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The Planck Mass and the Chandrasekhar Limit

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The Planck Mass and the Chandrasekhar Limit The Planck mass and the Chandrasekhar limit ͒ David Garfinklea Department of Physics, Oakland University, Rochester, Michigan 48309 ͑Received 6 November 2008; accepted 10 March 2009͒ The Planck mass is often assumed to play a role only at the extremely high energy scales where quantum gravity becomes important. However, this mass plays a role in any physical system that involves gravity, quantum mechanics, and relativity. We examine the role of the Planck mass in determining the maximum mass of white dwarf stars. © 2009 American Association of Physics Teachers. ͓DOI: 10.1119/1.3110884͔ I. INTRODUCTION ery part of the gas is in equilibrium and thus must have zero net force on it. In particular, consider a small disk of the gas The Planck mass, length, and time can be constructed with base area A and thickness dr at a distance r from the from the Newton gravitational constant G, the Planck con- center of the ball. Denote the mass of the disk by m , and the ប 1 d stant , and the speed of light c. In particular, the Planck mass of all the gas at radii smaller than r by M͑r͒. Newton mass, MP is given by showed that the disk would be gravitationally attracted by បc the gas at radii smaller than r, but unaffected by the gas at M = ͱ . ͑1͒ radii larger than r, and the force on the disk would be the P G same as if all the mass at radii smaller than r were concen- The Planck mass is very small on the human scale because trated at the center. In other words, the inward force due to Ϸ −8 MP 10 kg. It is quite large on the scale of elementary gravity on the disk has a magnitude / Ϸ 19 particles because MP mH 10 , where mH is the mass of a GM͑r͒m d ͑ ͒ hydrogen atom. This large ratio is often referred to as the Fin = 2 . 2 “hierarchy problem;” the “problem” is that such a large ratio r seems unnatural and thus requires an explanation. Because ␳ The volume of the disk is Adr, and so we have md = Adr, present day particle accelerators probe energies correspond- where ␳ is the density of the gas at radius r. Thus using Eq. 2 ing to a few hundred mHc , present day accelerators are very ͑2͒ we find 2 far from the much higher scale of MPc , the energy scale at GM͑r͒␳Adr which quantum gravity becomes important. F = . ͑3͒ One might expect that the Planck mass occurs only in very in r2 exotic, highly speculative, and completely theoretical areas of physics. However, because the Planck mass is constructed If gravity is pulling the disk inward, then why does it not fall G ប c and the whole gas ball collapse? Because pressure differ- from only , , and , it occurs in cases where gravity, ͑ ͒ quantum mechanics, and relativity all play an essential role. ences push the disk outward. Let P r be the pressure of the One such problem is the Chandrasekhar limit, the maximum gas as a function of radius. Then the outward force exerted mass of a white dwarf star.2 All stars are held in equilibrium on the inner surface of the disk is P͑r͒A. This force is almost by the opposing forces of self-gravity and pressure. Thus but not exactly balanced by the inward force of P͑r+dr͒A. gravity plays a role in the properties of stars. In white dwarf The net outward force on the disk due to pressure is stars the pressure is supplied by the motion of electrons due dP ͑ ͒ ͑ ͒ ͑ ͒ to the uncertainty principle and the Pauli exclusion principle. Fout = P r A − P r + dr A =−A dr. 4 Thus quantum mechanics plays a role. In the most extreme dr white dwarf stars, those nearing the maximum possible mass, Because the inward force and outward force must cancel and the electrons are moving so fast that their speeds are compa- thus must have equal magnitudes, it follows that Fin=Fout, rable to the speed of light, and thus relativity plays an essen- and therefore tial role. In this paper we will give a simple derivation of the Chan- dP GM͑r͒␳ =− . ͑5͒ drasekhar limit and show its connection with the Planck dr r2 mass. We define N ϵM /m and call this large dimension- h P H ͑ ͒ less number the hierarchy number. Nature also presents us As we increase r to r+dr, M r increases because the mass with another large dimensionless number: the maximum contained within that radius is augmented by a thin shell of number of nucleons ͑protons and neutrons͒ in a white dwarf radius r and thickness dr. Such a shell has volume 4␲r2dr, star, which we will call the Chandrasekhar number and de- and because the density at that radius is ␳, it follows that Ϸ 3 M͑r+dr͒−M͑r͒=4␲␳r2dr,or note as NC. We can show that NC 0.774Nh. We will show how this result comes about and specifically the roles of G, dM͑r͒ ប, and c in the calculation. =4␲r2␳. ͑6͒ dr II. GRAVITATION Suppose that the matter that makes up the star is so simple that the pressure is a function only of the density ͑and not the We start with some 19th century physics—the equilibrium temperature͒. If we know the equation of state P͑␳͒,wecan configurations for spherical balls of self-gravitating gas.3 Ev- rewrite Eq. ͑5͒ as 683 Am. J. Phys. 77 ͑8͒, August 2009 http://aapt.org/ajp © 2009 American Association of Physics Teachers 683 dP d␳ GM͑r͒␳ N ␳ ͑ ͒ ͑ ͒ =− 2 . 7 = . 12 d␳ dr r V 2mH We then multiply Eq. ͑7͒ by r2 /͑G␳͒, differentiate with re- Hence, we can write Eq. ͑11͒ as ͑ ͒ 2Uץ spect to r, and use Eq. 6 to obtain dP V = . ͑13͒ V2ץ d r2 dP d␳ d␳ ␳ 1 ͩ ͪ + ␳ =0. ͑8͒ 4␲Gr2 dr ␳ d␳ dr ͑ ͒ ϰ −1/3 From Eq. 9 it follows that pmax V and therefore that ץ Equation ͑8͒ allows us to make a mathematical model of the pmax 1 pmax ␳ =− . ͑14͒ V 3 Vץ star. Choose a value of the central density c and integrate Eq. ͑8͒ outward until the pressure vanishes at the surface of the star. Complete the model by assuming vacuum at all radii We apply Eq. ͑13͒ to Eq. ͑10͒ and use Eq. ͑14͒ and straight- ␳ larger than this stellar surface. Each value of c yields one forward algebra and obtain stellar model. dP 1 d⑀ = p4ͯ ͯ . ͑15͒ d␳ 9␲2ប3␳ dp p=pmax III. QUANTUM MECHANICS IV. SPECIAL RELATIVITY Matter in a white dwarf consists of electrons and nuclei, ͑ ͒ ͑ ͒ / ␳ mostly of carbon and oxygen. Because the matter is electri- Equation 15 with pmax given by Eq. 9 gives dP d ,if cally neutral, there is one proton for every electron. Also we have an expression for ⑀͑p͒. In nonrelativistic mechanics ⑀ 2 /͑ ͒ because the number of neutrons equals the number of pro- =p 2me , where me denotes the mass of the electron. In tons for both carbon and oxygen nuclei, there is one neutron relativistic mechanics for each electron. Usually the pressure of a gas depends on ⑀ ͱ 2 2 2 4 ͑ ͒ its temperature. However, the pressure at low temperatures is = p c + mec , 16 a purely quantum mechanical effect that is independent of and thus temperature. Recall that the uncertainty principle implies that 2 particles confined to a box must have a momentum. Because d⑀ c p = . ͑17͒ faster moving particles have larger energy, it takes work to dp ͱp2c2 + m2c4 make the box smaller. This work is due to the pressure. e Consider N electrons along with N protons and N neutrons We can gain insight by expressing Eq. ͑15͒ in terms of combined in carbon and oxygen nuclei confined to a cubical dimensionless quantities that have factored out the relevant box of side L. The electrons are free particles except for the physical scales. We define the dimensionless quantity X as confining walls of the box. Because of the Pauli exclusion p principle, each state can have no more than two electrons. X = max . ͑18͒ Thus for N electrons the ground state energy U is obtained mec / by putting the N electrons in the N 2 states of lowest energy. It follows from Eq. ͑9͒ that As shown in standard textbooks on statistical mechanics, the ␳ maximum momentum of an electron in the ground state pmax X3 = , ͑19͒ ͑which is the Fermi momentum͒ is given by ˜␳ ប ␳ ͑ ␲2 ͒1/3 ͑ ͒ where the constant ˜ is given by pmax = 3 N , 9 L 2 m ˜␳ = H , ͑20͒ ␲2 ␭3 and the ground state energy is given by 3 e 3 p ␭ 1 L max and e is the Compton wavelength of the electron given by U = ͩ ͪ ͵ ⑀͑p͒p2dp, ͑10͒ ␲2 ប ប 0 ␭ = . ͑21͒ e m c where ⑀͑p͒ is the energy of a single particle. e What we want to calculate is not U itself, but dP/d␳. It is straightforward to see the physical meaning of the scale -V, it follows that ˜␳: By the uncertainty principle, we expect electrons to acץ/Uץ−=Because P quire relativistic speeds at number densities on the order of 2Uץ dP =− . ͑11͒ one electron per cubic Compton wavelength. Because there -V are two nucleons per electron, this number density of elecץ␳ץ d␳ trons occurs at a mass density ˜␳ that is of the order of m per ͑ 3 ͒ H Here V=L is the volume of the box.
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