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# STATISTICAL PHYSICS Physics 115

UCI

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This 10 page Class Notes was uploaded by Carlotta Dare DVM on Saturday September 12, 2015. The Class Notes belongs to Physics 115 at University of California - Irvine taught by Clare Yu in Fall. Since its upload, it has received 45 views. For similar materials see /class/201919/physics-115-university-of-california-irvine in Physics 2 at University of California - Irvine.

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Date Created: 09/12/15

Physics 115A Statistical Physics Prof Clare Yu email cyu uciedu phone 949 824 6216 Of ce 210E RH Spring 2009 LECTURE 1 Introduction So far your physics courses have concentrated on what happens to one object or a few objects given an external potential and perhaps the interactions between objects For example7 Newton7s second law F ma refers to the mass of the object and its acceleration In quantum mechanics7 one starts with Schroedinger7s equation Hz E1 and solves it to nd the wavefunction 1 which describes a particle But if you look around you7 the world has more than a few particles and objects The air you breathe and the coffee you drink has lots and lots of atoms and molecules Now you might think that if we can describe each atom or molecule with what we know from classical mechanics7 quantum mechanics7 and electromagnetism7 we can just scale up and describe 1023 particles That7s like saying that if you can cook dinner for 3 people7 then just scale up the recipe and feed the world The reason we cant just take our solution for the single particle problem and multiply by the number of particles in a liquid or a gas is that the particles interact with one another7 ie7 they apply a force on each other They see the potential produced by other particles This makes things really complicated Suppose we have a system of N interacting particles Using Newton7s equations7 we would write 112 125 dt 7 m 1752 213 Wig51752 Mm 1 2 where E is the force on the 2th particle produced by the jth particle Vi is the total potential energy of the 2th particle assuming that all the forces are conservative We must also specify the initial conditions7 ie7 the initial positions and velocities of all the particles Then we must solve the N coupled second order partial differential equations This is quite a daunting task Its true that we now have powerful computers and molecular dynamics simulations carry out such tasks7 but they can only handle a few thousand particles and track their motions for perhaps a 10 8 seconds So we throw up our hands This is not the way to go Fortunately there is a better way We can use the fact that there are large numbers of particles to apply a statistical analysis to the situation We are usually not interested in the detailed microscopics of a liquid or a gas Instead we are usually interested in certain macroscopic quantities like 0 pressure temperature entropy speci c heat electrical conductivity 0 magnetic and electric susceptibilities 0 etc These concepts don7t make sense for one atom or even a few particles For example what is the temperature of one atom That doesn7t make sense But if we have a whole bucket of atoms then temperature makes sense Since we are not concerned with the detailed behavior of each and every particle we can use statistical methods to extract information about a system with N gt 1 interacting particles For example we may be interested in the average energy of the system rather than the energy of each particle Statistical methods work best when there are large numbers of particles Think of polling Typically we have on the order of a mole7s worth of particles A mole is the number of atoms contained in 12 grams of carbon 12 A carbon 12 atom has 6 protons and 6 neutrons for a total 12 nucleons lts atomic weight is 12 So a mole has 6022 X1023 atoms This is called Avogadro7s number There are two basic approaches to describing the properties of a large number of particles Thermodynamics This is essentially a postulational approach or a deductive approach On the basis of a few general assumptions general relationships between and among macroscopic parameters are deduced No reference is made to the microscopic details of the system Statistical Mechanics This is an inductive approach With only a few assumptions relations between the macroscopic parameters of a system are induced from a con sideration of the microscopic interactions This approach yields all of the thermo dynamic results plus a good deal more information about the system of interacting particles Probability We will talk more about probability in the discussion section If we ip a coin7 the chance of heads is 12 and the chance of tails is 12 The probability of an outcome is the number of ways of getting that outcome divided by the total number of outcomes We often write for the probability that an outcome is 2 So if we ip 2 coins7 then Pheadhead147 and Phead7tail12 because there are a total of 4 possible outcomes and 2 ways of getting one coin heads and one coin tails Binary Model Let7s work out a simple example of a binary model Rather than heads or tails7 think of a line of spins which can be up or down The spins correspond to magnetic moments im m corresponds to T and 7m corresponds to 1 Suppose we have N sites7 each with a spin On each site the spin can be up or down The total number of arrangements is 2 X 2 X 2 X 22 2N So the total number of states or outcomes is 2N We can denote a state by mum 2 Now suppose we want the number of con gurations with n up sites7 regardless of what order they7re in The number of down sites will be N n7 since the total number of sites is N The number of such con gurations is N nN 7 72 To see where this comes from7 suppose we have N spins of which n are up and N n are down How many ways are there to arrange them Recall that the number of ways to arrange N objects with one object on each site or in each box is 3 o the rst place can be occupied by any one of the N objects 0 the second place can be occupied by any one of the N 1 remaining objects etc until the last place can be occupied by the last 1 object So there are NN71N721N 4 con gurations In these con gurations the n up spins can be arranged in any of 72 ways Also the remaining N n sites have down spins which can be arranged in any of N 7 72 ways The spins are regarded as distinguishable So there are 2 ways to have TT TT and TT They look the same7 but if one were red and the other blue7 you would see the di erence red7 blue and blue7 red We are diViding out this overcounting that7s what the denominator nN 7 n is for So the probability of a state with n up spins and N n down spins is N Pn T7 N m 7 total number of states N 1 nN 7n 2 N 5 3 Now suppose that the probability of a site getting T is p and the probability of getting 1 is q 1 7 p So far we have considered up and down to be equally probable so p q 12 But what if p a q This might be caused by an external magnetic eld which biases the spins one way or another If we have 2 spins the probability of getting 2 up spins is p2 Recall that the probability of ipping a coin twice and getting heads both times is 14 If we want n up spins and N n down spins then the probability is PW 17 N number of arrangements X p qun N N n 7n 6 nN 7 n p q When p q 12 this reduces to our previous result Equation 6 is called the binomial distribution because the prefactor NnN 7 n is the coe icient in the binomial expansion N a bN 7 WNLiWaN b 7 Normalization Averages Second Moment Lets go back to our general considerations of probability distribution functions Let be the probability distribution for the probability that an outcome is zi If we sum over all possible outcomes one of them is bound to happen so 2 1 lt8 In other words the probability distribution function is normalized to one One can also have a probability distribution function Pz of a continuous variable z For example suppose you leave a meter stick outside on the ground Pzdz could be the probability that the rst raindrop to hit the stick will strike between position z and z dz Pz is called the probability density The normalization condition is dzPz 1 9 Note that fdzPz is the area under the curve The normalization sets this area equal to 1 The average of some quantity z is dP ltgtM f dzPz 10 Averages can be denoted by lt z gt or T For a discrete quantity 22 22 11 E If the probability is normalized7 then the denominator is 1 For example7 suppose NA is the number of people With age Ai Then the average age is 2230 N 142 142 Z 2320 WAD f dA NA A f M MA 12 The probability that a person is A old is NM PA 7 fdA AMA 13 Notice that this satis es the normalization condition More generally7 if f is a function of 7 then the average value of f is f Mow we 7 MW lt14 One of the more useful functions concerns the deviation of m from the mean T Am m 7 T 15 Then 16 A more useful quantity is the square of the deviation from the mean W xii m 72m72 P42 17 This is known as the second moment of m about its mean The rst moment is just T In general the nth moment of m about its mean is given by f mm 7 7 n 18 as so f MW lt gt Two other terms that are often used are the most probable value of X and the median The most probable value of m is the maximum of The median mmed is the value of m such that half the values of m are greater than mmd ie7 m gt m5d7 and half the values 5 of m are less than and ie7 m lt mmed In terms of the area under the curve7 the median is the vertical dividing line such that half the area lies to the left of the median and half to the right of the median Gaussian Distribution and Central Limit Theorem One of most useful distributions is the Gaussian distribution This is sometimes called the bell curve which is well known as the ideal grade distribution PX I I I I I I I I I I I I H x The formula for a Gaussian distribution is 1 x 27w where u E is the mean The coef cient is set so that the normalization condition is satis ed 02 Amy m if 20 is the width of the distribution There is a 68 chance that 70 S m S a One obtains this by integrating Pm from 70 to 0 0 is sometimes called the rootimeanisquare rms deViation or the standard deViation You will have the opportunity to check some of these assertions in your homework For a Gaussian distribution7 all the higher moments can be expressed in terms of the rst and second moments For example m 7 T4 3W 7 T There is one other interesting aspect of Gaussian distributions Its called the central limit theorem It can be stated in terms of a random walk in one dimension A drunk starts at a lamp post and staggers back and forth along the one dimensional sidewalk Let w3ds be the probability that a step length lies in the range between 3 and 3 d3 No matter what the probability distribution 103 for each step may be7 as long as the steps are statistically independent and 103 falls o rapidly enough as 3 a 007 the total displacement m will be distributed according to the Gaussian law if the number of steps N is suf ciently large This is called the central limit theorem It is probably the most famous theorem in mathematical probability theory The generality of the result also accounts for the fact that so many phenomena in nature eg7 errors in measurement obey approximately a Gaussian distribution Pd e z f gt2202d 19 ln fact7 if we go back to the binomial distribution given in eq 67 then in the limit of large N7 this becomes the Gaussian distribution see Reif 15 In this case7 m represents the net magnetization7 ie7 the difference between the number of up and down spins NT 7 Ni The average magnetization M p 7 WV 21 and the mean square deviation is 02 4Npq 22 More on the Random Walk Reference Howard C Berg7 Random Walks in Biology The random walk is an important concept Let us go back to the drunk starting at the lamp post and stumbling either to the right or the left Consider an ensemble of M drunks Let be the position of the 2th drunk after N steps Suppose everyone starts at the origin so that N 0 0 for all 239 Let the step size be of xed length 3 Then N 71i 3 23 The average displacement from the origin is 1 M TAN M 2N 21 M Z N 71i 3 24 So the mean displacement is zero because on average7 half of the displacements are negative and half are positive But this does not mean that all the drunks are sitting at the origin after N steps It is useful to look at the spread in positions A convenient measure of spreading is the root mean square rms displacement 2N Here we average the square of the displacement rather than the displacement itself Since the square of a negative or a positive number is positive7 the result will be positive x Nz N71i2sziN7132 25 Then we compute the mean com M Z 71i 23N 7132 N71s2 26 So Pm 01 32 22 232 and PN N32 So the rms displacement PN W3 27 The rms displacement is a good measure of the typical distance that the drunk reaches Notice that it goes as the square root of the number of steps lf 739 is the time between successive steps then it takes a time t NT to take N steps So W g N is lt28 Notice that the rms displacement goes as square root of the time t A random walk is What particles execute When the di use Think of a opening a bottle of perfume The perfume molecules di use through the air We can de ne a di usion co ef cient D by 2 9 D 29 2739 The factor of 2 is by convention Then in one dimension 2 A t 2Dt 7a W 30 For 2 dimensions if the X and y directions are independent 2 720 05 W 4Dt 31 Similarly for 3 dimensions 720 P05 y2t 2 205 6Dt 32 Notice that in all dimensions the mean square displacement is linear in the number of steps N and in the time t The di usion coef cient D characterizes the migration of particles in a given kind of medium at a given temperature In general it depends on the size of the particle the structure of the medium and the absolute temperature For a small molecule in water at room temperature D 2 10 5 cm2sec 8 If we go back to the drunk who is taking steps of size 3 to the right or to the left then we can use the binomial distribution Let p be the probability to take a step to the right and q 1 7 p be the probability to take a step to the left After taking N steps the mean number m of steps to the right is see Reif Eq 144 N l Z PTZ1N711N 7110 The mean number of steps to the left is Let the displacement be Am m3 n1 7 n23 35 The mean displacement is Reif Eq 146 M m9 W1 7 23 Np 7 q3 36 So if p q 12 then the mean displacement M 0 The displacement squared is Amy Am232 m 7 m 37 The dispersion of the net displacement or the mean square displacement is Ax Am232 4Npq32 38 where we used Reif Eq 1412 If p q 12 then the mean square displacement Amy N32 which is what we got before The rms displacement is m2 21qu W3 39 If p q 12 then the rms displacement is A xNs which is what we got before Notice again that the rms displacement goes as the square root of the number of steps Again the characteristic or typical displacement is given by Amy N W3 Poisson Distribution Ref C Kittel Thermal Physics It is worth mentioning another widely used dis tribution namely the Poisson distribution This distribution is concerned with the occurrence of small numbers of objects in random sampling processes For example if on average there is one bad penny in 1000 what is the probability that no bad pennies will be found in a given sample of 100 pennies The problem was rst consided and solved by Poisson 1837 in a study of the role of luck in criminal and civil trials in France So suppose that there are N trials and the probability for a desired outcome is p The probability that there will be 72 desired outcomes is ltngtn ltngt PM n 40 where the mean number of desired events is Np You will derive the Poisson distribution in your homework starting from the binomial distribution Quantum Mechanics Since you are just starting your quantum mechanics course lets just go over what you need to know for starters Mostly its just notation and jargon Particles or systems of particles are described by a wavefunction 71 The wavefunction is a function of the coordinates of the particles or of their momenta but not both You can7t specify both the momentum and position of a particle because of the Heisenberg uncertainty principle AmApz i h 41 So the wavefunction of a particle is written 7007 The wavefunction is determined by Schroedinger7s equation H22 E21 42 where H is the Hamiltonian and E is the energy The Hamiltonian is the sum of the kinetic and potential energies You solve Schroedinger7s equation to determine the wave function Schroedinger7s equation can be written as a matrix equation or as a second order di erential equation Often the wavefunction solutions are labeled by de nite val ues of energy momentum etc These values can only have discrete values ie they are quantized They are called quantum numbers We can label the di erent wavefunction solutions by their quantum numbers Quantum numbers correspond to conserved quanti ties like energy momentum angular momentum spin angular momentum etc Perhaps the most familiar example of this is the electronic orbitals in the hydrogen atom The electron orbitals like the 1s and 2p states have de nite discrete energies and de nite values of the orbital angular momentum For example the energy of the 1s state in hydrogen is 1 Rydberg or 136 eV The electron in this state can be described by a wavefunction 715 The quantum numbers of this state are n 1 and L 0 n 1 means that it has energy E1 136 eV and L is the orbital angular momentum 3 is the spectroscopic notation for L 0 Solving the Schroedinger equation for the hydrogen atom yields many wavefunctions or orbitals 1s 2s 2px Zpy 2pz 3s etc Another example is a particle in a box

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