1 00:00:02,110 --> 00:00:04,480 The following content is provided under a Creative 2 00:00:04,480 --> 00:00:05,870 Commons license. 3 00:00:05,870 --> 00:00:08,080 Your support will help MIT OpenCourseWare 4 00:00:08,080 --> 00:00:12,170 continue to offer high quality educational resources for free. 5 00:00:12,170 --> 00:00:14,710 To make a donation or to view additional materials 6 00:00:14,710 --> 00:00:18,670 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:18,670 --> 00:00:19,870 at ocw.mit.edu. 8 00:00:23,342 --> 00:00:25,300 PROFESSOR YEN-JIE LEE: Welcome back, everybody, 9 00:00:25,300 --> 00:00:28,810 to 8.03 Today, we are going to continue 10 00:00:28,810 --> 00:00:31,870 the discussion of symmetry matrix, 11 00:00:31,870 --> 00:00:33,970 which we started last time. 12 00:00:33,970 --> 00:00:37,460 And this is what we have been doing. 13 00:00:37,460 --> 00:00:39,370 OK? 14 00:00:39,370 --> 00:00:41,050 So the thing which we have been doing 15 00:00:41,050 --> 00:00:45,480 is to solve the normal mode frequencies of component 16 00:00:45,480 --> 00:00:48,490 systems by looking at equation of motion, 17 00:00:48,490 --> 00:00:51,850 MX double dot equal to minus KX. 18 00:00:51,850 --> 00:00:56,740 And in the end of the day, what we are doing is really 19 00:00:56,740 --> 00:00:59,910 to solve Eigenvalue problem-- 20 00:00:59,910 --> 00:01:02,830 M minus one K matrix Eigenvalue problem. 21 00:01:02,830 --> 00:01:07,770 And we have been exercising this several times 22 00:01:07,770 --> 00:01:10,570 the last few lectures. 23 00:01:10,570 --> 00:01:12,617 And the M minus one K matrix says 24 00:01:12,617 --> 00:01:16,540 you're describing how each component in the system 25 00:01:16,540 --> 00:01:18,110 interact with each other. 26 00:01:18,110 --> 00:01:18,610 OK? 27 00:01:18,610 --> 00:01:21,760 So that's actually what we are trying to do, 28 00:01:21,760 --> 00:01:24,310 to solve the normal modes. 29 00:01:24,310 --> 00:01:28,420 So we have been making progress, and we 30 00:01:28,420 --> 00:01:31,330 are increasing the number of coupled oscillators 31 00:01:31,330 --> 00:01:32,630 as a function of time. 32 00:01:32,630 --> 00:01:35,230 And now we finally arrive at our limit, which 33 00:01:35,230 --> 00:01:36,990 is infinite system, right? 34 00:01:36,990 --> 00:01:39,310 So basically, what we have been discussing 35 00:01:39,310 --> 00:01:43,960 is a special kind of infinite system, which 36 00:01:43,960 --> 00:01:47,470 actually satisfies translation symmetry, right? 37 00:01:47,470 --> 00:01:51,730 So in general, we don't know how to solve infinite system. 38 00:01:51,730 --> 00:01:53,490 If this system is really complicated, 39 00:01:53,490 --> 00:01:57,730 no symmetry, then who knows how to solve this, right? 40 00:01:57,730 --> 00:02:01,330 But very luckily, in 8.03 we have 41 00:02:01,330 --> 00:02:06,340 started to get a highly symmetrical infinite system, 42 00:02:06,340 --> 00:02:09,580 and in this case, it's translation symmetry. 43 00:02:09,580 --> 00:02:13,390 And that is really pretty nice, and we can actually 44 00:02:13,390 --> 00:02:18,490 use this example to learn an interesting fact, which 45 00:02:18,490 --> 00:02:22,380 we can see from the physical system we discuss here. 46 00:02:22,380 --> 00:02:26,440 So, one thing which we have been discussing 47 00:02:26,440 --> 00:02:30,420 is the space translation symmetry matrix. 48 00:02:30,420 --> 00:02:35,800 As a reminder, S matrix, as defined here. 49 00:02:35,800 --> 00:02:39,460 Basically, if you have a vector A, which 50 00:02:39,460 --> 00:02:43,910 describes the amplitude of an individual component 51 00:02:43,910 --> 00:02:47,500 in a system, basically these vectors 52 00:02:47,500 --> 00:02:51,700 have Aj, Aj plus one, Aj plus two blah blah blah. 53 00:02:51,700 --> 00:02:58,520 All those amplitudes are included in this vector. 54 00:02:58,520 --> 00:03:01,310 And what does this S matrix do? 55 00:03:01,310 --> 00:03:03,790 It's actually the following-- 56 00:03:03,790 --> 00:03:08,200 so if A prime is equal to S times A, 57 00:03:08,200 --> 00:03:12,910 then the base component of A prime 58 00:03:12,910 --> 00:03:16,100 will be equal to Aj plus one. 59 00:03:16,100 --> 00:03:20,890 OK, so that's actually what this S matrix does to A vector. 60 00:03:20,890 --> 00:03:21,550 OK? 61 00:03:21,550 --> 00:03:25,230 So you can think about S metrics as an operator. 62 00:03:25,230 --> 00:03:28,840 It's actually picking up the Aj plus 63 00:03:28,840 --> 00:03:33,080 one component, and moving it to Aj in the new vector. 64 00:03:33,080 --> 00:03:33,580 OK? 65 00:03:33,580 --> 00:03:37,180 So that's actually what this S matrix does. 66 00:03:37,180 --> 00:03:42,790 And we were talking about the Eigenvector of S matrix. 67 00:03:42,790 --> 00:03:47,470 For example, if A is an Eigenvector of S matrix, 68 00:03:47,470 --> 00:03:52,000 then we have this relation, SA equal to beta A. Now, 69 00:03:52,000 --> 00:03:57,830 beta is actually the Eigenvalue of S matrix. 70 00:03:57,830 --> 00:03:58,600 OK? 71 00:03:58,600 --> 00:04:02,380 And then we also discussed last time, that means, 72 00:04:02,380 --> 00:04:05,680 based on this logical extension, Aj prime 73 00:04:05,680 --> 00:04:08,170 would be equal to Aj plus one based 74 00:04:08,170 --> 00:04:10,300 on the definition of S matrix. 75 00:04:10,300 --> 00:04:14,320 And now this is going to be equal to beta times Aj, right? 76 00:04:14,320 --> 00:04:21,450 Because we assume that that A vector is an Eigenvector of S 77 00:04:21,450 --> 00:04:21,950 matrix. 78 00:04:21,950 --> 00:04:23,050 OK? 79 00:04:23,050 --> 00:04:28,340 That means Aj will be equal to beta to the j A zero. 80 00:04:28,340 --> 00:04:29,320 Right? 81 00:04:29,320 --> 00:04:33,550 According to this relation. 82 00:04:33,550 --> 00:04:36,460 Therefore, we can conclude that Aj will be 83 00:04:36,460 --> 00:04:39,460 proportional to beta to the j. 84 00:04:39,460 --> 00:04:43,000 OK, beta is still some coefficient, 85 00:04:43,000 --> 00:04:48,020 which we have not determined, and it can be anything. 86 00:04:48,020 --> 00:04:49,180 OK? 87 00:04:49,180 --> 00:04:56,470 So, in order to consider a system of little masses, which 88 00:04:56,470 --> 00:04:58,760 are oscillating up and down instead 89 00:04:58,760 --> 00:05:05,580 of going in one direction forever or the amplitude 90 00:05:05,580 --> 00:05:08,770 grows exponentially. 91 00:05:08,770 --> 00:05:11,740 And we also don't want the system 92 00:05:11,740 --> 00:05:14,720 to have the amplitude go to infinity 93 00:05:14,720 --> 00:05:19,360 when we go to very very large j value or very small j value, 94 00:05:19,360 --> 00:05:22,633 therefore we limit our discussion 95 00:05:22,633 --> 00:05:28,440 in the case of beta equal to exponential ika. 96 00:05:28,440 --> 00:05:29,030 OK? 97 00:05:29,030 --> 00:05:35,570 In this case, the absolute value of beta is one. 98 00:05:35,570 --> 00:05:36,660 Right? 99 00:05:36,660 --> 00:05:40,160 OK, and that means-- 100 00:05:40,160 --> 00:05:44,990 OK, Aj is actually proportional to beta to Aj, right? 101 00:05:44,990 --> 00:05:49,040 So if you take a ratio of Aj plus one and the Aj, 102 00:05:49,040 --> 00:05:51,160 the ratio is beta. 103 00:05:51,160 --> 00:05:55,100 If beta Is not equal to one or its absolute value 104 00:05:55,100 --> 00:05:59,180 is not equal to one, then this Aj value 105 00:05:59,180 --> 00:06:02,540 is going to be increasing according 106 00:06:02,540 --> 00:06:04,220 to a power law, right? 107 00:06:04,220 --> 00:06:06,350 So the amplitude is going to be, whoa, 108 00:06:06,350 --> 00:06:08,930 going to a very, very large value, right? 109 00:06:08,930 --> 00:06:11,810 And that corresponds to some kind of physical system, 110 00:06:11,810 --> 00:06:14,060 but not corresponding to oscillation. 111 00:06:14,060 --> 00:06:15,710 OK? 112 00:06:15,710 --> 00:06:19,010 And if we do this and assume that beta 113 00:06:19,010 --> 00:06:23,120 is equal to exponential ika. 114 00:06:23,120 --> 00:06:28,220 OK, if we do this, what is going to happen is the following. 115 00:06:28,220 --> 00:06:32,390 So again, the ratio of Aj plus one and Aj 116 00:06:32,390 --> 00:06:39,410 is a fixed value beta, equal to exponential ika. 117 00:06:39,410 --> 00:06:42,980 But what it does is, instead of changing the amplitude, 118 00:06:42,980 --> 00:06:47,911 it's actually doing a rotation in the complex plane. 119 00:06:47,911 --> 00:06:48,410 OK? 120 00:06:48,410 --> 00:06:54,080 So if this is Aj, J plus one, J plus two, et cetera, 121 00:06:54,080 --> 00:06:57,850 as a function of say j, the variable, then what it does 122 00:06:57,850 --> 00:07:01,810 is that this operation multiplied by beta 123 00:07:01,810 --> 00:07:07,640 is really a rotation in the complex plane. 124 00:07:07,640 --> 00:07:11,240 And while we actually see in the physical system, 125 00:07:11,240 --> 00:07:15,620 it's actually a projection of this complex imaginary plane 126 00:07:15,620 --> 00:07:17,990 to the real axis. 127 00:07:17,990 --> 00:07:21,610 And that would give you a sine function and cosine function. 128 00:07:21,610 --> 00:07:25,760 So very interestingly, if we choose wisely, 129 00:07:25,760 --> 00:07:29,030 the beta to have absolute value of one, 130 00:07:29,030 --> 00:07:36,020 then that will give you a system which is actually oscillating 131 00:07:36,020 --> 00:07:37,850 up and down and the amplitude is actually 132 00:07:37,850 --> 00:07:40,730 confined within some value. 133 00:07:40,730 --> 00:07:43,340 OK, so that is actually quite interesting. 134 00:07:43,340 --> 00:07:46,760 The other thing which I would like to talk about is-- 135 00:07:46,760 --> 00:07:52,690 OK, I choose to have exponential ika as my beta. 136 00:07:52,690 --> 00:07:54,890 OK, why ka, right? 137 00:07:54,890 --> 00:07:56,620 It looks really strange here. 138 00:07:56,620 --> 00:08:00,140 Suddenly the k and and the a are coming to play, right? 139 00:08:00,140 --> 00:08:02,591 So what is actually the small a? 140 00:08:02,591 --> 00:08:09,440 The small a is actually the distance between little mass, 141 00:08:09,440 --> 00:08:11,360 just a reminder. 142 00:08:11,360 --> 00:08:13,520 So this is actually the length scale 143 00:08:13,520 --> 00:08:16,410 between all those little mass, right? 144 00:08:16,410 --> 00:08:21,950 Therefore, what I'm doing is to factorize out the length scale, 145 00:08:21,950 --> 00:08:25,620 and then we suddenly found that. 146 00:08:25,620 --> 00:08:31,520 OK, after I do this, I define beta equal to exponential ika 147 00:08:31,520 --> 00:08:33,980 Instead of exponential i theta, right? 148 00:08:33,980 --> 00:08:36,980 So you can also do exponential i theta, right? 149 00:08:36,980 --> 00:08:40,309 But instead, I gave theta a fancy name, 150 00:08:40,309 --> 00:08:42,850 which is k times a. a is actually 151 00:08:42,850 --> 00:08:45,940 the distance between mass. 152 00:08:45,940 --> 00:08:49,090 Then something interesting happens, because k suddenly 153 00:08:49,090 --> 00:08:51,000 also will have a meaning. 154 00:08:51,000 --> 00:09:00,660 It's actually the wave number of the resulting sine wave. 155 00:09:00,660 --> 00:09:01,160 OK? 156 00:09:01,160 --> 00:09:03,980 So that's actually why we actually 157 00:09:03,980 --> 00:09:08,050 factor out this a factor. 158 00:09:08,050 --> 00:09:10,340 Then, after that, we actually found out 159 00:09:10,340 --> 00:09:12,210 the amplitude would be proportional 160 00:09:12,210 --> 00:09:16,580 to exponential ijka, and j is actually 161 00:09:16,580 --> 00:09:20,480 just a label of a phase component in the system. 162 00:09:20,480 --> 00:09:22,330 OK. 163 00:09:22,330 --> 00:09:27,890 So, once we have solved the Eigenvalue problem 164 00:09:27,890 --> 00:09:32,360 for the symmetry matrix, S, as we discussed before, 165 00:09:32,360 --> 00:09:35,720 if you look at the slides, OK? 166 00:09:35,720 --> 00:09:40,430 If S and the M minus one K matrix, they commute. 167 00:09:40,430 --> 00:09:41,720 OK? 168 00:09:41,720 --> 00:09:45,860 Commutes means that you can actually change S and and the M 169 00:09:45,860 --> 00:09:48,950 minus one matrix, you can swap them, OK, 170 00:09:48,950 --> 00:09:50,735 when you multiply them together. 171 00:09:50,735 --> 00:09:51,740 OK? 172 00:09:51,740 --> 00:09:55,190 If you can swap S and M minus one K matrix, 173 00:09:55,190 --> 00:09:57,680 that means they commute. 174 00:09:57,680 --> 00:10:00,740 And our conclusion from last time 175 00:10:00,740 --> 00:10:06,230 is that they will share the same Eigenvectors. 176 00:10:06,230 --> 00:10:07,130 OK? 177 00:10:07,130 --> 00:10:09,910 Of course, not necessarily the same Eigenvalue, 178 00:10:09,910 --> 00:10:11,990 but they share the same Eigenvector. 179 00:10:11,990 --> 00:10:13,400 So that's great news! 180 00:10:13,400 --> 00:10:17,330 Because instead of solving M minus one K matrix, which 181 00:10:17,330 --> 00:10:19,820 can be really complicated; depends 182 00:10:19,820 --> 00:10:23,490 on what kind of physical system you are talking about. 183 00:10:23,490 --> 00:10:28,060 I can solve S matrix Eigenvalue problem. 184 00:10:28,060 --> 00:10:29,310 OK? 185 00:10:29,310 --> 00:10:34,560 And then this Eigenvector, which I just found here, 186 00:10:34,560 --> 00:10:41,240 is going to be the Eigenvector of M minus one K matrix. 187 00:10:41,240 --> 00:10:45,110 And that's actually really making things much easier, 188 00:10:45,110 --> 00:10:49,460 because now instead of solving Eigenvalue problem of M 189 00:10:49,460 --> 00:10:52,210 minus one K matrix, what I am doing 190 00:10:52,210 --> 00:10:55,550 is just multiplying M minus one K times A, 191 00:10:55,550 --> 00:11:00,740 then you can actually obtain the normal mode frequency omega. 192 00:11:00,740 --> 00:11:01,310 OK? 193 00:11:01,310 --> 00:11:03,750 So that's the issue of the great excitement. 194 00:11:03,750 --> 00:11:05,210 What does that mean? 195 00:11:05,210 --> 00:11:09,370 That means, if you have all kinds of systems, which 196 00:11:09,370 --> 00:11:13,200 are translation symmetric-- you can have a line, 197 00:11:13,200 --> 00:11:17,960 you have however many people together, 198 00:11:17,960 --> 00:11:20,540 whatever a system which is so on here. 199 00:11:20,540 --> 00:11:25,330 They are all going to have the same Eigenvector. 200 00:11:25,330 --> 00:11:25,830 You see? 201 00:11:25,830 --> 00:11:27,290 You already know how they are going 202 00:11:27,290 --> 00:11:30,020 to interact with each other, and what is actually 203 00:11:30,020 --> 00:11:35,010 the amplitude as a function of j, which is the location label. 204 00:11:35,010 --> 00:11:35,850 OK? 205 00:11:35,850 --> 00:11:39,160 So that's actually really wonderful. 206 00:11:39,160 --> 00:11:43,040 So, in order to help you with understanding 207 00:11:43,040 --> 00:11:46,490 of this system some more, we are going 208 00:11:46,490 --> 00:11:51,290 to discuss another system which is actually also 209 00:11:51,290 --> 00:11:52,870 very interesting. 210 00:11:52,870 --> 00:11:55,330 It's actually a spring-- 211 00:11:55,330 --> 00:11:59,540 OK, last time we discussed a spring and mass system, right? 212 00:11:59,540 --> 00:12:01,920 And then we solved it together. 213 00:12:01,920 --> 00:12:05,210 And this time we are going to solve a system which 214 00:12:05,210 --> 00:12:08,060 is made of mass and strings. 215 00:12:08,060 --> 00:12:08,900 Ok? 216 00:12:08,900 --> 00:12:11,680 So that may actually copy-- 217 00:12:11,680 --> 00:12:13,640 OK, let me actually introduce you 218 00:12:13,640 --> 00:12:17,770 to this new system we are going to talk about today. 219 00:12:17,770 --> 00:12:23,060 It has many little mass here, from left hand side 220 00:12:23,060 --> 00:12:25,610 of the universe and right hand side of the universe. 221 00:12:25,610 --> 00:12:28,740 Take forever to actually construct this system. 222 00:12:28,740 --> 00:12:29,430 OK? 223 00:12:29,430 --> 00:12:33,860 Then my student actually carefully link them by strings, 224 00:12:33,860 --> 00:12:38,750 and we make sure that the string tension, OK, is actually 225 00:12:38,750 --> 00:12:41,630 a fixed value, which is T. OK. 226 00:12:41,630 --> 00:12:44,090 T is actually the string tension. 227 00:12:44,090 --> 00:12:47,300 And of course, as what we discussed before, 228 00:12:47,300 --> 00:12:50,570 the space or the distance between little mass 229 00:12:50,570 --> 00:12:53,240 is actually A. OK? 230 00:12:53,240 --> 00:12:54,860 So that's, again, the same length 231 00:12:54,860 --> 00:12:57,080 scale, which we were using. 232 00:12:57,080 --> 00:12:59,560 And finally, in order to describe 233 00:12:59,560 --> 00:13:04,740 all those little masses in the system, 234 00:13:04,740 --> 00:13:10,020 I label them as j, j plus one, j plus two, et cetera, et cetera. 235 00:13:10,020 --> 00:13:11,270 OK. 236 00:13:11,270 --> 00:13:15,470 So, the question we are asking is, 237 00:13:15,470 --> 00:13:20,190 what would be the resulting motion of this system, right? 238 00:13:20,190 --> 00:13:24,650 So what we can do is what we have done last time, right? 239 00:13:24,650 --> 00:13:28,820 We take jth object in this system, 240 00:13:28,820 --> 00:13:34,520 and we look at the force diagram and this M minus one-- 241 00:13:34,520 --> 00:13:37,240 to write down the equational motion and also the 242 00:13:37,240 --> 00:13:39,650 and M minus one K metrics. 243 00:13:39,650 --> 00:13:49,130 So what is actually the force diagram of j's component? 244 00:13:52,010 --> 00:13:52,510 OK. 245 00:13:52,510 --> 00:13:54,495 If I take this-- 246 00:13:54,495 --> 00:14:02,200 so first, I take my jth mass, and it's connected 247 00:14:02,200 --> 00:14:05,560 to two strings, right? 248 00:14:05,560 --> 00:14:10,150 So there are two strings connected to jth mass. 249 00:14:10,150 --> 00:14:10,960 OK? 250 00:14:10,960 --> 00:14:13,810 And of course, left hand side you 251 00:14:13,810 --> 00:14:17,470 have another mass, right hand side you have another option. 252 00:14:17,470 --> 00:14:19,660 OK? 253 00:14:19,660 --> 00:14:24,380 And I know the tension of this string. 254 00:14:24,380 --> 00:14:27,160 Its actually a fixed value, and the string 255 00:14:27,160 --> 00:14:31,510 tension is fixed at the T. OK? 256 00:14:31,510 --> 00:14:33,460 So, in order to describe this system, 257 00:14:33,460 --> 00:14:35,960 I need to find my coordinate system, right? 258 00:14:35,960 --> 00:14:36,810 As usual. 259 00:14:36,810 --> 00:14:38,560 So, what is actually the coordinate system 260 00:14:38,560 --> 00:14:40,010 I'm going to use.? 261 00:14:40,010 --> 00:14:42,520 So I need to define horizontal direction 262 00:14:42,520 --> 00:14:48,460 to be x, and the vertical direction to be y. 263 00:14:48,460 --> 00:14:52,560 Therefore, I cannot express this little mass to be-- 264 00:14:52,560 --> 00:14:58,580 the position of jth little mass as Xj and the Yj. 265 00:15:01,540 --> 00:15:03,040 OK? 266 00:15:03,040 --> 00:15:07,350 We can do the same thing for the left-hand side mass 267 00:15:07,350 --> 00:15:16,567 is Xa minus one, Yj minus one, and Xj plus one Yj plus one. 268 00:15:19,250 --> 00:15:22,870 So to simplify the discussion, what I'm going to assume 269 00:15:22,870 --> 00:15:28,930 is that all those little masses can only move up and down, OK, 270 00:15:28,930 --> 00:15:30,580 instead of back and forth. 271 00:15:30,580 --> 00:15:33,940 OK, so only one direction is allowed, 272 00:15:33,940 --> 00:15:38,520 and also I assume that the up and down motion 273 00:15:38,520 --> 00:15:42,640 of all those little masses is really really very small. 274 00:15:42,640 --> 00:15:46,450 So I can use small angle approximation. 275 00:15:46,450 --> 00:15:49,160 OK, so therefore I can-- 276 00:15:49,160 --> 00:15:57,040 this is essentially the zero of Y axis, when I actually-- 277 00:15:57,040 --> 00:15:59,350 before I actually move-- 278 00:15:59,350 --> 00:16:02,720 wind up with the mass away from the equilibrium position. 279 00:16:02,720 --> 00:16:07,720 So when the string mass system is at rest, OK, 280 00:16:07,720 --> 00:16:14,800 not really moving, then all the mass are at Y equal to zero. 281 00:16:14,800 --> 00:16:16,420 OK? 282 00:16:16,420 --> 00:16:23,880 So now, I actually move this mass to Yj, OK? 283 00:16:23,880 --> 00:16:28,960 Then, apparently there are two forces acting on this mass. 284 00:16:28,960 --> 00:16:31,060 The left hand side is string force, 285 00:16:31,060 --> 00:16:32,740 and the right hand side is string force. 286 00:16:32,740 --> 00:16:38,300 And the magnitude of the force is actually T. OK? 287 00:16:38,300 --> 00:16:42,590 So, in order to help us with solving this problem. 288 00:16:42,590 --> 00:16:44,360 I would define two angles-- 289 00:16:44,360 --> 00:16:50,140 the left hand side angle to be theta one, and the right hand 290 00:16:50,140 --> 00:16:52,540 side angle to be theta two. 291 00:16:52,540 --> 00:16:56,320 Then I can now write down the equation of motion 292 00:16:56,320 --> 00:16:59,800 in the horizontal direction, and then in the vertical direction. 293 00:16:59,800 --> 00:17:01,960 OK? 294 00:17:01,960 --> 00:17:02,710 All right. 295 00:17:02,710 --> 00:17:06,280 Since I have assumed that all of those masses 296 00:17:06,280 --> 00:17:10,089 can only move up and down, now I mean-- and also 297 00:17:10,089 --> 00:17:13,880 the displacement with respect to the equilibrium position, 298 00:17:13,880 --> 00:17:18,400 which is Y equal to zero, is really small, compared to, 299 00:17:18,400 --> 00:17:21,400 for example, the length scale a. 300 00:17:21,400 --> 00:17:21,970 OK? 301 00:17:21,970 --> 00:17:25,569 So therefore, I can write that condition explicitly. 302 00:17:25,569 --> 00:17:27,310 So I have a condition. 303 00:17:27,310 --> 00:17:32,680 I assume that Yj is actually much, much smaller than a, 304 00:17:32,680 --> 00:17:36,670 which is the distance between those masses. 305 00:17:36,670 --> 00:17:41,530 And that means theta one and theta two 306 00:17:41,530 --> 00:17:47,320 are going to be much, much smaller than one. 307 00:17:47,320 --> 00:17:47,860 OK? 308 00:17:47,860 --> 00:17:49,930 So that's actually given us a chance 309 00:17:49,930 --> 00:17:53,240 to use small angle approximation. 310 00:17:53,240 --> 00:17:55,660 So based on this force diagram, I 311 00:17:55,660 --> 00:17:59,070 can now write down the two equations of motion; 312 00:17:59,070 --> 00:18:01,220 one in the horizontal direction, the other one 313 00:18:01,220 --> 00:18:02,960 in the vertical direction. 314 00:18:02,960 --> 00:18:09,670 So what I'm going to get in the horizontal direction 315 00:18:09,670 --> 00:18:14,530 is M X j double dot. 316 00:18:14,530 --> 00:18:16,000 These will be equal to-- 317 00:18:16,000 --> 00:18:17,480 OK, there are two forces. 318 00:18:17,480 --> 00:18:17,980 Right? 319 00:18:17,980 --> 00:18:19,990 Horizontal direction, I would need 320 00:18:19,990 --> 00:18:23,320 to calculate the projection to the X direction, 321 00:18:23,320 --> 00:18:25,310 therefore the left hand side force 322 00:18:25,310 --> 00:18:30,970 will give you minus T cosine theta one, 323 00:18:30,970 --> 00:18:33,620 and the right hand side of this string force 324 00:18:33,620 --> 00:18:40,500 is going to give you plus T cosine theta two. 325 00:18:40,500 --> 00:18:42,160 OK? 326 00:18:42,160 --> 00:18:48,020 And then in the vertical direction, what I'm 327 00:18:48,020 --> 00:18:51,750 going to get is Myj double dot. 328 00:18:51,750 --> 00:18:57,210 This is equal to minus T sine theta y. 329 00:18:57,210 --> 00:19:01,220 Now I'm doing the projection in the y 330 00:19:01,220 --> 00:19:10,030 axis in the vertical direction, and minus T sine theta two. 331 00:19:10,030 --> 00:19:11,510 OK. 332 00:19:11,510 --> 00:19:14,960 And since I have this condition, all the mass 333 00:19:14,960 --> 00:19:20,810 can only move up and down, and also the displacement 334 00:19:20,810 --> 00:19:25,820 is much, much smaller than a; therefore, I 335 00:19:25,820 --> 00:19:28,080 have the small angle approximation. 336 00:19:28,080 --> 00:19:33,440 Cosine theta is roughly equal to one, 337 00:19:33,440 --> 00:19:38,700 and the sine theta is roughly equal to theta. 338 00:19:38,700 --> 00:19:39,410 OK? 339 00:19:39,410 --> 00:19:42,440 And I would call this the equation number one, 340 00:19:42,440 --> 00:19:45,590 and the second equation in the vertical direction 341 00:19:45,590 --> 00:19:48,920 to be equation number two. 342 00:19:48,920 --> 00:19:50,900 OK. 343 00:19:50,900 --> 00:19:54,710 So, up to here, everything is essentially exact, 344 00:19:54,710 --> 00:19:58,040 and now I would like to make a small angle approximation 345 00:19:58,040 --> 00:19:59,570 and see what will happen. 346 00:19:59,570 --> 00:20:01,860 And now equation number one will be 347 00:20:01,860 --> 00:20:07,820 called MXa double dot equal to minus T plus T, 348 00:20:07,820 --> 00:20:10,580 and this is equal to zero. 349 00:20:10,580 --> 00:20:15,710 OK, so that means we will not have horizontal direction 350 00:20:15,710 --> 00:20:16,400 acceleration. 351 00:20:16,400 --> 00:20:18,980 Therefore, in the horizontal direction, 352 00:20:18,980 --> 00:20:21,740 there will be no acceleration and therefore 353 00:20:21,740 --> 00:20:26,900 no movement in the X direction, or horizontal direction. 354 00:20:26,900 --> 00:20:31,190 OK, and now I can take a look at the vertical direction. 355 00:20:31,190 --> 00:20:32,630 OK? 356 00:20:32,630 --> 00:20:38,180 So basically, what I am going to get is MYj double dot, 357 00:20:38,180 --> 00:20:41,670 and this will be equal to minus T. Now, 358 00:20:41,670 --> 00:20:47,450 sine theta one will be roughly equal to theta one. 359 00:20:47,450 --> 00:20:49,460 So what is actually theta one? 360 00:20:49,460 --> 00:20:52,110 Theta one is going to be-- 361 00:20:52,110 --> 00:21:00,010 OK, so this is actually the Yj minus Yj minus one. 362 00:21:00,010 --> 00:21:01,169 Right? 363 00:21:01,169 --> 00:21:02,710 So now that's actually the difference 364 00:21:02,710 --> 00:21:10,140 between the amplitude of the displacement of the mass j, 365 00:21:10,140 --> 00:21:13,030 and the displacement of mass j minus one. 366 00:21:13,030 --> 00:21:13,770 Right? 367 00:21:13,770 --> 00:21:16,980 Divided by a, I get theta one. 368 00:21:16,980 --> 00:21:17,940 Right? 369 00:21:17,940 --> 00:21:20,480 Therefore, the first sine theta one, 370 00:21:20,480 --> 00:21:29,430 will become Yj minus Yj minus one, divided by a. 371 00:21:29,430 --> 00:21:30,820 OK? 372 00:21:30,820 --> 00:21:34,250 And of course, you can do the same thing for the sine theta 373 00:21:34,250 --> 00:21:35,320 two, right? 374 00:21:35,320 --> 00:21:38,735 Which is actually just the theta two. 375 00:21:38,735 --> 00:21:40,360 Then, basically, what I am going to get 376 00:21:40,360 --> 00:21:51,330 is minus Tyj minus Yj plus one divided by a. 377 00:21:51,330 --> 00:21:53,110 OK. 378 00:21:53,110 --> 00:21:54,128 Any questions? 379 00:21:56,880 --> 00:21:57,500 OK. 380 00:21:57,500 --> 00:21:59,360 I hope everybody is following it. 381 00:21:59,360 --> 00:22:03,740 OK, so now I can actually simplify equation number two, 382 00:22:03,740 --> 00:22:07,850 and basically, what I am going to get is MYj 383 00:22:07,850 --> 00:22:12,670 double dot will be equal to minus T over a. 384 00:22:12,670 --> 00:22:17,700 Basically, I take T over a out of it again. 385 00:22:17,700 --> 00:22:23,180 And I also collect all the terms related to Yj minus one, 386 00:22:23,180 --> 00:22:30,850 minus two Yj, plus Yj plus one. 387 00:22:30,850 --> 00:22:34,690 OK, I just ask you to rewrite equation number two 388 00:22:34,690 --> 00:22:40,800 in a form which we like more. 389 00:22:40,800 --> 00:22:45,110 OK, so that is actually the equation of motion, so from now 390 00:22:45,110 --> 00:22:47,670 on, I am going to ignore all the motion 391 00:22:47,670 --> 00:22:51,920 in the horizontal direction, because in this small angle 392 00:22:51,920 --> 00:22:54,545 approximation we have shown you that there 393 00:22:54,545 --> 00:22:58,700 will be no acceleration in the x direction, right? 394 00:22:58,700 --> 00:23:04,520 So now it's actually getting a step forward again. 395 00:23:04,520 --> 00:23:06,470 Basically, we have the equation of motion, 396 00:23:06,470 --> 00:23:09,240 and what is usually the next step? 397 00:23:09,240 --> 00:23:12,510 The next step is to write down what matrix? 398 00:23:12,510 --> 00:23:14,384 Anybody can help me? 399 00:23:14,384 --> 00:23:15,960 STUDENT 1: M minus one k matrix. 400 00:23:15,960 --> 00:23:18,085 PROFESSOR YEN-JIE LEE: M minus one k matrix, right? 401 00:23:18,085 --> 00:23:21,080 So actually, as usual, we actually follow the procedure. 402 00:23:21,080 --> 00:23:25,020 Now I would like to write down the m minus one k matrix. 403 00:23:25,020 --> 00:23:28,960 OK, so before I do that what, I will define-- 404 00:23:28,960 --> 00:23:35,600 I will actually assume my normal mode has this functional form, 405 00:23:35,600 --> 00:23:43,850 yj is equal to the real part of aj exponential i omega t 406 00:23:43,850 --> 00:23:45,500 plus phi. 407 00:23:45,500 --> 00:23:48,200 So basically, that tells you that all the components 408 00:23:48,200 --> 00:23:51,710 are oscillating at the same frequency, omega, 409 00:23:51,710 --> 00:23:53,990 and the same phase, phi. 410 00:23:53,990 --> 00:23:55,338 Yes. 411 00:23:55,338 --> 00:23:58,124 STUDENT 2: When there's a scenario-- 412 00:23:58,124 --> 00:23:59,290 PROFESSOR YEN-JIE LEE: Yeah? 413 00:23:59,290 --> 00:24:01,266 STUDENT 2: [INAUDIBLE] also [INAUDIBLE].. 414 00:24:01,266 --> 00:24:02,972 PROFESSOR YEN-JIE LEE: This one? 415 00:24:02,972 --> 00:24:07,578 STUDENT 2: No, there should be one [INAUDIBLE].. 416 00:24:07,578 --> 00:24:08,744 PROFESSOR YEN-JIE LEE: Here? 417 00:24:08,744 --> 00:24:09,225 STUDENT 2: Yeah. 418 00:24:09,225 --> 00:24:10,308 PROFESSOR YEN-JIE LEE: Ah. 419 00:24:10,308 --> 00:24:12,524 OK, maybe I made a mistake somewhere. 420 00:24:22,690 --> 00:24:26,880 Yeah I think it should be plus, right? 421 00:24:26,880 --> 00:24:27,380 OK. 422 00:24:30,380 --> 00:24:31,117 All right. 423 00:24:31,117 --> 00:24:31,950 Thank you very much. 424 00:24:34,750 --> 00:24:36,850 So now we have all the ingredients 425 00:24:36,850 --> 00:24:41,500 and we assume that it has a normal mode of aj, yj 426 00:24:41,500 --> 00:24:43,130 in this functional form. 427 00:24:43,130 --> 00:24:44,350 And then now I can-- 428 00:24:44,350 --> 00:24:47,858 the next step is to get m minus one k matrix. 429 00:24:47,858 --> 00:24:49,100 All right. 430 00:24:49,100 --> 00:24:50,850 So what is actually m metrics? 431 00:24:50,850 --> 00:24:53,050 M matrix is really really straightforward. 432 00:24:53,050 --> 00:25:00,940 It's m, m, m in the diagonal terms, 433 00:25:00,940 --> 00:25:02,960 and all the rest of the terms are zero. 434 00:25:02,960 --> 00:25:04,130 All right. 435 00:25:04,130 --> 00:25:05,980 And those, of course, you can also 436 00:25:05,980 --> 00:25:09,940 write down k matrix, right? 437 00:25:09,940 --> 00:25:14,004 So the k matrix will be equal to-- 438 00:25:14,004 --> 00:25:17,020 there are many terms, and in the middle 439 00:25:17,020 --> 00:25:25,180 you have minus t over a, two t over a, minus t over a, 440 00:25:25,180 --> 00:25:27,780 and all the rest of the terms are zero. 441 00:25:27,780 --> 00:25:31,030 And of course these patterns go on and on. 442 00:25:31,030 --> 00:25:37,780 Minus t over a, two t over A, minus t over a and zeros, et 443 00:25:37,780 --> 00:25:38,740 cetera et cetera. 444 00:25:38,740 --> 00:25:43,790 And this pattern is going to go on forever, because this 445 00:25:43,790 --> 00:25:47,650 is actually infinitly long matrix, infinite times 446 00:25:47,650 --> 00:25:49,540 infinitely long matrix. 447 00:25:53,200 --> 00:25:58,490 OK, so once we have this, we can now write down 448 00:25:58,490 --> 00:26:01,916 the m minus one k matrix. 449 00:26:01,916 --> 00:26:04,845 What is actually the m minus one k matrix? 450 00:26:04,845 --> 00:26:10,840 It has a similar structure to k matrix, right? 451 00:26:10,840 --> 00:26:12,600 All those are zeros-- 452 00:26:12,600 --> 00:26:14,520 OK, all those are the other values, 453 00:26:14,520 --> 00:26:24,380 but it has a fixed structure minus t over ma, two t over ma, 454 00:26:24,380 --> 00:26:29,660 minus t over ma, and then zeros. 455 00:26:29,660 --> 00:26:35,960 And this will go on forever in the diagonal term and also 456 00:26:35,960 --> 00:26:37,640 the next two diagonal terms. 457 00:26:37,640 --> 00:26:41,520 And all the rest of the terms. are zero. 458 00:26:41,520 --> 00:26:44,155 OK, Any questions? 459 00:26:48,770 --> 00:26:50,750 OK, so that's really nice. 460 00:26:50,750 --> 00:26:53,240 Now we have our m minus one k matrix, 461 00:26:53,240 --> 00:26:54,980 and the good news is that you don't 462 00:26:54,980 --> 00:26:58,790 have to solve m minus one k matrix's Eigenvalue again, 463 00:26:58,790 --> 00:26:59,360 right? 464 00:26:59,360 --> 00:27:01,880 Because we have solved the Eigenvalue problem 465 00:27:01,880 --> 00:27:05,540 of s matrix, therefore what is our left over 466 00:27:05,540 --> 00:27:11,280 is to multiply m minus one k by a, right? 467 00:27:11,280 --> 00:27:19,130 a is actually one of the Eigenvectors of s matrix. 468 00:27:19,130 --> 00:27:23,900 OK, so I am going to multiply that for you, 469 00:27:23,900 --> 00:27:30,010 and now we calculate m minus one k equal to omega square a, 470 00:27:30,010 --> 00:27:34,480 then I can get omega square out of this calculation. 471 00:27:34,480 --> 00:27:36,740 OK? 472 00:27:36,740 --> 00:27:39,920 If I again focus on this term. 473 00:27:42,830 --> 00:27:44,770 OK, so basically, what I am going to get 474 00:27:44,770 --> 00:27:51,880 is, right hand side, I have omega square aj, OK? 475 00:27:51,880 --> 00:27:55,420 Now that's actually from the right hand side, OK? 476 00:27:55,420 --> 00:28:00,940 And left hand side m minus one k times a, what I'm going to get 477 00:28:00,940 --> 00:28:16,641 is t over ma minus aj minus one plus two aj minus aj plus one. 478 00:28:16,641 --> 00:28:17,140 All right? 479 00:28:17,140 --> 00:28:20,350 Because if you take this term-- 480 00:28:20,350 --> 00:28:26,680 this term is actually in the exact diagonal 481 00:28:26,680 --> 00:28:30,310 of this m minus one k matrix, therefore this matches with j, 482 00:28:30,310 --> 00:28:36,270 and this will match with j minus one; match with j plus one. 483 00:28:36,270 --> 00:28:40,750 OK, therefore, if you multiply m minus one k and the a, you get 484 00:28:40,750 --> 00:28:46,510 this result. OK? 485 00:28:46,510 --> 00:28:53,810 And we also know that aj is proportional to exponential 486 00:28:53,810 --> 00:28:55,870 ijka, right? 487 00:28:55,870 --> 00:28:59,430 Therefore I can take aj out of this 488 00:28:59,430 --> 00:29:07,930 and basically I get t over ma aj minus exponential minus 489 00:29:07,930 --> 00:29:11,640 ika plus two. 490 00:29:11,640 --> 00:29:19,350 Because I take aj out of this bracket, OK. 491 00:29:19,350 --> 00:29:23,150 And minus exponential ika. 492 00:29:26,710 --> 00:29:27,620 OK. 493 00:29:27,620 --> 00:29:30,830 Now I actually can cancel aj. 494 00:29:30,830 --> 00:29:33,200 Basically, what I get is, omega square 495 00:29:33,200 --> 00:29:43,200 will be equal to t over ma two minus exponential ika 496 00:29:43,200 --> 00:29:45,884 plus exponential minus ika. 497 00:29:50,730 --> 00:29:57,161 And that will be equal to two t over ma. 498 00:29:57,161 --> 00:29:57,660 OK. 499 00:30:00,330 --> 00:30:07,200 One minus-- OK, so exponential ika plus exponential minus ika, 500 00:30:07,200 --> 00:30:12,960 you are going to get two cosine ka, all right? 501 00:30:12,960 --> 00:30:16,594 Therefore, you get one minus cosine ka. 502 00:30:20,797 --> 00:30:22,200 OK. 503 00:30:22,200 --> 00:30:31,770 I define omega zero to be square root of t over ma, 504 00:30:31,770 --> 00:30:33,870 just to make my life easier. 505 00:30:33,870 --> 00:30:35,180 OK? 506 00:30:35,180 --> 00:30:39,270 Then, what is going to happen is that I 507 00:30:39,270 --> 00:30:48,560 will have omega square equal to two 508 00:30:48,560 --> 00:30:56,450 omega zero square one minus cosine ka. 509 00:30:56,450 --> 00:30:57,650 Any questions? 510 00:31:02,510 --> 00:31:09,600 OK, of course if you like, you can also rewrite this as four 511 00:31:09,600 --> 00:31:21,800 omega zero square sine square ka divided by two. 512 00:31:21,800 --> 00:31:25,230 OK, so if you like. 513 00:31:25,230 --> 00:31:28,960 OK, so look at what we have done. 514 00:31:28,960 --> 00:31:33,280 We studied a highly symmetric system, 515 00:31:33,280 --> 00:31:35,800 which is as you were shown in the slide. 516 00:31:35,800 --> 00:31:41,050 OK, basically you satisfy the space translation symmetry. 517 00:31:41,050 --> 00:31:41,880 OK? 518 00:31:41,880 --> 00:31:44,080 Now what we have been doing is to derive 519 00:31:44,080 --> 00:31:47,830 the equation of motion, make use of the small angle 520 00:31:47,830 --> 00:31:51,670 approximation, then you will be able to find that, 521 00:31:51,670 --> 00:31:55,390 OK, only the y direction is actually 522 00:31:55,390 --> 00:31:57,850 moving as a function of time. 523 00:31:57,850 --> 00:32:02,110 Therefore, based on this derivation of m minus one k 524 00:32:02,110 --> 00:32:06,920 matrix, I arrive, and also based on the equation of motion, 525 00:32:06,920 --> 00:32:09,280 which I derived from the first diagram, 526 00:32:09,280 --> 00:32:12,190 I get this m minus one k matrix. 527 00:32:12,190 --> 00:32:16,305 And since we know that m minus one k matrix and s 528 00:32:16,305 --> 00:32:18,780 matrix will share this Eigenvector, 529 00:32:18,780 --> 00:32:24,220 I can multiply m minus one k matrix, and I come back to a. 530 00:32:24,220 --> 00:32:27,340 Then I will be able to solve the functional 531 00:32:27,340 --> 00:32:33,970 form of omega square, and that is actually given here. 532 00:32:33,970 --> 00:32:38,320 Omega square is equal to two omega zero square y 533 00:32:38,320 --> 00:32:40,240 minus cosine ka. 534 00:32:40,240 --> 00:32:41,860 OK? 535 00:32:41,860 --> 00:32:49,710 And is this actually telling you that omega is a function of k. 536 00:32:49,710 --> 00:32:51,540 OK. 537 00:32:51,540 --> 00:32:52,884 What is k? 538 00:32:52,884 --> 00:33:01,730 k is actually the wave number and omega is actually 539 00:33:01,730 --> 00:33:06,610 the angular frequency of the normal modes, right? 540 00:33:06,610 --> 00:33:10,665 So that means what we were talking about-- 541 00:33:10,665 --> 00:33:20,960 that means if we fix the wavelength or the wave number, 542 00:33:20,960 --> 00:33:27,601 k, then there will be a corresponding omega. 543 00:33:27,601 --> 00:33:28,100 OK? 544 00:33:28,100 --> 00:33:31,040 If you fix the wavelengths you are talking about, 545 00:33:31,040 --> 00:33:37,460 then the omega is also fixed by this omega of k function. 546 00:33:37,460 --> 00:33:41,080 OK, now we actually call it dispersion relation. 547 00:33:50,700 --> 00:33:54,300 This term may not mean much to you now, 548 00:33:54,300 --> 00:33:57,650 but later in the discussion, you will find, aha! 549 00:33:57,650 --> 00:34:00,960 It really makes sense, and that we will talk about dispersion 550 00:34:00,960 --> 00:34:03,315 in the later lectures. 551 00:34:06,160 --> 00:34:11,010 OK, so the conclusion from here is that, basically, 552 00:34:11,010 --> 00:34:17,000 if I have this distance that satisfies this translation 553 00:34:17,000 --> 00:34:25,010 symmetry, then what it tells us is that the normal modes, what 554 00:34:25,010 --> 00:34:31,260 looks like some kind of sinusoidal function, 555 00:34:31,260 --> 00:34:32,550 as we discussed last time. 556 00:34:35,139 --> 00:34:39,219 And also this-- 557 00:34:39,219 --> 00:34:44,260 OK, so this is actually the amplitude, what 558 00:34:44,260 --> 00:34:47,530 I am drawing here, this curve. 559 00:34:47,530 --> 00:34:53,800 And all those masses are only moving up and down, OK? 560 00:34:53,800 --> 00:34:55,250 As a function of time. 561 00:34:55,250 --> 00:35:00,040 And this is aj, and that is the oscillation frequency, which 562 00:35:00,040 --> 00:35:03,190 is actually the frequency of moving up and down, 563 00:35:03,190 --> 00:35:05,790 this kind of motion, is actually omega. 564 00:35:08,510 --> 00:35:15,140 And also, we learned that omega is equal to omega of k. 565 00:35:15,140 --> 00:35:19,030 And that is actually decided by the length, 566 00:35:19,030 --> 00:35:25,190 and how distorted is this normal mode-- 567 00:35:25,190 --> 00:35:27,110 the shape of the normal mode? 568 00:35:27,110 --> 00:35:31,190 And this is actually determined by the k, which is actually 569 00:35:31,190 --> 00:35:36,770 the wave number, and of course you can also get the wavelength 570 00:35:36,770 --> 00:35:38,810 from two pi over k. 571 00:35:38,810 --> 00:35:41,180 OK? 572 00:35:41,180 --> 00:35:44,930 In short, if you give it a specific wave 573 00:35:44,930 --> 00:35:49,370 number or wavelength, than the oscillation frequency 574 00:35:49,370 --> 00:35:53,830 is already fixed because of the equation of motion, 575 00:35:53,830 --> 00:35:57,200 which we did, right, from the first diagram. 576 00:35:57,200 --> 00:35:58,388 Any questions? 577 00:36:04,250 --> 00:36:05,980 OK. 578 00:36:05,980 --> 00:36:08,080 The last point which I would like to remind you 579 00:36:08,080 --> 00:36:12,730 is that, at this point, since we are talking 580 00:36:12,730 --> 00:36:17,110 about infinitely long systems, therefore 581 00:36:17,110 --> 00:36:22,226 all possible k are allowed. 582 00:36:22,226 --> 00:36:23,020 Right? 583 00:36:23,020 --> 00:36:25,400 Because, basically, you have an infinite number 584 00:36:25,400 --> 00:36:28,370 of coupled oscillators, and therefore you 585 00:36:28,370 --> 00:36:31,280 have an infinite number of normal modes. 586 00:36:31,280 --> 00:36:38,030 So all possible cases are allowed, and that actually 587 00:36:38,030 --> 00:36:41,300 because we have even an infinitely long system. 588 00:36:41,300 --> 00:36:45,080 After the break, which we will take a five minute break, 589 00:36:45,080 --> 00:36:48,950 we will discuss how to use infinitely long systems 590 00:36:48,950 --> 00:36:53,340 to actually understand a finite system. 591 00:36:53,340 --> 00:36:56,840 So you will see that, actually I can use, now, 592 00:36:56,840 --> 00:36:59,750 this space translation symmetry, and to solve, 593 00:36:59,750 --> 00:37:02,690 in general, infinitely long systems. 594 00:37:02,690 --> 00:37:05,815 And I can actually even go back to find to a finite system 595 00:37:05,815 --> 00:37:07,860 and see what we can get from there. 596 00:37:07,860 --> 00:37:08,360 OK? 597 00:37:08,360 --> 00:37:12,920 So we will be back at 12:20. 598 00:37:12,920 --> 00:37:18,370 If you have any questions, I will be here 599 00:37:18,370 --> 00:37:20,530 OK, welcome back, everybody. 600 00:37:20,530 --> 00:37:23,590 So we will continue the discussion. 601 00:37:23,590 --> 00:37:27,680 So there were a few questions asked during the break. 602 00:37:27,680 --> 00:37:30,730 So, the first question is related to how we actually 603 00:37:30,730 --> 00:37:32,870 arrive at this equation. 604 00:37:32,870 --> 00:37:35,620 And that is actually because-- 605 00:37:35,620 --> 00:37:39,180 OK, two t over ma is actually really happening 606 00:37:39,180 --> 00:37:40,570 in a diagonal term. 607 00:37:40,570 --> 00:37:43,370 Therefore, if you multiply m minus one k 608 00:37:43,370 --> 00:37:48,200 matrix and A matrix, which is actually shown there, 609 00:37:48,200 --> 00:37:51,750 then you will get this term, minus t 610 00:37:51,750 --> 00:37:54,770 over ma multiplied by aj minus one 611 00:37:54,770 --> 00:37:59,860 plus two t over ma times aj price minus t 612 00:37:59,860 --> 00:38:02,560 over ma times aj plus one. 613 00:38:02,560 --> 00:38:06,070 And that is actually why we can arrive at this expression. 614 00:38:06,070 --> 00:38:06,610 OK? 615 00:38:06,610 --> 00:38:08,990 Then what happens afterward is that we 616 00:38:08,990 --> 00:38:13,180 found that aj can be factorized out, and they cancel. 617 00:38:13,180 --> 00:38:17,560 And then now, my solution depends now upon the amplitude, 618 00:38:17,560 --> 00:38:23,560 and still omega is actually dependent on the k value, which 619 00:38:23,560 --> 00:38:24,540 we actually choose. 620 00:38:24,540 --> 00:38:25,780 OK. 621 00:38:25,780 --> 00:38:32,370 The second question is, why do I say k is the wave number? 622 00:38:32,370 --> 00:38:34,390 OK, where is that coming from? 623 00:38:34,390 --> 00:38:36,220 So that is because-- 624 00:38:36,220 --> 00:38:40,250 OK, so aj is proportional to exponential ijka. 625 00:38:40,250 --> 00:38:40,750 OK? 626 00:38:40,750 --> 00:38:43,060 It has a fancy name. 627 00:38:43,060 --> 00:38:48,900 If I take the real part, OK, as we 628 00:38:48,900 --> 00:38:51,370 did when we went to the description 629 00:38:51,370 --> 00:38:55,840 of physical systems, then you get cosine jka. 630 00:38:55,840 --> 00:38:57,070 OK? 631 00:38:57,070 --> 00:39:00,730 And j times a is actually-- 632 00:39:00,730 --> 00:39:02,470 j is actually a label, right? 633 00:39:02,470 --> 00:39:05,260 Labeling which mass I am talking about. 634 00:39:05,260 --> 00:39:07,960 A is actually the distance between all those masses. 635 00:39:07,960 --> 00:39:14,440 j times a will give you the x location of the mass. 636 00:39:14,440 --> 00:39:19,970 So j times a is actually the the x position of the mass. 637 00:39:19,970 --> 00:39:20,470 OK? 638 00:39:20,470 --> 00:39:26,800 Therefore, if you accept that, this becomes cosine kx, 639 00:39:26,800 --> 00:39:28,690 and from there you will see immediately 640 00:39:28,690 --> 00:39:32,580 that k has a meaning, which is actually the wave number. 641 00:39:32,580 --> 00:39:34,630 OK. 642 00:39:34,630 --> 00:39:37,240 All right, is that? 643 00:39:37,240 --> 00:39:42,430 OK, so that was the questions raised, 644 00:39:42,430 --> 00:39:45,130 which I can quickly explain. 645 00:39:45,130 --> 00:39:47,960 So, what I am going to do now is that-- 646 00:39:47,960 --> 00:39:52,120 OK, we have solved, in general, an infinitely long system. 647 00:39:52,120 --> 00:39:56,000 What are actually the resulting normal modes 648 00:39:56,000 --> 00:39:57,470 of infinitely long systems? 649 00:39:57,470 --> 00:40:02,830 It has an infinite number of normal modes, 650 00:40:02,830 --> 00:40:06,970 and we will wonder if I can actually 651 00:40:06,970 --> 00:40:08,980 borrow this infinitely long system 652 00:40:08,980 --> 00:40:14,680 and solve finite systems to see if I can arrive at the solution 653 00:40:14,680 --> 00:40:15,890 really quickly. 654 00:40:15,890 --> 00:40:16,620 OK? 655 00:40:16,620 --> 00:40:20,320 So the answer is actually yes. 656 00:40:20,320 --> 00:40:25,210 So if I consider a finite system that looks like this; 657 00:40:25,210 --> 00:40:32,510 so I have many, many little masses on this system 658 00:40:32,510 --> 00:40:35,620 and they are connected to each other 659 00:40:35,620 --> 00:40:41,290 by the center strings, which I prepared before, OK? 660 00:40:41,290 --> 00:40:44,830 And I call this the position in the y 661 00:40:44,830 --> 00:40:51,310 direction of this object y1, and then the next object y2, y3, 662 00:40:51,310 --> 00:40:52,330 etc. 663 00:40:52,330 --> 00:40:56,670 And I have an object in this system 664 00:40:56,670 --> 00:41:03,310 and both ends of the string are fixed on the wall. 665 00:41:03,310 --> 00:41:05,380 OK? 666 00:41:05,380 --> 00:41:10,270 So I can actually now argue that the infinitely long system 667 00:41:10,270 --> 00:41:13,540 can help us with the understanding 668 00:41:13,540 --> 00:41:15,460 of this finite system. 669 00:41:15,460 --> 00:41:17,150 Why is that? 670 00:41:17,150 --> 00:41:20,860 That is because, now, I can assume that, huh, this 671 00:41:20,860 --> 00:41:24,870 is actually just part of an infinitely long system. 672 00:41:24,870 --> 00:41:26,440 All right. 673 00:41:26,440 --> 00:41:29,710 So I construct my infinitely long system, 674 00:41:29,710 --> 00:41:40,270 and now I nail the yth mass, I nail the y n plus one mass, 675 00:41:40,270 --> 00:41:43,030 and I fix that so that it cannot move, OK? 676 00:41:43,030 --> 00:41:46,720 So it's still an infinitely long system, 677 00:41:46,720 --> 00:41:52,090 but there are two interesting boundary conditions 678 00:41:52,090 --> 00:41:57,481 at j equal to zero and j equal to n plus one. 679 00:41:57,481 --> 00:41:57,980 OK. 680 00:41:57,980 --> 00:42:00,340 What are the two boundary conditions? 681 00:42:08,880 --> 00:42:15,260 The first one is y zero equal to zero. 682 00:42:15,260 --> 00:42:16,610 OK? 683 00:42:16,610 --> 00:42:24,060 And the second condition is y n plus one equal to zero. 684 00:42:24,060 --> 00:42:25,520 OK? 685 00:42:25,520 --> 00:42:27,280 So there are two boundary conditions, 686 00:42:27,280 --> 00:42:31,180 so basically what I'm looking at is still an infinitely long 687 00:42:31,180 --> 00:42:36,800 system, but I require y zero and y n plus one 688 00:42:36,800 --> 00:42:39,610 to satisfy these two conditions. 689 00:42:39,610 --> 00:42:40,550 OK? 690 00:42:40,550 --> 00:42:44,090 And we will find that, huh, with this procedure, 691 00:42:44,090 --> 00:42:48,165 we can also solve this finite number of couple oscillators. 692 00:42:48,165 --> 00:42:52,610 The problem, in this case we, have coupled oscillators. 693 00:42:52,610 --> 00:42:54,317 OK? 694 00:42:54,317 --> 00:42:56,150 So the first thing which I would like to say 695 00:42:56,150 --> 00:43:01,190 is, based on the functional form, the functional 696 00:43:01,190 --> 00:43:05,660 form of omega square, now this is equal to four omega 697 00:43:05,660 --> 00:43:09,380 zero square sine square ka over two. 698 00:43:09,380 --> 00:43:10,220 OK? 699 00:43:10,220 --> 00:43:15,260 What we actually have is that omega k is 700 00:43:15,260 --> 00:43:19,670 equal to omega minus k. 701 00:43:19,670 --> 00:43:20,720 OK? 702 00:43:20,720 --> 00:43:24,920 So both of them will give you the same angular frequency. 703 00:43:24,920 --> 00:43:29,250 OK, therefore, what does that mean? 704 00:43:29,250 --> 00:43:32,080 This means that linear combination 705 00:43:32,080 --> 00:43:39,260 of exponential ijka and the exponential minus ijka-- 706 00:43:42,250 --> 00:43:46,545 OK, linear combination of these two vectors-- 707 00:43:49,240 --> 00:43:54,980 is also an Eigenvector of m minus one k metrics. 708 00:43:54,980 --> 00:43:58,290 OK, so you can do linear combination 709 00:43:58,290 --> 00:44:00,995 of these two vectors. 710 00:44:06,390 --> 00:44:13,380 OK, so if we do that, now I can guess my solution 711 00:44:13,380 --> 00:44:23,340 will be like yj equal to real part of exponential i omega t 712 00:44:23,340 --> 00:44:26,100 plus phi. 713 00:44:26,100 --> 00:44:30,500 I can now have a linear combination of exponential ijka 714 00:44:30,500 --> 00:44:33,240 and the exponential minus ijka. 715 00:44:33,240 --> 00:44:39,670 Basically, I have alpha exponential ijka 716 00:44:39,670 --> 00:44:43,600 plus beta exponential minus ijka. 717 00:44:48,400 --> 00:44:50,120 OK? 718 00:44:50,120 --> 00:44:52,580 And I would like to determine why that's actually 719 00:44:52,580 --> 00:44:57,800 alpha and beta which actually satisfy these boundary 720 00:44:57,800 --> 00:45:01,010 conditions, one and two. 721 00:45:01,010 --> 00:45:06,290 OK, so now I can use the first boundary condition, 722 00:45:06,290 --> 00:45:09,740 y zero equal to zero, right? 723 00:45:09,740 --> 00:45:11,810 So j equal to zero. 724 00:45:11,810 --> 00:45:16,510 Therefore, basically, what I get is, when j is equal to zero, 725 00:45:16,510 --> 00:45:21,060 then this is actually one and this is actually one, right? 726 00:45:21,060 --> 00:45:27,530 And this actually gives you y zero equal to zero. 727 00:45:27,530 --> 00:45:31,160 If y zero is equal to 0 at all times, 728 00:45:31,160 --> 00:45:36,320 no matter what t as you give it to this system, then basically 729 00:45:36,320 --> 00:45:41,060 you have alpha plus beta equal to zero, right? 730 00:45:41,060 --> 00:45:43,580 Because j is equal to zero. 731 00:45:43,580 --> 00:45:49,085 So you have alpha plus beta, and that has to be equal to zero. 732 00:45:49,085 --> 00:45:52,850 Therefore, you can conclude that alpha is equal to minus beta. 733 00:45:55,980 --> 00:46:00,000 And I've reused the second boundary condition, 734 00:46:00,000 --> 00:46:05,270 y n plus one equal to zero, because I nailed this mass 735 00:46:05,270 --> 00:46:08,510 and then fixed that so that it cannot move. 736 00:46:08,510 --> 00:46:13,810 Then basically, what you get is y n plus one is equal to zero, 737 00:46:13,810 --> 00:46:21,510 then basically you have alpha exponential i n 738 00:46:21,510 --> 00:46:24,735 plus one ka plus-- 739 00:46:28,350 --> 00:46:32,460 okay, so beta is your equal to minus alpha, right? 740 00:46:32,460 --> 00:46:44,530 So basically, you can get minus exponential minus i 741 00:46:44,530 --> 00:46:49,000 n plus one ka, right? 742 00:46:49,000 --> 00:46:51,640 Multiplied by alpha. 743 00:46:51,640 --> 00:46:56,720 And now this is actually equal to zero. 744 00:46:56,720 --> 00:46:59,150 Now we have the choice. 745 00:46:59,150 --> 00:47:02,690 We can actually set alpha to be equal to zero, 746 00:47:02,690 --> 00:47:05,300 but if I set alpha to be equal to zero, 747 00:47:05,300 --> 00:47:08,030 then beta is also zero. 748 00:47:08,030 --> 00:47:10,040 Then I have zero everywhere, right? 749 00:47:10,040 --> 00:47:11,810 Then there's no oscillation. 750 00:47:11,810 --> 00:47:15,450 And that's not fun, right? 751 00:47:15,450 --> 00:47:16,260 OK. 752 00:47:16,260 --> 00:47:18,790 Therefore, what I'm going to set is actually 753 00:47:18,790 --> 00:47:21,030 the second turn equal to zero. 754 00:47:21,030 --> 00:47:21,530 OK. 755 00:47:21,530 --> 00:47:24,020 The second turn, I can actually simplify 756 00:47:24,020 --> 00:47:33,980 that to be two i sine n plus one ka. 757 00:47:33,980 --> 00:47:37,380 And now this is actually equal to zero. 758 00:47:37,380 --> 00:47:38,290 OK. 759 00:47:38,290 --> 00:47:42,150 What is actually the condition of this thing equal to zero? 760 00:47:42,150 --> 00:47:46,640 Basically, n plus 1 is actually a given number, 761 00:47:46,640 --> 00:47:52,040 a is actually the distance between those masses, 762 00:47:52,040 --> 00:47:56,690 therefore, what I can actually change is the k value. 763 00:47:56,690 --> 00:47:57,530 Right? 764 00:47:57,530 --> 00:48:00,500 So I can now solve this condition, 765 00:48:00,500 --> 00:48:03,080 and I will conclude that k will have 766 00:48:03,080 --> 00:48:09,800 to be equal to n times pi divided by N plus one. 767 00:48:09,800 --> 00:48:12,500 I hope you can see it. 768 00:48:12,500 --> 00:48:24,550 Where small n is equal to one, two, three, until capital N. 769 00:48:24,550 --> 00:48:27,500 So what does that mean? 770 00:48:27,500 --> 00:48:29,450 This means that-- 771 00:48:29,450 --> 00:48:33,730 OK, originally, before I introduced the boundary 772 00:48:33,730 --> 00:48:38,080 condition, this system is infinitely long, OK, 773 00:48:38,080 --> 00:48:45,000 and it has an infinite number of normal modes, right? 774 00:48:45,000 --> 00:48:47,880 But once I introduced these boundary conditions, 775 00:48:47,880 --> 00:48:53,760 which I actually require y zero equal to zero, 776 00:48:53,760 --> 00:48:57,080 because I fixed this point on the wall. 777 00:48:57,080 --> 00:49:00,660 Y m plus one equal to zero because I fixed, also, 778 00:49:00,660 --> 00:49:02,910 that point on the wall. 779 00:49:02,910 --> 00:49:05,220 Something really happened. 780 00:49:05,220 --> 00:49:14,960 Now it actually gives us, first, the shape of the system when it 781 00:49:14,960 --> 00:49:16,470 is actually in the normal mode. 782 00:49:16,470 --> 00:49:18,840 Basically, the shape-- what I mean here 783 00:49:18,840 --> 00:49:22,350 is the amplitude as a function of j, OK? 784 00:49:22,350 --> 00:49:24,600 That's actually what I mean by shape, OK? 785 00:49:24,600 --> 00:49:29,430 The shape is now like a sine function. 786 00:49:29,430 --> 00:49:32,670 That's the first thing which we get from here. 787 00:49:32,670 --> 00:49:35,540 The second thing which we get here 788 00:49:35,540 --> 00:49:41,750 is that now, the k values are not arbitrary anymore. 789 00:49:41,750 --> 00:49:47,290 The k values are equal to n pi over n plus one. 790 00:49:47,290 --> 00:49:51,910 And the the small n is actually equal for one, two, three, 791 00:49:51,910 --> 00:49:57,280 until N. So now, once you actually fix this two point, 792 00:49:57,280 --> 00:50:05,370 you actually have only how many normal modes? 793 00:50:05,370 --> 00:50:07,990 N normal modes! 794 00:50:07,990 --> 00:50:08,830 Right? 795 00:50:08,830 --> 00:50:12,880 So what I want to tell you is that, in general, 796 00:50:12,880 --> 00:50:16,690 the sinusoidal shape is actually fixed already 797 00:50:16,690 --> 00:50:20,440 by this translation symmetry argument. 798 00:50:20,440 --> 00:50:21,240 OK. 799 00:50:21,240 --> 00:50:26,560 And once we nail both sides-- 800 00:50:26,560 --> 00:50:29,680 actually, we also restrict ourselves 801 00:50:29,680 --> 00:50:37,360 to the discussion of only a few k which actually satisfy 802 00:50:37,360 --> 00:50:38,730 the boundary conditions. 803 00:50:38,730 --> 00:50:43,150 And if I plot all those normal modes as a function of i, 804 00:50:43,150 --> 00:50:46,780 basically what you can see from here, 805 00:50:46,780 --> 00:50:50,800 you can see if I have n equals to one and capital 806 00:50:50,800 --> 00:50:54,560 N equal to four in this case. 807 00:50:54,560 --> 00:50:56,620 OK, so I have four-- 808 00:50:56,620 --> 00:51:00,200 basically, I'm going to have four normal modes. 809 00:51:00,200 --> 00:51:06,510 The first one will be like a really long wavelength 810 00:51:06,510 --> 00:51:09,730 one, when n is equal to one. 811 00:51:09,730 --> 00:51:14,350 And if I increase the small n value 812 00:51:14,350 --> 00:51:17,110 so that the k becomes bigger, then you 813 00:51:17,110 --> 00:51:23,330 can see that there is more distortion when this system is 814 00:51:23,330 --> 00:51:24,910 in one of the normal modes. 815 00:51:24,910 --> 00:51:30,440 And this shape is actually going to be oscillating up and down, 816 00:51:30,440 --> 00:51:31,170 instead of-- 817 00:51:31,170 --> 00:51:34,230 OK, so all those points are only moving up and down, right? 818 00:51:34,230 --> 00:51:35,140 Just a reminder. 819 00:51:35,140 --> 00:51:35,790 OK? 820 00:51:35,790 --> 00:51:41,170 And why do we have all those cases? 821 00:51:41,170 --> 00:51:45,300 Because of the boundary conditions. 822 00:51:45,300 --> 00:51:46,871 OK, any questions? 823 00:51:50,720 --> 00:51:54,150 OK, so I have several other cases, 824 00:51:54,150 --> 00:51:57,720 which is open end and the closed end, and also 825 00:51:57,720 --> 00:52:01,320 the driven and the coupled oscillator examples. 826 00:52:01,320 --> 00:52:04,190 Also in the lecture notes, but unfortunately, we 827 00:52:04,190 --> 00:52:06,050 are will not be able to go over them, 828 00:52:06,050 --> 00:52:08,240 but I think they are very, very detailed, 829 00:52:08,240 --> 00:52:10,400 the notes in the lecture notes. 830 00:52:10,400 --> 00:52:11,360 OK. 831 00:52:11,360 --> 00:52:14,980 So, let me-- before I move on to the discussion 832 00:52:14,980 --> 00:52:17,900 of continuous systems, OK, I would 833 00:52:17,900 --> 00:52:23,400 like to discuss with you what we have learned so far. 834 00:52:23,400 --> 00:52:27,200 So what we have learned is that, if I have a symmetry which 835 00:52:27,200 --> 00:52:29,810 is a translation symmetry, and plus, 836 00:52:29,810 --> 00:52:34,730 we only limit ourselves in the discussion of oscillation. 837 00:52:34,730 --> 00:52:39,530 OK, in other words, we limit the amplitude 838 00:52:39,530 --> 00:52:42,680 so that it doesn't explode at the edge of the universe. 839 00:52:42,680 --> 00:52:43,640 OK. 840 00:52:43,640 --> 00:52:46,760 And I will give you a beta value which 841 00:52:46,760 --> 00:52:52,160 is the functional form of exponential ika, 842 00:52:52,160 --> 00:52:57,290 and equation of motion can be derived from the first diagram. 843 00:52:57,290 --> 00:53:01,070 Once we entered the equation of motion, 844 00:53:01,070 --> 00:53:03,320 we can get m minus one k matrix, then 845 00:53:03,320 --> 00:53:07,630 we can derive omega square from this expression. 846 00:53:07,630 --> 00:53:12,860 And finally, we actually can simplify everything and then 847 00:53:12,860 --> 00:53:15,470 get the dispersion relation omega 848 00:53:15,470 --> 00:53:20,050 equal to omega k, which is a function of k, the wave number. 849 00:53:22,640 --> 00:53:25,850 Before we actually introduce boundary conditions 850 00:53:25,850 --> 00:53:30,420 to go from an infinitely long system to a finite system, 851 00:53:30,420 --> 00:53:33,610 all the k values are allowed. 852 00:53:33,610 --> 00:53:36,290 Once you introduce boundary conditions, 853 00:53:36,290 --> 00:53:42,050 you find that you only have a limited number of normal modes. 854 00:53:42,050 --> 00:53:46,460 Second, the k value not continuous at any value 855 00:53:46,460 --> 00:53:48,410 any more, it becomes discrete. 856 00:53:48,410 --> 00:53:53,710 And only n values allowed from this exercise. 857 00:53:53,710 --> 00:53:57,410 And finally, what is actually the most general solution 858 00:53:57,410 --> 00:53:59,690 is actually the linear combination 859 00:53:59,690 --> 00:54:03,760 of all those normal modes, which we show here. 860 00:54:03,760 --> 00:54:08,760 And what is actually the ratio between all those normal modes? 861 00:54:08,760 --> 00:54:11,480 All those free coefficients are determined 862 00:54:11,480 --> 00:54:14,220 by initial conditions if you are given. 863 00:54:14,220 --> 00:54:15,650 OK? 864 00:54:15,650 --> 00:54:18,485 So that's actually what we have learned so far. 865 00:54:22,160 --> 00:54:29,240 And now I would like to make a leap of faith 866 00:54:29,240 --> 00:54:30,340 to see what happens. 867 00:54:30,340 --> 00:54:34,460 OK, what we are going to do is to introduce you 868 00:54:34,460 --> 00:54:39,950 to a continuous infinite number of coupled oscillators. 869 00:54:39,950 --> 00:54:40,850 OK? 870 00:54:40,850 --> 00:54:41,970 So what does that mean? 871 00:54:41,970 --> 00:54:44,900 What I am going to do is to go from this-- 872 00:54:44,900 --> 00:54:55,080 so I have t and then a t, a lot of string and mass system, 873 00:54:55,080 --> 00:54:56,700 et cetera, et cetera, but I would 874 00:54:56,700 --> 00:55:06,510 like to go from there to just an infinitely long string 875 00:55:06,510 --> 00:55:15,110 with string tension t and some kind of density or mass. 876 00:55:15,110 --> 00:55:16,240 OK? 877 00:55:16,240 --> 00:55:19,830 We like to make it continuous to see what will happen. 878 00:55:19,830 --> 00:55:21,690 OK? 879 00:55:21,690 --> 00:55:30,710 So, just a reminder of the jth term of the m minus one 880 00:55:30,710 --> 00:55:32,250 k matrix operation. 881 00:55:32,250 --> 00:55:37,890 Basically, we have m minus one k times a, the j's term 882 00:55:37,890 --> 00:55:40,180 of m minus one k a. 883 00:55:40,180 --> 00:55:45,540 That is actually given by omega square aj. 884 00:55:45,540 --> 00:55:54,480 This is equal to t over ma minus aj minus one 885 00:55:54,480 --> 00:56:01,060 plus two aj minus aj plus one? 886 00:56:01,060 --> 00:56:01,680 OK. 887 00:56:01,680 --> 00:56:05,510 So this is actually just a copy of that formula here. 888 00:56:05,510 --> 00:56:10,570 OK, so if I make it continuous-- 889 00:56:16,360 --> 00:56:19,400 OK, so that means what I'm going to get 890 00:56:19,400 --> 00:56:25,060 is omega square a but evaluated at position 891 00:56:25,060 --> 00:56:30,340 x, where x is actually equal to j times a. 892 00:56:30,340 --> 00:56:32,200 OK? 893 00:56:32,200 --> 00:56:37,750 And this will be equal to t over ma 894 00:56:37,750 --> 00:56:50,067 minus a, evaluated at x minus a plus two a x minus a x plus a. 895 00:56:53,730 --> 00:56:59,100 Now I am going to make this a very, very small, right? 896 00:56:59,100 --> 00:57:02,350 So that, when I make a very, very small, 897 00:57:02,350 --> 00:57:06,240 then it becomes a very, very continuous system. 898 00:57:06,240 --> 00:57:07,530 OK? 899 00:57:07,530 --> 00:57:09,900 So what I'm going to do is-- 900 00:57:09,900 --> 00:57:14,030 I can now make a go to zero. 901 00:57:16,680 --> 00:57:25,470 I can now use Taylor's series fx plus delta x-- 902 00:57:25,470 --> 00:57:29,880 and just a reminder, if you do a Taylor expansion of this, 903 00:57:29,880 --> 00:57:39,640 you are going to get f of x plus delta x f prime x plus one 904 00:57:39,640 --> 00:57:46,080 over two factorial delta x square f double prime x. 905 00:57:49,349 --> 00:57:53,400 OK, so that means now, I can actually 906 00:57:53,400 --> 00:57:59,670 do a Taylor expansion of a x minus a and then a x plus a. 907 00:57:59,670 --> 00:58:05,230 So what I'm going to get is like this. 908 00:58:05,230 --> 00:58:11,770 So a x minus a will become a of x 909 00:58:11,770 --> 00:58:22,240 minus a a prime x plus one over two a square a double prime x. 910 00:58:26,048 --> 00:58:29,500 I can also do the same thing to do a Taylor expansion 911 00:58:29,500 --> 00:58:33,380 for a x plus a. 912 00:58:33,380 --> 00:58:44,960 a x plus a will be equal to a of x plus a prime x plus one 913 00:58:44,960 --> 00:58:51,355 over two a square a double prime x. 914 00:58:51,355 --> 00:58:51,855 OK. 915 00:58:59,620 --> 00:59:02,545 Once I have done this, basically, then, I 916 00:59:02,545 --> 00:59:11,315 can calculate minus a x minus a plus two a 917 00:59:11,315 --> 00:59:17,360 of x minus a x plus a. 918 00:59:17,360 --> 00:59:22,630 OK, I'm just taking the middle term and copying it there. 919 00:59:22,630 --> 00:59:24,170 OK? 920 00:59:24,170 --> 00:59:30,320 If I use, now, this expression, OK, then basically I 921 00:59:30,320 --> 00:59:36,200 will see that, OK, a of x terms actually cancel, right? 922 00:59:36,200 --> 00:59:40,790 Because I have two a of x from these two terms, 923 00:59:40,790 --> 00:59:44,670 and they cancel with this two a x. 924 00:59:44,670 --> 00:59:45,951 OK? 925 00:59:45,951 --> 00:59:50,600 And also, a prime terms, also cancel, right? 926 00:59:50,600 --> 00:59:59,780 Because you have a x minus and a x plus a, therefore, 927 00:59:59,780 --> 01:00:01,790 the x prime turns cancel here. 928 01:00:01,790 --> 01:00:05,010 This is a minus sign, this is a plus sign. 929 01:00:05,010 --> 01:00:05,790 You see? 930 01:00:05,790 --> 01:00:11,550 So therefore, what is actually left over is all those terms. 931 01:00:11,550 --> 01:00:24,060 This is going to give you a double prime x a squared, 932 01:00:24,060 --> 01:00:28,390 plus many other higher order terms, right? 933 01:00:28,390 --> 01:00:31,560 Because those are actually not completed, 934 01:00:31,560 --> 01:00:34,030 we have many, many higher order terms. 935 01:00:34,030 --> 01:00:35,230 OK? 936 01:00:35,230 --> 01:00:39,640 In the limit of x go to zero-- 937 01:00:39,640 --> 01:00:40,398 Yes. 938 01:00:40,398 --> 01:00:43,624 STUDENT: Isn't the xa prime [INAUDIBLE]?? 939 01:00:43,624 --> 01:00:44,790 PROFESSOR YEN-JIE LEE: Yeah. 940 01:00:44,790 --> 01:00:46,860 Oh yeah, you are right. 941 01:00:46,860 --> 01:00:49,280 Thank you very much. 942 01:00:49,280 --> 01:00:50,920 OK, thank you for that. 943 01:00:50,920 --> 01:00:54,370 So there should be a minus sign in front of it, OK. 944 01:00:54,370 --> 01:00:59,470 So basically what we have is a double prime x times a square 945 01:00:59,470 --> 01:01:10,570 plus some higher order turn of a cubed, et cetera, et cetera. 946 01:01:10,570 --> 01:01:16,540 OK, since we are talking about the limit of a goes to zero, 947 01:01:16,540 --> 01:01:20,670 we can safely ignore all the higher order terms. 948 01:01:20,670 --> 01:01:22,120 OK? 949 01:01:22,120 --> 01:01:32,550 So now, if we go back to this equation, omega square a 950 01:01:32,550 --> 01:01:40,840 will become, basically, t over m minus t over ma 951 01:01:40,840 --> 01:01:46,234 a double prime x times a squared. 952 01:01:51,180 --> 01:01:54,900 OK, and then plus some higher order terms. 953 01:01:54,900 --> 01:02:02,850 OK, now I can define rho l to be equal to m divided by a. 954 01:02:02,850 --> 01:02:03,460 OK. 955 01:02:03,460 --> 01:02:07,500 And I will I will have to be very careful when 956 01:02:07,500 --> 01:02:11,490 I go to the continuous limit, OK, I also 957 01:02:11,490 --> 01:02:17,670 do not want to make this system infinitely massive, right? 958 01:02:17,670 --> 01:02:20,700 Therefore, I would like to fix the rho l when 959 01:02:20,700 --> 01:02:24,030 I go to the continuous limit. 960 01:02:24,030 --> 01:02:28,980 So basically what I am doing is that I cut this system in half 961 01:02:28,980 --> 01:02:32,880 so that a becomes smaller and the mass also because smaller, 962 01:02:32,880 --> 01:02:37,320 so that rho l stays as a constant, OK, 963 01:02:37,320 --> 01:02:39,350 when I go to a continuous limit. 964 01:02:39,350 --> 01:02:43,640 OK, so what I'm going to get is omega square a 965 01:02:43,640 --> 01:02:54,010 will be equal to minus t over rho l a double prime x. 966 01:02:54,010 --> 01:02:56,730 And to write it explicitly, this is actually 967 01:02:56,730 --> 01:03:01,620 equal to minus t over rho l. 968 01:03:01,620 --> 01:03:06,730 I'll just square a partial x squared. 969 01:03:06,730 --> 01:03:07,230 OK? 970 01:03:07,230 --> 01:03:11,160 Because each prime is actually the differentiation 971 01:03:11,160 --> 01:03:14,580 which is spread to x, right? 972 01:03:14,580 --> 01:03:15,270 OK. 973 01:03:15,270 --> 01:03:17,220 Don't forget-- what is actually this? 974 01:03:17,220 --> 01:03:21,530 This is actually m minus one ka, right? 975 01:03:21,530 --> 01:03:23,788 Equal to omega square a. 976 01:03:27,180 --> 01:03:32,330 And that this is actually just partial 977 01:03:32,330 --> 01:03:39,860 square a partial t square. 978 01:03:39,860 --> 01:03:40,360 Right? 979 01:03:40,360 --> 01:03:42,850 Because this is actually m minus one k matrix. 980 01:03:42,850 --> 01:03:49,256 It's actually originally-- going back to the original equation, 981 01:03:49,256 --> 01:03:55,330 we are actually solving m x double dot equal 982 01:03:55,330 --> 01:03:57,328 to minus kx problem, right? 983 01:04:00,610 --> 01:04:02,970 So there should be a minus sign there as well. 984 01:04:02,970 --> 01:04:03,830 Right? 985 01:04:03,830 --> 01:04:08,400 So mx double dot equal to minus kx, 986 01:04:08,400 --> 01:04:14,490 therefore x double dot will be equal to minus m minus one kx. 987 01:04:14,490 --> 01:04:15,330 Right? 988 01:04:15,330 --> 01:04:21,176 So minus m minus one k matrix will be equal to x double dot, 989 01:04:21,176 --> 01:04:21,950 right? 990 01:04:21,950 --> 01:04:25,460 Therefore, what this is actually x double dot? 991 01:04:25,460 --> 01:04:29,600 It's basically-- in the current presentation, 992 01:04:29,600 --> 01:04:33,570 it's actually just partial square a, partial t square. 993 01:04:33,570 --> 01:04:34,650 OK? 994 01:04:34,650 --> 01:04:37,470 Can everybody accept this? 995 01:04:37,470 --> 01:04:40,440 OK, so now we have some sensibility 996 01:04:40,440 --> 01:04:43,400 going from a discrete system, which 997 01:04:43,400 --> 01:04:46,560 you have a length scale of a, to a continuous system, 998 01:04:46,560 --> 01:04:48,120 because a goes to zero. 999 01:04:48,120 --> 01:04:52,400 By a certain time, I fix the ratio of m and a so 1000 01:04:52,400 --> 01:04:56,910 that the system doesn't grow too infinitely massive. 1001 01:04:56,910 --> 01:05:01,530 OK, so if I do this, then in short, you get this equation. 1002 01:05:01,530 --> 01:05:05,220 Partial square a partial t square, 1003 01:05:05,220 --> 01:05:09,660 and that is equal to t over rho l partial square 1004 01:05:09,660 --> 01:05:13,740 a partial x square. 1005 01:05:13,740 --> 01:05:14,550 OK? 1006 01:05:14,550 --> 01:05:24,190 I can now define vp as equal to square root of t over rho l. 1007 01:05:24,190 --> 01:05:24,930 OK? 1008 01:05:24,930 --> 01:05:29,750 And this will become the p square partial square 1009 01:05:29,750 --> 01:05:31,800 a partial x square. 1010 01:05:34,850 --> 01:05:37,310 What is this? 1011 01:05:37,310 --> 01:05:40,460 This is what equation? 1012 01:05:40,460 --> 01:05:43,580 Wave equation! 1013 01:05:43,580 --> 01:05:45,440 OK, you see that now? 1014 01:05:45,440 --> 01:05:49,250 After all the hard work, OK, going 1015 01:05:49,250 --> 01:05:52,940 from infinity long systems, discrete systems, 1016 01:05:52,940 --> 01:05:54,810 then go to a continuous limit, we 1017 01:05:54,810 --> 01:05:59,060 discovered the wave equation. 1018 01:05:59,060 --> 01:06:02,990 This is probably the most important equation 1019 01:06:02,990 --> 01:06:07,070 you actually learn, until now. 1020 01:06:07,070 --> 01:06:11,120 More important than f equal to ma, right? 1021 01:06:11,120 --> 01:06:13,390 Because you can listen to my lecture, 1022 01:06:13,390 --> 01:06:16,850 even, if this equation didn't exist, right? 1023 01:06:16,850 --> 01:06:20,220 Then I can not propagate a sound wave to your ear. 1024 01:06:20,220 --> 01:06:22,100 Right? 1025 01:06:22,100 --> 01:06:24,690 And you cannot even see the black board, 1026 01:06:24,690 --> 01:06:26,810 because the electromagnetic wave, 1027 01:06:26,810 --> 01:06:30,050 which I will show in a later lecture, 1028 01:06:30,050 --> 01:06:34,670 also kind of satisfies the same function or form. 1029 01:06:34,670 --> 01:06:37,280 So I think this is an achievement and a highlight 1030 01:06:37,280 --> 01:06:38,980 of today's class. 1031 01:06:38,980 --> 01:06:42,110 We actually realize now, what we have 1032 01:06:42,110 --> 01:06:45,980 been doing is really solving something 1033 01:06:45,980 --> 01:06:48,920 related to the wave equation. 1034 01:06:48,920 --> 01:06:53,180 And next time, what I would like to actually discuss with you 1035 01:06:53,180 --> 01:06:56,070 is the solution of this wave equation. 1036 01:06:56,070 --> 01:06:58,070 So here, I have-- 1037 01:06:58,070 --> 01:06:59,930 again, last time you have seen this. 1038 01:06:59,930 --> 01:07:02,765 This is a coupled oscillator system. 1039 01:07:02,765 --> 01:07:05,420 It has 72 components. 1040 01:07:05,420 --> 01:07:08,810 As a physicist, that's actually equal to an infinite number 1041 01:07:08,810 --> 01:07:11,090 of coupled oscillators, OK. 1042 01:07:11,090 --> 01:07:12,560 That's good enough. 1043 01:07:12,560 --> 01:07:18,590 And you can see that, if I do this, 1044 01:07:18,590 --> 01:07:21,210 it does something really strange, right? 1045 01:07:21,210 --> 01:07:27,380 You see a progressing wave going back and forth, 1046 01:07:27,380 --> 01:07:30,290 and it disappears because of friction. 1047 01:07:30,290 --> 01:07:31,850 OK? 1048 01:07:31,850 --> 01:07:35,270 If I can construct something closer 1049 01:07:35,270 --> 01:07:39,290 to a perfect coupled system without friction, 1050 01:07:39,290 --> 01:07:42,470 what is going to happen is that this wave 1051 01:07:42,470 --> 01:07:47,800 is going to be there bouncing back and forth forever. 1052 01:07:47,800 --> 01:07:49,790 And that actually can be understood 1053 01:07:49,790 --> 01:07:52,260 by the wave equation. 1054 01:07:52,260 --> 01:07:58,810 And also, if I oscillate this system at a fixed frequency, 1055 01:07:58,810 --> 01:08:02,380 you will see that these become standing waves. 1056 01:08:04,950 --> 01:08:07,780 And of course I can I do crazy things, 1057 01:08:07,780 --> 01:08:11,070 I can oscillate and stop it and it becomes really, really 1058 01:08:11,070 --> 01:08:13,050 complicated motion. 1059 01:08:13,050 --> 01:08:15,360 And of course you are welcome to come here and play 1060 01:08:15,360 --> 01:08:16,500 after the class. 1061 01:08:16,500 --> 01:08:20,370 And next time, on Thursday, we are 1062 01:08:20,370 --> 01:08:24,540 going to talk about the solution to this equation 1063 01:08:24,540 --> 01:08:28,710 and how to understand all kinds of fancy motion 1064 01:08:28,710 --> 01:08:32,220 this system can do, given by the nature. 1065 01:08:32,220 --> 01:08:34,210 Thank you very much.