1 00:00:00,060 --> 00:00:01,780 The following content is provided 2 00:00:01,780 --> 00:00:04,019 under a Creative Commons license. 3 00:00:04,019 --> 00:00:06,870 Your support will help MIT OpenCourseWare continue 4 00:00:06,870 --> 00:00:10,730 to offer high quality educational resources for free. 5 00:00:10,730 --> 00:00:13,330 To make a donation or view additional materials 6 00:00:13,330 --> 00:00:17,217 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:17,217 --> 00:00:17,842 at ocw.mit.edu. 8 00:00:20,560 --> 00:00:22,610 MEHRAN KARDAR: You decide at first look 9 00:00:22,610 --> 00:00:28,830 at the our simple system, which was the ideal gas. 10 00:00:33,930 --> 00:00:38,250 And imagine that we have this gas contained 11 00:00:38,250 --> 00:00:44,630 in a box of volume V that contains N particles. 12 00:00:44,630 --> 00:00:48,970 And it's completely isolated from the rest of the universe. 13 00:00:48,970 --> 00:00:55,240 So you can know the amount of energy that it has. 14 00:00:55,240 --> 00:00:58,770 So the macroscopic description of the system 15 00:00:58,770 --> 00:01:02,015 consists of these three numbers, E, V, and N. 16 00:01:02,015 --> 00:01:07,540 And this was the characteristics of a microcanonical ensemble 17 00:01:07,540 --> 00:01:10,600 in which there was no exchange of heat or work. 18 00:01:10,600 --> 00:01:13,540 And therefore, energy was conserved. 19 00:01:13,540 --> 00:01:17,710 And our task was somehow to characterize the probability 20 00:01:17,710 --> 00:01:21,370 to find the system in some microstate. 21 00:01:21,370 --> 00:01:26,410 Now if you have N particles in the system, 22 00:01:26,410 --> 00:01:30,180 there is at the microscopic level, some microstate that 23 00:01:30,180 --> 00:01:36,520 consists of a description of all of the positions of momenta. 24 00:01:36,520 --> 00:01:39,690 And there is Hamiltonian that governs 25 00:01:39,690 --> 00:01:43,850 how that microstate evolves as a function of time. 26 00:01:43,850 --> 00:01:49,660 And for the case of ideal gas, the particles don't interact. 27 00:01:49,660 --> 00:01:55,070 So the Hamiltonian can be written as the sum of n terms 28 00:01:55,070 --> 00:01:59,680 that describe essentially the energy of the end 29 00:01:59,680 --> 00:02:04,770 particle composed of its kinetic energy. 30 00:02:04,770 --> 00:02:08,490 And the term that is really just confining the particle 31 00:02:08,490 --> 00:02:10,240 is in this box. 32 00:02:10,240 --> 00:02:12,990 And so the volume of the box is contained. 33 00:02:12,990 --> 00:02:14,510 Let's say that in this potential, 34 00:02:14,510 --> 00:02:19,050 that it's zero inside the box and infinity outside. 35 00:02:19,050 --> 00:02:25,900 And we said, OK, so given that I know what the energy, volume, 36 00:02:25,900 --> 00:02:29,910 number of particles are, what's the chance 37 00:02:29,910 --> 00:02:35,350 that I will find the system in some particular microstate? 38 00:02:35,350 --> 00:02:38,490 And the answer was that, obviously, you 39 00:02:38,490 --> 00:02:45,450 will have to put zero if the particles are outside box. 40 00:02:50,320 --> 00:02:55,520 Or if the energy, which is really just the kinetic energy, 41 00:02:55,520 --> 00:03:00,560 does not much the energy that we know is in the system, 42 00:03:00,560 --> 00:03:06,030 we sum over i of P i squared over 2m 43 00:03:06,030 --> 00:03:10,071 is not equal to the energy of the system. 44 00:03:14,790 --> 00:03:18,170 Otherwise, we say that if the microstate corresponds 45 00:03:18,170 --> 00:03:22,010 to exactly the right amount of energy, then 46 00:03:22,010 --> 00:03:24,710 I have no reason to exclude it. 47 00:03:24,710 --> 00:03:26,960 And just like saying that the dice can 48 00:03:26,960 --> 00:03:29,400 have six possible faces, you would 49 00:03:29,400 --> 00:03:33,650 assign all of those possible phases equal probability. 50 00:03:33,650 --> 00:03:36,050 I will give all of the microstates that 51 00:03:36,050 --> 00:03:37,720 don't conflict with the conditions 52 00:03:37,720 --> 00:03:40,800 that I have set out the same probability. 53 00:03:40,800 --> 00:03:43,400 I will call that probability 1 over 54 00:03:43,400 --> 00:03:47,530 some overall constant omega. 55 00:03:47,530 --> 00:03:49,140 And so this is one otherwise. 56 00:03:56,780 --> 00:03:59,590 So then the next statement is well 57 00:03:59,590 --> 00:04:04,980 what is this number omega that you have put it? 58 00:04:04,980 --> 00:04:07,960 And how do we determine it? 59 00:04:07,960 --> 00:04:12,100 Well, we know that this P is a probability so 60 00:04:12,100 --> 00:04:15,640 that if I were to integrate over the entirety of the face 61 00:04:15,640 --> 00:04:25,480 space of this probability, the answer should be 1. 62 00:04:29,580 --> 00:04:32,700 So that means this omega, which is 63 00:04:32,700 --> 00:04:36,250 a function of these parameters that I set out from the outside 64 00:04:36,250 --> 00:04:39,190 to describe the microstate, should 65 00:04:39,190 --> 00:04:53,660 be obtained by integrating over all q and p of this collection 66 00:04:53,660 --> 00:04:57,620 of 1s and 0s that I have out here. 67 00:04:57,620 --> 00:05:02,330 So I think this box of 1s and 0s, put it here, 68 00:05:02,330 --> 00:05:03,060 and I integrate. 69 00:05:07,630 --> 00:05:09,940 So what do I get? 70 00:05:09,940 --> 00:05:13,850 Well, the integration over the q's is easy. 71 00:05:13,850 --> 00:05:17,890 The places that I get 1 are when the q's are inside the box. 72 00:05:17,890 --> 00:05:22,280 So each one of them will give me a factor of V. 73 00:05:22,280 --> 00:05:24,290 And there are N of them. 74 00:05:24,290 --> 00:05:28,490 So I would get V to the N. 75 00:05:28,490 --> 00:05:32,090 The integrations over momenta essentially 76 00:05:32,090 --> 00:05:36,960 have to do with seeing whether or not the condition sum 77 00:05:36,960 --> 00:05:42,730 over i Pi squared over 2m equals to E is satisfied or not. 78 00:05:42,730 --> 00:05:46,232 So this I can write also as sum over i 79 00:05:46,232 --> 00:05:55,180 P i squared equals to 2mE, which I can write as R squared. 80 00:05:55,180 --> 00:06:00,690 And essentially, in this momentum space, 81 00:06:00,690 --> 00:06:05,200 I have to make sure that the sum of the components of all 82 00:06:05,200 --> 00:06:09,320 of the momenta squared add up to this R. squared, 83 00:06:09,320 --> 00:06:11,950 which as we discussed last time, is 84 00:06:11,950 --> 00:06:29,620 the surface of hypershpere in 3N dimensions of radius 85 00:06:29,620 --> 00:06:31,370 R, which is square root of 2mE. 86 00:06:35,850 --> 00:06:40,350 So I have to integrate over all of these momenta. 87 00:06:40,350 --> 00:06:42,820 And most of the time I will get 0, 88 00:06:42,820 --> 00:06:47,850 except when I heat the surface of this sphere. 89 00:06:47,850 --> 00:06:50,500 There's kind of a little bit of singularity 90 00:06:50,500 --> 00:06:53,010 here because you have a probability that there's 0, 91 00:06:53,010 --> 00:06:57,960 except at the very sharp interval, and then 0 again. 92 00:06:57,960 --> 00:07:02,660 So it's kind of like a delta function, which is maybe 93 00:07:02,660 --> 00:07:05,930 a little bit hard to deal with. 94 00:07:05,930 --> 00:07:09,860 So sometimes we will generalize this 95 00:07:09,860 --> 00:07:13,900 by adding a little bit be delta E here. 96 00:07:13,900 --> 00:07:18,130 So let's say that the energy does not have to be exactly E, 97 00:07:18,130 --> 00:07:20,970 but E minus plus a little bit, so 98 00:07:20,970 --> 00:07:25,850 that when we look at this surface in three 99 00:07:25,850 --> 00:07:28,520 n dimensional space-- let's say this was two 100 00:07:28,520 --> 00:07:31,220 dimensional space-- rather than having 101 00:07:31,220 --> 00:07:34,940 to deal with an exact boundary, we 102 00:07:34,940 --> 00:07:42,720 have kind of smoothed that out into an interval that 103 00:07:42,720 --> 00:07:47,340 has some kind of a thickness R, that presumably is related 104 00:07:47,340 --> 00:07:49,690 to this delta E that I put up there. 105 00:07:49,690 --> 00:07:51,860 Turns out it doesn't really make any difference. 106 00:07:51,860 --> 00:07:53,970 The reason it doesn't make any difference I 107 00:07:53,970 --> 00:07:56,960 will tell you shortly. 108 00:07:56,960 --> 00:07:59,750 But now when I'm integrating over all of these 109 00:07:59,750 --> 00:08:03,360 P's-- so there's P. There's another P. This could be P1. 110 00:08:03,360 --> 00:08:04,650 This could be P2. 111 00:08:04,650 --> 00:08:06,730 And there are different components. 112 00:08:06,730 --> 00:08:13,530 I then get 0, except when I hit this interval 113 00:08:13,530 --> 00:08:16,150 around the surface of this hypersphere. 114 00:08:16,150 --> 00:08:22,280 So what do I get as a result of the integration over this 3N 115 00:08:22,280 --> 00:08:24,110 dimensional space? 116 00:08:24,110 --> 00:08:29,250 I will get the volume of this element, which 117 00:08:29,250 --> 00:08:32,950 is composed of the surface area, which 118 00:08:32,950 --> 00:08:37,380 has some kind of a solid angle in 3N dimensions. 119 00:08:37,380 --> 00:08:41,429 The radius raised to the power of dimension 120 00:08:41,429 --> 00:08:45,330 minus 1, because it's a surface. 121 00:08:45,330 --> 00:08:49,620 And then if I want to really include a delta R 122 00:08:49,620 --> 00:08:54,960 to make it into a volume, this would be the appropriate volume 123 00:08:54,960 --> 00:08:58,640 of this interval in momentum space. 124 00:08:58,640 --> 00:08:59,395 Yes. 125 00:08:59,395 --> 00:09:01,135 AUDIENCE: Just to clarify, you're 126 00:09:01,135 --> 00:09:02,910 asserting that there's no potential 127 00:09:02,910 --> 00:09:05,782 inside the [INAUDIBLE] that comes from the hard walls. 128 00:09:05,782 --> 00:09:06,740 MEHRAN KARDAR: Correct. 129 00:09:06,740 --> 00:09:10,490 We can elaborate on that later on. 130 00:09:10,490 --> 00:09:13,710 But for the description of the ideal gas without potential, 131 00:09:13,710 --> 00:09:16,300 like in the box, I have said that potential 132 00:09:16,300 --> 00:09:19,610 to be just 0 infinity. 133 00:09:19,610 --> 00:09:20,110 OK? 134 00:09:24,170 --> 00:09:26,350 OK, so fine. 135 00:09:26,350 --> 00:09:28,080 So this is the description. 136 00:09:28,080 --> 00:09:29,620 There was one thing that I needed 137 00:09:29,620 --> 00:09:41,550 to tell you, which is the d, dimension, of solid angle, 138 00:09:41,550 --> 00:09:47,020 which is 2pi to the d over 2 divided 139 00:09:47,020 --> 00:09:51,100 by d over 2 minus 1 factorial So again, 140 00:09:51,100 --> 00:09:55,330 in two dimensions, such as the picture that I drew over here, 141 00:09:55,330 --> 00:10:00,210 the circumference of a circle would be 2 pi r. 142 00:10:00,210 --> 00:10:03,065 So this s sub 2-- right there'd be 2 pi-- 143 00:10:03,065 --> 00:10:06,380 and you can show that it is 2 pi divided 144 00:10:06,380 --> 00:10:09,490 by 0 factorial, which is 1. 145 00:10:09,490 --> 00:10:13,850 In three dimensions it should give you 4 pi r squared. 146 00:10:13,850 --> 00:10:15,340 Kind of looks strange because you 147 00:10:15,340 --> 00:10:20,090 get 2 pi to the 3/2 divided by 1/2 half factorial. 148 00:10:20,090 --> 00:10:22,530 But the 1/2 factorial is in fact root 2 149 00:10:22,530 --> 00:10:25,670 over pi-- root pi over 2. 150 00:10:25,670 --> 00:10:27,670 And so this will work out fine. 151 00:10:30,340 --> 00:10:33,840 Again, the definition of this factorial in general 152 00:10:33,840 --> 00:10:36,280 is through the gamma function and an integral 153 00:10:36,280 --> 00:10:38,270 that we saw already. 154 00:10:38,270 --> 00:10:43,910 And factorial is the integral 0 to infinity, dx, x to the n 155 00:10:43,910 --> 00:10:46,680 into the minus x. 156 00:10:46,680 --> 00:10:49,990 Now, the thing is that this is a quantity that 157 00:10:49,990 --> 00:10:55,990 for large values of dimension grows exponentially v E d. 158 00:10:55,990 --> 00:11:00,750 So what I claim is that if I take the log of this surface 159 00:11:00,750 --> 00:11:06,610 area and take the limit that d is much larger than 1, 160 00:11:06,610 --> 00:11:09,280 the quantity that I will get-- well, let's 161 00:11:09,280 --> 00:11:10,340 take the log of this. 162 00:11:10,340 --> 00:11:13,390 I will get log of 2. 163 00:11:13,390 --> 00:11:23,840 I will get d over 2 log pi and minus the log 164 00:11:23,840 --> 00:11:26,680 of this large factorial. 165 00:11:26,680 --> 00:11:30,650 And the log of the factorial I will use Sterling's formula. 166 00:11:30,650 --> 00:11:32,870 I will ignore in that large limit 167 00:11:32,870 --> 00:11:36,920 the difference between d over 2 and d over 2 minus 1. 168 00:11:36,920 --> 00:11:38,530 Or actually I guess at the beginning 169 00:11:38,530 --> 00:11:46,380 I may even write it d over 2 minus 1 log of d over 2 minus 1 170 00:11:46,380 --> 00:11:49,337 plus d over 2 minus 1. 171 00:11:53,000 --> 00:11:56,340 Now if I'm in this limit of large d again, 172 00:11:56,340 --> 00:12:00,210 I can ignore the 1s. 173 00:12:00,210 --> 00:12:06,020 And I can ignore the log 2 with respect to this d over 2. 174 00:12:06,020 --> 00:12:10,070 And so the answer in this limit is in fact proportional 175 00:12:10,070 --> 00:12:11,950 to d over 2. 176 00:12:11,950 --> 00:12:14,390 And I have the log. 177 00:12:14,390 --> 00:12:16,550 I have pi. 178 00:12:16,550 --> 00:12:21,290 I have this d over 2 that will carry in the denominator. 179 00:12:21,290 --> 00:12:26,460 And then this d over 2 times 1 I can write as d over 2 log e. 180 00:12:26,460 --> 00:12:30,520 And so this is the answer we get. 181 00:12:30,520 --> 00:12:38,590 So you can see that the answer is exponentially large 182 00:12:38,590 --> 00:12:41,530 if I were to again write s of d. 183 00:12:41,530 --> 00:12:46,500 S of d grows like an exponential in d. 184 00:12:46,500 --> 00:12:49,930 OK, so what do I conclude from that? 185 00:12:49,930 --> 00:12:59,740 I conclude that s over Kd-- and we said that the entropy, 186 00:12:59,740 --> 00:13:05,080 we can regard as the entropy of this probability distribution. 187 00:13:05,080 --> 00:13:08,065 So that's going to give me the log of this omega. 188 00:13:12,050 --> 00:13:16,460 And log off this omega, we'll get a factor 189 00:13:16,460 --> 00:13:18,970 from this v to the n. 190 00:13:18,970 --> 00:13:24,190 So I will get N log V. I will get 191 00:13:24,190 --> 00:13:30,130 a factor from log of S of 3N. 192 00:13:30,130 --> 00:13:35,260 I figured out what that log was in the limit 193 00:13:35,260 --> 00:13:37,110 of large dimensions. 194 00:13:37,110 --> 00:13:46,340 So I essentially have 3N over 2 because my d is now roughly 3N. 195 00:13:46,340 --> 00:13:48,980 It's in fact exactly 3N, sorry. 196 00:13:48,980 --> 00:13:53,645 I have the log of 2 pi e. 197 00:13:53,645 --> 00:13:56,900 For d I have 3N. 198 00:13:56,900 --> 00:14:00,940 And then I actually have also from here 199 00:14:00,940 --> 00:14:09,600 a 3N log R, which I can write as 3N over 2 log of R squared. 200 00:14:09,600 --> 00:14:13,506 And my R squared is 2mE. 201 00:14:13,506 --> 00:14:14,464 The figure I have here. 202 00:14:17,430 --> 00:14:24,950 And then you say, OK, we added this delta R. 203 00:14:24,950 --> 00:14:27,250 But now you can see that I can also 204 00:14:27,250 --> 00:14:33,230 ignore this delta R, because everything else that I have 205 00:14:33,230 --> 00:14:37,240 in this expression is something that grows radially 206 00:14:37,240 --> 00:14:43,400 with N. What's the worse that I can do for delta R? 207 00:14:43,400 --> 00:14:49,910 I could make delta R even as big as the entirety of this volume. 208 00:14:49,910 --> 00:14:52,680 And then the typical volume would 209 00:14:52,680 --> 00:14:57,530 be of the order of the energy-- sorry, the typical value of R 210 00:14:57,530 --> 00:15:00,220 would be like the square root of energy. 211 00:15:00,220 --> 00:15:02,310 So here I would have to put this log 212 00:15:02,310 --> 00:15:04,680 of the square root of the energy. 213 00:15:04,680 --> 00:15:07,495 And log of a square roots of an extensive quantity 214 00:15:07,495 --> 00:15:10,400 is much less than the extensive quantity. 215 00:15:10,400 --> 00:15:12,470 I can ignore it. 216 00:15:12,470 --> 00:15:17,870 And actually this reminds me, that some 35 years ago when 217 00:15:17,870 --> 00:15:22,680 I was taking this course, from Professor Felix Villiers, 218 00:15:22,680 --> 00:15:25,370 he said that he had gone to lunch. 219 00:15:25,370 --> 00:15:30,515 And he had gotten to this very beautiful, large orange. 220 00:15:30,515 --> 00:15:32,780 And he was excited. 221 00:15:32,780 --> 00:15:36,700 And he opened up the orange, and it was all skin. 222 00:15:36,700 --> 00:15:39,500 And there was just a little bit in the middle. 223 00:15:39,500 --> 00:15:41,560 He was saying it is like this. 224 00:15:41,560 --> 00:15:43,370 It's all in the surface. 225 00:15:43,370 --> 00:15:48,560 So if Professor Villiers had an orange in 3N dimension, 226 00:15:48,560 --> 00:15:52,040 he would have exponentially hard time extracting an orange. 227 00:15:55,070 --> 00:16:00,880 So this is our formula for the entropy of this gas. 228 00:16:00,880 --> 00:16:04,110 Essentially the extensive parts, n log v 229 00:16:04,110 --> 00:16:07,590 and something that depends on n log E. 230 00:16:07,590 --> 00:16:10,100 And that's really all we need to figure out 231 00:16:10,100 --> 00:16:12,950 all of the thermodynamic properties, 232 00:16:12,950 --> 00:16:18,540 because we said that we can construct-- 233 00:16:18,540 --> 00:16:25,710 that's in thermodynamics-- dE is TdS minus PdV plus YdN 234 00:16:25,710 --> 00:16:29,440 in the case of a gas. 235 00:16:29,440 --> 00:16:32,950 And so we can rearrange that to dS 236 00:16:32,950 --> 00:16:43,410 be dE over T plus P over T dV minus Y over T dN. 237 00:16:43,410 --> 00:16:46,630 And the first thing that we see is 238 00:16:46,630 --> 00:16:50,090 by taking the derivative of S with respect to the quantities 239 00:16:50,090 --> 00:16:53,010 that we have established, E, V, and N, 240 00:16:53,010 --> 00:16:57,780 we should be able to read off appropriate quantities. 241 00:16:57,780 --> 00:17:01,040 And in particular, let's say 1 over T 242 00:17:01,040 --> 00:17:06,819 would be dS by dE of constant v and n. 243 00:17:06,819 --> 00:17:10,480 S will be proportional to kB. 244 00:17:10,480 --> 00:17:14,829 And then they dependents of this object on E 245 00:17:14,829 --> 00:17:19,599 only appears on this log E. Except 246 00:17:19,599 --> 00:17:23,210 that there's a factor of 3N over 2 out front. 247 00:17:23,210 --> 00:17:30,110 And the derivative of log E with respect to E is 1 over E. 248 00:17:30,110 --> 00:17:32,820 So I can certainly immediately rearrange 249 00:17:32,820 --> 00:17:37,720 this and to get that the energy is 3/2 N k 250 00:17:37,720 --> 00:17:46,160 T in this system of ideal point particles in three dimensions. 251 00:17:46,160 --> 00:17:51,067 And then the pressure, P over T, is 252 00:17:51,067 --> 00:17:54,380 the S by dV at constant e and n. 253 00:17:54,380 --> 00:17:56,520 And it's again, kB. 254 00:17:56,520 --> 00:18:00,850 The only dependence on V is through this N log V. 255 00:18:00,850 --> 00:18:06,365 So I will get a factor of N over V, which I can rearrange to PV 256 00:18:06,365 --> 00:18:12,150 is N kB T by the ideal gas law. 257 00:18:12,150 --> 00:18:15,530 And in principle, the next step would 258 00:18:15,530 --> 00:18:17,670 be to calculate the chemical potential. 259 00:18:17,670 --> 00:18:21,800 But we will leave that for the time being 260 00:18:21,800 --> 00:18:23,830 for reasons that will become apparent. 261 00:18:34,260 --> 00:18:46,480 Now, one thing to note is that what you have postulated here, 262 00:18:46,480 --> 00:18:51,640 right at the beginning, is much, much, more information 263 00:18:51,640 --> 00:18:56,390 than what we extracted here about thermodynamic properties. 264 00:18:56,390 --> 00:19:00,760 It's a statement about a joint probability distribution 265 00:19:00,760 --> 00:19:04,160 in this six N dimensional face space. 266 00:19:04,160 --> 00:19:08,210 So it has huge amount of information. 267 00:19:08,210 --> 00:19:11,290 Just to show you part of it, let's take 268 00:19:11,290 --> 00:19:14,670 a note at the following. 269 00:19:14,670 --> 00:19:19,820 What it is a probability as a function of all 270 00:19:19,820 --> 00:19:23,540 coordinates and momenta across your system. 271 00:19:23,540 --> 00:19:26,470 But let me ask a specific question. 272 00:19:26,470 --> 00:19:31,420 I can ask what's the probability that some particular particle-- 273 00:19:31,420 --> 00:19:36,160 say particle number one-- has a momentum P1. 274 00:19:36,160 --> 00:19:37,990 It's the only question that I care 275 00:19:37,990 --> 00:19:42,460 to ask about this huge amount of degrees of freedom 276 00:19:42,460 --> 00:19:45,090 that are encoded in P of mu. 277 00:19:45,090 --> 00:19:48,070 And so what do I do if I don't really care about all 278 00:19:48,070 --> 00:19:51,690 of the other degrees of freedom is I will integrate them. 279 00:19:51,690 --> 00:19:57,810 So I don't really care where particle number one is located. 280 00:19:57,810 --> 00:20:00,470 I didn't ask where it is in the box. 281 00:20:00,470 --> 00:20:05,750 I don't really care where part because numbers two through N 282 00:20:05,750 --> 00:20:08,680 are located or which momenta they have. 283 00:20:08,680 --> 00:20:12,410 So I integrate over all of those things 284 00:20:12,410 --> 00:20:17,200 of the full joint probability, which 285 00:20:17,200 --> 00:20:20,161 depends on the entirety of the face space. 286 00:20:23,460 --> 00:20:25,140 Fine, you say, OK. 287 00:20:25,140 --> 00:20:29,450 This joint probability actually has a very simple form. 288 00:20:29,450 --> 00:20:35,670 It is 1 over this omega E, V, and N, 289 00:20:35,670 --> 00:20:38,990 multiplying either 1 or 0. 290 00:20:38,990 --> 00:20:45,210 So I have to integrate over all of these q1's, all 291 00:20:45,210 --> 00:20:49,020 of these qi P i. 292 00:20:51,630 --> 00:20:56,420 Of 1 over omega or 0 over omega, this delta 293 00:20:56,420 --> 00:20:58,890 like function that we put in a box 294 00:20:58,890 --> 00:21:02,070 up there-- so this is this delta function that 295 00:21:02,070 --> 00:21:05,850 says that the particular should be inside the box. 296 00:21:05,850 --> 00:21:07,690 And the sum of the momenta should 297 00:21:07,690 --> 00:21:09,900 be on the surface of this hypershpere. 298 00:21:14,760 --> 00:21:18,390 Now, let's do these integrations. 299 00:21:18,390 --> 00:21:19,555 Let's do it here. 300 00:21:19,555 --> 00:21:22,020 I may need space. 301 00:21:22,020 --> 00:21:24,830 The integration over Q1 is very simple. 302 00:21:24,830 --> 00:21:28,490 It will give me a factor of V. 303 00:21:28,490 --> 00:21:31,505 I have this omega E, V, N, in the denominator. 304 00:21:35,320 --> 00:21:41,340 And I claim that the numerator is simply the following. 305 00:21:41,340 --> 00:21:50,550 Omega E minus P1 squared over 2m V N minus 1. 306 00:21:50,550 --> 00:21:52,250 Why? 307 00:21:52,250 --> 00:21:59,230 Because what I need to do over here in terms of integrations 308 00:21:59,230 --> 00:22:03,320 is pretty much what I would have to integrate over here 309 00:22:03,320 --> 00:22:08,040 that gave rise to that surface and all of those factors 310 00:22:08,040 --> 00:22:09,900 with one exception. 311 00:22:09,900 --> 00:22:14,540 First of all, I integrated over one particle already, 312 00:22:14,540 --> 00:22:18,520 so the coordinate momenta here that I'm integrating pertains 313 00:22:18,520 --> 00:22:21,280 to the remaining N minus 1. 314 00:22:21,280 --> 00:22:25,470 Hence, the omega pertains to N minus 1. 315 00:22:25,470 --> 00:22:27,900 It's still in the same box of volume V. 316 00:22:27,900 --> 00:22:31,250 So V, the other argument, is the same. 317 00:22:31,250 --> 00:22:33,530 But the energy is changed. 318 00:22:33,530 --> 00:22:34,030 Why? 319 00:22:34,030 --> 00:22:37,430 Because I told you how much momentum 320 00:22:37,430 --> 00:22:40,460 I want the first particle to carry. 321 00:22:40,460 --> 00:22:42,520 So given the knowledge that I'm looking 322 00:22:42,520 --> 00:22:47,030 at the probability of the first particle having momentum P1, 323 00:22:47,030 --> 00:22:50,810 then I know that the remainder of the energy 324 00:22:50,810 --> 00:22:54,570 should be shared among the momenta of all the remaining 325 00:22:54,570 --> 00:22:57,650 N minus 1 particles. 326 00:22:57,650 --> 00:23:03,580 So I have already calculated these omegas up here. 327 00:23:03,580 --> 00:23:06,710 All I need to do is to substitute them over here. 328 00:23:06,710 --> 00:23:09,500 And I will get this probability. 329 00:23:09,500 --> 00:23:16,000 So first of all, let's check that the volume part cancels. 330 00:23:16,000 --> 00:23:19,090 I have one factor a volume here. 331 00:23:19,090 --> 00:23:23,540 Each of my omegas is in fact proportional to V to the N. 332 00:23:23,540 --> 00:23:26,080 So the denominator has V to the N. 333 00:23:26,080 --> 00:23:29,060 The numerator has a V to the N minus 1. 334 00:23:29,060 --> 00:23:32,930 And all of the V's would cancel out. 335 00:23:32,930 --> 00:23:39,900 So the interesting thing really comes from these solid angle 336 00:23:39,900 --> 00:23:42,720 and radius parts. 337 00:23:42,720 --> 00:23:47,240 The solid angle is a ratio of-- let's write the denominator. 338 00:23:47,240 --> 00:23:48,300 It's easier. 339 00:23:48,300 --> 00:23:54,130 It is 2 pi to the 3N over 2 divided by 3N over 2 340 00:23:54,130 --> 00:23:56,200 minus one factorial. 341 00:23:56,200 --> 00:24:00,670 The numerator would be 2 pi to the 3 N and minus 1 342 00:24:00,670 --> 00:24:09,745 over 2 divided by 3 N minus 1 over 2 minus 1 factorial. 343 00:24:13,200 --> 00:24:18,620 And then I have these ratio of the radii. 344 00:24:18,620 --> 00:24:34,980 In the denominator I have 2mE to the power of 3 N minus 1 over 2 345 00:24:34,980 --> 00:24:36,130 minus 1. 346 00:24:36,130 --> 00:24:48,770 So minus 3N-- it is 3N minus 1 over 2. 347 00:24:48,770 --> 00:24:52,010 Same thing that we have been calculating so far. 348 00:24:52,010 --> 00:24:59,490 And in the numerator it is 2m E minus P1 squared over 2m. 349 00:24:59,490 --> 00:25:05,350 So I will factor out the E. I have 1 minus P1 squared 350 00:25:05,350 --> 00:25:09,380 over 2m E. The whole thing raised to something 351 00:25:09,380 --> 00:25:14,338 that is 3 N minus 1 minus 1 over 2. 352 00:25:25,954 --> 00:25:31,300 Now, the most important part of this 353 00:25:31,300 --> 00:25:39,360 is the fact that the dependence on P1 appears as follows. 354 00:25:39,360 --> 00:25:45,320 I have this factor of 1 minus P1 squared over 2m E. 355 00:25:45,320 --> 00:25:48,400 That's the one place that P1, the momentum 356 00:25:48,400 --> 00:25:51,760 of the particle that I'm interested appears. 357 00:25:51,760 --> 00:25:54,280 And it raised to the huge problem, 358 00:25:54,280 --> 00:25:57,080 which is of the order of 3N over 2. 359 00:25:57,080 --> 00:25:59,160 It is likely less. 360 00:25:59,160 --> 00:26:01,290 But it really doesn't make any difference 361 00:26:01,290 --> 00:26:06,050 whether I write 3N over 2, 3N over minus 1 over 2, et cetera. 362 00:26:06,050 --> 00:26:08,720 Really, ultimately, what I will have 363 00:26:08,720 --> 00:26:13,260 is 1 minus the very small number, because presumably 364 00:26:13,260 --> 00:26:15,920 the energy of one part they can is 365 00:26:15,920 --> 00:26:19,030 less than the energy of the entirety of the particle. 366 00:26:19,030 --> 00:26:22,670 So this is something that is order of 1 out of N raised 367 00:26:22,670 --> 00:26:25,420 to something that is order of N. So that's 368 00:26:25,420 --> 00:26:28,520 where an exponentiation will come into play. 369 00:26:28,520 --> 00:26:31,680 And then there's a whole bunch of other factors 370 00:26:31,680 --> 00:26:36,320 that if I don't make any mistake I can try to write down. 371 00:26:36,320 --> 00:26:39,300 There is the 2s certainly cancel when 372 00:26:39,300 --> 00:26:42,480 I look at the factors of pi. 373 00:26:42,480 --> 00:26:45,670 The denominator with respect to the numerator 374 00:26:45,670 --> 00:26:49,780 has an additional factor of pi to the 3/2. 375 00:26:49,780 --> 00:26:51,430 In fact, I will have a whole bunch 376 00:26:51,430 --> 00:26:54,840 of things that are raised to the powers of 3/2. 377 00:26:54,840 --> 00:26:57,980 I also have this 2mE that compared 378 00:26:57,980 --> 00:27:01,420 to the 2mE that comes out front has 379 00:27:01,420 --> 00:27:03,940 an additional factor of 3/2. 380 00:27:03,940 --> 00:27:06,220 So let's put all of them together. 381 00:27:06,220 --> 00:27:09,760 2 pi mE raised to the power of 3/2. 382 00:27:13,260 --> 00:27:18,230 And then I have the ratio of these factorials. 383 00:27:18,230 --> 00:27:22,350 And again, the factorial that I have in the denominator 384 00:27:22,350 --> 00:27:27,940 has one and a half times more or 3/2 times more 385 00:27:27,940 --> 00:27:31,420 than what is in the numerator. 386 00:27:31,420 --> 00:27:34,480 Roughly it is something like the ratio 387 00:27:34,480 --> 00:27:39,390 of 3 N over 2 factorial divided by 3 388 00:27:39,390 --> 00:27:42,065 N minus 1 over 2 factorial. 389 00:27:46,870 --> 00:27:50,390 And I claim that, say, N factorial compared 390 00:27:50,390 --> 00:27:55,430 to N minus 1 factorial is larger by a factor of N. 391 00:27:55,430 --> 00:27:59,990 If I go between N factorial N minus 2 factorial is 392 00:27:59,990 --> 00:28:02,960 a factor that is roughly N squared. 393 00:28:02,960 --> 00:28:08,810 Now this does not shift either by 1 or by 2, but by 1 and 1/2. 394 00:28:08,810 --> 00:28:12,250 And if you go through Sterling formula, et cetera, 395 00:28:12,250 --> 00:28:16,650 you can convince yourself that this is roughly 3 N over 2 396 00:28:16,650 --> 00:28:19,335 to the power of 1 and 1/2-- 3/2. 397 00:28:21,930 --> 00:28:28,180 And so once you do all of your arrangements, what do you get? 398 00:28:28,180 --> 00:28:32,210 1 minus a small quantity raised to a huge power, 399 00:28:32,210 --> 00:28:34,580 that's the definition of the exponential. 400 00:28:34,580 --> 00:28:41,650 So I get exponential of minus P1 squared over 2m. 401 00:28:41,650 --> 00:28:46,060 And the factor that multiplies it is E. 402 00:28:46,060 --> 00:28:50,260 And then I have 3N over 2. 403 00:28:55,660 --> 00:29:02,160 And again, if I have not made any mistake 404 00:29:02,160 --> 00:29:05,890 and I'm careful with all of the other factors that remain, 405 00:29:05,890 --> 00:29:12,340 I have here 2 pi m E. And this E also 406 00:29:12,340 --> 00:29:17,040 gets multiplied by the inverse of 3N over 2. 407 00:29:17,040 --> 00:29:20,890 So I will have this replaced by 2E over 3N. 408 00:29:29,140 --> 00:29:34,650 So statement number one, this assignment 409 00:29:34,650 --> 00:29:37,500 of probabilities according to just throwing the dice 410 00:29:37,500 --> 00:29:41,970 and saying that everything that has the same right energy 411 00:29:41,970 --> 00:29:45,080 is equally likely is equivalent to looking 412 00:29:45,080 --> 00:29:47,680 at one of the particles and stating 413 00:29:47,680 --> 00:29:50,455 that the momentum of that part again is Gaussian distributed. 414 00:29:53,660 --> 00:30:00,200 Secondly, you can check that this combination, 2E divided 415 00:30:00,200 --> 00:30:03,330 by 3N is the same thing as kT. 416 00:30:03,330 --> 00:30:05,360 So essentially this you could also 417 00:30:05,360 --> 00:30:09,030 if you want to replace 1 over kT. 418 00:30:09,030 --> 00:30:13,120 And you would get the more familiar kind 419 00:30:13,120 --> 00:30:16,370 of Maxwell type of distribution for the momentum 420 00:30:16,370 --> 00:30:20,320 of a single particle in an ideal gas. 421 00:30:20,320 --> 00:30:23,540 And again, since everything that we did 422 00:30:23,540 --> 00:30:26,760 was consistent with the laws of probability, 423 00:30:26,760 --> 00:30:30,200 if we did not mix up the orders of N, et cetera, 424 00:30:30,200 --> 00:30:32,860 the answer should be properly normalized. 425 00:30:32,860 --> 00:30:38,520 And indeed, you can check that this is the three dimensional 426 00:30:38,520 --> 00:30:43,490 normalization that you would require for this gas. 427 00:30:43,490 --> 00:30:49,790 So the statement of saying that everything is allowed 428 00:30:49,790 --> 00:30:56,220 is equally likely is a huge statement in space 429 00:30:56,220 --> 00:30:58,390 of possible configurations. 430 00:30:58,390 --> 00:31:01,980 On the one hand, it gives you macroscopic information. 431 00:31:01,980 --> 00:31:04,580 On the other hand, it retains a huge amount 432 00:31:04,580 --> 00:31:06,640 of microscopic information. 433 00:31:06,640 --> 00:31:08,340 The parts of it that are relevant, 434 00:31:08,340 --> 00:31:10,257 you can try to extract this here. 435 00:31:13,361 --> 00:31:13,860 OK? 436 00:31:22,680 --> 00:31:25,820 So those were the successes. 437 00:31:25,820 --> 00:31:29,470 Question is why didn't I calculate for you 438 00:31:29,470 --> 00:31:31,840 this u over T? 439 00:31:31,840 --> 00:31:35,300 It is because this expression as we wrote down 440 00:31:35,300 --> 00:31:39,450 has a glaring problem with it, which in order 441 00:31:39,450 --> 00:31:43,870 to make it explicit, we will look at mixing entropies. 442 00:31:52,490 --> 00:31:54,580 So the idea is this is as follows. 443 00:31:54,580 --> 00:31:59,585 Let's imagine that we start with two gases. 444 00:32:03,780 --> 00:32:11,080 Initially, I have N1 particles of one type in volume 1. 445 00:32:11,080 --> 00:32:16,640 And I have N2 particles of another type in volume 2. 446 00:32:16,640 --> 00:32:20,600 And for simplicity I will assume that both of them 447 00:32:20,600 --> 00:32:23,430 are of the same temperature. 448 00:32:23,430 --> 00:32:27,590 So this is my initial state. 449 00:32:27,590 --> 00:32:30,720 And then I remove the partition. 450 00:32:30,720 --> 00:32:33,770 And I come up with this situation 451 00:32:33,770 --> 00:32:38,950 where the particles are mixed. 452 00:32:38,950 --> 00:32:44,070 So the particles of type 1 could be either way. 453 00:32:44,070 --> 00:32:48,100 Particles of type 2 could be in either place. 454 00:32:48,100 --> 00:32:55,715 And let's say I have a box of some toxic gas here. 455 00:32:55,715 --> 00:32:57,510 And I remove the lid. 456 00:32:57,510 --> 00:33:00,080 And it will get mixed in the room. 457 00:33:00,080 --> 00:33:03,740 It's certainly an irreversible situation 458 00:33:03,740 --> 00:33:07,075 where is an increase of entropy i associated with that. 459 00:33:07,075 --> 00:33:10,310 And we can calculate that increase of entropy, 460 00:33:10,310 --> 00:33:14,400 because we know what the expression for entropy is. 461 00:33:14,400 --> 00:33:18,060 So what we have to do is to compare the entropy initially. 462 00:33:20,930 --> 00:33:23,315 So this is the initial entropy. 463 00:33:23,315 --> 00:33:26,430 And I calculate everything in units of kB 464 00:33:26,430 --> 00:33:30,610 so I don't have to write kB all over the place. 465 00:33:30,610 --> 00:33:33,940 For particle number one, what do I have? 466 00:33:33,940 --> 00:33:36,030 I have N1 log V1. 467 00:33:40,240 --> 00:33:50,100 And then I have a contribution, which is 3 N 1 over 2. 468 00:33:50,100 --> 00:33:53,970 But I notice that whatever appears here is really 469 00:33:53,970 --> 00:33:59,690 only a function of E over N. E over N 470 00:33:59,690 --> 00:34:02,970 is really only a function of temperature. 471 00:34:02,970 --> 00:34:11,840 So this is something that I can call a sigma of T over here. 472 00:34:11,840 --> 00:34:21,139 And the contribution of box 2 is N2 log V plus 3N 2 over 2. 473 00:34:21,139 --> 00:34:28,179 This-- huh-- let's say that they are-- 474 00:34:28,179 --> 00:34:30,000 we ignore the difference in masses. 475 00:34:30,000 --> 00:34:33,840 You could potentially have here sigma 1, sigma 2. 476 00:34:33,840 --> 00:34:37,330 It really doesn't make any difference. 477 00:34:37,330 --> 00:34:45,780 The final state, what do we have? 478 00:34:45,780 --> 00:34:50,179 Essentially, the one thing that changed 479 00:34:50,179 --> 00:34:53,770 is that the N1 particles now are occupying the box of volume 480 00:34:53,770 --> 00:35:00,420 V. So if call V to the V1 plus V2, what we have 481 00:35:00,420 --> 00:35:10,620 is that we have N1 log of V plus N2 log of V. 482 00:35:10,620 --> 00:35:15,240 My claim is that all of these other factors 483 00:35:15,240 --> 00:35:16,500 really stay the same. 484 00:35:22,920 --> 00:35:28,150 Because essentially what is happening in these expressions 485 00:35:28,150 --> 00:35:35,900 are various ratios of E over N. And by stating that initially I 486 00:35:35,900 --> 00:35:38,480 had the things at the same temperature, 487 00:35:38,480 --> 00:35:44,850 what I had effectively stated was that E1 over N1 488 00:35:44,850 --> 00:35:48,430 is the same thing as E2 over N2. 489 00:35:48,430 --> 00:35:51,640 I guess in the ideal gas case this E over N 490 00:35:51,640 --> 00:35:54,350 is the same thing as 3/2 kT. 491 00:36:01,530 --> 00:36:07,180 But if I have a ratio such as this, 492 00:36:07,180 --> 00:36:14,330 that is also the same as E1 plus E2 divided by N1 plus N2. 493 00:36:14,330 --> 00:36:19,090 This is a simple manipulation of fractions that I can make. 494 00:36:19,090 --> 00:36:26,510 And E1 plus E2 over N1 plus N2, by the same kinds of arguments, 495 00:36:26,510 --> 00:36:29,220 would give me the final temperature. 496 00:36:29,220 --> 00:36:35,060 So what I have to compute is that the final temperature 497 00:36:35,060 --> 00:36:37,210 is the same thing as the initial temperature. 498 00:36:37,210 --> 00:36:40,320 Essentially, in this mixing of the ideal gases, 499 00:36:40,320 --> 00:36:42,880 temperature does not change. 500 00:36:42,880 --> 00:36:48,870 So basically, these factors of sigma 501 00:36:48,870 --> 00:36:51,630 are the same before and after. 502 00:36:51,630 --> 00:36:56,630 And so when we calculate the increase in entropy, Sf 503 00:36:56,630 --> 00:37:00,690 minus Si, really the contribution 504 00:37:00,690 --> 00:37:05,720 that you care about comes from these volume factors. 505 00:37:05,720 --> 00:37:10,620 And really the statement is that in one particle currently 506 00:37:10,620 --> 00:37:13,190 are occupying a volume of size V, 507 00:37:13,190 --> 00:37:15,990 whereas previously they were in V1. 508 00:37:15,990 --> 00:37:17,760 And similarly for the N2 particles. 509 00:37:22,670 --> 00:37:26,550 And if you have more of these particles, more of these boxes, 510 00:37:26,550 --> 00:37:28,980 you could see how the general expression 511 00:37:28,980 --> 00:37:32,550 for the mixing entropy goes. 512 00:37:32,550 --> 00:37:34,930 And so that's fine. 513 00:37:34,930 --> 00:37:38,110 V is certainly greater than V1 or V2. 514 00:37:38,110 --> 00:37:40,940 Each of these logs gives you a positive contribution. 515 00:37:40,940 --> 00:37:43,045 There's an increase in entropy as we expect. 516 00:37:48,080 --> 00:37:52,430 Now, there is the following difficulty however. 517 00:37:57,830 --> 00:38:04,050 What if the gases are identical-- are the same? 518 00:38:11,180 --> 00:38:14,930 We definitely have to do this if I take a box of methane here 519 00:38:14,930 --> 00:38:18,615 and I open it, we all know that something has happened. 520 00:38:18,615 --> 00:38:21,920 There is an irreversible process that has occured. 521 00:38:21,920 --> 00:38:25,020 But if the box-- I have essentially 522 00:38:25,020 --> 00:38:27,310 taken the air in this room, put it in this box, 523 00:38:27,310 --> 00:38:30,632 whether I open the lid or not open the lid, 524 00:38:30,632 --> 00:38:32,710 it doesn't make any difference. 525 00:38:32,710 --> 00:38:35,250 There is no additional work that I 526 00:38:35,250 --> 00:38:38,300 have to do in order to close or open the lid. 527 00:38:38,300 --> 00:38:42,430 Is no there no increase of entropy one way or the other. 528 00:38:42,430 --> 00:38:45,450 Whereas if I look at this expression, 529 00:38:45,450 --> 00:38:48,530 this expression only depends on the final volume 530 00:38:48,530 --> 00:38:50,860 and the initial volumes, and says 531 00:38:50,860 --> 00:38:53,690 that there should an increase in entropy 532 00:38:53,690 --> 00:38:57,650 when we know that there shouldn't be. 533 00:38:57,650 --> 00:39:02,500 And of course, the resolution for that 534 00:39:02,500 --> 00:39:03,780 is something like this. 535 00:39:06,390 --> 00:39:15,280 That if I look at my two boxes-- and I said maybe one of them 536 00:39:15,280 --> 00:39:19,420 is a box that contains methane. 537 00:39:19,420 --> 00:39:22,000 Let's call it A. And the other is 538 00:39:22,000 --> 00:39:27,820 the room that contains the air. 539 00:39:27,820 --> 00:39:31,480 Now this situation where all of the methane 540 00:39:31,480 --> 00:39:35,750 is in the box and the oxygen freely floating in the room 541 00:39:35,750 --> 00:39:39,820 is certainly different from a configuration 542 00:39:39,820 --> 00:39:44,140 where I exchange these two and the methane is here 543 00:39:44,140 --> 00:39:48,148 and the oxygen went into the box. 544 00:39:48,148 --> 00:39:50,340 They're different configurations. 545 00:39:50,340 --> 00:39:54,320 You can definitely tell them apart. 546 00:39:54,320 --> 00:40:01,570 Whereas if I do the same thing, but the box and outside 547 00:40:01,570 --> 00:40:11,710 contain the same entity, and the same entity is, let's say, 548 00:40:11,710 --> 00:40:16,520 oxygen, then how can you tell apart these two configurations? 549 00:40:16,520 --> 00:40:19,740 And so the meaning of-- yes. 550 00:40:19,740 --> 00:40:21,192 AUDIENCE: Are you thinking quantum 551 00:40:21,192 --> 00:40:22,644 mechanically or classically. 552 00:40:22,644 --> 00:40:25,064 Classically we can tell them apart, right? 553 00:40:27,980 --> 00:40:29,820 MEHRAN KARDAR: This is currently I 554 00:40:29,820 --> 00:40:32,810 am making a macroscopic statement. 555 00:40:32,810 --> 00:40:36,410 Now when I get to the distinction of microstates 556 00:40:36,410 --> 00:40:40,590 we have to-- so I was very careful in saying 557 00:40:40,590 --> 00:40:43,210 whether or not you could tell apart 558 00:40:43,210 --> 00:40:45,970 whether it is methane or oxygen. 559 00:40:45,970 --> 00:40:48,960 So this was a very macroscopic statement as to 560 00:40:48,960 --> 00:40:50,570 whether or not you can distinguish 561 00:40:50,570 --> 00:40:53,660 this circumstance versus that circumstance. 562 00:40:53,660 --> 00:40:57,090 So as far as our senses of this macroscopic process 563 00:40:57,090 --> 00:41:02,940 is concerned, these two cases have to be treated differently. 564 00:41:02,940 --> 00:41:08,560 Now, what we have calculated here for these factors 565 00:41:08,560 --> 00:41:12,130 are some volume of phase space. 566 00:41:12,130 --> 00:41:15,190 And where in the evening you might 567 00:41:15,190 --> 00:41:18,790 say that following this procedure 568 00:41:18,790 --> 00:41:23,690 you counted these as two distinct cases. 569 00:41:23,690 --> 00:41:26,730 In this case, these were two distinct cases. 570 00:41:26,730 --> 00:41:30,050 But here, you can't really tell them apart. 571 00:41:30,050 --> 00:41:31,990 So if you can't tell them apart, you 572 00:41:31,990 --> 00:41:34,580 shouldn't call them two distinct cases. 573 00:41:34,580 --> 00:41:39,950 You have over counted phase space by a factor of two here. 574 00:41:39,950 --> 00:41:43,620 And here, I just looked at two particles. 575 00:41:43,620 --> 00:41:47,080 If I have N particles, I have over 576 00:41:47,080 --> 00:41:50,980 counted the phase space of identical particles 577 00:41:50,980 --> 00:41:56,110 by all possible permutations of n objects, it is n factorial. 578 00:41:56,110 --> 00:42:03,680 So there is an over counting of phase space 579 00:42:03,680 --> 00:42:19,020 or configurations of N identical particles 580 00:42:19,020 --> 00:42:24,300 by a factor of N factorial. 581 00:42:31,210 --> 00:42:38,390 I.e., when we said that particle number one can be anywhere 582 00:42:38,390 --> 00:42:42,060 in the box, particle number two can be anywhere 583 00:42:42,060 --> 00:42:46,250 in the box, all the way to particle number n, 584 00:42:46,250 --> 00:42:50,370 well, in fact, I can't tell which each is which. 585 00:42:50,370 --> 00:42:53,060 If I can't tell which particle is which, 586 00:42:53,060 --> 00:42:57,195 I have to divide by the number of permutations and factors. 587 00:42:59,950 --> 00:43:02,105 Now, as somebody was asking the question, 588 00:43:02,105 --> 00:43:06,950 as you were asking the question, classically, 589 00:43:06,950 --> 00:43:11,240 if I write a computer program that 590 00:43:11,240 --> 00:43:14,620 looks at the trajectories of N particles 591 00:43:14,620 --> 00:43:18,780 in the gas in this room, classically, your computer 592 00:43:18,780 --> 00:43:22,670 would always know the particle that started over here 593 00:43:22,670 --> 00:43:24,430 after many collisions or whatever 594 00:43:24,430 --> 00:43:28,960 is the particle that ended up somewhere else. 595 00:43:28,960 --> 00:43:33,010 So if you ask the computer, the computer 596 00:43:33,010 --> 00:43:37,600 can certainly distinguish these classical trajectories. 597 00:43:37,600 --> 00:43:42,180 And then it is kind of strange to say that, well, I 598 00:43:42,180 --> 00:43:44,860 have to divide by N factorial because all of these 599 00:43:44,860 --> 00:43:45,760 are identical. 600 00:43:45,760 --> 00:43:48,950 Again, classically these particles 601 00:43:48,950 --> 00:43:51,070 are following specific trajectories. 602 00:43:51,070 --> 00:43:54,170 And you know where in phase space they are. 603 00:43:54,170 --> 00:43:58,530 Whereas quantum mechanically, you can't tell that apart. 604 00:43:58,530 --> 00:44:02,670 So quantum mechanically, as we will describe later, rather 605 00:44:02,670 --> 00:44:04,820 than classical statistical mechanics-- when 606 00:44:04,820 --> 00:44:07,880 we do quantum statistical mechanics-- if you have 607 00:44:07,880 --> 00:44:10,580 identical particles, you have to write down 608 00:44:10,580 --> 00:44:13,820 of a wave function that is either symmetric or 609 00:44:13,820 --> 00:44:16,960 anti-symmetric under the exchange of particles. 610 00:44:16,960 --> 00:44:19,500 And when we do eventually the calculations 611 00:44:19,500 --> 00:44:23,070 for these factors of 1 over N factorial 612 00:44:23,070 --> 00:44:26,070 will emerge very naturally. 613 00:44:26,070 --> 00:44:31,070 So I think different people have different perspectives. 614 00:44:31,070 --> 00:44:34,540 My own perspective is that this factor really 615 00:44:34,540 --> 00:44:39,130 is due to the quantum origin of identity. 616 00:44:39,130 --> 00:44:42,290 And classically, you have to sort of fudge it and put it 617 00:44:42,290 --> 00:44:43,670 over there. 618 00:44:43,670 --> 00:44:46,710 But some people say that really it's a matter of measurements. 619 00:44:46,710 --> 00:44:50,090 And if you can't really tell A and B sufficiently apart, 620 00:44:50,090 --> 00:44:51,800 then you don't know. 621 00:44:51,800 --> 00:44:53,890 I always go back to the computer. 622 00:44:53,890 --> 00:44:56,260 And say, well, the computer can tell. 623 00:44:56,260 --> 00:44:58,990 But it's kind of immaterial at this stage. 624 00:44:58,990 --> 00:45:03,390 It's obvious that for all practical purposes for things 625 00:45:03,390 --> 00:45:08,040 that are identical you have to divide by this factor. 626 00:45:08,040 --> 00:45:12,210 So what happens if you divide by that factor? 627 00:45:12,210 --> 00:45:17,180 So I have changed all of my calculations now. 628 00:45:17,180 --> 00:45:24,040 So when I do the log of-- previously I had V to the N. 629 00:45:24,040 --> 00:45:29,300 And it gave me N log V. Now, I have log of V 630 00:45:29,300 --> 00:45:32,550 to the N divided by N factorial. 631 00:45:32,550 --> 00:45:36,780 So I will get my Sterling's approximation additional factor 632 00:45:36,780 --> 00:45:42,300 of minus N log N plus N, which I can sort of absorb here 633 00:45:42,300 --> 00:45:43,110 in this fashion. 634 00:45:48,825 --> 00:45:53,450 Now you say, well, having done that, you have to first 635 00:45:53,450 --> 00:45:57,610 of all show me that you fixed the case of this change 636 00:45:57,610 --> 00:46:00,960 in entropy for identical particles, 637 00:46:00,960 --> 00:46:04,380 but also you should show me that the previous case where 638 00:46:04,380 --> 00:46:07,570 we know there has to be an increase in entropy just 639 00:46:07,570 --> 00:46:12,640 because of the gas being different that that is not 640 00:46:12,640 --> 00:46:15,530 changed because of this modification that you make. 641 00:46:15,530 --> 00:46:17,280 So let's check that. 642 00:46:17,280 --> 00:46:25,490 So for distinct gases, what would be the generalization 643 00:46:25,490 --> 00:46:29,648 of this form Sf minus Si divided by kV? 644 00:46:35,310 --> 00:46:38,820 Well, what happens here? 645 00:46:38,820 --> 00:46:51,000 In the case of the final object, I have to divide N1 log of V. 646 00:46:51,000 --> 00:46:58,640 But that V really becomes V divided by N1, because 647 00:46:58,640 --> 00:47:02,920 in the volume of size V, I have N1 oxygen 648 00:47:02,920 --> 00:47:04,630 that I can't tell apart. 649 00:47:04,630 --> 00:47:08,630 So I divide by the N1 factorial for the oxygens. 650 00:47:08,630 --> 00:47:13,020 And then I have N2 methanes that I can't tell apart 651 00:47:13,020 --> 00:47:18,560 in that volume, so I divide by essentially N2 factorial that 652 00:47:18,560 --> 00:47:20,840 goes over there. 653 00:47:20,840 --> 00:47:30,450 The initial change is over here I would have N1 log of V1 654 00:47:30,450 --> 00:47:32,440 over N1. 655 00:47:32,440 --> 00:47:41,980 And here I would have had N2 log of V2 over N2. 656 00:47:41,980 --> 00:47:46,100 So every one of these expressions that was previously 657 00:47:46,100 --> 00:47:51,870 log V, and I had four of them, gets changed. 658 00:47:51,870 --> 00:47:57,150 But they get change precisely in a manner that this N1 log of N1 659 00:47:57,150 --> 00:48:00,530 here cancels this N1 log of and N1 here. 660 00:48:00,530 --> 00:48:05,140 This N2 log of N2 here cancels this N2 log of N2 here. 661 00:48:05,140 --> 00:48:08,320 So the delta S that I get is precisely the same thing 662 00:48:08,320 --> 00:48:09,960 as I had before. 663 00:48:09,960 --> 00:48:18,230 I will get N1 log of V over V1 plus N2 log of V over V2. 664 00:48:21,230 --> 00:48:27,240 So this division, because the oxygens were 665 00:48:27,240 --> 00:48:29,370 identical to themselves and methanes 666 00:48:29,370 --> 00:48:31,670 were identical to themselves, does not 667 00:48:31,670 --> 00:48:37,110 change the mixing entropy of oxygen and nitrogen. 668 00:48:37,110 --> 00:48:40,520 But let's say that both gases are the same. 669 00:48:40,520 --> 00:48:42,380 They're both oxygen. 670 00:48:42,380 --> 00:48:43,130 Then what happens? 671 00:48:50,890 --> 00:48:55,500 Now, in the final state, I have a box. 672 00:48:55,500 --> 00:48:59,110 It has a N1 plus N2 particles that are all oxygen. 673 00:48:59,110 --> 00:49:01,370 I can't tell them apart. 674 00:49:01,370 --> 00:49:07,140 So the contribution from the phase space 675 00:49:07,140 --> 00:49:14,450 would be N1 plus N2 log of the volume divided by N1 plus N2 676 00:49:14,450 --> 00:49:15,240 factorial. 677 00:49:15,240 --> 00:49:18,460 That ultimately will give me a factor of N1 plus N2 here. 678 00:49:22,550 --> 00:49:24,800 The initial entropy is exactly the one 679 00:49:24,800 --> 00:49:26,706 that I calculated before. 680 00:49:26,706 --> 00:49:32,040 For the line above, I have N1 log of V1 681 00:49:32,040 --> 00:49:37,157 over N1 minus N2 log of V2 over N2. 682 00:49:45,440 --> 00:49:48,840 Now certainly, I still expect to see some mixing entropy 683 00:49:48,840 --> 00:49:53,870 if I have a box of oxygen that is at very low pressure 684 00:49:53,870 --> 00:49:56,340 and is very dilute, and I open it 685 00:49:56,340 --> 00:49:59,185 into this room, which is at much higher pressure 686 00:49:59,185 --> 00:50:01,600 and is much more dense. 687 00:50:01,600 --> 00:50:03,520 So really, the case where I don't 688 00:50:03,520 --> 00:50:06,210 expect to see any change in entropy 689 00:50:06,210 --> 00:50:09,505 is when the two boxes have the same density. 690 00:50:12,680 --> 00:50:14,860 And hence, when I mix them, I would also 691 00:50:14,860 --> 00:50:16,255 have exactly the same density. 692 00:50:21,170 --> 00:50:26,090 And you can see that, therefore, all of these factors that 693 00:50:26,090 --> 00:50:30,370 are in the log are of the inverse of the same density. 694 00:50:30,370 --> 00:50:33,320 And there's N1 plus N2 of them that's positive. 695 00:50:33,320 --> 00:50:36,100 And N1 plus N2 of them that is negative. 696 00:50:36,100 --> 00:50:38,830 So the answer in this case, as long 697 00:50:38,830 --> 00:50:44,020 as I try to mix identical particles of the same density, 698 00:50:44,020 --> 00:50:46,340 if I include this correction to the phase 699 00:50:46,340 --> 00:50:48,586 space of identical particles, the answer 700 00:50:48,586 --> 00:50:54,438 will be [? 0. ?] Yes? 701 00:50:54,438 --> 00:50:58,422 AUDIENCE: Question, [INAUDIBLE] in terms 702 00:50:58,422 --> 00:51:04,398 of the revolution of the [INAUDIBLE] 703 00:51:04,398 --> 00:51:09,400 there is no transition [INAUDIBLE] 704 00:51:09,400 --> 00:51:13,388 so that your temporary, and say like, oxygen and nitrogen can 705 00:51:13,388 --> 00:51:15,778 catch a molecule, put it in a [? aspertometer. ?] 706 00:51:15,778 --> 00:51:18,168 and have different isotopes. 707 00:51:18,168 --> 00:51:20,668 You can take like closed isotopes of oxygen 708 00:51:20,668 --> 00:51:22,604 and still tell them apart. 709 00:51:22,604 --> 00:51:25,992 But this is like their continuous way 710 00:51:25,992 --> 00:51:28,684 of choosing a pair of gases which 711 00:51:28,684 --> 00:51:31,980 would be arbitrarily closed in atomic mass. 712 00:51:31,980 --> 00:51:34,500 MEHRAN KARDAR: So, as I said, there 713 00:51:34,500 --> 00:51:38,030 are alternative explanations that I've heard. 714 00:51:38,030 --> 00:51:40,530 And that's precisely one of them. 715 00:51:40,530 --> 00:51:44,620 And my counter is that what we are putting here 716 00:51:44,620 --> 00:51:47,670 is the volume of phase space. 717 00:51:47,670 --> 00:51:52,180 And to me that has a very specific meaning. 718 00:51:52,180 --> 00:51:55,120 That is there's a set of coordinates and momenta 719 00:51:55,120 --> 00:51:58,960 that are moving according to Hamiltonian trajectories. 720 00:51:58,960 --> 00:52:03,690 And in principle, there is a computer nature 721 00:52:03,690 --> 00:52:06,060 that is following these trajectories, 722 00:52:06,060 --> 00:52:08,800 or I can actually put them on the computer. 723 00:52:08,800 --> 00:52:11,920 And then no matter how long I run 724 00:52:11,920 --> 00:52:14,980 and they're identical oxygen molecules, 725 00:52:14,980 --> 00:52:18,010 I start with number one here, numbers two here. 726 00:52:18,010 --> 00:52:21,420 The computer will say that this is the trajectory of number one 727 00:52:21,420 --> 00:52:23,760 and this is the trajectory of numbers two. 728 00:52:23,760 --> 00:52:28,030 So unless I change my definition of phase space 729 00:52:28,030 --> 00:52:32,710 and how I am calculating things, I run into this paradox. 730 00:52:32,710 --> 00:52:36,050 So what you're saying is forget about that. 731 00:52:36,050 --> 00:52:39,720 It's just can tell isotopes apart or something like that. 732 00:52:39,720 --> 00:52:41,140 And I'm saying that that's fine. 733 00:52:41,140 --> 00:52:43,480 That's perspective, but it has nothing 734 00:52:43,480 --> 00:52:45,091 to do with phase space counting. 735 00:52:51,300 --> 00:52:51,800 OK? 736 00:52:54,560 --> 00:52:57,100 Fine, now, why didn't I calculate this? 737 00:52:57,100 --> 00:53:02,220 It was also for the same reason, because we 738 00:53:02,220 --> 00:53:07,460 expect to have quantities that are extensive 739 00:53:07,460 --> 00:53:11,362 and quantities that are intensive. 740 00:53:11,362 --> 00:53:13,950 And therefore, if I were to, for example, 741 00:53:13,950 --> 00:53:16,630 calculate this object, that it should be something 742 00:53:16,630 --> 00:53:18,672 that is intensive. 743 00:53:18,672 --> 00:53:22,420 Now the problem is that if I take a derivative with respect 744 00:53:22,420 --> 00:53:26,470 N, I have log V. And log V is clearly 745 00:53:26,470 --> 00:53:30,590 something that does not grow proportionately to size 746 00:53:30,590 --> 00:53:33,890 but grows proportionately to size logarithmically. 747 00:53:33,890 --> 00:53:36,640 So if I make volume twice as big, 748 00:53:36,640 --> 00:53:40,590 I will get an additional factor of log 2 here contribution 749 00:53:40,590 --> 00:53:42,470 to the chemical potential. 750 00:53:42,470 --> 00:53:45,500 And that does not make sense. 751 00:53:45,500 --> 00:53:52,520 But when I do this identity, then this V becomes V over N. 752 00:53:52,520 --> 00:53:56,270 And then everything becomes nicely intensive. 753 00:53:56,270 --> 00:54:02,520 So if I allowed now to replace this V over N, 754 00:54:02,520 --> 00:54:14,774 then I can calculate V over T as dS by dN at constant E and V. 755 00:54:14,774 --> 00:54:17,470 And so then essentially I will get 756 00:54:17,470 --> 00:54:22,880 to drop the factor of log N that comes in front, so I will get 757 00:54:22,880 --> 00:54:28,840 kT log of V over N. And then I would 758 00:54:28,840 --> 00:54:40,060 have 3/2 log of something, which I can put together as 4 pi N 759 00:54:40,060 --> 00:54:45,030 E over 3N raised to the 3/2 power. 760 00:54:49,730 --> 00:54:53,980 And you can see that there were these E's 761 00:54:53,980 --> 00:54:57,500 from Sterling's approximation up there that 762 00:54:57,500 --> 00:55:01,265 got dropped here, because you can also take derivative 763 00:55:01,265 --> 00:55:04,260 with respect to the N's that are inside. 764 00:55:04,260 --> 00:55:07,310 And you can check that the function of derivatives 765 00:55:07,310 --> 00:55:09,320 with respects to the N's that are inside 766 00:55:09,320 --> 00:55:11,415 is precisely to get rid of those factors. 767 00:55:15,841 --> 00:55:16,340 OK? 768 00:55:23,850 --> 00:55:26,610 Now, there is still one other thing 769 00:55:26,610 --> 00:55:30,950 that is not wrong, but kind of like jarring 770 00:55:30,950 --> 00:55:32,430 about the expressions that they've 771 00:55:32,430 --> 00:55:39,280 had so far in that right from the beginning, 772 00:55:39,280 --> 00:55:43,510 I said that you can certainly calculate entropies out 773 00:55:43,510 --> 00:55:49,710 of probabilities as minus log of P average if you like. 774 00:55:49,710 --> 00:55:51,610 But it makes sense only if you're 775 00:55:51,610 --> 00:55:55,170 dealing with discrete variables, because when you're 776 00:55:55,170 --> 00:55:57,220 dealing with continual variables and you 777 00:55:57,220 --> 00:55:59,870 have a probability density. 778 00:55:59,870 --> 00:56:01,590 And the probability density depends 779 00:56:01,590 --> 00:56:03,490 on the units of measurement. 780 00:56:03,490 --> 00:56:06,240 And if you were to change measurement from meters 781 00:56:06,240 --> 00:56:08,580 to centimeters or something else, 782 00:56:08,580 --> 00:56:13,460 then there will be changes in the probability densities, 783 00:56:13,460 --> 00:56:17,750 which would then modify the various factors over here. 784 00:56:17,750 --> 00:56:22,080 And that's really also is reflected ultimately 785 00:56:22,080 --> 00:56:24,720 in the fact that these combinations of terms 786 00:56:24,720 --> 00:56:28,890 that I have written here have dimensions. 787 00:56:28,890 --> 00:56:32,250 And it is kind of, again, jarring 788 00:56:32,250 --> 00:56:36,450 to have expressions inside the logarithm or in the exponential 789 00:56:36,450 --> 00:56:39,550 that our not dimensionless. 790 00:56:39,550 --> 00:56:43,480 So it would be good if we had some way 791 00:56:43,480 --> 00:56:46,500 of making all of these dimensionless. 792 00:56:46,500 --> 00:56:50,570 And you say, well, really the origin of it 793 00:56:50,570 --> 00:56:53,430 is all the way back here, when I was 794 00:56:53,430 --> 00:56:56,050 calculating volumes in phase space. 795 00:56:56,050 --> 00:56:59,060 And volumes in phase space have dimensions. 796 00:56:59,060 --> 00:57:03,560 And that dimensions of pq raised to the 3N power 797 00:57:03,560 --> 00:57:07,240 really survives all the way down here. 798 00:57:07,240 --> 00:57:12,150 So I can say, OK, I choose some quantity 799 00:57:12,150 --> 00:57:15,775 as a reference that has the right dimensions of the product 800 00:57:15,775 --> 00:57:18,690 of p and q, which is an action. 801 00:57:18,690 --> 00:57:22,790 And I divide all of my measurements 802 00:57:22,790 --> 00:57:28,140 by that reference unit, so that, for example, here I 803 00:57:28,140 --> 00:57:32,130 have 3N factors of this. 804 00:57:32,130 --> 00:57:34,120 Or let's say each one of them is 3. 805 00:57:34,120 --> 00:57:39,720 I divide by some quantity that has units of action . 806 00:57:39,720 --> 00:57:41,680 And then I will be set. 807 00:57:41,680 --> 00:57:47,008 So basically, the units of this h is the product of p and q. 808 00:57:51,160 --> 00:57:56,050 Now, at this point we have no way 809 00:57:56,050 --> 00:57:59,330 of choosing some h as opposed to another h. 810 00:57:59,330 --> 00:58:04,070 And so by adding that factor, we can make things look nicer. 811 00:58:04,070 --> 00:58:08,610 But then things are undefined after this factor of h. 812 00:58:08,610 --> 00:58:11,150 When we do quantum mechanics, another thing 813 00:58:11,150 --> 00:58:14,640 that quantum mechanics does is to provide us 814 00:58:14,640 --> 00:58:18,670 with precisely [? age ?] of Planck's constant as a measure 815 00:58:18,670 --> 00:58:21,900 of these kinds of integrations. 816 00:58:21,900 --> 00:58:25,390 So when we eventually you go to calculate, say, 817 00:58:25,390 --> 00:58:28,900 the ideal gas or any other mechanic system 818 00:58:28,900 --> 00:58:32,180 that involves p and q in quantum mechanics, 819 00:58:32,180 --> 00:58:34,400 then the phase space becomes discretized. 820 00:58:34,400 --> 00:58:39,030 You would have-- The appropriate description 821 00:58:39,030 --> 00:58:42,730 would have energies that are discretized corresponding 822 00:58:42,730 --> 00:58:46,650 to various other discretization that are eventually 823 00:58:46,650 --> 00:58:52,670 the equivalent to dividing by this Planck's constant. 824 00:58:52,670 --> 00:58:56,400 Ultimately, I will have additionally 825 00:58:56,400 --> 00:59:01,600 a factor of h squared appearing here. 826 00:59:01,600 --> 00:59:05,920 And it will make everything nicely that much [? less. ?] 827 00:59:05,920 --> 00:59:16,700 None of these other quantities that I mentioned calculated 828 00:59:16,700 --> 00:59:18,572 would be affected by this. 829 00:59:28,990 --> 00:59:31,250 So essentially, what I'm saying is 830 00:59:31,250 --> 00:59:38,340 that you are going to use a measure for phase 831 00:59:38,340 --> 00:59:43,960 space of identical particles. 832 00:59:51,310 --> 00:59:59,270 Previously we had a product, d cubed Pi, d cubed Qi. 833 00:59:59,270 --> 01:00:02,430 This is what we were integrating and requiring 834 01:00:02,430 --> 01:00:06,540 that this integration will give us [INAUDIBLE]. 835 01:00:06,540 --> 01:00:11,950 Now, we will change this to divide by this N factorial, 836 01:00:11,950 --> 01:00:14,920 if the particles are identical. 837 01:00:14,920 --> 01:00:21,560 And we divide by h to the 3N because 838 01:00:21,560 --> 01:00:25,386 of the number of pairs of pq that appear in this. 839 01:00:30,850 --> 01:00:35,020 The justification to come when we ultimately do quantum study. 840 01:00:40,597 --> 01:00:41,180 Any questions? 841 01:00:59,640 --> 01:01:04,780 So I said that this prescription when 842 01:01:04,780 --> 01:01:08,480 we look at a system at complete isolation, 843 01:01:08,480 --> 01:01:12,360 and therefore, specify fully its energy 844 01:01:12,360 --> 01:01:16,300 is the microcanonical ensemble, as opposed 845 01:01:16,300 --> 01:01:27,440 to the canonical ensemble, whereas the set 846 01:01:27,440 --> 01:01:32,145 of microscopic parameters that you identified 847 01:01:32,145 --> 01:01:39,400 with your system, you replace the energy with temperature. 848 01:01:39,400 --> 01:01:42,630 So in general, let's say there will 849 01:01:42,630 --> 01:01:48,590 be some bunch of displacements, x, that give you 850 01:01:48,590 --> 01:01:51,490 the work content to the system. 851 01:01:51,490 --> 01:01:54,550 Just like we fixed over there the volume and the number 852 01:01:54,550 --> 01:02:00,420 of particles, let's say that all of the work parameters, 853 01:02:00,420 --> 01:02:03,130 such as x microscopically, we will 854 01:02:03,130 --> 01:02:06,130 fix in this canonical ensemble. 855 01:02:06,130 --> 01:02:09,040 So however, the ensemble is one in which 856 01:02:09,040 --> 01:02:12,640 the energy is not specified. 857 01:02:12,640 --> 01:02:15,550 And so how do I imagine that I can 858 01:02:15,550 --> 01:02:18,120 maintain a system at temperature T? 859 01:02:18,120 --> 01:02:21,740 Well, if this room is at some particular temperature, 860 01:02:21,740 --> 01:02:25,150 I assume that smaller objects that I put in this room 861 01:02:25,150 --> 01:02:27,480 will come to the same temperature. 862 01:02:27,480 --> 01:02:31,260 So the general prescription for beginning something other 863 01:02:31,260 --> 01:02:34,860 than temperature T is to put it in contact 864 01:02:34,860 --> 01:02:37,550 with something that is much bigger. 865 01:02:37,550 --> 01:02:39,390 So let's call this to be a reservoir. 866 01:02:43,330 --> 01:02:48,420 And we put our system, which we assume to be smaller, 867 01:02:48,420 --> 01:02:49,580 in contact with it. 868 01:02:49,580 --> 01:02:53,810 And we allow it to exchange heat with the reservoir. 869 01:02:58,570 --> 01:03:04,600 Now I did this way of managing the system 870 01:03:04,600 --> 01:03:07,750 to come to a temperature T, which 871 01:03:07,750 --> 01:03:09,990 is the characteristic of a big reservoir. 872 01:03:09,990 --> 01:03:11,980 Imagine that you have a lake. 873 01:03:11,980 --> 01:03:15,390 And you put your gas or something else inside the lake. 874 01:03:15,390 --> 01:03:21,140 And it will equilibrate to the temperature of the lake. 875 01:03:21,140 --> 01:03:23,900 I will assume that the two of them, 876 01:03:23,900 --> 01:03:26,610 the system and the reservoir, just 877 01:03:26,610 --> 01:03:30,450 for the purpose of my being able to do some computation, 878 01:03:30,450 --> 01:03:33,230 are isolated from the rest of the universe, 879 01:03:33,230 --> 01:03:38,750 so that the system plus reservoir is microcanonical. 880 01:03:44,400 --> 01:03:52,620 And the sum total of their energies is sum E total. 881 01:03:59,320 --> 01:04:05,120 So now this system still is something like a gas. 882 01:04:05,120 --> 01:04:08,970 It's a has a huge number of potential degrees of freedom. 883 01:04:08,970 --> 01:04:12,270 And these potential number of degrees of freedom 884 01:04:12,270 --> 01:04:16,020 can be captured through the microstate of the system, 885 01:04:16,020 --> 01:04:17,660 u sub s. 886 01:04:17,660 --> 01:04:21,390 And similarly, the water particle in the lake 887 01:04:21,390 --> 01:04:22,600 have their own state. 888 01:04:22,600 --> 01:04:25,330 An there's some microstate that describes 889 01:04:25,330 --> 01:04:28,880 the positions and the momenta of all of the particles 890 01:04:28,880 --> 01:04:32,150 that are in the lake. 891 01:04:32,150 --> 01:04:32,940 Yes? 892 01:04:32,940 --> 01:04:35,400 AUDIENCE: When you're writing the set of particles used 893 01:04:35,400 --> 01:04:37,750 to describe it, why don't you write N? 894 01:04:37,750 --> 01:04:39,874 Since it said the number of particles in the system 895 01:04:39,874 --> 01:04:40,560 is not fixed. 896 01:04:40,560 --> 01:04:43,800 MEHRAN KARDAR: Yes, so I did want to [INAUDIBLE] 897 01:04:43,800 --> 01:04:48,660 but in principle I could add N. I wanted to be kind of general. 898 01:04:48,660 --> 01:04:51,800 If you like X, [? it ?] is allowed 899 01:04:51,800 --> 01:05:04,210 to include chemical work type of an X. 900 01:05:04,210 --> 01:05:09,020 So what do I know? 901 01:05:09,020 --> 01:05:14,940 I know that there is also if I want to describe microstates 902 01:05:14,940 --> 01:05:17,260 and their revolution, I need to specify 903 01:05:17,260 --> 01:05:21,260 that there's a Hamiltonian that governs 904 01:05:21,260 --> 01:05:23,780 the evolution of these microstates. 905 01:05:23,780 --> 01:05:25,930 And presumably there's a Hamiltonian 906 01:05:25,930 --> 01:05:31,610 that describes the evolution of the reservoir microstate. 907 01:05:31,610 --> 01:05:35,910 And so presumably the allowed microstates 908 01:05:35,910 --> 01:05:40,180 are ones in which E total is made up 909 01:05:40,180 --> 01:05:44,090 of the energy of the system plus the energy of the reservoir. 910 01:05:53,350 --> 01:06:04,300 So because the whole thing is the microcanonical, 911 01:06:04,300 --> 01:06:07,810 I can assign a probability, a joint probability, 912 01:06:07,810 --> 01:06:15,070 to finding some particular mu s, mu r combination, 913 01:06:15,070 --> 01:06:18,040 just like we were doing over there. 914 01:06:18,040 --> 01:06:21,180 You would say that essentially this 915 01:06:21,180 --> 01:06:25,120 is a combination of these 1s and 0s. 916 01:06:25,120 --> 01:06:30,940 So it is 0 if H of-- again, for simplicity I 917 01:06:30,940 --> 01:06:38,030 drop the s on the system-- H of mu s plus H 918 01:06:38,030 --> 01:06:43,660 reservoir of mu reservoir is not equal to E total. 919 01:06:43,660 --> 01:06:52,595 And it is 1 over some omega of reservoir in the system 920 01:06:52,595 --> 01:06:53,095 otherwise. 921 01:06:58,980 --> 01:07:02,010 So this is just again, throwing the dice, 922 01:07:02,010 --> 01:07:05,860 saying that it has so many possible configurations, given 923 01:07:05,860 --> 01:07:08,210 that I know what the total energy is. 924 01:07:08,210 --> 01:07:13,290 All the ones that are consistent with that are allowed. 925 01:07:13,290 --> 01:07:17,150 Which is to say, I don't really care about the lake. 926 01:07:17,150 --> 01:07:21,720 All I care is about to the states of my gas, 927 01:07:21,720 --> 01:07:24,550 and say, OK, no problem, if I have the joint probability 928 01:07:24,550 --> 01:07:27,950 distribution just like I did over here, 929 01:07:27,950 --> 01:07:29,980 I get rid of all of the degrees of freedom 930 01:07:29,980 --> 01:07:32,190 that I'm not interested in. 931 01:07:32,190 --> 01:07:37,370 So if I'm interested only in the states of the system, 932 01:07:37,370 --> 01:07:41,680 I sum over or integrate over-- so this would be a sum. 933 01:07:41,680 --> 01:07:44,200 This would be an integration, whatever-- 934 01:07:44,200 --> 01:07:46,085 of the joint probability distribution. 935 01:07:50,230 --> 01:07:54,330 Now actually follow the steps that I had over here 936 01:07:54,330 --> 01:07:58,040 when we were looking at the momentum of a gas particle. 937 01:07:58,040 --> 01:08:00,940 I say that what I have over here, 938 01:08:00,940 --> 01:08:04,165 this probability see is this 1 over omega R, 939 01:08:04,165 --> 01:08:09,110 S. This is a function that is either 1 or 0. 940 01:08:09,110 --> 01:08:13,660 And then I have so sum over all configurations 941 01:08:13,660 --> 01:08:16,120 of the reservoir. 942 01:08:16,120 --> 01:08:23,050 But given that I said what the microstate of the system 943 01:08:23,050 --> 01:08:30,735 is, then I know that the reservoir has to take energy 944 01:08:30,735 --> 01:08:35,149 in total minus the amount the microstates has taken. 945 01:08:35,149 --> 01:08:39,040 And I'm summing over all of the microstates that 946 01:08:39,040 --> 01:08:41,830 are consistent with the requirement 947 01:08:41,830 --> 01:08:44,569 that the energy in the reservoir is 948 01:08:44,569 --> 01:08:48,050 E total minus H of microstate. 949 01:08:48,050 --> 01:08:50,910 So what that is that the omega that I 950 01:08:50,910 --> 01:08:54,390 have for the reservoir-- and I don't know what it is, 951 01:08:54,390 --> 01:08:59,010 but whatever it is, evaluated at the total energy 952 01:08:59,010 --> 01:09:03,250 minus the energy that is taken outside the microstate. 953 01:09:03,250 --> 01:09:07,390 So again, exactly the reason why this became E minus Pi 954 01:09:07,390 --> 01:09:08,670 squared over 2. 955 01:09:08,670 --> 01:09:14,000 This becomes E total minus H of mu S. Except 956 01:09:14,000 --> 01:09:16,640 that I don't know either what this is or what this is. 957 01:09:20,200 --> 01:09:22,790 Actually, I don't really even care about this 958 01:09:22,790 --> 01:09:26,695 because all of the H dependents on microstate dependents 959 01:09:26,695 --> 01:09:28,359 is in the numerator. 960 01:09:28,359 --> 01:09:31,890 So I write that as proportional to exponential. 961 01:09:31,890 --> 01:09:36,060 And the log of omega is the entropy. 962 01:09:36,060 --> 01:09:40,470 So I have the entropy of the [INAUDIBLE] in units of kB, 963 01:09:40,470 --> 01:09:54,980 evaluated at the argument that this E total minus H of mu S. 964 01:09:54,980 --> 01:09:57,370 So my statement is that when I look 965 01:09:57,370 --> 01:10:02,030 at the entropy of the reservoir as a function of E 966 01:10:02,030 --> 01:10:06,650 total minus the energy that is taken out 967 01:10:06,650 --> 01:10:11,490 by the system, my construction I assume 968 01:10:11,490 --> 01:10:16,440 that I'm putting a small volume of gas in contact 969 01:10:16,440 --> 01:10:19,150 with a huge lake. 970 01:10:19,150 --> 01:10:24,520 So this total energy is overwhelmingly larger 971 01:10:24,520 --> 01:10:28,230 than the amount of energy that the system can occupy. 972 01:10:28,230 --> 01:10:32,530 So I can make a Taylor expansion of this quantity 973 01:10:32,530 --> 01:10:39,920 and say that this is S R of E total minus the derivative of S 974 01:10:39,920 --> 01:10:41,290 with respect to its energy. 975 01:10:41,290 --> 01:10:44,310 So the derivative of the S reservoir 976 01:10:44,310 --> 01:10:47,990 with respect to the energy of the reservoir times 977 01:10:47,990 --> 01:10:54,010 H of the microstate and presumably higher order terms 978 01:10:54,010 --> 01:10:55,090 that are negligible. 979 01:10:58,340 --> 01:11:02,090 Now the next thing that is important about the reservoir 980 01:11:02,090 --> 01:11:04,150 is you have this huge lake. 981 01:11:04,150 --> 01:11:07,850 Let's say it's exactly at some temperature of 30 degrees. 982 01:11:07,850 --> 01:11:11,030 And you take some small amount of energy from it 983 01:11:11,030 --> 01:11:12,610 to put in the system. 984 01:11:12,610 --> 01:11:14,910 The temperature of the lake should not change. 985 01:11:14,910 --> 01:11:17,590 So that's the definition of the reservoir. 986 01:11:17,590 --> 01:11:22,930 It's a system that is so big that for the range of energies 987 01:11:22,930 --> 01:11:26,420 that we are considering, this S by dE 988 01:11:26,420 --> 01:11:30,706 is 1 over the temperature that characterizes the reservoir. 989 01:11:34,600 --> 01:11:38,070 So just like here, but eventually 990 01:11:38,070 --> 01:11:41,520 the answer that we got was something 991 01:11:41,520 --> 01:11:44,830 like the energy of the particle divided by kT. 992 01:11:44,830 --> 01:11:48,940 Once I exponentiate, I find that the probability 993 01:11:48,940 --> 01:11:54,370 to find the system in some microstate is proportional to E 994 01:11:54,370 --> 01:12:01,250 to the minus of the energy of that microstate divided by kT. 995 01:12:04,330 --> 01:12:06,570 And of course, there's a bunch of other things 996 01:12:06,570 --> 01:12:11,206 that I have to eventually put into a normalization that 997 01:12:11,206 --> 01:12:11,705 will cause. 998 01:12:15,390 --> 01:12:21,310 So in the canonical prescription you sort of replace 999 01:12:21,310 --> 01:12:23,550 this throwing of the dice and saying 1000 01:12:23,550 --> 01:12:26,630 that everything is equivalent to saying that well, 1001 01:12:26,630 --> 01:12:30,750 each microstates can have some particular energy. 1002 01:12:30,750 --> 01:12:34,800 And the probabilities are partitioned according 1003 01:12:34,800 --> 01:12:39,480 to the Boltzmann weights of these energies. 1004 01:12:39,480 --> 01:12:43,650 And clearly this quantity, Z, the normalization 1005 01:12:43,650 --> 01:12:48,630 is obtained by integrating over the entire space 1006 01:12:48,630 --> 01:12:50,870 of microstates, or summing over them 1007 01:12:50,870 --> 01:12:54,865 if they are discrete of this factor of E to the minus beta 1008 01:12:54,865 --> 01:13:01,370 H of mu S. And we'll use this notation beta 1 over kT 1009 01:13:01,370 --> 01:13:02,837 sometimes for simplicity. 1010 01:13:28,200 --> 01:13:33,660 Now, the thing is that thermodynamically, we 1011 01:13:33,660 --> 01:13:38,520 said that you can choose any set of parameters, 1012 01:13:38,520 --> 01:13:42,070 as long as they are independent, to describe 1013 01:13:42,070 --> 01:13:45,510 the macroscopic equilibrium state of the system. 1014 01:13:45,510 --> 01:13:50,670 So what we did in the microcanonical ensemble is 1015 01:13:50,670 --> 01:13:54,330 we specified a number of things, such as energy. 1016 01:13:54,330 --> 01:13:56,805 And we derived the other things, such as temperature. 1017 01:13:59,360 --> 01:14:03,750 So here, in the canonical ensemble, 1018 01:14:03,750 --> 01:14:07,349 we have stated what the temperature of the system is. 1019 01:14:07,349 --> 01:14:08,390 Well, then what happened? 1020 01:14:11,950 --> 01:14:16,060 On one hand, maybe we have to worry because energy 1021 01:14:16,060 --> 01:14:19,580 is constantly being exchanged with the reservoir. 1022 01:14:19,580 --> 01:14:25,770 And so the energy of the system does not have a specific value. 1023 01:14:25,770 --> 01:14:28,420 There's a probability for it. 1024 01:14:28,420 --> 01:14:36,740 So probability of system having energy 1025 01:14:36,740 --> 01:14:44,940 epsilon-- it doesn't have a fixed energy. 1026 01:14:44,940 --> 01:14:47,770 There is a probability that it should have energy. 1027 01:14:47,770 --> 01:14:49,850 And this probability, let's say we indicate 1028 01:14:49,850 --> 01:14:55,610 with P epsilon given that we know what temperature is. 1029 01:14:55,610 --> 01:14:58,920 Well, on one hand we have this factor of E 1030 01:14:58,920 --> 01:15:01,980 to the minus epsilon over kT. 1031 01:15:01,980 --> 01:15:03,560 That comes from the [INAUDIBLE]. 1032 01:15:06,710 --> 01:15:10,660 But there isn't a single state that has that energy. 1033 01:15:10,660 --> 01:15:14,340 There's a whole bunch of other states 1034 01:15:14,340 --> 01:15:18,770 of the system that have that energy. 1035 01:15:18,770 --> 01:15:23,740 So as I scan the microstates, there 1036 01:15:23,740 --> 01:15:28,800 will be a huge number of them, omega of epsilon in number, 1037 01:15:28,800 --> 01:15:30,740 that have this right energy. 1038 01:15:30,740 --> 01:15:33,600 And so that's the probability of the energy. 1039 01:15:33,600 --> 01:15:39,710 And I can write this as E to the minus 1 1040 01:15:39,710 --> 01:15:42,610 over kT that I've called beta. 1041 01:15:42,610 --> 01:15:44,730 I have epsilon. 1042 01:15:44,730 --> 01:15:47,387 And then the log of omega that I take 1043 01:15:47,387 --> 01:15:52,580 in the numerator is S divided by Kb. 1044 01:15:52,580 --> 01:15:57,620 I can take that Kb here and write this as T S of epsilon. 1045 01:15:57,620 --> 01:16:00,440 And so this kind of should remind you 1046 01:16:00,440 --> 01:16:02,135 of something like a free energy. 1047 01:16:06,620 --> 01:16:14,980 But it tells you is that this probability 1048 01:16:14,980 --> 01:16:22,310 to have some particular energy is some kind of [? a form. ?] 1049 01:16:22,310 --> 01:16:25,300 Now note that again for something like a gas 1050 01:16:25,300 --> 01:16:30,360 or whatever, we expect typical values of both the energy 1051 01:16:30,360 --> 01:16:33,000 and entropy to be quantities that 1052 01:16:33,000 --> 01:16:35,700 are proportional to the size of the system. 1053 01:16:35,700 --> 01:16:39,470 As the size of the system becomes exponentially large, 1054 01:16:39,470 --> 01:16:42,140 we would expect that this probability would 1055 01:16:42,140 --> 01:16:45,800 be one of those things that has portions that 1056 01:16:45,800 --> 01:16:49,190 let's say are exponentially larger than any other portion. 1057 01:16:52,440 --> 01:16:55,720 There will be a factor of E to the minus N something 1058 01:16:55,720 --> 01:16:58,870 that will really peak up, let's say, the extremum 1059 01:16:58,870 --> 01:17:02,800 and make the extremum overwhelmingly more likely 1060 01:17:02,800 --> 01:17:05,770 than other places. 1061 01:17:05,770 --> 01:17:08,450 Let's try to quantify that a little bit better. 1062 01:17:10,990 --> 01:17:14,130 Once we have a probability, we can also 1063 01:17:14,130 --> 01:17:16,920 start calculating averages. 1064 01:17:16,920 --> 01:17:22,566 So let's define what the average energy of the system is. 1065 01:17:22,566 --> 01:17:25,590 The average energy of the system is 1066 01:17:25,590 --> 01:17:30,380 obtained by summing over all microstates. 1067 01:17:30,380 --> 01:17:34,020 The energy of that microstate, the probability 1068 01:17:34,020 --> 01:17:40,580 of that microstate, which is E to the minus beta H 1069 01:17:40,580 --> 01:17:44,900 microstate divided by the partition function, which 1070 01:17:44,900 --> 01:17:48,400 is the sum-- OK, the normalization, 1071 01:17:48,400 --> 01:17:51,760 which we will call the partition function, which 1072 01:17:51,760 --> 01:17:54,074 is the sum over all of these microstates. 1073 01:17:57,890 --> 01:18:03,290 Now this is something that we've already seen. 1074 01:18:03,290 --> 01:18:08,180 If I look at this expression in the denominator 1075 01:18:08,180 --> 01:18:13,610 that we call Z and has a name, which is the partition 1076 01:18:13,610 --> 01:18:19,530 function, then it's certainly a function of beta. 1077 01:18:19,530 --> 01:18:24,610 If I take a derivative of Z with respect to beta, what happens 1078 01:18:24,610 --> 01:18:28,050 I'll bring down a factor of H over here. 1079 01:18:28,050 --> 01:18:33,610 So the numerator up to a sine is the derivative 1080 01:18:33,610 --> 01:18:36,330 of Z with respect to beta. 1081 01:18:36,330 --> 01:18:41,580 And the denominator is 1 over Z. And so this is none other 1082 01:18:41,580 --> 01:18:47,090 than minus the log Z with respect to beta. 1083 01:18:47,090 --> 01:18:52,900 So OK, fine, so the mean value of this probability 1084 01:18:52,900 --> 01:18:57,890 is given by some expression such as this. 1085 01:18:57,890 --> 01:19:07,850 Well, you can see that if I were to repeat this process 1086 01:19:07,850 --> 01:19:11,980 and rather than taking one derivative, 1087 01:19:11,980 --> 01:19:27,320 I will take n derivatives and then divide by 1 over Z. 1088 01:19:27,320 --> 01:19:33,000 Each time I do that, I will bring down a factor of H. 1089 01:19:33,000 --> 01:19:40,740 So this is going to give me the average of H to the N. 1090 01:19:40,740 --> 01:19:43,490 The end moment of this probability distribution 1091 01:19:43,490 --> 01:19:46,890 of energy is obtainable by this procedure. 1092 01:19:50,920 --> 01:19:56,220 So now you recognize, oh, I've seen things like such as this. 1093 01:19:56,220 --> 01:19:59,560 So clearly this partition function 1094 01:19:59,560 --> 01:20:03,080 is something that generates the moments 1095 01:20:03,080 --> 01:20:05,920 by taking subsequent derivatives. 1096 01:20:05,920 --> 01:20:09,360 I can generate different moments of this distribution. 1097 01:20:11,880 --> 01:20:13,570 But then there was something else 1098 01:20:13,570 --> 01:20:16,040 that maybe this should remind you, 1099 01:20:16,040 --> 01:20:19,450 which is that if there's a quantity that generates 1100 01:20:19,450 --> 01:20:24,470 moments, then its log generates cumulants. 1101 01:20:24,470 --> 01:20:31,590 So you would say, OK, the nth cumulant 1102 01:20:31,590 --> 01:20:36,580 should be obtainable up to this factor of minus 1 to the n, 1103 01:20:36,580 --> 01:20:41,630 as the nth derivative with respect to the beta of logs. 1104 01:20:45,887 --> 01:20:48,560 And it's very easy to check that indeed 1105 01:20:48,560 --> 01:20:50,970 if I were to take two derivatives, 1106 01:20:50,970 --> 01:20:53,190 I will get the expectation value of H 1107 01:20:53,190 --> 01:20:58,080 squared minus the average of H squared, et cetera. 1108 01:20:58,080 --> 01:21:04,460 But the point is that clearly this to log Z 1109 01:21:04,460 --> 01:21:08,730 is, again, something that is extensive. 1110 01:21:08,730 --> 01:21:12,600 Another way of getting the normalization-- I 1111 01:21:12,600 --> 01:21:15,600 guess I forgot to put this 1 over Z here. 1112 01:21:21,990 --> 01:21:25,670 So now it is a perfectly normalized object. 1113 01:21:25,670 --> 01:21:32,660 So another way to get z would be to look 1114 01:21:32,660 --> 01:21:34,810 at the normalization of the probability. 1115 01:21:34,810 --> 01:21:39,570 I could integrate over epsilon this factor of E to the minus 1116 01:21:39,570 --> 01:21:45,580 beta epsilon minus T S of epsilon. 1117 01:21:45,580 --> 01:21:51,710 And that would give me Z. Now, again 1118 01:21:51,710 --> 01:21:54,750 the quantities that appear in the exponent, energy-- 1119 01:21:54,750 --> 01:21:57,010 entropy, their difference, free energy-- 1120 01:21:57,010 --> 01:21:59,180 are quantities that are extensive. 1121 01:21:59,180 --> 01:22:02,850 So this Z is going to be dominated again 1122 01:22:02,850 --> 01:22:04,950 by where this peak is. 1123 01:22:04,950 --> 01:22:08,370 And therefore, log of Z will be proportional to log 1124 01:22:08,370 --> 01:22:10,170 of what we have over here. 1125 01:22:10,170 --> 01:22:12,660 And it be an extensive quantity. 1126 01:22:12,660 --> 01:22:16,720 So ultimately, my statement is that this log of Z 1127 01:22:16,720 --> 01:22:23,510 is something that is order of N. 1128 01:22:23,510 --> 01:22:26,560 So we are, again, kind of reminiscent 1129 01:22:26,560 --> 01:22:29,290 of the central limit theorem. 1130 01:22:29,290 --> 01:22:33,860 In a situation where we have a probability distribution, 1131 01:22:33,860 --> 01:22:37,540 at large N, in which all of the cumulants 1132 01:22:37,540 --> 01:22:42,150 are proportional to N. The mean is proportional to N. 1133 01:22:42,150 --> 01:22:45,690 The variance is proportional to N. All of the cumulants 1134 01:22:45,690 --> 01:22:48,010 are proportional to N, which means 1135 01:22:48,010 --> 01:22:51,420 that essentially the extent of the fluctuations that you have 1136 01:22:51,420 --> 01:23:00,380 over here are going to go the order of the square root of N. 1137 01:23:00,380 --> 01:23:04,020 So the bridge, the thing that again allows us, 1138 01:23:04,020 --> 01:23:09,130 while we have in principle in the expression that we have 1139 01:23:09,130 --> 01:23:13,540 said, a variable energy for the system. 1140 01:23:13,540 --> 01:23:18,530 In fact, in the limit of things becoming extensive, 1141 01:23:18,530 --> 01:23:23,960 I know where that energy is, up to fluctuations 1142 01:23:23,960 --> 01:23:25,680 or up to uncertainty that is only 1143 01:23:25,680 --> 01:23:28,060 of the order of square root of N. 1144 01:23:28,060 --> 01:23:32,580 And so the relative uncertainty will vanish as the N goes 1145 01:23:32,580 --> 01:23:34,790 to infinity, limit is approached. 1146 01:23:34,790 --> 01:23:36,570 So although again we have something 1147 01:23:36,570 --> 01:23:39,410 that is in principle probabilistic, 1148 01:23:39,410 --> 01:23:41,990 again, in the thermodynamic sense 1149 01:23:41,990 --> 01:23:46,740 we can identify uniquely an energy for our system as, 1150 01:23:46,740 --> 01:23:49,440 let's say, the mean value or the most likely value. 1151 01:23:49,440 --> 01:23:53,250 They're all the same thing of the order of 1 over N. 1152 01:23:53,250 --> 01:23:58,680 And again, to be more precise, the variance 1153 01:23:58,680 --> 01:24:05,050 is clearly the second derivative of log Z. 1 derivative of a log 1154 01:24:05,050 --> 01:24:07,560 Z is going to give me the energy. 1155 01:24:07,560 --> 01:24:10,483 So this is going to be d by d beta up 1156 01:24:10,483 --> 01:24:14,000 to a minus sign of the energy or the expectation 1157 01:24:14,000 --> 01:24:17,820 value of the Hamiltonian, which we identified 1158 01:24:17,820 --> 01:24:19,900 as the energy of the system. 1159 01:24:19,900 --> 01:24:21,615 The derivative with respect to beta, 1160 01:24:21,615 --> 01:24:25,320 I can write as kB T squared. 1161 01:24:25,320 --> 01:24:28,530 The derivative of energy with respect to T, 1162 01:24:28,530 --> 01:24:32,110 everything here we are doing that conditions of no work. 1163 01:24:32,110 --> 01:24:37,030 So the variance is in fact kB T squared, 1164 01:24:37,030 --> 01:24:40,880 the heat capacity of the system. 1165 01:24:40,880 --> 01:24:45,300 So the extent that these fluctuations squared 1166 01:24:45,300 --> 01:24:50,680 is kB T squared times the heat capacity the system. 1167 01:24:50,680 --> 01:24:53,790 OK, so next time what we will do is 1168 01:24:53,790 --> 01:24:57,490 we will calculate the results for the ideal gas. 1169 01:24:57,490 --> 01:24:59,790 First thing, the canonical ensemble 1170 01:24:59,790 --> 01:25:02,990 to show that we get exactly the same macroscopic 1171 01:25:02,990 --> 01:25:05,620 and microscopic descriptions. 1172 01:25:05,620 --> 01:25:09,460 And then we look at other ensembles. 1173 01:25:09,460 --> 01:25:11,250 And that will conclude the segment 1174 01:25:11,250 --> 01:25:13,700 that we have on statistical mechanics 1175 01:25:13,700 --> 01:25:16,387 of non-interacting systems.