bert0.txt

 Good afternoon everybody. Even though we are in a flood, here we can start the lecture today. Welcome back. And the title of... Oh my god, it's so hard to see that. This is the realness of this one. Perfect. Okay, so we can start with this lecture. So, which was the state of the art at the end of the last lecture? The idea was this one. Now we have a thermodynamic potential that we want to use, which is the Landau free energy, the GF. And the GL, the free energy is the F, the L must be energy, okay, minus here 0 and then in general if you have an extended body you have to carry out an integral over the volume of H dot m in the column. This is also called the Zeeman term because in some sense it describes the interaction of the magnetic dipoles with an external field or a field which is also a field felt by each type. And then the problem is here. We have to place some reasonable interposed for this family interaction. And the basic interaction is probably the energy. This comes from the magnetic world. Now the problem is, which are now the energy contributions that we have to take into account when you write down the expression for the energy. We start from the very beginning. You know that the origin of firm magnetism is exchange interactions. So let's start now with exchange interaction. You know many things about exchange interaction, of course. And we are used to see Eisenberg Hamiltonian, which is Hamiltonian, which means Eisenberg is an operator which acts on wave function. and especially that's on the skin part of the waist. When you describe it, you remember the story. Describing the interaction, which is in the end something describing Coulomb. In the end, you end up with an amygdala which does not have the contour of this wheel. Not on the back of the waist. The consequence of the fact that we are dealing with So, particle in the quantum limits, but in the end, we have the Newtonian. And the Newtonian acts on the spins, and this is the form, the general form that we are used to see. So, it's the sum, in case you have a lattice, over i and j, which are describing now the different points in the lattice. of terms like minus j, where j is the exchange constant, by si dot sj. Si and sj are operators of it. You are supposed to use the polymathic system, but we will see that in case of micromanagement, we use sort of We use now not really speed but vectors, which are now if you want a vector that has only the expectation value of sx, y and z. So we are referring now to a sort of classical approximation. Even though the exchange is not a classical. So in this description, J is written according to the general formalism of the Blonde-Elland Expression. So according to this definition, 2 times J is the difference between the singlet and and 12 states. Then 2 times j. Which means that j is equal to the difference between the energy for the state and the state variable. The reason of this factor is just connected to this way of writing this sum. This sum over i and j, what does it mean? It means that in this sum, you have count this twice. ij and ji, 2, 1, and 1, 2. in all the 12 samples, J must be R, the difference between the energy in the singlest and the tarpest state. But this is just a problem connected with this particular definition. We will see, I'm not sure today, but probably tomorrow, that another situation we have to Think about this definition when you write something that is considered a definition of J by the law. But for this slide, where we have this Hamiltonian, which is written for a lattice, where you have I and J, which are now an index, which are a lattice. OK, slide this. J is defined according to this definition. Now moving to classical speed. In the micro-magnetism program, the issue is essentially this one. For each cell to measure the . Each square represents itself. Inside each cell, I want to take here, the speed are . Why? Because locally, you know, that makes changes . I will try to represent this in a way that for each wave, each volume, if it is rolling up, I have to give this one to a spin, then to another spin, and they are all parallel. Okay? So at the end of the day, A equals C, so the same thing, one cell and then A equals one. Okay? So let's go back now to the Eisenberg algorithm and we jump from the Eisenberg to this So we start with some sort of classical speed, from vectors. And I'm using right now here this definition of the reduced moment, which is mode m, which is the magnetization divided by the saturation magnetization. Is that according to what I was telling you, you now have a small ball in which all the spins are essentially parallel, okay? This means that locally the magnetization is equal to the saturation magnetization. What does it mean? What does it mean? It means that in general, if you measure a hysteresis loop, you have a curve like this. Now the exact shape depends on many things. The shape of your body, the direction of application of the material. The material which is the opposite of your body. your body. But in principle you have a saturation magnetization, this one, MS. And during the loop you have the... okay, it's not exactly... the equivalent, the magnetization during the loop is not exactly MS. Because here, what is happening? You are breaking your body into domains. But in this situation, the cell volume is so small that inside due to exchange interaction, you typically have the oldest piece of the particle. magnetization and the small volume is always the same. You could explain from exchange. Remember our calculation of the I choose the small sphere was right smaller than the exchange. In this small sphere, all the speeds are the same. In the approach of magnetization, it is something like that. If you consider a small volume inside this volume, all the spins are parallel, which means that locally the magnetization equals the saturation magnetization. Maximum magnetization that you can have. Just to make an example, you take hydrogen. How many bond magnetons do you have in the atom? 2.2. So this means 2.249 for each atom. This is the magnetic moment of the atom. Before, we have an isolated atom, but in a crystal it's 2.2. You have to have a certain configuration to be full. But in terms of what is happening in a crystal, we have 2.2. So, if you take a small volume of iron in the end, you will find that the magnetic moment is 2.2 by the number of atoms, so they will be smaller. Because all the things are small. An extended body is not like that. It has the range, so in the end it will not correspond to the situation. But, for a small volume, if you have A, which is much smaller than the exchange length, Okay, this is our approach. So something that we are used to do is to define this reduced magnetic moment or reduced moment which is just the magnetization divided by saturation magnetization. And so which is the magnitude of this reduced moment? The magnitude is 1. It's a unit vector. The unit vector is small m. The unit vector was direction cosine, so the relevant quantities here, x and y and mz represent the direction cosine of this unit vector. So, let's try now to move on. If you want to calculate the exchange energy starting from some pseudo vector representing now the student, And from this expression you are dealing with typically a scalar problem between two vectors. So clearly the angle between the two vectors is involved in this calculation. Now the question is how big this angle can be. And the answer is very simple. Why? Because the extreme horizon is so strong that you cannot admit a rotation just by a 45 degree. when you jump from one side to the other. That's important. Okay? The energy curve should be different. But what is happening in practice is that usually you just have a very small angle when you are moving from one left point to the nearest one. And the problem is that j is very small. So exchange this price. It's probably for magnets so that's a small angle can be, I'll say, affordable for the system in terms of the energy. Okay? So this means that if I'm thinking about this situation, with this m i and m j representing now the direction of the speed, this single vector that I'm using, or the direction of the local reduced moment, same story. In the end of the day, what you see here is a minus one side between the spin and magnetic moment, so that I'm getting the minus of the heat. There's no problem with this. So imagine now that you have two magnetic moments, or two spins, these are the unit vectors representing the direction, so they are separated by this distance which is rij and the angle between them is this phiij. So phiij must be very small, much smaller than one. The reason is that the energy, both otherwise would be not sustainable for the system, not affordable. So when you jump from this operator to the energy, and it's okay in the idea of jumping from the to a way for calculating the energy for the state for different orientation of the spins. So you have again this sum over i and j, okay, the magnitude, okay, of the spins is the same, which is s, so you take s squared out of the sum, and you're left with what? Essentially in this color problem you have just the cosine of phi ij. But now phi ij is more so you can expand this part cosine of phi ij using the Taylor series so you get that for the cosine of phi ij it will be 1 minus phi ij squared over 2. This one is giving you just a constant term which is represented by this constant here and you are learning something which is given by minus by this minus means plus And this is exactly the expression that you have to calculate. So J square divided by 2 the sum of pi i J square. Now, we are really bridging from the Hamiltonian of the Heisenberg operator to Watt, to the description of micromagnetism, in which we want to treat what typically our problem will be to find out m is the function of the position. So it's a vector field our variable. So we want to know in a condition of a applied field at a given temperature which is now the arrangement of the spin, the expression of this vector field. So, with this in mind, we would like to express this pi ij squared as a function of this vector field, capital M or small m, as a function of r, which is exactly what we are planning to use here. So how can we do that? So it's a very simple approximation, but you can have a look here, So let me now plot again the Mi in this position, this is phi ij. So if I take now the difference between these two unit vectors, okay, the difference between these two unit vectors can be used as a reasonable approximation for the phi ij. And remember, okay, it's the difference which is exactly this small vector here, and in the end is something which is, if you can see it in different ways, something which for small values of the angle is giving you a very nice approximation of phi j. So instead of phi j you write down m i minus m j. And now we move towards this This region of the story in which we want to use this vector field, which is our variable, something that we want to know is why don't we use other variables? They should be difference and m i, tangent of v j. We can use m i as a vector. Okay, so okay, we are doing that, but m i is what we want. So I want to jump from this difference to something involving now the vector of z. What I can do is to say, okay, the difference between this vector and the other one can be calculated as what? Rij, which multiplies the gradient of this M. This example is written here. I calculate the difference of this M and Rij by the gradient of this M. And this is the general expression that must be calculated. calculated. We will see an example, we will make an exercise for a cubic map example. But okay, in the general approach of the story, independently of the Latin, we can replace phi ij with rij dot degree n. And now let's move back to the expression for the exchange interaction energy which is okay J s squared divided by 2 the sum over i and j of r i j dot the gradient of n square. It turns out that if typically you take an r i j a the lattice meters of the energy, because we use this. It turns out that we will see, especially in the case of a cubic lattice, that you can express this form of the energy using this integral, okay? So I'm jumping from the sum to the integral, which is the normal approach where when you're dealing with the integral over the space of the big cow in the volume and you see that the components have the gradient of mx which is the x component of m squared plus the gradient of my squared plus the gradient of mz squared multiplied by a divided by 2. is this A? Which is the new character of this formula. A is called exchange stiffness. Because it gives you a way for measuring the strength. The iron, the A, the iron, the energy associated with the exchange for a given configuration of the sphere. So the A contains the information of the lattice and of the J. And as a matter of fact, the expression for a cubic lattice, as we will see in Y, is this one. Two times J, S squared by Z, where Z is the number of sides in the unit cell of a cubic lattice, divided by A. So you see, you have J, you have S, and you have A. and it represents now the strength of the exchange interaction. This is the expression for a cubic lattice, for an hexagonal closed packed lattice. You see a different expression here, okay, so it depends now on the structure. So there's a general framework so that in the end of the day, when you have to put inside the F here exchange interaction energy you are supposed to use this information, this form here, this interval. Let's comment about this interval. So what is there? So you immediately see that, according to this expression, there is a strong contribution to have you can exchange the energy only if you have some gradients of energy on zero which means that if you have a system with energy which is beautiful which means all the spin functions there is no extra energy force because this is the ground state if all spin are parallel the system is in the ground state for exchange energy I am not saying it is the ground state for the system at all same energy, if all the spin are parallel this will represent the minimum energy. And as soon as you create some field between the spin, clearly you have some extra energy flux coming from the design of the angle of y and j. And this corresponds to the fact that in this formula the gradient of mx and y and mz are not null, and in the end you integrate something which is non-zero and you get an extra energy cross. Okay, so the take-home method is this is the expression, a divided by two b is the integral, and you have some exchange energy if you have some gradient of nx and y and mz. Otherwise, if the gradients are zero, all the spins are parallel, and there is no additional energy cross. energy from the system of the ground. That's the fundamental framework in which we are moving. In order to better understand what is going on here let's go for the solution of an exercise. Which is this one. Find the expression for exchange energy in a simple cubic lattice. Which is the simplest case in which we can work today. So let's start from the beginning. So you remember that in our expression it was written like this, j s squared divided by 2 and the sum over i and j of the cosine pi ij. So let's first express pi ij squared and the reason is that you remember that in the end of the it was something like this. At the end, after the expansion in the Taylor series from the formula. Okay, now, find a j-square. Put it to what we have seen is m i minus m j-square. And so what does it mean? I'm considering now the magnitude of the difference between the vectors. So how can I calculate the magnitude of this difference? I just take what? the magnitude of the vector Is the difference which has the nx component which is delta mx The y component which is delta my and z component which is delta mz So in the end of the day it will be delta mx squared plus delta my squared plus delta mz squared now, how can I calculate delta mx now and talking about what the delta of of a scalar mx, which is the x-component. So this is something that you can easily calculate is rij dot the gradient of mx. The gradient of the scalar number is easier to see and calculate than the gradient of the x-component. The same goes through also for the delta m y. Again you have r i j dot what the gradient of m y and the same for delta m z. r i j dot the gradient of m z. Which you can summarize with this formula is r i j dot lambda applied to m. This is the vector field. This is the formulation for the the color field mx, my and mz, but you can summarize this formula using this equation within the brackets. Now, okay, this is the energy, now the problem is to carry out this summation, okay? We want to carry out the summation and we take now a simple cubic lapse. So let's sit down on a specific lattice point. Of course, now we have our reference system X, Y, and Z. So we are sitting here, this is high field, and now we are considering what the nearest because it's an exchange interaction, and as you know, exchange interaction falls down very rapidly when you move far away. With exchange interaction, invariant exchange depends on the superposition of which function, but in any case, exchange interaction is only accumulated by distance. So what we can say is that, okay, essentially I have a nearest neighbor which is here, another one which is there, so this is for plus a, the other one is for minus a. And then I will have other four contributions for the nearest neighbor in a cubic light distance. So let's start for these two guys, which are at plus a and minus a along the x-axis. And let's see which is their contribution. So let me say that for this specific choice, I have calculate now this rij dot the gradient of fx. But in this state, for this one for instance, rij becomes mod a by ux, where ux is the unique vector of the x-axis. So considering that you have this one, now you have to calculate the dot product, the product of Rij and the gradient of Mx. Of course, if you write down the gradient of Mx, this will be derivative of Mx with respect to x by Ux plus partial derivative of Mx with with respect to y. u y plus the partial derivative of m x with respect to z by u k. But when you take the dot product, the product in this one and that one, we're just selecting this component, okay? Just this one, not the other two. Just this one. So you find that this bracket here is giving you for this specific choice of the ij, it's giving you this contribution, a by the partial derivative of mx with respect to x. Just this ux is selecting this one, not the other one. Then you can say, okay, but I also have the other one, which is for Rij, which is equal to minus a by ux. What is happening here is essentially the same story, okay? It's just a minus sign, but as you are taking now the square, you're giving you exactly the same contribution, but the reason why is that factor 2 appearing here, okay? And now we see what is going on, because now we have to move on to the second part. Here you have what? Rij dot the gradient of my, but Rij is always like this, a by ux, okay? And now you're considering what? with the gradient of m y, but again the fact that i j is along x is selecting the partial derivative of m y with respect to x. And that's the reason why you see the appearance of this partial derivative, the derivative of m y with respect to x. Here, we're selecting the partial derivative of m x with respect to x, but again here, it's selecting the partial derivative of my with respect to a. And now we move to the last term here which is rij dot the gradient of mz, same story, okay, you are selecting what? The partial derivative of mz with respect to a. And of course there is a factor of two also for these two contribution. We have plus a and minus a. This one. This is the first line which correspond to the two first angles along the x-axis. Now we have to repeat the calculation for the other nearest angles along the y-axis. Of course you will have this one and the other one. Now, Rij becomes what? A by Uy, unit vector in the y direction. And what is the difference right now? The calculation is essentially the same, but right now the fact of having here something which is directly along y is selecting all the partial derivatives with respect to y. So you see that the structure is very similar, but this second row, which is describing now the torrential angle along the y-direction, is now showing the partial derivative with plus by 2 y. And the reason is this one. The r i j that I am now considering in the second row is directed along y. And when you move to the third goal, of course, you have now something which is placed here, and there, plus a and minus a again, in this case, r i j, a by u z. So clearly you're selecting the partial derivative in the gradient with respect to z, which is the third goal that you see here. a partial derivative of nx with respect to z, same for my and mz. Okay, this is really the summation over a j when you are seated on one specific high. And now when you look at this, you can see my row or my column. you look at this expression not by who but by code what do you recognize what is the sum of these three terms You see the component of the gradient of what? Of Mx. And here in the second column you see the component of the gradient of My. third of the column you have to do the following. So you can rewrite everything using now the gradient of mx, my and mz. In this matrix by column not by rule. And it turns out that this solution can be written, solution over j. Now you can write it simply two times and then you can write down the word essentially. Two times a, the gradient of mx squared plus the gradient of my squared plus the gradient of M z squared. You see? Now we see the appearance of the gradient of M x and y and M z is a routine. It's a general expression for the exchange energy written for a vector field which is mod M. So this is exactly, ok. Here is a squared, ah, because I have to take the square. Now I can just rearrange these things and especially I have to move from a sum to an integral. Okay, so how can I do that? Essentially our expression J sqrt divided by 2 was coming from the addition of the isometric Hamiltonian, which is the summation. So essentially, I just calculate an integral over the volume, and I divide by the volume of the unit cell. The volume of the unit cell is a to the cube, a cube, which is here. So this means that in the end, you're left with this expression. So you would have j s squared divided by a. Now, it's just a historical reason, but people decided that, okay, let's do a different thing. So, let's write the pre-factor as A divided by 2. Choice. It's a historical choice. Sometimes you can, but we can't really understand why people decide to put it on half or not. The first one who decided A is decided to put it on half, and historically it's there, and we are not changing it. So the prefactor is written as a divided by 2 and a takes this form 2 times j and squared divided by 3. But this is true for a cubic. For a cubic. For a simple cubic lattice it's like this. If you have now a PCC, you have two hand-held superior cells. Because we have the one for a simple cubic, plus the center of the cube. on the FCC you have 4 atoms, because in each phase you have a contribution of 4 atoms, which is a scale of 30 years, and then you get 4, and then the general expression. Numbers, which are relevant here. Your number manual does not design the vector master very well. A, essentially for typical ferromagnets, cobalt for instance, or thermo-mobiles, they are quite different in terms of energy. Not to be, but to be. They are sharing the fact that they are made of cobalt, and thermalloy is made of nickel and iron. So, 3D ferromagnets. For typical 3D ferromagnets, the order of magnitude of A is the order of tens or 5 joules per meter. 5 joules per meter. Tens, this is 31, this is 10, but the order of magnitude is 10, 20, 30 something. The order of 10, 5 joules. This is permalloy. P-Y. It's an alloy. It's widely used in magnets. thermalloy. It's made by one. It's a very soft material. It can have different composition. In general, it's nickel X, iron Y. Something which is widely used for this is nickel 20, iron 80. But you can have different composition. It's a nickel iron alloy. Depending on the ratio between nickel and iron, you change a lot the saturation magnetization and change a lot the amount of the . Okay, the message here is this is exchange interaction, it's described by this integral and the free factor has an order of magnitude of 10, 5 joules per meter. Good. But this is just the first ingredient we are considering. This is not the unique thing. What's next? I suppose the next contribution is coming from anisotropy. Okay, that's the general framing. We have already seen shape and energy. What's the origin of shape and energy? Which is the physical interaction giving rise to shape and energy. Dark-color interaction. You take a body, it's a magnetic body some shape, and you discover that when you write down the magnetostatic energy, which is a polar interaction, but calculated over the whole door, taking into account all the contributions, all the different interactions, you discover that this magnetostatic energy depends now on the wall, depends on the orientation of the diametrization with respect to some peculiar axis of the earth. For a cylinder, for instance, we have found that the mesiatic is set along the middle direction, the set direction. Then for a sphere, for example, and for a film we discovered that the magnetization want to be in the plane or shape of the lithotripy for minimizing dipolar energy. But this is not the unique source of my photography. Now the source of my photography which is less connected to the world, with the crystal in our project, has two sources, one is the German Interaction, the Alibaba Interaction, the Korn Interaction, and the calculation of the Navisat energy we found the code which was containing a tensor lambda, and I told you okay. Usually, the vistas are left apart are considered inside the energy. Anyway in this slide I'm just describing the general formalism of anisotropy energy. In general you say that there is an anisotropy term or anisotropy energy contribution when the energy of your magnetic body depends on the orientation of the magnetism with respect to some specific or peculiar aspect. So having the magnetism point in this direction, that direction, makes it difficult to work on the totalism. In this case, we are dealing with the isomorphism, the total magnetic answer. And how can you describe now this anisotopy? Essentially you just use a reference system so that you can identify the direction by the polar and antimotor angle theta and phi. And so you say that again you can introduce and use a moment which is giving you point by point the rotation of the whole volume. Okay and you say okay now I'm dealing with anisotopy if I have a function which is a capital F and it's not a p which is a function of m it's an energy which depends on the orientation of that you can also introduce something which is the energy density just by dividing by the volume and a very practical way for representing the disorder p is even by the energy surfaces which are now a plot and pretty poor giving you an idea of the anisotropy energy. How can you build up this form? Essentially you say, okay, in the reference system I consider now a transitive direction which is identified by a theta and a fine angle and now the point in this anisotropy energy surface is placed at the distance distance from the origin so that OA is proportional to the anisotropy energy density. Which means that in this case for instance the maximum anisotropy energy is found for the magnetization pointing along the z-axis and the minimum is for the magnetization pointing in the xy plane. You see this distance is smaller than the other one. So the way you build up this surface is quite simple okay. A point in the surface for a given direction as a distance to the origin which is proportional to the value of the anisotropy energy for that specific direction of the magnetization. Of course looking at this representation can easily be seen by hybrid inspection if you are dealing with some easy axes also hard axes. So an easy axis will be what? It will be a local minimum. The point where you see that we have a minimal distance from the origin. And the narrowed axis is the maximum. So that looking at this z-axis here. What is that? The Nisi axis or the Narda axis? The Z-axis, the direction. Narda, okay? In that direction, you see there is a sort of delineation, fusion of the anisotopy energy surface, okay? So this is an R-axis radio system. It's a really visual way of seeing the form of anisotropy. And for instance, here it's not really clear, but you see that it looks like you have a sort of symmetry, well, rotational symmetry around there, which means that probably this is sort of a unilateral. Now it's time to have a break from coffee, and then we'll start with the different forms of anisotropy. Okay, let's move on with the first example of anisotropy energy for magnetism. Uniancial anisotropy. So we say that we have a unilateral anisotropy, well the anisotropy energy just depends on the angle formed with an axis. But it doesn't depend on the wheel orientation. Imagine that the easy axle, the axle reference axis of the system is a vertical one, so that configuration holds the other one, okay? What makes the difference is just the polar angle of theta. And, say, plus theta or minus theta, same, but also theta plus pi. These two are completely different. What makes the difference is just the angle from the axis. with no real, uh, original impact on the orientation. That, or the other one, that is the idea of unilateral anisotropy. Which means that this function must be unchanged upon rotation around the anisotropic axis, and this must be an even function of MZ. There, for the other one, they have Mz which is opposite, but the magnitude is the same, and they share the same magnitude. Okay? Good. Now, how can you easily describe this kind of anisotropic? So clearly, it must be an even function of Mz. Okay? So you have different choices. it could be just a function of the equal to theta, or it could be just a function of one minus the equal to theta, which means the sum of the square root of the angle theta. What does it mean? It's just a matter of taste. You want to write it as a function of the projection of the x, y, and z, which is the sum of the square root of the x, y, and z, or just in terms of the magnitude of the projection of the z-axis. Whatever is the choice of the equation. there is no physical message on this. Traditionally in many textbooks the choice is that of using the sine square of the angle of the finger. It's one possible choice so that for the uniaxial anisotropy you write down an expression which is in agreement with these two constraints here. So the typical expression is k0 constant You can need it or sometimes you suppress it in calculation because it's just a constant. Okay, k0 plus k1 sine squared of the n of theta. And then you move to k2 sine to the fourth of theta. And then you have k3 minus the sine of 6 of the n of theta. Okay? In such a way that you expect that plus m to the n minus m. They do not change the undercurrent. In most of the cases, in all our calculations in this course, when we will be using this adiabatic form, the uniaxial adiabatic energy, we will be using just the first term, which is k1 sine squared of n. So let's see two difficult situations that we can propose in practical application. The first possibility is that you have k1 which is even than 0, 0, which is positive. Clearly if k1 is positive, okay, it turns out that for theta equal to 0, and so you are along the z-axis, you have to move k0 to the top. And when you move to theta equal to pi divided by 2, you have k0 plus k1. This means that we have now these two loads of this red line which represents what? It's a cross-section of the surface and the isotope energy. And the red is a cross-section of the surface and the isotope energy in the plane X and Z. Here you have just k0 and here you have k0 plus k1. If k1 is positive along the x direction, you have a larger anisotropy energy than in case of the direction of the magnetization which points along z. So clearly you are in a situation of an easy axis. The easy axis the Z axis, okay? Because it's the directional magnetization for which you find a minimum value of the anisotropy energy, which corresponds to the minimum distance of the point of the surface anisotropy energy with respect to the point. Okay. And where is the other possible situation for K1 which is lower than zero? So in this case you have again along the z-axis you just have k0 and along the x-axis you have what? You have k0 plus k1 but k1 is negative and so you have a lower energy with respect to the case of the lightning position along the z-axis. Okay? Again, you see a lobe, but the lobe is along the z-axis here. And then the situation of the easy plane and the isotopy. What does it mean? That all the directions in the x-y plane are equivalent, and they correspond to minimum. You can more easily appreciate the story. I don't really get why does the mean of the energy certainly compute like the day ok so how does p1 influence this let me build up this one X and in general this is the angle theta. So, M is for the distance. Theta equals to zero means that the energy is the unit here, the small f. F and isotherm for zero is equal to just k0. And F anisotropy for pi over 2 is equal to k0 plus k1. So I'm starting from k0, and here I have k0 plus k1. Now if you just plot this function, and I will show you plot using Matlab in the next slide you will see now the load. That's not fantastic same square. We'll pronounce it this. You get something like this. And in the other situation, the other way around, because you have k0 plus k1, but k1 is negative, so instead of being here, you are here. So you have something here. You see my point? Okay. positive easy axis. A0 negative is k1 greater than zero. k1 lower than zero. A0 positive means that you have an easy axis. A0 negative means that the xy plane will be an easy plane. And this one will be an hard axis. You can easily get this point, looking at the 3D representation of the circle. That's the case of K1, which is larger than 0, it's positive. You see that you have a minimum here, because of the z-axis, and the maximum is in the equatorial plane, in the xy plane. When you move now to the K1, which is lower than 0, which is negative, you see that the minima are in the xy plane. The easy problem you can do with a machine. Okay, but so far this is just a mathematical representation. I'm not saying which is the origin, but you have already seen a possible source of this type of algorithm. You tell me which type of algorithm you have already seen, which is the reaction. The spheroids, the oblate spheroid, for instance, or the prolate spheroid. The prolate spheroid is elongated along the z-axis, which has now an easy axis along the z, which is bigger on the left. It's just like a microscopic element, it's shaped like a z-axis. This is just a mathematical representation of the story, that the physical object is fine, clean, and is completely, perfectly, also, as I mentioned, described by the linear geometry. It's just mathematical, it's just a way for expressing how to treat now the anisotopy. And as I was telling you, this is the easiest way for including anisotopy energy in your calculation especially in analytical calculation it's quite simple okay that's the first exercise the first example of the story of the thermal dynamics it's an exercise which is asking you to calculate the anisotropic rate of the system This is now the idea. Let's assume that you have an easy axis and isotopy, which means the system described by a surface and isotopy energy like this, with an easy axis which is pointing in this direction. What does it mean that you have an easy axis? It means that the system will like to create magnetization along this axis. Because in this way, it minimizes the magnetostatic energy, or the energy in general. But now, let's imagine that you want to drive the magnetization out of this axis, in the plane, in the xy plane, which of course is not the natural position for the magnetization. Magnetization wants to be aligned with the xy axis, and due to the presence of z-magnetization, energy, the rational energy, sooner or later the magnetization will arrive to the earth. Okay? How it happens in practice is something that I'm not commenting about right now. There will be a strange story. It's not like this motion. You know, there, what does the magnetic field do? but due to the fact that sooner or later, so there are also these situations that describe the identity of the LHG machine, sooner or later the mycoplasm will align to the external field. Now the question is how big is the external field that you have to apply in order to be able to saturate your body in the direction which is particularly meaningful. You see my problem? The problem. This is the definition of the anisotropy field. HA, the field needed to saturate the magnetization along the R-dominant. And Slinger K, which is the field that has to apply Pertweeber to Z in order to drive magnetization along X. How can you use now all the information that you have so far? You should say, okay, everything is described by the thermodynamic potential, in particular by the low-doubled free energy. So let's write down now the special problem, the low-doubled free energy of my system. Okay, and then let's find now the different equilibrium states for the different observations. Okay. You see here that I'm writing down a simplified version of the story which applies to a situation in which typically we don't have exchange interaction software. In which we don't have an access of exchange and of course we have exchange but we don't have an additional contribution to exchange interaction. You remember that the expression for the exchange energy was a divided by two integral of the gradient of mx squared plus the gradient of my squared and so on. The message is that this is zero in case of uniform magnetic speciation. Well, this is feasible. In which condition you find the uniform magnitude? Typically in small bodies. The reason is quite simple. You will see it later, but it is very important. The exchange interaction tends to be obviously in a line. This is not very convenient, but the size of the body is large. because there are other things that can be done to the system, that's a very key domain. But if the body is very, very small, it's more convenient to keep this, I think, to avoid the skin alive, instead of creating domains very, very small. We will not stay here, we will sit. So, for a small body, it's true that you typically have a single domain. So that's the tool that you are dealing with a single domain configuration. So in a single domain configuration, you write down the gl, and now the small gl represents the energy per unit volume, okay? Press the jout in the cube. The depth of the group, the small gl is the depth. And you see two contributions here. k1 sine square of theta, k0. It's a constant. I want to minimize this and it came out. Minus the Z-minus term that's coming from the magnetic wolf. Minus mu0 ms dot h. Okay, this is the sketch of the story. Let's assume now that you have the z which is the easy axis and now the angle theta which represents the angle between the magnetization and the easy axis and then what? The applied field which is h. According to our definition, it's a trickle-driven easy axis. Now the question is, which is the value of h which is large enough to drive the magnetization out of the way. For a single domain configuration. All spins are powerless or with a so-called coherent rotation. What does it mean? Inside my body I see that all the spins move coherently, they just rotate. Okay. Okay. So in terms now of the angles, nearly the sine square of theta is this angle, and in the Z-mult energy term, I have to put one. The angle between the magnetization and the field that could be written as pi divided by two minus Z. So instead of the cosine of pi divided by 2 minus theta, I can write down just the sine of eta. And this is the expression of my GL. In this case, it's a very simple problem because in general, the GL is a function. What does it mean? It's a function of another function because in principle the GL is a function of the magnetization. the magnetization, the magnetization is a vector field, so it's a function of the principle. It's an un-mathematical problem, but in this case it's very, very simple. The unique coordinate is the angle. Okay? Just feel. And, of course, that's another story. Let's stop this consideration. So the unique coordinate is the angle. How can you now work with this function? We know that irreversible processes moves towards the minimum. So what happens? Let's now read the magnetization prompting like this. We apply a field like that. Let's look for the minimum. So there would be something which is irreversible that finally drive the magnetization to be impassive to the planet. But this process happens if you minimize the load up here. So let's find the minima for the field which is applied to the base. You start with this condition, it's not the minimum, it might be position, but it takes in order to find the minimum for the new energy landscape that you have to change because you introduced this field, pointing to this function. Okay? So let's find out the minimum, the possible minimum. And now it becomes a very simple mathematical problem, it's just a problem that's used here, of, say, physical engineering. It's just the minimization of a single value function. So you calculate now the stationary point, you take the derivative of sine square of the angle theta, which means the time sine by the cosine of the angle theta, and then you calculate the rigidity of this Zeeman term that gives you the cross-agonal angle. And immediately you realize that you have two families, two of them, both arising from the fact that the cross-agonal theta is equal to zero. And of course this means that you have now two possibilities, pi plus pi over two, or minus pi over two. So these are the two possible stationary points. Sorry, not me. I'm being there just stationary points. And then you are left with the other family, which is coming from this bracket here. So when two k1 sine of theta minus zero ms by h is equal to zero, you have the second family. And the second family is expressed by this. So now you have some stationary points, but the stationary point is not the minimum, it's the maximum. So now you have to verify under which condition this final is . The possible final state for the condition . To do that, you calculate the second derivative of g with respect to theta, and okay, Here, it's something like k1 the sine of 2 theta, so the derivative is just k1, with the cosine of 2 theta, and you have a factor of 2, which is a field here, okay? And then you have, of course, the derivative of the cosine, minus cosine, which is plus. And now you have this second derivative. Now, what shall we do? We just check for each family under which condition they represent some minimum. For instance, we start with, I don't know, theta equal to minus pi over 2. What does it mean? So let's start in this situation. This is the z-axis. And I don't know, that could be for instance the x-axis, I don't know, or y, x. Okay, when I say theta equal to minus pi over 2, Now it's just a matter of, so we are applying the field in this direction, hx, this h, it has just a component along x, so essentially when you have that theta is equal to minus pi over 2 to apply to the in this definition. So you are asking yourself, this condition, which is theta equal to minus pi over 2, is it really the minimum and for which values of the field that you apply? That's not it. What you do is that you take this expression and you put inside theta equal to minus pi over 2. When you do this, you find here the cosine of minus I calculate here the derivative of g L relative to theta, for theta equal to minus pi over 2. And this gives rise to what? So essentially is the cosine of minus pi is minus 1, which means minus two times k1 and then I have here the sine of theta which is the sine of minus pi divided by two which is again minus one this leads to minus zero by ms by h. ms because it's saturation magnetization according to what we know okay it's a single domain inter-regulation . So 2.24 magneton per atom of iron, for instance. OK. And now the question is, in order to find a minimum, this quantity must be positive. So you say, OK, this is the condition. I want to find the condition. And the condition is for what? It's for H. That's my problem. So for which values of h, this is positive. Okay? And this link, the family h, must be lower than minus 2 times k1 divided by a real zero. Only for h which is smaller than this value, which is negative, I can find that this condition represents what? a minimum, an equilibrium point of my system. What does it mean? It means that, of course, for having the magnetization point, I have to apply a field which is negative, lower than zero, and a magnitude larger than this expression, only in this case. So, if you want to build up now the Stoic loop of your system. What do I mean by Stoic loop? It's a typical measurement that you can carry out in a magnetometer, a DSM, a screen, whatever you want. So you measure what? How you measure the mx component of the magnetization in the x direction function of H ms. I apply the field in this direction, I measure the component and magnetization the same way. So theta equal to minus pi over 2. What does it mean? That the magnetization must be equal to minus ms. Or if you want to plot down down something which is easy. Right by saturation by imposition, which is the small m mx so that it will be bounded between minus one and one. See my point? It's the normal argumentation. So, when theta is equal to minus pi over two, you have that mx is equal to minus one. And you know that this minus 1 is possible, so it's really an equilibrium configuration here, just for hx, which is lower than this value. Okay? Let me call this minus HA. It means that HA is two times k1 divided by minus 0 ms. Only in this region of the applied field you have an equilibrium configuration. Okay? And then there is something which changes. Okay, now you can repeat this kind of motion and you discover immediately that the other solution which correspond to theta equal to plus pi over two is a minimum so it's a equilibrium point if the applied field is larger than two times k1 divided by mu zero mass So you are here. And the other solution is this one. This is a minimum only for fields applied along x, which are at this interval from minus HA, 2 plus HA. Okay? Does it make sense? Yes. Why? I apply a strong field in this direction, the magnetization saturates in this direction. And after HA, of course, once you saturate, you cannot reach a larger value for the magnetization. The maximum value is ms. So you cannot do more than 2.2 volt magneton per atom of hydrogen. No way. On the other side, same story, but you just change the sign. You start to break with a large field. And in between for hx equal to zero, what happens? mx is equal to zero. Where is m? Z. Because z is the easy axis. Okay? When you are here, M is pointing along the z axis. So an exit is equal to zero Then you apply a field in this direction and there is a rotation like that You apply the field in the other direction and there is a rotation like this Okay, and it's a coherent rotation in small Small bodies so that you can treat it as a macro speed. That's another word which is widely used It's a macrospeed. It's a correlation speed which are linked by the exchange. So that they use as a unit. They use a unique speed, as a macrospeed. They rotate as a unique vector which has a large magnetic moment because it's sound. It's a macrospeed. But that's the basic behavior and that's the definition of the anisotropy. So HA in the end is 2 times K1 by mu 0 and S. Of course, it indicates the strength of the anisotropy. At the numerator you have K1, K1 which represents volts. and the green factor in the first step of the reaction isotopy. What about the units? K1 is expressed in the information system of units in which? Which is the unit of measurement? K1. The energy per unit volume. The unicorn, okay? Small GL is an energy per unicorn. K1 is an energy per your spirit. It's expressed in joule per unit. The million order of magnitude for a robot is the unit of 10 to the fifth. 10 to 5, okay? For a Bermaloy, for instance, 4 quarter minus more than, depends now, Bermaloy is a soft Bermaloy. So you can easily obtain the magnetization of the alloy. And that's the definition of anisotropy. Yeah let's move on. Any questions on this? I guess it's like H. Sorry? The rate of measure of the... The field, you measure it in amps per meter. So H is the field that you apply. If you have a coil, H is given by what? a coil with the density of twice m, number of times m, and i is the current, h is m by 5. So we measure it in amps per meter. With the standard, the solution of the magnetic field. So essentially this experiment is something that we do. Every time my students draw a film, the first thing they have to do is go to the DSM and measure the anisotopy. How can you do that? You can measure something like this. You see for different orientation of the sample, you start with the film, which is the response, you carry out the interpolation, the anisotopy, the film and so on. every time, every day in any magnetic laboratory, magnetism laboratory. Laboratory is not magnetic, magnetism laboratory. Okay, now let's move on. Magnetocrystalline ion is of the, what does it mean? So now we are talking about the different sources of magnetosorbic. I'm telling you anisotopy, in actual anisotopy it's just a mathematical description, but this is the source. Typically magnetocrystalline anisotopy is telling you there there is a difference behavior magnetic body when you apply the field along different crystal orientation. So you carry out the measurement. You see that along the 001 family, the 110, the 111 family in Ireland, can you find the difference? This is pretty mentally what you can do. And you see here a sort of summary of what you can measure, for instance, in DCC Ireland, which is the stable phase of Ireland. So this is the magnetization curve like this, okay? So you apply H in one direction and you measure the component of the magnetization along the same direction. And you see that the behavior is very different. Look, when you apply the field along the 100, you see that there is a step increase here, okay, of the magnetization then you tend to saturation. But there is a steep increase of the magnetization here. And now it depends on the system, but I can tell you that if you take a single crystal, made of iron, crystal of energy, or crystal of physical energy, and you carry out this measurement, you just see a stack, okay? Just one stack. Boom! So the versatility could be less than one person at least. It would be half field, it is . And you have one . Any kind of population now creates a complex pattern of domain . It's that function, OK? And you start to . But if you take the same crystal and you rotate it, So then you apply the field along the in the plane. Of course, I don't think it's a mass membership or something, but in the plane. And you discover that that's a different. You need a much stronger field in order to do a saturation. And indeed, you see here that, OK, with a first step here and there, it takes a while for reaching the saturation. So you immediately realize that in this system, an easy axis to respond to each iteration. . . But in that iteration, as soon as you apply the feed, you immediately, boom, you jump in. You can't . Essentially, you find a system that's configuration, if you reverse the theorem, which was essentially the experiment carried out here, immediately move to the opposite direction but it's done in the very very small and while for the 1 1 1 you have to move up to 6 10 to the minus 2 in this unit in test okay okay 6 to the minus 2 It's not the one blue square. Ah, sorry. That's something which is important. The units. You know that in principle, we should build the international system of measurement, which says that age is measured in F and milk is measured in F. Good. Fine. The problem is that we have to deal with some system in which the coercive freedom creates one person needs ten to the minus one. Instead of saying every time ten to the minus one, we are using this. The conversion is this one. This is one Tesla. It was bought in 1024 U.S. sorry, Gauss, and h is m per meter. Okay, now typically one m, so when you come from this to the other one, you have to divide by a factor which is on the order of 8, okay, and the order of magnitude. So, one amp per meter is on the order of 1 divided by 80. First step, sorry. First step, okay. Just because sometimes you will find these different theories. Okay, the picture is different when you move to nickel for instance. Nickel is an FCC. And looking at this picture, which is the experimental result, you see the easy axis for nickel. The one, one, one. See the other way around. When you move to cobalt, which is HCP, again it's different because typically the easy axis is the one which has the C axis. The diagonal and the C axis is the axis of the display and the C axis is the axis of the display. So the method is for different crystal structure we have different which are represented here. You see for example iron, the easy axis correspond to for instance the direction or the or the . You see this minimum place here on the axis. For nickel of course the y, z and x axis correspond to what? to the arc axis. So it's a different story. And the minima are along the axis of the diagonal of the cube. Now the problem is which is the origin of this magneto-crystalline anisotropic? There are two origins which are usually called single ion anisotropics and two ion anisotropics. The first one is single island isotopy which corresponds to something with given C is spinomic interaction. Because you know what spinomic interaction is. So let's see the origin according to this model. So essentially spinomic interaction energy is described by water in a unit which is A by L dot O. L is the operator for the orbital and the momentum and S is the T. And depending now on the sign of this A or alpha, depends now, I'm using A, alpha in other slides I will explain later, there is a three-type of K. But you can have situations in which if f is parallel to S, you have the minimum or the maximum energy But the basic story is that it depends now on the energy or the relative orientation of the orbital angular momentum and the speed of the object. Now, what happens in the so-called weak mass-penolobics interaction beam? It's that you can think that in many systems the direction, the preferential direction of air is fixed by water, by the crystal fist. the crystal fields that you took mind for instance, the octa-hebra, in the tree, like this one, the lowest energy states for these states are T2g states, instead of the EG. This is the rationale for the sample that you have seen. So essentially, the T2G wave function, they display some lobes along the diagonal, okay? So that when you look at these structures, these are orbitals which are centered here, okay? And for instance, this could be the oxygen atoms with the lobes along the axis. So this wave function avoids the superposition charges that the union wire could own to the partial gradient of the wave. So in the end of the day, in the octahedral symmetry, you find that the crystal field selects some orbital which correspond to some values of the angular momentum. Now what happens? Let's imagine that my left arm represents the orbital angular momentum, which is pointing to the shield. And now this represents the speed. The speed, imagine that I'll take positive. This is the minimum energy state, this one. And now you apply a field. You're presenting the field. Your arm pointing there. What is happening? So the field is applied like that. And now I have my... This is the field, this is my spin. So it's the problem that we have seen before, okay? So the spin or the magnetization, they tend to align to the applied field, okay? See what is going on. But it turns out that in this case, you create what? An angle here. Which means an increase in... The energy which is liquid is being created in the large. This is going, the energy is being created by the crystal field. This is the spin, ok? Which is oriented like this. But now you apply a field. The field tends to tilt. The spin, we can be talking about the magnetization, which is the problem we have seen before. The magnetization is given by the spin. The magnetization is given in 3D metabyte for the spin angle. the angular momentum. The orbital angular momentum is 0. So you act with the field on the magnet position, but you are acting on the spin direction. And if you create the field, you create both an axis of spin of the energy. What does it mean? It means that if you want to magnetize your body in this direction, you can do all this by applying a particular you are creating an excess cost. So in principle the best condition is this one. So the final total ends of the system will have a contribution that will be minimized when magnetization is parallel to this factor. And that's the origin for magnetization It's a dynamic state. It's a spirochic interaction. S is fixed like this up here. S is something on which S is connected to the dynamic condition. Different magnetization and dynamic condition require different energy properties. That's all for today. I hope I could finish with you. Bye. See you tomorrow morning. Thank you.