1 00:00:00,150 --> 00:00:03,450 So I come back to this lecture about -- potentials. 2 00:00:04,080 --> 00:00:09,720 So in the previous lecture, I have just given you the solution to this for General Maxwell's equations 3 00:00:10,050 --> 00:00:13,020 and the solutions are these -- potentials. 4 00:00:13,890 --> 00:00:20,700 But in reality, it's very difficult to find out that these are really the solutions and that this solution 5 00:00:20,700 --> 00:00:21,540 here is true. 6 00:00:22,680 --> 00:00:29,850 So this lecture is optional, and this is because you will not really learn new physics here, because 7 00:00:29,850 --> 00:00:33,270 I already told you that these are the solutions, we just have to verify it. 8 00:00:33,930 --> 00:00:39,640 And furthermore, this proof is rather difficult, mathematically speaking. 9 00:00:40,320 --> 00:00:45,960 So if you're not really that interested in the mathematics, you can go ahead and skip this video and 10 00:00:45,960 --> 00:00:47,040 go to the next video. 11 00:00:47,940 --> 00:00:53,370 But I think it's really worth the effort to go through this video because you'll learn some new cool 12 00:00:53,370 --> 00:00:54,810 tricks about mathematics. 13 00:00:54,810 --> 00:01:01,110 And we will also learn about the so-called Delta distribution, which is a very important concept in 14 00:01:01,110 --> 00:01:02,240 theoretical physics. 15 00:01:03,930 --> 00:01:06,920 So I will only prove one half of this part. 16 00:01:06,930 --> 00:01:14,820 I will only prove that this electrostatic potential with this -- time here solves this Maxwell 17 00:01:14,820 --> 00:01:15,560 equation here. 18 00:01:16,290 --> 00:01:21,900 And then afterwards, if you want, you could verify also that the vector potential solves the other 19 00:01:21,900 --> 00:01:22,860 Maxwell equation. 20 00:01:23,100 --> 00:01:26,880 But the steps on the method in terms of the mathematics, they are very similar. 21 00:01:28,170 --> 00:01:29,700 OK, so let us get started. 22 00:01:30,390 --> 00:01:36,090 So first of all, we must realize that this large class operator acting on the electrostatic potential 23 00:01:36,510 --> 00:01:38,870 is just the divergence of the gradient. 24 00:01:39,420 --> 00:01:46,000 And so we go step by step and first calculate the gradient of fire before we calculate the plaza operator. 25 00:01:47,190 --> 00:01:50,940 Now, since we have here this are ministers, other aren't here as well. 26 00:01:50,940 --> 00:01:57,570 And always we only consider distance might introduce a new variable, which just is Capital R, which 27 00:01:57,570 --> 00:01:59,030 describes this distance here. 28 00:01:59,790 --> 00:02:06,000 And of course, if I don't write it bold, not as a vector, then I mean this absolute value. 29 00:02:08,550 --> 00:02:11,640 So let us calculate the gradient of the electrostatic potential. 30 00:02:12,360 --> 00:02:17,670 So this is a pretty factor and we can also switch around the integration and the differentiation. 31 00:02:18,000 --> 00:02:21,710 So we just have to differentiate this integration to you. 32 00:02:21,720 --> 00:02:23,470 We have to calculate the gradient. 33 00:02:24,690 --> 00:02:27,570 Now, of course, here we have to are dependent terms. 34 00:02:27,570 --> 00:02:36,300 We have to charge density, which depends on R because we have here in this time argument the R and 35 00:02:36,300 --> 00:02:40,740 then we have here another R, so it's a quotient. 36 00:02:40,740 --> 00:02:45,270 Or you could also say a product of two functions that are dependent. 37 00:02:46,470 --> 00:02:53,250 Now another comment, this KNOBLER operator here, this gradient is typically with respect to this position 38 00:02:53,250 --> 00:02:57,210 vector, but since they are linear to each other, it's the same. 39 00:02:57,210 --> 00:03:03,240 If we also write it here with respect to this capital R so we can just differentiate with respect to 40 00:03:03,240 --> 00:03:04,110 this R here. 41 00:03:04,920 --> 00:03:06,060 OK, let's do this. 42 00:03:06,330 --> 00:03:12,360 So we have here this product of these two are dependent functions, so we get a sum of two terms. 43 00:03:13,350 --> 00:03:20,790 The first term here is just this one here times the derivative of one over R, so the derivative of 44 00:03:20,790 --> 00:03:25,470 one of our is of course one of our square then the minus sign. 45 00:03:25,860 --> 00:03:31,110 And we also need to consider the inner derivative, which is the gradient of our. 46 00:03:32,840 --> 00:03:39,560 Then the other term, we know we have to apply the naval operators to discharge density and just have 47 00:03:39,560 --> 00:03:44,330 to multiply by these other functions, so we have to gradient over divided by R. 48 00:03:45,900 --> 00:03:52,650 Now we have to calculate what is the gradient of our I think you know this already, but here I have 49 00:03:52,650 --> 00:03:53,820 written it down once again. 50 00:03:54,390 --> 00:03:59,810 So a comment you could do this much more elegantly if you would use spherical coordinates. 51 00:04:00,180 --> 00:04:07,170 But this in itself is also a bit difficult because then you first have to learn about spherical coordinates. 52 00:04:07,560 --> 00:04:12,230 So in principle, you can also do it in Cartesian coordinates, as you can see here. 53 00:04:12,690 --> 00:04:20,220 But then you need several steps and you need to be very, very careful that you do not miss any any 54 00:04:20,520 --> 00:04:24,450 in a derivative so that you do not miss this term here, for example. 55 00:04:25,050 --> 00:04:29,390 And you will see that this is just are divided by ah, OK. 56 00:04:29,400 --> 00:04:31,310 This means we have now this term here. 57 00:04:31,620 --> 00:04:35,520 So we still need this one, the gradient of the charge density. 58 00:04:36,780 --> 00:04:42,840 And to you we have to gradient with respect to capital R and this coordinate enters this -- time 59 00:04:42,840 --> 00:04:43,480 argument. 60 00:04:44,370 --> 00:04:50,520 So what we have to do is we have to first calculate the outer derivative which is charged density divided 61 00:04:50,550 --> 00:04:52,170 with respect to this argument. 62 00:04:52,650 --> 00:04:59,460 And then we have to multiply and calculate the inner derivative, which is the gradient of this argument. 63 00:05:00,560 --> 00:05:07,130 So we have here robots, there's just a shorthand notation, as you typically use it in physics, and 64 00:05:07,130 --> 00:05:12,410 then for the for the inner derivative, we have the gradient of the argument, which is the gradient 65 00:05:12,410 --> 00:05:17,240 of tea, which gives zero and then the gradient of minus are divided by. 66 00:05:18,380 --> 00:05:27,260 So this in a derivative is the gradient of ah and then times minus one divided by C and the gradient 67 00:05:27,260 --> 00:05:29,150 of are we know already we have it here. 68 00:05:29,540 --> 00:05:37,580 This is the vector are divided by the absolute value of R, so this is our final result for this one 69 00:05:37,580 --> 00:05:37,850 here. 70 00:05:39,020 --> 00:05:45,370 So now let's take these two terms here, this one and that one and put them into our our gradient here. 71 00:05:45,830 --> 00:05:48,800 And this is what we get for the gradient of fire. 72 00:05:49,490 --> 00:05:52,540 We get two terms and they look kind of similar. 73 00:05:52,550 --> 00:05:58,080 But here we have a dependent's one of our square until one of our to the power of three. 74 00:05:58,460 --> 00:06:02,000 And here we get a row until we get the time derivative of row. 75 00:06:04,090 --> 00:06:09,110 So this is what we have, but what we want to calculate is the plant's operator. 76 00:06:09,520 --> 00:06:12,960 So now we need to calculate the divergence of this term. 77 00:06:14,470 --> 00:06:21,190 So we have now these two terms and we need to calculate the divergence of both of these terms. 78 00:06:22,570 --> 00:06:30,430 And now, of course, you can see we have here three functions, Row R and R Square, and here also 79 00:06:30,430 --> 00:06:32,760 three functions that all depend on R. 80 00:06:33,550 --> 00:06:42,400 So we need to use the product rule many times and let us start by calculating these two terms and write 81 00:06:42,400 --> 00:06:42,880 them down. 82 00:06:43,300 --> 00:06:47,210 First of all, I used to the product rule here. 83 00:06:47,530 --> 00:06:50,470 So I write down real dot, for example. 84 00:06:50,470 --> 00:06:54,620 rawData times the divergence of dysfunction here. 85 00:06:54,820 --> 00:07:03,400 So that's done divergence of this one plus the other way around this function, then the scale of products 86 00:07:03,700 --> 00:07:11,140 times the gradient of row dot, and then similarly I do it for this term to get these two terms. 87 00:07:12,650 --> 00:07:19,910 But here we are not finished, I want to go one step further, I have to apply the product rule also 88 00:07:20,060 --> 00:07:26,930 to this term here and to this term, because here we have also the product of two are dependant functions, 89 00:07:27,740 --> 00:07:29,600 so we get even more terms. 90 00:07:30,560 --> 00:07:35,090 So, for example, from this one here, we get these two terms. 91 00:07:36,450 --> 00:07:40,920 And then from this one here, we get these two terms. 92 00:07:43,720 --> 00:07:49,760 Now, this is what we have for our the plant operator and this is what we get for the integrity. 93 00:07:50,260 --> 00:07:54,370 So now we need to simplify all of these six terms. 94 00:07:55,220 --> 00:07:57,310 So let's get started with the first term. 95 00:07:57,850 --> 00:08:03,880 Of course, here the only difficulty is calculating the divergence of R, but actually it's it's not 96 00:08:03,880 --> 00:08:10,780 at all difficult because you can imagine the operator is a vector of the partial derivatives and you 97 00:08:10,780 --> 00:08:14,730 calculate the scale of product with this position vector itself. 98 00:08:14,980 --> 00:08:24,130 So you calculate basically derivative of respect to X of X, which is one plus derivative of Y with 99 00:08:24,130 --> 00:08:30,030 respect to Y, which is one plus the derivative of Z with respect to Z, which is also one. 100 00:08:30,820 --> 00:08:33,340 So we get for this term here three. 101 00:08:34,390 --> 00:08:39,460 So the answer is three times robots divided by C are square. 102 00:08:41,320 --> 00:08:45,150 Now the second term you can see already highlighted that's red. 103 00:08:45,160 --> 00:08:48,160 So we will get a similar functional dependence. 104 00:08:49,390 --> 00:08:50,440 This a bit more difficult. 105 00:08:50,620 --> 00:08:57,490 So here we have to calculate the gradient of one of our square, which is a bit similar to what we previously 106 00:08:57,490 --> 00:08:59,970 did when we already calculated. 107 00:09:00,820 --> 00:09:07,450 So what we what we get here is first we have to construct the outer derivative, which is our to the 108 00:09:07,450 --> 00:09:08,380 power of three. 109 00:09:08,560 --> 00:09:14,560 So one divided by R to the poor three times minus two, which comes from this two here. 110 00:09:15,070 --> 00:09:19,970 But then please do not forget about the inner derivative, which is the gradient of arm. 111 00:09:21,590 --> 00:09:28,330 Now the gradient of are we have also already calculated this was the arbiter divided by R so this is 112 00:09:28,330 --> 00:09:29,440 what we get in total. 113 00:09:29,440 --> 00:09:35,770 And now you can see when we only look at these vectors here, our thoughts are, is of course the absolute 114 00:09:35,770 --> 00:09:42,280 value of our square so we can get rid of the Spectre's here and just right our square and we divide 115 00:09:42,280 --> 00:09:44,310 by our to the power of four. 116 00:09:44,650 --> 00:09:48,280 So we end up with one divided by R to the power of two. 117 00:09:48,940 --> 00:09:54,670 And then all of these three factors, of course, and you can see it's exactly the same dependence here. 118 00:09:54,820 --> 00:09:58,180 But here we have this term three times and here we have it minus two times. 119 00:09:58,860 --> 00:10:00,640 So these two will simplify already. 120 00:10:02,740 --> 00:10:06,640 Now here we get a different turn here. 121 00:10:06,650 --> 00:10:08,860 We have to calculate the gradient of DOT. 122 00:10:09,220 --> 00:10:13,570 And this is, of course, very, very simple because on the previous slide, we have already calculated 123 00:10:13,870 --> 00:10:15,060 the gradient of row. 124 00:10:15,790 --> 00:10:19,420 And so we can just take our solution from the gradient of row. 125 00:10:19,690 --> 00:10:23,460 But instead of getting here what we get now, roadable dot. 126 00:10:24,610 --> 00:10:26,530 But besides that, it's the same thing. 127 00:10:26,530 --> 00:10:33,130 And we need to consider here also this in a derivative, which is, of course, are divided by R and 128 00:10:33,130 --> 00:10:35,290 then we get here also this minus one A. 129 00:10:36,460 --> 00:10:43,150 So if you don't remember how it is done and if you can't do it yourself, just skip a few minutes back 130 00:10:43,150 --> 00:10:46,440 where we have already calculated this gradient of row. 131 00:10:47,800 --> 00:10:49,510 Now here we get the same thing. 132 00:10:49,510 --> 00:10:53,800 We get again, our dot are where we can just ride our square. 133 00:10:54,160 --> 00:10:57,910 And here in the denominator we have our to the power of two times. 134 00:10:57,910 --> 00:10:58,270 Ah. 135 00:10:58,630 --> 00:11:02,590 So we end up with one of our and then these three factors here. 136 00:11:03,460 --> 00:11:05,350 So this is a different dependent's. 137 00:11:05,350 --> 00:11:10,240 So we cannot cancel this term with the other two two terms. 138 00:11:12,300 --> 00:11:19,800 Now, here we get another term here we have to calculate again the divergence of our which is three. 139 00:11:19,800 --> 00:11:21,210 So this one is very simple. 140 00:11:23,670 --> 00:11:30,640 Here we have to calculate the Dow and sorry, the gradient of one of our to the power of three. 141 00:11:31,200 --> 00:11:38,580 So if we know what the gradient of one of our square is, we also know what the gradient of one of our 142 00:11:38,580 --> 00:11:39,780 to the power of three is. 143 00:11:40,680 --> 00:11:49,500 We just get here are to depart four instead of three, and here we get minus three instead of minus 144 00:11:49,500 --> 00:11:53,920 two, which comes, of course, from these three instead of this two. 145 00:11:54,810 --> 00:12:00,330 And then again, we have the derivative, which is the gradient of art, which simplifies to this one 146 00:12:00,330 --> 00:12:00,630 here. 147 00:12:01,080 --> 00:12:07,660 So as a total result here, we get minus three times row divided by R to the point of three. 148 00:12:07,980 --> 00:12:09,870 So you see, it's again very similar here. 149 00:12:09,870 --> 00:12:14,100 We get to minus three instead of two, minus two, and here the three instead of the two. 150 00:12:15,750 --> 00:12:19,530 And see that it's the same functional dependence as this one here. 151 00:12:19,560 --> 00:12:22,800 And in fact, these two terms even cancel out. 152 00:12:23,530 --> 00:12:29,130 So that's really cool, which means we do not have these two terms here anymore because they are exactly 153 00:12:29,130 --> 00:12:31,130 the same for the last term. 154 00:12:31,290 --> 00:12:38,490 We have something here are times the gradient of row and we already know what the gradient of row is. 155 00:12:39,180 --> 00:12:43,770 This is, of course, this minus one of a sea times. 156 00:12:43,770 --> 00:12:48,860 Roedad times are divided by are we have calculated this on the previous slide. 157 00:12:49,590 --> 00:12:52,460 So again, we get our times are, which is our square. 158 00:12:52,830 --> 00:12:57,460 So we get here from this arg to power for only are to the power of two. 159 00:12:58,560 --> 00:13:05,000 And as you can see, this term once again has the same functional dependence as these two terms here. 160 00:13:06,030 --> 00:13:12,950 And furthermore, you can see that all of these terms compensate and that all of these blue terms compensate. 161 00:13:13,800 --> 00:13:19,920 So all of these difficult looking derivation letters to one single term. 162 00:13:21,550 --> 00:13:30,310 So we have that the entire rent is equal to minus roadable dots divided by sea squared times are. 163 00:13:31,120 --> 00:13:37,450 So we have that our the class operator is the integral of this expression here. 164 00:13:40,280 --> 00:13:48,110 Now, what we can do next is we can write down this road double dots as the second order, time to time 165 00:13:48,110 --> 00:13:49,770 derivative, which. 166 00:13:50,690 --> 00:13:53,200 Yeah, second order a time derivative of rope. 167 00:13:54,020 --> 00:13:59,020 And we can then switch around the derivative and integration. 168 00:13:59,660 --> 00:14:03,050 So we write down the second or the time derivative. 169 00:14:04,250 --> 00:14:12,360 Of this integral of roll over are we can do this because these are here is not time dependent and Rose 170 00:14:12,360 --> 00:14:15,350 was the only time dependent thing in his intake around here. 171 00:14:18,360 --> 00:14:25,890 Now, another thing is that I have used this square here in the Integral and have put it outside of 172 00:14:25,890 --> 00:14:26,630 this term here. 173 00:14:27,630 --> 00:14:33,380 And so here you see, I have already introduced these two brackets to highlight this term here. 174 00:14:33,930 --> 00:14:41,490 And if you remember really well what this term was, this was our electrostatic potential from the beginning 175 00:14:42,030 --> 00:14:42,780 to look here. 176 00:14:42,960 --> 00:14:48,400 If you're introduced as are you get the same thing for this term and for this term here. 177 00:14:49,080 --> 00:14:56,160 So what we have here is one of a C Square, the second order time derivative of the electrostatic potential. 178 00:14:57,880 --> 00:15:04,420 So what we wanted to show is that this class operator is equal to one of us square the second or third 179 00:15:04,420 --> 00:15:06,670 time derivative of five, which is this one here. 180 00:15:07,450 --> 00:15:10,050 But there is this term missing here. 181 00:15:10,480 --> 00:15:10,780 Here. 182 00:15:10,790 --> 00:15:15,160 We wanted to also show that we have another term, which is minus one of Epsilon zero. 183 00:15:16,030 --> 00:15:23,830 So I think given the fact that this whole derivation was kind of difficult, it's a good first step 184 00:15:24,040 --> 00:15:25,090 that we have to as a result. 185 00:15:25,480 --> 00:15:27,640 But then again, where is this term? 186 00:15:28,000 --> 00:15:29,590 It cannot cannot be missing. 187 00:15:31,030 --> 00:15:36,630 And the reason why it is missing is that we made a small mistake in the derivation. 188 00:15:37,420 --> 00:15:46,080 So our derivation is only valid when are equal or is only valid when R is not equal to zero. 189 00:15:46,870 --> 00:15:50,860 This is because we divide by art in many different locations. 190 00:15:51,930 --> 00:16:00,420 So whenever we have divided by hour or derivation is in fact wrong, and we need to be much more careful 191 00:16:00,420 --> 00:16:03,030 for the case where ours is equal to zero. 192 00:16:04,570 --> 00:16:11,260 So now, um, what what this means is that this expression here is only true if R is different from 193 00:16:11,260 --> 00:16:11,760 zero. 194 00:16:12,100 --> 00:16:17,560 So this means if R is equal to zero, which is, of course the case, and we also have to integrate 195 00:16:17,860 --> 00:16:23,290 it, this means we get here additional terms that we have been lacking so far. 196 00:16:25,570 --> 00:16:33,460 And so let's go back to this slide here where I have written down all of these different terms. 197 00:16:33,790 --> 00:16:40,330 And I want to tell you which of the terms gives us additional results that we have been missing so far. 198 00:16:41,080 --> 00:16:48,910 And it turns out that all of these four terms here give us nothing new except for this single term here, 199 00:16:48,910 --> 00:16:50,660 which corresponds to these two terms. 200 00:16:51,400 --> 00:16:57,220 So this term for our equals zero gives us another thing and let's look at what it is. 201 00:16:58,690 --> 00:17:06,040 So first of all, we need to, again, remind ourselves that when we straightforwardly calculate this 202 00:17:06,040 --> 00:17:11,440 divergence of this function here, we get these two terms and we get that at zero. 203 00:17:12,160 --> 00:17:16,090 And it's zero for every are different from zero. 204 00:17:17,910 --> 00:17:24,540 But we can also do a different thing, we can just take this term here, so not this role, but this 205 00:17:24,540 --> 00:17:30,900 divergence of are divided by the power of three and we integrate over the whole three dimensional space. 206 00:17:31,800 --> 00:17:38,850 And if we want to calculate this integral, we can use Goussis Theorem as we did before, and transformed 207 00:17:38,850 --> 00:17:45,090 this integration to a surface integral where we integrate not over the divergence, but just over this 208 00:17:45,090 --> 00:17:48,240 function here of our divided by our to the power of three. 209 00:17:49,110 --> 00:17:51,620 And then this one here is this. 210 00:17:52,420 --> 00:17:52,690 Yeah. 211 00:17:52,800 --> 00:18:01,110 Just this element, which is which is a surface and its orientation is normal to the surface. 212 00:18:01,830 --> 00:18:03,880 So it's always in radial direction. 213 00:18:03,900 --> 00:18:09,330 So this one here is actually parallel to our if we consider a sphere. 214 00:18:10,050 --> 00:18:15,600 So what we are doing here is we want to calculate over the whole three dimensional space, which is 215 00:18:15,600 --> 00:18:22,590 infinitely large, and then he on the right hand side, we must integrate over the surface of the three 216 00:18:22,590 --> 00:18:29,100 dimensional space, which is not really possible, of course, because what is the surface of something 217 00:18:29,100 --> 00:18:30,140 infinitely large? 218 00:18:30,930 --> 00:18:32,220 So we must use a trick here. 219 00:18:32,520 --> 00:18:40,620 We must consider a finite sphere at first and what the radius are, and then we must take the limit 220 00:18:40,620 --> 00:18:42,920 of this radius which goes to infinity. 221 00:18:43,230 --> 00:18:48,840 And if we do this, we get an infinitely large sphere, which is of course equal to this whole three 222 00:18:48,840 --> 00:18:49,740 dimensional space. 223 00:18:51,320 --> 00:18:59,880 OK, so we have transformed this one here to this one, which is with just me, which just means that 224 00:18:59,880 --> 00:19:07,860 we take a sphere, we integrate over the surface of the sphere with Radius R and afterwards we make 225 00:19:07,860 --> 00:19:09,540 the radius infinitely large. 226 00:19:10,500 --> 00:19:18,480 And then here, since we have used the sphere, now this surface element here, this orientation is 227 00:19:18,480 --> 00:19:26,490 parallel to our so we get to your unit vector R and then this is given by our square synesthete ADT 228 00:19:26,490 --> 00:19:27,450 to define. 229 00:19:30,630 --> 00:19:34,840 So now let's use this relation here that these two are parallel to each other. 230 00:19:35,190 --> 00:19:41,280 So this just gives us ah, so we get ah to the power of three divided by ah to the power of three. 231 00:19:42,390 --> 00:19:43,980 And that's the whole trick here. 232 00:19:44,800 --> 00:19:48,510 It means that our integration's is not dependent on. 233 00:19:48,510 --> 00:19:56,640 Ah, so this means it doesn't matter how large our sphere is now we can just take the limit ah to infinity 234 00:19:56,940 --> 00:20:00,560 and it will change nothing that will just give us this term here. 235 00:20:01,890 --> 00:20:04,840 And when we calculate this term, this is really easy for us. 236 00:20:04,850 --> 00:20:09,740 First, we can solve this integration with respect to PHI because there is no fi here. 237 00:20:10,170 --> 00:20:16,500 So we just get the function or we just get to the difference of this integration boundary. 238 00:20:16,530 --> 00:20:23,640 So we get two PI times this integration and this integration we can solve by. 239 00:20:24,090 --> 00:20:29,970 Yeah, calculating minus cosine theta from the uh from the boundaries. 240 00:20:30,210 --> 00:20:31,370 So zero to pi. 241 00:20:32,280 --> 00:20:38,190 And so if you do this you get minus minus one, minus minus one, so you get two. 242 00:20:38,730 --> 00:20:43,440 And so our total integral here is equal to four pi. 243 00:20:45,190 --> 00:20:46,960 So what does this mean? 244 00:20:47,380 --> 00:20:56,530 It means that this this term, he had divergence of are divided by R to the power of three is zero everywhere 245 00:20:56,530 --> 00:20:59,070 besides the position are equal to zero. 246 00:21:00,370 --> 00:21:07,180 But if we integrate this term over the whole three dimensional space, which basically means we add 247 00:21:07,180 --> 00:21:12,830 up this expression here at every possible position, it gives us four pi. 248 00:21:14,080 --> 00:21:16,630 So this means at the position R equals zero. 249 00:21:16,810 --> 00:21:24,400 There must be a very large signal or a very large term that gives rise to this integration that gives 250 00:21:24,400 --> 00:21:25,130 for PI. 251 00:21:26,710 --> 00:21:30,310 And so this is something that has bothered physicists for a long time. 252 00:21:30,310 --> 00:21:36,760 And it's really hard to comprehend because it's a bit counterintuitive that you just have a single point 253 00:21:36,760 --> 00:21:40,360 of which you integrate and you get some finite signal. 254 00:21:41,920 --> 00:21:50,170 But this has led to the introduction of the so-called Delta distribution function and this Delta distribution 255 00:21:50,170 --> 00:21:58,420 function you can write here as Delta are, and in very simple terms, which are not really correct. 256 00:21:58,420 --> 00:22:09,310 You could say that the Delta function is zero for everything different from R equals zero and it's infinitely 257 00:22:09,310 --> 00:22:11,960 large for our equals zero. 258 00:22:12,850 --> 00:22:16,540 So this means you really get only a signal from this point. 259 00:22:16,540 --> 00:22:19,980 R equals zero by this definition is not really correct. 260 00:22:20,530 --> 00:22:27,370 The real definition is that when you write it in an integral, you can get rid of the integer and just 261 00:22:27,370 --> 00:22:30,240 take this other function here that you're right as a product. 262 00:22:30,580 --> 00:22:37,480 So you take this function here, which is just for PI and take the argument are at the position zero, 263 00:22:38,080 --> 00:22:42,570 which in this case just gives for PI because there is no argument. 264 00:22:42,580 --> 00:22:44,190 Ah, OK. 265 00:22:44,200 --> 00:22:51,910 This may sound a bit confusing to you maybe, but on the next, on one of the next slides I will show 266 00:22:51,910 --> 00:22:53,640 you a bit more about the Delta distribution. 267 00:22:53,650 --> 00:23:00,490 Then this may become clearer, but the important thing is we can write down that this divergence of 268 00:23:00,490 --> 00:23:05,860 R divided by R to typeof three is given by four PI times the Delta function. 269 00:23:06,460 --> 00:23:15,550 And this also includes our previous results that for a position where R is different from zero, then 270 00:23:15,550 --> 00:23:17,230 this whole divergence is zero. 271 00:23:17,950 --> 00:23:23,230 And it also includes that when we integrate over this term, we do not get zero. 272 00:23:23,440 --> 00:23:26,730 We get a finite term, which is four pi in this case. 273 00:23:27,610 --> 00:23:35,320 So we get our new term here, which is four PI Delta times zero, which you must not forget because 274 00:23:35,320 --> 00:23:36,850 here we have to zero also. 275 00:23:38,270 --> 00:23:44,520 It's now since this Delta distribution is a bit, yeah, difficult to imagine and a bit counterintuitive, 276 00:23:44,520 --> 00:23:49,680 I think here I have prepared one more slide that I want to show you. 277 00:23:50,130 --> 00:23:57,840 And there exists some kind of continuous transformation of this function here that approximates the 278 00:23:57,840 --> 00:24:02,400 Delta distribution very well if you take the limit of eight to zero. 279 00:24:03,360 --> 00:24:10,320 So this function here is a Gorshin function, which is basically x e to the power of minus X square. 280 00:24:11,180 --> 00:24:16,650 And you have here in this argument also some parameter aid and also here you have a perimeter. 281 00:24:17,990 --> 00:24:24,260 And now if you take this parameter eight to zero, you will see that this prefecture becomes infinitely 282 00:24:24,260 --> 00:24:29,840 large and also that this Gorshin function becomes infinitely narrow. 283 00:24:30,620 --> 00:24:33,890 So those are not Desharnais animation on Wikipedia about this. 284 00:24:34,430 --> 00:24:41,060 So you can see here for smaller values of a this peak in the middle becomes larger and larger. 285 00:24:41,510 --> 00:24:47,590 And the yeah, the broadness of the Delta function also becomes more narrow. 286 00:24:48,410 --> 00:24:55,370 And the important thing is that when you take the limit to infinity, you will see that the value of 287 00:24:55,370 --> 00:24:57,470 this function here becomes zero. 288 00:24:57,680 --> 00:25:05,330 It becomes zero everywhere except for the position X equal to zero, and therefore the value will become 289 00:25:05,330 --> 00:25:06,470 infinitely large. 290 00:25:07,430 --> 00:25:14,840 And also, this may be hard to see, but maybe you can imagine this when you integrate over this area 291 00:25:14,840 --> 00:25:19,940 here below this curve, this will give you the same value for every value of a. 292 00:25:20,870 --> 00:25:27,320 So when you have some some final value of A, you can just calculate the integral and it will give you 293 00:25:27,320 --> 00:25:27,830 a number. 294 00:25:28,880 --> 00:25:34,010 Now, if you make a smaller and smaller, this this number stays the same. 295 00:25:34,700 --> 00:25:41,360 And so this means also for the limit of eight to zero, where you could say that this function is zero 296 00:25:41,360 --> 00:25:47,010 everywhere and it's infinitely large at zero, he will still get the same area. 297 00:25:47,960 --> 00:25:55,700 And so this is the whole important thing about this data distribution, that when you integrate a function 298 00:25:55,700 --> 00:26:01,690 times this Delta function or Delta distribution, then you will get a finite value. 299 00:26:02,270 --> 00:26:07,610 And in fact, you will get the value of this function at the position of zero. 300 00:26:08,810 --> 00:26:14,690 And if this data function is not centered around zero, as in this case, where it's centered around 301 00:26:14,840 --> 00:26:19,340 X zero, then you will get the function at the position X zero. 302 00:26:21,430 --> 00:26:28,360 So in our case, this is very similar here, we have an integral role like this and we have shown that 303 00:26:28,360 --> 00:26:30,760 we can transform it to this expression here. 304 00:26:31,570 --> 00:26:38,320 So this means when we integrate over this function, which is just four PI times Delta distribution, 305 00:26:38,680 --> 00:26:41,880 we get this function at the position are equals zero. 306 00:26:42,310 --> 00:26:47,130 And since this does not have any dependence, we just get for PI. 307 00:26:48,010 --> 00:26:54,740 And so this means we can really express this divergence here in terms of this Delta function. 308 00:26:57,350 --> 00:27:04,600 OK, that was really difficult, but this is only one of two occasions where we will use the Delta distribution 309 00:27:04,600 --> 00:27:05,900 throughout this whole course. 310 00:27:06,340 --> 00:27:12,130 So this is why I decided to not spend too much time on the Delta distribution, because for our purposes, 311 00:27:12,130 --> 00:27:13,740 it's not that important. 312 00:27:14,350 --> 00:27:21,250 We could say it's for us just a mathematical trick to write down the things in an elegant way. 313 00:27:22,170 --> 00:27:28,140 But what's important is that, you know, what this Delta function does, and this is when you integrate 314 00:27:28,140 --> 00:27:31,020 over overhead, you get a finite value. 315 00:27:32,430 --> 00:27:38,850 OK, so we have now here over the plant operator and we have transformed our integrations and have found 316 00:27:38,850 --> 00:27:45,690 that it's actually not just a single term, but that there is another term which becomes only important 317 00:27:45,690 --> 00:27:47,310 when R is equal to zero. 318 00:27:49,110 --> 00:27:55,050 So this means when we write it down like this, we get to hear this new term, previously we only had 319 00:27:55,050 --> 00:27:59,430 this term and this term transformed to this term here. 320 00:27:59,460 --> 00:28:00,600 We have shown this already. 321 00:28:01,680 --> 00:28:07,140 Now, the other term here, we have this Delta distribution. 322 00:28:08,270 --> 00:28:12,560 And on the previous slide, we have already learned what the standard distribution does. 323 00:28:13,630 --> 00:28:24,460 It sets the the argument are to this argument are with this dash here, so this means as the integral 324 00:28:24,460 --> 00:28:34,840 here, we get this function, which is for pay roll at or for the case where R is equal to this are 325 00:28:34,840 --> 00:28:35,210 here. 326 00:28:35,770 --> 00:28:44,260 So this means we get no here an hour and we get here a zero, which means we have finally got rid of 327 00:28:44,260 --> 00:28:45,550 the retardation effect. 328 00:28:46,300 --> 00:28:54,580 So this means this integral here is for Pirjo and the arguments are R and T now these four pieces here, 329 00:28:54,580 --> 00:29:03,490 they cancel out and we just get minus one of Epsilon zero zero of our N t and this is exactly this term 330 00:29:03,490 --> 00:29:04,690 here that was missing. 331 00:29:06,130 --> 00:29:12,340 Sort of all this means after this long time, after this difficult lecture, we have finally proven 332 00:29:12,700 --> 00:29:18,120 that indeed this -- scalar potential solves this Maxwell equation. 333 00:29:19,180 --> 00:29:21,640 And now I told you, it is already in the beginning. 334 00:29:22,090 --> 00:29:26,130 You can do a very similar derivation for the vector potential. 335 00:29:27,190 --> 00:29:29,260 So the mathematics is really similar. 336 00:29:29,620 --> 00:29:35,860 But again, it's quite an effort, maybe even more, because this is here a vector and this was only 337 00:29:35,860 --> 00:29:36,400 a scalar. 338 00:29:36,970 --> 00:29:41,500 But I think if you really take your time, you could do it yourself. 339 00:29:41,500 --> 00:29:45,520 If you follow along the steps that I have just shown you for this Caillat potential.