Conformal Invariance Be Found with a Green’S Function G(~X, ~Y) Suitable to the Region, Φ(~X) = D3yg(~X, ~Y)Ρ(~Y)

Home , Conformal symmetry, Invariant (physics)

204. Last Latexed: April 25, 2017 at 9:45 Joel A. Shapiro

∂2 ∂2 ∂2 rule shows that 2 = + + , of the same form as in terms of ∇ ∂u2 ∂v2 ∂w2 (x, y, z). Of course it might be even better to make a change of variables to spherical coordinates, where we would see that the r dependence factors out, and actually gives us the differential equations in θ and φ which would in general provide the decomposition of Φ into spherical harmonics, and for Chapter 18 a conducting sphere would tell us Φ is independent of the angles. The solution of a linear partial differential equation with sources in some region, with boundary conditions, for example the Poisson equation, can Conformal Invariance be found with a Green’s function G(~x, ~y) suitable to the region, Φ(~x) = d3yG(~x, ~y)ρ(~y). The invariance under symmetries is reflected in the Green’s Rfunction. For example, invariance under translations tells us G(~x, ~y)= G(~x ~y), and invariance under the full rotation group tells us that is actually− a 2 At the beginning of the semester we motivated our investigation of symme- function only of (~x ~y) . −2 tries by illustrating that, given differential equations which were symmetric, To argue that was rotationally invariant, we showed that the form ∇ the solutions had to transform into each other under the symmetries as a rep- did not change under the rotation of coordinates (x, y, z) (u,v,w), but −→R resentation of the symmetry. The first illustrations considered Schrödinger the form does change under (x, y, z) (r,θ,φ). This change is to be con- equations with symmetric potentials, such as the electrons in the spheri- sidered a change of variables rather→ than a symmetry of the physics, and cally symmetric potential of an atom, having wavefunctions transforming we may still write Φ(r,θ,φ)= µ(r′, θ′,φ′) G ((r,θ,φ), (r′, θ′,φ′)) ρ(r′, θ′,φ′) under the rotation group. The Laplacian is invariant under both rotations with the correct measure µ = Rr2dr sin(θ) dθ dφ and a suitably modified G. and translations, but the source of the potential may be taken as invariant This would be true, though probably not useful, even if there is no symmetry rotationally but not translationally. Under translations, we might make a in the physics. This kind of transformation is considered a passive one, cor- connection of the wave function at ~x for an atom with a nucleus at ~y with responding only to a change of description rather than an actual change of the wave function at ~x +~a for an atom with a nucleus at ~y +~a, as the physics the physical situation under which the physics is invariant, that is, an active is translationally invariant if we translate both the point of evaluation and symmetry transformation. the boundary conditions, or sources. This kind of coordinate independence plays a crucial role in general rel- Thus the way symmetries act on solutions depend on both the differen- ativity, where we make a big point of quantities such as 2 and Φ having tial operator having the symmetry and the boundary conditions or sources physical meaning independent of the coordinates used to∇ express them.∇ This having the symmetry. The electric field of a point charge at the center of a leads to the ideas of co- and contra-variant tensors, and in particular of the conducting cube would not have a full rotational symmetry, but would be- metric tensor, which expresses the distance ds between two positions xµ and have as a representation of the symmetry group of the cube. It is obvious xµ + dxµ as 2 2 2 2 ∂ ∂ ∂ 2 µ ν that the differential operator = + + is symmetric under (ds) = gµνdx dx ∇ ∂x2 ∂y2 ∂z2 x x, under x y, and under x z, which generate the group. For a µ ↔ − ↔ ↔ where x can be any set of generalized coordinates, and ds might refer to the spherical conductor with a charge at the center, we should have O(3) sym- Minkowski rather than the Euclidean length. Under a change of coordinates, metry, but how do we know 2 is rotationally invariant, when it seems to µ ′ ν µ ′ ν ∇ x x , the relation of dx to dx is given by the partial derivative depend on a choice of three axes? One way is to perform a change of vari- matrix,→ so the metric tensor g′ transforms on each index with that matrix, ables (x, y, z) (u,v,w) where R is an orthogonal matrix. Then the chain and g dxµ dxν is an invariant object. −→R µν

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µν µν µ ρ σ If we define g to be the inverse matrix to gµν, i.e. g gνρ = δρ , and if we whenever ηρσdx dx = 0. An antisymmetric Lσρ clearly satisfies that, and define g := det g , (or det( g ) for Minkowski space), then we can show gives a Lorentz transformation, but also clearly L = η or Lµ = δµ sat- µν − µν σρ σρ ν ν that the invariant volume element is given by √g dxµ and the laplacian isfies this condition as well. Choosing lightlike dxρ’s with opposite spacelike (or d’Alembertian) by Q components shows a symmetric Lσρ must have L0j = 0, and then we are 2 i j left with L00 ~a + L a a = 0 for any ~a, so the only symmetric L which | | ij µν 2 1 ∂ µν ∂ survives is η. = µ g √g ν . ∇ √g ∂x ∂x Notice that this is precisely a change in the scale of the metric, gµν µ ∂ → hgµν. As a differential operator, this is D = ix ∂xµ , which commutes with Now if we start with ordinary Euclidean or Minkowski space with carte- µ µ ν ∂ the Lorentz transformations Λ = x L ν , but not with the momenta sian coordinates, g = δ or η respectively, there are changes of coor- ν µ ∂x µν µν µν P = i ∂ , as [D,P ]= ∂ = iP . dinates which leave the form of g invariant, which include the transla- µ ∂xµ µ ∂xµ µ µν Thus we may add D to the Poincaréalgebra to get an 11 dimensional tions, rotations in space, and Lorentz transformations, which suggest physics Lie algebra, with Lorentz generators M , translations P , and dilatations is invariant as well. The postulates of special relativity suggest that for µν µ D, and commutation relations physics to be invariant, these are the correct set of symmetry transformations. Leaving the form invariant means η is a fixed specified matrix, and [Mµν , Mρσ] = iηµρMνσ iηνρMµσ iηµσMνρ + iηνσMµρ (18.1) the Poincarétransformations leave (ds)2 = η dxµdxν invariant. But the − − µν [M ,P ] = iη P iη P (18.2) only physical law Einstein’s postulate about the invariant speed of light re- µν ρ µρ ν − νρ µ 2 quires is that for light in vacuum, ds = 0. Thus a change in coordinates [Pµ,Pν] = 0 (18.3) ′ ′ ρ ρ ρ for which gµν(x ) = h(x )gµν (x ) would still have that piece of physics un- [Pµ,D] = iPµ (18.4) 2 changed, though different observers would not agree on the mean ds lifetimes [Mµν ,D] = 0 (18.5) of muons. If we consider, in flat Minkowski space, an infinitesimal transformation As differential operators these may be represented by xµ x′ µ = f µ(x), and ask that two infinitesimally separated points with 2 → ′ 2 ∂ ds = 0 also have (ds ) = 0, we need M = iη xα (18.6) µν − µα ∂xν ∂f µ ∂f ν ∂ η dxρ dxσ = 0 whenever η dxρ dxσ =0. P = i (18.7) µν ∂xρ ∂xσ ρσ µ ∂xµ

µ µ ∂ ∂f µ D = ix (18.8) Writing ∂xρ =Λ ρ this is ∂xµ

ρ σ µ ν ρ σ ηρσdx dx = 0 implies ηµνΛ ρΛ σ dx dx =0. There is, however, another transformation which preserves the null di- µ µ xµ rections. Consider the inversion I : x y = x2 , for which the Jacobian When we were looking for Poincar´einvariance, we insisted that the two ∂yµ δµ x xµ 7→ matrix =Λµ = ν 2 ν , so sides be equal even if they were not zero, and that gave rise to the pseudo- ∂xν ν x2 − (x2)2 orthogonality of Λ, but here the requirement is less restrictive and harder µ to interpret. But if we consider only an inﬁnitesimal transformation Λ = 1 2 2 ρ η Λµ Λν = η δµx 2xµ x δνx 2xν x δµ + L ρ, the right hand side is µν ρ σ (x2)4 µν ρ − ρ σ − σ ρ µ 1 2 2 2 2 ηρσ ρ σ ρ σ ρ σ = 2 4 ηρσ(x ) 4xσxρx +4x xρ xσ = 2 2 ηρσ + ηµσLµ + ηρνLν dx dx =0=(Lσρ + Lρσ) dx dx =0 (x ) − (x ) 618: Last Latexed: April 25, 2017 at 9:45 207 208. Last Latexed: April 25, 2017 at 9:45 Joel A. Shapiro

µ ν ρ σ ρ σ d+1 ′ ρ so ηµν Λ ρΛ σdx dx vanishes whenever ηµν dx dx does. λ Fµν (x ) satisfy the same Maxwell’s equations with the modified source As D gives an infinitesimal scale change and I inverts, IDI = D. Note term j (xρ)= λd+2j (x′ ρ). Thus we may say that electromagnetism is scale − µ µ I acts as a Lorentz scalar, so [I, Mµν] = 0. But I does not commute with invariant. Pµ, and in fact gives us something new and exciting. That electromagnetism should be scale invariant should be expected be- Of course I is highly singular on the light cone, and is certainly not an cause the theory does not have any parameters with dimensions of length. infinitesimal transformation, but I2 = 1I, so I times an infinitesimal generator But a hint of a more surprising symmetry is familiar from the method of times I is an infinitesimal generator. Let images. In electrostatics, if Φ(~x) is a solution of Poisson’s equation 2Φ(~x)= 2 p ~x∇ ρ(~x), might we find a solution with Ψ(~x)=(x ) Φ(~y) where ~y = x2 ? Kµ := IPµI. (18.9) ∂ψ 2 p−1 2 p ∂yj ∂Φ µ µ −ib Kµ −ib Pµ = 2pxi (x ) Φ(~y)+(x ) Then e = Ie I maps ∂xi ∂xi ∂yj xµ xµ xµ + x2bµ (xµ + x2bµ) x2 2 p−1 2 p δij 2xixj ∂Φ µ µ = 2pxi(x ) Φ(~y)+(x ) x + b = 2 2 4 → x2 → x2 x2 → (xµ + x2bµ) x − x ∂yj 2 2 p−1 2 2 p−2 (xµ + x2bµ) ψ = 6p(x ) +4p(p 1)x (x ) Φ(~y) = xµ + bν (x2δµ 2xµx ) ∇ − i 1+2b xν + x2b2 ≈ ν − ν ν 2 p−1 δij 2xixj ∂Φ +4pxi(x ) 2 4 so as a differential operator, x − x ∂yj

2 p 2xiδij 6xj +2δijxi (2xixj)(4xi) ∂Φ 2 µ µ ∂ +(x ) − 4 4 + 6 Kν = i(x δν 2x xν) µ . (18.10) x − x x ∂yj − ∂x 2 2 p δij 2xixj δik 2xixk ∂ Φ From the definition, we see that +(x ) 2 4 2 4 x − x x − x ∂yj ∂yk −1 −2 ∂Φ −2 ∂Φ [Kµ,Kν] = I[Pµ,Pν]I =0 (18.11) = 2p(2p + 1)(x2)p Φ(~y) 4p(x2)p x 2(x2)p x − j ∂y − j ∂y [Mµν ,Kρ] = [Mµν ,IPρI]= I[Mµν ,Pρ]I = iηµρIPν I iηνρIPµ I j j − 2 = iη K iη K (18.12) 2 p−2 ∂ Φ µρ ν − νρ µ +(x ) δjk ∂yj ∂yk [Kµ,D] = [IPµ I,D]= IPµ ID DIPµ I = IPµ D I + IDPµ I − − (x2)−5/2 2 Φ(~y). = I [P ,D] I = iI P I = iK (18.13) − y − µ − µ − µ p=−→1/2 ∇ [K ,P ] = 2iη D 2iM (18.14) µ ν µν − µν So we see that a solution to the Laplace equation, transformed by inversion, So we see that by adding in the special conformal transformations we have a is also a solution, and in fact even the Poisson equation with a source ρ is 15 dimensional Lie algebra called the conformal symmetry group. a solution for a new ρ which transforms as a density ought (with fixed total charge). If we consider an arbitrary coordinate transformation 18.1 Maxwell’s Equations ′ ′ ′ ′ rµ r µ(rν), g dr µdr ν = g drµdrν → µν µν ρ Electromagnetic fields are described by the 4-vector Aµ(x ) and the field the metric tensor in the new coordinates is ρ strength F (x ) which is its antisymmetrized derivative. Under a scale trans- ρ σ µν ′ ∂x ∂x ρ ′ ρ ρ ′ ρ d ′ ρ ′ ρ formation x x = λx , new fields A (x ) = λ A (x ) and F (x ) = gµν = ′ ′ gρσ. → µ µ µν ∂x µ ∂x ν 618: Last Latexed: April 25, 2017 at 9:45 209 210. Last Latexed: April 25, 2017 at 9:45 Joel A. Shapiro

′ ′ ~ ~ If the new gµν(r ) has the same form, up to a symmetry of the theory, as class of flows is that of irrotational flow, for which V = 0, and so if the ′ ∇ × gµν, we can view r r not as a change of variables of a fixed physical system, region is simply connected, we may define a velocity potential Φ for which but as a map from→ one system to another under a symmetry transformation. V~ = ~ Φ. Then if the fluid is also incompressible, so that dρ/dt = 0 and ∇ Now in general gµν, as a symmetric D D matrix has D(D+1)/2 indepen- by continuity ~ ρV~ = 0, we have that Φ again satisfies Laplace’s equation. × ∇· dent elements, and it is unlikely that an arbitrary transformation will produce So the same tool, finding analytic maps, is useful for fluid flow, at least for such a change. But in two dimensions, if we treat r = (x, y) z = x + iy nonviscous irrotational incompressible fluids in two dimensions. ′ ∼ and r = (u, v) w = u + iv, something very special happens if the map The idea of conformal equivalence has many applications, one dear to me (x, y) (u, v) is∼ a complex analytic function z w, for which the Cauchy- in particular is in string theory. The states of string theory are described Riemann→ conditions → not as in field theory, with particle paths (Feynman diagrams) in space- ∂u ∂v ∂u ∂v ∂x ∂y ∂y ∂x times, but as mappings of the world-sheet, that is the world-surface of a = , = , and also the inverse = , = ∂x ∂y ∂y −∂x ∂u ∂v ∂u −∂v one-dimensional string traveling through time, into to full (possibly 10 or 26 dimensional) full space-time. Just as the length of a path embedding hold. Then ρ σ in space is independent of the parameterization of that path, the area of ′ ∂x ∂x g = g = A ρA σg , the world sheet may be expressed as an integral over two parameters, but µν ∂x′ µ ∂x′ ν ρσ µ ν ρσ it is independent of coordinate transformations of those parameters, and ρ ρ ∂x a b ∂x ∂x hence conformally invariant. As a particular case, the analogue of a one-loop where Aµ = ′ = with a = , b = . Now if we started off ∂x µ b a ∂u ∂v Feynman diagram for a closed string is a mapping of a torus into the full − ′ T T 2 2 in Euclidean space, with gµν = δµν , we have g = AgA = AA =(a b )1I, space-time. A torus is a two-dimensional surface so can be parameterized − and we see that we have just a dilation, so if our physics is scale invariant, by two variables wj, but, at least if embedded in 3-D Euclidean space, it is an arbitrary analytic transformation is a symmetry. not a flat surface, so gjk = δ . A two-dimensional surface may have several w 6 ij Consider an electrostatics problem of finding the electric potential inside curvatures when considered as embedded in three dimensions, for example, a container with specified potentials at each point of the wall. Let us consider on the inside of the torus, a path circling the hole is curved outside the this problem in two dimensions, which is physical if we really have a long dough of the donut but the circle of each cut you make if you slice it into cylinder with our 2-D space as cross section and we are looking for solutions three portions is curved around the dough. But these curvatures are not uniform in the third dimension. Then inside the container we have Laplace’s intrinsic to the surface. For example, a cylindrical surface is flat, in the sense equation that you can make it from a flat piece of paper, but a sphere or donut is not. 2 2 ∂ φ ∂ φ 2 1 ∂ µν ∂Φ So the intrinsic curvature of a cylinder is zero, of a sphere is positive, and of 2 + 2 =0= Φ= µ gw √gw ν , ∂x ∂y ∇ √gw ∂w ∂w a donut is positive in some places (further from the center) and negative in where the first expression assumes cartesian coordinates (x, y), but the second others. µν µν µν is good in any coordinates. If gw is proportional to δ , it is δ /√gw, as But as a two dimensional surface has only one curvature degree of free- gw := det gw µν. But then we see that, even though gw is a function of ~r, the dom, and that can be modified by a position-dependent scale transformation, 2 ∂2φ ∂2φ term between the derivatives is not, and Φ=0 2 + 2 = 0. any two dimensional surface is conformally equivalent, locally, to a flat sur- ∇ ⇔ ∂u ∂v So if we solve the electrostatic problem in any region with any boundary face. Of course globally there may be problems. For the torus, we have potential specified, and if we can map this region with an analytic function periodic conditions in both parameters, so the conformal mapping is to a into another region, we have solved the problem for this new region as well. parallelogram with opposite sides identified. A rectangle will not work in This has an application to fluid flow as well. Consider the motion of a general, because the na¨ıve map from the rectangle into the torus might not fluid in two dimensions, with a velocity V~ (~r) at each point. An important preserve the angles given by the torus’ metric (Note that conformal maps 618: Last Latexed: April 25, 2017 at 9:45 211 212. Last Latexed: April 25, 2017 at 9:45 Joel A. Shapiro keep angles unchanged, being locally just a scale transformation).

So all (simple) toruses are conformally equivalent to some parallelogram, with some ratio of one side to the other and some speciﬁc angle. Considering the edges as complex numbers, this may be restated as one complex parameter τ which is the ratio of one edge to the other. But notice that changing which edge is in the denominator is equivalent to mapping τ 1/τ, and also, because the parallelogram represents periodic boundary conditions,→ − we may consider this a period lattice, and note that replacing the numerator edge by the numerator edge plus the denominator edge, (or τ τ + 1) simply changes to an equivalent unit cell on the lattice. So our→ theory is invariant under the modular group (see homework 4, problem 1).

In field theory we sum over Feynman graphs integrating over positions of all vertices, using the functional integral interpretation of quantum mechan- ics. In string theory we should integrate over all conformally inequivalent toruses, or roughly speaking, over the two dimensional parameter τ. But we should not be integrating over the equivalent configurations described by the Modular group, so we need to factor out the modular equivalent configurations. Thus the correct integration over loop parameters for the closed string is to integrate over one “fundamental region” of the modular group.

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