting particles at a nonzerotemperature T, presentednextin the full glory of traditional notations consistent with the rest of this page.
- (2008-10-07)
The Hamiltonian energy E is not proportional to the inertial mass M. Here's the great formula which I obtained many years ago by pushing to their logicalconclusion some of the arguments presented in the 1969 addenda to the1967doctoral dissertation of Abdelmalek Guessous. I've kept this result of my younger self in pectore for far too long (I first read Guessous' work in 1973).
E = M c 2 N k T We define the inertial mass (M) of a relativisticsystem of N point-masses as the ratio of its total momentum p to the velocity v of its center-of-mass:
p = Mv
It's not obvious that thedynamical momentum p is actually proportional to the velocity v so that M turns out to be simply a scalar quantity !
The description of a moving object of nonzero size always takes placeat constant time in the frame K of the observer. The events which are part of such a description are simultaneous in K but are usually not simultaneous in the rest frame (K) of the object. That viewpoint has been a basic tenet ofSpecialRelativity ever since Einstein showed in excruciating detailshow it explains theLorentz-Fitzgerald contraction,which makes the volume V of a moving solidappear smaller than its volume at rest V0 :
V = V0 1 v/c - (2008-10-13)
Local temperature is higher on the outer parts of a rotating body. - (2005-06-20)
Each unit of area at the surface of a black body radiates a total powerproportional to the fourth power of its thermodynamic temperature. This law was discovered experimentally in 1879, by the Slovene-Austrian physicistJoseph (Jozef) Stefan (1835-1893). It would be justified theoretically in 1884, by Stefan's most famous student: Ludwig Boltzmann (1844-1906).
The energy density (in J/m3 or Pa) of the thermalradiation inside an oven of thermodynamic temperature T (in K) isgiven by the following relation:
[ 4 / c ] T 4 = [ 7.566 10-16 Pa/K4 ] T 4
On the other hand, each unit of area at the surface of a black bodyradiates away a power proportional to the fourth power of its temperature T. The coefficient of proportionality is Stefan's constant (which is also known as the Stefan-Boltzmann constant). Namely:
= ( 2 5 k 4) / ( 15 h 3 c 3 ) = 5.6704 10-8 W/m2/K4
Those two statements are related. The latter can be derived from the former using the following argument,based on geometrical optics, which merely assumes thatradiation escapes at a speed equal to Einstein'sconstant (c).
One of the best physical approximations to an element of the surfaceof a "black" body is a small opening in the wall of a large cavity ("oven"). Indeed, any light entering such an opening will never be reflected directly. Whatever comes in is "absorbed", whatever comes out bears no relation whatsoevertoany feature of what was recently absorbed... The thing isblack in this precise physical sense.
Luminosity of a Spherical Blackbody L = 4 T4 R2 To a good enough approximation, this formula relates the surface temperature T of a star of radiant absolute luminosity L to its radius R.
- (2005-06-19)
Several arguments would place anupper bound on temperature... Several arguments have been proposed which would put a theoretical maximumto the thermodynamic temperature scale. This has been [abusively] touted as a "fourth law" of thermodynamics. Some arguments are obsolete, others are still debated within the latest contextof thestandard model of particle physics:
In 1973, D.C. Kelly argued that no temperature could ever exceed a limitof a trillion kelvins or so, because when particles are heated up, very highkinetic energies will be used to create new particle-antiparticle pairs ratherthan further contribute to an increase in the velocities of existing particles.Thus, adding energy will increase the total number of particles rather than thetemperature.
This quantum argument is predated by a semi-classical guess,featuring a rough quantitative agreement: In 1952, French physicist Yves Rocard (father of Michel Rocard, who was France's prime minister from 1988 to 1991) had argued that the density of electromagnetic energy ought not to exceed by muchits value at the surface of a "classical electron" (a uniformly charged sphere with a radius of about 2.81794 fm). Stefan's law would then imply an upper limit for temperatureon the order of what has since been dubbed "Rocard's temperature", namely:
3.4423K
One process seems capable of generating temperatureswell above Rocard's temperature: theexplosionof a black hole viaHawking radiation.
Rocard's temperature would be that of a black hole of about8 1011 kg, which is much too smallto be created by the gravitational collapse of a star.Such a black hole could only be a "primordial" black hole,resulting from the hypotheticalcollapse of "original" irregularities (shortly after the big bang). Yet, the discussion below shows that a black hole whose temperature isRocard's temperature would radiate away its energy for a very long time:about 64 million times the age of the present Universe... It gets hotter as its gets smaller and older.
As the derivation we gave forStefan's Lawwas based on geometrical optics,it does not apply in the immediate vicinityof a black hole (space curvature is importantand wavelengths need not be much shorter than the sizes involved). A careful analysis would show that a Schwarzschild black hole absorbsphotons as if it was a massless black sphere (around which space is "flat") with aradius equal to a = 33 MG/c (about 2.6 times the Schwarzschild radius). Thus, it emits like a black body of that size (obeying Stefan's law). Its power output is:
P = ( 4a 2) T 4 = 108 M 2 G 2 T 4/ c 4
Using for T the temperature of a Schwarzschild black hole of mass M:
P = 9 h c6 20480 M2 2 G2 As this entails a mass lossinversely proportional to the square of the mass, the cube of the black hole's mass decreases at a constant rateof 27 / 20480 (in natural units). The black hole will thus evaporate completely after a time proportionalto the cube of its mass, the coefficient of proportionality being about 5.96 s per cubic kilogram. A black hole of 2 million tonnes (2 kg)would therefore have a lifetime about equal tothe age of the Universe (15 billion years).
Hawking thus showed that his first hunch was not quite right: a black hole's area maydecrease steadilybecause of the radiation which does carry entropy away. The only absolute law is that, in this process like in any other,the total entropy of the Universe can only increase. There is little doubt that Hawking's computations are validdown to masses as small as a fraction of a gram. However, it must be invalid for masses of the order of the Planck mass (about 0.02 mg),as the computed temperature would otherwise be such that a "typical" photonwould carry away an energy kT equivalent to the black hole's total energy.
In his wonderful 1977 book The First Three Minutes, 1979 Nobel laureateSteven Weinberggives credit to R. Hagedorn, of the CERN laboratory in Geneva,for coming up with the idea of a maximum temperature in particle physics when anunlimited number of hadron species is allowed. Weinberg quotes work on the subjectby a number of theorists including Kerson Huang (of MIT) and himself, and statesthe "surprisingly low" maximum temperature of 2 000 000 000 000 K for the limit based on this idea...
However, in an afterword to the 1993 edition of his book,Weinberg points out that the "asymptotically free" theory of stronginteractions made the idea obsolete: a much hotter Universe would "simply"behave as a gas of quarks, leptons, and photons (until unknown territory is foundin the vicinity of Planck's temperature).
In spite of these and other difficulties, there may be a maximum temperaturewell below Planck's temperature which is not attainable by any means, includingblack hole explosions: One guess is that newly created particles could form a hot shell aroundthe black hole which could radiate energy back into the black hole. The black hole would thus lose energy at a lesser rate,and would appear cooler and/or larger to a distant observer. The "fourth law" is not dead yet...
- (2005-06-20)
Like all perfect absorbers, black holes radiate with blackbody spectra. The much-celebrated story of this fundamental discoverystarts with the original remark by Stephen W. Hawking (in November 1970) thatthe surface area of a black hole can never decrease. This law suggested that surface area is to a black hole what entropy is to any other physical object.
Jacob Bekenstein (1947-2015) was then a postgraduate student working at Princeton under John Wheeler. He was the first to take this physical analogy seriously, before all themathematical evidence was in. Following Wheeler, Bekenstein remarked that black holes swallow the entropy ofwhatever falls into them. If thesecond law of thermodynamics is to holdin a Universe containing black holes, some entropy must be assigned to black holes. Bekenstein suggested that the entropy of a black hole was, in fact,proportional to its surface area...
At first, Hawking was upset by Bekenstein's "misuse" of his discovery,because it seemed obvious that anything having an entropy wouldalso have a temperature, and that anything having a temperature would radiate awaysome of its energy. Since black holes were thought to be unable to let anythingescape (including radiation) they could not have a temperature or anentropy. Or so it seemed for a while... However, in 1973, Hawking himself made acclaimed calculations confirming Bekenstein's hunch:
The entropy (S) of a black hole is proportional
tothe area (A) of its event horizon.S = k 2 c 3 ( ¼ A ) h G What Stephen Hawking discovered is that quantum effects allow blackholes to radiate (and force them to do so). One ofseveralexplanatory pictures is based on the steady creations and anihilations ofparticle/antiparticle pairs in the vacuum, close to a black hole... Occasionally, a newly-born particle falls into the black hole before recombiningwith its sister, which then flies away as if it had beendirectly emitted by the black hole. The work spentin separating the particle from its antiparticle comesfrom the black hole itself, which thus lost an energy equal to the mass-energy ofthe emitted particle.
For a massive enough black hole, Stephen Hawking found the corresponding radiation spectrumto be that of a perfect blackbody having a temperature proportional to the surface gravity g of the black hole (which is constant over the entire horizon of a stationary black hole). In 1976,Bill Unruhgeneralized this proportionality between any gravitional field g (or acceleration) and the temperature T of an associated heat bath.
Temperature T is proportional to (surface) gravity g. kT = g h / (42c) [ Any coherent units ] T = g / 2 [ In natural units ] In the case of the simplest black hole (established by Karl Schwarzschild as early as 1915) g is c/2R, where R is the Schwarzschild radius equal to 2MG/c for a black hole of mass M. In SI units, kT is about 1.694 / M.
Temperature of a Schwarzschild black hole : kT = c 3(h/2) /(4G M) [ Any coherent units ] T = 1 / M [ In natural units ] Rationalized Natural System of Units c = 1 [Einstein's constant ] h = 2 [Planck's constant ] G = 1 / 4 [gravitational constant ] o = 1 [magnetic constant ] k = 1 [Boltzmann's constant ] - (2005-06-13)
Z is a sum over all possible quantum states: Z() = exp ( E)
E is the energy of each state. = 1/kT is related to temperature. A simplified didactic model: N independent spins
Consider a large number (N) of paramagnetic ions whose locations in a crystalare sufficiently distant so interactions between them can be neglected. The whole thing is submitted to a uniform magnetic inductionB. Using the simplifying assumption that each ion behaves like an electron,its magnetic moment can only be measured to be aligned with the external fieldor directly opposite to it... Thus, each eigenvalue E of the [magnetic] energy of the ions can be expressed interms of N two-valued quantum numbers i = 1
E = B
i =1
i
is a constant(it would be equal to Bohr's magneton for electrons). As the partition function introduced above involves exponentials of such sums, whichare products of elementary factors, the entire sum boils down to:
Z() = exp ( E) = [ exp ( B)+ exp ( B) ] N