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Study Guide #3 - Thermal Physics

by: Hazel Medina

Study Guide #3 - Thermal Physics Physics 60

Marketplace > University of California - Irvine > Physics 2 > Physics 60 > Study Guide 3 Thermal Physics
Hazel Medina
GPA 3.0
Thermal Physics
Feng, J.

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Hi, everyone! Here's the last set of study guide notes. If you feel like you haven't quite yet absorbed earlier material, please download the Midterm #1 and Midterm #2 study guides, which cover the...
Thermal Physics
Feng, J.
Study Guide
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This 6 page Study Guide was uploaded by Hazel Medina on Friday December 4, 2015. The Study Guide belongs to Physics 60 at University of California - Irvine taught by Feng, J. in Fall 2015. Since its upload, it has received 19 views. For similar materials see Thermal Physics in Physics 2 at University of California - Irvine.


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Date Created: 12/04/15
Physics 60 Exam date December 7 2015 Study Guide 3 Ch 7 Quantum Statistic Ch 71 The Gibbs Factor 0 Consider a system that can exchange particles with its environment NOW 39PS139P52 QR52QR51 e5RSZSRs1k I Using the thermodynamic identity dSR 1TdUR PdVR udNR we find SR52 SR51 391TE52 Esl uNs2 pNsl o This means that p52psl elE52uN52kTeEsluNslkT This exponential is the Gibbs factor Gibbs factor e39ES39 NS kT 39 135 1 Ze39ES39 NSW W Z is the grand partition function Z ZS e39lElsl39WsnHn If there are two types of particles Gibbs factor e39lElsl39MNAlsl BNBisllkT Ch 72 Bosons and Fermions 0 Quantum statistics study of dense systems in which two or more identical particles have reasonable chance of wanting to occupy the same singleparticle state Remember 2 1NZlN 0 Some particles can occupy the same space and others can t Bosons particles that can share a state with another similar particle I Has integer number of spins O 1 2 Fermions particles that cannot share a state with another particle of the same type I Pauli exclusion principle rule that two identical fermions cannot occupy the same state I Has halfinteger spins 12 32 0 When the number of available singleparticle states is much greater than the number of particles then 21 gtgt N For an ideal gas the singleparticle partition function is 21 VZintVQ I vQ hV 27tka3 0 Because 21 gtgt N and Z ZlNN then VN gtgt va 3 The Distribution Functions 0 When a system violates 21 gtgt N then we can use the Gibbs factor Consider a singleparticle state of a system whose energy is occupied by e and consists of n particles I This means that 39Pn 1Ze39quot 39 kT 1Ze39quot 39 kT o For fermions Z 1 e39 39 kT Occupancy 77 average number of particles in the state 17 Zn n39Pn O39PO 1421 e39 39 kT1 aleW For fermions 1e 39 kT 1 this equals the FermiDirac distribution mm which goes to zero when 6 gtgt 1 and goes to one when 6 ltlt 1 meaning that states with energy much less than u tend to be occupied o For bosons Z 11 e39 39 kT For the boson if x E e ukT then 77 Zn ne39 x Z 1ZZnooxe39 x 1ZGZ6X For bosons 1e BE 0 Using Boltzmann statistics we find 39Ps 1Zle39 kT This means that BoItzmann N39Ps NZle39 kT ellHe39d e39 39 kT 39E39WkT 1 this is equal to the BoseEinstein distribution Ch 73 Degenerate Fermi Gases 0 Consider a gas of fermions at a very low level Think of conduction electrons inside a chunk of metal in a temperature low enough for Boltzmann statistics to apple to an ideal gas so that VN ltlt vQ I At room temperature vQ hV 27tka3 43 nm3 but in a typical metal there is about one conduction electron per atom which means the volume per conduction electron is roughly that of an atom 02 nm3 st Zero Temperature 0 At T 0 the FermiDirac distribution becomes a step function Singleparticle states with energy 1 is occupied and states with energy greater than u are unoccupied I This means that for this situation 1 is the Fermi energy eF uTO I A gas is degenerate if it is a gas of fermions that is so cold that nearly all states below 6 are occupied while nearly all states above 6 are unoccupied o The value of 6 is determined by total number of present electrons Assuming the electrons are free particles confined in a box of volume V L3 I Remember wavelength and momentum An 2Ln and pn MN hn2L o For the 3D box this applies in the x y and 2 directions This means that e 1322mh28mL2nX2 ny2 n22 0 Consider a large sphere with center at the octant since n values are positive then only mind the electrons in the first octant This means that the Fermi energy 1 eF is the energy of the particles on boundary of the sphere which means 6 hznmaX28mL2 339 Small Nonzero Temperatures 0 We cannot calculate heat capacity using the approximation T 0 because the particles acquire energy of about kT But in degenerate electron gas most electrons cannot acquire this energy because they may jump into already occupied space going against the FermiDirac distribution I Only electrons that can acquire some thermal energy are those within kT of the Fermi energy which can jump into unoccupied states above 6 This means that additional energy oc number of affected electrons x energy acquired by each and so additional energy oc NkT x kT or additional energy oc NkT2 I The total energy of a degenerate Fermi gas where T ltlt 6 is U 35NeF n24NkT2eF 0 Therefore heat capacity is CV oUoTV thNsz26F 339 The Density of States 0 Remember that 6 h2n28mL2 and n 8mL2h212e12 This means that dn 8mLZh21212e12de I From this we can find that the energy integral for a Fermi gas at zero temperature is U If elTt28mL2h23 2Ede We can get the density of states ge the number of singleparticle states per unit energy 36 n8m3 22h3V 3N2EF32E I 3N2eF32E implies that ge depends on N but note that the N dependence is canceled by 6 o For an electron gas at zero temperature N fOEFgede fT is a nonzero then N fowge FDede f0 ge1eeukf 1 Total energy is U If 6516 FDede L ege1e 39 1 339 The Sommerfeld Expansion 0 Sommerfeld expansion method used to evaluate integrals involved with finding chemical potential and total energy of a free electron gas in the limit kT ltlt 6 Start with N fooo ge FDede gofo00 elZ FDede 00 32 I Integrating by parts we get N 23goe32 FDe o 23g0f0 e d FDdede d FDde ddee 39 kT 1 1 1kTeXex 12 X e ukT o N 23gof 1eXex 12e32dx kT I Using the Taylor series about 6 u we expand e32 and keep the first few terms and get e3 2 H 326 uu1238e um o The integral becomes N 23gof ooeXex 12l32 32kau12 38XkT2p3912 dx Using u z eF we find ueF 1 HZ12kTEF2 All of this proves that U 35NeF HZ4NkT2EF Ch 74 Blackbody Radiation 0 Now consider the electromagnetic radiation inside a box at a given temperature 339 The Ultraviolet Catastrophe 0 Treat electromagnetic radiation as a continuous quotfieldquot permeating all space In a box this field is like a combination of various standing wave patterns each standing wave is a harmonic oscillator with frequency f cA and average thermal energy 212kT Ultraviolet catastrophe the electromagnetic field should be infinite because the total number of oscillators is infinite but this cannot be observed experimentally 339 The Planck Distribution o In quantum mechanics a harmonic oscillator is allowed energy of En 0 hf 2hf The partition function of a single oscillator is Z 1 e39Bhf e39ZBhf 11 e39Bhf Average energy is E 1ZoZoB hfehfkT 1 The average number of units of energy in the oscillator is found through Planck s distribution 7 1ehfkT 1 I Note that it requires oscillator energies are quantized 0 v Photons o Photons particles representing units of energy in the electromagnetic field They are bosons and the pattern of their field is given by the BoseEinstein distribution 1735 1e e39WkT 1 where e hf I This requires that u 0 for the photons o This means that oFoNTV 0 at equilibrium 0 When a photon y is created or destroyed by an electron e lt gt e y then He 6 uv at equilibrium This means the chemical potential for photons is zero so the BoseEinstein distribution reduces to the Planck distribution 339 Summing over Modes 0 Planck constant tells how many photons are in any single quotmodequot quotsingleparticle state of an electromagnetic field 0 To calculate the total number of photons inside a box and the total energy of those photons start with a box of length L The wavelength is A 2Ln and the momentum is p hn2L I Because photons are ultrarelativistic particles their energies are given by e pc hcn2L instead of e p22m 0 This means that U 2an Zny ane pue 2nxnynzhCnL1ehcn2LkT 1 With the upper limit n being infinity this integration is U fooo dn fen2 d6 fen2 dwnzsin6hcnL1eh quot2LkT 1 The Planck Spectrum 0 With L3 V and e hcn2L the integral over n can become UV f0w8ne3hc3e kT 1de If x ekT then this become UV 87tkT4hc3f0 x3eX 1dx which is still proportional to x 282 or e 282kT I Wien s law higher temperatures tend to give higher photon energies The integrand is the spectrum or energy density per unit photon energy ule 8nhcl3lleseekT 1 v Total Energy 0 Total energy density is UV 8n5kT415hc3 339 Entropy of a Photon Gas 0 For a box of thermal photons with volume V specific heat CV oUoTV 4aT3 a 8n5k4V15hc3 o The absolute entropy is ST L CVT T dT 4af0TT 2dT 43aT3 32n545VkThc3k N 8nVkThc3f0 xZex 1dx v The Cosmic Background Radiation 0 Radiation that fills the entire observable universe has an almost perfect thermal spectrum at a temperature of 273K Photons making up the cosmic background radiation have low energies with the spectrum peaking at e 382kT 661O394 eV which corresponds to wavelengths of about a millimeter in the far infrared 339 Photons Escaping through a Hole 0 Consider a photon gas in a box with a hole that lets some photons out All photon travel at the same speed in a vacuum regardless of wavelength Photons escape in time interval dt from shell with thickness cdt of radius R I This means volume of chunk of the shell RdG x Rsinedw x cdt O 90 o The energy density of this chunk is UV 8n515kt4hc3 This means that energy in the chunk UVcdtR2sin6de Probability of escape AcosG4TtR2 Energy escaping from chunk AcosG4TtUVcdtsin6d6dD Integrating the energy escaping from the chunk with 6 from 0 to TtZ and D from 0 to 27 you get total energy escaping A4UVcdt I Power per unit area c4UV 2n515kt4h3c2 0T4 0 StefanBoltzmann constant 0 2n5k415h3c2 5671O398 WmZK4 339 Radiation from Other Objects o Blackbody radiation photons emitted by any nonreflecting quotblackquot surface at temperature T Consider a blackbody a box of photon gas from earlier examples I Total power emitted by the blackbody should be the same as that emitted by the hole or else it would violate the second law of thermodynamic Emissivity e the fraction of photons absorbed at some given wavelength and therefore the fraction emitted I Power oeAT4 339 The Sun and the Earth 0 Solar constant amount of solar radiation received by the Earth 1370 Wm2 o Luminosity total energy output Luminosity of the sun 391026 W Using T luminosityoA14 we find the surface temperature to be 5800K I Spectrum of sunlight peaks at photon energy of e 282T 141 eV solar constant HR2 4TER20T4 Ch 75 Debye Theory of Solids o Einstein model each atom is treated as independent threedimensional harmonic oscillators Heat capacity of the Einstein model CV 3NkekT2e kTe kT 12 I N is the number of atoms and e hf is the universal size of the units of energy for the identical oscillators I When kT gtgt 6 heat capacity approaches a constant value of 3Nk which agrees with the equipartition theorem 0 When kT z e the heat capacity approaches zero as temperature goes to zero 0 However Einstein model doesn t work for crystals where atoms oscillating influence each other Mechanical oscillations aka sound waves behave like light waves I Three key differences between sound waves and light waves are 1 Sound waves travel slower than light and sound wave speed cS depends on stiffness and density of the material through which it passes 2 Light waves must be transversely polarized but sound waves can also be longitudinal which means is has three polarizations 3 Light waves can arbitrarily short wavelengths but sounds waves in solids must have wavelengths longer than twice the atomic spacing Equally spaced energy level of each mode of oscillation e hf thA thnZL where L is the magnitude of the vector in nspace and when this mode is in equilibrium at temperature T the number of energy units is p 1eekT 1 these energy units are phonons 39 nmax 6Nn13 0 With Debye s approximation and knowing U 32716213an E pe we find u an2f0quotm quotth2Ln3ehcquot2L 1dn With x thnZLkT then the upper limit is xmax hcsnmaXZLkT hCSZkT6NTV13 E TDT I TD is the Debye temperature I This becomes U 9NkT4TD3fOTDT x3eX 1dx which becomes U 3NkT when TID gtgt T o When TltltTD u 3n45NkT4TD3 which means cV 12n45TTD3Nk when TltltTD You can get heat capacity with CV 9NkTTD3fOTDT x4eXeX 12dx Ch 76 BoseEinstein Condensation 0 Now we will find the distribution of quotordinaryquot bosons such as atoms with integer spin known as the BoseEinstein condensation Atoms confined to a box of volume V have ground state of 60 3h28mL2 I The BoseEinstein distribution is No 1e 39 kT 1 o This yields N0 kTeo u when N0 gtgt 1 N 2a S 1elelslukT 1 Using the function ge ZxE2nmh232VE we find the integral of N to be N fooo ge1e 39 kT 1de 0 We find N 26122nkach232V or H 0527h22nmNV23 Nexcited TTc32N TltTc and the rest of the atoms must be in ground state which means No N Nexcled 1 TTc N when T lt T I Tc is the BoseEinstein condensation and this transition temperature Tc is called the condensation temperature the groundstate atoms are called the condensate Ch 8 System of Interacting Particles Ch 82 The Ising Model of a Ferromagnet o Ferromagnet a material with neighboring dipoles aligning parallel to each together even in the absence of an external field If they align antiparallel they are antiferromagnet At the Curie temperature the net magnetization becomes zero when there s no external feed Domains parts of metal that are magnetized o Ising model simplified model of a magnet consisting of a single domain within a ferromagnet neglecting longrange magnetic interactions between dipoles and can only point wither parallel or antiparallel The total energy of the system is U Zneighbormg palTS U Sis Exact Solution in One Dimension 0 We can find 2NcoshBeN391 z 2coshBeN With this we find 7 4553an NetanhBe 339 The Mean Field Approximation o Es XMLgm Sneighbor en E is the average alignment of the neighbors A system becomes magnetized below Tc at ch ne


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