Hi everybody, it's Sam at Mixed Up Craft. Thank you for watching my tutorial today. I'm gonna be showing you how to make this. It's kind of an explosion box in terms of when we open it, it all pops out. But this time it's a triangular shape with this curved closure on the top. Now I saw this picture on the internet just in Google Images. And as you can see, it's in Chinese. And yeah, I think I've got a pretty good representation of that. You can see, I think this one is a little bit longer. But I'm really pleased with the outcome. So that's the inspiration for today's project. And then I finished it off with a little happy birthday sentiment and some little faceted embellishments there. And then this real life plastic butterfly on the top. It's using the nature's grace papers so they've got the butterflies in. So they just seem to work really well. And I was looking back at my tutorials and I seem to use a lot of butterflies. I mean, I do like them, but it's not something that, you know, I guess I go, well, I guess I do really like because I use them all the time and I'm using them again now. So anyway, right. So I've got another sheet from this paper pack which is here. So it's that one there. Use this load. Got about four sheets left now. And it's just beautiful. Really like that one. Again, I'll share all the links in the video description below on my blog. So I've already done my little sentiment. You don't need to watch me do that. So that's all ready to go. I've got my paper. And these are all the butterflies I've got. I brought these from Walmart and they come with a backing with like a silver. So they're basically the butterflies printed twice. But I mean, it does look pretty real when it's not got the glare of the light on it. I think they're brilliant. And I've had these for ages. I brought these for early last year and they've just been in the straw and I had a little mooch around my craft room and thought, oh, I haven't used these. Let's get them out. So yeah, there you go. So I'm gonna be choosing another one of those. I might put the little yellow one on there maybe. Okay, so this is what you need. So it's just a one sheet wonder, as they say. And this is 10 by 12. Okay, so I'm gonna flip it over and do the scoring on this side because you will be able to see it better when I lift it up. But along the 10-inch side, you want to score at three inches all the way down. Push that in a bit further, a little bit up. And at seven, all the way down, okay. Then rotate your card onto, so you can see there are those two, all right. Then what we're gonna create is these score lines. So we're creating a box just in the middle, but you don't wanna score on these sides. So what you wanna do is lie in the 12-inch position with your ruler.  Like so, and then you wanna make sure it's lined up. I'll just come up here, it's lined up with the four and a half at the bottom. So I've got a marker here in black, just here. You may just pick it out as a little black marker, which is at three inches. So I can just now count one, two, three, four, five, six, seven, eight, which is then four inches. And then four, which will give me the half. So one, two, three, four. And I can see there it lines up perfectly with my ruler. Okay, so with that in mind, you then just want to score the middle part. So from the score lines that you just created, you just wanna join them up in the middle. You don't wanna score on the top or the bottom. And then the next one, move along and score it seven and a half.  I've got another marker at the bottom here, which I've popped in at six inches. So again, I can count one, two, three, four, five, six, seven, eight, and then four, one, two, three, four, and just move my ruler there so it's nice and flush. And then again, I nearly went there, I did the wrong bit, just score in the middle. And it will fit perfectly into the little score line, well, or track. Okay, so you will now see that I've got these two long score lines, which were the ones that you done along the 10 inch side, the three and the seven inch. And then you've got these ones in the middle, which are along the 12 inch side, which is four and a half and seven and a half. Okay, so that's all the scoring of the scoreboard, get rid of that. Then what you want to do is along the 12 inch side, so your ruler will line up perfectly along there. What you want to do is with a pencil, is just mark the middle points, the six inches. Okay, and again, on the other 12 inch side, just mark at six. And then just, you're not going to draw a line, but just make sure that that is nice and straight.  And it should go through the middle of that square in the middle, in the center. Then what you want to do is flip it so that you've got the 10 inch side this way. And from the square, so here's the square in the middle, up from each side, up from each corner of the square, you wanna put a little pencil mark at three and a quarter. So just flip it on this side. So I've got my square in the center here, just pop in my ruler at a three and a quarter there, and then put a little pencil mark. Then go along to this side, three and a quarter, a little pencil mark. And then again, the corner down here, three and a quarter, and here, three and a quarter. You can rub the pencil marks out in a minute, you'll probably find you will cut some of them off anyway, but you can see that I've just put little crosses. And each of these crosses is three and a quarter from the four corners of the square in the middle. Ignore if you can faintly see two lines there, because I did go and score there, but what you can do if you do do that, if you're scoring like I have, is just flip it over, and wherever you've got the lump of the score line, which is on this side for me, get your bone folder and just rub over it. And on this side, it's completely disappeared, and you can only slightly see it on the white side. You can also use a rubber, but I find a bone, a bone folder much better. Okay, so now what you need is, let me grab, so you need some circular piece. Again, I'm using, my faithful seem to be using these, I'm on a bit of a roll now with these at the moment. So what you want to do is, with your, whatever this is that you're using, a plate, any circular piece, but it needs to be able to meet the two crosses side by side here. And you just want to bring that up. So in my case, I'm going with the inner circle of this until it meets with those crosses. And then all I'm going to do is, with a pencil is just do a little arch. So you can see there what I've just done. And it will just come up to the very edge of that 12 inches. I mean, if you're using an oval shape, you can do, but sometimes you might be, an oval is obviously, I'll do it on here, like that. If you go more over this side, then you're going to have a wonky curve.  Whereas with a circle, it's the same all the way around, it doesn't matter. So try and use a circle because it does, it will make life easier. Just get rid of that quickly before I forget. Okay, then do the same on the other end. So again, like so, okay. And then you want to do the other side. So you're going to go in this way now. And connect it up that way. And it will give you the exact same arch. This is what I mean, if you're using an oval, the likelihood of getting that exact is hard to do. Whereas with a circle, you won't have any problems. And then again, on this side, like so. Now, as I said before, if you're new to paper grafting or you haven't done this kind of stuff before, then do it all in pencil first because you can rub it out. But what you can do is if you are already confident with using and doing this kind of thing, what we're now going to do is we're going to score over this pencil mark.  But now you're just literally going over that. And you're joining that score line that we've just done then with the two score lines either side here. Then again, flip it around and do it. And you're doing it on the inside one because the outside one we're going to cut, it's just the inside, like so. Okay? This is all the preparation. Now you're probably going to end up cutting away the outer pencil line. But just while I'm here, I'm just going to rub out the inner one because that one would be seen once it's open. Like so. Then what you want to do is another, again, you can free cut this, you can freehand to cut this. There's a lot of, this is the easiest way I've found to make this little kind of box. So what you want to do now is with your ruler and your pencil, very lightly, because again, this is just going to be a cut line.  Again, do it on the other side. This template you will be able to find on my blog. Okay, so you can revert to the template, like so. Just do that again on the other side, like so. So you can see now, we've got this kind of odd shape.  So that six inch marker we're going to join to these corners here. This is the side of this inside box. So I'm going to draw it in pencil just so that you guys can see it.  You can see what I've done. So that's that inside square, the side of it. You just want to join up to that six inch and then again, do a pencil mark to here and there. But now we're going to score over these. That's what I mean. You don't have to do that pencil mark, but if you find it will help you better than go for it because you can just rub it out. So now I'm going to go and score over all of those pencil lines, okay? And then just rub them back out again. Okay, so now the only pencil marks that you should be left with are these big ones here and the curves at each end. All the rest should be score lines. So now you can just rub your scissors and literally go around and cut that all out. Before we burnish it, it's just easier to do it when it's all nice and flat. So just very neatly with your neatest cutting, you want to work your way around those arches. Okay, so that is the shape you should have, okay? It's kind of like a Christmas decoration, isn't it? One of those fancy ones. And then you should have your square score lines in the middle, two long score lines here and then triangular score lines here and here and these little kind of, I guess it's like a cat's eye, the inside of a cat's eye, that kind of shape. So now we can go and burnish everything. So just carefully and then that inside one. So now when we go to, so we fold those ones in a minute because you don't want to fold across here. So just fold, pinch in your little, you know, whatever this shape is. You just want to kind of help it into place like so. And then again with this side here like so. Then fold your triangles. So you want to fold in and just create nice score lines there. Okay, so now what we want to bring this in, this piece here needs to be folded. Now you can do another score line here but I've just found a bit of a faff. It's much easier if you just basically bring the triangle part up, okay, this side. Then the side with the curved piece, bring that over and you want it to line up perfectly. So this corner here should line up with the top of your triangle, which mine does. Now you can see inside, this has kind of got this shape. If you line it on its side, trying to do it so you can see it, so like so, okay. And then just fold it down on one side just so you can burnish the end. So now I've just formed a, so again, this is the triangle, this is the top with the arch. You're bringing it so that it meets up at the top there and then on that piece inside just pinch it like so. Okay, so now it gives you that pop effect. And by doing it by pinching it and not burnishing it, it gives it more of a bouncy feel. So again, I'm just gonna go around to that next side. Always make sure that this all lines up before you pinch it into place. So now I can see if I bring that together, I've got a really nice corner, okay. So again, on this side, just bring those up. Okay, so now you can see it's all coming together. Then you should now just be able, because we've just burnished those pieces, pop those on top and there is your gift box. So we're gonna put a Velcro dot on that in a minute, but then once they open it, you want it to pop open. So it's like that pop box I've done, but it's just obviously a different shape. So let me grab my Velcro. So this is the, I use the brand Velcro. So you get your hook and loop on separate pieces. So that's the hook. So just take one of them off and cut one of the loops off. So again, I'll share all the links, but these are the 16 millimeter ones that I get off of Amazon. There you get loads. You can use magnets. If you want, you could just use ribbon to tie it up. Got a stutter coming on. Okay, so just pop them together and remove the backing off of one of the sides. It doesn't matter which. Then choose if you've got preference to the lid. I mean, they've both got butterflies on mine, but I think I prefer this one because it's got all three of them on. So that's gonna be the top. So you want to put your Velcro or whatever it is you're using. You may want to put a little hole punch through this bit and feed the ribbon through. And then when I bring it together, make sure your corners are all married up and neat and then stick it down. So I'll leave that there for a minute. Then I've got my little sentiment, which I'd already put my backing on. So that will just go. So I don't want it to cover the butterflies. Maybe I'll put this a little bit higher up and slightly over to that side this time. There we go. And you can still see all those lovely little butterflies. And then let's choose a butterfly. So, see that's a good one as well. That looks quite nice against that. I think I'm gonna go for that one. That was quick. I thought I would have taken a bit longer. I'm quite indecisive when it comes to things like that. I sit here talking to myself for ages. So I'm just refolding them because sometimes they've got dodgy where they've been obviously through the factory, but they've kind of bent them in the wrong part. So, because I've peeled these off what was on, they were on their original sticker I need to add. And because these are plastic, again, you wanna pop the red tape because it will stick better. And then that one is gonna go. I'm gonna pop that one there this time. There you go. But I think they look really real. I'm looking at it down through the camera now and I think they look so real. So there you go. So that is that. So it's, again, I guess I don't find it hard because I'm so used to doing all this, but I don't think it is particularly hard. It's just all about precision, getting all those score lines right, getting those pencil lines correct. But now when I open this one up, I just carefully, because I haven't lifted that off yet, just go over that Velcro, the dots on each side. But you can fill that now with treats, put some nice jewelry in there or something and wrap it in tissue if you're putting a gift. So just if it is loose, it doesn't fall out, but it holds itself up on the sides and I just think they're really nice. And as always, I can now, because these are traveling with me. So both of these are just going to go completely flat and be sandwiched in between a paper pack or something to keep them all nice and safe. But yeah, there you have it. So there's so quick and easy to do if you've got a party and you want to do these as little favours. They're great to do. Have a little convey about going with them all. So there you have it. So if you've enjoyed today, please hit the thumbs up and subscribe to my channel to see more. Thanks for watching, bye.


Hi, I'm Zor. Welcome to Unisor Education. Today I would like to continue talking about circles. But apparently the further we move down the geometry road, the more objects we are familiar with. And circles actually are very interesting objects and there are many very nice theorems about circles. And as you remember, the main point is basically developing your creativity, your logic, your abilities to rigorously think about certain statements which you are making. And circles is just adding one more object into this language, into this universe of different objects which we are talking about, which we are using, just as a brain exercise if you wish.  Most of them are about circles but obviously some other elements which we have already learned will be involved in this. So this is about circles, mini theorems, it's number two. I have more actually. Alright, so let's just go one by one. Measure of an angle formed by two quarts, okay, so we have two quarts which are intersecting inside a circle. So let's call it a b a k is equal to half the sum of measures of central angles. Okay, so we have these angles and we are interested in measuring this particular angle. So this is the center of a circle and the theorem states that any one of these angles, and they are, by the way, congruent to each other because they are vertical, so any one of those is equal to half the sum of these two central angles. So this is the center of a circle. So this particular arc and this particular arc, I think I better take it to this way.  And these two arcs support these two vertical angles. So the statement of this theorem is that the measure of any angle which is formed by two intersecting quarts let's say this angle is actually half of the measure of corresponding central angles which are supported by the same arcs. So if you take this arc and its central angle, this arc and its central angle, then the half sum of these two central angles is equal to this. Alright, now how can we prove it? You probably should remember that if you have an inscribed angle, then we have proved that it's half the measure of a corresponding central angle which is supported by the same arc. So we will use this theorem in this particular case. So we know about inscribed angle. Now this is a more complicated picture of two quarts actually crossing each other, but the theorem is, you know, in a way it's very, very much in tune with this one. So in this case we have two different arcs which support the angle. So half the sum would be the measure. Now how can we do it? Alright, let's do it this way. Let me wipe out the central angles and switch to inscribed angles. It would be easier. And then we use the fact that the central is twice as big as inscribed. Alright, so let's just connect these two things, these two points. Let's call this point M where two quarts intersect each other. Now AB prime B is an inscribed angle. And what we can say about this particular angle which we are talking about, AMB. We know that angle AMB is exterior to a triangle AB prime M. See this is an exterior angle. And that's why it's equal to a sum of two angles not supplemental with it. So it's equal to sum of AB prime B plus B prime AMB prime angle B prime AM or A prime. Doesn't really matter. So sum of this angle and this angle is equal to this one which we are looking to be measured. But now let's just think about AB prime B is an inscribed angle which is supported by this arc AB. And B prime A A prime B prime A A prime is actually an inscribed angle which is supported by this arc. So that's why this angle is equal to one half of, let's call this point O, this is a center. So angle A OB, the central corresponding angle which is supported by the same arc. And this angle is equal to one half of a corresponding central angle which is supported by this arc which is B prime O, A prime. Or B prime O, B prime. Same thing. And that's exactly what needs to be proven. This particular angle which is formed by two chords intersecting each other and supported by these two arcs is equal to half the sum of the corresponding central angle supported by the same arcs. That's it. So this is just a tiny bit more complicated case than just inscribed angles. All I was using is the theorem that the exterior angle of a triangle is equal to sum of interior angles not supplemented with it. Now these are two chords which are intersecting inside a circle. And that's why you have half sum of the corresponding angles, central angles supported by the same arcs. Now what if these chords do not cross each other inside a circle and they are intersecting outside? So this is A, A prime. This is B prime. And this is the angle which we would like to measure. So right now, actually the theorem is exactly the same but instead of edging two central angles which are supported by the same arcs, you really have to subtract. And here is Y. So let's forget for a second about the center. Let's just connect these two points. And again, this is an exterior angle to A prime and B triangle which means it's equal to sum of this angle plus this angle. So the angle A A prime B, this one, is equal to a sum of angles A and B plus this one A prime B D prime. Now the angle we are looking for is this one. Just solve this particular equation for A and B. So you will have A and B equals to the difference between angle A A prime B minus angle A prime B B prime. Well, this actually concludes the whole proof because A A prime B, A A prime B is an inscribed angle which is supported by this arc which means, I will not even write it down, which means it's equal to half of the central angle which is supported by the same arc. Now the angle A prime B B prime, A prime B D prime is inscribed which is supported by this arc and again it's equal to half of the central angle which is supported by the same arc. So our angle which we are looking for A and B is equal to half the difference between the corresponding central angles which are supported by the same arcs which are cut out from the circle. In a way it's really very similar to the previous theorems. If chords are intersecting insides is half the sum of the central angles which are supported by the same arcs. If they are intersecting outside then it's half the difference between the corresponding central angle. So you might actually remember it this way. Alright, so these are two very much related theorems to each other. They're all based on the fact that the measure of an inscribed angle is always half of the corresponding central which is supported by the same arc. These are just small complications to that particular theorems. They're used much much rarer but again our purpose is not to provide you with any practical knowledge just to exercise the brain. Okay let's go further. Angle between a tangent and a chord originated in a point where okay here it is. You have a circle, you have a tangent and you have a chord which has an endpoint exactly where the tangent touches the circle. Now as you remember the radius which is drawn in the same point where the tangent touches the circle is perpendicular to this tangent. That's the main property of the tangent. Now what's necessary to prove is if you put letters a and b and o here okay. Alright, so this is the arc which corresponds to this particular chord. So the theorem states that this particular angle is measured as half of the central angle which is supported by the same arc related to the chord. So you have a chord and in the endpoint you have a tangent and you are measuring the angle between the chord and the tangent. Now this particular arc is related to the chord and the theorem states that this angle is half of this angle. Now how can that be proven? Well again let's just use the fact that the radius is perpendicular to the tangent which means that this particular angle which we are looking for well let's call it alpha. This alpha is equal to 90 degree minus beta. That's obvious because this is the perpendicular. Now a o b is isosceles triangle because these are two radiuses so this is also beta. Okay now what can we say after that? If I will draw a perpendicular to a chord let's call it c then angle a o c a o c it's the right triangle if this is beta so this one is 90 minus beta which is alpha right because alpha is 90 minus beta. So angle a o c is equal to alpha well obviously a o c is half of a o b because these are equal congruent right triangles which means this is also an angle alpha and that's basically concludes the proof that this particular angle is equal to half the central angle which is supported by the same. This is alpha and this is alpha so this is two alpha and that's why it's twice as big as this one and the proof. As you see these proofs are really very very easy as long as you understand just a small thing for instance just draw an extra line which connects the couple of points or draw a perpendicular to a chord in this particular case and everything actually immediately becomes you know quite obvious because the whole proof was in one line or two lines so that's really very easy but what's really interesting is you have to know what additional construction what line to draw what perpendicular to drop which two points to connect to really make the whole theorem quite obvious so that's the most important part of it. Next among all chords that contain certain point inside a circle the shortest is the one perpendicular to a radius that contains this point okay so you have a circle this is the center and this is a chosen point inside. Now what this particular theorem says is that this chord it's not a good drawing I'm sorry I think this would be better okay so this is the radius this is AE this is all let's say M so and this is the point P through which all the chords are actually drawn so if you take any other chord let's say A prime B prime which is not perpendicular to all P it would be longer than AP so the shortest chord which is which contains a particular point inside a circle is the one which is perpendicular to a radius which goes through this point in this case it's OM so P is a given point and to draw a shortest chord through this point you have to draw a radius and then the perpendicular chord to this radius okay now let's think about how to prove it okay I'll probably change slightly the letter so it will correspond to whatever I have on the on the web so this is a point M okay AB the chord perpendicular that's right it's perpendicular to OM and A prime B prime is another chord which is not perpendicular to to OM so A prime B prime is not perpendicular to OM okay now what we do is we drop a perpendicular from the center to this new chord okay let's call this an M prime so OM is perpendicular to AB OM prime is perpendicular to A prime B prime now we know that um the radius which is perpendicular to a chord divides it in half it's a perpendicular bisector so basically A prime M prime is congruent to M prime B prime as well as AM is congruent to MB OM prime is perpendicular to this chord or M is perpendicular to this chord okay fine that's done what's next now let's consider the triangle OM prime M it's a right triangle and OM is um a hypotenuse which means OM is greater than OM prime because OM prime is a characters and OM is a hypotenuse now so what kind of a picture we have here we have two chords one of them is closer to the center than another and as we have learned uh from one of the previous lectures one of the previous theorems actually I think it was mini theorem one the closer the chord to the center the longer it is so A prime B prime is closer to the center than OM because OM prime is is smaller than OM that's why A prime B prime therefore A prime B prime is greater than AB so you just have to remember that previous theorem that the the shorter the distance from a chord to a center the distance being perpendicular obviously the longer the chord is and what element of a circle has the shortest distance to the center well obviously a diameter so if you draw a diameter doesn't really matter whether it's from this point or not diameter has zero lengths to the center that's the shortest as it can be and therefore diameter is the longest that's the longest chord okay that that's end for this particular problem now what was interesting in this particular problem what was important to to to lead us to a conclusion well number one a small construction which we made just a perpendicular to A prime B prime and what's also important is always remember what was before because the whole building of mathematics is built from the ground up so you know certain things and you can use it on the next floor in this case we were using this theorem that the closer chord to the center the longer it is okay theorem number five given the circle was a center O and chord AB center O and chord AB okay on equal distance on both sides from the midpoint M of this chord so this is the midpoint and probably this is as we know a perpendicular because the perpendicular bisector is actually crossing midpoint of the chord and goes through the center okay so on equal lengths from both sides of M inside a circle we choose two points C and D on equal side on equal lengths let's say this and this C and D we choose two points okay at each of these points we draw perpendicular to chord AB from C we call it CE and from D we call it DF okay prove that segments CE and DF are congruent so these two are supposed to be congruent all right so how can we prove that that let me first do a simpler task let's say our chord is not just any chord but a diameter it would be a little bit easier in this case now from the diameter on the diameter we have two points on equal distance from the center by the way M and the center O are coinciding right now right we draw perpendicular now it's quite obvious that CE and DF are congruent because CM and MG are congruent by the premise of the theorem by the condition because we actually use the same distance on both sides of M and these are radiuses so we have two right triangles which have congruent calculus and hypotenuse and that's why they are congruent and that's why GE and DF and CER congruent to each other so that's easy now how can we reduce this to this well simply let's draw a diameter parallel to a chord so diameter is parallel to the chord now it's obvious that um CG let's call these K and L CG L K is rectangle Y because these are parallel to each other these are parallel to each other because they're perpendicular to the same line AB and obviously since they are perpendicular these are right angles so CGL K is rectangle which means that this piece is equal to this piece now this part is exactly like this because these two segments obviously are congruent to each other because again these are all parallel lines these are equal so these are equal so basically we have reduced our problem to two simpler ones first that these particular pieces of the segments are congruent to each other because these are opposite sides of a rectangle and these two pieces are congruent to each other because we could just prove it for that case uh second to go so some of these and some of these would be congruent to each other so what's the lesson to to learn from this particular task well the complicated task can be reduced to two simpler ones right so by drawing the the diameter parallel to a chord i have basically broken two big segments into different into two parts each and prove the theorem for these two parts separately and then just sum them up with some sum them together okay next two chords are perpendicular to diameter okay so you have diameter C C prime so we have chord AB no AA prime sorry AA prime and let me check that's exactly what this is is AC A prime now that's the other way around i'm sorry it's not this diameter they're supposed to be perpendicular not parallel my fault diameter is this C C prime right AC A prime B prime C prime and B now it's correct okay so two chords are perpendicular to the end this is the center consider a segment that can take connects midpoints of AB prime and A prime B this is a midpoint and this is the midpoint prove that it's also uh perpendicular to the same diameter okay that's quite interesting actually there are many ways to prove this particular theorem but here is what i suggest let's just consider this diameter as an axis of symmetry is the picture symmetrical well let's just think about it since a diameter perpendicular to a chord is a perpendicular bisector A and A prime are symmetrical relative to C C prime same thing DB prime are symmetrical now if these points are symmetrical and these points are symmetrical then the segments which connect these points are symmetrical as well therefore their midpoints are symmetrical as well and since they are symmetrical the line which connects them is perpendicular to the diameter diameter is actually a perpendicular bisector of this particular segment which connects the midpoints so i have a approach from a symmetry standpoint that's an interesting approach i guarantee that it can be actually solved differently using some congruent triangles and basically using the same principle of symmetry but behind the scene and not really mentioning this uh because what is the symmetry symmetry is perpendicular and equal in length right that's what A and A prime symmetrical means they are on the same perpendicular to C C prime and on equal distance now from this you can obviously derive for instance congruence of this triangle and this triangle to right triangles and stuff like this so basically it's the same thing i have just shortened the way to explain all these problems by using a little bit more complex concept of symmetry but anyway it's quite quite a legitimate way to to to prove the theorem as long as you know that the corresponding points all are symmetrical to each other so what have i used if one point is symmetrical to another point and the third point symmetrical to the fourth point then the segment which is connecting which is connecting these points are symmetrical as well which is kind of a trivial symmetrical problem all right we can argue about rigorousness of this particular proof but it's kind of obvious that it can be made rigorous if we will just open up the concept of symmetry and prove each particular component of that concept the perpendicularity and the equal distance etc all right next given a circle with center o and chord ab continue ab by bc equal to radius so this is the radius this is the radius and we continue to bc so the bc is equal to radius well whenever you see something like this the isosceles triangle you know it just comes to my mind immediately as soon as you have these two segments equal with the common vertex most likely something will be necessary to consider like o bc as an isosceles triangle but anyway let's see whether i'm right or wrong i know it's just my intuition draw a second from c through a center of a circle c through a center a second okay so we are crossing the whole thing all right i was almost right this second intersects a circle at point g which is further from c and e which is closer to c prove that angle a od a od which is a central angle is three times greater than acg than acb all right how can we prove that so this is supposed to be three times greater than this well as i was saying o bc is isosceles triangle and that's why this is these two angles are equal to each other so that's obvious right okay what's next well recall that if you have two quads which are intersecting outside of a circle then the measure of this angle is half the difference between central angles which are supporting these two arcs one arc is ag another arc is ve so basically let's call this angle alpha so alpha can be measured as one half of a central angle which is angle a od minus central angle b oe but it's also alpha right i mean half of alpha i'm sorry i see something is not exactly right half of alpha so again this angle is measured as half the difference between two central angles one central angle is our a od and another central angle is same alpha because they are in in the isosceles triangle and two angles at base well let's just look at this obviously if you multiply it by two you will have two alpha equals to a od minus alpha from which three alpha is equal to a od and that's exactly what's necessary to prove simple isn't it so all you have to know is the theorem which was learned before about this angle between two quads intersecting outside of a circle measured as half of the central angle supported by the same arcs okay let's end of it and the last problem i have in this lecture given a circle with the center o and point m outside of this circle okay circle with the center o and point m outside of this circle all right consider all seconds that contain point m and intersect circle in two points okay this is one in the second one closer to m and one further so let's say this is p and q so the p is closer intersection point with the circle to m and q is further all right prove that a second that contains a center which is this second let's call this p prime q prime and these will be p and q so it's necessary to prove that among all the different seconds which are originated at m the second which goes through a circle has the closest point has the closer point of intersection closest among all other and the further point of intersection furthest among all other so m q is greater than m q prime but m p is less than m p prime so that's what's necessary to prove m q is greater than m q prime and m p is less than m p prime so that's what's necessary to prove now how can we prove it well um remember the triangle inequality if you have three points then the sum of distances along the angular line is greater than the straight line which connects these two points this is called triangle um inequality we will use in this particular case how very simple let's just connect these two points think of the line at the segment m q since all q and all q prime are two radiuses m q is equal to m o plus and instead of this radius i will use this all q prime this plus this now obviously since this is angular connection between m and q prime and m q prime is a straight line this would be greater than m q prime because of a triangle inequality so that's how we prove the first one now the second one is very similar o p prime m o p prime plus p prime m is greater than o m which can be represented as o p plus p m but all p prime and o p are congruent to each other so we can take out from this inequality and we will have that p prime m is greater than p m p prime m is greater than p m and that's the second one which is necessary to prove that's it okay that concludes my series of mini theorems number two number two there will be others dedicated to circles so as i'm saying there will be other mini theorems there will be other problems which are also related to circles and circles are really quite an interesting culture there are many very interesting problems to train your curiosity which i can provide everything will be obviously on the website on the unisor.com you are encouraged to spend as much time as you want on this website and obviously very important for parents who want their children to well succeed in developing of their creativity and other great qualities of the character please don't hesitate enroll your children into different programs let them take exam control them control their progress control their educational experience so everybody's welcome to unisor.com that concludes my lecture for today thank you very much


to where we were last time. We worked out a general relationship for a rotation operator representing an axis angle form acting on an arbitrary vector u. There's three terms involving trigonometric functions, and then vector cross products and dot products and so on. In a special case in which an angle is very small, this simplifies and becomes just u, which is the identity term, plus a small correction, which is the angle theta times the axis crossed into the vector u. Now there's another approach to small angle rotations, which starts from just a rotation matrix as a model of matrix. If it says small angle rotation, the matrix R has to be close to the identity, so we write the correction term as epsilon times A. The epsilon here is just a reminder that this term is so small that if we wanted to, we could absorb the epsilon into the definition of A, and then we just have to think of A as a small matrix. But I split up the epsilon just as, really for psychological reasons, to remind you that that term is small. And then at the end of the hour last time, I showed that if you oppose the requirement that R be refoggable, R transpose R is identity, it follows that this matrix A is anti-symmetric. So the rule here is that the small correction matrix for a small angle rotation is an anti-symmetric matrix superimposed on the identity, of course, which is what holds for zero angles. All right, thank you. Excuse me? Should there be a minus sign to pretend as? Minus sign, excuse me, yes, this would be symmetric. This is minus A transpose indeed. The anti-symmetric matrices. Now, so let's talk about anti-symmetric matrices. These are three by three matrices, let's write them out, write them out. The S has zeroes on the diagonal because it's anti-symmetric. In the one-two position, let's put a coefficient, I'll call minus A3, and then the opposite side, we have to have plus A3. In the upper right, let's put what we call A2 and then minus A2 opposite that. In the remaining slide, minus A1 and then A1 here. Plus A1 to the other side. So this is, as you can see, is an arbitrary anti-symmetric matrix expressed in terms of three real parameters, A1, A2 and A3. Let's write this this way as A1 times the matrix which has the components zero zero zero zero zero minus one zero one zero plus A2 times A2 times the matrix, zero zero one zero zero zero minus one zero zero plus A3 times the matrix which is zero minus one zero, one zero zero and zero zero zero zero like this. And if I do this, then you see that these three matrices I've written here, a form of basis of three by three matrices that span the space of all possible a's and a's of matrices. And we give it some of these three pieces of chalk here. And that these lowercase a's are the expansion coefficients. Now let's write these three matrices, give them names. I'll call the first one J1 and the script J. The second one is J2 and the third one is called J3. When I write ten notes on a computer, I use a sense of response for these matrices because that's just a standard of matrices. Then the blackboard I'll write in the script J's here. But anyway, remember that these are two matrices. So this linear combination here, it's conventional to write it in a particular way. I should write it as A vector dotted into J vector. In other words, here A is a vector of numbers, three numbers. And J is a vector of, quote unquote, a vector of three by three matrices. This is similar to the notation A dot sigma that you're, which you're used to in quantum mechanics, or a sigma with power matrices. Here A is a vector of numbers and sigma is a quote unquote vector of two by two matrices. So it's the same type of notation. When you use this, you have to remember that the J is really a vector of matrices. Anyway, this is just notation. This shows that an arbitrary anti-symmetric matrix can be associated with a three vector, A, if you, quote unquote, dot an entry, is a vector of basis matrices. And it's a nice notation for anti-symmetric matrices. This also introduces for us these matrices script J, which I'll be telling you about. Actually, I intended to do this on another board so you can see if I'm covering it up. But if you can kind of hold your, hold your attention to that. Before I cover it up, let me mention, well, since I started here, maybe I'll finish this before. Let me just mention some facts about this. You can see these J matrices are constants. They just have ones and zeros and minus points in them. And in fact, let me list the properties, the properties of the J matrices here. There's really two properties. This one comes in two versions. The first one is that if you look at the components of one of these J matrices called a J i, and we take the J k component of the other, the answer is that it's a Levy-Chibbier symbol by epsilon i J k with, as it turns out, a minus sign in front. And to check that, you just need to look at the ones and zeros. There's 27 numbers, so they turn out to be this. It's just another way of writing an epsilon symbol, actually. There's a close variation on this property, which is the following. Maybe I'll derive it before I write it down. Derivation is this, is that if we take a J, let's take a dotted into J. This is, of course, a anti-symmetric matrix. And let's multiply that out into an arbitrary vector u, the matrix multiplication times a vector. Let's take the i component of this and work it out. This is the same thing as the matrix a dotted into J, the ij component of that times uj, for I'm using the summation convention. That's the same thing as a sub k times J sub k, the ij component of that times u sub j. Just by using what we mean by a dot j, it's this. But then, except that the Js are related to the epsilon symbol like this, this becomes equal to minus a sub k times epsilon kij times uj. And now let me swap the k and the i, swap two indices in the epsilon, which gives a change in sign.  If that in turn is the same thing as the cross product a cross u, the i component of that. So the i component at the top is equal to the i component at the bottom, so the two vectors are equal. And this gives me property 1d, which I want to write out, which is that matrix and symmetric matrix a dotted into j by matrix multiplication after that vector u is the same thing as a cross u. And this is useful because it gives a matrix notation for the cross product, which it turns out is quite useful in the application. So these are properties 1a and 1b. Now, there's a second set of properties called 2a and 2d. These are commutation relations. If we compute the commutation relations of two of these j matrices, ji and jj, which you do just by doing the matrix multiplication, you'll find this is epsilon ijk times j sub k. This, of course, reminds you of the commutation relations of the Pauli matrices. And see, we're getting similar linear combinations here and here, although these j matrices are coming out of classical rotations and by themselves don't have anything to do with quantum mechanics. In fact, they really only have to do with Euclidean geometry in three-dimensional space. In that sense, they don't have anything to do with classical mechanics either. It's just cheap geometry, as well. Anyway, this is the commutation relations, which I'll let you verify by multiplying the matrices. Now, there's a variation on this, which I'll call 2d, and it has this for it. If we take a dotted into j, this is one particular anti-symmetric matrix parameterized by the three-vector a, and then take another anti-symmetric matrix parameterized by a different three-vector column b. Well, it's easy to show that the commutator of two anti-symmetric matrices is anti-symmetric. So the result has to be expressible in terms of the dot product that's summing three-vector into the j matrices. And the question is, what is that three-vector? The answer is it is the cross product of a with b. It's a cross b, not j. So these are the main properties of the j matrices for four years. All right, so this is a way of talking about j matrices. By the way, if we go back to this earlier expression, we have for a... I'll call this an infinitesimal rotation. The rotation is an infinitesimal. The angle is, however, it's a small angle. That's what I mean by an infinitesimal rotation. The rotation itself is an identity plus a small direction. Let's take this, and now let's use this property number 1b here, which allows us to re-express the cross product in terms of these j matrices. So this says that this line, this box here is equal to, first of all, there's the u, which I'll copy. And then for this cross product, we divide it this way. It's theta times n hat dotted inches for j. That's a matrix. Multiply u. This, of course, is the same thing as the identity plus theta times n hat dotted into j. The whole thing, multiply u. And the result is that the near-identity rotation matrix can be a one-off, down-ended bar. It can be written. These are also trying to reduce us. It can be written this way is that r of n hat comma theta is equal to the identity plus theta times n hat dotted into square root of j. And there's extra terms. This is only the leading term in theta. And this applies when theta is small. If you neglect the higher order terms, it's an approximation with theta is small. So this is another way of writing near-identity rotation matrix now in terms of these three matrices. All right.  Next, let me make some comments about the commutativity of the rotation matrices. Let's say we've got two rotation matrices, r1 and r2, which are related, which are defined in terms of different axes and different angles. So the first one has an axis n1 theta 1 and the second one has an axis n hat 2 theta 2. These are just two arbitrary rotations now. I just want to make a very simple point at this point, which is that r1, r2 is not equal to r2, r1 in general. In general. And to believe this, all you have to do is just try an example. So if you, for example, multiply your rotation about the x-axis and then one about the y-axis by pi over 2 is something simple and then do it in the other order. I'll actually do this for yourselves. In your spare time, your co-op is spare time. And if you do, you'll see the answers are not the same. So this is a way of saying the rotation is not the same in general. In the language of root theory, we say that s of 3 is not a brilliant root and this terminology is used pretty much these days. So I'll just tell you what it means. Not a brilliant would imply the root just means non-inheritance. That's all it means. So the statement is the rotation root, which is the, in this case, the root root. These are the proper rotations. It's essentially a non-inheritance root. It just means that general rotations don't include it. If I may go back to this expression down here at the bottom of the board and we're live here for a small angle rotation it's expressed in terms of the axis of the rotation dotted into these J matrices multiplied by the angle of the rotation. You can see this is the first term of a Taylor series expansion of the rotation matrix in terms of powers of the angle. And there's a question about what the higher order terms are. As it turns out, it's actually pretty easy to get the higher order terms. And I'll now show you how we do this. Because this result of knowledge in the first word of term, it turns out that's enough to get all the other terms without too much trouble. So here's how this works. We'll lock this up and we'll get this involved. Here's what we do. We take a rotation in axis angle form and we look for differential equations in terms of the angle theta. Actually before I proceed, let me do go back to this earlier remark about the commutativity of rotations. Because there is an exception to this. There's an exception when rotations do commute. And that's when the axes are the same. Now this is geometrically clear. If I have an axis in that like this, and then you do a rotation by the right-handed rule, it's like the rotations. It's pretty clear that the rotations about a fixed axis just add, the angles add. So in other words, you have r of in hat comma theta one at the same time, r of in hat comma theta two. Or the point is these are the same axes. This is the same thing as r of the axis in hat theta one plus theta two. The angles just add. And this is the same thing if you do this in the other order. You won't play another order. So rotation matrices do commute if the axes are the same. In general, they don't commute because the two axes in general are not the same. Now using that fact, I can work out a useful expression for the derivative of the rotation matrix of axis angle four. It's actually pretty easy. So write this using the definition of derivative. So limit is epsilon goes to zero of r of in hat comma theta plus epsilon minus r of in hat comma theta divided by epsilon. The first term in this numerator here is the same thing as r of in hat comma epsilon multiplied times r of in hat comma theta because that's the same axis. We have about the same axis so the angles just add. And so you see the entire numerator has a factor of r of in hat comma theta which I can take out to the right. So this thing turns into the limit as epsilon goes to zero of r of in hat comma epsilon minus identity divided by epsilon times r of in hat comma theta.  And we know what that is. Now we're going to go to the bottom of the board it's identity plus the angle times in hat dot j. Now here I call it the angle theta and here I'm calling it epsilon but the result is in fact off the identity divided by epsilon what's left over is just the in hat dot j. So this whole thing is in hat dot of j times r of in hat comma theta. In hat dot j is independent of theta. So this is a differential equation which is easy to solve. The solution is this. This is r of in hat comma theta. There's an initial condition by the way which is r of in hat comma zero as if the identity of the angle is zero to the rotation matrix is the identity. And so this initial condition the solution is that it's the exponential of theta times in hat dotted into script j. This is an important result it's the exponential form of rotation, these are probably proper rotations now. Proper rotations in axis angle form. And so if we expand this out in a Taylor series we get the identity plus theta times in hat dotted into script j plus theta squared over two factorial times in hat dotted into script j squared plus dot by dot. It's easy to write down the general term and this confirms what I said in every minute of the go that this earlier result we got can be extended up we can extend this into the Taylor series and write down the general term it's just an exponential series and this is what it is here. So this is an important result. All right, that's the exponential form for matrices, rotation matrices in axis angle form. Next slide please. Next slide please. Another result and the story on this begins with the cross product. Let's take the cross product of two vectors, I'll call them a and u like this. There is a fact about proper rotations that the cross product transforms as a vector. What that means in present context is that if you take the cross product of these two vectors and then rotate the result like this by some rotation this is the proper rotation now. In fact I'll almost exclusively be talking about proper rotations quite a bit of time. We'll come back to improper ones when we want to discuss parity but for proper rotation we rotate the cross product and the answer it turns out is the same as if you rotate the vectors first and then take the cross product second. So this is the same thing as r a that vector cross with the vector r u. Now I'm not going to prove this formula rather I'll leave it as an exercise for you but when you do this exercise what you'll find is the answer is only true if the rotation r is proper if the rotation r is improper you get a minus sign on the other side. If you wanted to handle both cases what you would do is you would insert a sign here which is the determinant of r which is of course equal to plus or minus one with a plus sign being the proper ones and the minus one being improper ones. One of the ways it's saying this is that if you take the cross product of two vectors what you get is a pseudo the difference between vectors and pseudo vectors is how they transform under parities an example of an improper rotation to change sign under improper rotations whereas under ordinary rotations they don't. Now I only want to talk about proper rotations so I'll leave that sign out and we just have a plus sign on the right hand side. Alright Now I want to use this to derive a useful result and I'm going to use this notation which I got covered up unfortunately the notation for the cross product which is property 1B here that A cross with U is A dot, J over over 4 and U A dot, J is the matrix now it's a matrix multiplication and so let's use that notation to rewrite this so on the left hand side what we've got is we're rotating A cross U is the same thing as A dotted into J multiplied by U A dotted into J is the matrix so now you can see this is a product of two matrices multiply by U and on the right hand side we can write this as R A which is a vector dotted into the matrix excuse me vector of matrices 3 matrices J like this multiply R A like this so this is again a matrix multiplication this matrix runs down one you can do this in a sense as a kind of a commutation rule because if I've got R multiply times this anti-symmetric matrix and I want to pull the R through it what I have to do is replace the A there by R A then the R appears into the side and it drops part of the solve for the A anyway in fact this is where this is going it's a kind of a commutation relation or a conjugation relation involving anti-symmetric matrices to bring this up more clearly let's replace let's write U equal to R inverse of another vector let's call V because if we do this then this equation becomes R times A dotted into J times R inverse multiplying vector V it's equal to on the right hand side R A which is a vector dotted into J which gives us a matrix which multiplies R U you see R U is running backwards and U is equal to V which is V over the right hand side like this V is arbitrary so we'll strike it in both sides and to summarize this we get an equation that says that R times A dotted into J times R inverse is equal to R A vector dotted into J we unbox that because that's a simple formula we'll even give this a name called the adjoin formula not an official name it's just my name but this is in fact related to the adjoin representation of the group which is why I call it that I won't take you into that but it is useful to have a name for this there are several ways of viewing this result A dotted into J is an architecture of anti-symmetric matrix which is a thing parameterized in terms of this three vector A based on matrices J now it's easy to show that if you take an anti-symmetric matrix and you conjugate by a rotation you get another anti-symmetric matrix and so that therefore is also expressible in terms of some three vector dotted into J and the question is what is the new three vector and the answer is just the rotated version of the old three vector so contributing anti-symmetric matrices is equivalent to rotating the corresponding that's the meaning of this equation that leads to another useful result let's let A here be equal to Theta times Inhat Axis times the angle so to write this out we just have R times Theta Inhat dotted into J times R inverse for the supremacists here this equals to Theta times R Inhat dotted into J now a lot of way to exponentiate both sides on the left hand side we've got an expression that looks like this it's either the A times B times A minus 1 it's that exponential you had something like this in the homework problem a couple of weeks ago this is the same thing as A times B A inverse it's the same as conjugating you're conjugating exponential it's the same thing as exponentiating the conjugated matrix the power series each term is just a power conjugating the powers like conjugating each factor so this left hand side is the same thing as R times E to the Theta Inhat dotted into J times R inverse I think it will avoid some confusion here instead of writing R I call this R zero let me put a zero subscript from this R that appears here just to distinguish it from another rotation matrix which is coming up R zero is in the rotation let's just put a zero on it so this is all following from this exaggerated formula so what you can see here is in the middle you've got a rotation matrix and that's the axis angle form that's being conjugated by a rotation matrix R zero on the right hand side is another rotation matrix the axis is rotated and so to summarize this the formula can be written this way is that R zero times R of Inhat comma Theta times other zero inverse is equal to R of R zero Inhat comma Theta let me talk about that result which is another important result this is essentially an exponentiated version of the adjoint formula and so I'll also call it so it's related in such a simple way to the adjoint formula let's call this an exponentiated version of it but let's think about what it means it says if you take a rotation in axis angle form and you conjugate by second rotation of fixed rotation R zero you get a new rotation and the question is what happens to the axis in the angle and the first thing it says an angle doesn't change it's a new rotation of the same angle but the axis has been rotated by the rotation into use of the conjugation now one of the ways of saying this in words which helps you remember it is to say that the axis of a rotation transforms as a vector under rotations and the angle of rotation transforms as a scalar it doesn't change this is all logical it's a simple meaning but this is one way of looking at it now alright so that's the exponentiated version of the adjoint I'm going to apply it in just a moment in fact I'm going to apply it next so far we've been parameterizing rotations in axis angle form you can see there's really three parameters here there's certainly one parameter in axis angle but a unit vector is equivalent to a polynomial on a unit sphere and that requires two angles so if you wonder in terms of angles you see it's all got three angles applied in this parameterization so the basic rule is that the parameterized rotation requires three parameters the set of rotations is a three dimensional space this is one of the common parameterizations of rotations another common parameterization and I want to tell you about that now so the basic idea of boiler angles works like this let's start with our coordinate system x, y, z are an aerosol frame like this this has a unit vector so-called e hat i i equals one, two, three means x, y, and z let's suppose we've got some fixed rotation of r that we're just getting the point here is to parameterize it let's give it a fixed rotation of r let's let r act on these old base perspectives and give us ourselves new ones which we call prime products so this means there's a new axis, a set of axes and we're trying to sketch the meaningless best I can let's say this is x prime, this is y prime this is z prime this gives us two frames an old frame and a rotate frame now I'm going to take it as geometrically obvious let's specify the orientations of all three axes of the rotated frame that's equivalent to specifying rotation r so to parameterize the rotation r we need to parameterize the orientations of the three prime axes well there's three of them let's simplify just starting with z axes so let me draw this picture again except I'll I'll permit the x prime and y prime axes here here's the old axes x, y, and z here's the z prime axes like this now by our rotation matrix r that's all the old axes in the new ones so in particular it maps the z axis of the z prime axes let's parameterize the z prime axes by the spherical coordinates we want the orientation of the z prime axes let's parameterize it by as usual spherical angles let's call it polar angle beta polar angle alpha like this so those are two angles giving us the orientation of the z prime axes now allow me to define another rotation which is not the same as our fixed one we started with, I'll call it r1 but it's one subscript and I just distinguish it from the old one so r1 is defined this way it's equal to r of z hat prime alpha times r of y hat prime beta and the reason this is interesting is because it turns out that r1 also maps the old z axis of the new z prime axes and to see that all we have to do is just draw pictures so if I take the old axes x, y, and z like this here's the original z prime grid vector sticking up like that of the z prime axes now let's begin by rotation about the y axis by angle beta, this is the right hand rule you see around the y axis you do the rotation on the right first and then you move from right to left and you order the operators to apply so this is going to take this z vector and swing it out of the xz plane by an angle of beta like this okay then we follow that by rotation about the z axis by an angle alpha let me look this up so we're going to do this by an angle alpha and that's going to take this vector to the column as you can see and bring the vector over to a new orientation like this in which the angle alpha is the usual azimuthal angle of the new vector and the angle beta is this one so that coincides with the usual spherical angles of the new z prime axes so by drawing a picture like this what we see is at the rotation r1 also maps the z axis the z axis and the z prime axes it does the same thing as our original matrix r for given rotation r now does this imply that these two rotations are the same oh no because both of them have the same effect on the z axis they put the z prime axes in the right orientation but nothing says that they're going to get the y prime axes right and generally don't so r1 and r differ from one another because they don't do the same thing with the x and y prime axes however if the z prime axes is right then the x and y prime axes can only be wrong by some rotation about the z prime direction so if we take this r1 and follow it by a rotation about the z prime direction by some angle is called a gamma if you choose gamma right the y prime axes are right so there must be some angle gamma such that r is equal to r of z hat prime gamma times r1 if you write out r1 this is the same thing as r of z hat prime gamma times r of z hat of alpha times r of y comma beta and this gives us the well-arranged parameterization r will now write as r alpha theta gamma z equal to this r is starting with and you can even see what the range is on these three angles have to be. First of all alpha and beta are easy to interpret because they're the polar angles of the z prime axes so the alpha is the azimuthal angle and it has to be in the range of 0 and 2 pi and beta is the polar angle so it has to be in the range of 0 and pi to see what it is to make the south pole but when the gamma is the rotation that gets the x and y prime axes right inside their x prime y prime plane the general requires an angle of the range of 0 and 2 pi and so these are the ranges on the polar angles such that if alpha and pi and gamma are in those ranges you cover all possible rotations well the way that this is written out is that this is all right but the way this is written out it's not the most convenient form for the polar angles because the rotation the given rotation is now being expressed as a product of three rotations but you can see it's a mixture of old and new axes there's two old axes z and y there's the new axes, the prime axes there and it's more convenient to write it purely in terms of the old axes yes how do you know you don't but the point is we want to be able to parameterize rotation in terms of three angles so if you gave me an arbitrary rotation I could actually work this out and find out what extra angle needed to be opposed to how you were about to do this I think what you're driving at is that the geometrical interpretation of gamma is not as clear as it is about the beta and that's certainly true it's related to the line of nodes between the planes think about that but right now let's just say there is if there exists an angle of gamma now so we'll fix this up in the following way let's take this first expression here so here's our R at alpha beta gamma and it's equal to R of z hat gamma times R1 I'll write it this way R1 times R1 inverse times R of z hat prime gamma times R1 write it like that the reason I do that is because now the last three factors are a conjugation of a rotation by a second rotation of R1 well here's our nice formula of the above which we've just worked out I call it R0 here and I'm calling it R1 there and it's actually R0 there corresponds to R1 inverse here as you can see there's an inverse on the left hand side but the result is but now it's got to be the inverse rotated R1 inverse applied to z hat prime that's the axis that's what this is equal to well here's what R1 did to z so R1 inverse applied to z hat prime takes us back to the original unrotated axis z this is the same thing as our z hat gamma so you see we've gone from the primed axis to the un-primed axis by this conjugation and so the final result is in fact what I've done is I've taken this same here and moved it over to the other side and it's sort of doing the drop to prime so the result is this is an R of alpha, beta, gamma which is a R of z hat, comma, alpha R of y hat, comma, beta times R of z hat, comma, gamma the product of three rotations that look like this and that's the Euler angle privatization this is the most convenient form for the Euler angle privatization of the privatization axis now this is sometimes called the ZYZ convention for Euler angles and in addition I'm using the active point of view on rotations which as I mentioned last time is following throughout this whole this whole subject you've probably seen Euler angles in your course in classical mechanics and they are that you they almost always use a ZXZ convention and they also use the passive point of view so there's details are different but the basic idea is the same it's convenient to use the ZYZ convention in one mechanics because I actually think we started doing this because the conventions we have for which of the which of the holy matrices are real and which ones are imaginary and why are these imaginary reasons for that alright so that's the story of Euler angles alright now one last topic in classical rotations before we move on to quantum rotations in quantum mechanics and this is to examine the question of the sensitivity of rotations in more detail let's write down two rotations let's say r1 and r2 which have different axes and angles in one theta 1 and in two theta 2 and if we write these in exponential form we can write the first one as e to the a1 and even second one as e to the a2 or a1 and a2 or any symmetric matrices a1 is equal to 1 times n hat 1 dotted in the script J and a2 is the theta 2 times n hat 2 dotted in the script J now I mentioned earlier that r1 times r2 is not in general the same thing as r2 times r1 let me introduce a matrix which is r1 times r2 times r1 inverse times r2 inverse and call this matrix C just to give it an A first thing to notice is that if r1 and r2 do commute then I can bring the r1 inverse past the r2 here and it cancels out with r1 and likewise what's left over is r2 times r2 inverse and that cancels the result is that C is equal to the identity if r1 and r2 is equal to r2 and r1 it can happen to commute and if they don't commute then this is in some sense a measure of the amount by which they don't commute now I need to tell you this I hesitate to tell you this in general literature the C is called a commutator the reason I hesitate is because it's not the commutator that you're used to in quantum mechanics it is however related to the commutator that you're used to and I'll show you now how that's solved it's particularly interesting to look at this this matrix C in a special case in which the angles theta one and theta two are small so let's do a Taylor series expansion of these four factors that empowers the angles what happens to C when the angles are small maybe before I do that whom you mentioned that in the book in the Socorite book he does an example of how he considers the first rotation of the x-axis about some angle and then about the y-axis by different angle and then reverses the order goes back and x and back and y and what he shows is that the final result is the rotation of the z-axis well that's just what this is doing going one direction or another this is a slight generalization what he does in the book all right, anyway if we expand these out into series then for r one we've got the identity plus a one plus one half a one squared plus dot dot dot that's the exponential series and then for r two it's identity plus a two plus one half a two squared plus dot dot dot and then for r one inverse it's identity minus a one plus one half a one squared plus dot dot dot and then for r two inverse it's identity minus a two plus one half a two squared plus dot dot dot writing out the exponential series which I carry out in the second order of the backward order now you multiply these four things together obviously what you get in both order is the identity and then in the first order what you get is just the sum of the first four order terms here which is a one plus a two which is zero so we just write this as one plus zero this c thing vanishes in the first order when the angles are small if you want to get a the next correction in other words if you're interested in how these two matrices don't come in with each other you've got to see how it is to see the difference from the identity and if you want to do that you have to go to the second order that's why I expanded these series here to the second order so now we go through and connect up all the second order terms here's some algebra in that there's six second order terms I'll let you do it what you find is that the second order when the slope clears is actually the commutator of the matrices a1 and a2 in the usual sense of order mechanics and then there's higher order terms this is the second order in fact plugging in these specific values of a1 and a2 in terms of the axes and angles this becomes the identity plus theta one times theta two here impact one dotted into j with the commutator of impact two dotted into j plus higher order terms now we'll go back to the properties of j matrices which are summarized over here what should the property 2 be the commutator of two these anisometric matrices is expressed in terms of the cross product of course of all the vectors so the result is in power series this matrix c is the identity plus theta one times theta two times impact one cross with impact two dotted into j plus higher order terms that's something I'll just post that result because we'll need it in a little while okay so that proved the efficiency of this process in little big pieces of chalk okay so that's all I want to say about classical rotations it's really just the this isn't even classical mechanics it's just the geometry of three-dimensional space it's three-dimensional Euclidean geometry it's really all the same now however I want to turn to the question of rotation operators in quantum mechanics let's begin by supposing we have a quantum system and so there's some altered space for the system and in fact at the beginning we're not going to be very specific about what the system is because there's a lot of conclusions that can be drawn without having to do that so we're not going to say necessarily whether it's a spin system or whether there's orbital degrees of freedom or multi-particle or going through this thing or any of those things will actually won't matter some quantum system and what we like to do is to make a reasonable definition of what we'll call rotation operators which we'll think of as being parameterized by the classical rotations r so u is an operator that acts on this orbital space the fact that it's parameterized by the classical the classical rotation r will look this way it's a given a classical rotation even associated with a corresponding operator u of r and in some sense what u of r will do to make our quantum system before I go on we need to think a little bit about physically about what it means to rotate a quantum system if you have like the 2p state of a hydrogen atom that's a certain wave function and you want to ask what does it mean to rotate it well you can't go in with a wrench and turn an electron in a hydrogen atom so in that sense what does it mean to rotate it but I need to point out or remind you that in spite of the language we use all the time about we say the wave function of the electron the truth is what the wave function represents is the statistical results of measurements on an ensemble of identity-prepared systems that describes the properties of an ensemble and not so much of a single system the ensemble itself is prepared by some preparation apparatus which for simplicity may assume can also come by density operators but let's say you have apparatus that pairs with your state so one way to define the rotated state is to say that it's the state produced by the rotated preparation apparatus and that's certainly clear because you can always rotate your Stern-Gerlach apparatus for example alright so that's one point of view on rotation operators in quantum mechanics another point of view is that rotation operators in quantum mechanics oftentimes arise as a result of interactions or as the time evolution of specific systems this is most notably true in the case of spin systems and magnetic fields which give rise to rotations and spins we'll look at that in detail after a while but in any case this gives you at least some idea about what rotation operators in quantum mechanics mean from a physical standpoint right now we want to address the general question about how can we take a classical rotation and in general reduce an operator that represents those rotations so to do this we're going to make some postulates of demands on what this operators U of R should satisfy these are reasonable demands the first one is is that the operator U of R is unitary this follows from the requirement that symmetry operation should preserve probabilities if you rotate a system you don't expect particles to disappear so it should be unitary the second simple requirement is that the unitary operator corresponding to the identity should be the identity operator in quantum mechanics and the third requirement which is less trivial is that if we take the unitary operator corresponding to a product of rotations it should correspond to the product of the unitary operator like this and these are the requirements we'll impose now if we find a set of unitary operators that satisfy these three postulates then what we say is if we have a representation as the magic word here is used frequently in quantum mechanics more exactly this mapping that takes you from the classical rotations to the quantum rotations is a representation of classical rotations by means of the unitary operators which act on the over space that means that these unitary operators reproduce the multiplication law and that's the idea as it turns out these three postulates I've written down that are in general too strong and we can't actually meet them we'll see this comes up it has to do with spin limit half systems but you can almost meet these requirements and when you're done you'll learn some extra things about spin limit half systems this is an extra minus one phase that comes when you rotate an electron by 360 degrees we'll come back to that in just a little while anyway, these are the postulates and now the problem is to work on the consequences we'll stop here


Cartagena. Very, very tempting to go in there, but I'm only two to three days away from Valencia now. And despite the fact I've got no milk, I've got no cigarettes, I've really got no energy left whatsoever. We are pressing on regardless. Terrible decision, but I blame the sting. Valos, which is directly upwind from us, and frankly the tack ink is rather tiresome. More old dragimore, she's not very close-winded. And you can see where as close-hold as we possibly could be. Well that's what I'm trying to do, is keep as close-hold as possible and drive in towards the land and see whether or not the land has any effect on the wind direction and maybe we can steal a few degrees by staying very close to the land. And if not, maybe there's a dead band of air very close to the land as well. If there is a dead band of air, I'm going to put the engine on and drive because that will save me hours of tacking. Frankly, life is too short right now for hours of tacking. The MG Velocity made good is the vector of your velocity directly towards, in this case, the wind. Which also happens to be in line with my waypoint. So you can see we're only going at two knots directly towards the waypoint that I want to go at. The actual speed over the ground is five knots. And that's the problem with going upwind is that you have to sail a lot of miles to get a short distance. Here we can see tack, tack, tack. Here's where I played going further in short and it paid off a little bit. There's a few degrees advantage there. I'm probably going to pay for them now though. So there was no dead band nowhere where it was smooth enough to get any advantage from the engine. And so this is what we're going to do now until we get through here and around this waypoint here. I might see if there's any shortcuts to be taken. Impending collision, I'm trying to line you up with the victim. And don't be fooled by the way my boat is pointing where it's pointing and where it's going. Two different things, but it looks like he's maneuvered. You don't have to avoid everything by two miles, three miles. The ocean's not big enough for that. I did give him a call and he did say he was monitoring me. I found yet another cape. This one was particularly difficult so I'm glad to see the back of that. And over in the distance you can see the high-rised nightmare that is the Costa Blanca. And over there somewhere is Ibiza and Mallorca. But we're not going there today.


There's been a virtual explosion of biomedical knowledge over the last century, and we're teaching different material in medical schools. However, our approach to teaching that material really hasn't changed over that time period. The foundational facts of medicine are taught in large lecture halls, the veritable sage on the stage. I'm a big believer that human beings are in a room together. They should be interacting with each other. If you have 300 people in the room, at least in my mind, one of the worst things you could do is one person lecture to the other 299 people. Students have little opportunity to really engage with the material, so they struggle with understanding its relevance, and they really are unprepared then for their bedside instruction when they have to take the material that they've learned and apply it to the patients that they're taking care of. Medical school is filled with facts, but facts are fragments of knowledge. We have the opportunity in these early years of medical school to create a foundational context for these facts so that the students will be excited about combining their knowledge of biology with the diseases they will learn about in their patients. This curriculum was created to make the material come alive for the students and to inspire them to really master the knowledge they need to take care of their future patients. His tracheal and bronchial epithelial self. By engaging the students through patient stories and using the technology we have today to animate and narrate those stories. We have the power to create these new teaching tools that really bring the basic sciences to life. We can show the pre-clinical medical students the links between basic microbiology and the infections they'll see in their patients. This frees up the lecture time for truly interactive sessions where the students work together as part of a medical team in the classroom where they integrate the basic science foundational facts with the clinical information, skills that they will need as they enter their clinical training. Students have been involved in every step of this process. We've been working with the faculty and giving constant feedback to make sure that the material is clear, compelling, and relevant. It's really rewarding to have this opportunity to impact the education of other medical students all across the country. Two days later I get a call. This also lets us bring the basic scientists into the classroom and allows the basic scientists and the clinicians to really collaborate in teaching the medical students. The learning is more fun. The material is more engaging. It's more efficiently delivered and it will stay with the students into their futures.