1 00:00:07,395 --> 00:00:12,590 PROFESSOR: Having looked at the quizzes in a preliminary 2 00:00:12,590 --> 00:00:16,500 fashion, I come to the conclusion that some notes on 3 00:00:16,500 --> 00:00:19,880 tensors would be useful for the remainder of the term. 4 00:00:19,880 --> 00:00:25,300 So at great pain and personal sacrifice I will endeavor to 5 00:00:25,300 --> 00:00:26,550 [INAUDIBLE] 6 00:00:32,292 --> 00:00:37,110 All right, let me remind you of where we were a week ago-- 7 00:00:37,110 --> 00:00:42,010 before a slight unpleasantness intervened-- 8 00:00:42,010 --> 00:00:47,640 and we had just begun to look at the properties of a very 9 00:00:47,640 --> 00:00:50,460 useful surface, the representation quadric. 10 00:00:59,270 --> 00:01:04,400 And we said that to define it we would take the elements of 11 00:01:04,400 --> 00:01:08,777 a second rank tensor, something of the form A11, 12 00:01:08,777 --> 00:01:21,060 A12, A13, A21, A22, A23, A31, A32, A33. 13 00:01:21,060 --> 00:01:28,110 And we'd use those elements as coefficients in a second rank, 14 00:01:28,110 --> 00:01:38,650 a quadratic equation, of the form Aij Xi Xj equals 1. 15 00:01:38,650 --> 00:01:41,960 OK, so the Xi and Xj are the coordinates in our 16 00:01:41,960 --> 00:01:43,650 three-dimensional space. 17 00:01:43,650 --> 00:01:50,990 And the set of coordinates that satisfy this equation-- 18 00:01:50,990 --> 00:01:53,320 setting the right hand side equal to a constant-- 19 00:01:53,320 --> 00:01:56,770 will define some surface in space. 20 00:01:56,770 --> 00:02:00,780 And it's going to be a surface that is a quadratic form, so 21 00:02:00,780 --> 00:02:06,450 it is going to be one of four different surfaces that can be 22 00:02:06,450 --> 00:02:08,300 defined by such an equation. 23 00:02:08,300 --> 00:02:14,930 One would be an ellipsoid, and in general this would be an 24 00:02:14,930 --> 00:02:18,970 ellipsoid oriented in an arbitrary fashion with respect 25 00:02:18,970 --> 00:02:21,470 to the reference axes. 26 00:02:21,470 --> 00:02:25,460 And it would have a general shape, so the three principle 27 00:02:25,460 --> 00:02:31,630 axes of the ellipsoid would be of different lengths. 28 00:02:31,630 --> 00:02:35,405 Then we saw we could also get a hyperboloid of one sheet. 29 00:02:41,470 --> 00:02:43,545 One sheet means one surface-- 30 00:02:46,770 --> 00:02:48,430 continuous surface. 31 00:02:48,430 --> 00:02:54,010 This was the surface that had an hourglass like shape that 32 00:02:54,010 --> 00:02:56,960 pinched down in the middle, and the cross-section, in 33 00:02:56,960 --> 00:02:58,990 general, would be an ellipsoid. 34 00:02:58,990 --> 00:03:06,130 And the asymptotes to this surface would define a 35 00:03:06,130 --> 00:03:10,770 boundary between directions in which the radius was real, 36 00:03:10,770 --> 00:03:13,790 which meant a positive value of the property, and a 37 00:03:13,790 --> 00:03:18,100 direction in which the radius was imaginary, even though 38 00:03:18,100 --> 00:03:19,620 that may boggle the mind. 39 00:03:19,620 --> 00:03:22,460 And if you take an imaginary quantity and square it, you're 40 00:03:22,460 --> 00:03:23,850 going to get a negative number. 41 00:03:23,850 --> 00:03:28,725 So this surface would describe a property that was positive 42 00:03:28,725 --> 00:03:32,720 in some directions and negative in magnitude over 43 00:03:32,720 --> 00:03:34,370 another range of directions. 44 00:03:34,370 --> 00:03:37,905 Then the third surface was a hyperboloid of two sheets. 45 00:03:41,740 --> 00:03:46,860 And this was a surface that looked like two [INAUDIBLE] 46 00:03:46,860 --> 00:03:49,490 with an elliptical cross-section that were nose 47 00:03:49,490 --> 00:03:51,920 to nose, but not touching the origin. 48 00:03:51,920 --> 00:03:55,090 And in this range of directions between the 49 00:03:55,090 --> 00:04:00,660 asymptote to those two lobes the radius was imaginary, the 50 00:04:00,660 --> 00:04:03,170 property negative and then there were a range of 51 00:04:03,170 --> 00:04:08,660 directions that would intersect the surface of 52 00:04:08,660 --> 00:04:11,870 either of these two sheets in there in those directions the 53 00:04:11,870 --> 00:04:13,590 property would be positive. 54 00:04:13,590 --> 00:04:14,880 Then the fourth one-- 55 00:04:14,880 --> 00:04:18,459 which is encountered only rarely, but as we pointed out 56 00:04:18,459 --> 00:04:20,269 does exist in the case of thermal 57 00:04:20,269 --> 00:04:21,899 expansion as a property-- 58 00:04:21,899 --> 00:04:23,400 and this would be an imaginary ellipsoid. 59 00:04:29,380 --> 00:04:32,710 This would be an ellipsoid in which the distance from the 60 00:04:32,710 --> 00:04:36,540 origin out to the surface was everywhere imaginary. 61 00:04:36,540 --> 00:04:40,030 And this would be a property then that was negative in all 62 00:04:40,030 --> 00:04:41,770 directions in space. 63 00:04:41,770 --> 00:04:45,020 And that doesn't seem to be a physically realizable thing, 64 00:04:45,020 --> 00:04:48,920 we pointed out that for the linear thermal expansion 65 00:04:48,920 --> 00:04:53,140 coefficient there are a small number of very unusual 66 00:04:53,140 --> 00:04:56,750 materials that when you heat them up contract uniformly in 67 00:04:56,750 --> 00:04:58,110 all directions. 68 00:04:58,110 --> 00:05:01,370 Very remarkable property, but that does indeed, thankfully, 69 00:05:01,370 --> 00:05:06,240 provide me with an example of an imaginary ellipsoid quadric 70 00:05:06,240 --> 00:05:10,100 that does represent, indeed, a realizable physical property. 71 00:05:12,690 --> 00:05:17,400 OK, this surface, we said, had two rather remarkable 72 00:05:17,400 --> 00:05:19,930 properties. 73 00:05:19,930 --> 00:05:26,130 The first property was that if we looked at the radius from 74 00:05:26,130 --> 00:05:31,380 the origin out to the surface of the quadric, a property of 75 00:05:31,380 --> 00:05:38,310 the quadric is that the radius out in some direction-- 76 00:05:38,310 --> 00:05:40,190 that would be defined in terms of the 77 00:05:40,190 --> 00:05:42,270 three direction cosines-- 78 00:05:42,270 --> 00:05:48,640 the radius has the property that it is given by 1 over the 79 00:05:48,640 --> 00:05:50,130 square root of the value of the 80 00:05:50,130 --> 00:05:51,720 property in that direction. 81 00:05:51,720 --> 00:05:55,140 Or, alternatively, the value of the property in the 82 00:05:55,140 --> 00:05:57,880 direction is 1 over the magnitude 83 00:05:57,880 --> 00:06:00,510 of the radius squared. 84 00:06:00,510 --> 00:06:04,000 OK, so this is what gives us the meaning of the imaginary 85 00:06:04,000 --> 00:06:07,550 radii, the imaginary radii 1 squared would give you a 86 00:06:07,550 --> 00:06:09,080 negative property. 87 00:06:09,080 --> 00:06:12,660 But again, I point out, it's obvious here-- but we have to 88 00:06:12,660 --> 00:06:13,910 keep it in mind-- 89 00:06:13,910 --> 00:06:19,110 that the value of the property and the magnitude of the 90 00:06:19,110 --> 00:06:23,140 radius are related in a reciprocal fashion, not only 91 00:06:23,140 --> 00:06:25,850 reciprocal, but reciprocal of the square. 92 00:06:25,850 --> 00:06:32,530 So if this is the quadric A, a polar quad of the value of the 93 00:06:32,530 --> 00:06:37,050 property A as a function of direction would have its 94 00:06:37,050 --> 00:06:41,450 maximum value along the minimum principal axis and, 95 00:06:41,450 --> 00:06:44,750 correspondingly, the minimum value of the property along 96 00:06:44,750 --> 00:06:47,520 the maximum radius of the quadric. 97 00:06:47,520 --> 00:06:50,480 So that is sort of counter-intuitive, you think 98 00:06:50,480 --> 00:06:53,520 big radius, big value of the property, no, goes not only as 99 00:06:53,520 --> 00:06:57,500 the reciprocal, but as the reciprocal of the square. 100 00:06:57,500 --> 00:07:00,030 So the value of the property with direction is not a 101 00:07:00,030 --> 00:07:04,250 quadratic form any longer, it's based on a quadratic 102 00:07:04,250 --> 00:07:06,350 equation, but when you square the radius 103 00:07:06,350 --> 00:07:07,780 it's no longer quadratic. 104 00:07:11,510 --> 00:07:13,960 Looks as though we have the value of the property as a 105 00:07:13,960 --> 00:07:16,830 function of direction, but looks as though we've lost 106 00:07:16,830 --> 00:07:20,610 information about the direction of 107 00:07:20,610 --> 00:07:24,030 the resulting vector. 108 00:07:24,030 --> 00:07:26,320 The direction that we're talking about is always the 109 00:07:26,320 --> 00:07:30,050 direction in which we're applying the generalized 110 00:07:30,050 --> 00:07:32,820 force, the electric field, or the temperature gradient, or 111 00:07:32,820 --> 00:07:34,320 the magnetic field. 112 00:07:34,320 --> 00:07:38,190 But the direction of the thing that happens is defined by the 113 00:07:38,190 --> 00:07:40,610 basic equation that gives us the tensor. 114 00:07:40,610 --> 00:07:45,630 But as I demonstrated with you last time, that information is 115 00:07:45,630 --> 00:07:48,320 in the quadric also, and this is the so-called 116 00:07:48,320 --> 00:07:49,810 radius-normal property. 117 00:07:53,580 --> 00:07:59,090 Which doesn't involve exactly what it sounds like, but what 118 00:07:59,090 --> 00:08:02,200 it tells us to do is a way of determining the direction of 119 00:08:02,200 --> 00:08:06,560 what happens is to go out in a particular direction, along 120 00:08:06,560 --> 00:08:09,450 some radius that will intersect the surface of the 121 00:08:09,450 --> 00:08:12,010 quadric at some point. 122 00:08:12,010 --> 00:08:16,110 And then at the point where the directional vector 123 00:08:16,110 --> 00:08:25,210 emerges, construct a vector that is normal to the surface. 124 00:08:25,210 --> 00:08:28,060 And so if this then were the direction of an applied 125 00:08:28,060 --> 00:08:33,940 electric field, the direction to the-- 126 00:08:33,940 --> 00:08:37,650 normal to the surface of the quadric where that direction 127 00:08:37,650 --> 00:08:38,450 intersected-- 128 00:08:38,450 --> 00:08:42,110 the surface of the representation quadric-- 129 00:08:42,110 --> 00:08:45,450 this would be in the case of conductivity the direction in 130 00:08:45,450 --> 00:08:46,700 which the current flows. 131 00:08:49,620 --> 00:08:54,330 So everything that you care to know about the anisotropy of 132 00:08:54,330 --> 00:08:57,480 the property that's described by a given tensor, and about 133 00:08:57,480 --> 00:08:59,760 the direction of what happens-- 134 00:08:59,760 --> 00:09:02,770 the generalized displacement as you apply a vector-- 135 00:09:02,770 --> 00:09:04,370 is contained within the quadric. 136 00:09:04,370 --> 00:09:10,580 But one caveat, this holds only if 137 00:09:10,580 --> 00:09:11,830 the tensor is symmetric. 138 00:09:18,240 --> 00:09:21,340 And I think this is where we finished up just before the 139 00:09:21,340 --> 00:09:27,630 quiz, only if Aij equals Aji does the radius-normal 140 00:09:27,630 --> 00:09:28,880 property work. 141 00:09:31,850 --> 00:09:34,130 OK, any comments or questions on this? 142 00:09:41,600 --> 00:09:47,500 OK, let me now turn to a practical question of 143 00:09:47,500 --> 00:09:49,640 interpretation. 144 00:09:49,640 --> 00:09:53,810 If you were indeed to measure a physical property as a 145 00:09:53,810 --> 00:09:59,270 function of direction, and you get a tensor that is a general 146 00:09:59,270 --> 00:10:02,315 tensor, A11, A12, A13. 147 00:10:02,315 --> 00:10:07,555 A21, A22, A23. 148 00:10:07,555 --> 00:10:13,340 A31, A32, A33. 149 00:10:13,340 --> 00:10:14,590 All the numbers are non-zero. 150 00:10:16,930 --> 00:10:22,630 You know that the diagonal values in the tensor are going 151 00:10:22,630 --> 00:10:25,440 to give you the value of the property along X1. 152 00:10:30,650 --> 00:10:33,860 Or, in general, the diagonal terms give you the value of 153 00:10:33,860 --> 00:10:36,030 the property along Xi. 154 00:10:36,030 --> 00:10:39,835 Aii gives you the property along Xi. 155 00:10:42,400 --> 00:10:47,310 You have not the foggiest idea, if the quadric has a 156 00:10:47,310 --> 00:10:51,790 general orientation, what the maximum and minimum values of 157 00:10:51,790 --> 00:10:54,880 the property might be, and these might concern you. 158 00:10:54,880 --> 00:10:59,180 Given the tensor that you've determined, what are the 159 00:10:59,180 --> 00:11:01,720 largest values, what is the largest value of the property, 160 00:11:01,720 --> 00:11:04,890 what is the smallest value of the property? 161 00:11:04,890 --> 00:11:09,030 And if the crystal has any symmetry these directions 162 00:11:09,030 --> 00:11:13,480 might be the direction of symmetry axes in the crystal. 163 00:11:13,480 --> 00:11:15,870 So for many reasons you might want to know the maximum and 164 00:11:15,870 --> 00:11:18,120 minimum value of the property. 165 00:11:18,120 --> 00:11:21,300 And then for some applications for a real chunk of crystal 166 00:11:21,300 --> 00:11:25,350 that you've examined relative to an arbitrary set of axes, 167 00:11:25,350 --> 00:11:28,700 you might want to ask how should I orient that crystal 168 00:11:28,700 --> 00:11:32,330 so that I can cut a rod from it that has, let's say, the 169 00:11:32,330 --> 00:11:36,730 maximum or minimum thermal conductivity. 170 00:11:36,730 --> 00:11:39,710 If you're using a piece of ceramic let's say, as a probe 171 00:11:39,710 --> 00:11:43,260 into a furnace, you wouldn't want that probe to conduct 172 00:11:43,260 --> 00:11:46,390 much heat, so you'd like the minimum thermal expansion 173 00:11:46,390 --> 00:11:48,870 coefficient, for example. 174 00:11:48,870 --> 00:11:54,430 If you were making a window out of a transparent single 175 00:11:54,430 --> 00:11:57,810 crystal, it'd be a very small window, but you might want the 176 00:11:57,810 --> 00:12:02,360 smallest thermal conductivity normal to the surface, and 177 00:12:02,360 --> 00:12:04,690 therefore you might want to make your single crystal 178 00:12:04,690 --> 00:12:07,280 window oriented in a particular direction. 179 00:12:07,280 --> 00:12:10,950 So I hope I've convinced you with enough straw questions 180 00:12:10,950 --> 00:12:13,390 that we can knock down easily that, yes, this would be an 181 00:12:13,390 --> 00:12:16,340 interesting thing to know. 182 00:12:16,340 --> 00:12:19,490 So how can we do this? 183 00:12:19,490 --> 00:12:22,100 Let me show you some simple geometry that let's us set up 184 00:12:22,100 --> 00:12:26,990 an equation for finding these directions right away. 185 00:12:26,990 --> 00:12:32,820 And it'll be based on the following observation, that 186 00:12:32,820 --> 00:12:38,610 when the direction of, let's say, the applied electric 187 00:12:38,610 --> 00:12:44,040 field is a long one of the principal axes, then and only 188 00:12:44,040 --> 00:12:50,750 then is the direction of the generalized displacement 189 00:12:50,750 --> 00:12:55,020 exactly parallel to the radius vector. 190 00:12:55,020 --> 00:12:57,390 We go off in any other direction other than a 191 00:12:57,390 --> 00:13:00,090 principal axis, the generalized displacement is 192 00:13:00,090 --> 00:13:03,190 not parallel to the radius vector out to the surface of 193 00:13:03,190 --> 00:13:04,440 the quadric. 194 00:13:09,970 --> 00:13:22,480 We look at the equation of the property A11 X1 plus A12 X2 195 00:13:22,480 --> 00:13:26,030 plus A13 X3. 196 00:13:26,030 --> 00:13:33,505 This is the X1 component of the generalized displacement. 197 00:13:37,840 --> 00:13:46,860 Similarly, A21 X1 plus A22 X2 plus A23 X3 is going to be the 198 00:13:46,860 --> 00:13:48,110 X2 component. 199 00:13:51,000 --> 00:13:59,710 And A31 X1 plus A32 X2 plus A33 times X3, guess what? 200 00:13:59,710 --> 00:14:04,140 That's going to be the X3 component of the displacement. 201 00:14:04,140 --> 00:14:09,880 Now we said that these X's give us the direction of the 202 00:14:09,880 --> 00:14:14,180 displacement as we let the coordinates of that point X1, 203 00:14:14,180 --> 00:14:18,370 X2, X3 roam around the surface of the quadric. 204 00:14:18,370 --> 00:14:22,500 When we land at a point on the surface of the quadric which 205 00:14:22,500 --> 00:14:27,490 is where a principal axis emerges, then, and only then, 206 00:14:27,490 --> 00:14:29,830 is the resulting vector parallel 207 00:14:29,830 --> 00:14:31,570 to the applied vector. 208 00:14:31,570 --> 00:14:40,140 And this means each component of the resulting vector has to 209 00:14:40,140 --> 00:14:44,490 be proportional to a vector out to the point at that 210 00:14:44,490 --> 00:14:47,260 location, X1, X2, X3. 211 00:14:47,260 --> 00:14:50,360 So what I'm saying then is that when we are on a 212 00:14:50,360 --> 00:14:55,130 principal axis the X1 component of what happens is 213 00:14:55,130 --> 00:14:59,490 going to be proportional to X1. 214 00:14:59,490 --> 00:15:03,320 And the X2 component of the vector that happens has to be 215 00:15:03,320 --> 00:15:06,420 proportional to X2, and this is the radial vector out to 216 00:15:06,420 --> 00:15:07,230 that point. 217 00:15:07,230 --> 00:15:10,940 And, similarly, the X3 component of what happens has 218 00:15:10,940 --> 00:15:16,470 to be parallel to the coordinate X3, the vector out 219 00:15:16,470 --> 00:15:19,840 to the surface of the quadric along X3. 220 00:15:19,840 --> 00:15:24,670 So let me make this an equation, now, by putting in a 221 00:15:24,670 --> 00:15:28,430 constant for the unknown proportionality constant. 222 00:15:28,430 --> 00:15:32,930 And what I'll do, just for arbitrary reasons, is call 223 00:15:32,930 --> 00:15:37,850 that proportionality constant "lambda." So this, then, is a 224 00:15:37,850 --> 00:15:43,245 set of equations that will give me, if I solve for X1, 225 00:15:43,245 --> 00:15:49,240 X2, and X3, the coordinates of one of the three points that 226 00:15:49,240 --> 00:15:53,140 sit out on the surface of the quadric at the location where 227 00:15:53,140 --> 00:15:56,840 a principal axis emerges. 228 00:15:56,840 --> 00:15:59,970 So let me rearrange these equations to a 229 00:15:59,970 --> 00:16:02,410 form that I can solve. 230 00:16:02,410 --> 00:16:07,400 And I'll make the equations homogeneous by bringing lambda 231 00:16:07,400 --> 00:16:12,510 X1 over to the left hand side of the equation, and let me 232 00:16:12,510 --> 00:16:14,740 point out that this is the same lambda 233 00:16:14,740 --> 00:16:16,730 in all three equations. 234 00:16:16,730 --> 00:16:20,340 There's a proportionality constant between the radial 235 00:16:20,340 --> 00:16:28,140 vector out to the surface of the quadric and each component 236 00:16:28,140 --> 00:16:30,210 of the vector that results. 237 00:16:30,210 --> 00:16:38,020 So my set of equations will be A11 minus lambda X1 plus A12 238 00:16:38,020 --> 00:16:45,850 times X2 plus A13 times X3, and that's now equal to zero 239 00:16:45,850 --> 00:16:46,800 on the right. 240 00:16:46,800 --> 00:16:51,670 Next equation would be A21 X1-- 241 00:16:51,670 --> 00:16:56,590 and notice I'm not combining A12 and A21 into numerically 242 00:16:56,590 --> 00:16:59,520 the same constant, representing it by the same 243 00:16:59,520 --> 00:17:01,990 quantity, I'm just leaving them be separate and 244 00:17:01,990 --> 00:17:03,830 independent at this point. 245 00:17:03,830 --> 00:17:07,900 And then the middle term here is A22 minus lambda times X2 246 00:17:07,900 --> 00:17:12,599 plus A23 times X3 equals 0. 247 00:17:12,599 --> 00:17:15,060 And the third equation, you can anticipate how that turns 248 00:17:15,060 --> 00:17:24,940 out, it's A31 X1 plus A32 times X2 plus A33 minus 249 00:17:24,940 --> 00:17:31,560 lambda, and that's equal to zero. 250 00:17:31,560 --> 00:17:38,220 So this, then, is a set of linear equations and they're 251 00:17:38,220 --> 00:17:39,710 called homogeneous equations. 252 00:17:47,270 --> 00:17:50,980 And that has solved our problem for us, all we have to 253 00:17:50,980 --> 00:17:54,210 do is solve for X1, X2, X3. 254 00:17:54,210 --> 00:18:00,140 Except that if the nine coefficients Aij are all 255 00:18:00,140 --> 00:18:05,360 arbitrary and lambda is a constant this set of equations 256 00:18:05,360 --> 00:18:13,850 has only one solution, and that solution is X1 equals X2 257 00:18:13,850 --> 00:18:18,530 equals X3 equals 0, and that indeed will wipe out every one 258 00:18:18,530 --> 00:18:19,630 of these lines. 259 00:18:19,630 --> 00:18:21,820 And that is not a very interesting solution. 260 00:18:24,380 --> 00:18:31,350 The only case in which this is not the only solution is that 261 00:18:31,350 --> 00:18:34,470 if the condition that the determinant of the 262 00:18:34,470 --> 00:18:40,710 coefficients is equal to zero, then we can find a real set of 263 00:18:40,710 --> 00:18:42,220 X1, X2, X3. 264 00:18:42,220 --> 00:18:46,070 So the condition that a solution exists is that A11 265 00:18:46,070 --> 00:19:01,046 minus lambda A12 A13, and A21 A22 minus lambda A23, A31 A32 266 00:19:01,046 --> 00:19:05,590 A33 minus lambda, this determinant has 267 00:19:05,590 --> 00:19:06,840 to be equal to zero. 268 00:19:12,090 --> 00:19:14,540 All right, so we know how to expand the determinant, we'll 269 00:19:14,540 --> 00:19:17,750 do a little number crunching, and I'm not going to write it 270 00:19:17,750 --> 00:19:23,450 down explicitly, but we'll have a term A11 minus lambda 271 00:19:23,450 --> 00:19:28,690 and this will be among other terms in the expansion A22 272 00:19:28,690 --> 00:19:35,870 minus lambda A23 A32 A33 minus lambda and then there'll be 273 00:19:35,870 --> 00:19:39,960 another determinant that involves this term A12 plus 274 00:19:39,960 --> 00:19:43,980 its cofactor plus A13 times its cofactor. 275 00:19:43,980 --> 00:19:46,560 Unfortunately none of these terms are zero in general, so 276 00:19:46,560 --> 00:19:48,900 I'm going to have three terms here. 277 00:19:48,900 --> 00:19:52,830 If I expand this term, I'm going to get something in A11 278 00:19:52,830 --> 00:20:01,590 minus lambda A22 minus lambda times A33 minus lambda and a 279 00:20:01,590 --> 00:20:04,640 bunch of other terms that I won't bother to write down 280 00:20:04,640 --> 00:20:05,890 explicitly. 281 00:20:08,070 --> 00:20:14,360 The only point I want to draw at this particular juncture is 282 00:20:14,360 --> 00:20:19,300 to note that this is a third rank equation. 283 00:20:23,160 --> 00:20:24,410 In lambda. 284 00:20:26,940 --> 00:20:28,190 It's a cubic equation. 285 00:20:31,380 --> 00:20:33,880 And what this means is that there are 286 00:20:33,880 --> 00:20:35,150 going to be three roots. 287 00:20:40,080 --> 00:20:44,000 And lets call them lambda 1, lambda 2, and lambda 3. 288 00:20:48,180 --> 00:20:49,810 And that's the way it should be. 289 00:20:49,810 --> 00:20:55,010 There are three principle axes, so we should have three 290 00:20:55,010 --> 00:20:59,050 values of the constant lambda that we could put into this 291 00:20:59,050 --> 00:21:02,720 equation that will let us solve explicitly for the 292 00:21:02,720 --> 00:21:06,670 coordinates X1, X2, and X3, which sits on the surface of 293 00:21:06,670 --> 00:21:09,920 the quadric where our principal axis pokes out. 294 00:21:09,920 --> 00:21:13,000 Three different principal axes, so this should be three 295 00:21:13,000 --> 00:21:18,770 solutions to the third rank equation that 296 00:21:18,770 --> 00:21:20,020 we've defined here. 297 00:21:23,090 --> 00:21:27,120 OK I think you've all seen this sort of problem before. 298 00:21:27,120 --> 00:21:30,435 The lambdas are called characteristic values. 299 00:21:38,930 --> 00:21:42,640 Or, on the basis that things sound more impressive in 300 00:21:42,640 --> 00:21:52,770 German, these are called the eigenvalues, 301 00:21:52,770 --> 00:21:55,210 meaning the same thing. 302 00:21:55,210 --> 00:21:59,450 So using nothing more than the properties of the quadric and 303 00:21:59,450 --> 00:22:03,040 some linear algebra we have stumbled headlong into an 304 00:22:03,040 --> 00:22:06,150 eigenvalue problem, which is a well known sort of 305 00:22:06,150 --> 00:22:09,700 mathematical problem associated with a large number 306 00:22:09,700 --> 00:22:12,980 of physical situations and a lot to do with physical 307 00:22:12,980 --> 00:22:16,520 properties in particular. 308 00:22:16,520 --> 00:22:19,570 OK, now I have a question for you. 309 00:22:19,570 --> 00:22:22,180 How we going to solve this silly thing? 310 00:22:22,180 --> 00:22:22,950 I know-- 311 00:22:22,950 --> 00:22:27,260 well the answer is I poke it into my laptop, and out comes 312 00:22:27,260 --> 00:22:28,210 the answer. 313 00:22:28,210 --> 00:22:31,450 But that's not satisfying, we should know how to do this if 314 00:22:31,450 --> 00:22:35,834 our batteries were dead and we needed the answer in a hurry. 315 00:22:35,834 --> 00:22:38,530 Well, I know how to solve a second rank question. 316 00:22:38,530 --> 00:22:45,500 If I have equation of the form ax squared plus bx plus c 317 00:22:45,500 --> 00:22:49,250 equals 0, I know the solution to that. 318 00:22:49,250 --> 00:22:52,800 It's etched indelibly in my memory and it says that x 319 00:22:52,800 --> 00:22:57,080 equals minus b plus or minus the square root of b squared 320 00:22:57,080 --> 00:23:01,190 minus 4ac all over 2a. 321 00:23:01,190 --> 00:23:02,450 Impressed you, didn't I? 322 00:23:02,450 --> 00:23:05,210 You know why I remember that? 323 00:23:05,210 --> 00:23:11,030 I learned high school algebra from a throwback-- 324 00:23:11,030 --> 00:23:15,050 a teacher who was a throwback to the Elizabethan period. 325 00:23:15,050 --> 00:23:18,770 You know how when people go to a remote island they suddenly 326 00:23:18,770 --> 00:23:21,250 find an example of a species that 327 00:23:21,250 --> 00:23:23,700 everyone thought was extinct? 328 00:23:23,700 --> 00:23:30,150 Well, to secondary education [? Ms. Dourier ?] 329 00:23:30,150 --> 00:23:34,410 was a species that had long gone extinct, but happened to 330 00:23:34,410 --> 00:23:37,820 be preserved at the high school that I attended. 331 00:23:37,820 --> 00:23:38,750 [? Ms. Dourier ?] 332 00:23:38,750 --> 00:23:44,600 was a very tall woman, ramrod straight, with a pile of snow 333 00:23:44,600 --> 00:23:46,950 white hair on the top of her head. 334 00:23:46,950 --> 00:23:51,090 She invariably dressed in a long, black skirt, and a white 335 00:23:51,090 --> 00:23:55,910 blouse that had puffy sleeves, and a collar with frilly 336 00:23:55,910 --> 00:23:59,930 things on the top, and, invariably, a black ribbon 337 00:23:59,930 --> 00:24:03,820 around the frilly lace collar top. 338 00:24:03,820 --> 00:24:09,530 Her mode of enlightened education was to have all of 339 00:24:09,530 --> 00:24:14,170 the students in the class have a notebook that was open. 340 00:24:14,170 --> 00:24:20,160 And one hapless member of the class was called upon to solve 341 00:24:20,160 --> 00:24:24,310 a problem in real time, and as the student recited we all 342 00:24:24,310 --> 00:24:27,120 wrote down what the student was saying in our notebooks. 343 00:24:27,120 --> 00:24:28,960 All the while [? Ms. Dourier, ?] 344 00:24:28,960 --> 00:24:32,730 holding a ruler like a swagger stick, strode up and down the 345 00:24:32,730 --> 00:24:37,080 aisles, and woe be unto the poor student who faltered 346 00:24:37,080 --> 00:24:40,660 slightly, let alone get something wrong. 347 00:24:40,660 --> 00:24:45,780 And that would bring a slap of the ruler down on the desk and 348 00:24:45,780 --> 00:24:48,450 a sigh of disgust, and then, all right, 349 00:24:48,450 --> 00:24:50,100 you take it, Audrey. 350 00:24:50,100 --> 00:24:53,240 And then Audrey would begin to recite until she screwed up 351 00:24:53,240 --> 00:24:56,710 and we would all copy in our notebooks. 352 00:24:56,710 --> 00:25:04,920 That tyrant so terrorized and intimidated a bunch of kids in 353 00:25:04,920 --> 00:25:09,760 a redneck, working class high school that we 354 00:25:09,760 --> 00:25:11,220 really learned algebra. 355 00:25:11,220 --> 00:25:16,960 So to this very day, as a reflex action, minus b plus or 356 00:25:16,960 --> 00:25:21,140 minus the square root of b squared minus 4ac all over 2a. 357 00:25:21,140 --> 00:25:24,410 But I don't have the foggiest idea how to solve a third 358 00:25:24,410 --> 00:25:27,050 order equation until I look it up. 359 00:25:29,560 --> 00:25:33,390 You probably never had to do it, so let me, for your 360 00:25:33,390 --> 00:25:36,370 general education and amusement, pass around a sheet 361 00:25:36,370 --> 00:25:41,975 that tells you how to solve a third rank equation. 362 00:25:46,520 --> 00:25:51,290 OK the first step is to convert an equation in, let's 363 00:25:51,290 --> 00:25:57,020 say, y squared plus y cubed plus py squared plus qy plus r 364 00:25:57,020 --> 00:26:01,010 equals 0 to a so-called normal form where you get rid of one 365 00:26:01,010 --> 00:26:02,770 of the terms. 366 00:26:02,770 --> 00:26:07,980 So if you make a substitution of variables and let y be 367 00:26:07,980 --> 00:26:12,590 replaced by x minus the coefficient p/3, then you get 368 00:26:12,590 --> 00:26:17,440 an equation that has a cubic term, linear 369 00:26:17,440 --> 00:26:20,040 term x, and a constant. 370 00:26:20,040 --> 00:26:26,370 And the constants a and b are combinations of p and q and r 371 00:26:26,370 --> 00:26:28,670 in the original equation. 372 00:26:28,670 --> 00:26:32,090 And then the solutions, not well known to be sure, are 373 00:26:32,090 --> 00:26:36,470 there's a first solution x is equal to a plus b, and then, 374 00:26:36,470 --> 00:26:43,720 something messy, x2 and x3 are minus 1/2 of a plus b plus or 375 00:26:43,720 --> 00:26:46,560 minus an imaginary term-- those are the 376 00:26:46,560 --> 00:26:48,100 second and third roots-- 377 00:26:48,100 --> 00:26:51,610 i times the square root of 3 over 2a minus b where A and B 378 00:26:51,610 --> 00:26:55,420 are complicated functions of a and b. 379 00:26:55,420 --> 00:26:57,275 Capital A and capital B are functions of little 380 00:26:57,275 --> 00:26:58,690 a and little b. 381 00:26:58,690 --> 00:27:02,990 And then if the coefficients in the original equation are 382 00:27:02,990 --> 00:27:06,070 real, then you get one real and two 383 00:27:06,070 --> 00:27:08,070 conjugated imaginary roots. 384 00:27:08,070 --> 00:27:11,920 If this combination of terms b squared and a squared are 385 00:27:11,920 --> 00:27:18,200 greater than 0, it is b squared over 4 plus a cubed 386 00:27:18,200 --> 00:27:19,850 over 27 equals 0. 387 00:27:19,850 --> 00:27:22,520 If it's that, then you get three real roots, two 388 00:27:22,520 --> 00:27:23,820 of which are equal. 389 00:27:23,820 --> 00:27:26,590 In the most general case that we would be encountering here 390 00:27:26,590 --> 00:27:32,260 is b squared over 4 plus a cubed over 27 is less than 0, 391 00:27:32,260 --> 00:27:35,300 and then there are three real unequal roots. 392 00:27:38,100 --> 00:27:44,570 Unfortunately, if b squared over 4 plus a cubed over 27 is 393 00:27:44,570 --> 00:27:49,150 negative that term appears inside of a square root sign 394 00:27:49,150 --> 00:27:51,870 in the solutions. 395 00:27:51,870 --> 00:27:55,500 So the only case that we'd really be interested in 396 00:27:55,500 --> 00:27:57,430 doesn't work for the solution. 397 00:27:57,430 --> 00:27:59,760 But fortunately there's another solution and that's 398 00:27:59,760 --> 00:28:02,930 given at the bottom of the page, and if you really wanted 399 00:28:02,930 --> 00:28:07,910 to solve a third rank equation by hand, these solutions would 400 00:28:07,910 --> 00:28:10,560 do it for you. 401 00:28:10,560 --> 00:28:13,750 But I think you'd much prefer to have your computer solve 402 00:28:13,750 --> 00:28:17,890 the equations for you, but let me show you a way of doing 403 00:28:17,890 --> 00:28:25,090 this without any computer, without solving any equations. 404 00:28:25,090 --> 00:28:31,550 And it is a very clever method of successive approximations 405 00:28:31,550 --> 00:28:33,600 that's based on the properties of the quadric. 406 00:28:38,470 --> 00:28:43,560 So let's suppose we have a tensor, and that tensor has 407 00:28:43,560 --> 00:28:50,640 some set of coefficients Aij, all of which are non-zero. 408 00:28:50,640 --> 00:28:53,630 And I'll assume although this will work for other quadratic 409 00:28:53,630 --> 00:28:59,690 surfaces that the quadric has the shape of an ellipsoid. 410 00:29:05,294 --> 00:29:09,400 All right let me now pick some direction at random, actually 411 00:29:09,400 --> 00:29:11,840 I don't have to pick it at random, but I'll show you what 412 00:29:11,840 --> 00:29:15,160 the shrewd first guess would be. 413 00:29:15,160 --> 00:29:19,380 So let's say we picked this direction. 414 00:29:19,380 --> 00:29:25,650 And let us find the direction if this is-- 415 00:29:25,650 --> 00:29:28,850 let's do it in terms of current and conductivity. 416 00:29:28,850 --> 00:29:34,110 Let's let this be the first guess for the applied field. 417 00:29:34,110 --> 00:29:37,690 That's clearly not going to be one of the principal axes, and 418 00:29:37,690 --> 00:29:40,250 let this be the first resulting direction of 419 00:29:40,250 --> 00:29:41,500 current flow J1. 420 00:29:44,190 --> 00:29:51,600 So what I'm finding then is a J sub i in terms of an Aij 421 00:29:51,600 --> 00:29:56,060 times an E sub J, and this is my first result, and this was 422 00:29:56,060 --> 00:29:58,590 my first assumption. 423 00:29:58,590 --> 00:30:05,380 Let me now let the direction of J be taken as the direction 424 00:30:05,380 --> 00:30:09,150 of the applied field for a second guess. 425 00:30:09,150 --> 00:30:12,130 And we could normalize to a unit vector, but we don't even 426 00:30:12,130 --> 00:30:13,020 have to do that. 427 00:30:13,020 --> 00:30:16,710 So let's simply say that my second guess for the electric 428 00:30:16,710 --> 00:30:19,900 field that I hope will point along one principal axis is 429 00:30:19,900 --> 00:30:23,390 E2, and E2 I'll take as identical-- 430 00:30:23,390 --> 00:30:24,460 let's say proportional-- 431 00:30:24,460 --> 00:30:28,370 to J1 from my original guess. 432 00:30:28,370 --> 00:30:32,516 So this then would be E2, the second guess. 433 00:30:32,516 --> 00:30:34,900 And if I find the direction to the normal-- 434 00:30:34,900 --> 00:30:37,410 to the quadric in that direction, this would be my 435 00:30:37,410 --> 00:30:44,070 second result for the current flow J, I should have put this 436 00:30:44,070 --> 00:30:46,990 in parentheses to indicate that these are not components 437 00:30:46,990 --> 00:30:49,140 of E and J. 438 00:30:49,140 --> 00:30:51,330 And you can see what's happening, look at where I am, 439 00:30:51,330 --> 00:30:55,650 I started out here after just one iteration I have defined 440 00:30:55,650 --> 00:30:58,800 this as the potential direction of one of the 441 00:30:58,800 --> 00:31:00,480 principal axes. 442 00:31:00,480 --> 00:31:02,540 So I'm going to find E1-- 443 00:31:02,540 --> 00:31:03,910 the direction of the field-- 444 00:31:03,910 --> 00:31:08,955 E1 that comes from sigma ij times Ji, I'll take my second 445 00:31:08,955 --> 00:31:19,250 guess E2 either as identical to J1 and this is going to 446 00:31:19,250 --> 00:31:24,920 give me then a new value for my second iteration, this 447 00:31:24,920 --> 00:31:27,730 would be J2. 448 00:31:27,730 --> 00:31:34,330 And this process is going to converge very, very rapidly on 449 00:31:34,330 --> 00:31:35,730 the shortest principal axis. 450 00:31:45,050 --> 00:31:47,940 As you can see in two shots I'm pretty close to being 451 00:31:47,940 --> 00:31:51,100 parallel to this direction, which would be X2 in my 452 00:31:51,100 --> 00:31:52,350 illustration. 453 00:31:54,285 --> 00:31:57,410 Now if I wanted to-- if I wanted to see if I was close 454 00:31:57,410 --> 00:31:58,420 to convergence-- 455 00:31:58,420 --> 00:32:02,330 this is going to be awkward because if I don't normalize 456 00:32:02,330 --> 00:32:06,030 the magnitude of the resulting vector, J is going to get 457 00:32:06,030 --> 00:32:07,710 larger and larger and larger. 458 00:32:07,710 --> 00:32:11,060 So periodically I would want to normalize-- 459 00:32:11,060 --> 00:32:13,770 take the magnitude of J, divide that into the 460 00:32:13,770 --> 00:32:16,060 components, and then I'd have a unit vector. 461 00:32:16,060 --> 00:32:20,740 If I wrote a computer program to do this, I would do the 462 00:32:20,740 --> 00:32:24,300 normalization each time to test convergence. 463 00:32:24,300 --> 00:32:29,580 Now, this is my kind of procedure, because if I'm 464 00:32:29,580 --> 00:32:33,470 doing this by hand and I screw up and I make a mistake and my 465 00:32:33,470 --> 00:32:37,190 answer is thrown off a little bit, if I continue to iterate, 466 00:32:37,190 --> 00:32:42,940 the thing will proceed to continue to converge to the 467 00:32:42,940 --> 00:32:46,310 shortest principal axis. 468 00:32:46,310 --> 00:32:48,990 So I can make a mistake and it will correct itself and come 469 00:32:48,990 --> 00:32:51,160 back again, and that's my kind of solution. 470 00:32:51,160 --> 00:32:52,020 Yeah, Jason? 471 00:32:52,020 --> 00:32:53,500 AUDIENCE: It's going to give you the minimum value of the 472 00:32:53,500 --> 00:32:54,980 property, right? 473 00:32:54,980 --> 00:32:55,924 PROFESSOR: No. 474 00:32:55,924 --> 00:32:59,000 It's going to give you the minimum principal axis, that's 475 00:32:59,000 --> 00:33:02,620 going to be the maximum value of the property. 476 00:33:02,620 --> 00:33:05,520 So that's one out of three, hey, that's not bad. 477 00:33:05,520 --> 00:33:08,010 What do I do for the others? 478 00:33:08,010 --> 00:33:16,650 Let me tell you without proof to find the 479 00:33:16,650 --> 00:33:18,215 maximum principal axis. 480 00:33:25,760 --> 00:33:28,470 And that would be the minimum value of the property. 481 00:33:40,260 --> 00:33:43,530 What you would do is exactly the same procedure, and you 482 00:33:43,530 --> 00:33:56,060 would operate on not the tensor, but on-- using as a 483 00:33:56,060 --> 00:33:58,270 matrix of coefficients-- 484 00:33:58,270 --> 00:34:02,435 the matrix of the tensor inverted. 485 00:34:08,280 --> 00:34:14,040 And I assume you know how to find the inverse of a matrix, 486 00:34:14,040 --> 00:34:17,110 except that you don't have to even find the inverse, the 487 00:34:17,110 --> 00:34:19,920 inverse is going to be a collection of functions of the 488 00:34:19,920 --> 00:34:22,610 original elements divided by the determinant. 489 00:34:22,610 --> 00:34:25,000 You don't have to worry about normalizing, all that's 490 00:34:25,000 --> 00:34:27,245 important is the relative value of these coefficients. 491 00:34:27,245 --> 00:34:28,090 Yes, sir? 492 00:34:28,090 --> 00:34:30,719 AUDIENCE: If your tensor is symmetric, can't you say that 493 00:34:30,719 --> 00:34:32,870 they'll just [INAUDIBLE] on to another? 494 00:34:32,870 --> 00:34:34,310 PROFESSOR: That's what you're going to do. 495 00:34:34,310 --> 00:34:36,505 But when you know just one, that won't work. 496 00:34:36,505 --> 00:34:38,822 The other two are floating around somewhere and you don't 497 00:34:38,822 --> 00:34:39,870 know in what direction. 498 00:34:39,870 --> 00:34:45,340 So if you do this, then you have two out of the three and 499 00:34:45,340 --> 00:34:49,590 now very shrewdly, as you point out, if we have two of 500 00:34:49,590 --> 00:34:53,969 the principal axes and the tensor is symmetric, then we 501 00:34:53,969 --> 00:34:56,910 can automatically get the direction of the third. 502 00:34:56,910 --> 00:35:02,370 If it's not symmetric, then you have to use the set of 503 00:35:02,370 --> 00:35:06,360 equations, and you have three principal axes as 504 00:35:06,360 --> 00:35:10,140 eigenvectors, as they're called in eigenvalue problems. 505 00:35:10,140 --> 00:35:13,440 And they do not have to be orthogonal, but you have the 506 00:35:13,440 --> 00:35:17,740 components in a Cartesian coordinate system so you can 507 00:35:17,740 --> 00:35:20,836 find the angle between those axes. 508 00:35:20,836 --> 00:35:21,250 OK? 509 00:35:21,250 --> 00:35:22,480 Yes, sir. 510 00:35:22,480 --> 00:35:25,286 AUDIENCE: With respect to whether the matrix is 511 00:35:25,286 --> 00:35:27,360 symmetrical or is not symmetric, 512 00:35:27,360 --> 00:35:29,520 the quadric is always-- 513 00:35:29,520 --> 00:35:32,400 has three principal axes at right angles right? 514 00:35:32,400 --> 00:35:36,720 PROFESSOR: Only if the tensor is symmetric. 515 00:35:36,720 --> 00:35:39,130 Only if the tensor is symmetric. 516 00:35:39,130 --> 00:35:44,340 And let me put your question aside for about five minutes, 517 00:35:44,340 --> 00:35:47,640 and I want to take an overview of what we've learned about 518 00:35:47,640 --> 00:35:51,160 the geometry of the quadric and the symmetry restrictions 519 00:35:51,160 --> 00:35:54,460 that we can impose on the tensor in a formal fashion, 520 00:35:54,460 --> 00:35:57,100 and see how they compare and how one allows an 521 00:35:57,100 --> 00:35:59,600 interpretation of the other. 522 00:35:59,600 --> 00:36:02,110 So I'm going to answer your question, I'd like to wait for 523 00:36:02,110 --> 00:36:04,880 about three minutes. 524 00:36:04,880 --> 00:36:07,520 Other questions? 525 00:36:07,520 --> 00:36:10,930 OK, again this is all very symbolic, I'm not giving you 526 00:36:10,930 --> 00:36:13,370 an explicit answer, I can't do it. 527 00:36:13,370 --> 00:36:14,960 But what I can do you-- 528 00:36:14,960 --> 00:36:16,980 do for you is, ha ha, do to you. 529 00:36:16,980 --> 00:36:19,870 I almost slipped and said that, what I can't do to you, 530 00:36:19,870 --> 00:36:24,800 is give you a problem that asks you to solve for the 531 00:36:24,800 --> 00:36:28,560 principal axes, for example of a property tensor. 532 00:36:28,560 --> 00:36:31,955 I will be merciful and perhaps not have all nine coefficients 533 00:36:31,955 --> 00:36:34,320 non-zero, I will have one of them zero, or 534 00:36:34,320 --> 00:36:36,590 maybe two of them zero. 535 00:36:36,590 --> 00:36:42,190 So again to try this it is very instructive and it will 536 00:36:42,190 --> 00:36:45,200 cement what the individual steps that I performed here 537 00:36:45,200 --> 00:36:48,820 actually involve. 538 00:36:48,820 --> 00:36:50,940 Now, a question that I thought you were going to ask about 539 00:36:50,940 --> 00:36:57,640 principal axes is that in order to do what I did here I 540 00:36:57,640 --> 00:37:02,280 am using the radius-normal property, and the 541 00:37:02,280 --> 00:37:08,580 radius-normal property only works for a symmetric tensor. 542 00:37:08,580 --> 00:37:12,850 So sounds like I'm swindling you, except that if I use this 543 00:37:12,850 --> 00:37:19,910 procedure, the radius-normal won't actually be identically 544 00:37:19,910 --> 00:37:24,390 parallel to the generalized displacement. 545 00:37:24,390 --> 00:37:27,960 But it's not going to be terribly different from it, 546 00:37:27,960 --> 00:37:30,600 and I'm going to get something that again will converge 547 00:37:30,600 --> 00:37:32,940 towards the shortest principal axis. 548 00:37:32,940 --> 00:37:36,750 And once I'm close to the principal axis it is true-- 549 00:37:36,750 --> 00:37:38,500 symmetric or not-- 550 00:37:38,500 --> 00:37:43,880 that the only three directions before which the generalized 551 00:37:43,880 --> 00:37:46,570 displacement is parallel to the radius vector are the 552 00:37:46,570 --> 00:37:48,990 principal axes. 553 00:37:48,990 --> 00:37:52,280 But anyway this is a dicey situation if the tensor is not 554 00:37:52,280 --> 00:37:56,530 symmetric, and it probably comes as a great relief to 555 00:37:56,530 --> 00:38:00,710 learn that there's really only one property tensor of second 556 00:38:00,710 --> 00:38:04,060 rank that's known for sure to be non-symmetric. 557 00:38:04,060 --> 00:38:06,440 So in most cases-- 558 00:38:06,440 --> 00:38:09,560 99 out of 100, if not more-- we will be dealing with 559 00:38:09,560 --> 00:38:12,350 property tensors that are symmetric, so we could use 560 00:38:12,350 --> 00:38:13,600 this procedure. 561 00:38:16,240 --> 00:38:19,730 So again this property converges remarkably well, and 562 00:38:19,730 --> 00:38:23,230 as I say it has the admirable quality of being 563 00:38:23,230 --> 00:38:26,040 self-correcting if you make a mistake, if you're grinding 564 00:38:26,040 --> 00:38:27,410 this out by hand. 565 00:38:27,410 --> 00:38:32,110 But a computer program that you write that just simply 566 00:38:32,110 --> 00:38:35,600 takes the direction of J, normalizes it to a unit 567 00:38:35,600 --> 00:38:40,420 vector, applies that as the generalized force, finds a 568 00:38:40,420 --> 00:38:44,020 quote "J" as a second iteration, normalizes that, 569 00:38:44,020 --> 00:38:47,400 and then you can check to whatever degree of convergence 570 00:38:47,400 --> 00:38:49,330 you wish to have in your solution. 571 00:38:49,330 --> 00:38:52,320 And it's a simple matter to set up a program to do this. 572 00:38:56,487 --> 00:39:02,390 All right, I think we still have some time, so let me now 573 00:39:02,390 --> 00:39:04,650 put everything together. 574 00:39:04,650 --> 00:39:12,630 And look at what we've seen of the geometric properties of 575 00:39:12,630 --> 00:39:20,080 the quadric, and the tensor arrays that we found for 576 00:39:20,080 --> 00:39:22,340 different symmetries. 577 00:39:22,340 --> 00:39:29,520 For triclinic crystals, for which the recommended 578 00:39:29,520 --> 00:39:32,940 procedure is to leave by the nearest exit and work on 579 00:39:32,940 --> 00:39:40,340 something different instead, you would have nine elements 580 00:39:40,340 --> 00:39:46,210 altogether that are necessary to define the property. 581 00:39:53,830 --> 00:39:57,690 I'll discuss this in terms of tensors that 582 00:39:57,690 --> 00:40:00,870 gives you an ellipsoid. 583 00:40:00,870 --> 00:40:05,050 The argument is not different for hyperboloids of one or two 584 00:40:05,050 --> 00:40:08,640 sheets, and we can't draw imaginary ellipsoids, but an 585 00:40:08,640 --> 00:40:12,510 ellipsoid is a much easier thing to draw. 586 00:40:12,510 --> 00:40:15,910 OK, triclinic symmetries, either one or one bar, nine 587 00:40:15,910 --> 00:40:17,160 elements in the tensor. 588 00:40:21,700 --> 00:40:28,010 No constraints whatsoever on the shape of the quadric or on 589 00:40:28,010 --> 00:40:32,150 its orientation relative to our coordinate system. 590 00:40:32,150 --> 00:40:36,960 So, let's total up now looking at the quadric how many 591 00:40:36,960 --> 00:40:39,650 degrees of freedom there are in this situation. 592 00:40:39,650 --> 00:40:46,710 We have three principal axes, so those are 593 00:40:46,710 --> 00:40:48,910 three degrees of freedom. 594 00:40:48,910 --> 00:40:54,360 The orientation of the quadric is arbitrary, so there are 595 00:40:54,360 --> 00:40:57,080 three orientational degrees of freedom. 596 00:41:08,790 --> 00:41:11,635 And that comes out to six. 597 00:41:14,360 --> 00:41:16,320 Supposedly nine degrees of freedom, what 598 00:41:16,320 --> 00:41:19,190 are the other three? 599 00:41:19,190 --> 00:41:22,340 Again, if this is a general tensor, which is 600 00:41:22,340 --> 00:41:24,840 non-symmetric, the eigenvectors-- 601 00:41:27,720 --> 00:41:32,330 the principal axes that you have to choose to squeeze this 602 00:41:32,330 --> 00:41:34,550 thing into a diagonal form-- 603 00:41:34,550 --> 00:41:37,360 become non-orthogonal. 604 00:41:37,360 --> 00:41:38,280 OK? 605 00:41:38,280 --> 00:41:43,860 So you have another three degrees of freedom that 606 00:41:43,860 --> 00:41:49,160 specify the mutual directions of the eigenvectors-- 607 00:41:53,690 --> 00:41:55,140 the angles between them. 608 00:41:57,970 --> 00:41:58,380 OK? 609 00:41:58,380 --> 00:42:02,720 The directions in which you have to pick your coordinate 610 00:42:02,720 --> 00:42:10,080 system, X1, X2, and X3, that force this thing into a 611 00:42:10,080 --> 00:42:13,520 diagonal form is going to involve a coordinate system in 612 00:42:13,520 --> 00:42:18,820 which these three angles are not 90 degrees and are fixed 613 00:42:18,820 --> 00:42:20,640 if you're going to squeeze this thing 614 00:42:20,640 --> 00:42:21,895 into a diagonal form. 615 00:42:26,690 --> 00:42:27,300 And that's it. 616 00:42:27,300 --> 00:42:30,595 So we add these three interaxial angles in, they're 617 00:42:30,595 --> 00:42:34,150 a total of nine degrees of freedom for a general non 618 00:42:34,150 --> 00:42:35,400 symmetric tensor. 619 00:42:41,860 --> 00:42:45,800 To convince you that these interaxial angles for the 620 00:42:45,800 --> 00:42:49,410 eigenvectors really are variables, let me tell you 621 00:42:49,410 --> 00:42:53,400 something that we may prove later as a recreational 622 00:42:53,400 --> 00:42:56,300 exercise, or I may give it to you on a problem set, it's not 623 00:42:56,300 --> 00:42:57,600 difficult to prove. 624 00:42:57,600 --> 00:43:06,470 And that is the result that a symmetric tensor remains 625 00:43:06,470 --> 00:43:13,660 symmetric for any arbitrary transformation of axes. 626 00:43:16,720 --> 00:43:22,320 That is from one Cartesian coordinate system to another. 627 00:43:48,888 --> 00:43:55,250 Now a very salutary effect of this is that the tensor has 628 00:43:55,250 --> 00:43:56,500 nine elements. 629 00:44:06,900 --> 00:44:08,620 And if the tensor is symmetric-- 630 00:44:08,620 --> 00:44:10,860 if you want to transform that tensor-- 631 00:44:10,860 --> 00:44:14,870 you only have to do the off-diagonal terms, because 632 00:44:14,870 --> 00:44:17,750 whatever it turns into when you change the axes, these 633 00:44:17,750 --> 00:44:20,060 off-diagonal terms are going to be equal to the 634 00:44:20,060 --> 00:44:22,746 off-diagonal terms in the new tensor. 635 00:44:22,746 --> 00:44:23,170 OK? 636 00:44:23,170 --> 00:44:25,040 So this means that only six-- 637 00:44:25,040 --> 00:44:27,810 if the tensor was symmetric in one coordinate system, only 638 00:44:27,810 --> 00:44:31,750 six elements have to be transformed. 639 00:44:31,750 --> 00:44:33,870 Actually, it's better than that, you don't have to do 640 00:44:33,870 --> 00:44:36,770 six, you only have to do five. 641 00:44:36,770 --> 00:44:41,770 And this is the last diagonal term that can be obtained from 642 00:44:41,770 --> 00:44:43,020 the trace of the tensor. 643 00:44:50,460 --> 00:44:55,620 And that would be the trace of the tensor T prime after 644 00:44:55,620 --> 00:45:03,040 transformation because A11 prime plus A22 prime plus A33 645 00:45:03,040 --> 00:45:07,360 prime has to be equal to the original trace T, which was 646 00:45:07,360 --> 00:45:11,680 A11 plus A22 plus A33 in the original system. 647 00:45:14,960 --> 00:45:17,480 So if the tensor is symmetric, you really have to crank 648 00:45:17,480 --> 00:45:22,200 through a transformation for only five of the nine terms. 649 00:45:22,200 --> 00:45:25,150 Now what is the relevance of this to what I just said? 650 00:45:25,150 --> 00:45:31,057 If we have the tensor diagonalized, to a new form 651 00:45:31,057 --> 00:45:41,380 A11, 0, 0, 0, A22 prime, 0, 0, 0, A33 prime, that is a 652 00:45:41,380 --> 00:45:45,180 special case of a symmetric tensor. 653 00:45:45,180 --> 00:45:48,450 So if you decide to go from the coordinate system that 654 00:45:48,450 --> 00:45:52,270 produced this diagonalized form to some other coordinate 655 00:45:52,270 --> 00:45:55,130 system, the new tensor that you're going to get is going 656 00:45:55,130 --> 00:45:56,530 to be symmetric. 657 00:45:56,530 --> 00:46:00,225 But the thing was not originally symmetric, so how 658 00:46:00,225 --> 00:46:02,010 can that be? 659 00:46:02,010 --> 00:46:05,140 The answer is you can put it in diagonal form only in a 660 00:46:05,140 --> 00:46:07,740 non-Cartesian coordinate system. 661 00:46:07,740 --> 00:46:12,150 And, therefore, that is evidence that 662 00:46:12,150 --> 00:46:14,540 the reference axes-- 663 00:46:14,540 --> 00:46:17,680 the principal axes-- cannot be orthogonal. 664 00:46:17,680 --> 00:46:20,600 Otherwise you could take the diagonalized tensor and crank 665 00:46:20,600 --> 00:46:23,690 it back to some arbitrary Cartesian coordinate, and 666 00:46:23,690 --> 00:46:26,080 suddenly it would go to a symmetric tensor when it was 667 00:46:26,080 --> 00:46:28,674 not symmetric to begin with. 668 00:46:28,674 --> 00:46:29,500 OK? 669 00:46:29,500 --> 00:46:30,820 QED. 670 00:46:30,820 --> 00:46:33,950 So you can diagonalize a general tensor only if you 671 00:46:33,950 --> 00:46:37,940 take the axes along the eigenvectors, which cannot be 672 00:46:37,940 --> 00:46:39,960 orthogonal. 673 00:46:39,960 --> 00:46:42,940 OK, I saw you raise your hand, I wasn't ignoring you. 674 00:46:42,940 --> 00:46:43,695 That was it? 675 00:46:43,695 --> 00:46:44,455 Good. 676 00:46:44,455 --> 00:46:45,425 AUDIENCE: I wanted to make sure that Cartesian 677 00:46:45,425 --> 00:46:46,675 [INAUDIBLE] 678 00:46:49,310 --> 00:46:51,610 PROFESSOR: It's always nice to see the class one step ahead 679 00:46:51,610 --> 00:46:52,860 of what you're doing. 680 00:46:56,786 --> 00:47:01,290 All right, let's look at a couple of more crystal systems 681 00:47:01,290 --> 00:47:05,705 in the form of tensors in those systems and show that 682 00:47:05,705 --> 00:47:09,680 that in fact does correspond to the geometric constraints 683 00:47:09,680 --> 00:47:11,070 on the quadric as well. 684 00:47:11,070 --> 00:47:12,320 For monoclinic crystals-- 685 00:47:16,530 --> 00:47:22,780 and this is symmetry 2, symmetry M, and symmetry 2/M. 686 00:47:22,780 --> 00:47:25,560 The form of the tensor when we took the coordinate system 687 00:47:25,560 --> 00:47:31,080 along symmetry elements was A11, A12, 0-- 688 00:47:31,080 --> 00:47:33,230 terms with a single three vanished-- 689 00:47:33,230 --> 00:47:39,648 A21, A22, 0, 0, 0, A33. 690 00:47:42,980 --> 00:47:48,130 So this was the case where X3 was along the two-fold axis 691 00:47:48,130 --> 00:47:50,420 and or perpendicular to the mirror point. 692 00:47:55,690 --> 00:48:01,440 OK, number of independent variables to describe this 693 00:48:01,440 --> 00:48:02,810 property is five. 694 00:48:06,150 --> 00:48:09,600 And if we look at this in terms of a constraint and 695 00:48:09,600 --> 00:48:18,530 shape in orientation of an ellipsoidal quadric, says that 696 00:48:18,530 --> 00:48:21,540 one of the principal axes, if the quadric is to remain 697 00:48:21,540 --> 00:48:28,690 invariant, has to be along the two-fold axis and or 698 00:48:28,690 --> 00:48:32,700 perpendicular to a plane that contains the mirror plane, if 699 00:48:32,700 --> 00:48:36,220 a mirror plane is also present. 700 00:48:36,220 --> 00:48:38,810 So what are the degrees of freedom here? 701 00:48:38,810 --> 00:48:42,640 This has to be along one reference axis, x3. 702 00:48:42,640 --> 00:48:47,570 And, indeed, this form of the tensor occurred only when the 703 00:48:47,570 --> 00:48:52,730 two-fold axes were parallel, 2 X3. 704 00:48:52,730 --> 00:48:55,900 And this ellipsoid then was left with 705 00:48:55,900 --> 00:48:57,880 one degree of freedom. 706 00:48:57,880 --> 00:49:02,950 If this is the direction of the reference axis X2, we have 707 00:49:02,950 --> 00:49:05,040 a degree of freedom in-- 708 00:49:05,040 --> 00:49:08,330 shouldn't have called these X1 and X2, these need not be the 709 00:49:08,330 --> 00:49:10,050 coordinate systems. 710 00:49:10,050 --> 00:49:14,260 Looking down on the quadric along the two-fold axis, the 711 00:49:14,260 --> 00:49:17,310 section of the quadric perpendicular to the two-fold 712 00:49:17,310 --> 00:49:23,370 axis can have a general shape, and it can have one degree of 713 00:49:23,370 --> 00:49:28,670 freedom in the orientation of the principal axis relative to 714 00:49:28,670 --> 00:49:30,370 the coordinate system. 715 00:49:30,370 --> 00:49:34,480 So we have three parameters to specify the principal axes, 716 00:49:34,480 --> 00:49:35,960 since this is a general quadric. 717 00:49:39,180 --> 00:49:57,590 We have one degree of freedom in orientation, and that adds 718 00:49:57,590 --> 00:50:00,425 up to four, but we said five. 719 00:50:03,180 --> 00:50:04,260 So what's going on here? 720 00:50:04,260 --> 00:50:07,180 Again this is a question of where the eigenvectors point. 721 00:50:10,770 --> 00:50:15,910 If this term is not equal to this term, in order to force 722 00:50:15,910 --> 00:50:20,240 the tensor into a diagonal form, you have to pick 723 00:50:20,240 --> 00:50:25,350 eigenvectors in the X1, X2 plane, which will force the 724 00:50:25,350 --> 00:50:27,740 tensor into a symmetric form. 725 00:50:27,740 --> 00:50:31,810 That is, to make this term equal to this term. 726 00:50:31,810 --> 00:50:35,150 It's only a pair of axes involved, so there's only one 727 00:50:35,150 --> 00:50:36,420 angle between these two 728 00:50:36,420 --> 00:50:38,320 eigenvectors, which is a variable. 729 00:50:38,320 --> 00:50:42,930 So for a nonsymmetric tensor plus one angle between the 730 00:50:42,930 --> 00:50:44,180 eigenvectors. 731 00:50:56,220 --> 00:50:58,150 So that comes out, a-ha, five. 732 00:51:04,380 --> 00:51:07,880 And again the way to see that is I will never get the tensor 733 00:51:07,880 --> 00:51:18,420 into a diagonal form making all the off-diagonal terms 734 00:51:18,420 --> 00:51:22,560 zero when I'm starting with a tensor which is not symmetric, 735 00:51:22,560 --> 00:51:25,960 and I know that a symmetric tensor, even in the special 736 00:51:25,960 --> 00:51:29,080 case where the off-diagonal terms are zero, is going to go 737 00:51:29,080 --> 00:51:33,860 to a symmetric tensor in another coordinate system. 738 00:51:33,860 --> 00:51:35,270 So I know that there has to be an 739 00:51:35,270 --> 00:51:37,680 additional degree of freedom. 740 00:51:37,680 --> 00:51:41,290 OK, the rest come very, very easily, so let me take just a 741 00:51:41,290 --> 00:51:44,270 couple of minutes to finish up this stage of the discussion, 742 00:51:44,270 --> 00:51:47,080 and we will shortly go on to something different. 743 00:51:51,310 --> 00:51:53,240 The next step up in symmetry is orthorhombic. 744 00:52:00,070 --> 00:52:06,890 And if we pick arc-- and this could be 222, 2MM or 2/M 2/M 745 00:52:06,890 --> 00:52:19,680 2/M. We found that when we took X1, X2, and X3 along 746 00:52:19,680 --> 00:52:29,416 two-fold axes that the tensor took a diagonal form A11, 0, 747 00:52:29,416 --> 00:52:35,665 0, 0, A22, 0, 0, 0, A33. 748 00:52:38,370 --> 00:52:44,310 That says that the quadric if it's an ellipsoid-- 749 00:52:44,310 --> 00:52:47,690 or any other quadratic form-- 750 00:52:47,690 --> 00:52:52,330 has all three of its principal axes along the reference 751 00:52:52,330 --> 00:52:55,230 system X1, X2, X3. 752 00:52:55,230 --> 00:52:58,120 So the only degree of freedoms here-- and things are again 753 00:52:58,120 --> 00:53:02,410 starting to set up like supercooled water-- 754 00:53:02,410 --> 00:53:06,065 the only degrees of freedom are the three principal axes. 755 00:53:11,960 --> 00:53:14,065 And that's it, the orientation is fixed. 756 00:53:20,030 --> 00:53:25,400 And, moreover, the tensor is diagonal in this reference 757 00:53:25,400 --> 00:53:29,180 system-- it's going to be symmetric in this reference 758 00:53:29,180 --> 00:53:32,850 system, it's going to remain symmetric in any other 759 00:53:32,850 --> 00:53:34,060 coordinate system. 760 00:53:34,060 --> 00:53:40,040 So the tensor is always symmetric, the eigenvectors 761 00:53:40,040 --> 00:53:48,300 then are always orthogonal, even if the coordinate system 762 00:53:48,300 --> 00:53:49,700 stays Cartesian. 763 00:53:49,700 --> 00:53:52,260 And there are three variables, three degrees of freedom. 764 00:54:02,520 --> 00:54:05,210 And finally two remaining cases can 765 00:54:05,210 --> 00:54:08,550 be disposed of quickly. 766 00:54:08,550 --> 00:54:11,510 We saw that for an arbitrary theta that 767 00:54:11,510 --> 00:54:13,220 is a symmetry element-- 768 00:54:16,940 --> 00:54:18,110 so let me not say a symmetry-- 769 00:54:18,110 --> 00:54:20,245 for an arbitrary rotation about X3. 770 00:54:25,970 --> 00:54:34,560 This was the ingenious little proof in which we transform 771 00:54:34,560 --> 00:54:41,470 the tensor by an arbitrary rotation theta about X3. 772 00:54:41,470 --> 00:54:48,370 We found that the form of the tensor was A11, A12, 0, and 773 00:54:48,370 --> 00:54:52,380 now I get to use tensor notation by calling this term 774 00:54:52,380 --> 00:54:57,810 A12 again and not the proper indices A21, so I'll put 775 00:54:57,810 --> 00:54:59,190 quotation marks. 776 00:54:59,190 --> 00:55:07,170 "A22" was equal to "A11." 777 00:55:07,170 --> 00:55:11,640 So going to an arbitrary rotation theta this must be 778 00:55:11,640 --> 00:55:14,670 the form of the tensor we discovered, and this will 779 00:55:14,670 --> 00:55:16,810 cover then fourfold axes, three-fold 780 00:55:16,810 --> 00:55:19,220 axes, and sixfold axes. 781 00:55:19,220 --> 00:55:23,570 These two elements are constrained to have the two 782 00:55:23,570 --> 00:55:27,410 values, these two are constrained to have the same 783 00:55:27,410 --> 00:55:29,650 values, so the tensor is always symmetric. 784 00:55:33,080 --> 00:55:35,820 Symmetric relative to these axes, symmetric in all 785 00:55:35,820 --> 00:55:40,470 coordinates, and any choice of coordinate system, then. 786 00:55:55,280 --> 00:55:59,870 And that means that if the quadric has this property, 787 00:55:59,870 --> 00:56:04,080 it's going to have one principal axis along X3, it's 788 00:56:04,080 --> 00:56:11,120 going to be circular in the plane that contains X1 and X2, 789 00:56:11,120 --> 00:56:20,160 so there are only, count them up, there are only three 790 00:56:20,160 --> 00:56:21,410 degrees of freedom. 791 00:56:29,670 --> 00:56:32,790 In terms of tensor elements, these degrees of 792 00:56:32,790 --> 00:56:37,400 freedom are A1, A12-- 793 00:56:37,400 --> 00:56:41,430 in the diagonalized form-- and A33. 794 00:56:41,430 --> 00:56:46,540 And there are three degrees of freedom in terms of the 795 00:56:46,540 --> 00:56:48,540 independent elements as well. 796 00:56:48,540 --> 00:56:53,850 So this is for anything that involves rotational symmetry 797 00:56:53,850 --> 00:56:56,540 other than 180 degrees. 798 00:56:56,540 --> 00:57:07,420 And, finally, for cubic crystals, if a symmetry 799 00:57:07,420 --> 00:57:11,150 operation other than 180 degrees gives us this form of 800 00:57:11,150 --> 00:57:14,960 the tensor, two off-diagonal terms are zero, two diagonal 801 00:57:14,960 --> 00:57:16,100 terms equal. 802 00:57:16,100 --> 00:57:21,140 If we impose this along all three reference axes, then the 803 00:57:21,140 --> 00:57:24,540 form of the tensor is going to go to a diagonal form with all 804 00:57:24,540 --> 00:57:28,440 diagonal terms equal. 805 00:57:31,730 --> 00:57:39,720 The quadric that corresponds to that form is a sphere that 806 00:57:39,720 --> 00:57:44,640 says that there's one quantity, one degree of 807 00:57:44,640 --> 00:57:47,720 freedom in the form of the tensor. 808 00:57:54,410 --> 00:57:56,200 There are zero degrees of 809 00:57:56,200 --> 00:57:58,120 orientational degree of freedom. 810 00:58:11,430 --> 00:58:14,310 Thus the spherical quadrics stay spherical for any 811 00:58:14,310 --> 00:58:15,540 orientation. 812 00:58:15,540 --> 00:58:19,120 So again, one independent element in the tensor, just 813 00:58:19,120 --> 00:58:21,150 one degree of freedom in the quadric-- 814 00:58:21,150 --> 00:58:22,530 namely its radius-- 815 00:58:22,530 --> 00:58:25,510 and, again, the formal constraints that we obtained 816 00:58:25,510 --> 00:58:31,700 by symmetry transformations agree with those that are the 817 00:58:31,700 --> 00:58:34,090 same degrees of freedom when you want the quadric to 818 00:58:34,090 --> 00:58:36,050 coincide with the axes. 819 00:58:36,050 --> 00:58:36,844 Yes? 820 00:58:36,844 --> 00:58:39,925 AUDIENCE: Speaking of the circle and the exponents to 821 00:58:39,925 --> 00:58:43,960 play only if A12 equals zero? 822 00:58:43,960 --> 00:58:48,450 PROFESSOR: These two were equal, but non-zero. 823 00:58:48,450 --> 00:58:51,400 These two are equal, this one was independent. 824 00:58:51,400 --> 00:58:53,590 And that would be for the specific case 825 00:58:53,590 --> 00:58:55,660 of a fourfold axis. 826 00:58:55,660 --> 00:58:59,300 Now if we put a fourfold axis along X1, that's going to 827 00:58:59,300 --> 00:59:02,910 involve these two being equal and the 828 00:59:02,910 --> 00:59:07,400 non-zero elements become-- 829 00:59:07,400 --> 00:59:08,350 let's see-- 830 00:59:08,350 --> 00:59:16,718 along X2 it would be A13 and A23. 831 00:59:16,718 --> 00:59:17,968 No. 832 00:59:21,200 --> 00:59:26,680 OK, we put the four-fold axis along X2, then this one would 833 00:59:26,680 --> 00:59:40,440 be A22, A11, and A13 would have to be equal, and 21 and 834 00:59:40,440 --> 00:59:48,650 12 would have to be zero, 23 and 32 would have to be equal. 835 00:59:48,650 --> 00:59:51,110 I think that's the way it'll go, and these 836 00:59:51,110 --> 00:59:53,440 three have to be zero. 837 00:59:53,440 --> 00:59:56,210 So when you impose all three constraints, here we've had 838 00:59:56,210 --> 01:00:00,060 these two equal, now we have these two equal, put the 839 01:00:00,060 --> 01:00:01,860 four-fold access in the next-- 840 01:00:01,860 --> 01:00:03,880 in the remaining direction, and again it has to be 841 01:00:03,880 --> 01:00:08,290 diagonal, and pairwise all of the off-diagonal terms will be 842 01:00:08,290 --> 01:00:09,748 required to be zero. 843 01:00:09,748 --> 01:00:15,130 AUDIENCE: I mean, when you diagonalize your matrix the 844 01:00:15,130 --> 01:00:20,820 diagonal term are going to be all different in this case if 845 01:00:20,820 --> 01:00:25,430 A12 is different from zero. 846 01:00:25,430 --> 01:00:27,860 PROFESSOR: This is not a final result, this is an 847 01:00:27,860 --> 01:00:29,270 intermediate step. 848 01:00:29,270 --> 01:00:33,230 So we impose this constraint, plus this constraint, and 849 01:00:33,230 --> 01:00:35,270 that's actually going to give us all the qualities we're 850 01:00:35,270 --> 01:00:38,105 going to get, but we'd want to do the same thing for a 851 01:00:38,105 --> 01:00:42,600 four-fold axis along X3. 852 01:00:42,600 --> 01:00:46,890 You put them all together, the fact that these are zero will 853 01:00:46,890 --> 01:00:48,750 wipe out all of the off-diagonal 854 01:00:48,750 --> 01:00:51,880 terms, and these things-- 855 01:00:51,880 --> 01:00:53,420 for another choice of axes, these two 856 01:00:53,420 --> 01:00:54,880 would have to be zero. 857 01:00:54,880 --> 01:00:58,790 And for another choice of the orientation of the four-fold 858 01:00:58,790 --> 01:01:01,010 axis, this and this would have to be equal. 859 01:01:01,010 --> 01:01:03,900 And, finally, this and this would have to be equal. 860 01:01:03,900 --> 01:01:06,780 So this is what we found, in fact, when we cranked through 861 01:01:06,780 --> 01:01:11,050 the symmetry transformations, and this is what we would get 862 01:01:11,050 --> 01:01:14,640 when we said that the quadric has to have a shape that 863 01:01:14,640 --> 01:01:17,912 conforms to cubic symmetric. 864 01:01:17,912 --> 01:01:22,268 AUDIENCE: My question was in the case of the arbitrary 865 01:01:22,268 --> 01:01:24,700 relation of those three. 866 01:01:24,700 --> 01:01:28,859 All three degrees of freedom, are they, in fact, the three 867 01:01:28,859 --> 01:01:30,109 principal axes? 868 01:01:32,840 --> 01:01:34,040 PROFESSOR: For this case here? 869 01:01:34,040 --> 01:01:36,010 AUDIENCE: Yes. 870 01:01:36,010 --> 01:01:38,870 PROFESSOR: For a symmetric tensor, they are 871 01:01:38,870 --> 01:01:41,190 two principal axes. 872 01:01:41,190 --> 01:01:41,650 OK? 873 01:01:41,650 --> 01:01:45,490 If the tensor is nonsymmetric, then these two don't have to 874 01:01:45,490 --> 01:01:49,640 be equal any longer, and they give you the 875 01:01:49,640 --> 01:01:50,890 extra degree of freedom. 876 01:01:53,640 --> 01:01:54,890 OK? 877 01:02:00,640 --> 01:02:01,890 AUDIENCE: What are the three degrees of 878 01:02:01,890 --> 01:02:03,140 freedom in this case? 879 01:02:17,640 --> 01:02:17,990 PROFESSOR: I just want you to know I 880 01:02:17,990 --> 01:02:19,240 appreciate your question. 881 01:02:23,336 --> 01:02:24,797 AUDIENCE: May I make a suggestion? 882 01:02:24,797 --> 01:02:25,771 PROFESSOR: Yeah. 883 01:02:25,771 --> 01:02:27,719 AUDIENCE: I think you need two degrees of freedom to specify 884 01:02:27,719 --> 01:02:30,641 the first one, and since it's symmetric and since it's 885 01:02:30,641 --> 01:02:33,439 orthogonal you'd only need one more to find the second one 886 01:02:33,439 --> 01:02:35,260 and then another to divide the third. 887 01:02:35,260 --> 01:02:37,525 So that's three degrees of freedom to find three 888 01:02:37,525 --> 01:02:39,430 principal axes in an orthogonal system. 889 01:02:39,430 --> 01:02:41,603 PROFESSOR: Not sure I like that, because this has to be 890 01:02:41,603 --> 01:02:46,395 the shape of the quadric for an arbitrary rotation angle. 891 01:02:56,359 --> 01:02:58,545 AUDIENCE: Does that angle between X1 and X2-- does it 892 01:02:58,545 --> 01:03:00,335 have to be 90 degrees if it's not connected? 893 01:03:00,335 --> 01:03:03,814 PROFESSOR: That's correct, that's correct. 894 01:03:03,814 --> 01:03:09,030 OK, let's not tie up the whole audience, let me-- 895 01:03:09,030 --> 01:03:11,620 let's talk about this and I'll think about it too. 896 01:03:11,620 --> 01:03:15,160 But you make a very, very good point, but let's not settle it 897 01:03:15,160 --> 01:03:16,820 in real time. 898 01:03:16,820 --> 01:03:19,930 We'll throw everybody out, we'll close the door, and you 899 01:03:19,930 --> 01:03:25,010 can come back and see who won the scuffle, OK? 900 01:03:25,010 --> 01:03:27,050 OK, so let's take our break, I think you're more 901 01:03:27,050 --> 01:03:28,300 than ready for it.