Lecture 8, the outline

Transcription

1 Lecture, the outline loose end: Debye theory of solids more remarks on the first order hase transition. Bose Einstein condensation as a first order hase transition 4He as Bose Einstein liquid Lecturer: Anna Liniacka Date: //

2 Energy, entroy, F and ressure of hoton gas We have calculated the energy density of the hoton gas in equilibrium 4 4 h c U h c 4 What is the entroy? d du du U d d 4 d h c In rocesses at a constant entroy which hoton gas undergoes we must have thus τ =const Helmholtz free energy F=U στ = 4 U 4 h c ressure of the hoton gas = F 4 U 4 h c heat caacity = C U Heat caacity of a boson gas ~ τ remember ~ τ for degenerate fermion gas and constant for ideal classical gas Lecturer: Anna Liniacka Date: // 4 h c Note here the ressure U/ relation which is different from what we were getting for non relativistic articles. For non relativistic non interacting articles we were getting =/ U/. This relation deends on the energy momentum relation for the gas article. It is different for relativistic articles hotons. note also that G=F+= U/+U/ = in agreement with what we think the chemical otential should be.

3 Phonons in solids, Debye theory The energy of elastic wave in a solid should be quantized in the same way as an electromagnetic wave in the cavity is quantized. The wave is essentially lattice deformation, and the quantum of the energy of such a wave is called a honon. thermal average number of honons of given frequency : <N> e h As elastic wave (differently then EM wave ) can be both transversal and longitudinal there are three ossible olarizations. And there is a limit on the maximal number of all the modes = N, where N is the number of articles in the system. As before : n nv L, n n n x y n z we will now n_max =n_d such that L." modes where "v" is the velocity of sound in the solid. we will assume here that it is indeendent on frequency and olarization, useful but not totally true aroximation = Lecturer: Anna Liniacka Date: // 4 n max n D n dn N n D energy of all honon modes: U N 6N n D n D n dn h n e h! n" # $ Phonons can be treated like hotons. There are three differences honons can have not olarization states, and there is a maximal number of honon modes= N, number of articles in the system ( when each article roduces modes oscillation with ossible olarizations ), and the velocity of honons (sound) is different than that of lights. We call the maximal occuied honon mode n_d ( n Debye)

4 - calculate U, 4 Energy and heat caacity of the solid n D U n dn h n e h! n" # $ x D dx 4 h v x e $ x introducing x h if temerature is small comared to Debye teme rature θ we can set the uer limit of the integral to the infinity, and : x D n D hv U 4 4 =N h v k B U C 4 N n % nhv x D k B ' ( ) % L % L 6 N & hv Lecturer: Anna Liniacka Date: // ' * L % 6 N d x n D 6N + hv x k B e x. / We would like now to calculate energy of the honon gas in analogy what we have calculated for the hoton gas. We would like also to calculate the heat caacity, we hoe it will come similar to the one of the hoton gas, thus deending on the temerature to the third ower. ur roblem is that now we have to do the integral to the finite n_d instead of infinity. We are not able to do it, so we will just secify here the conditions in which we can aroximate the integral with the one u to infinity. This is ossible if n_d is big, thus the maximal frequency which can be occuied is large comared to temerature, which means that temerature is small comared what we call Debye temerature. The Debye temerature is roortional to concentration of articles to / ower. We get now the same result for secific heat as for the hoton gas, valid for small temeratures in case of honons. k B4 4

5 For non metallic solids Heat caacity of solids C N 4 k B4 at low temerature the heat caacity is roortional to the third ower of temerature. This was verified in the temerature range u to K with great accuracy. For metallic substances, the degenerate electron gas contributes to the heat caacity C tot C el6 C vib 7 4 where A= N N k B4 F 6 A 9 C tot 7 N 6 F A Both effects are confirmed exerimentally with high accuracy Lecturer: Anna Liniacka Date: //

6 ?? Phase equilibrium : hases ( ) ( Phase transitions ( :,( :,( Deciding on which hase is N, and what is the hase ( ) comosition If we consider hase_+hase_ at constant ressure > minimum of G =Nµ. if we consider hase_+hase_ at constant volume > minimum of F. First order hase transitions: entroy er article ("s") is non continuos=> latent heat and volume er article is tyically not continuos : ( < v < v = : ( : > < =? G? C s < s = : >,NA D F ( Lecturer: Anna Liniacka Date: // < F ( = G ( < G ( Entroy and volume discontinuities result in heat caacity at constant ressure being infinite at hase transition and comresability being infinite? E? C at hase transition,na NA B

7 G R F F T.4 G L N liquid c.44 N.9 c gas.96.9 F G,H S G H, g l F G g l F H F H I J K liquid liquid+gas for < critical reassure First order hase transitions L M L c latent heat g,h I K gas σ or heat l, H G,N Lecturer: Anna Liniacka Date: // Q,N liquid F G g U I liquid + gas for T < critical temerature T NP F N = T G gas,n l U J v gi v l Q R Here examle G comes from an der Waals equation of state. Isotherm (for temerature lower than critical temerature) and isobar for ressure below the critical ressure are for a "tyical" first order hase transition.> water > water vaor.

8 Chemical otential of the boson gas Consider boson gas at low temerature. If we have the system with N bosons in total we must have : N f W BE s s N ex Q P Q ex W sq P X Y Q Q if we choose the energy scale in such a way that the lowest energy level has energy =, then the occuancy of the th level must be : absolute activity must be < so that N_ >, thus the chemical otential must be <. For τ >, the factors ex (ε/τ) will be very big, thus we can assume, that all states with energy> have much less occuancy than N_. Assume: N N Z P Q x lets find x. It has to be retty small at low temerature, so the system can hold a large number of articles. N lim 9 ex Q P Q ex x Q [ Lecturer: Anna Liniacka Date: // x x N Thus : 9 P Q N Here we are trying to find what is the chemical otential of the boson gas. We will consider here the occuancy of the lowest energy state, we gauge it such so the lowest energy =. We see first of all that the chemical otential has to be negative At small temeratures the factor (ε+ µ )/τ which governs the occuancy of the orbitals will be very big. Thus we might exect that the occuancy of all orbitals will be small excet of the one with zero energy, which will have to hold most of the articles of the system. From the condition that zero orbital holds all the articles of the system we are trying to find now what the chemical otential must be.

9 d _ N \ s Number of bosons in the ground and excited states Lets calculate now the number of bosons in the excited states. The total number of bosons is: s f ] BE \ s ] ^ ex N ` \ sa (.. )e f.. dn g h f D g (.. ) dg ] ^ ex Lecturer: Anna Liniacka Date: // b c ` ` \sa ] ^ ex we used to aroximate the sum over states with the integral over the energy and density of states D i j dn di j i k m h k 4l However, this aroximation is only good if there is no state which holds much more articles then any other, which is not true for the ground state, with energy=, which can have all the articles at low temerature. We will use the aroximation for all excited states: N b c ` ` m D ] d] ex ] ^ _ N ex o N /.6 n q N /.6 n q n we will erform this integral assuming µ=, we get in this case: N ex / m " h For concentrations bigger than.6 n_q there is a substantial fraction of articles in ground orbital Einstein condensations. As n_q goes down with τ this henomenon m r s occurs for boson "gases" at low temerature n Q 4 n.6 " /.6 n q q h The energy density of states is the same we calculated for fermions, but is is factor of two smaller so sin degrees of freedom. We calculate searately number of articles in ground and excited states, as integral aroximation cannot be used for the ground state. For excited states we use the aroximation of zero chemical otential, which is valid at relatively small temeratures.

10 l Calculations, reminder of energy density of states We calculated the energy density of states for a free article obeying Schroedinger equation, residing in a box of volume =L. π h ε ( ) n, n, n = n x + ny + n x y z z =t ml su v h n s N i j m L 4l nx N i j 6 nx j z D N exc m ] ] ^ d] ex = 4 m _h m i m h { m k 6l D ] ] d] ex _ x ex x d x _ m y D i j N b c 4 h dn di ` ` m m m j i k m h D ] d] ex ] ^ Lecturer: Anna Liniacka Date: // h ] ex ] } _ w k 4l N ` N exc d] = 4 ~.6.6 n q

11 / N ex o N /.6 n q N ex / N.6 n q n / Bose Einstein gas equation of state..6 n q N / n.7 m N h n " For τ < τ_c N_ex< N and articles start to move to the ground state. Lets discuss how it haens. Consider Bose gas at constant temerature and to which we are adding articles. For small concentrations Bose gas behaves like the ideal gas and N_ex=N. Chemical otential is negative and aroaches with increasing N. If it is close to, excited states cannot hold any more articles=> there is a maximal number number of articles excited states can hold namely: N ex /.6 m h " We exress conditions in which N_ is coma rable with N in terms of critical temerature " Lecturer: Anna Liniacka Date: // N N_ex. h n " normal gas characteristic of hase transition o m "isotherm" condensate+gas " N c. m h Ν % ln n n Q, m r s q h We have calculated here the maximum number of articles the excited states can hold, for a given temerature and volume. Why is it maximum? Because we assumed the chemical otential maximal available for the boson gas, equal to zero. What would haen if we add to the boson gas more articles than excited states can hold? They have to fall to the zero energy state. The lot of Nex as a function of N for a constant temerature and volume has a characteristics of first order hase transition lot.. Imagine you do the same for the water vaor (gas) and lot N_gas vs N. " n Q

12 ˆ ˆ z s Energy, entroy, ressure We can treat all the articles in the ground state as a condensed hase of the Bose Einstein gas. They have different roerties from all articles in excited states, for examle zero energy and zero entroy. Lets calculate roerties of such a gas for equilibrium conditions of condensed and normal hase, namely µ. The energy will be that of all articles in the excited states: U m ] ƒ D ] ] d] ex _ U d ˆ _ du ƒ m h du ƒ ƒ d d ƒ Helmholtz free energy F=U στ = ˆ F ˆ U ˆ ƒ m h 4 m ƒ ƒ Š C ressure does not deend the volume! x m h m d x ex x _ Š Cdƒ m Lecturer: Anna Liniacka Date: // h Š C =/ σ Τ/ D h ˆ = / U ƒ s dn d 4 C m h r m h Š C σ = / U/T r 4q The condensed hase will have zero volume, zero entroy, and chemical otential close to zero. Lets calculate roerties of normal hase in equilibrium with condensed hase, thus for zero chemical otential. We calculate here the energy of "normal hase", the entroy, free energy and ressure, and observe that the ressure does not deend on volume. As the "normal hase" is in equilibrium with condensed hase this will be as well the ressure of condensed hase. We can calculate as well G=F+ = /U+/U =, in agreement with what we assumed about the chemical otential.

13 Isotherms and hase diagrams ˆ ƒ m h Bose condensate+gas τ=const Š C =Nτ/ Bose "gas" For constant and N,hase coexistence critical N condensate ƒ Bose "gas" c. m h τ As ressure does not deend on volume for condensate gas equilibrium we can comress the system to volume, corresonding to all articles in the ground state. This corresonds to infinite τ_c and thus Lecturer: Anna Liniacka Date: // N ex N c for c Œ coexistence curve has to fulfill Clausius C. equation: s gq s c v gq v c R U We suly /U in heat and and u with U. =/ U is the work done in exansion by the condensate gas c. L = / U m h N L Ž σ = / U/T v " Here we draw isotherm for condensed "normal gas" hase. Remember that here U and σ is the energy and the entroy of the normal gas hase (cause condensed hase has zero energy and entroy) and that the ressure of the normal hase in equilibrium with condensed hase reresents "hase equilibrium ressure" and, if we calculate the derivative of it over temerature we will can use Clausius Claeyron equation to calculate the latent heat on moving the system from condensed to normal hase. This latent heat is L=/U according to our calculations, where U is the energy of the normal hase. Thus /U must be the work done by the condensed hase while decomressing from volume to volume. This work should be = / U/*=/U thus we get a consistent result here. Condensate normal hase coexistence ends at critical temerature, above which nearly all articles are in the excited states normal hase.

14 ˆ Heat caacity of condensate "gas" in equilibrium What is the heat caacity of condensate gas in equilibrium? C / C C_ U Heat / / critical τ C ƒ C ˆ τ C U N Lecturer: Anna Liniacka Date: // U / = _ du=τdσ d m h C and U= / However, at constant, can change from to anything like Nτ/ without any change of temerature. So C / / ƒ m h Š C U _ m h C We rove here that the heat caacity at constant ressure is infinite at hase transition.

15 Helium 4He as Bose gas We can calculate critical (or Einstein) temerature for 4He, which is a bosonic atom with m=4u c. m h N " We take N= Avogadro number and = molar volume at atmosheric ressure= 7.6 cm^ and get T_c =. K C_ "critical" Τ. Κ P condensate. Τ Aarently, there is an interesting behavior of the heat caacity of Helium close to what should be the condensation temerature resembling of what we would exect from forming the condensate We should not forget that Helium becomes liquid at T=4 K at atm so the condensed hase is in equilibrium with liquid, not only gas. However, liquid helium has much larger entroy and volume er article than exected for the liquid with aroriate tye of inter atomic interactions we are often justified to use gas aroximation Lecturer: Anna Liniacka Date: //

16 ž Suerfluidity Liquid 4He in condensate hase shows roerty of suer fluidity > very little friction to flow. The onset of suer fluidity occurs at the same temerature at which jum in secific heat occurs. iscosity is due to energy transfer between the moving body and the liquid, or between various layers of the liquid. To transfer energy to the Bose Einstein condensate one has to move articles from the zero energy state. iscosity can be due to the excitations from the ground state by interactions with the moving body. Consider a body moving in the condensate with velocity. We can consider the excitations of the condensate as "excitons", quasi articles like honons. The body will interact with them. We will show that a certain minimal velocity is needed for a article to emit "exciton". M M i œ h k M f α h,h k energy momentum conservation i k M š M i h k iž h k i œ M i hk i cosÿ œ h ž min œ Lecturer: Anna Liniacka Date: // k M š œ M h k M f M f M h k M h k For honons this is just the seed of sound, thus article moving with a lower seed will feel no resistance. We show here that to interact with the condensate the moving article must have some minimal velocity. Particles with lower velocity are not able to emit "excitons" > excitations of the condesate to the higher state. Excitons are a model of transferring energy into a gas of non interacting articles in a zero energy state. In reality articles of the gas are interacting to some extent, and one would think that the direct interaction with the moving body due to inter atomic forces should be enough to move articles "one by one" from the ground state and induce some kind of viscosity, and soil suer fluidity. However, the same attractive inter atomic forces hel to stabilize to condensate and kee the article in the lower energy state once it is there. Thus in the resence of inter atomic forces also the excitations of the condesate should be "global" > in the form of the excitons sound waves for examle

17 R Summary, L First order hase transitions are characterized by dis continuos entroy er article= latent heat, and dis continuos volume er article. This results in infinite comresability at hase transition and infinite heat caacity at constant ressure > flat region in aroriate diagrams G C Q T, N,N,N Bose Einstein condensate = articles in the ground state, can be treated as a hase of the Bose gas. It has zero energy, zero volume, zero entroy, and zero chemical otential. It can coexist with the normal hase for temeratures below Einstein ( or critical) temerature: c. m h N " For temeratures below this substantial number of articles is in the ground state N ex o N /.6 n q n / c " valid for τ<τ_c Pressure of the Bose gas at τ<τ_c does not deend on the volume, and secific heat at constant is infinite, while C_ has a marked tilt at critical temerature. Bose Einstein condensation + interaction between articles lead to suerfluidity=> no resistance to motion against the condensate for low velocities. Lecturer: Anna Liniacka Date: //