 Come to the second session of the day and the last session of our same lecture on LSS. Okay. So today's lecture is going to be mostly on the blackboard and it's going to deal with the Zeldowich approximation and redshift space distortions. So before I switch over to the blackboard, I just want to say this very simple thing about the Zeldowich approximation so you get a sense for what one is trying to do and here the idea is to follow particles or fluid elements. Okay, so you switch from the Eulerian picture which we have been using until now where we ask at a fixed location in space, what is the density and velocity of a fluid? The fluid could be baryons, dark matter, photons, etc. So from this picture now because we are now interested in late time growth of structure where radiation does not play a role anymore and it's mostly non-relativistic matter, dark and baryons. So it becomes useful to think in the Lagrangian, from the Lagrangian point of view where I take fluid elements and I follow their motion along. Okay, so this one of the first such calculations in the Lagrangian picture was by Zeldowich. It is probably the oldest non-linear approximation in cosmology. It goes all the way back to 1970. So this movie which is, is it not playing? So this movie was made by a former PSD student of mine who just set up the rules that Zeldowich prescribed on a computer. And the interesting thing here is that you can see that this approximation is producing visually something that looks very similar to the images and movies I was showing you before from full n body simulations. And there are very interesting differences between this and full n body which maybe we can chat about as we go along. For the timing just take away the fact that this is a very simple prescription which is basically straight line motion. In the sense that once I give my particles some kicks initially, their velocities do not change direction in this approximation. Okay, so it's clearly wrong because gravity will make particles change the direction of their velocity. But in the Zeldowich approximation, you say let particles move in the direction of velocity which they had initially. What changes is the magnitude of their velocity. And the magnitude changes according to linear perturbation theory. Okay, so this is the Zeldowich approximation. And remarkably it produces something very, very interesting. So there's just some motivation to tell you that it's a useful thing to follow the mathematics and ask what is happening. So let me stop sharing, I hope because I'm not seeing that, yes, okay. So camera is on me. So let's start with discussing the Lagrangian picture and the Zeldowich approximation. Okay, so in the Lagrangian picture as I said, you think of particle motion. Or when I say particles now, you should always convert in your mind to fluid elements. The particle motion from its initial location to its final location through gravitational evolution can be thought of as a mapping in space from the initial position to the final position. So it's sort of like a coordinate transformation. It's not really a coordinate transformation. It's a map which takes you from the initial position q of a fluid element to its final position x at some time t. So this final position depends on where you started from. So it's a function of q and it depends on where, at what time you want to study the problem. So it depends on the conformal time eta. And I can now just mathematically say that x of q and eta, let me just re-parameterize this in terms of the initial q, plus a displacement field. So there is nothing fancy going on here. I just introduced some new notation where I've brought in this displacement field, which tells me how much the particle has moved. And you can appreciate that if I know the displacement field for every particle in my volume, then I have at a later time eta, then I have solved the problem of structure formation, right? So that's the goal, because then I will know where it's starting from some initial locations where every fluid element has gone. And if I can calculate this analytically, that's the goal. And it would solve the entire problem. So this is the idea of thinking in these lines. Now, of course, the question is how do you determine the displacement field psi? So to do this, or at least to try to approximate it, we need to know what the equations of motion are for these fluid elements. And this is not difficult because in this approximation that we are working in, which is late-time sub-horizon density perturbations. Newtonian gravity is an excellent thing to assume. So all relativistic effects have gone away. Remember yesterday we saw the relevant equations in this late-time sub-hubble limit were the continuity equation, the usual continuity equation, the usual Euler equation, and the usual Poisson equation. So I didn't actually need to know anything about relativistic cosmology in order to write down these equations. The only non-trivial term was the expansion of the universe, which also you can actually absorb into the definition of time. So you can always rearrange things so that that is not an issue. But basically the gravity sector is purely Newtonian, apart from the expansion. And even if not, even if you start with GR, what you can do is to calculate the geodesic equation obeyed by the positions of these fluid elements. Again in the same approximations. And this geodesic equation or the equation of motion will just look like Newton's law, except that there will be a drag force coming from or a drag term coming from the expansion of the universe. So this is the same notation that I have been using until now, except that now this x is the position of a fluid element as a function of time. Okay, so this is equal to the negative x gradient of the gravitational potential evaluated at the solution of the problem. Evaluated at the position that I'm talking about. So every fluid element has a position along this trajectory. And the equation of motion says that it's acceleration. The acceleration that it feels at this location is given by this equation. There on the right hand side, the source term contains the Eulerian gradient. It's the gradient with respect to x of the gravitational potential, which is a field pervading all of space. And I evaluate that field at the location of the particle. So that is the gravitational force experienced by the fluid element. Fantastic. So this is your equation of motion which you would like to solve. And of course, because we are talking about gravity, so the x Laplacian of phi is then given. By three halves, omega m, h squared times delta. Where this delta is the density contrast, it's delta rho over rho, which is again an Eulerian quantity. So at the location of the particle, I have to put down a little volume element in Eulerian space and that gives me a density of all the particles that happen to have reached there at that point, in that little volume element. And that density is the delta that appears here. So this is an Eulerian equation. And you need it to specify the force which is experienced by any fluid element. Okay, so it's a mixture of things. But you are still following the trajectory of an individual fluid element. Everyone with me? Okay, good. So now let's come back to this mapping that we wrote down and make a crucial assumption that the mapping is one to one. Meaning that if I give you a q and I specify the time eta, there is a unique x or vice versa. If I give you that at time eta, a fluid element has reached x, then it must have come from a unique q. Okay, so it's a bijective mapping. And this automatically excludes shell crossing or multi-streaming. So for people who are not around in the discussion session, multi-streaming can happen in phase space where I start with in one plus one d. I can start with initial conditions that look like this. And this will evolve in general to a situation that looks like that. Which means that at a particular value of x, I can have multiple velocities appearing, okay? And it also tells me that at a particular value of x. So here, these are Lagrangian fluid elements, which were initially, so you can think of this as q, this is initial, and this is final, okay? So in terms of q, there is one fluid element at every location of q. But in terms of x, these fluid elements have been displaced such that there is a fluid element here, there's a fluid element here, there's a fluid element here, okay? So there are multiple fluid elements sitting at the same location x. So if I just give you x at time t in this picture, you cannot tell me uniquely which q that particular x came from, it depends on which of these you're talking about. So we are assuming that this situation does not happen, that the mapping actually is one to one. So we are not talking about the shell cross regime, okay? So this is a crucial assumption. So if we do this, that we assume the mapping to be one to one, then you can imagine what happens if I take a position q and I put a little infinitesimal volume around it. This entire infinitesimal volume will evolve, and it will become some other infinitesimal volume around x, okay? So the particles contained in this little infinitesimal volume will define a unique infinitesimal volume at late times. So particle number in this sense will be conserved. If I follow each of these particles starting from a particular infinitesimal volume in the initial conditions. So this conservation of number, or if you think of these fake particles as having the same mass, this is conservation of mass. So this conservation of mass I can write in this way. Let's say that the density field in the initial conditions was perfectly uniform, or very, very close to being perfectly uniform. This is a safe assumption because even if you put in the fact that the initial conditions were slightly perturbed, these small perturbations are not really relevant for the equation that we are writing down right now, okay? So let's start with just saying that the density field of matter was perfectly uniform in the initial conditions. So then the mass contained in this infinitesimal volume d3q, initially, is just rho bar times d3q. And this, due to mass conservation, will become some density times d3x when I evaluate things at the location x at a time eta, okay? So this is just mass conservation. And now rho of m, I'm working in co-moving coordinates. So powers of a cancel out. So go back and think about why this happens. But both q and x are co-moving. So the scale factor dependence has gone away. There is a difference between x and q because things have moved, okay? But I'm thinking of this motion in terms of motion on the co-moving grid, yeah? In the perfectly homogeneous and isotropic universe, I would have no motion. Because my fluid element would just maintain its x, or maintain its q. Now, because the universe is perturbed, my fluid elements will move on the co-moving grid. And that is the motion that we are tracking, okay? So for this reason, powers of a will never appear in this expression. And this rho m will just be rho bar, the same rho bar that appears here, times 1 plus delta times d3x. Because that's how I define the density contrast at late times. To be a bit careful here, this is delta at time eta evaluated at the position where the initial q has reached, okay? So this is again the same delta actually that appears in the Poisson equation. It's exactly the same thing, all right? I'm being careful because when I was learning this, this was a point of confusion for me, okay? So I thought I'll just spend a few seconds on this. So let's use this and move forwards. This now simplifies to saying 1 plus delta. So now that I know what delta is, I won't keep writing this full dependence. I'll just write 1 plus delta times this d3x. I can write this as d3q. Again, please don't yell at me for equating infinitesimal quantities like this. We are physicists, so we know what we are talking about, hopefully. Okay, so this is one way of writing the expression for mass conservation, fine, yeah? So now let's put on a slightly different hack, where we are almost mathematicians and we say that we have changed the initial coordinates q into some final location x. So I think of this as a mathematical transformation from q to x. So if I did that, then what I would find is that my infinitesimal volume element d3x is given by the magnitude of the determinant of this transformation matrix times d3q due to change of variables. The chain rule and this determinant I will call j. This is a definition which is given by this. And I hope this notation is okay. I'm taking a matrix of transformations dxi by dqj and I'm taking the determinant of that matrix. So that is the standard chain rule. So now I can look at this equation, d3x is j, sorry, sorry, this is d3q, okay? So d3x is j d3q and 1 plus delta d3x is d3q. So j must be 1 over 1 plus delta. So I can think of the density contrast of the Eulerian set of distribution particles as related to the Jacobian of the transformation from the initial positions to the final positions. And this is just coming from mass conservation, okay? So this is one way of, so this is due to mass conservation. And mathematically, if I just take this definition, then I can also think of this j and use the fact that I've parameterized x in terms of this displacement field to write j in a slightly different form, which is also useful. So this becomes the determinant of the identity plus the q tensor derivative of psi. I'll also just write this in component language, so that it's clear. So this is determinant of the Kronecker delta plus d psi i by dqj, okay? And if you don't see why this is, just go back to the definition, okay? X is q plus psi, I have to take a derivative with respect to q. And each coordinate has to be differentiated with respect to every coordinate of q. Each x coordinate, each coordinate of x has to be differentiated with respect to every coordinate of q. That's what gives me the three by three matrix, and I take its determinant, okay? Good. So these are two different ways of writing j. I will also commonly use this notation. So let me write it down here. So d psi i by dqj, I will write as psi i comma j, okay? So the comma means a derivative with respect to q, not with respect to x. When I want to differentiate with respect to x, I will explicitly say it, okay? So this is nice because I can now go back to the equation of motion, okay? Which we kind of forgot about for a few seconds. So in the equation of motion, I can plug in this stuff, right? And I can also try to get rid of this phi here. Because now I have an expression for delta through this j. I can replace occurrences of delta whenever I see them in terms of j. And j itself, I also know how to write in terms of the displacement field, okay? So if I'm somehow, I know that this phi is determined by delta. Delta is very simply related to j. And j itself depends on the displacement field. My goal is to find the evolution of the displacement field. Now on the right-hand side, I have phi here. On the left-hand side, I have only psi because this d by d eta will kill the q. And it will differentiate the psi. So psi will remain. So the left-hand side is a differential operator acting on psi. But the right-hand side has the gravitational potential. So I need to get rid of it and bring in a dependence on psi. And that's how I propose to do it by relating phi to delta to j to psi, okay? And there is a straightforward but slightly long way of doing it. So I will not derive it, but I'll tell you what I'm doing. I take the x divergence of this equation, okay? I take the gradient operator with respect to x and apply it to this equation. On the right-hand side, I will just get the x laplacian of phi, which immediately gets replaced with delta. On the left-hand side, I have to be careful because I'm taking the x divergence of a quantity which depends on q. But that's not a problem because I know how to apply the chain rule to derivatives as well, okay? So I will just use this transformation law again to convert my x divergence into a q divergence. So without giving you further details, I will scan these notes and post them so you can work through it. Even here, I've not really given a lot of details, but I've given some, okay? So let me just write down the final equation. Is it okay if I erase all of this because I need a little bit of space? Yeah? So we can derive what is called, what can be called the master equation for the Lagrangian evolution. So I have j, I've manipulated it, okay? It's not exactly what comes out. I've manipulated it to make it a little simple. So this is the identity plus the q tensor derivative of psi. It's inverse, and then I take the ijth component of this. So I'm mixing tensor and component language. Then I take the q gradient component j. So this is like ij and then j here. So this is summed Einstein convention. And then here I have d2 psi i by d eta squared plus h d psi i by d eta. So this i is also now summed, okay? So everything is summed, and the left-hand side becomes a scalar. And on the right-hand side, because of this business of the x Laplacian, I have 3 halves omega matter h squared times j minus 1. And the reason this happened is because the Laplacian will give me a delta here. But j is 1 over 1 plus delta. So I can write delta in place of j and I'll get a 1 over j minus 1, which I manipulated to get a j minus 1 here and then I pulled the j over there, okay? So that's the thing that happened. So this is the most important equation for the purposes of this stock, okay? Okay, so of course because of our strong assumption of one to one mapping between q and psi, this equation will not always hold. It will break down once multi-streaming or shell crossing happens. Okay, so that's always a restriction you have to keep in mind. The usual approach to this problem now is to perturbatively expand in powers of psi. This is different from perturbatively expanding in powers of delta, which we were doing in the Eulerian picture. And the reason it is different and that you can expect strong differences is because psi is a displacement. So small displacements can actually create, even conceptually, they can create very large differences in densities. Because if I have particles here and here, at any given time, if my displacements are small, you can see that in a few time steps, I can end up with a situation where a lot of particles have collected together and given me very large densities. So, perturbing in psi has the potential to be a very, very useful tool for accessing the non-linear regime. Because large densities can naturally occur. Okay, that's the expectation without any calculations even. All right, so basically we are not assuming that delta is small, but we will assume that psi is a small displacement. And of course, because of this strong assumption of being pre shell crossing, it has the same problems that we had when we wrote down the fluid equations for the Eulerian equations, right? And then we tried to perturbatively expand them and wrote down these second order non-linear equations, etc. So those equations are also valid only up to shell crossing. These also have exactly the same problems. Okay, so those problems have not gone away. But for what it's worth, this is a different way of doing things. And there is this issue that it may become interesting. Okay, so now let's do the first order perturbation in this, explicitly, because this teaches us some very interesting things. So let's look at the linear solution. By the way, I should just mention, perturbative expansion of this in psi, this is called Lagrangian perturbation theory. So it's an alternative to the usual perturbation theory. Okay, so let's look at the linear solution for Lagrangian perturbation theory. On the left-hand side of the master equation, you see that there is already an explicit power of psi. See, this is the full equation, okay? It's completely non-linear, it's very non-linear actually, because I have to take the inverse of this matrix. Okay, so take the inverse is a very non-linear operation, and then I have these psi's here, and then I have j's appearing. And j has the determinant of the same matrix again. Okay, so it's a highly, highly non-linear problem. So now if I look to linearize this equation, what I can realize is that on the left-hand side, there is psi which is linearly appearing explicitly here. So if I want to only retain linear powers of psi, in whatever else appears here, I can set psi to zero, okay, on the left-hand side. So meaning that in the definition of j, maybe I should not erase the definition of j. Okay, so just one second. j is determinant of delta ij plus psi i comma j, okay? Is there a question? So in the expression for j on the left-hand side, if I set psi to zero, then I just have the determinant of the identity which is one. So j becomes one. Here I'll just set psi to zero. So this just becomes the identity. The inverse of the identity is the identity. So this whole thing doesn't do anything, okay? It's just the identity matrix acting on the rest of the thing. So it's just going to take a trace over this entire expression. Okay, so the left-hand side becomes very simple. Where did it go? Okay, I haven't written it yet. But the left-hand side will become simple for this reason. On the right-hand side, I have a j minus one. And I want to expand j to linear order in psi. So there's a cute trick to do this, to recognize that j is a determinant. And there's a relationship between the determinant of a matrix and the trace of that matrix, okay? So just remember that log of determinant of a is trace of log of a. If you don't know how to prove this, it's actually very easy. First prove it for a diagonal matrix, which is extremely simple. And then argue that both determinant and trace are rotationally invariant. So I can rotate to any other matrix that I care about. And this expression will still be true. Okay, it's very straightforward. But because this is true, I also have that in the language of Taylor expansions of matrices, if I have an expression like if, okay, so, sorry. So if my a is identity plus a small matrix epsilon. Epsilon is a three by three matrix, but it is composed of very tiny elements. Then the determinant of one plus epsilon. This you can show will just be because of this expression. It will turn out to be one, the number one, plus the trace of epsilon. So please prove this if you've never done it before, it's very easy. Okay, you have to take the exponential of this expression. So I'll get that a here and e raised to trace log a and then just do Taylor expansions appropriately, okay? Using this relation. You'll also need the Taylor expansion for log of one plus epsilon. Which again will be related to something, okay? So figure it out, okay? So using this, what you will find is that on the right-hand side, j can be replaced as the number one plus the q divergence of the displacement field psi at linear order. So j in its full glory is this determinant, but at linear order, it is one plus the q divergence of psi, all right? And what you can also appreciate here is that because here I just said j to one. And I said this thing to zero, so I just have identity. The identity ij is contracting the j and the i here. So this is also giving me a q divergence of psi, okay? So now the whole equation becomes an equation for the q divergence of psi at linear order. That's point number one. Point number two, again coming back to this expression. We have another expression for j, which is that it is equal to one over one plus delta. So now if I think of Taylor expanding, if I write delta as a linear quantity, okay, if I assume that delta is also small in this linear limit, if that is true. Then this one plus delta, so one over one plus delta. If I think of this as one over one plus delta linear, first order. This is approximately one minus delta linear, right? So it also tells you that, so I should put a one here, because this is only true at linear order. So this also tells you that the q divergence of psi one is actually just equal to minus delta one, the linear theory delta, okay? This is what this Lagrangian picture is telling us. But in order to be sure, you have to also check one more thing, which is that you now have a claim that the q divergence of psi at linear order is the linear theory delta. But then it should evolve like the linear theory delta also in order to be consistent. And we have an equation of motion for the q divergence of psi, okay? From this linear order calculation. So at linear order, this thing becomes d2 by d eta squared of the q divergence of psi, psi one, plus h d by d eta of q divergence of psi one, minus three halves omega matter h squared times the q divergence of psi one equal to zero. Have you seen this expression before? People who did the first exercise might know it. I think I maybe flashed it also, okay? This is exactly the differential equation you get in the subhubble regime in during matter domination, okay? Or well after matter radiation equality, it may be matter or lambda domination. But this is the equation obeyed by the linear theory delta, okay? So things are self consistent. The equation obeyed by the q divergence of psi is identical to the equation obeyed by linear theory delta. And in fact, just from looking at the structure of the Jacobian also, you are seeing that the q divergence of psi one is just the negative of the linear theory delta, okay? So this is very nice that the lowest order, so lowest order LPT, it recovers the lowest order. Standard perturbation theory, which is an Eulerian perturbation theory, okay? So they agree at this order. If you assume that delta is small, okay? And this we will discuss in a second. But continuing with this, everything there, okay? So let's continue with this. Now remember we were making this approximation in addition to not having shell crossing, we were also saying let's set, let's set vorticity to zero. So curls of velocities we can set to zero, okay? And this is still true because as long as you are not in the shell crossed regime, and you did not have any sources for vorticity to begin with, you can again prove in this picture as well that vorticity will not evolve. It will decay. I mean it will evolve, but it will not grow, it will decay. So it is still valid to set vorticity to zero, which means that because we are writing down everything in terms of this displacement picture that the q curl of psi can be set to zero, okay? And one way of thinking about this is that the peculiar velocity in this case is nothing but d psi by d eta. So if you're setting vorticity to zero, it's because you're setting the curl of psi to zero, okay? So now the linear growing mode, for example, in this picture. Because again, this has two solutions, one will grow, one will decay. So let's focus on the growing mode. So the linear growing mode I can write, this is a vector psi. This is minus of the d1 that we have seen before times the q gradient of some scalar, I hope the difference between my scalars and vectors is clear. I have introduced a scalar now to describe the quantity whose gradient is the displacement field psi. Because if psi has zero curl, then it is the gradient of a scalar. That scalar I'm calling psi one, this other psi, not vector psi, okay? Yeah, I've done this in the notes, so I'm following it. But you could use better notation perhaps, okay? So then we also know from standard perturbation theory that delta one of q and eta, this is, I can write q here because actually delta one should be evaluated at x and eta, but then I'm doing linear perturbation theory. So x is q plus psi, psi is itself a linear perturbation. So if I evaluate delta one at x, it is effectively the same as evaluating it at q, okay? Because the corrections to this will be higher order in the perturbations, which I'm throwing away. So this I can just replace with q here. And this is just equal to d1 times delta one initial of q, okay? Whatever the initial conditions were. So for example, these were Gauss and random field in the case that we were discussing. So then this psi one, because delta one is related to the divergence of psi, right? So the q divergence of this will give you the q laplacian of this scalar potential, okay? And also you know that delta initial is the q laplacian or the x laplacian, but in linear theory it's the q laplacian of the gravitational potential. So this psi is actually the gravitational potential. It is at least proportional to the gravitational potential, okay? So you can argue this out. So because, so then the q laplacian of psi this is equal to delta, delta one. And that means that this psi is just proportional to the initial gravitational potential, okay? It's the initial one so I can just not even worry about perturbations, okay? So this gives you the full solution at linear order for x, which is q minus d one of eta times the q gradient of psi one of q, okay? All the time dependence is in eta, is in the growth factor. And all the dependence on q is in this potential psi one. And the velocities of these fluid elements are given by negative d one over d eta times again the q gradient of psi one. I just take a time derivative of this, so it only affects the growth factor. This is usually written as in this way, what happened to this sign? This should be consistent. So this becomes a plus h times f times psi one, okay? Did I do this correctly? Yes, that's right. So this negative sign, what I have done is I have multiplied and divided by a d one, okay? So when I divide by d one, I'll define this f here. So f is d log d one by d log a. So I had a d d one by d eta. I have changed variables from eta to a. So I pick up a conformal Hubble parameter. And then I multiplied and divided by d one. So this became a d log d one by whatever. And then the d one that I multiplied by the minus d one grad q psi one has become a psi one, the vector psi one, okay? So my peculiar motion is given by the displacement vector itself. Up to a multiplicative factor, which depends on time. And this is what I was saying before, that the direction of motion, see this vector psi one, its direction is completely determined by the initial conditions. Its direction is not evolving. And this direction propagates into defining the direction of the velocities of these fluid elements. So in this sense, we have straight line motion. The magnitude of the vector velocity of course keeps changing, and it changes proportionately to this factor, okay? So one very interesting thing for observational cosmology is to try to constrain the value of this factor. Because this factor depends not just on the growth itself, but the rate of growth. And the rate of growth can be very different in different cosmological models. Lambda CDM versus dark energy or GR versus non-GR, okay, even. So this is a very interesting quantity track observation. Good, so this is all linear Lagrangian perturbation theory. And this allows us to very easily set up what is called the Zeldovich approximation. Because what Zeldovich did is to basically do all this calculation. And say, eventually, let's assume that the full displacement field psi, which would actually be psi one plus psi two plus psi n, nth order Lagrangian perturbation theory. Just replace this with the linear solution, psi one, okay? It's as simple looking as that. Just say that the full displacement field is given by the linear solution. Why is this not a trivial statement? It's not trivial, because in all of the previous calculation, in order to solve for all of this stuff, I made the assumption. So when I wanted to relate this psi one to the initial density, I assumed that the initial density is small, okay? And that's what allowed me to calculate the solution for psi one. But now, having calculated the solution, I'm going to say, let the full displacement be given by that solution. But the full displacement is related to the full density in a non-linear way. Through j is equal to one over one plus delta, okay? So this now is the equation that I need to worry about. Where this j is then also equal to the determinant of delta ij plus now psi i comma j, where this is the full solution. So I am not going to perturbatively expand the left-hand side and then I'll see what happens, okay? And you'll see that immediately that something very interesting happens, okay? So let's do this calculation. So now I have one plus delta, which is one over j, and this is one over this determinant, okay? I'm not Taylor expanding anything yet, I'm just writing this down. And seeing what happens. So now this quantity psi i comma j in the linear limit. So in the Zeldowich approximation, so under Zeldowich, this psi i comma j is psi one i comma j. And this is very easy because I know the full solution. So this is minus d one of eta times this derivative of this scalar psi comma ij, okay? This scalar psi, which is proportional to the initial gravitational perturbation. And I take two derivatives of it, okay? So this is just del two psi one by del qi del qj just to be completely explicit. This thing is a tensor. It's a symmetric three-dimensional tensor. It's called the tidal tensor. So strictly speaking, it's the linear tidal tensor. Why tidal tensor? Because it is doing exactly what a tidal tensor is supposed to do. The gradient of the gravitational potential is the gravitational field. And tidal effects are obtained by differentiating the gravitational field. Spatial variations of the gravitational field are what lead to tides on the earth, for example, right? The fact that the gravitational field at one end of the earth is different from that on the other end of the earth due to the influence of the moon and the sun. So this is calculating exactly that, the spatial derivative of the gravitational field. And it turns out to be a symmetric three by three tensor, okay? So now I have the determinant of something which has the identity matrix plus this linear tidal tensor. So I need to know how to deal with this. So a nice way of dealing with it is to say, just recognize that all of this depends on q, right? So I have to know which q I'm talking about. So let's pick a particular q and calculate the tidal tensor there. So I will get a three by three matrix. It's a real symmetric matrix, so I can diagonalize this matrix, okay? So I can locally diagonalize the tidal tensor. And let's say that in this diagonal reference frame, so in the eigenframe of the tidal tensor, let's say that this i1, ij, this is a diagonal matrix with eigenvalues lambda 1 of q, lambda 2 of q, and lambda 3 of q, okay? So it will have real eigenvalues because it's a real symmetric matrix. And without loss of generality, I can adjust, I can orient my axes so that lambda 1 is bigger than or equal to lambda 2 is bigger than or equal to lambda 3. There's nothing wrong with making this choice. So this means, okay, this I can always do. But this means that this factor here is going to be something very interesting because of this relation here, okay? So what am I going to get? I will get 1 now in full glory, 1 plus delta evaluated at x of q eta comma eta. This is going to be equal to 1 over in the diagonal frame. See, I have the identity plus now in the diagonal frame. In the eigenframe, I have identity plus a diagonal matrix, right? So identity plus the diagonal matrix is itself a diagonal matrix. So the determinant of this full diagonal matrix is just the product of the diagonal elements. So what are the diagonal elements? For each of them, the diagonal elements will be 1 plus the contribution that came from the corresponding eigenvalue of this object, okay? That eigenvalue is given by lambda, which will go and sit here with this minus sign. So I will get the product from j equals 1 to 3 of 1 minus d1 eta times lambda j of q. So 1 minus d times lambda 1, 1 minus d times lambda 2, 1 minus d times lambda 3, all right? Why is this interesting? Can anyone already guess why this is interesting? In the context of everything we've been talking about. Suppose I say that lambda 1 is some small positive number. It has to be small because I assume that the initial conditions are small. But I fix it to be some small positive number. Let's say it's plus 10 to the minus 5. What happens as a function of time to the density? I can't hear. It will become the dominant term you're saying. But keep progressing forwards in time. What will happen to the density eventually? It will become infinite, right? So I started with something that looked like perturbation theory, in which there was some small quantity, a small displacement. But it is actually leading to infinities in the density. So this is the thing that tells you that the Zeldowich approximation already is enough, small displacements are enough to produce very, very large departures in the density contrast. Okay, so there is a certain language associated with this. How am I doing for time? Half an hour. Fantastic, okay? So there is a language associated with this. It's called the language of the cosmic web. Because all of this cosmic web stuff is inspired from the Zeldowich approximation. So this lambda one, which is the largest local eigenvalue of the tidal tensor. If you think about what it's, okay, let me write this down. This corresponds to the direction E1 of maximum compression. Let's think about what this means. So if lambda one is positive, I'm saying that lambda one is the largest of the three eigenvalues. If it is a positive number, the other two eigenvalues, whatever they are, are less than this. And lambda one, we've assumed it's a positive for a second, and it is the largest of the three. So it is the one which is going to give you the most amount of compression. Right? Because it is the one which is going to give the largest increase in density. It, and it is the one which will cause the density to go to infinity for the first time, okay? Of course, it could happen that lambda one itself is negative. So if all three eigenvalues are negative and lambda one is the largest of them, right? So if these are the lambdas, and this is lambda equal to zero, I could have lambda one here and lambda two here and lambda three here. So all of them are negative and lambda one is also negative. If all of them are negative, then this beast is very well behaved, because it always remains finite. And in fact, because all of them have this negative sign, this whole thing will just keep decreasing as a function of time, right? So it will always be finite and a decreasing function of time. So in that case, lambda one would be the direction. So basically this maximum compression could actually be an expansion, which is diluting the density, okay, if lambda one is negative. But if lambda one is positive, then this E1 will collapse first. In the sense that this particular direction will be the one which goes to zero radius, okay? Or it will create a density which is infinite, because this denominator is going to zero, that particular factor corresponding to lambda one, all right? So this is, so for example, when this happens, when D1 times lambda one becomes equal to one. And there is nothing to tell you that you can't say that oh, lambda one is small, so this will not happen. If you wait long enough, it will happen, okay? Because it is finite, lambda one is not zero. So this will happen at some point in the future, okay? So this kind of structure, Zeldowic called a pancake. Because what one is saying is that in, it's a three dimensional distribution of these fluid elements. And along one direction, they are becoming singular. They're all coming together. So it's like a sheet or a pancake is forming, okay? And the other two directions are still doing something. Now, if lambda two is positive, this means that lambda one and lambda two are positive, because lambda one is bigger than lambda two. So that means lambda one is bigger than equal to lambda two is bigger than zero. Then the direction E2, so then of course lambda one, that direction will collapse, it will go to infinite density. But imagine that you have a way of progressing beyond infinite density, okay? You have some rule to say that when these particles go through the same location at the same time, with some rule, I will allow them to keep going forward. So the ones that are going this way, they will all appear and I will get an infinite density here. But then this one which came here, I will just let it keep going. And this one which came from here, I will let it keep going there, okay? So I can mathematically extend the evolution beyond the formation of the first such singularity, first such caustic as it is called. If I do that, and if my lambda two is positive, then I will find that at some point in the future, there will be another zero in the denominator, which will be given by one minus D1 times lambda two being zero, right? Because lambda two is the next number which is controlling the factors here. So let me just write this again. This is one over one minus D1 lambda one, one minus D1 lambda two, one minus D1 lambda three. Of course, now whatever I'm saying is all really approximate because you have to have a way of progressing beyond the first infinity, okay? So you have to say that something happens and makes things regular again, and then you ask what happens next. So if you do it like that, you will find that the E2 direction collapses next. And if I now think about what this is describing, I already had the E1 direction collapsing, so I had a sheet forming. Now in an additional direction, I am reaching a singularity. So instead of a two dimensional distribution, I end up with a one dimensional distribution, which is just a line. It's like all the particles now fall into a line or a filament. And by extension, if lambda three is then I play the same trick again. I allow things to progress beyond the second singularity. Then if lambda three is zero, then E3 will collapse the third. And now all three directions have been compressed. And I will form what is called a node or a knot or a cluster, okay? And if lambda one is negative, that means all of them are negative, then all directions expand and I form a void, okay? So this is the usual language of the Cosmic Web, which is even used today. Although we don't really think that this is the way structure formation proceeds because we now think that structure formation proceeds hierarchically with small structures forming first and larger structures forming later. It is very interesting though that something that looks like the cosmic web produced by Zeldovich is in fact present. We have seen it already in these simulations, right? And then you can also go back and look at the movies. The lines that trace out the filaments, etc. They are already in place at very high red ships, okay? So this hierarchical structure formation is very interesting to understand because it has very deep connections with what Zeldovich was talking about, even though it is not exactly the same, okay? All right. So there is a section here on how the Zeldovich approximation is used to set initial conditions for simulations, which is there in the notes, but I will skip it for this discussion because I want to really talk about red ship space distortions next, okay? So this is what I will say for the Zeldovich approximation on its own. If there are any questions, I can take them now, yes. You don't have shell crossing, you mean? Because shell crossing, you can think of shell crossing as the time when the first of these axes collapses, okay? In this particular approximation, that's when the first shell crossing will happen. So in the case of the voids in this approximation, not in the full Lagrangian perturbation theory or in the Lagrangian theory, but in the Zeldovich approximation, you can think of shell crossing as happening when the first axis collapses. So if lambda 1 is negative, then no axis will ever collapse and shell crossing will never happen. It will just keep increasing other way around, right? The least negative will expand the fastest. Lambda 1 gives you the direction of maximum compression, right? So the maximum compression is the opposite of maximum expansion. The last one will give you maximum expansion if it is negative. Lambda 3 will give you the fastest rate of dilution. I mean, just if you can work it out. There's another one? Yeah, yeah. So that's what I'm saying. This you have to take with a pinch of salt because it depends on how you choose to evolve the fluid elements later. Okay? So it's a hand-waving thing that I've done here. In reality, of course, neither the Zeldovich approximation nor this business of pancakes to clusters is exactly correct. So you have to solve everything in the presence of gravity. Okay? So there will not be any problem because the gravitational evolution will give you how to move the trajectories of these particles. Okay? So you would have to do something here, but whatever you do is fake. Okay? So I'm saying let's not worry too much about it. Anything else? Yeah? Larger size can do the same thing. And if you want to study larger size, you could, in principle, study the next orders in perturbation theory. So that's how one organizes this calculation. And you can ask what is the largest typical size that you expect? This is a very nice question, actually. I'll say, can I repeat the question? Yes. Okay. So the question is, I said that small size can lead to large densities, which is true. We have seen this. But large size can also lead to large densities, of course. So now the question is how large can psi typically, how large is psi expected to become? Okay? So you could formally study Lagrangian perturbation theory and ask this question order by order. Do you see things converging or not? Or you can do a slightly simpler calculation, at least for order of magnitude estimates, which is to evaluate the velocity power spectrum in linear theory. Which, because velocity is now it's erased, but remember, in linear theory, velocities of fluid elements are given by psi, linear. So the power spectrum of velocities, which roughly looks like u dot u in k space. So u k, u k prime. This will be related to psi 1 k, psi 1 k prime. I'm taking a dot product here to get a scalar power spectrum. You can even put a star here to make things consistent. Calculate this thing. It's a nice exercise to do. It's just a linear theory calculation. Everything you need to know is there, because this psi 1, the divergence of this psi 1, is the density, linear density. So in k space, it's like k dot psi 1 is related to the linear density. So in fact, let me just write down this expression, because then I was going to write it when I was doing the simulation IC, but I'm not doing that. So psi 1 of q, I can think of as the Fourier transform or the Fourier inverse of i k by k squared times delta 1 k. And where k is the Fourier conjugate of q now, because I'm doing everything in the initial conditions. This quantity I can use to calculate this power spectrum. So this psi 1 k is this thing that's written down inside. Calculate the power spectrum of this. Unlike the matter power spectrum, which has this problem, that the dimensionless matter power spectrum, it has increasing power. So if I calculate k cubed p delta in linear theory by 2 pi squared for normalization and I plot it as a function of log k, I will find that this power keeps increasing. This is not the same for the velocity power spectrum or the power spectrum of psi. I will find that the velocity power spectrum is very well behaved, it actually falls, and I can integrate this quantity. Integrating the power spectrum gives me a measure of the total variance of velocity in the initial conditions, because the power spectrum is the variance of the Gaussian random field at individual modes, and the integral of this is the sum of the variances of all the modes. So it's the total variance in the initial velocities. This number, if you calculate, will turn out to be the square of roughly 10 megaparsec over h squared in Hubble law language, because velocities can be converted into distances. So if I ask, given the typical velocities I get from here, how far can a typical particle travel? This number, the variance of this number will be something like 10 megaparsec squared. So the idea is that this linear theory calculation tells you that typical particles will not be displaced by more than about 10 megaparsec or so in the age of the universe, a randomly chosen particle. This is what allows you to think that displacements overall are probably going to remain small. You may never actually have to deal with displacements which are of the order of the size of the universe. That never happens because you're dealing with non-relativistic initial conditions. That's it. All right. I assume there is a question. Yes. So the question is, was an or base assumption that of bijectivity, so even the infinite density condition shouldn't be allowed under this model that along them cross over? Infinite density should not be allowed in this model. Yes. I mean, yes, this signals that something is going wrong. But it is also true that the presence of caustics is not unknown in physics, right? Because in optics, for example, you do see caustics and it is not that you cannot treat them. You cannot describe them. This is, of course, an idealized model. It's perfectly collisionless, dark matter. We are just studying what happens to this system. So in the real world, what you expect is that gravity will change things because here gravity has not been accounted for correctly. So the correct accounting of gravity could change at least where the locations of these caustics happen. But caustics do occur in this model. Now, if you really care about these things, you have to ask what is observable. So then things depend on what exactly you are observing. So in a galaxy cluster, for example, you will not directly be observing the dark matter. You'll be observing astrophysical quantities like gas, et cetera, which will not have caustics, but it may have shocks. So there will be places where densities become high and the basic theory here may still be useful. So it is not that just because you see a singularity, you should throw away everything. I don't know if this answers the question because I didn't fully understand the gist. So in the time remaining, let me tell you how the Zeldowich approximation can be related to the notion of redshift space distortions. So what are redshift space distortions? This is a slight change of gears, but we'll see. We'll come back to all of this language in very little wine. So imagine now from an observational point of view. So now I leave aside my theory. I'm an observer. I'm sitting on Earth and I'm observing galaxies in the sky. I have shown you all these nice-looking images from SDSS where they plotted redshift and they converted it. They were using redshift as a proxy for distance because Hubble and LeMethra told us so. So now you have to think about what it means to measure a redshift. What can create a redshift? If you think about the way Hubble and LeMethra created their plots, they always converted their measured redshifts into velocities. Why was that? Because their thinking at the time, or at least Hubble's thinking at the time when he created his plot, was that redshifts must be related to Doppler motions. So I should just multiply and Doppler effect in the non-relativistic limit tells me that the redshift is V over C. So if I take my redshift and I multiplied with speed of light, I'll get the Doppler velocity that is the speed of this thing which is moving. So that's my interpretation. Then LeMethra and then other people figured out that, ah, actually if you think in relativistic terms, the thing that you should really be saying is that the universe is expanding and the photon propagation is what leads to a change in the wavelength. So that is the origin of the cosmological redshift. But because of the FLRW space time, that cosmological redshift has a linear relation with the distance. It is still true that if you take C times Z, it is a proxy for some kind of velocity, but it's not really a velocity, it is actually a proxy for distance. And Hubble's law, Hubble-LeMethra law said that CZ is H0D, or HD. D is your estimate of distance. So writing this in this form, I can estimate distance by measuring the redshift and knowing Hubble's parameter. Suppose I now say that I will do this for all the galaxies that I observe. In a homogeneous and expanding universe for nearby galaxies, this would give you the proper distance to these galaxies. But the universe is not homogeneous, right? That's what we've been studying so far. There are perturbations everywhere, which means that these galaxies need not be fundamental observers who always see the same constant CMB around them. They can have motions relative to the co-moving grid, which is used to define the FLRW geometry. So these peculiar motions of these galaxies will create Doppler shifts, even if you are sitting in a reference frame very close to the galaxy and co-moving with the fundamental grid. So on top of the fact that the photons have taken a long time to reach you and therefore expansion has increased their wavelengths, there will be an additional small effect, because even if I remove the effect of this expansion and I go to the reference frame of the, I go to the location of the galaxy, if I sit in the reference frame where the CMB is evolving uniformly, then there will still be a peculiar motion of that galaxy relative to me, and that peculiar motion will cause a Doppler effect in the redshift, in the measured spectrum. So that Doppler effect has to be added to the total cosmological effect that one is usually talking about when dealing with Hubble's law. So in pictorial terms, if I am an observer here and there is a galaxy here and this galaxy, so I am an observer at a co-moving distance x and therefore a proper distance A x and imagine that it's a nearby galaxy, so distances, I don't really care about whether it's luminosity or angular diameter distance, they're all the same. And this galaxy is moving relative to the co-moving grid with a peculiar motion U and just for convenience to change units, I will divide the velocity by the Hubble parameter. So this is the vector U over H. The component of this vector, let me draw it a little bit better, so the component of this vector along my line of sight to this galaxy, let me call this U parallel over H. It's a scalar quantity. This creates an extra Doppler effect in the measured spectrum coming from this galaxy. So I measure a spectrum which has an observed redshift z observed and I want to use this z observed to convert into a measure of distance of this galaxy from me. So what will I do? I will just multiply this by C and I will divide it by H and I will call this the observed distance. Let's call it S. But what is this quantity? The numerator is actually the original, is the effect of Hubble flow which will be an A times H times AX because this is the proper distance multiplied by the Hubble parameter. This is the actual cosmic, the cosmological Hubble's law applied because of cosmological expansion. So this is what the thing would have been without peculiar motion but then in the numerator, I also have an additional effect in the redshift which is due to the peculiar velocity which is U parallel in units where C is one. So I have to, sorry, C is taken on this side. So this is just U parallel because the Doppler effect will be V over C or U over C. So this quantity then gives me A times X plus, have I done this correctly? I want to actually write A times S here. I want a co-moving distance to be defined. So let me call it AS. So this is A times X plus U parallel over H. So this, if I just reorganize in vector notation, the vector distance to this galaxy will be written in terms of the actual co-moving vector to this galaxy from my location plus a term which depends on U parallel and the direction of this second term is only along the line of sight because the effect of this extra Doppler shift is only along the line of sight. It's only that component which matters. So when I write it in vector form, I will have, so let's say that this is the Z axis. It's convenient because Z is also the label we use for redshift, so I can call this the Z axis without loss of generality. So I have an effect along the line of sight and the effect is just U dot Z hat divided by H. U dot Z hat is U parallel. All right? So why is this nice? Because X, going back to what we were doing before, I can think in Lagrangian terms and I can write this X as Q plus psi of Q and eta plus this effect, U dot Z hat, Z hat by H. And I will just club these two terms together and I will define this thing in square brackets as a redshift space displacement vector, psi s of Q and eta. Sorry, there should be an A H here everywhere because I divided by an extra power of H and this A H is nothing but script H. So I have a script H here. So now I can do some nice things with this because what I have done is to rewrite my position of my interesting tracers in terms of the initial positions of those tracers and a displacement field, psi s. So everything that I understood about the Zeldovich approximation and all the tricks that we used in analyzing the displacement field there can also be applied here with the additional contribution of the velocities taken into account properly. So let's do this calculation which is very nice. Are people following this or are you lost? If you are lost, ask me some kind of question so that I understand why you are lost. Or if it's clear, then I can keep moving, okay? I can ask a question. Yeah, go ahead. What is s? Yes, s is the magnitude of s is the distance that I would infer from the observed redshift by simply applying Hubble's law, the Hubble-Limitri law. So Hubble-Limitri law would tell me that c times z is h times distance. So I have written this in a convention such that this s is the co-moving distance that I would infer from such a calculation. Then I recognize that the observed redshift is not just the application of Hubble's law to the co-moving position of the object, but it has an additional contribution due to a local Doppler motion. So I add that in and then from here to here I have just vectorized the equation by saying that there is that galaxy in that direction and the quantity which is affecting the redshift is something that is pointing along the line of sight. So that is why this z is pointing along line of sight and its amplitude is determined by the dot product of the local velocity with the line of sight because I only need u parallel here, all right? Hope that also clarified things for other people. And now let's, we need to know what is the displacement field here, right? Let's assume the Zeldovich approximation for that displacement field. So u z a for psi. And now we know what to do because psi is then psi 1 which is the Fourier transform of i k by k squared delta 1. By the way, if you are not following why this i k by k squared came, it's very simple. Remember we said that the vorticity is zero. So this vector psi is the gradient of a scalar. That scalar is just the gravitational potential, okay? So the gravitational potential solves the Poisson equation which is Laplacian of that gravitational potential is delta. So in Fourier space, the gravitational potential is just delta divided by k squared, okay? So this part is the gravitational potential and this i k is the gradient of that gravitational potential, okay? Which is the quantity which determines the displacement field. So this is just this, okay? And we also know that in this Zeldovich approximation, the velocity is h times f times psi 1 again, okay? And this is super nice because now we have everything. We have psi which is replaced with psi 1 and u is also related to psi 1. It's linearly proportional to psi 1 with some factors. So let's track everything. So now my Fourier space, so in Fourier space, my psi s of k, this will be psi 1 of k, this approximation because I'm using the Zeldovich approximation here. So psi 1 of k plus just do this on your own, it's very easy. It is f times z hat, only f because the h gets cancelled out, okay? So f times z hat times psi 1 of k dotted with z hat, okay? And now I do these manipulations. I replace psi 1 of k with this quantity here, okay? i k by k squared times delta 1. So this will have i k dependence here. This will have a z hat. This will have a k dot z hat, okay? So I track everything. And then you can show that this will be equal to i delta 1 k by k squared. This will come out. And inside, sorry, this will be a k because 1 power of k will cancel. So check this as a homework. And inside, I will have the unit vector along k plus the unit vector along z multiplied by f, multiplied by the dot product of these unit vectors. Very, very simple, okay? Everything is determined by the initial conditions, but modulated by all of these vectors and these amplitudes of k, okay? So from here, this is the displacement field. And if I again assume whatever I was assuming about linear theory, if I assume that this displacement field corresponds to a change in density, which is also linear, then the redshift space density, delta s, will be related to the q divergence of this displacement field for the same reason that the linear theory density field was related to the q divergence of the real space displacement field. Okay? So this delta s, if I treat it in perturbative language, then this will be equal to minus the q divergence of the redshift space displacement field, okay? So this delta s in k space, the goal is to calculate the power spectrum in k space, okay? The macro power spectrum in redshift space. This is called the redshift space picture where I have used all distances as inferred from observations using actual redshifts. So then the positions of galaxies that I get will be affected by these peculiar motions. I can put down a grid and I can calculate a density field of these positions. I can Fourier transform it. I can calculate the power spectrum. What is the power spectrum that I will get and how will it be related to the actual power spectrum? So what I will get is this delta s of k will be minus ik in linear theory, okay? So minus ik dot psi s, because it is the Fourier transform of this, okay? And now I know an expression for psi s. I take it from here and this will turn out to be delta 1k. This is the real space linear field into 1 plus f times k hat dot z hat squared, okay? Why did this square come? There was one z hat dot k hat, which came from the fact that the Doppler effect involved the magnitude of the parallel component, right? So that's what led to this z hat dot k hat. And the second one came because there is a divergence of the displacement field appearing when you calculate the density. So there is a dot product with k appearing there and the direction of this effect is in the line of sight anyway, okay? So there are two powers of this that appear. So if I define a quantity mu, it is similar to that p dot k that was there earlier. It is now nothing to do with photons. It has to do with the direction of the line of sight. So if this is defined in this way, then my delta s of k is just delta 1k times 1 plus f mu squared. This is showing you something very, very interesting. There is now a dependence of these Fourier modes on the line of sight direction through the appearance of this mu, which is a dot product of k with the line of sight. So when I take the power spectrum of this, power spectrum of this object, what I will get is the power spectrum of linear theory, the usual one, times 1 plus f mu squared, the whole squared. So something that I didn't discuss earlier is that what we expect for the Gaussian random field in the standard assumptions is that not only is it statistically homogeneous, but it is also statistically isotropic. That statistical isotropy tells you that the quantity k that appears here, that the linear theory power spectrum, I had written down generally as a function of the vector k, but due to statistical isotropy, that linear theory power spectrum can only depend on the magnitude of k. So that is the usual assumption. What the Zeldowich approximation and redshift space calculation is showing you is that in redshift space, this isotropy assumption is explicitly broken because I have an explicit dependence on mu here. So this is one important result. This calculation was first done by Nick Kaiser in 1987. So for example, if I can now do various things with this, I can try to test whether this isotropy exists. This is all in linear theory, so it only holds at large scales. I will not have time to talk about what happens at small scales. But for example, I can calculate the average over mu squared, average over angles. And then what I get is what is called the monopole, or let me just call it the monopole. And this will turn out to be the linear theory times a very specific factor which involves f. So this will be 1 plus 2 thirds f plus 1 fifth f squared, if I just integrate over all mu. And so you see that there is a clear dependence on f now, which is interesting, because it allows you to do cosmology. So I just want to end with... Okay, I'm 10 minutes over. Let me just show you a power spectrum and stop. So this is what this difference in power spectra looks like. So in blue is the linear theory. So it is the power spectrum in a simulation. So it is not really linear theory. It agrees with linear theory at very small values of k. But it is in real space, the blue one. The red one is in redshift space. So in a simulation, I can always move my particles according to their velocities and I can define a redshift space field and I can calculate its power spectrum. And what you can see here is that at these small k values, which are the large scales where linear theory holds, I've taken the ratio of the redshift space power spectrum to the real space power spectrum. And I've compared it with this horizontal line, which is this factor, set beta to one. And it's that factor, one plus two-thirds f plus one-fifth f squared. So set beta to f, sorry. So you can see that the theory works exceedingly well at very large scales. And at small scales or large values of k, there is a departure. These departures are also very interesting. Maybe we can have a question on this in the discussion. So I think stop here. I have given you a flavor for the various techniques that go into understanding the evolution of large-scale structure. I haven't covered everything and several important things that I have missed, but I hope this gives you enough motivation to study the field. So thank you. Thank you. Okay, let me just say that this afternoon we have a seminar by Mustafa Amin about dark matter. And he agreed to give a short introduction since we do not have a specific lecture on dark matter this year. He is going to give some general remarks and then talks about the more specific research that he is doing. So the total seminar would be about one and a half hours. And those who are online need to register to get a new link and passcode. Okay, thanks. Let's go for lunch.