One of these is about recognizing that the health of humans is dependent or is closely linked to that of animals and the environment where they interact.  So there are a few examples where gender is important for example in exposure to diseases. For example where women are in the kitchen cooking and maybe they are expected to slaughter for example chicken and exposed to pathogens that are likely in a chicken meat. For example Campylobactas and Salmonera among others. The other one is recognizing that women are more prone to certain diseases. We saw in the case of Listeriosis for example in South Africa where majority of those that were affected were women. The other thing is the role of gender in targeting interventions. Cooperating gender in one health just makes it easier for countries to achieve health but also to meet several other sustainable development goals. This is the HEAL project which here is one of the partners. So this project operates in Ethiopia, Kenya and Somalia. And it involves professionals from the health domain, animals and environment working together with the local community. So this project operates through what they call one health unit and through this unit the community is actively engaged. So the needs that they have are captured, well captured. JEDA is strongly part of it. Tools like for example the training manuals, the setting of time for meetings and interactions during these one health units. It considers the gender. One investment is of course research evidence generation. We need more data for example the gender segregated data to begin to understand what are some of these factors that would warrant further work to bring out the gender element in interventions. The other one is trying to see which roles are being defined by gender and how do those roles for example increase or decrease the risk of exposure, for example to important zoonotic diseases. Also what are some of the factors that will limit women's participation or women's contribution in ensuring that their health is good, the health of their families and communities is good and also the health of the animals that support their life loads and that of the environment is well taken care of. So the other one is making sure that gender is captured during the project life cycle. These ideas of thinking about gender can only come clear if you include gender specialists in these projects and one important thing is ensuring that there is budget to have gender considered in the project.


In the last lecture, I worked out the energy levels for a particle, a chart particle, moving in the uniform magnetic field. And I summarized some of the calculations here on the board. The main result is the energy levels, which depend on two quantum numbers, M, which is a Landau quantum number, and then P3, which is the momentum parallel to the magnetic field. The particle loses a free particle in that direction. And the Landau level corresponds to a harmonic oscillator with an energy of 1.5 h part omega. Interestingly enough, the Q and the P of the harmonic oscillator are basically the X and Y components of the velocity, which don't commute with each other. So the harmonic oscillator itself is really just the X squared plus the Y squared times M over 2 as kinetic energy. This is the quantized values of the perpendicular kinetic energy of the particle. In fact, the whole energy is kinetic energy. Anyway, that was the main result of the last lecture. And in this lecture, what I'd like to do is to start by working on the, not just the magnetic levels, but also the energy hiding functions. So, let me start with that thing over here. And it worked with that. So the strategy is going to be to choose a complete set of commuting observables, a CSCO. And for starters, we can take the, for starters, I mean, notice that the perpendicular Hamiltonian and the parallel Hamiltonian commute with each other. I want to keep looking at this. I'll have to show you the summary for the last time. The reason they do is that, you know, there's three degrees of freedom. One, which is, basically, the guiding center position X and Y, the Q and the P. The two degree of freedom is the X and Y components of velocity. And the three degree of freedom is the Z and PZ in the parallel direction. So the reason that, the reason that H perpetuates parallel commute with each other is that H perpetually involves number two degrees of freedom whereas H parallel involves number three. They commute with each other because the commutation relations are Q, I, P, J is I, H bar, delta, I, J. So the commutators are zero for different degrees of freedom. So the Hamiltonian is the sum of two terms that commute with each other. And so we can take these two terms H perpetuate parallel with being two of our members in the commuting observables, let's write this set like this. Now, actually, as it turns out, it's still an H per, excuse me, it's still an H parallel. But it's actually better to use P3, the parallel component of momentum. This is really just one dimensional free particle. And the idea here is as it shows, momentum is preferable because an energy is degenerate. You know, free particle of what I mentioned, you've got two waves traveling in opposite directions have the same energy. That's resolved if you use the momentum as a fun number. So allow me to do that and replace H parallel just by P3 like this. That gives us two commuting observables. But we need another one, and it's pretty clear that we do because the number one degree of freedom is not represented by H parallel only per. That's the x and y that might be certain positions. So if we want to find the third observable which commutes with the first two, it obviously can be any function of Q1 and P1. For that matter, it could be called x and y is the same thing apart from these constants. So it's a guiding center position, some function of n. And the simplest choice is just to take it to be x. We can make it y. Let's make it x just to give it a choice. So let's take a complete set of commuting observables, and we'll take it to be x H parallel to P3. And our strategy of them will be to find a set of simultaneous eigenfunctions of those observables. All right. So what we're going to look for then is to solve a set of equations like this. We've got wave functions from these sides, x, y, and z. And we want, first of all, that x acting on this, the operator, brings out an eigenvalue we call x. And you can see that they're distinguished from the operator from the eigenvalues. So let me use my usual trick of putting hats on any operator where I think there's going to be a confusion between the operator and an eigenvalue or some other C number. So anyway, this is the eigenvalue eigenfunction equation for the operator x hat. Likewise, when that h-perf acting on the same side will bring out an e-perf perpendicular energy and it will have a P3 hat acting on the side will bring out a P3 eigenvalue acting on the side, so these are three simultaneous eigenvalue equations that have to be satisfied. Now, so let's work on these. The h-perf is the most complicated of all. And the P3 and x hat are simpler. So let's do the simple ones first. Let's work on P3 hat first. P3 hat is from, it's from, importantly down here, it's the mass times BZ. And BZ, the velocity, is expressed in terms of the canonical momentum and vector potential by this equation. So we'll put those together. P3 hat then is equal to the mass times BZ, I'll put a hat on BZ. And that's the same thing as momentum PZ plus E over C times the vector potential AC, which in general ends at x, y, z. And so this eigenvalue equation, the third eigenvalue equation looks like this now. It replaces PZ by the operator minus i h bar BZ. If we get an equation like this, minus i h bar BZ plus E over C times ACZ, which is the function of x, y, and z, acting on psi is equal to eigenvalue P3 times psi of x, y, and z. Now, what makes this equation hard to solve is the vector potential component ACZ. And so we ask ourselves, is it possible to choose a gauge? Let's try to choose a gauge now to simplify the equations. Can we choose a gauge such that ACZ is equal to zero and then be done? Well, let's write out the equations, the components of B equals del cross A. We take the z component, and we have B, which is the same as the z component. It's at dA and y with respect to x minus partial A to the respect of y, which is right up the full equations. Zero, which is Bx, is equal to partial AC with respect to y minus partial A to y with respect to z. And zero, which is By, which is the third component, is equal to partial A to x with respect to z minus partial AC with respect to x, just writing on the curl. Now, if AC is equal to zero, if we provisionally guess that we can do that, then these two terms here go away. And you can see that what's left is that dA, y with respect to z is equal to zero, and also dA, x with respect to z is equal to zero. So these equations imply that Ax is a function of only a maximum y now, and it also is the same as true for Ay is only a function of x and y. The AC is equal to zero. This means, as you can see, that the entire vector potential is now independent of the z. So partial of the whole vector potential with respect to z is equal to zero. This means that the vector potential is invariant in the translations in the z direction. This is one of the symmetries of the magnetic field. The magnetic field is uniform in the z direction like this. It obviously has a symmetry, a translational symmetry in the z direction. If you move it up and down, it doesn't change anything. You can also move it to the right and left or backwards and forward. And that doesn't change anything either. The magnetic field has a translational symmetry in all three directions. It also has a rotational symmetry about the z-axis. Only about the z-axis, not about the z-axis. Anyway, this magnetic field is uniform in the z direction, and I feel that's four obvious symmetries. So now one of them is reproducing the vector potential, making it the z-translational symmetry. So let's do that. Let's make this assumption, and if we do that, things will suffer because this term goes away. And also, if we do that in this term goes away, we'll see that P3, the operator, is the same as the Pz operator. So let's just change the notation under the same thing. So let's just take P3 hat and just replace it by Pz. That's one of this. And likewise, do the same for the eigenvalue. We'll replace that by Pz. So I'll take this, place this just by Pz. And so now the equation is easy to solve, and gives you the psi of x to y and z is equal to e to the i Pz times z over h bar times an arbitrary function of x and y. Because we're running the x and y derivatives here, this is the general solution for that equation. And so the z dependence is determined, and it turns into a, as you see, a plane wave in the z direction. A logical sense of motion in the z direction is that of a free particle. All right. Anyway, this takes care of the third eigenvalue, which is the P3 equation. Now, the next most easiest one to work with is the x equation. So let's do that. I'll try to start it here. I'm going to shift to my board. So the x, I'll remind you, the x, the guiding center x, is defined as the coordinate x minus dy over omega. And dy is given in terms of the canonical element of the second potential by the standard formula. So in terms of operators, this says that the guiding center operator x hat is equal to the z particle position x hat minus p y hat over omega. And that's the same thing as particle position x hat minus one over little omega times p y hat plus university a y, which we now know is a function only of x and y. And that's an expression for the x operator. I'm putting x at everything where I want to distinguish an operator from the number. And so the equation x hat psi is equal to x psi, the eigenvalue equation, turns into a differential equation, which I can write out like this. It becomes x minus one over little omega. And then we've got minus psi h bar dy, that's equal to p y here, plus university a of x of y. The whole thing multiplies our psi of x, y, z and gives us eigenvalue of x times psi of x, y, z. Now, however, psi is already determined as a phase factor in z direction times an unknown function of an x and y, of x and y variables. We're going to determine this pi function by using these other two eigenvalue conditions. And in particular, by plugging in this expression for the inside, at this point, even to the inside and two sides of this equation, you can see there aren't any z derivatives or anything else that depends on z on either side of this equation. Thus, the phase factor here, the i pz over h bar, pzz over h bar, just cancels. And so, let's look at this up to another board here. Oh, yeah. So it just goes to cancel and ends up being just an equation of y. That's right. I want to do one more thing. Now, having done that, the still is not exactly an easy-looking equation. The main reason is, again, because of this vector potential here. But in addition, there's also this x over here. So before proceeding, let's ask whether we can take this vector potential and use it to cancel out that x. This is there on the other side. What do you think that happens? Is that a supposed to be a x? That's supposed to be a x, yes. A x. No, excuse me. It's a y. Because this is the dy, which came from dy, yes. All right, so can we do that? Well, if we do, if those two terms cancel, then the x here is a minus sign and one over omega and a plus sign here. So what we have to get is a condition that says that x is equal to d over m omega c times ay. Or, obviously, ay is equal to mc over e times omega, which is eb over mc times x. And mc is canceled and eb is canceled. And so this leads to ay is equal to x times b. We do that. Now, if ay is equal to x times b, then this term in the curl equation is just equal to b. So b equals b, and therefore, ea x dy must be equal to zero. And so the logical choice of a x is to take a x equals to zero. And if we do this, of course, these are now, they are functions of x and y, they're really only a function of x. If we do this, we now have a specific to completely determine the age here. And there's a pretty good choice of age. And you can see that the effective potential now has only a y component, and the y component depends only on x. This means that now not only is the effective potential independent with c, but it's also independent of y. So it's translationally invariant also in the y direction, both the y and the z directions. It is not, however, translationally invariant in the x direction. And it better not be because if it were, it would have to be zero. I mean, we're at the constant. If it were constant, it wouldn't the magnetic field would be zero. And the magnetic field was not zero. So there's a moral in this, which is the effective potential can never have all the symmetries of the magnetic field. But anyway, as soon as you set it up, so it has the y and z translationally invariants, but not the x. All right. Anyway, this is not giving us a complete choice of age. And we can now solve this x equation. And so it's going to look like this. It's pretty easy to solve. So it's going to be, I just need to clean this stuff up here. The x in the area will cancel out. And the rest of it is like I get an IH bar over elevator ddy from here to here. And that's almost going to be pi. And that's a y. And that's equal to capital X times pi. That's a y. Capital X is behind the value of the finding certain position of it. And this is easy to solve. It says a pi of x and y is equal to e to the minus i m omega times capital X times y h bar times the arbitrary function of x. So as you see by taking care of the x hat equation we've now gotten the unknown eigenfunction down to just one variable x. And so this leaves us finally with just h per. So let's now write down the h per equation. So h per as an operator is one half m times px plus e over cax plus one over 2m px plus e over cax plus one over 2m py plus e over cay. Of course we've just decided that ax is zero so that goes out. And px here is going to turn into minus i h bar dpx. As far as e over cay y is concerned eay is up here. It's x times b. So this is eb over c times x. Which if I multiply and divide by the mass turns into m omega times x. That's the impartive position x. As far as py is concerned that's minus i h bar ddy. But this is going to act on the full function psi because the z variables don't appear anywhere in h parts. So the z face factor will just drop out. So what we wanted to let it act on is the function phi. The function phi is the face factor of y times f of x. Now the y derivatives do appear. They appear right there. They're going to act on this face factor and bring down a constant. As you can see y is going to be minus i h bar times c. This is the number that's going to turn into a constant factor which is going to be m omega times capital X. Just to multiplication by it. I think it's going to be minus m omega capital X. Right. So after that's been done then there's going to be a common face on the equation h per phi is equal to e per phi. So it's all there's a common face factor. This y dependent face factor is in both sides and we can cancel that. So what's left over when we clean all this up is this. It's minus h bar squared over 2 m times b squared over the x squared. So the equation just in the unknown function f is c. Then what we get is m omega squared over 2 times x minus capital X squared over x. And we get this e squared over x. Now this equation is the eigenvalue equation for a one-dimensional harmonic oscillator. You can see there's a short one-dimensional trigger equation. Here's the kinetic energy. Here's the potential energy. Frequency of omega. The only thing that's a little bit funny about it is the margin has been shifted from x equals 0 over the x equals capital X which is the eigenvalue of the Geithung center operator. So we can write down the energies in the sleuth and the eigenfunctions right away. The energy is e p perpendicular to the distance less than half h bar omega. That's something we do already. And the energy eigenfunctions f of x are going to be able to let me call the harmonic oscillator eigenfunctions just use it in. This is a useful harmony polynomial exponential. But it's evaluated in position x minus x. And so we can now put the whole solution together and we have psi, the wave function which is characterized by the quantum numbers capital X to land on a quantum number in the momentum in the z direction which is a function of x, y and z is equal to e to the i p z z over h bar to the z direction minus i m omega capital X times y over h bar to the y direction times u n of x minus capital X for the X variable. And that's the solution. So those are the wave functions. These are the simultaneous eigenfunctions of these three commuting operators which we chose to set. And because the following isn't a function of those operators in particular just the last two of them p dot energies the total energies are on the level plus the perpendicular energies. Yes? Our argument for why we needed a third of the determinants of the whole thing was that in z I just filled two, three we still had x and y the initial position but we used the pictures of x and now y so we still have another degree of freedom that's our argument to. Well this system has three degrees of freedom. You can see that in the original x, y and z which are just the coordinates of the particle but then we also transformed them to these q's. There was three q, p pairs. It's the same number of degrees of freedom. But in the Hamiltonian there's only two degrees of freedom that appear in number two and number three degree of freedom. The guidance in a position is x and y don't appear because the energy doesn't depend on those parameters. It depends where the plane is from the circle. So in order to find an eigenfunction we need to have a third, we need a third operator between two first two. And it has to be a function of the guidance in the coordinates because the other two would take care of degrees of freedom in number two and three of the guidance in the variables in number one. Does that make sense? Yes. However there was a lot of arbitrariness in the choice of this operator. It could have been any function of the number one variable. It could have been y because it was a function of x and y. It still would have been u to the remaining two. This is a way of saying there's lots of ways of writing down energy eigenfunctions that have the same energy eigenvalue. Energy eigenvalue depends only on the last two fundamental numbers to get into the d z. So this is a way of saying this is a highly degenerate system. Given energy is as many different rates of writing down about energy eigenfunctions. The reason it's so degenerate is because it's highly symmetric. If you include physical effects to break the symmetry, the realistic thing they feel is don't go on forever in x and y. They have to end somewhere in a solid. There's going to be an end of the solid or there's also going to be potentials or electric potentials or something. That will break the symmetry and then the Hamiltonian will defend an x and y. And that will change things. But anyway this is the simplified model. Okay if we take a thought, let's take this way function and sketch it to see what it looks like and try to get some ideas what it means. If I notice the z and y dependence from space factors if you square them and they drop out then all it's left is the only possibility of writing function. If we plot this in the x, y plane and kind of ignore the z direction we can put in a dotted line here for where a particle position is equal to a writing function. Let's also take n equals 0 which is the ground state of the Landau motion then u of n of x and u of n is Gaussian but it's going to be centered on it at x equals x. So you've got a Gaussian that looks like this. It's set to extend out in the y direction. It continues around the y. If I put the z direction it would be a kind of a Gaussian wall that would be out here just in the x, y plane. It's like a Gaussian ground in range which is centered at x equals x. This doesn't look at all like classical motion, classical motion in a circle in the x, y plane situated somewhere like this. What's the relationship between this energy eigenfunction in this class? We might have expected the quantum wave function to be concentrated in a circle. Something of that sort. It doesn't look like that. Why is that? Because we made this an eigenfunction of the operator x hat, the 90 centered position. But the 90 centered y position does not commute with x. In fact, they're like a Q and a B apart from the scaling factor or an x and a B. And so by constructing a state in which the value of x hat from the observable x is known precisely, it means that the value of the course plane y of the observable is completely unknown. And that's why it has to stretch out to the plus and minus infinity in the y direction. It's the uncertainty principle that's forcing this on us. If we had chosen y, not the y, 90 centered y as the set of x as the first observable, then we would have gotten the amount of range going in the other direction. Because then y would have been known that x would be completely unknown. There's a homework problem in which you don't lose either y or x, but you use angular momentum. And if you do that, you find the y function is rotationally varying around the origin. It still doesn't look like the classical motion because the circle in the classical motion doesn't have to be in the origin. It can be anywhere. But anyway, at least it doesn't go to infinity. So anyway, these are all aspects of this problem. There are ways of constructing energy eigenfunctions that do look like the classical motion, but I don't think I want to go into it. Here's another curious thing about this problem. It's that the x-y plane, if I talk about the particle coordinates x and y, of course it's just an ordinary configuration space, and the x-y variables can be used so they can be measured in simultaneous state. If, however, we talk about the guiding center x-y plane, those are operators that don't commute. And in fact, the commentator is i h part times the constant, so they look like an x and a p. In other words, the guiding center x-y plane is really more like a phase space, and it is like a configuration space. And by the way, the same thing is true for the vx and vy variables, which don't commute either. They're like another phase space in a different dimension. In fact, the classical orbit is a circle in velocity space. This is actually like the circular orbit of a harmonic oscillator in the phase space in the classical picture. Well, in any case, to go back to the guiding center variables, x and y, it's like in some respects, it's like a phase space. There's one thing we know about a phase space, which is that there's a concept of a quant cell. Quant cell is a cell that has an area, which is a 2 pi h bar, and it's the area that supports a single quantum state. This is a roughly classical rule. Well, the the x and y variables aren't quite q's and p's. There's one of them, one of the real omega involved in them. So a quant cell in the guiding center plane is going to be a cell in which delta q1 has delta q2. That's the phase space area as equal to 2 pi h bar. So let's talk about quant cells. This is what this would be the condition for it. However, from this formula here, delta q and delta p are square root of omega times delta x delta y. So this is the same thing as m omega times delta x delta y. That's the same thing as m omega is equal to bb over c delta x delta y. And b times delta x delta y is the magnetic flux through the area of the x and y plane. And so what we find is that the flux through a single quant cell v delta x delta y is equal to 2 pi h bar over here times c over e. Which is more usually written this way. It's hc over e. hc over e is the unit of magnetic flux. It's called the flux quantum. And this is actually an important concept in various condensed matter applications. Fun home-effects are where you have not just one particle in magnetic fields. And the quantum states the ground state of the land model fills up the area with one state per quant cell. But the quant cell is an area of the x y plane that captures a single flux of magnetic flux. Which is hc over e. It's a very common concept. The terminology is a little bit bad because the magnetic flux is not quantized. This is just merely a natural unit of magnetic flux that emerges in problems involving magnetic fields. But magnetic flux can take on any value whatsoever. That's the meaning of the flux quantum hc over e. Now that's all I want to say about the charged particle in a uniform magnetic field. I'd like to turn to another problem involving magnetic fields which I guess I'll do here. And this is the hard enough home-effect which is interesting because it challenges our understanding of the global negative-effect potentials in quantum mechanics. The hard enough home-effect may be a fact based on about 1961 so I'm sort of surprising nobody thought of this earlier but there's in a sense there's nothing much to it it's just solving it. Surely nobody invents a new Schrodinger equation or any new physics. It's just solving a Schrodinger equation in some standard way but the results were surprising to some people anyway. Anyway, the situation the situation at all is that of a charged particle moving in the field of a long solenoid a long solenoid a long solenoid is a standard problem in electromagnetic theory so I assume you've seen it already. Let's take the solenoid looking like this it actually is supposed to and ideally it extends to infinity in both directions let's take the z direction just pointing up and put coordinates x and y in this I'll also use I'll also use cylindrical coordinates when it's coming in this is rho, phi, and z where rho is the radius of the x-y plane square root of x squared plus a y squared but in any case the solenoid is supposed to have a current sheet a singular current sheet applied in the azimuthal direction around the solenoid and the result is if you get a magnetic field a constant uniform magnetic field with zero magnetic field outside so the magnetic field is equal to let's just say magnitude b times z hat if the radius is less than the radius of the solenoid let's call that a as the radius of the solenoid in fact while we're making another picture we look at this from above we look at the solenoid from above so the x-y plane is going to be hard to know let's say the radius is a like this so if you're inside the magnetic field is constant if you're outside the magnetic field is zero now in classical mechanics a charged particle moving in the exterior region of the solenoid would not feel the solenoid at all because there's no magnetic field there so there's no force in the particle what's interesting is that there is an effect in the quantum mechanics as we'll see in just a moment to look at this quantum mechanics of course we're running any vector potential because that's what appears in the Schrodinger equation so let's find the vector potential for this magnetic field we'll do this in this manner it turns out that you can find the vector potential which is purely in the 5 direction and we get the vector potential by doing an integral around a closed loop with some radius around the solenoid so let's integrate around some loop a.dl that's equal to the radius the circumference of the loop which is 2 pi a rho times 8 pi however by so stair this is also equal to the magnetic flux it's the integral of b.da over the certainly closed surface and that's the same as the magnetic flux which is enclosed by the loop the magnetic flux which is enclosed by the loop depends on whether the loop is inside the solenoid when it's outside these dotted lines are two examples of loops that we're thinking of here like this if the loop is inside the solenoid then it picks up the flux which is inside the loop so in that case what we get is the area of the loop which is pi over the square times the magnetic field and if the loop is outside the solenoid then we pick up the magnetic flux from the entire solenoid with radius a but nothing beyond that because there's no magnetic field beyond that so in this case it just turns into pi a squared times b for a rho greater than a so dividing by 2 pi rho it's solved that after the 10th hole we find a5 is equal to b over 2 times rho if rho is less than a then b over 2 times a squared over rho if rho is greater than a so we can make a plot it's a function of rho a sub rho we draw in here the radius rho equals a out to in the interior solution the a5 is proportional to rho so it's a straight line like this and then the exterior solution that falls off is 1 over rho so it decays like this if we plot similarly the magnetic field on the same on the same chart same diagram it's constant interior and zero outside so although the magnetic field is zero outside the solenoid the vector potential is non-zero and the vector potential appears in the Schrodinger equation so the question is is that going to affect the quantum mechanics of particles which are outside the solenoid to make this a little more kind of we can imagine that the solenoid is like a hard wall that the particles can't get through so we've got a quantum mechanical problem if I look at it from above we've got a quantum mechanical problem in which the wave function is equal to zero it's a boundary condition in the edge of the solenoid and we're going to be interested in solving the Schrodinger equation in the exterior region we don't care what happens in the interior because the particles don't get in there okay so let's think about doing that well from one point of view you might be tempted to say um in this exterior region we have a non-zero vector potential it's true this is what I just worked out here but this is just one choice of gauge obviously if you have a zero magnetic field the possible choice of gauge is just eight and zero because the current zero is zero so when we do a gauge transformation to get rid of this vector potential end up with eight and zero and then in the exterior region we ought to have just the Schrodinger equation for a free particle and the magnetic field wouldn't make any difference well let's try to do that so the vector potential we've got is this one here this is my component this is B A squared over 2 times rho times pi hat both of my neurons are generated by pi because then pi A squared B is the magnetic flux inside the solenoid so it's pi over 2 pi times rho times pi hat that's what's supposed to be one over rho one over rho times pi hat as we say this vector potential is pure gauge you can get high gauge transformation that can be eliminated that means it must be equal to the gradient of some scalar F what is that scalar F well the scalar F is proportional to the S of the angle phi don't confuse the magnetic flux with the S of the angle phi if you compute the gradient of phi then the coordinates will find it's equal to pi hat divided by rho which is precisely what we have here and so the F here which is the of which A is the gradient to summarize this we can write the vector potential in the exterior region is actually equal to the gradient of the scalar which is the flux of the solenoid divided by 2 pi times the S of the angle phi you'll recall that when you do a gauge transformation on a wave function it goes into psi times e to the i q over h bar c times the gauge scalar F the gauge scalar F is going to be flux over 2 pi times the S of the angle phi and so the result that's not right now so the result is in this attempt to gauge away the vector potential in the exterior region what we end up with is a new wave function which is related to the old wave function by a phase factor that goes like the azimuthal angle specifically it's this the new wave function of the transformation is this e to the i q over h bar c times the function F is where did I write it down the function F is the flux phi over 2 pi times times angle phi h bar c times angle phi flux capital phi times angle phi q over 2 pi h bar c is the same thing as q over h c which is one over what I'll call pi zero where pi zero is the same flux one that we saw in the charge particle in the uniform field this can be written this way psi times e to the i capital phi over phi naught times the azimuthal angle phi like this trouble is you see is that when you go completely around the phi over phi naught is not an injury but this is how we do it in phi yes let me hear that gauge transformation this is the vector bullet inside well I'm glossing over that because I'm saying suppose we just want to solve the Schrodinger equation in the exterior region so I'm supposing let's just try to do a gauge transformation here to get rid of it if you do that what you find is this is the new wave function it's related to the whole wave function times the factor which is the flux in the solenoid divided by the flux prime times the azimuthal angle the new wave function is is discontinuous that's the problem it's discontinuous in the solenoid of its infinite you've got problems with that so let me not go into the problems of the discontinuities of the wave function but this is a reflection of the fact that it's not going to work to gauge away the vector potential in the exterior region even though the magnetic field is zero out there another way of saying it is that if I were able to do a gauge transformation if I were able to do the vector potential in the exterior region and obviously the integral the volume integral of a dot dL would be equal to zero because if a was equal to zero this would certainly be equal to zero but that can't be right because this integral must be equal to the flux which is contained in the solenoid so the result is is you cannot gauge away the vector potential in the exterior region whether it's a hole in its face that's really what makes the difference about being able to do this let me not follow this any further except just to say that the result of this is that there's a phase shift in effect on going around the solenoid compared to the free particle solutions and the phase shift is precisely this that are reflected by two pi going around the solenoid the phase shift is this is the ratio of the flux to the flux one which is in the presence of that vector potential compared to solutions, free particle solutions when there's no vector potential differ by this phase shift going around the solenoid now let me show you how this might have an effect physically it does have an effect physically this is maybe the most dramatic physical effect way of describing the effect of this let's consider a double splitting experiment where we have, let's say, holdways coming here like this two holes and you know what will happen is you get waves radiating from the holes and then you've got a screen down stream from this and as you know you'll get an interference pattern here and the crest are the places where the phase difference coming from the two holes is an an integral of two pi so that the two waves are in phase with one another now let's modify the experiment by putting an arm of bone solenoid right there behind the screen it's part that has never seen the solenoid it's never come closer it's just passing through the holes but because of the presence of the solenoid the solutions in the presence of the solenoid have this extra phase factor a relative phase going around the two sides of the solenoid compared to the three particle solutions and what that means is the relative phase of these two waves is precisely this amount as we say by the flux part of pi naught times 2 pi that's the phase shift the extra phase shift is induced between these two waves and so what happens is the interference pattern moves to the right and the left depending on the flux in the solenoid and if you make this time dependent by changing the current in this we can move this interference pattern back and forth now the point of this is it's not physically observable so this is a physical effect which appears in quantum it's a purely quantum effect it's nothing to know in this classically which occurs in a situation in which charged particles never enter a region where there's a non-zero magnetic field and so this is what Harnoff and Fong pointed out in 1961 and some people said oh it's no big deal it's just a solution to the Schrodinger equation and other people said oh this is the greatest thing so anyway I think there's still kind of a skincey attitude about the Harnoff-Fong effect and it certainly is interesting and this is what it predicts now by the way the Harnoff-Fong effect has been actually experimentally verified as usual in all kind of experiments the real experiments are have to be practical reasons that it's so much different but the Harnoff-Fong effect has been verified experimentally so there's no question that this sort of thing actually really does happen I'm sorry yes would you finish the Harnoff-Fong solution thing you said so this is wrong and now you're saying that it's right no what I said was you see if it were possible to gauge the wave of active potential in the exterior region the Schrodinger equation in the exterior region would be that in a free particle not at all in which case the interference pattern would be just what you always get for what you've always learned but the point is if I put the solenoid in there and turn it on so I've got a magnetic flux then there's an extra phase shift between these two waves so then computing where the peaks occur you have to include an extra shift now it's a relative shift between the two waves it's a relative phase shift and then we might like to think that there's spherical waves coming out of here and these spherical waves of course are the lines of constant phase but who defines the phase it's easy enough that there's no vector potential around you just say for A equals 0 these are the spherical waves where the peaks need to be IKR coming out of here the IKR or R but in the presence of the vector potential then you have to specify a gauge and the phase of the wave function depends on the gauge it changes if you do a gauge transformation so what the wave fronts are depends on your phase convention as weird as it sounds that's true but the phases themselves are not measurable what's measurable are phase differences and the phase difference between these two routes even in the presence of the solenoid depends only on the flux in the solenoid and the flux in the solenoid depends on B not on A so it's a gauge and variant quantity so what's physically measurable is gauge and variant so although the vector potential plays a part of no role in quantum physics and it does in classical physics it still remains true that the gauge convention is still just a convention and that the real physics doesn't depend on that convention the real physics there is a phase shift a relative phase shift between waves only on the flux the flux in the solenoid well so at a time I didn't I didn't really have time to go into what else I wanted to talk about which is how this appears in a pathological formulation let me just say that in a pathological formulation you know you have a phase in the path integral it's either the I over H bar it's a integral over time with a Lagrange let's say T0 and T1 and this is integrated over the path space in the X and T for the real impact integral this is a propagator K and the Lagrangian has two terms it has a kinetic energy which is M times M times X dot 1 over 2 times X dot squared for the path and then it has another term which is Q over C times X dot dot of the vector potential we can call this let's say called 0 which is the free particle contribution of the Lagrangian let's call this LLM which is the magnetic contribution of the Lagrangian so in comparison with the path integral for a free particle of course, let the solenoid in the center and a hard wall on the particles in a sense is only free in the exterior region but we can accommodate that as a path integral just by taking paths that have never interacted with the region so let's look at that here in the path space for integral so then all we've got here is just a free particle of Lagrangian plus a magnetic part of the Lagrangian but the magnetic part of the Lagrangian turns out is is independent of the path up to a it's largely independent of the path I'm sure you want to know about that let's say that here's our hard-on-cold solenoid let's say here's a position x0 here's a position x1 and we've got a path that connects the two of them let's define the magnetic action as being the integral along the path of the magnetic contribution of the Lagrangian let's call this aM it's a magnetic it's a function of the path and it's a t this is the this is q over c times the integral of t0 to t1 of a dotted into the velocity x dot dt well, this turns into a y integral it's q over c the integral from x0 to x1 simply a dot dx and so you see the time where amortization drops out and the magnetic contribution of the action does not depend on how fast a particle is passing actually, it depends on even less than that because if I have two paths let's call them p1 and p2 such that they can be continuously deforming to one another then the magnetic action along these two paths is the same because if I go along one path and back along the other that gives me a loop integral which is equal to the magnetic plus through the enclosed region but this is entirely an exterior region and the magnetic field is 0 so if you stay in the exterior region and you continuously deform one path into another you don't change the magnetic action and so the magnetic action for paths of that sort can be taken out of the path then well, not quite because there are paths for which the magnetic action is not the same it's called as p3 these are paths that relative to the initial path go around the solenoid in fact, if we deform this path continuously we can turn it into this partially follow p1 and make an excursion over here, go around, come back and then follow p1 again so we go backwards and forwards like this if we do this, then the magnetic action along p3 is equal to the magnetic action along p1 plus q over c times the flux in the solenoid and so the result is in the path integral you can break paths up, you can take the homotopy classes these are the classes that indicate how many times it goes around the solenoid and all paths that belong to the same homotopy class give the same magnetic action and that can actually be taken out of the path integral anyway, the result is the path integral becomes a free particle path integral taken over or taken over paths belonging to the single homotopy class multiplied by a phase factor which is precise on this hard-on-bone phase exactly this here times an integer so it becomes e, i, n, 5, or 5, 0 and this is the phase factor that distinguishes one homotopy class from the other anyway, our doing this becomes obvious that the propagators in the presence of the solenoid is not the same as it was in the absence of the solenoid okay, I think that's all so just to remind you that the class this week can all be on code there's a few more homotopes here remember that we're going to say


Where in the US do you want to be during a nuclear apocalypse? While you obviously want to avoid these cities, go ahead and add these places to the list too. Why? They're connected to our own nuclear infrastructure. Also, don't forget you don't want to live near a nuclear power plant. Also also, I hope you still remember your state capitals, because they're likely targets too according to FEMA. With one exception, Bismarck, North Dakota, you go Bismarck. So, where do you want to be? Eastern Oregon and Western Kansas are amongst the safer bets, and maybe even Western Texas if you're in a pinch.


way more so than I thought. Are we here? Oh yeah. What's up everybody and welcome back here to the channel. We are here in Japan for a little Asetakorsa Matsuri. So the Drift Matsuri did end over in Japan. So we are here today to have a little fun. We got a pretty big server, 18 cars here and we're going to go send it here and see if we can't get some tandems. I gotta make sure nobody's coming in. Oh there's a line of cars coming down the stretch. So we're trying to throw ourselves in with all these first time out here on the track. We are in an S14 and that uh, that car's going everywhere. So I haven't driven this car yet, so I'm very not comfortable in it. So I'm gonna take a lap or two to get used to it and we got a slew of cars on us. So try and grab as many doors as we can here for today. Apparently we're drifting this. Oh and I'm already getting bumped just like in the actual Matsuri. So we're gonna try and go up here and send it in here. Oh too much. We're in the dirt. Oh we didn't crash anybody though. We're good. It's kind of a crazy session with all these cars here. We lost it. The S14 probably not my favorite car to drive. Would have preferred to be in a JZX100 but they were all taken. We got a line going on. So just basically a random thrasher lobby here.  Look at Rex. Somebody else is wrecking and everybody's off into the dirt. We gotta get a good run here. Next one's gonna be perfect. Right there on the door though. Try not to hit him. Now I know some people did say that the Tando buddies cars are very easy to drift and it's the only server out right now. A lot of people in it doing this track with you know a lot of people. So that's why we're in them. There we go. This is better. Oh geez don't say that. Oh we dead. Oh man. Oh we are very dead. Rip. Let's go back to pits. Drop our tires a little bit more and then we're gonna go back out again.  Let's cut into that chaser right there. See if we can't run a little run by ourselves and see if we can't get a handle on this S14. Sounds good. I like how it sounds. Oh my that car came out of nowhere. See this guy wants to go. Oh okay. That yellow car is not very friendly right now so see if we can't throw a good one ourselves. There we go. So we can do it. I love the backfires on the pop side of this car.  There we go. So we did get a single solid run. So we're just gonna try and hook up with some more people. Try and throw some tandems. We're gonna wait for some people to show up and then we're gonna set it sideways and see what we can do with some other cars. We'll get out of this man's way. Come up here and wait for a second. I know there should be a lot more people coming so we're gonna wait and see if we can't hook up with a couple more people. Just messy it a little bit. There we go. A little bit closer. I see smoke. We're gonna throw ourselves in the mix. I mean all right he can pass me. I'm gonna straight line that a little bit. I think he went out. He went off. I was trying to follow him and he went off grass. There we go.  Try and get some tandems without wrecking the car. It's not gonna work when we go in the grass like that though. All right now we here we go. All right let's try not to screw this one up. We got bumped but we good. We got bumped but we're still sliding. We didn't lose it. Look her pop. That man's pulling wheelies up there. This is just as crazy as the actual amount series look. Oh sorry my dude would love to get my chance to go and drift these roads myself. But right now we're just doing it here on a set of Corsa. Way more so than I thought. Are we here? Oh yeah. I think he lost it. He did. That was good until he lost it. So much fun just to get in a random thrasher lobby and have some fun and just just send it. Now we got a line of cars kind of getting pulled on right now. Trying to be smooth. That man lost it. Oh I don't think he knew I was there. And then he decided to spin me out. Well he's put out first but you know you can't be mad. Online session. Definitely very fun. Very crazy. Oh huge understeer. All right so we're gonna try and get our own good entry. The clutch kick. Probably too soon. Yeah way too soon. So we're gonna straight line approach this. Well that man's doing hydro stuff in the dirt. Oh man that could have been bad. Oh we got cars wrecking everywhere. We got massive carnage on this lap. Bumped from behind by the skyline. Just bumped me into that guy. Some people don't really care. Oh sorry dude.  Man that was a bad move. But people on this server are kind of a ruthless it looks like. And they're just trying to throw doors without you know any regard if they hit you or whatnot. Oh I'm sorry dude. Can't tell if that was intentional. We're just bashing today. Luckily this is not a like real cars because then this could be very bad. Another 180 following us. I don't know what kind of line that was. That was super unorthodox. I think we're gonna end it there. So if you guys like to see you want to see more of it subscribe do appreciate all the support and as always I'll be a rabbit. I'll see you guys on the track.


What's going on everyone? Sam here, late in sports cards. We are ripping 2022 Panina Origins football hobby 8 box half case break number 26 Pick your team style. I'm gonna read off our list of our names and our teams Brian H Jets, Evan C Patriots, Gregory M Jags Giants, Jason L Steelers, Jeff N Panthers, Bangles, Texans, Jeffrey M Cardinals, Jeremy H Chiefs, Jeremy T with the Ravens Raiders, Joe M Falcons Keith B with the Bears, Louise S Broncos Titans, Manny V Saints, Michael S Cowboys, Nanod M Browns Rams Dolphins, Nicholas S football team, Nick H with the Packers, Roger W Colts, Russell P Seahawks, Sam L Chargers Niners, Scott V Vikings Eagles, Timothy M Bucks, Todd C, Bills, and Will W with the Lions. Good luck everybody. Let's do it. Brad does a lot of the behind the scenes stuff right now, Bryce. Yeah, I ripped up good amount of overtime early. It's pretty solid actually, Grady. I mean, there's supers in there, reds, golds, all that kind of stuff, two guaranteed autos. A couple guys are projected lottery picks for this upcoming draft, the Thompson Twins. So it's their first like on card autos, which is pretty cool. Matt Jones to 299 and Matt Ryan to 99. I appreciate that laser beam, Stefan Diggs, rookie George Pickens, Wandaille Robinson, 24 out of 75. Okay, this is kind of wild. Okay, we've got Wow, one of 10. Kenny Pickett, Desmond Ritter, Malik Willis, boom. Steelers, Falcons, Titans, geez, dude, Falcons is Joe. Steelers is, that's kind of wild, dude. This is the first one of these I've ever seen. Steelers, Jason L, and then Titans is Luis S, all on card, all on card. That's kind of sick, dude. One out of 10. So yeah, this is probably, this is too big to random right now. So we'll get in contact with the three owners of this and see what they want to do. That is nasty right there. That is so sick, dude. We'll get that right into a mag for you guys. That's, I guess I have not seen that yet. And then Alec Pierce, 67 out of 99, going to the Colts. Nice one there. You'll take it, Joe. Yeah, I'll just give it to Joe. Yeah, the fact it's on card is kind of wild, dude. Yeah, we'll get in contact with the three owners, Joe, Luis, and Jason, and see what they want to do, whether it's random, random it, if someone wants to buy the other owners out, or you guys want to sell it together. So a couple of options there. There's Aaron Rodgers to 75, Chris Carson to 99, Keenan Allen, and Mitchell Chabrisky. Our first patch is going to be nice. Kenneth Walker, 36 out of 99. Three color patch going to the, that is Seahawks. We've got Aiden Hutchinson, six out of 49. Nice one there. Yeah, I got you right after this card. Sean Peerstrong Jr. Patriots. Rookie Jersey Auto. We've got a Wisconsin Timber Rattlers that I minorly went with the Diamondbacks Jersey. So I decided to throw that one on. I'm going to take a picture of it really quick, dude. Yeah. That's what you thought it was. There you go, Sean. Yeah, I love this hat, dude. Got it for like crazy sale in Cooperstown. To 199, Cooper Cup, and Daven Cook to 299. Daniel Jones, and Joe Burrow. Oh yeah, Stefan, there's personal trip tonight. Yeah, after group breaks. Got nice. Dak Prescott, 16 out of 49. Hometown Roots, Player Warn. Yeah, I meant to answer that. I'm sorry, Stefan. We've got George Pickens on card Rookie Auto. Going to the Steelers, and Tyquan Thor and Rookie Jersey Auto for the Patriots. Nice one there. Heather Lockett to 199 for the Seahawks. Zeke Elliott to 299 Cowboys. Aaron Rogers and Derek Stingley. Alright, some good stuff, man. Caleb was telling me about what we've got. Sky Moore, 75 out of 99 for the Chiefs on the Jumbo Patch. Damien Pierce on card Rookie Auto for the Texans. And EZE for the Dolphins. Eric Azucanma. Nice one there. Next one up. This is Google. Justin Jefferson to 299. And Justin Jefferson to 299. Justin Jefferson. Go for that rainbow. Josh Jacobs. Travis E.T.N. on your base. Wandaille Robinson, 1 out of 49. There's a good chance, Bryce. There's definitely a good chance it can replace the book. Nice Christian Watson, 45 out of 99 for the Packers. And Xamir White, Raiders, Rookie Jersey Auto. I always pick your team, Michael, so it's a little bit different per team. So anywhere from like 200 to probably like 20 would be my guess, a 20 up to 200. Derek Carr to 99. Jersey number four out of 99. Nice. And Saquon Barkley to 299. We've got a Barkley base. And John Mechie. Max Crosby, nice one there, 35 out of 49. And Trayvon Walker, Rookie Jersey Auto going to the Jags. Nice one there, Jacksonville. And then we've got a Bailey Zappy, 42 out of 49, four colored RPA. Congrats, Patriots. Nice one there for the Pats. That's kind of sick. Two to go. And Zappy. Sean's boy. Sean's favorite NFL player right there. Mitch Trubisky, 10 out of 75. Jersey numbered. Cam Akers to 199. Rams. Tony O'Gibson, Michael Pittman Jr. Deontay Johnson, hometown roots, 20 out of 99. Jersey for the Steelers. That is Ruskin, Florida. Sam Howell, on-card rookie auto not numbered for the Commanders. And Belish Jones Jr. Rookie Jersey Auto going to the Bears. Dub Bears. Two to go. Or one to go, I should say. Last box. So yeah. No, I think there's a book in here. Yeah, the triple did not replace the book. That's good to know, actually. Heck yeah. Ryan Tannehelton, 199. And Remandre Stevenson to 299. Yeah, I'll tell you right now, Joe. I think it's pretty stacked right now. Yeah, there's a lot. There's a pretty good cue. Pretty good cue for sure. Jerome Ford Rookie. That's Greg Dualchich. Rookie Auto, 30 out of 99 for the Broncos. Carson Strong, Rookie Jersey Auto going to the Eagles. Got John Mechie, 68 out of 99 for the Texans. And then we've got going to the Cowboys. Five or six out of 25. That is Jalen Tolbert. Nasty three colored patch on that. Congrats, Cowboys. Nice last box mojo for America's team. We've got, who's got Dallas? Let's get in that card. Dallas is Michael S. Congrats, Michael. That'll do it for the break, guys. We'll start off with this right here. Jalen Tolbert. I'm going to let Sean go ahead and package that one up as well. And we'll get into our recap. John Mechie Texans. Deontay Johnson Steelers. Max Crosby Raiders. Wondale Robinson Giants. Skymore Chiefs. Dak Prescott Cowboys. Kenneth Walker Seahawks. Wondale Robinson Giants. Greg Dualchich for the Broncos now on to Autos. Carson Strong, Eagles. Bielish Jones Bears. Sam Howe, Commanders. Bailey Zappie for the Patriots. Got Trayvon Walker-Jaxx. Christian Watson Packers. Zameer White for the Raiders. Eric Isaacman Dolphins. Damian Pierce Texans. Tyquan Thor and Patriots. Alec Pierce for the Colts. George Pickens Steelers. Aiden Hutchinson-Lions. Pierce Strong Jr. Patriots. And then the hit of the break. Triple, Kenny Pickett, Desmond Renner, Malik Willis, 1 out of 10. That'll do it for the break, guys. Thanks again for the fill. Next up, we've got Origins Tier.


Hi everyone, Joe for Jasper's casebreaks.com coming at you with another flawless baseball random number block randomizer Featuring those teams right there Mets Phillies Pirates Cubs twins Giants A's Diamondbacks and Rockies That's how it works. If you're not familiar with it big. Thanks to this group. This group's familiar with it Thank you. There are the numbers right there. Let's roll it randomize it names and numbers one and three four times one two three and four For four we got Jeremy down to David one of three four times for the numbers one two three and four nine down to five Excuse me losing my voice It's the end of the week Breaking a lot talking a lot All right Jeremy with nine Stephen with four David with seven Peter with eight David with three and zero Michael with six Shay with two Jeremy with one and David with five Let's order these numerically and finally in the next video will be the flawless break itself We'll see you for that Jasper's casebreaks.com


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Hi everyone, Joe for JaspiesCaseBreaks.com. We did it. We knocked out tonight's Tuesday nights all spots sold promo All we had to do was knock out these four things to give away that much break credit So we did the mosaic basketball pick your team seven plus any fillers Which was the break itself Then the the random team and then the filler for the random team We did the jaspies four box high end baseball mixer. We didn't have to do any fillers for that So that was just straight up We did nt collegiate football plus the two number blocks and only the flawless number Collegiate number blocks counted so there are two of those so you'll see it on the list in just a second And then we did two fillers and then the break itself for the three box first off the line football mixer Excellent work ladies and gentlemen. I think from from the beginning of the day I said listen if we want to get this done We're going to start hammering these breaks early and often to get us to Money'sville right and we did it so I appreciate that. Thank you for for digging deep and making this happen Hopefully you got some nice hits. If not fingers crossed on some money If you don't get the money and didn't get nice hits. I appreciate you trying. Thank you. We'll get them next time Here's the big list right there, which I've been keeping track of on the break schedule page on the 12 7 promo tab So there is the flawless collegiate There was the mosaic basketball hobby pack There's the other flawless collegiate There's the number block for the nt collegiate the other number block for nt collegiate The nt collegiate break itself the people who bought spots straight up There's the nfl solo pack 162. There's the other mba Mosaic hobby pack number 11 Which sold out the random team for picker team seven and then that unlocked picker team seven So everyone's included in there.  Everybody's in there, too Then there was the four box high-end baseball mixer as a divisional no fillers So that's a total of 258 names. That's not bad For all for the amount of break credit that we're giving away So 258 names New dice new list And name on top gets $2,000 in break credit to jaspyscasebrakes.com. The next five will get a hundred dollars of break credit each After 11 times good luck one two three four five six seven eight nine 10 And 11 then final time after 11 I need like a drum roll feature here. You can drum roll on your desk or something like that Drum roll drum roll drum roll drum roll drum roll Thanks everybody here on this list for giving this a go. I appreciate it Hopefully we've got you some nice hits. If not, listen, we'll get them next time. Like I said Thank you. Thank you. Thank you. I appreciate it. Appreciate you all Thanks for digging deep. I was happy to give this money away All right, so we're getting close to the top So right at spot seven just on the outside kevin. Sorry but thank you For getting in next five Up to spot two from six to two hundred dollars of break credit going your way So a little bit of money going back to david will tommy david and mark And that's what the extra spot that you want to mark nice and shon mattock is on top $2,000 of break credit. You're on the top and the bottom $2,000 of break credit to jaspy's case breaks dot com going your way So congrats shon $2,000 of break credit mark david tommy will and david 100 dollars each going your way in fact david i'm going to send you $200 all in one Perfect time to use this break credit for tomorrow's brand new releases on wednesday december 8th 2021 We've got spectra football and panini one and one basketball Feel free to use that there i'm joe for jaspy's case breaks dot com. Thanks for watching Thanks for breaking with us and i'll see you next time for the next giveaway. Bye. Bye


So we've made connections between enthalpy and heat, in particular enthalpy is the same thing as heat when we do something at constant pressure. And we motivated that conversion of heat into enthalpy in part by saying that it's more convenient to use enthalpy as a state function rather than heat which is a path function. So now we begin to get the real benefit of using enthalpy as a state function because it will allow us to calculate the enthalpy change for processes that we can't actually physically perform or at least not perform easily. So let me show you what I mean by that with a few examples. As the first relatively simple example, let's suppose the chemical process we're interested in is the decomposition of iron III oxide, Fe2O3, decomposing that into pure solid iron and oxygen gas. So if I balance that reaction, two irons and three halves of an oxygen II molecule balance this Fe2O3. So let's call that reaction A. So reaction A that I'm interested in, maybe I want to know what is the delta H for that reaction. That's not the kind of thing that we can necessarily terribly easily measure in a calorimeter because it's a little difficult to force that reaction to happen. Iron II, iron III oxide, that's essentially rust. Spontaneously iron and oxygen will combine and form Fe2O3 but it's a little more difficult to make it go the other way around. But what that suggests is we can, if I just take and call this reaction B, if I flip this reaction around and say iron solid and oxygen gas forming Fe2O3, the reverse of the reaction I was originally interested in, I can make that reaction happen without much difficulty. If I take some iron, it'll rust very slowly in the presence of oxygen but if I shave the iron down into iron filing, so it's got a lot more surface area, perhaps give it a little spark to initiate this combustion process, I can cause iron II pretty quickly combust and form Fe2O3. So that's the type of thing I could do in a calorimeter. So I can measure delta H for that reaction relatively easily but since we know that enthalpy is a state function, the difference in enthalpy between these products in this reaction is exactly the negative of the enthalpy change between these products and these reactants. So what that means is delta H for reaction A is the negative of delta H for reaction B. So like I said, that's a relatively simple example. Perhaps we can't do the reaction we're interested in but we can do the reverse reaction, just throw a negative sign in front of the enthalpy and that makes us able to use a reaction that we can measure the properties of more easily. As a slightly more involved example, let's take, let's say now that what we're interested in is hydrogen peroxide. So let's say that what we're interested in doing is talking about the, let's talk about the formation of hydrogen peroxide, so there's a reaction that's fairly easy to balance. Hydrogen peroxide could be formed by combining H2 and O2 gas together. So let's say that's the reaction that we're interested in. That reaction looks like something we might be able to do in a calorimeter. Take hydrogen, combust it in the presence of oxygen and perhaps we would form hydrogen peroxide except in fact we wouldn't. What actually happens if you put hydrogen gas in the presence of oxygen gas and react them is it will combust, hydrogen will burn in oxygen, but what you form is the more stable water, let's say water liquid product. So balancing this reaction, hydrogen and only half an oxygen will form water liquid. It doesn't react to form hydrogen peroxide, it reacts to form water instead. So let's say this is reaction number one, not quite, so that's the reaction that nature will choose to have happen if we try to combust hydrogen in a calorimeter in order to measure the enthalpy change for this process. But we know some other things about hydrogen peroxide. We know that if we take hydrogen peroxide it's not terribly difficult to get hydrogen peroxide to decompose. As you know if you've ever poured hydrogen peroxide on a cut it bubbles and it's a very good oxidizing agent. It will reduce relatively easily. So this reaction is one that it's not crazy to consider causing to happen. Hydrogen peroxide decomposing to form water and giving off some oxygen gas. So between reaction number one and reaction number two we have all the pieces we need to understand this reaction. If I draw those out maybe graphically, let's say if I start with a mole of oxygen and a mole of hydrogen, the things that I want to combine to form H2O2, I can combine the mole of hydrogen and half of a mole of oxygen to form a mole of H2O and I've got left over a half of a mole of an O2. Half of a mole of O2 didn't react. So that's what we call reaction number one. But water and a half a mole of O2, those are exactly the products in this reaction number two. So as products hydrogen and peroxide decomposing into water and one half an O2, that's reaction number two. So again remembering that enthalpy is a state function, the delta H for this process converting H2O2 directly into H2O2, it doesn't matter whether we take the direct path or whether we take the indirect path of first converting it to H2 and a half an oxygen and then converting it further to H2O2, which is the reverse of reaction two, it doesn't matter which of these two paths I take, they're both going to have the same enthalpy. So the delta H of this reaction that I'm interested in is the delta H of step one and not delta H for step two but the reverse of that, so a negative delta H of step two. So very often if there's a chemical reaction that we can't literally perform in the laboratory, we can still find out something about the thermodynamics of that reaction, the thermochemistry by breaking it down into reactions that we can either measure the properties of or themselves can be broken down in other ways to measure the properties of. So the key observation here is something called Hess's law. Any time we do something like this or something like this, what we've used is Hess's law which says that we can treat chemical reactions essentially algebraically. If I turn a reaction around backwards, I just flip the sign of the delta H or indeed the delta U if I'm using energies instead of enthalpies. If I add two reactions together, then I add their enthalpies together. If I flip one around and then add them together, I can treat them algebraically, add them together but one of them having a negative sign. So I can combine reactions algebraically and their enthalpy changes for those reactions combine algebraically in the same way. So we could go through many other examples. If I double a reaction, I double its enthalpy. If I add three different reactions one after another, I just add the three enthalpies together. So that's essentially what Hess's law says is you combine the enthalpies algebraically the same way that you treat the chemical reactions if you can form a reaction that you're interested in by combining simpler chemical reactions. So it turns out this is especially useful when we use Hess's law to combine formation reactions and that's what we'll consider in the next lecture.


So let's say for example, I'm a nice little coffee Food truck sort of person known as pony express and one of the things that I want to know is what is the maximum possible? Revenue I can make on a given day if I sell coffee cups Coffee lattes and mochas at these different prices The problem is we're talking about physical goods and that means that I have certain limitations Obviously, I'm not gonna buy an infinite amount of coffee cups I'm only gonna have five hundred coffee cups. So collectively I can only sell five hundred But then I only have the materials For say a premium latte or a premium mocha I only have enough materials to total sort of make 350 those and on top of that I only have enough material to make at most a 125 mochas, so I can't just say oh, let's just sell mochas all the time So what can I do? Well, this is actually where the solver option can come into play because it allows us to have a little bit more control Than something like the goal seek So how do I do this?  I can set it to be the maximum the minimum or the value of but in our case, you know We'll leave it as max, but now we get into changing what variables. Well, this is actually where cups sold needs to come into play obviously This is the variables that are gonna change and so I hit D5 But as soon as I hit D9 Notice this changed that becomes very annoying because well, I need all three of these guys So what can I do? Well, I can hold down the control key. I can click on D9 And you see that it now highlights both of them. I can click on D13 and It's gonna select all of them So now we get into our constraints. This is where these can come into play So subject to what constraints in our case, we're just gonna go ahead and hit add Now the part I want to focus in on is no longer this side, but these Things over here like I mentioned. I'm limited by the number of cups I can sell so total cups can only be 500 so in this case total cups can be less than or equal to 500 now, I know what you're thinking you're all well, you know We should probably sell as many as we possibly can but for our sake Now I'm gonna go ahead and hit add Because the next one I want to deal with is my max premiums. So all of a sudden premium cups I can only sell 350 premium cups same kind of thing comes in G7 is less than or equal to 350 I'm gonna hit add one more time And this time we're actually talking about max mocha. I only have enough material to make 125 mochas. So the same thing I'm gonna come in establish that I only have This is the Variable the cell that will be constrained by 125 I'm gonna hit okay now all of a sudden I've built out my constraints and We could go through this song and dance of selecting a solving method Luckily for ours, we'll go ahead and just leave it at GRG nonlinear But you can see we have different options that we can play with Now what I have to do is I basically hit solve And it's gonna go through and you can see I get this nice little oh it found a solution. I need to sell a hundred and fifty normal cups and 225 premium lattes and 125 Premium mocha. So again, that was my cap. I can only sell that many mochas So then it became this sort of song and dance And so all of a sudden I can see that I can list those out and my total revenue suddenly is $918.75