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faculteit Wiskunde en Natuurwetenschappen

Nonrelativistic .

Bacheloronderzoek Wis- en Natuurkunde, deel 1: Wiskunde Augustus 2012 Student: P.L.Los Begeleider natuurkunde: D. Boer

Begeleider wiskunde: J. Top 1 Introduction

Symmetry is, in physics, the invariance of properties of a physical system under coordinate trans- formations of this system. This often can be used to deduce properties of the system. The main purpose of this bachelor’s thesis is, besides giving a general mathematical description of continuous symmetries and Lie groups, to provide the correct mathematical language to study some questions about different physical systems with nonrelativistic symmetries. One of these questions, which will be studied not in this thesis but in a continuation on it, is: What is the exact physical meaning of the central extension with a certain generator M of the Lie algebra of the Schr¨odingergroup? It has something to do with mass, but what does it mean exactly? Similar questions are about the physical meaning of subgoups of the Lorenz group and very spe- cial relativity and about examples of nonrelativistic physical systems and systems with Lifshitz symmetry. In this thesis we first will descibe what symmetry exactly is. Then we will consider continuous symmetries: what are they and how to describe them mathematically? This will bring us to the important concepts of topological group, and Lie algebra. Further we want to describe Noether’s theorem about the relation between symmetry and conser- vation laws. As an important physical example of a system with continuous symmetry, we will derive the maximal symmetry group of a free nonrelativistic point particle, which will be found to be the Schr¨odingergroup. This example is important in studying the question above about the physical meaning of a central extension of the Schr¨odingergroup. Finally we will discuss a list of some common groups and relations between them.

2 What is symmetry?

In physics, symmetry is invariance of certain properies of a system under certain transformations of this system. For example: the area of a rectangle (not a square) is invariant under every rota- tion of this rectangle, the rectangle itself is only symmetric under rotations of 180 degrees.

y y

x x

(x, y) −→ (x cos θ − y sin θ, y cos θ + x sin θ)

Rotation of a rectangle: The rectangle is not invariant under this rotation, but the area of the rectangle stays invariant. Many, but not all symmetries are geometrical symmetries, which means that the symmetry trans- formations are space-time coordinate transformations (for example space-time translations, rota- tions, scalings).

The different invertible symmetry transformations of a system form a group: The identity transformation is the identity element of this group, the composition of two invertible symmetry transformations is again an invertible symmetry transformation, and the inverse of a symmetry transformation is a symmetry transformation.

2 2.1 Continuous symmetry and Lie groups: What is continuous symmetry, and how can continuous symmetries be described by Lie groups? Continuous symmetry can be imagined as invariance under “very small” coordinate transforma- tions, in contrast to discrete symmetry such as for example reflection symmetry. More precisely: continuous symmetry is invariance under transformations which depend continu- 2 ously on a set of parameters. For example the translations (x, y) 7→ (x + α1, y + α2) in R depends continuous on the parameters α1 and α2. Like ordinary symmetry transformations form an ordinary group, the symmetry transformations of a continuous symmetry form a topological group: A topological group is a topological space in which the elemens form a group, where the compo- sition and the inversion map are continuous functions. Example: (simplified) gravity: Under which spatial coordinate transformations is the gravitational energy of an object invariant? Suppose the gravitational energy of an object is given by the following volume integral: R 3 E = − V (ρ(~x) ∗ g ∗ z)d~x , with ρ(x, y, z) > 0 the mass density, and g the gravitation constant. Then for a coordinate transformation of the form (x, y, z) → (x − ∆x, y − ∆y, z − ∆z) (a trans- ˜ R lation), we have for the gravitational energy in the transformed system: E = − V (˜ρ(x − ∆x, y − ∆y, z −∆z)∗g∗(z −∆z))d~x3 withρ ˜ the mass density function in the transformed system. Because the object itself is not changed, we haveρ ˜(x − ∆x, y − ∆y, z − ∆z) = ρ(x, y, z), and hence: Z E˜ = − (ρ(x, y, z) ∗ g ∗ (z − ∆z))d~x3 V Therefore, the gravitational energy is invariant under this transformation if and only if there is no translation in the z-direction. In formulas: ˜ R 3 E = E ⇐⇒ V (ρ(~x)∆z)d~x = 0 ⇐⇒ ∆z = 0. Hence this gravitational energy is invariant under translation transformations which leave z in- variant, these symmetry transformations depend continuously on the parameters ∆x and ∆y. The topological group of symmetry transformations of a continuous symmetry often is a Lie-group. A Lie group is a smooth manifold, from which the elements form a group where the composition and inversion map are not only continuous but even smooth functions.

2.2 Group Generators and the Lie Algebra A system with symmetry transformations which form a Lie group is invariant under “infinitesimal small transformations”. Such infinitesimal small transformations can be described by linearising the symmetry transformations of the system around the identity transformation. Suppose we ~ ~ have a coordinate transformation ~x → ξ(~x,α1, α2, ...), where ξ depends differentiable on the set of ~ parameters ~α = α1, α2,... , and ξ(~x,~0) = ~x is the identity transformation. We can linearise this transformation around the identity transformation, the linearized transformation ~x → ζ~(~x,~α) is ~ ~ ~ ∂ξ~(~x,~a) then given by: ζ(~x,~α) = ξ(~x, 0) + ( ∂~a |~a=~0)~α. The infinitesimal generators of the group are now given by:

n o ζ~(~x, 1, 0,..., 0) − ~x, ζ~(~x, 0, 1,..., 0) − ~x, . . . , ζ~(~x, 0, 0,..., 1) − ~x ( ) ∂ξ~(~x,~α) ∂ξ~(~x,~α) ∂ξ~(~x,~α) = ( | )(1, 0,..., 0), ( | )(0, 1,..., 0),..., ( | )(0, 0,..., 1) ∂α ~α=~0 ∂α ~α=~0 ∂α ~α=~0

This set of generators can be seen as a basis of a vectorspace V which is the tangent space to the Lie group in ~x. We now will give the definition of Lie algebra to be able to describe what the Lie algebra of a group is. A Lie algebra s is a vectorspace X over some field F together with some operation [., .]: X×X → X

3 which is called the Lie bracket and which satisfies the following properties: Bilinearity: [α~x+β~y, ~z] = α[~x,~z]+β[~y, ~z] and [~z, α~x+β~y] = α[~z, ~x]+β[~z, ~y] ∀α, β ∈ F, ∀~x,~y, ~z ∈ X; Alternating property: [~x,~x] = 0; Jacobi identity: [[~x,~y], ~z] + [[~y, ~z], ~x] + [[~z, ~x], ~y] = 0.

For a matrix group, the Lie algebra is given by the vectorspace spanned by the infinitesimal generators of the group, with the Lie bracket given by the commutator: [A, B] = AB − BA.

2.3 Example: Rotations (in 3 dimensions): As an example of a Lie group and its Lie algebra we will look at the group of rotations in 3 dimensions: An arbitrary rotation in 3 dimensions is given by the coordinate transformation: ~x → A~x, with A an orthogonal matrix with det(A) = 1. Which is: A ∈ SO(3) = {A ∈ GL(3) | AT A = I, det(A) = 1}. (GL(3) is the general linear group in 3 dimensions, which is: the group of all 3x3 invertible ma- trices, with matrix multiplication as group operation) The group of all rotations in 3 dimensions is a 3-dimensional Lie group. We want to linearize this group around the identity element to get the Lie algebra of this group. A rotation over an angle ϕz around the z-axis is given by:   cos ϕz − sin ϕz 0 sin ϕz cos ϕz 0 0 0 1

We can linearise this with respect to ϕz:

 cos ϕ − sin ϕ 0   1 −ϕ 0 ∂ z z z I + ϕ sin ϕ cos ϕ 0 | = ϕ 1 0 z ∂ϕ  z z  ϕz =0  z  z 0 0 1 0 0 1

The same we can do for rotations ϕx and ϕy around resp. the x- and the y-axis. Hence, lineari- sation of the rotations around the identity gives:   1 −ϕz ϕy  ϕz 1 −ϕx −ϕy ϕx 1

In - more or less - the same way, the infinitesimal generator of rotations around the z-axis is given by: cos ϕ − sin ϕ 0 0 −1 0 ∂ z z J = sin ϕ cos ϕ 0 | = 1 0 0 z ∂ϕ  z z  ϕz =0   z 0 0 1 0 0 0 And analogue the infintitesimal generators around the x- and the y-axis are given by:

0 0 0   0 0 1 Jx = 0 0 −1 ,Jy =  0 0 0 0 1 0 −1 0 0

The Lie algebra of the group of rotations is generated by Jx, Jy and Jz with the commutation relations:

4 [Jx,Jy] = JxJy − JyJx 0 0 0   0 0 1  0 0 1 0 0 0  = 0 0 −1  0 0 0 −  0 0 0 0 0 −1 0 1 0 −1 0 0 −1 0 0 0 1 0 0 −1 0 = 1 0 0 0 0 0

= Jz

Similarly: [Jy,Jz] = Jx and [Jz,Jx] = Jy.

3 Noether’s Theorem[24],[32],[12],[35] 3.1 Noether’s Theorem and its proof Noether’s Theorem is about the relation between symmetry and conservation laws. Intuitively, it makes sense that often it is possible to derive conservation laws from the continuous symmetry of a system: if there is something invariant under symmetry transformations, then there is something conservated. There are different versions of Noether’s Theorem. Here we will describe one of them. First we will look what Noether’s Theorem says, then we will give a proof of it, and finally we look at some examples where it is possibe to apply this theorem. Noether’s Theorem (in the here described form) says that invariance of the action of a particle under a continuous tranformation implies a conserved quantity of the system: The action of a particle is given by R L(~q, ~q,˙ t) where L (as a function of ~q, ~q˙ and t) is the La- grangian of the particle (which can be see as the difference between potential and kinetic energy), ˙ d~q and ~q(t) is the coordinates of the particle at time t. ~q = dt (in general, a dot above a function is used to be the total time derivative). Noether’s Theorem: If the Action of a particle R L(~q, ~q,˙ t) is symmetric under a coordinate transformation of the form ∆ = ξ ∂ +ζ~ ∂ +ζ~t ∂ , then ( ∂L ~q˙−L)ξ − ∂L ζ~ is constant (for ~q(t) a solution of the Euler-Lagrange ∂t ∂~q ∂~q˙ ∂~q˙ ∂~q˙ equations). Where ∆ = ξ ∂ + ζ~ ∂ + ζ~t ∂ is an infinitesimal transformation. ∂t ∂~q ∂~q˙ The invariance of the action under ∆ we can better describe in the following way: The action ˜ is invariant under a coordinate transformation T (ε):(t, ~q, ~q˙) → (t˜(ε), ~q˜(ε), ~q˙(ε)) which we can linearize by (t, ~q, ~q˙) → (t + εξ, ~q + εζ,~ ~q˙ + εζ~t). ˜ Because in the transformed system ~q˙(ε) has to be the derivative of ~q˜(ε) with respect to t˜(ε), we can derive that ζ~t has to depend on ζ~ and ξ in the following way:

5 ∂ ∂ ∂(~q + εζ~) | (~q˙ + εζ~t) = | ( ) ∂ε ε=0 ∂ε ε=0 ∂(t + εξ) ∂ ∂(~q + εζ~) ∂t = | ( ) ∂ε ε=0 ∂t ∂(t + εξ) ∂ ∂ζ~ 1 = | ((~q˙ + ε ) ) ∂ε ε=0 ∂t dξ 1 + ε dt ∂ ∂ζ~ dξ dξ ∂ζ~ = | (~q˙ + ε − ε ~q˙ − ε2 ), ∂ε ε=0 ∂t dt dt ∂t ∂ζ~ dξ ⇒ ζ~t = − ~q˙ ∂t dt Now we will give the proof of Noether’s theorem: We start with requiring the action invariant under ∆, which is (as described above): Z Z Z ˜ L(~q, ~q,˙ t)dt = T (ε)( L(~q, ~q,˙ t)dt) = L(~q˜(ε), ~q˙(ε), t˜(ε))dt˜(ε)

Differentiating with respect to t to get rid of the integral gives: ˜ d(t˜(ε)) L(~q, ~q,˙ t) = L(~q˜(ε), ~q˙(ε), t˜(ε)) dt Linearizing this with respect to ε, setting ε to zero, and substracting L(~q, ~q,˙ t) gives: ∂ d(t + εξ) 0 = (L(~q + εζ,~ ~q˙ + εζ~t, t + εξ) )| ∂ε dt ε=0 This we can work out to get something from which we can with use of the Euler-Lagrange equations derive Noether’s theorem: (This looks like a long derivation, but there does not happen much:) ∂ d(t + εξ) d(t + εξ) ∂ 0 = L(~q + εζ,~ ~q˙ + εζ~t, t + εξ) | + L(~q + εζ,~ ~q˙ + εζ~t, t + εξ)| ∂ε dt ε=0 dt ∂ε ε=0 dξ ∂L(~q, ~q,˙ t) ∂L(~q, ~q,˙ t) ∂L(~q, ~q,˙ t) = L(~q, ~q,˙ t) + ζ~ + ζ~t + ξ dt ∂~q ∂~q˙ ∂t dξ ∂L ∂L ∂ζ~ dξ ∂L = L + ζ~ + ( − ~q˙) + ξ dt ∂~q ∂~q˙ ∂t dt ∂t dξ ∂L ∂L ∂L ∂L ∂L ∂L ∂L ∂ζ~ dξ = L + ( + ~q˙ + ~q¨)ξ − ( ~q˙ + ~q¨)ξ + ζ~ + ( − ~q˙) dt ∂t ∂~q ∂~q˙ ∂~q ∂~q˙ ∂~q ∂~q˙ ∂t dt d(Lξ) ∂L ∂L ∂L ∂L ∂ζ~ dξ = − ( ~q˙ + ~q¨)ξ + ζ~ + ( − ~q˙) dt ∂~q ∂~q˙ ∂~q ∂~q˙ ∂t dt d(Lξ) ∂L ∂L d(~qξ˙ ) dξ ∂L ∂L ∂ζ~ dξ = − ~qξ˙ − ( − ~q˙ ) + ζ~ + ( − ~q˙) dt ∂~q ∂~q˙ dt dt ∂~q ∂q˙ ∂t dt d(Lξ) ∂L ∂L ∂L ∂ζ~ d(~qξ˙ ) = − ~qξ˙ + ζ~ + ( − ) dt ∂~q ∂~q ∂~q˙ ∂t dt d(Lξ) ∂L ∂L d ∂L d ∂L = − ~qξ˙ + ζ~ + ( (ζ~ − ~qξ˙ )) − (ζ~ − ~qξ˙ ) dt ∂~q ∂~q dt ∂~q˙ dt ∂~q˙ ∂L d ∂L ∂L d ∂L d ∂L ∂L = ζ~ − ζ~ − ~qξ˙ + ~qξ˙ + (Lξ + ζ~ − ~qξ˙ ) ∂~q dt ∂~q˙ ∂~q dt ∂~q˙ dt ∂~q˙ ∂~q˙ ∂L d ∂L d ∂L ∂L = ( − )(ζ~ − ~qξ˙ ) + ((L − ~q˙)ξ + ζ~) ∂~q dt ∂~q˙ dt ∂~q˙ ∂~q˙

6 By the Euler-Lagrange equations we have ∂L − d ∂L = 0. Substituting this in the last equation ∂~q dt ∂~q˙ gives: ⇒ 0 = d ((L − ∂L ~q˙)ξ + ∂L ζ~) dt ∂~q˙ ∂~q˙ ⇒ ( ∂L ~q˙ − L)ξ − ∂L ζ~ is constant. ∂~q˙ ∂~q˙ Therefore, if the action is invariant under the infinitesimal transformation ∆, which is the same as invariance under a transformation T (ε) which can be linearized by (t, ~q, ~q˙) → (t+εξ, ~q+εζ,~ ~q˙+εζ~t), then ( ∂L ~q˙ − L)ξ − ∂L ζ~ is constant (for all ~q(t) satisfying the Euler-Lagrange equations). ∂~q˙ ∂~q˙

3.2 Examples of Noether’s Theorem 1 ˙ 2 Suppose for a particle, the Lagrangian is given by L = 2 m|~q| − V (~q). If the action stays invariant under the following coordinate transformations, then we can use Noether’s Theorem to derive conserved quantities from this transformations: ⇒ For ∆ = ξ ∂ + ζ~ ∂ + ζ~t ∂ with ζ~ = 0 and ξ constant, we have by Noether’s theorem: ∂t ∂~q ∂~q˙ ∂L ~q˙ − L is constant. ∂~q˙ 1 ˙2 ⇒ 2 m~q − V (~q) is constant, ⇒ conservation of Energy. ⇒ For ∆ = ξ ∂ + ζ~ ∂ + ζ~t ∂ with ξ = 0 and ζ~ constant, we have by Noether: ∂t ∂~q ∂~q˙ ∂L is a constant quantity. ∂~q˙ ⇒ m~q˙ is constant, ⇒ conservation of momentum. ⇒ For ∆ = ξ ∂ + ζ~ ∂ + ζ~t ∂ with ξ = 0, ζi = −qj, ζj = qi for some i 6= j, and ζk = 0 ∀k 6∈ {i, j} ∂t ∂~q ∂~q˙ (using qi and ζi for the i-th component of ~q respectively ζ~), Noether’s theorem gives: ∂L ∂L − (−qj) − qi is constant. ∂q˙i ∂q˙j ⇒ m(q ˙iqj − q˙jqi) is constant. ⇒ conservation of angular momentum.

4 The maximal symmetry of a nonrelativistic free point particle

An example of an system which has Schr¨odinger-symmetry (At most points following the paper of Jahn and Shreedhar[22]; we provide more details compared to [22].) We want to describe what the maximal symmetry of a simple nonrelativistic free point particle is, which is: what the maximal group of symmetry transformations is under which such a particle remains invariant. Such a free point particle is completely described by its action. If the action of this particle is “form invariant” under a transformation, the particle itself is also invariant under this transformation. What we mean with “form invariant” will be explained more precisely below. In this section, we start with taking the action of a nonrelativistic free point particle form invariant under a continuous coordinate transformation (~x,t) → (ξ,~ τ), and then from this invariance derive constrains on ξ~(~x,t) and τ(~x,t), until we know the group of all transformations under which the particle remains invariant. And we will see that this is exactly the Schr¨odingergroup (see the list

7 of groups at the end of this thesis). dτ Because the transformation has to be invertible and continuous, dt has to be positive. The action of a nonrelativistic free point particle with mass m and position ~x(t) is given by:

Z tf 1 X S = m (x ˙ 2)dt 2 i t=ti i

dx (Wherex ˙ := dt , and xi is the i-th component of the vector x. In the rest of this section we will use sometimes this, and sometimes the vector notation. ) ti is the initial time of the system, and tf the final time. To derive the maximal symmetry group of this particle, we have to look under which coordinate transformations this action stays forminvariant. Which means: For which coordinate transforma- tions (~x,t) → (ξ,~ τ) do we have for the action: A = S + C for all ~x(t)?, with A the action in “the   τ(t )  2 transformed system”: A = R f 1 m P dξi dτ, and with C a free constant. τ=τ(ti) 2 i dτ Suppose the action is form invariant under the coordinate transformation (~x,t) → (ξ,~ τ), this means: ! ! Z tf  2 Z τ(tf )  2 1 X dxi 1 X ξi m dt = C + m dτ ∀~x(t) 2 dt 2 dτ t=ti i τ=τ(ti) i Which constrains on the coordinate transformation can we derive from this formula? First we want to get rid of the integrals in this formula. This we can do by transforming this formula into one with integrals over arbitrary intervals (a, b) in [ti, tf ]: Because the form invariance of the action has to hold for all possible paths ~x(t), we can choose ~x1(t) fixed, and choose ~x2(t) = ~x1(t) for t 6∈ [a, b], ~x2 arbitrary for t ∈ (a, b). For both of these paths ~x1(t) and ~x2(t) the action has to be form invariant. Because these paths differ only on the interval (a, b), we can compere the equations for both paths and end up with an equation with integrals over the interval (a, b) which we had chosen arbitrary:

Z b  2! 1 X dx2 m i dt 2 dt t=a i ! ! ! ! Z b  2 Z tf  2 Z tf  2 1 X dx1 1 X dx2 1 X dx1 = m i dt + m i dt − m i dt 2 dt 2 dt 2 dt t=a i t=ti i t=ti i ! ! ! Z b  2 Z τ(tf )  2 Z τ(tf )  2 1 X dx1 1 X dξ2 1 X dξ1 = m i dt + C + m i dτ − C − m i dτ 2 dt 2 dτ 2 2 dτ 1 t=a i τ2=τ(ti) i 2 τ1=τ(ti) i 1 ! ! ! ! Z τ( ~x1(b),b)  2 Z b  2 Z τ( ~x1(b),b)  2 1 X dξ2 1 X dx1 1 X dξ1 = m i dτ + m i dt − m i dτ 2 dτ 2 2 dt 2 dτ 1 τ2=τ( ~x1(a),a) i 2 t=a i τ1=τ( ~x1(a),a) i 1

Because the right hand term is an integral over (a, b) which does only depend on the path x1(t), which is fixed,

there exists a function F (~x,t) such that this right hand term is equal to F ( ~x1(b), b) − F ( ~x1(a), a) ∀a, b ⇒ ! Z τ( ~x1(b),b)  2 1 X dξ2 = m i dτ + (F ( ~x (b), b) − F ( ~x (a), a)) 2 dτ 2 1 1 τ2=τ( ~x1(a),a) i 2 Z b  2! 1 X dξ2 dτ2 = m i dt + (F ( ~x (b), b) − F ( ~x (a), a)) 2 dτ dt 2 2 t=a i 2 Z b  2! ! 1 X dξ2 dτ2 dF ( ~x2(t), t) = m i + dt 2 dτ dt dt t=a i 2

8 This does not depend on ~x1(t), hence we have for all ~x(t) on an arbitrary interval (a, b):

Z b  2! Z b  2! ! 1 X dxi 1 X dξi dτ dF (~x(t), t) m dt = m + dt, 2 dt 2 dτ dt dt t=a i t=a i ⇒ for an coodinate transformation which leaves the action form invariant we have:  2!  2! X dxi(t) X dξi(~x(t), t) dτ(~x(t), t) 2 dF (~x(t), t) = + ∀~x(t). dt dτ(~x(t), t) dt m dt i i In this equation we do not have integrals anymore. We want to simplify it such that we get rid of dξ the derivative dτ , and are left with only partial derivatives:  2!  2! X dxi(t) X dξi(~x(t), t) dτ(~x(t), t) 2 dF (~x(t), t) ⇒ = + ∀~x(t) dt dτ(~x(t), t) dt m dt i i  2 P dξi(~x(t),t) i dt 2 dF (~x(t), t) = + dτ(~x(t),t) m dt dt  2 P P ∂ξi dxj ∂ξ   i j ∂x dt + ∂t j 2 X ∂F dxj ∂F = dx +  +  P ∂τ j + ∂τ m ∂xj dt ∂t j ∂xj dt ∂t j

dτ Because dt 6= 0 we can write this as:  2   2      dx  P dxi  − 2 P ∂F dxi + ∂F P ∂τ dxi + ∂τ = P P ∂ξi j + ∂ξ i dt m i ∂xi dt ∂t i ∂xi dt ∂t i j ∂xj dt ∂t And this has to hold for all ~x(t). d~x(t) Because this equation holds for all ~x(t), we may choose dt arbitrary, and if we look at a fixed ~ d~x(t) time t, then ~x(t) and ξ does not depend on dt . Hence we can subdivide the left and the right side of this equation in parts with different powers d~x(t) of the different components of dt . This leads to a lot of conditions which the coordinate trans- formation has to satisfy to leave the particle invariant: dx 2  dx  Terms with ( dxi )2 j :(∀i, j) Left: dxi  ∂τ j . Right: 0. dt dt dt ∂xj dt ⇒ ∂τ = 0 ∀j. ∂xj Hence the time in the transformed system τ(~x,t) depends only on t and not on ~x ! dxi 2 Terms with ( dt ) : 2      2 Left: dxi  ∂τ  − 2 ∂F dxi ∂τ dxi . Right: P ∂ξ dxi . dt ∂t m ∂xi dt ∂xi dt j ∂xi dt  2 ∂τ = 0, so we find: ∂τ  = P ∂ξj (∀i). ∂xi ∂t j ∂xi dx dx dx Terms with dxi j , i 6= j: Left: − 2 ∂F dxi ∂τ j − 2 ∂F j ∂τ dxi . Right: dt dt m ∂xi dt ∂xj dt ∂xj dt ∂xi dt dx P 2 ∂ξk dxi ∂ξk j . k ∂xi dt ∂xj dt ⇒ − 1 ∂F ∂τ − ∂F ∂τ = P ∂ξk ∂ξk m ∂xi ∂xj ∂xj ∂xi k ∂xi ∂xj If we use here ∂τ = ∂τ = 0 we find: 0 = P ∂ξk ∂ξk (∀i, j : i 6= j). ∂xi ∂xj k ∂xi ∂xj      ∂ξ   ∂ξ  Terms with ( dxi )1: Left: − 2 ∂F dxi ∂τ − 2 ∂F  ∂τ dxi . Right: P 2 j dxi j . dt m ∂xi dt ∂t m ∂t ∂xi dt j ∂xi dt ∂t       Because ∂τ = 0 this is: − 1 ∂F ∂τ  = P ∂ξj ∂ξj (∀i). ∂xi m ∂xi ∂t j ∂xi ∂t  2 dxi 0 2 ∂F  ∂τ  P ∂ξj Terms with ( dt ) : Left − m ∂t ∂t . Right: j ∂t . ⇒

9  2 2 ∂F  ∂τ  P ∂ξj − m ∂t ∂t = j ∂t .

We now have found a lot of constraints on the symmetry transformation which we want to combine and simplify.  2   ∂τ P ∂ξl P ∂ξl ∂ξl By differentiating = (∀i) and = 0 (∀i 6= j) with respect to xk, ∂t l ∂xi l ∂xi ∂xj we find:

 2   2 2  X ∂ ξl ∂ξl X ∂ ξl ∂ξl ∂ ξl ∂ξl 2 = 0 (∀i, k) and + = 0 (∀i, j, k : i 6= j), ∂xi∂xk ∂xi ∂xi∂xk ∂xj ∂xj∂xk ∂xi l l Combining this we find:  2 2  X ∂ ξl ∂ξl ∂ ξl ∂ξl ⇒ + = 0 (∀i, j, k), ∂xi∂xk ∂xj ∂xj∂xk ∂xi l  2   2   2   2  X ∂ ξl ∂ξl X ∂ ξl ∂ξl X ∂ ξl ∂ξl X ∂ ξl ∂ξl ⇒ = − = − = ∂xi∂xk ∂xj ∂xj∂xk ∂xi ∂xk∂xj ∂xi ∂xi∂xj ∂xk l l l l  2   2   2  X ∂ ξl ∂ξl X ∂ ξl ∂ξl X ∂ ξl ∂ξl = = − = − (∀i, j, k) ∂xj∂xi ∂xk ∂xk∂xi ∂xj ∂xi∂xk ∂xj l l l And hence we can simplify this:  2  X ∂ ξl ∂ξl ⇒ = 0 (∀i, j, k) ∂xi∂xk ∂xj l

  ∂a1 ∂a1 ··· ~ ! ∂b1 ∂b2 ∂~a(b) ∂a2 ∂a2 ··· From here, sometimes the following notation is used: :=  ∂b1 ∂b2   ∂~b  . . .  . . ..

We now want to show that ξ~(~x,t) depends only linearly on the components of ~x:

!T ! ∂ξ~ ∂ξ~ ∂τ We already have: = ( )I, (With I the identity matrix) ∂~x ∂~x ∂t !T !! ∂ξ~ ∂ ∂ξ~ and: = O (∀i), (With O the null matrix) ∂~x ∂xi ∂~x !T !! ∂ξ~ ∂ξ~ ∂ ∂ξ~ ∂τ ⇒ + = ( )I (∀i), ∂~x ∂~x ∂xi ∂~x ∂t

!!  !T −1 ! ∂ξ~ ∂ ∂ξ~ 1 ∂ξ~ ∂ξ~ ⇒ + = = (∀i) ∂~x ∂x ∂~x  ∂τ ∂~x  ∂~x i ∂t ! ∂ ∂ξ~ ⇒ = O (∀i) ∂xi ∂~x

2 And hence ∂ ξk = 0 ∀i, j, k, therefore ξ~ depends only linearly on the different components of ~x! ∂xi∂xj To write this in matrix notation: ∂ξ~ ⇒ ∂~x = A for some matrix A which depends only on t ⇒ ξ~ = A~x + ζ~ for some function ζ~ of t. But for this matrix A we already have:

10 T T  ∂ξ~   ∂ξ~  ∂τ A A = ∂~x ∂~x = ( ∂t )I √A Therefore R(t) defined by R(t) := ∂τ is an orthogonal matrix (which will turn out to be the ∂t rotation matrix of the rotational part of the symmetry transformation).

We now go further with the other equations we got above by comparing powers of dxi :   dt First we use − 1 ∂F ∂τ  to show that R(t) is constant: m ∂xi ∂t         1 ∂F ∂τ X ∂ξl ∂ξl − = m ∂xi ∂t ∂xi ∂t l         ! ∂ 1 ∂F ∂τ ∂ X ∂ξl ∂ξl ⇒ − = ∂xj m ∂xi ∂t ∂xj ∂xi ∂t l    2  X ∂ξl ∂ ξl = ∂xi ∂t∂xj l   X ∂Alj = A = (AT A˙) li ∂t ij l  2    T 1 ∂ F ∂τ T ⇒ (A A˙)ij = − = (A A˙)ji m ∂xi∂xj ∂t

⇒ RT R˙ = R˙ T R 1 ∂ 1 ∂ 1 ⇒ O = I = (RT R) = (RT R˙ + R˙ T R) = RT R˙ 2 ∂t 2 ∂t 2 ⇒ R˙ = O ⇒ R is constant. Hence for the spatial part of the symmetry transformation we find: r ∂τ ⇒ ξ~ = R~x + ζ~ with R a constant orthogonal matrix, and ζ~ a function of t. ∂t ∂τ We now will derive an equation with ∂t from which we can derive an equation for τ(t), and using this finally we will find an equation for ζ(t).

We know already:       − 1 ∂F ∂τ  = P ∂ξl ∂ξl m ∂xi ∂t l ∂xi ∂t   2 ⇒ − 2 ∂F  ∂τ  = P ∂ξl m ∂t ∂t l ∂t  We can compare these equations by differentiating the one with respect to t and the other with respect to xi:      −m P ∂ξl ∂ξl   ∂ l ∂xi ∂t = ∂t ∂τ ∂2F  ∂t   2  = −m P ∂ξl ∂xi∂t  ∂ l ∂t = ∂τ  ∂xi  2   ∂t       ∂2ξ  ∂ξ   ∂ξ  ∂2ξ  l l l l  ∂ξl  ∂ξl   ∂t∂x ∂t ∂x ∂t2 2  P i i ∂xi ∂t ∂ τ = −m l  ∂τ + ∂τ − ∂τ 2 ∂t2   ∂t ∂t ( ∂t )     ∂ξl ∂ ∂ξl   P ∂t ∂xi ∂t = −m l ∂τ  ∂t

11    2      2 X ∂ξl ∂ ξl ∂τ X ∂ξl ∂ξl ∂ τ ⇒ 2 = 2 ∂xi ∂t ∂t ∂xi ∂t ∂t l l Or, writing this in matrix form:

∂2ξ~ ∂τ ∂ξ~ ∂2τ ⇒ RT = RT ∂t2 ∂t ∂t ∂t2 Because R is orthogonal, we can get rid of it:

∂2ξ~ ∂τ ∂2ξ~ ∂τ ∂ξ~ ∂2τ ∂ξ~ ∂2τ ⇒ = RRT = RRT = ∂t2 ∂t ∂t2 ∂t ∂t ∂t2 ∂t ∂t2 q ~ ∂τ ~ Substituting ξ by ∂t R~x + ζ:

q  q  ∂2 ∂τ R~x + ζ~ ∂ ∂τ R~x + ζ~ ∂t ∂τ ∂t ∂2τ = ∂t2 ∂t ∂t ∂t2  q   q  ∂2 ∂τ ∂ ∂τ 2 ∂t ¨ ∂τ ∂t ˙ ∂ τ ⇒ R~x + ζ~ = R~x + ζ~  ∂t2  ∂t  ∂t  ∂t2

 2 2   3  ∂ τ   2  ∂ τ 2 ∂ τ 2 3 ∂t ¨ ∂τ 2 ˙ ∂ τ ⇒  ∂t −  R~x + ζ~ = ∂t R~x + ζ~ q 1  q  2  ∂τ ∂τ 1+ 2   ∂t ∂τ ∂t 2 ∂t 4 ∂t 2 ∂t

By comparing powers of ~x, we find:

 2 2  3  ∂ τ  2 ∂ τ 2 ∂ τ 2 3 ∂t ∂τ 2 ∂ τ  ∂t −  = ∂t q 1 q 2  ∂τ ∂τ 1+ 2  ∂t ∂τ ∂t 2 ∂t 4 ∂t 2 ∂t

and: ¨∂τ ˙ ∂2τ ζ~ = ζ~ ∂t ∂t2 The first of these two equations we can simplify, the second we will use later to derive an equation for ζ~(t). From the first one we find:

 2 2 ∂ τ 2 ∂3τ ∂τ ∂t2 ∂2τ  ⇒ − = ∂t3 ∂t 2 ∂t2 2 ∂3τ ∂2τ ! 3 3 2 ⇒ ∂t − ∂t = 0 ∂τ 2 ∂τ ∂t ∂t The left hand side of this equation is exactly the Schwarzian derivative of τ. We will now look for which functions the Schwarzian derivative is zero:

4.1 For which real functions the Schwarzian derivative is zero? 2 ∂3f  ∂2f  ∂x3 3 ∂x2 The Schwarzian derivative of a function f is given by: ∂f − 2 ∂f . We want to look for which ∂x ∂x funcions this derivative is zero: f 000 3 f 00 2 − = 0 f 0 2 f 0

12 f 00 0 1 f 00 2 f 000 f 00 2 1 f 00 2 f 000f 0 − 3 (f 00)2 ⇒ − = − − = 2 = 0 f 0 2 f 0 f 0 f 0 2 f 0 f 0 First we will consider the case in which f 00 6= 0: f 00 0 1 2 Set g = f 0 , then: g − 2 g = 0 −2 ⇒ g(x) = with c1 a constant. x+c1 f 00 −2 ⇒ 0 = . f x+c1 0 c2 0 ⇒ f = 2 with c1 and c2 constants, (f 6= 0). (x+c1) −c2 ⇒ f = + c3 with c1, c2, c3 constants. x+c1 c2 αx+β + c3 = with α = c3γ, β = (c1c3 − c2)γ, δ = c1γ, γ arbitrary 6= 0. x+c1 γx+δ 2 2 2 2 ⇒ αδ − βγ = c1c3γ − c1c3γ + c2γ = c2γ . 2 ⇒ we can choose γ such that |c2γ | = 1 0 If f > 0 then c2 > 0 and we can choose γ such that αδ − βγ = 1. 0 If f < 0 then c2 < 0 and we can choose γ such that αδ − βγ = −1.

00 αx+β 0 In the case f = 0, we have: f = δ . Because of f 6= 0 we can choose α, β and γ such that αδ = 1 for f 0 > 0 and αδ = −1 for f < 0. Now we have: If the Schwarzian derivative of f is zero, then:

αx + β  αδ − βγ = 1 and f 0 > 0, f = , with γx + δ or αδ − βγ = −1 and f 0 < 0

In the reverse direction, if any function f is given by:

αx + β  αδ − βγ = 1 and f 0 > 0, f = , with γx + δ or αδ − βγ = −1 and f 0 < 0 then it is easy to see that the Schwarzian derivative of this function is zero: α αx + β f 0 = − γ γx + δ (γx + δ)2 α αx + β ⇒ f 00 = −2γ + 2γ2 (γx + δ)2 (γx + δ)3 −2γ = f 0 γx + δ 2γ2 −2γ ⇒ f 000 = f 0 − f 00 (γx + δ)2 γx + δ 2γ2 −2γ −2γ = f 0 − f 0 (γx + δ)2 γx + δ γx + δ 3 4γ2 = f 0 2 (γx + δ)2 3 f 00 2 = f 0 2 f 0 f 000 3 f 00 2 ⇒ − = 0 f 0 2 f 0

Hence for τ(t), the time part of the symmetry transformation, we now have:

αx + β ⇒ τ = , with αδ − βγ = 1 γx + δ

13 This we will use to derive an equation for ζ: ∂τ α αt + β ⇒ = − γ ∂t γt + δ (γt + δ)2 αγt − γαt + αδ − γβ = (γt + δ)2 1 = (γt + δ)2 r ∂τ 1 ⇒ = ∂t γt + δ ⇒ For ζ we get:     ∂ αt+β ∂2 αt+β ¨ γt+δ ˙ γt+δ ζ~ = ζ~ ∂t ∂t2     ∂ α − γ αt+β ¨ α αt + β ˙ γt+δ (γt+δ)2 ⇒ ζ~ − γ = ζ~ γt + δ (γt + δ)2 ∂t ¨ α αt + β  ˙  αt + β γα  ⇒ ζ~ − γ = ζ~ 2γ2 − 2 γt + δ (γt + δ)2 (γt + δ)3 (γt + δ)2 ¨ αt + β  ˙  αt + β γα  ⇒ ζ~ α − γ = 2ζ~ γ2 − γt + δ (γt + δ)2 γt + δ ¨ −2γ ˙ ⇒ ζ~ = ζ~ γt + δ ∂ 1 −2γ We know that = ∂t (γt + δ)2 (γt + δ)3 ˙ c ⇒ ζ~ = for some constant c (γt + δ)2 −c/γ ⇒ ζ~ = + d for some constants c and d γt + δ ~vt + ~a ⇒ ∃~a,~v : ζ~ = γt + δ

We now have for any coordinate transformation (~x,t) → (ξ,~ τ) under which the action is form in- variant: r ∂τ R~x + ~vt + ~a ξ~ = R~x + ζ~ = ∂t γt + δ and αt + β τ = γt + δ with R orthogonal and αδ − γβ = 1. This is exactly the Schr¨odinger Symmetry (see the list of symmetry groups in the next section).

5 A list of some groups:

In this section we will give a list of some common groups and give some relations between them. The infinitesimal generators and Lie algebras of these groups can be obtained in the same way we did in the example of SO(3) (see section 2.3 above). In composing this list, the following

14 papers/texts have been useful: [1], [2], [3], [4], [5], [13], [14], [15], [17], [18], [20], [25], [29], [30], [31].

5.1 Relativistic Symmetries: There are different relativistic symmetries, which are well known in physics. They are all sym- metries of the Minkowski spacetime. This is a fourdimensional vectorspace with three spatial dimensions and one time dimension, were distances are given by plus or minus(by convention) (∆t)2 − |∆~x|2.

Group: SO(3) ⊂ Lorentzgroup ⊂ Poincar´egroup ⊂ Conformal Group Infinitesimal generators: Ji Ji,Ki Ji,Ki,Pµ Ji,Ki,Pµ,D,Cµ Dimension: 3D 6D 10D 15D (Where ...⊂... means: ... is a subgroup of ... .)

Ji: Rotation Ki: Boost Pµ: Space-Time-Translation D: Dilatation/Scaling Cµ: Inversion/Expansion Here we use i(,j,k,l,...) for the different spatial dimensions {1, 2, 3} (with Ji we mean for example {Jx, Jy, Jz}), and µ(,ν,σ,ρ,...) goes over all dimensions {0, 1, 2, 3} (this is common in physics). Now we will look at this different groups in a litte more detail:

5.1.1 Rotation group SO(3) We have already given the rotation group in three dimensions as an example of a Lie group with a Lie algebra existing of Jx, with the commutation relations: [Jx,Jy] = Jz,[Jy,Jz] = Jx,[Jz,Jx] = Jy. In physics it is common to write this as: [Ji,Jj] = εijkJk Where εijk, known as the Levi Civita symbol, is an total antisymmetric tensor: ε123 = ε231 = ε312 = 1 and ε132 = ε213 = ε321 = −1. Because rotations are only spatial transformations, the relativistic rotation group is the same as the ordinary transformation group in three dimensions.

5.1.2 Lorenz group (not connected): The total Lorenz group is given by all isometries of Minkowski spacetime which leave the origin invariant. It is a matrix Lie group which preserves the length: t2 − x2 − y2 − z2 (or −t2 + x2 + y2 + z2, you can use either of them by convention). t 1 0 0 0  x µ ν 0 −1 0 0  In physics this length of ~x =   is often written as: ηµν x x where η =   y 0 0 −1 0  z 0 0 0 −1 where summation over µ and ν is intended. Sometimes this is described as the Generalized Orthogonal Group in 3+1 dimensions: O(1, 3), where 3+1 is about the three - signs and one + sign in the formula for the distance between points.

Infinitesimal generators of the group: - Spatial Rotations Ji -Boosts Ki

15 -Parity (or space inversion) -Time reversal (or time inversion) -Spacetime inversion The boosts are given by matrices which are very like the rotation matrices Ji, they can be seen as a sort of “rotation” between one space dimension and time (in stead of between to space dimensions). Therefore, the infinitesimal generators for rotations and boosts we can also write in one equation as: Mµν = (xµ∂ν −xν ∂µ), which can be explained in the same way like the infinitesimal transformation ∂ ∆ in the section about Noether’s theorem (∂x := ∂x , in the rest of this list of groups we will use this notation). The total Lorenz group consists of four connected components. This is because space inversion and time inversion, and also spacetime inversion are discrete symmetries.

5.1.3 Restricted Lorenz group (6D) If we take one of the four connected components of the Lorenz group, the one containing the iden- tity, we have again a Lie group, which is called the restricted Lorens group, sometimes denoted by: SO+(1, 3) It is no surprise that this group preserves orientation of space and direction of time. The infinitesimal generators of the group are now given by: Mµν = (xµ∂ν − xν ∂µ).

5.1.4 Poincar´egroup (10D): If we extend the Lorenz group with the group of translations, we get the Poincar´egroup, which gives the full symmetry of special relativity. It consists of all isometries of Minkowski spacetime. In the same way, the restricted Lorenz group extended with the translations will give the restricted Poincar´egroup. Infinitesimal generators of the group: Lorenz transformations (Rotations and Boosts): Mµν = (xµ∂ν − xν ∂µ) Translations: Pµ = −∂µ

5.1.5 (Relativistic) Conformal group (15D)[5]: We can extend the Poincar´egroup to the (relativistic) conformal group, which can be seen as the group of all angle-preserving transformations. The infinitesimal generators of this group are given by: Lorenz transformations (Rotations and Boosts): Mµν = (xµ∂ν − xν ∂µ) Translations: Pµ = −∂µ Dilatations: D = −xµ∂µ Special conformal transformations: 2 ν Kµ = (x ∂µ − 2xµxν ∂ ) Nonzero commutation relations: [D,Kµ] = −Kµ, [D,Pµ] = Pµ,

16 [Kµ,Pν ] = 2ηµν D − 2Mµν , [Kµ,Mνρ] = (ηµν Kρ − ηµρKν ), [Pρ,Mµν ] = (ηρµPν − ηρν Pµ), [Mµν ,Mρσ] = (ηνρMµσ + ηµσMνρ − ηµρMνσ − ηνσMµρ).

5.2 Non-Relativistic Symmtries: Group: Galile¨ıgroup ⊂ Schr¨odingergroup Infinitesimal generators: H,Pi,Ji,Ki H,Pi,Ji,Ki,D,C Dimension: 10D 12D

The infinitesimal generators are given by: H: Timetranslation (x, t) → (x, t + s) −∂t Pi: Spacetranslation (x, t) → (x + a, t) −ai∂xi Ji: Rotation (x, t) → (Ax, t) xk∂xj − xj∂xk Ki: Boost (x, t) → (x − vt, t) vit∂xi 2 D: Dilatation/Scaling (x, t) → (λx, λ t) 2t∂t + xi∂xi x t 2 C: Inversion/Expansion (x, t) → ( 1−κt , 1−κt ) t + txi∂xi

5.2.1 Galilean group (10D): The Galilean group is the group under which Newton mechanics is invariant. Generators of the group: -Time Translation (H, Hamiltonian, (x, t) → (x, t + s), −∂t ) -Space Translation (Pi,(x, t) → (x + a, t), −ai∂xi ) -Spatial Rotation (Lij = Jk,(x, t) → (Ax, t), −(xi∂j − xj∂i)) -Galilean Boost (Ki,(x, t) → (x − vt, t), vit∂xi ) Commutation relations: [Pi,Pj] = 0, [Ki,Kj] = 0, [H,Pi] = 0, [H,Ki] = −Pi, [Ki,Pi] = 0, [H,Lij] = 0, ([H, Ji] = 0), [Lij,Pk] = (δikPj − δjkPi), ([Ji,Pj] = ijkPk), [Lij,Kk] = (δikKj − δjkKi), ([Ji,Kj] = ijkKk), [Lij,Lkl] = (δikLjl − δjlLjkδjkLil − δjlLik), ([Ji,Jj] = ijkJk),

5.2.2 Galilean conformal group (12D): (Extension of the Galilean Group, Group contraction of (relativistic) Conformal Group (with m → 0)) Generators of the group: -Time Translation (H, Hamiltonian, (x, t) → (x, t + s), −∂t ) -Space Translation (Pi,(x, t) → (x + a, t), −ai∂xi ) -Spatial Rotation (Lij = Jk,(x, t) → (Ax, t), −(xi∂j − xj∂i)) -Galilean Boost (Ki,(x, t) → (x − vt, t), vit∂xi ) 2 -Dilatation/scaling (D,(x, t) → (λx, λ t), t∂t + xi∂xi ) x t 1 2 -Expansion/inversion (C,(x, t) → ( 1−κt , 1−κt ), 2 t + txi∂xi ) Commutation relations:

17 [Pi,Pj] = 0, [Ki,Kj] = 0, [H,Pi] = 0, [H,Ki] = −Pi, [Ki,Pi] = 0, [H,Lij] = 0, [Lij,Pk] = (δikPj − δjkPi), [Lij,Kk] = (δikKj − δjkKi), [Lij,Lkl] = (δikLjl − δjlLjkδjkLil − δjlLik), [H,C] = D, [C,D] = −2C, [H,D] = 2H, [H,Ki] = −Pi, [Pi,D] = Pi, [Ki,D] = −Ki, [Pi,C] = −Ki, [Lij,C] = 0, [Lij,D] = 0,

5.2.3 Schr¨odingergroup (12D): This is the symmetry group we obtained as the maximal symmetry group of a free nonrelativistic point particle (see section 4 above). It is also the symmetry group under which the Schrdinger equation is invariant. (The Schr¨odingergroup is not the same as the Galilean Conformal Group ...) -Time Translation (H, Hamiltonian, (x, t) → (x, t + s), −∂t ) -Space Translation (Pi,(x, t) → (x + a, t), −ai∂xi ) -Spatial Rotation (Lij = Jk,(x, t) → (Ax, t), −(xi∂j − xj∂i)) -Galilean Boost (Ki,(x, t) → (x − vt, t), vit∂xi ) 2 -Dilatation/scaling (D,(x, t) → (λx, λ t), 2t∂t + xi∂xi ) x t 2 -Expansion/inversion (C,(x, t) → ( 1−κt , 1−κt ), t + txi∂xi )

5.2.4 Extended schr¨odingergroup (13D): (Schr¨odingergroup with central extension M) -Phase Transformation (M, compare [30]. To me it remains a question what this transformation means physically, this question will be studied in a continuation of this thesis).

5.2.5 Lifshitz algebra: Generators of the group: -Time Translation (H, Hamiltonian, (x, t) → (x, t + s)) -Space Translation (Pi,(x, t) → (x + a, t)) -Spatial Rotation (Lij,(x, t) → (Ax, t)) -Dilatation/scaling (D,(x, t) → (λx, λzt)) Commutation relations: [Lij,D] = 0, [Pi,D] = Pi, [H,D] = ziH.

18 6 conclusions

Symmetry is, in physics, the invariance of properties of a physical system under coordinate trans- formations of this system, and often can be used to deduce properties of the system. In this thesis first is explained how to describe symmetry in a mathematical way (section 2). Con- tinuous symmetries can be described by topological groups. Often these groups are Lie groups, which can be linearized to get the corresponding Lie algebra (section 2.1, 2.2). This is clarified by the example of rotations in three dimensions (section 2.3). Noether’s theorem about the relation between symmetry and conservation laws is proved (section 3.1), and there are some examples given in which it is used (section 3.2). As an example, the derivation of the maximal symmetry group of a nonrelativistic point particle is given in detail. This symmetry group has turned out to be the Schr¨odingergroup (section 4). Finally there is given a list of some common groups and some of the relations between them (section 5). As a continuation to this thesis, some questions about different nonrelativistic physical systems will be studied: What does central extension with M of the Schr¨odingerLie algebra physically mean? Which systems have Lifshitz symmetry? Which subgroups does the Lorenz group have, and what are the implications of this subgroups for the study of Lorenz symmetry breaking?

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