Müller I. A history of thermodynamics. The doctrine of energy and entropy (Müller I. A history of thermodynamics. The doctrine of energy and entropy.pdf), страница 67

Therefore, ironically, thefirst white dwarf entered the literature as the dark companion of Sirius, alsocalled Sirius B. The first person to see it – in 1862 – as a dim spot was theastronomer Alvan Graham Clark (1832-1897). And in 1914, Walter SidneyAdams (1876-1956) succeeded to measure the spectrum of Sirius B, andcould thus conclude that its surface temperature is about 10000K, makingit white hot, considerably hotter than the sun. Given that fact, the star had tobe quite small in order to appear dim.

Thus it came to be called a whitedwarf. From the known distance the diameter could be estimated as2.7.107m, or 4% of the solar diameter. And yet, according to Eddington’smass-luminosity relation, cf. Chap. 7, Sirius A was twice as massive as thesun and, in order to be forced into the observed orbit by the companion,the companion had to have about the same mass as the sun. This meant thatthe average density had to be a fantastic 140000 times that of the sun, or200000 times that of water. One cm3 has a mass of 200kg!Any scepticism about such numbers was quickly silenced when – at thesuggestion of Eddington – Adams re-examined his spectroscopic data andfound the relativistic red shift of spectral lines which must be expected forthe intense gravitational field of a massive and compact white dwarf.

Morewhite dwarfs were discovered as time went on – despite their dimness – andsome are considerably denser and hotter even than Sirius B.Therefore, it was clear that no atoms could exist in a white dwarf, onlynuclei and electrons. And the nuclei must be fairly heavy nuclei, becauseastronomers have good reasons to believe that white dwarfs are old stars,which have essentially burned up their light-weight-fuel, the protons. Nowthey consist of many electrons and few nuclei of atoms of intermediatemass, like iron, which cannot serve as fuel for further combustion. If that isso, the large majority of all particles are electrons and the mean relativemolecular mass is µ/µo = 2, cf. Chap.

7.µ cc 2ln a d0kTstrongly degenerate Bose2µ ccln a !! 1kTstrongly degenerate Fermidegeneratenon-degenerate lna<<1exp(p22 µ ckT1) 1pz021 for 0 d 2µ kT (ln a µkTcc )0 else12cµcp21a exp( kT ) exp( 2 µ ckT ) B 1Maxwell distributionNon-relativisticµ cc 2!! 1kT22a exp( µkTcc ) exp( 2 µpckT )exp(112p2µ cc 2) B11 ( µpcc )2 )1p2exp[ kT ( 1 ( µcc 2 ) ) 1)] 1µ cc 21 for2pz0ª ln a º0 d d « µcc 2 » 1¬ kT ¼0 elsepµ ccMaxwell-Jüttner distribution1aµ cc 2kT2a exp( µkTccRelativisticpz0Planck distribution for p = hȞ/ccpexp kT11kTc1cpexp( kT) B11 for 0 d p d0 else1aln aµ cc 2 1kTcpa exp( kT)Ultra-relativisticTable 10.1 Equilibrium distribution function in a gas at rest, i.e. with UA=(c,0,0,0) for a degenerate relativistic gas and limit values for weak and strongdegeneration and for non-relativistic and ultra-relativistic caseWhite Dwarfs295The only remaining source of energy for a white dwarf is gravitationalcontraction, -- Helmholtz fashion.

That keeps the star hot in the centre,perhaps hot enough – a thousand times as hot as the sun – that it must beconsidered a relativistic gas. Note that the small electronic mass helps inc 2this respect, because the relativistic coldness µkTc is more than 103 timessmaller for electrons than for nuclei, or atoms at the same temperature.Now, large speeds make for small de Broglie wave lengths, so that quantumeffects should be small.

However, the large gravitational pressurecompresses the star to such a degree that even the small de Broglie wavelengths interfere and thus produce quantum degeneration. Therefore in awhite dwarf the electron gas can perhaps be both: a relativistic gas and aquantum gas. Chandrasekhar adopted this assumption as the basis for histheory of white dwarfs.

In this way he provided an application for Jüttner’sformulae.Thermal equation of state inside a white dwarfIn relativistic thermodynamics the conservation of mass is replaced by theconservation of the number of particles, and momentum and energy conservationare combined in a vector equation. We haveN A , A 0 and T AB ,B0 , whereAABN is the particle flux vector and T is the energy-momentum tensor.

TheAABequilibrium quantities n, e, and p are related to N and T as shown in thefollowing table.number densityn1c2U ANAenergy density1c2eIn a gas in equilibrium NdistributionAU AU B Tand TABABpressurep13( U AU B g AB )T AB1c2are moments of Jüttner’s equilibriumYF1a§ UApA ·¸¸ B 1kT©¹exp¨¨ so that we haveNAAÔp Fdp1dp 2 dp3poand T ABdp1dp 2 dp3cÔ p A p B F.po29610 Relativistic Thermodynamicsdp1dp 2 dp 3pwith poµ cc 1 p2µ c2 c2is the scalar element of momentumo3space, and 1/Y – or h – determines the cell of the phase space.For a strongly degenerate Fermi gas we thus have, cf. Table 10.1nx4 ʌ ( µ cc ) 3 Y ³ z 2 dzoandp13xc 4 ʌ ( µ cc ) 4 Y ³oz41 z2dz ,where x( kTlna ) 2 1 .

It follows that p depends only on n, not on T ! Anexplicit form of the relation – the thermal equation of state – can be obtained, if theintegrals are evaluated, so that x can be eliminated.If relativistic effects were ignored, the square root in the integrand for p wouldbe absent.Insert. 10.1Subramanyan Chandrasekhar (1910–1995)Chandrasekhar was an astrophysicist with a particular interest in whitedwarfs.

As Eddington did for normal stars, he argued that inside a whitedwarf the atoms are broken down into nuclei and electrons, so that there is alot of space for the particles to move in freely, -- even when the densitiesare as big as described above: If the total mass of the star is big enough,however, the free space between the particles can be squeezed out, as itwere. The electrons are then pushed together and the resulting compactcluster of electrons resists the gravitational pull. That equilibrium canpersist even when the white dwarf cools and becomes a red dwarf andeventually, a black one.

But not all stars can follow that course as we shallnow see.In part of his work Chandrasekhar assumed that the electron gas is astrongly degenerate relativistic Fermi gas.11 In that case it was fairly easy toconsider the limit of the ultimate white dwarf characterized by an infinitemass density at the centre and zero radius.

Surely no other star could bedenser and, presumably, have more mass. That ultimate white dwarf came11S. Chandrasekhar: “The maximum mass of ideal white dwarfs.” Astrophysical Journal 74(1931) p. 81.S. Chandrasekhar: “The highly collapsed configurations of a stellar mass, I and II.”Monthly Notices of the Royal Astronomical Society 91 (1931) p.

456 and 95 (1935) p.207.See also: S. Chandrasekhar: “An Introduction to the Study of Stellar Structure” Universityof Chicago Press (1939). This book is available in a Dover edition, first published in 1957.Subramanyan Chandrasekhar (1910–1995)297out to have a mass of approximately 1.4 solar masses, cf. Insert 10.2.

Thislimiting mass for white dwarfs became known as the Chandrasekhar limit.It was confirmed by observation in the sense that no white dwarf was everseen that has more than Chandrasekhar’s limit mass.The Chandrasekhar limitSince the mean value of the relative molecular mass is 2, by Insert 10.1 the massdensity and the pressure are given byȡAx 3pxB³dz0 1 z2z44ʌ( µ cc ) 3 Y3withA2 µowithB4ʌc ( µ cc ) 4 Y.3andTherefore the momentum balance reads, see Chap.

7dpdr ȡGMrr2wherer4 ʌ ³ ȡ( r c)r c2 dr c .oMrDifferentiation with respect to r and the use of the thermal equation of state, cf.Insert 10.1, provides1 d È 2 d 1 ( ȡ/A) 2/3Ø2ÉrÙr dr ÊdrÚ2È2Ø4ʌGA É1 ( ȡ/A) 2/3 1ÙÉÙBÚ Ê3/2.21/LNon-dimensionalization with the unknown central value ȡc of ȡ provides§·¨¸2/3¨¸1 dd1 ( ȡ/A)Ș2¨¸Ș 2 dȘ ¨ dȘ 1 ( ȡc /A) 2/3 ¸¨¸ĭȘ¹©where Ș(1 ( ȡc /A) 2/ 3 )rL§¨2¨ 1 ( ȡ/A) 2/31¨2/3 ( ȡc /A)1 ( ȡc /A) 2/3¨ 1¨ĭ 2 Ș©·¸¸¸¸¸¹3/2,is the dimensionless radius.We investigate the case that ȡc is infinite.

Presumably that assumptioncharacterizes the ultimate white dwarf in the sense that no other one could bedenser and have more mass. In that case it is easy to solve – numerically – thedifferential equation for the central values ĭ(0) = 1 and ĭ c(0) = 0 and one obtainsthe graph shown in Fig. 10.1. On the surface of the star, at r = R, we must haveȡ = 0, hence ĭ = 0. According to the figure, that value occurs for Ș = 6.9, so that Ris zero, but the mass is not. It can be calculated as follows:29810 Relativistic ThermodynamicsFig.