When I was seven years old, I started being a cashier, and my dad's store, Radio Electric Company, started without TV because it was in the 40s. We didn't have TV in Tuscaloosa until 1949, I think, and he had a TV set in his store and he could pick up little glimmerings of the Memphis station, but there was not one in Birmingham at that point, as I recall. When I graduated from high school, I went to the University of Alabama for two years and my girlfriend's father ran the public television station here, so I got a job at the television station. I took a cut in salary, I went from 75 cents an hour to 40 cents an hour, and the long run that paid off. I got married and we wanted to go to New Mexico. I had been to New Mexico when I was a boy scout to the Philmont Scout Ranch and I liked it out there and I was accepted to the University of New Mexico and finished in newspaper journalism, was my major, and then was drafted, joined the Navy and was a naval officer for four years, and then my mother died and I decided because I had the GI Bill, I might as well go to graduate school, I might as well go to Alabama and be here with my dad, who had lost his wife. So I came back here in 66 and got the master's degree in communications, and then I continued in public television after that. In 1975, moved to Hollywood to see if I could make it out there and my first job was as a production manager at the ABC Television Network. From there I got a job as a producer and continued and assigned myself a job of directing an episode one time, and continued in producing and directing. George was scarfing shrimp at this meeting and this guy says, hey George, the ocean call, they're running out of shrimp. The last producing I did was with Seinfeld, I was also the director for five years, and then after that I only wanted to direct, so I directed Ellen DeGeneres' television show, I did a short series of Mary Tyler Moore and that was good, and News Radio was a lot of fun. The cast was good and Seinfeld was terrific too. In 2002, Lloyd Singleton, the dean of the School of Communications, had asked me if I would come teach a class here, and so for 12 years I taught a film class at the University of Alabama. There are a number of people in the very first class that we taught here that are working in Hollywood. Joe Borden is a writer and Kate Turnip seeds a producer out there, and Alan McLeod is an actor and a director, and others were working in Hollywood in various and sundry types of jobs. I enjoyed the teaching very well. It was rewarding and doing the job I was doing in Hollywood was also very rewarding. It doesn't matter where you go to school or where you're from, you can succeed if you just work hard at it. Take what you know and get a job doing what you want to do, and everything else will work out. And then here's a kid from Tuscaloosa that did OK in Hollywood.


Stan, thank you. Thank you very much for the guide on this. It's very pleasant to be here. So my talk will be about interacting particle systems and particular property of these systems, which I will refer to as color position symmetry and interacting particle systems. So we'll be ASAP, ASAP, so totally symmetric simple exclusion process, not totally. And the generalization, which I will call it Stahastic 6 vertex model. And I will study their multicolor versions. And again, there are like, so one can say, so one will call them like multi species, multi type, and probably multi classes. So I will define what is this. And my talk will be, we'll have two parts. So first part will be certain peculiar property of the systems, the color position symmetry. So there will be no asymptotics. And then in the second part, I will discuss some new ways how to apply this property to asymptotic analysis of these systems. And in the first part, I will formulate this statement actually in like two times, just the same statement. But well, first in a pictorial way, which is probably easier to remember, and then a little bit more formally. All right, so let me start with pictures. And my basic objects will be n vertical lines, which are colored by n colors. And both positions and colors are numbered by integers from 1 to n. So these are positions like from 1 to 5. Let us say n colors are also from 1 to 5. Like each vertical line has one color. So we have a permutation of size 5 here. And we encode this picture. Yeah, so permutation is constructed in the following way. So this is color number 1. It stands on the second position. So that means that p of 1 equals 2. Like color number 3 is on the fifth position. So p of 3 equals 5. So there is a permutation, and there is a picture how we think about this permutation. So that's the basic object. Now I want to have some probability, some operations with these vertical lines. And so I introduce crosses between these two vertical lines. The two neighboring vertical lines might have a cross. And then colors start to somehow the color can jump from one vertical line to another vertical line at this cross. And these jumps happen with certain probabilities. And well, basically we have two options at each cross. So we have blue color here, red color here. And blue color can go either to the left or to the right. And the only thing that I need to define are probabilities with what probability it could go left and it can go right. And these probabilities will depend on two parameters, q and x. So q will be the main somehow quantization or asymmetric parameter. And x is just like some inhomogeneity, slightly less important. And the ways are defined in the following way. So let me recall that these colors, they have numbers, which are denoted by this inverse permutation. So these numbers, I can compare these two numbers between each other. So one of them is smaller, another is larger, and this is also important. So if I am in the situation when the smaller color in the left and the larger color on the right, then I have probability 1 minus x and x. This is my definition. And in the opposite case, when this one is small and this one is larger, I will have 1 minus q times x and q times x. So these probabilities are associated to each cross. So next, I want to consider these vertical lines with many crosses and certain configurations of these crosses. And I want to define just by statistical physics. So if I have just one cross, I define the weight of this picture. So I have 5, 2. And with some probability, 5 will be here, 2 will be here. Or with some probability, 5 will go here, and 2 will be here. And the weight of this cross will define the weight of such a picture if I have only one cross. And if I have several crosses like here, then I just multiply the weights of these crosses. So that's the very usual definition from statistical mechanics. So we have all possible configurations. Each configuration has a certain weight. And the configuration has a weight, which is a product of weights of all crosses in this configuration. So now I can already formulate this property, color position symmetry of which I will talk about. So this property tells us that the partition functions of the pictures that are represented here, they are all the same. So what is in these pictures? So the second and third pictures are just a rotation of 180 degrees. So just the same picture, just like if we rotate the head or something, they will go one to another. So the fact that they are partition functions are equal to each other. This is trivial. So the trivial relation is between the first and the second one. So let me try to explain how these two pictures are related. So first, let's see that the positions of crosses are just at the same positions. So it's just the same places, we have the same crosses. But something happens with colors. So here we have colors from 1 to 5 going from left to right. And then we fix some permutation of these colors on the top. And we sum everything that happens inside. So something inside has color scheme. But actually I sum all possible options inside. So I can see the partition functions. So the only two things that are fixed here is the top boundary and lower boundary. And I can see the sum of weights of all configurations with fixed boundary conditions. So here I have fixed boundary conditions as identity permutations here and some arbitrary permutations on the top. And here I reverse the directions of all vertical lines and fixed identity permutations from right to left on the top and inverse permutation on the bottom. And the claim is that these two pictures, when I sum all possible variants that can happen inside, they will have the same weight for any choice of this permutation pi. So but it's important that here I have identity permutations and here I have arbitrary permutation pi. So this is somehow a pictorial statement. So I will formulate it a little bit more precisely in three minutes. So it is due to Amir, Angel and Valkor in the case when all these parameters x's are equal to 1. And we recently generalized it to the case when these parameters x's, they can depend on a cross and they can vary. But the parameter q remains the same. So that's how to think about the statement in pictures. Now let me say just exactly the same story. But in terms of permutations, we will need all permutations of size n in this statement. But later, we will also consider infinite permutations when I have integers and I have bijection between integers. So this is an infinite permutation. And again for such permutations, I can think about it as a configuration of particles on the line. So I have colors for each integer. I have one color and one position. And permutation tells me, so if I have pi of i equals j, this tells me that color i is at position j. So this is one way to take into correspondence like permutations into colors. And the inverse permutation will correspond to the opposite case when pi of i equals j tells us that at position i, I have color j. So both these ways will be needed because as you have seen, both permutations and its inverse are present in this property. So this inverse changes the roles of colors and particles. And this is exactly the color positions and colors. And this is exactly the color position symmetry. So then I will need a special role is played by an identity permutation, which I will refer as packed initial configuration. So this is just like identity permutations. Color i stands at position i. Then I need to say what is the stochastic cross in this case, or this is like update in exclusion process. So if I have two particles and one with smaller color and one with larger color, so smaller color is black here and gray color is a larger one, I apply a day update for two neighboring positions. And with some probabilities, they do not switch. And with some probability, the particles switch. So x1 minus x. And here, if they go in the opposite order, so the smaller color is to the right, then they also with some probability switch with some not. But this depends now on q times x. This is the same definition as with stochastic crosses. And I will have notation that this update happens at positions z and z plus 1. So these two particles are at neighboring sites, z and z plus 1. And when I apply this update to some configuration, I obtain a random configuration with these probabilities. So this is notation for such an update at positions z plus 1 and in homogeneity parameter x. And parameter q will be fixed throughout the talk, and parameter x will be allowed to vary. So that's another definition of stochastic cross. And now the color position symmetry reads that if we take a finitely many such updates at positions z1, zk, arbitrary positions, arbitrary and homogeneity parameters, and we apply them in some order to identity permutation. So we get some random permutation as a result of this, so this is like finitely many random updates. Then we do the following. So we take the same updates, but we apply them in the opposite order again to identity permutation, but in the opposite order. So we start with this one, then this one, and then this one. So we get some other random permutation. And the claim is that these two random permutations are related to each other in a very simple way. We just take an inverse, like permutation inverse, of one of these distributions, and we obtain another distribution. Or in other words, so like for one, so if we move from z1 to zk, in this order we take color position permutation. And if we take it in the opposite order, we take position to color, somehow, way to encode the permutations. And these two random distributions will be the same. So this is somewhat somehow in a way surprising, because it seems that positions and colors play very different roles. That positions were updated only in neighboring positions. And well, it seems that in the definition of dynamics they participate in very different ways. But this statement shows that we just need to reverse the time, and then the roles of colors and positions are also reversed. So this statement, like the story, is that first it was proved for TASAP, so which corresponds to q equals 0 in the paper by Engel, Hall-Royd, and Romig. Then in the ASAP case for general q, but without these inhomogeneity parameters in the paper, Amer, Engel, and Valko. And we generalized these, so included these inhomogeneity parameters, which are also allowed to depend on, somehow, on time. So x1, xk, these are arbitrary sequence of numbers. Any questions? All right, so far this is just some property of this system, how to apply it to something asymptotic, probabilistic, because that's what one usually studies about all these processes, what happens with them in the large time limit. And probably the main point of my talk is that this statement actually helps to understand these systems. So let me probably move to, somehow, more widely known models of standard ASAP, or standard ASAP. So standard means that usually people have just two colors, which they call particles and holes. So this is standard interacting particle systems. Also sometimes there are actually three colors. And in this case, it's usually called first class particle, second class particle, and hole. So first class particle, somehow, it interacts with second class and with hole in the same way, so it doesn't distinguish between them. And all these systems can be obtained from our infinitely many colors case when we somehow become partially colorblind. So we can distinguish only whether the color is, for example, smaller than a. Then we say that all of this is just one color. If the integer associated to color is between a and b, we say that this is another color or another class. And if it's greater than b, then this is like a third class or hole. So if we do this, somehow, degeneration or coupling, we do not lose any information. And the definitions of our dynamics, well, they remain the same. So we just lose some, like the multi-color case, infinitely many color case, it contains all the information about these cases. So but symmetry theorem, the statement that I had, this statement is somehow inherently a multi-color statement. So because it involves this inverse permutation, it's not very clear what is inverse permutation if you just have three colors or two colors. So this is a multi-color statement. And the main point of this talk is that actually this, it's even like if we want to study a standard ASAP, it's convenient to have in mind this multi-color picture because it gives somehow new results. It seems to give a new result, which it's hard to understand how to get them without this multi-color picture. So OK, so there is just like a degeneration to this like two-color or three-color cases. And now let me define finally standard ASAP. So we have continuous time at each position z, well, like pair of positions z, z plus 1, we have independent Poisson process. Like for each z, there is independent Poisson process. And when we get a point from this process, or like it evolves, all these processes evolve in time, and when we get a point in this process, we try to apply, well, we apply update w, z, z plus 1 that I have defined earlier. So this is the definition of continuous time ASAP. And then we are interested what happens with this, like with these particles, we're interested what happens with this dynamics when time goes to infinity. So this type of question, it was pioneered by Harris, Liget, Rost, and others. And there are many, many questions that can be asked about this picture. So I will just concentrate on the results that I will actually need. So the most ardent version is just two colors. These are particles. These are holes. And initial configuration is that all particles from the left to zero, or all holes are to the right from zero. And let me introduce these random variables, which is just like occupation random variables. So if it positions z, we either have a particle or not. So this will be either 0 or 1. And of course, for various z, they are highly dependent. And they have very trivial joint distribution. And they encode this process. And the question is, what is known about these random variables as time goes to infinity? So quite a lot of things are known. So the exact result will be on the next slide. But the type of question that I want to recall is that what is the density of this particle process? So in time, we have these particles move from the left to the right. So they jump with average speed. So they want to jump to the right with probability 1. And to the left only with probability q. So generally, they go to the right with the speed 1 minus q. And then, so we have a region, like a trivial region, when we have density 0. We have a trivial region when we have density 1. So in this region, nothing has happened. But in a non-trivial region, so it turns out that we have a density, which is a linear function. So this is, well, by now, this is a classical result. But this is not by no means a trivial result. But in the limit, so this point is 1 minus q times t. This is minus 1 minus q times t. And the linear density between them. Sorry? Well, one dimension, this is only integers. OK, well, this exact number is q equals 1 over 2, I guess. Yeah, so it's 1 over 2 and then scaled by time. So this point is 1 minus q and this 1 minus q. But the exact formula for this function is like this. So for a general q case. So here is this statement about density. Actually, there are two statements. And they are like the proofs in this Angel-Virus and Binasi Foucais papers, so which are independently and simultaneously. I, as far as I understand, were published in 1997. So we are interested in these occupation variables. And we need to assume something about a point where we look for this occupation variable. And we assume that it grows linearly. So this constant is y. And if y is larger than 0, if y is smaller than it's 1, and in between this linear function. So the probability to find the first correlation function in the limit will be given by this expression. But then I also need to know what happens if I look for several neighboring sites. So I take this position m of t and also have m of t plus i. So some, finally, many positions nearby. And the statement is that, actually, these occupation variables, they become independent. So we have IID Bernoulli process for the window of find length. So this is called local equilibrium in ASAP. And I need it only for very particular and probably the simplest initial conditions. All right, so how this can be related to this multicolor statement. Let me start with an example, which can be easily done by other methods. And this example, basically, is due to Ferrari and Kipnis just to illustrate the simplest application of color position symmetry. So now I assume that I have three colors, which means I have particles, first class particles, second class particles, and holes. And let us start with a situation when I have just like one second class particle. So just like one second class particle, l equals 0. So first class particles go to the right, holes go to the left. But what happens with this unique second class particle? This is not very trivial. Well, if you just see this for the first time, because it can go to the right, it can go to the left. And the result is that actually it is uniform on the segment minus 1 minus qt, 1 minus qt. So it has like a large variety of options what can happen with it. So it can go to the right with large speed and can go to the left with large speed. And how to prove this? With the use of symmetry theorem, one can see this in the following way. So we apply, so we have continuous ASAP for multi-species model, multicolor model. And the question is where is the particle of color 0 is given by this event. That permutation of 0 is less than x. So this is the event that governs what happens with this second class particle. Because we made a coupling, everything else, like our packed initial condition, is coupled and gives exactly this initial condition. First class particle to the left, holes to the right, and just one second class particle. And this event is exactly what we are interested in, the position of second class particle after time t. And what happens with the right-hand side of our color position symmetry? So we need to make reverse of time, but it's easy to see that this is just the same process if we have IID continuous ASAP. And we need to look at the probability that position to color at 0 is less than x. But this is a very different event. This is the event that we have at 0 a particle of color less than x. So it seems that these two quantities are not very much related, but it's immediately, well, the theorem gives us this equality. And this event, so this event is about three-color system. So this is like a second class particle, holes and first class particles. And this event is just about two colors. So here we're interested whether color is less than x or greater than x. So this right-hand side is immediately given by the theorem from the previous slide by the first correlation function of standard ASAP. So some quantity related to three-color ASAP is immediately related to the quantity, which depends only on two colors. And about it, everything is, well, not everything, but what we need is known. And it's immediate that the limit from the previous slide, the limit is just like this density function. I need to subtract minus x because this will be step initial conditions shifted by x. And this will be density at minus x. So this particular example can be done by another very simple argument by Ferrari and Kipnis. This is just an illustration. But now we can play with this a little bit more and to produce a little bit less simple results. But just by the same logic. So let's say that we have several second-class particles and just a simulation. So if I have L equals 3, I have three second-class particles. And I look what happens with the largest second-class particle. So it will have this distribution. So it is more likely that it goes to the right than to the left. And one can write the formula in general. So if I have several second-class particles, so second-class particles, they do not switch order between themselves. So if I have a largest second-class particle, it always remains the largest. And these largest second-class particles will be further to the right. And the smallest will be further to the left. And there is some explicit formula for what is the limited distributions. And the proof, basically, I already told you the proof of this result just in the case of one second-class particle. So we look to this probability. And we also use the second part of this statement about IID Bernoulli distribution. Then it immediately gives. So one can see that here is something like from binomial distributions. So this is exactly because I have IID local equilibrium for standard ASAP. And in the case when k equals 1, so for the smallest particle, it was also proved in a recent paper by Coastal Science and Zell. But so far, so good. But one can see that I always should have something like step initial conditions. So the smallest first-class particles here, second-class here, holes here. And because my statement all starts from identity permutation. So what can I say about some other cases? But in fact, actually, I can say something. So let's go to the situation, which is almost as the one that I described. But now I have first-class particles here, then second-class particle, again first-class particle, and holes. So I just switched the order of first-class particle and second-class particle are switched in the initial conditions. And I have the same questions. What happens with the second-class particle with the unique second-class particle in this picture? So it is just like a very small perturbation, just like two-particles switch positions. But it turns out that asymptotic behavior feels this. And this is the answer. This is the answer. So the second-class particle, ASAP, will have this limiting distribution. So this is the limiting distribution that was before, uniform distribution. And I have some additional contribution. So this particle is more likely bit on the left compared to Ferrari-Keithness case, rather than be to the right of 0. So how to prove this with scalar position symmetry? It's, again, just like one more very simple trick. We first apply deterministic update to these two positions, 0, minus 1, and 0. And then we start IID process. And then it becomes important that I have the reversal of time in my abstract theorem. Because when I look to the right-hand side, I first need to do IID process. And then in the end, I need to do this deterministic swap between minus 1 and 0. And what I get here is this IID Bernoulli from standard ASAP plus this deterministic swap that I have to introduce because of the change of initial conditions. So this is basically complete proof in a way of these results. And I don't know how to get this result by any other standard techniques for general queue. So the proof with this multicolor statement is very easy and doesn't require basically any. So I already told you during a 40-minute talk the proof of this result. So note that this distribution depends on this very small local perturbation of initial conditions. It also depends on queue in a substantial way. So because for ASAP and ASAP, for quite many quantities, it's often the case that they just have the same limited object, just like to prove things for ASAP are much harder. But here the answer also depends. So these types of questions, they depend on a queue in a substantial way. In the limit, we get something that substantially depends on queue. So and like a general statement that can be proved by using this philosophy is that if we have step initial conditions outside of minus LL, so first class particles here, holes here, and then here we can have arbitrary collections of first class particles, second class particles, holes, and we can describe the position of any second class particle, like any one second class particle in the limit as time goes to infinity. And the recipe, so I haven't formulated like this like a complete recipe, but the recipe is that just we have IID Bernoulli and then apply these deterministic swaps, deterministic updates, which correspond to the initial conditions. So that's one application of color position symmetry. So let me also give another one. It will have like a similar flavor, but like the type of asymptotic, like the objects that appear asymptotics will be somewhat different. So I assume that I have, yeah, I think, yeah. So here I have TSEP, so Q equals 0. And again, I create some non-trivial initial conditions. So I have first class particles, which starts from minus L. And also I have first class particles from 1 to L. And I have a unique second class particle at 0, and everything else are holes. And now I, so if L is fixed, this is the case that I had before. So I have second class particles and some finite perturbation of initial conditions. But now I want to allow L to depend on time. And then, so in like TSEP and ASAP, so this is called shock. So we have like this distance is quite large, but then these first particles at some point becomes closer, closer, and they finally catches this particle. And at this point like appears like, yeah. So this is called shock, I think it's quite a good name. So like one wave reaches the other one that is generated by this first collection of particles. So that's what happens for large L. And now the question is what happens with this second class particle, which sometimes can be rephrased as like what happens where exactly is the position of the shock. So again, so this is the question about three colors. So I have two, second class particle, first class particle, and whole. And color position symmetry gives the statement that relates the position of the second class particle. So on the left, I have just like probability that this position of second class particle is less than x. This probability is just equals to probability that certain interval in standard TSEP, so this is the number of particles in interval between minus x minus L and minus x plus L. So this interval of length to L plus 1 contains more than L plus 1 particle. So some observable of this system, namely observable, is the distribution of the second class particle. It has exactly like exact relation with some quantity related just to standard TSEP. And because like standard TSEP, so there are many available tools to study it, these exact statements are somehow very useful because all these like exact results can be immediately translated to this new type of questions that turn out to be related to just like to standard TSEP or standard ASAP. So this is the exact relation. And in order to formulate the synthetic results, so I'm interested in the situation like just like one specific case when this L grows as time to the power 2 thirds with some constant. So L time grows to infinity, L grows to infinity, but of the scale 2 thirds. Then the position of second class particle, so the position of second class particles, it also has a scale T to the power 2 thirds. And the limited distribution will be the difference of two sections of every process. So this is the exact statement. And it appears just from the known asymptotic results of the theorem of Johansson essentially that every process appears as a fluctuations of height function for standard TSEP. So these non-trivial results, like it was like we require that are many like detrimental formulas and like all these computations. So it can be just like translated because of the exact relation. It can be just like translated to the result about this second class particle in the short in some particular limit. So this is just like examples of this applications. So there are some other ones, but my main goal here was just like to, well, to formulate these abstract statements for multi-color process and to try to convince you that even like if one studies like usual TSEP, usual ASAP, like two color, three color situations, it's useful to have in mind this multi-color case as well because it's hard for me to understand how one can get this statement without like this multi-color. So there should be a proof. So it's not like a miracle if there should be a proof, but how to formulate it, how to know that this statement exists and then how to prove this. This is all very, I guess this is all quite non-trivial if like one just walks with second class particles and first class particles and holes and without this multi-color statement. So I guess the lunch is soon, so I will stop here.


Soldiers of Alpha Company, 2nd Battalion, 34th Armored Regiment, 1st Armored Brigade Combat Team, 1st Infantry Division, performed gunnery tables at Konitop Range in Dresden Komorski, Poland on March 2nd, 2022. Master gunners and graders rate the soldiers on tasks that include calling the tower with the correct commands and firing at targets. The staff uses pop-up targets to test the soldiers and a powerful camera to grade the soldiers on their performance. The training began early in the day and continued into the night, providing them an opportunity to practice their weapon systems and prepare for their future Table 6 qualification. This training is part of the ongoing task of maintaining the standard and making sure soldiers are ready for any challenge they may face. For the 1-13th Mobile Public Affairs Attachment from Dresden Komorski, I'm Army Sergeant Elisabeth Melendez.


Hello, hello, Andrea Trowsky here with Dental L Tutoring. I am going to be showing you guys some loops today because I had a comment on YouTube the other day asking to show you guys some loops and why I like certain loops over other ones. Now, I'm not an expert because I have only tried three but I definitely love one more than the other ones and I'll tell you why. So first I'll show you guys the ones that I don't like but also keep in mind that everybody has their own opinions and just because I don't like something it doesn't mean that it won't work for you. So just to keep that in mind. So the first loops that I got were actually used because I was a student, I didn't have any money but I wanted loops. So I had only paid about $500 for these which is a great price and they did fine. Like I did have them for several years. I love them, no issues at all. The thing about loops though is you do have to be careful you do have to be careful with the magnification. So buying used loops make sure to try them on first to make sure that they are good for you. Now I'm lucky because I just need the lowest magnification possible. So those are a lot easier to buy than if you need a much higher number because not everybody needs a higher number, right? The rep that I was talking to said that most people only need the lowest possible unless you start to have eyesight issues and then you might need something more. So always ask about the magnification. You won't know what you need unless you see a rep to look at that for you because a rep will actually measure your magnification. They'll measure the working distance and there's other things too. But as I said before, when I was a student, I was a hygienist for maybe, actually sorry, I was not a student because I was a hygienist for maybe a couple months, but I wanted loops. So I didn't know my working distance, magnification, nothing, right? So okay, sorry guys to make a long story short here. These are the ones that I bought used. So as you guys can see, they're the flip up style. There's nothing wrong with that, but they're heavier. So if you put them on, okay, and you don't look very good either, right? But who cares? But they're a little bit heavier around the nose area. But if you're not using them, they're so easy to just flip up. So if you were talking to your patient about something, if they had a question instead of having to take them off, you could easily flip them up like this. So that does help. And to adjust them, there's a little knob here. So you see the like loop lenses? You can make them further apart closer together, depending on your face shape. So they're more customizable if that makes sense. But it's kind of like a one size fits all. So you can widen them. So these are a wider and then you can kind of put them closer together. So you look through like this. But the one thing that I don't like about them is over time, this area becomes quite loose. You could probably make it tighter, but I just didn't have the tools to do that, nor did I care to do that, but you probably could. But you see how this could become quite loose. So if you're working, this can go like literally, see you guys, sorry, see that? Up and down. So if you're working, you're cleaning teeth and then this could move and then you're thinking, okay, now I can't see. So I have to do one of these, right? So they do not stay in place over time. When I first had them, they stayed in place where I put them, because you just have to tighten them up for them to stay, right? But I was happy that I did have these still because I had lost my main loops when we had moved. I couldn't find them anywhere. And now that I have used loops for such a long time, not using them, I had a very hard time working. I felt I couldn't see things properly. I felt that I wasn't seeing the tarder. Who knows if I was, but I do love loops and I highly recommend them. They're a great investment. So these are the ones that I don't love, but they are cheaper and they work if you are starting out, okay? But if you have the money, let me show you guys the ones that I love and the ones that I currently use. They are by Oroscoptic, did I say that right? Here's my light, Oroscoptic, okay? Now I'm being honest and I'm sorry, but I had horrible customer service experience with them, okay? Would I buy from them again? No, because I did not appreciate their customer service. I had damaged my loops, totally my fault, but they fell on the floor and they were slightly bent. A couple of years ago, so I did contact my rep to send them in, have them fixed. They were still covered under warranty because I had only had them for two years or something like that, a year and a half even. Worst customer service ever. The rep was fine, so not him, but the fact that I had to go back and forth constantly, they got lost in the mail. The rep didn't really seem to care, so I had to contact FedEx, I had to do this, I had to do that, because these are expensive, you guys. Don't quote me, I don't remember, but I think I paid about $2,000, okay? But it comes with a light. So again, I would not buy Oroscoptic again because of the horrible customer service that I had to deal with, but these are my favorite loops. So when these stop working, knock on wood, I will look into something else, but I'll tell you why I like these ones. And I'm just saying that because you guys know that I am honest, I will let you guys know how it is. So what it is is you fit them on like this and the lenses stay on the safety glasses, which I love.  And I should turn on my light, although it might not be charged you guys here, I'll see if it's charged. We went on our honeymoon last week, so it's probably not old, it is on. I'm surprised because we went on our honeymoon last week and I haven't used them in about a week and a half. So anyway, so here's the light, you plug it in, you hold the light here and there you go. I love the light.  Oh, see it says low battery, makes sense because I haven't used it for like a week and a half. But I love these loops because the lenses are stuck on so they don't move. You see that you guys? Totally worth it. And with your more expensive loops, a rep will measure your magnification, the working distance, everything.  So that's hugely important. The nice thing about oroscopic loops is that they come in different colors who doesn't love that. Blue is my favorite color, so that's why I have the blue ones. And you can adjust your handles here. Or sorry, your handles. You can adjust them to fit properly, which is nice. And I do highly recommend getting a light. Yes, it's expensive, you guys, but it's worth it. Once you work with a light, you will not want anything else. I'm serious. When my light runs out of battery in the middle of a working day, I am sad, okay? Because then I have to turn on the overhead light and it's just not the same, okay? You need a light. That's the best thing that I can say. Yes, there are a lot of money. It's worth the investment. It is worth it. So just like I say tutoring is worth the investment to pass and have less stress. Loops with a light is worth the investment. And the nice thing is, with a lot of loops, if you don't like them, just take them back and you get a full refund. So it just kind of depends on the style you want. You can have the flip style like this or where the lenses are stuck on, which is the style that I prefer, okay? With a light. So if you guys are curious, I will leave the link for the ore scoptic on the bottom. These have been my favorite loops ever. I did try to order loops on Amazon because again, you have to know your magnification, but they looked nice. They, I don't know, they looked nice actually. A lot cheaper. I paid only $50, $60 or something, but the magnification wasn't right. So even though I knew my magnification, I tried them on and it was like, whoa, like I'm dizzy. I'm, this isn't gonna work for me. I bought the ones on Amazon when I had lost these ones and I didn't know if I would ever find them and I didn't wanna spend the money on a new pair if I found them, right? So I was still looking, still unpacking. So that's why I did that. So there's no harm in trying loops from Amazon. And the nice thing is you can take them back. They're a lot cheaper. I know some that love them.  Some love them. So it doesn't hurt to try because they're so much cheaper. But I love these ones. They were about $2,000, I'm pretty sure. If you're a student still in school, they typically offer payment plans with no interest. That's what I did at the time because I was in school to be a restorative hygienist. And that helped me. Plus at the conventions, they are always cheaper. Talk to your rep. Just talk about loops. They will measure you at the convention, everything. So I hope that helped you guys. If you have any questions, let me know. And I will see you guys in the next video.