Chapter 5 Symmetries and Conservation Laws At a very fundamental level, physics is about identifying patterns of order in Na- ture. The inception of the field starts arguably with Tycho Brahe (1546-1601) — the first modern experimental physicist — and Johannes Kepler (1571-1630) — the first modern theoretical physicist. In the 16th century, Brahe collected large amounts of astrophysical data about the location of planets and stars with groundbreaking accuracy — using an impressive telescope set up in a castle. Kepler pondered for years over Brahe’s long tables of numbers until he could finally iden- tify a pattern underlying planetary dynamics: this was summarized in the three laws of Kepler. Later on, Isaac Newton (1643-1727) referred to these achieve- ments through his famous quote: “If I have seen further it is only by standing on the shoulders of giants”. Since then, physics has always been about identifying patterns in numbers, in measurements. And a pattern is simply an indication of a repeating rule, a constant attribute in complexity, an underlying symmetry. In 1918, Emmy Noether (1882–1935) published a seminal work that clarified the deep relations between symmetries and conserved quantities in Nature. In a sense, this work organizes physics in a clear diagram and gives one a bird’s eye view of the myriad of branches of the field. Noether’s theorem, as it is called, can change the way one thinks about physics in general. It is profound yet simple. While one can study Noether’s theorem in the Newtonian formulation of me- chanics through cumbersome methods, the subject is an excellent demonstration of the power of the new formalism we developed, Lagrangian mechanics. In this sec- tion, we use the variational principle to develop a statement of Noether’s theorem. We then prove it and demonstrate its importance through examples. 189 190 CHAPTER 5. SYMMETRIES AND CONSERVATION LAWS Figure 5.1: Two particles on a rail. EXAMPLE 5-1: A simple example Let us start with a simple mechanics problem from an example we saw in Section 4.1. We have two particles, with masses m1 and m2, confined to moving along a horizontal frictionless rail as depicted in Figure 5. The action for the system is 1 1 S = dt m q˙2 + m q˙2 V (q q ) (5.1) 2 1 1 2 2 2 − 1 − 2 Z ✓ ◆ where we are considering some interaction between the particles described by a potential V (q q ) that depends only on the distance between the particles. 1 − 2 Let us consider a simple transformation of the coordinates given by q10 = q1 +C ,q20 = q2 + C (5.2) where C is some arbitrary constant. We then have q˙10 =˙q1 , q˙20 =˙q2 . (5.3) Hence, the kinetic terms in the action are unchanged under this transformation. Furthermore, we also have q0 q0 = q q (5.4) 1 − 2 1 − 2 5.1. INFINITESIMAL TRANSFORMATIONS 191 implying that the potential term is also unchanged. The action then preserves its overall structural form under the transformation 1 2 1 2 S dt m q˙ + m q˙ V (q0 q0 ) . (5.5) ! 2 1 01 2 2 02 − 1 − 2 Z ✓ ◆ This means that the equations of motion, written in the primed transformed co- ordinates, are identical to the ones written in the unprimed original coordinates. We then say that the transformation given by (5.2) is a symmetry of our system. Physically, we are simply saying that — since the interaction between the particles depends only on the distance between them — a constant shift of both coordinates leaves the dynamics una↵ected. It is also useful to consider an infinitesimal version of such a transformation. Let us assume that the constant C is small, C ✏; and we write ! q0 q δq = ✏ (5.6) k − k ⌘ k for k =1, 2. We then say that δqk = ✏ is a symmetry of our system. To make these ideas more useful, we want to extend this example by considering a general class of interesting transformations that we may want to consider. 5.1 Infinitesimal transformations There are two useful types of infinitesimal transformations: direct and indirect ones. 5.1.1 Direct transformations A direct transformation deforms the degrees of freedom of a setup directly: δq (t)=q0 (t) q (t) ∆q (t, q) . (5.7) k k − k ⌘ k We use the notation ∆to distinguish a direct transformation. Note that ∆qk(t, q) is possibly a function of time and all of the degrees of freedom in the problem. In the previous example, we had the special case where ∆qk(t, q) = constant. But it need not be so. Figure 5.2(a) depicts a direct transformation: it is an arbitrary shift in the qk’s. Note also that ∆qk(t, q) is a small deformation. 192 CHAPTER 5. SYMMETRIES AND CONSERVATION LAWS Figure 5.2: The two types of transformations considered: direct on the left, indirect on the right. 5.1.2 Indirect transformations In contrast, an indirect transformation a↵ects the degrees of freedom indirectly — through the transformation of the time coordinate: δt(t, q) t0 t. (5.8) ⌘ − Note again that the shift in time can depend on time and the degrees of freedom as well! It is again assumed to be small. This is however not the end of the story since the degrees of freedom depend on time and will get a↵ected as well — indirectly dqk qk(t)=qk(t0 δt) qk(t0) δt qk(t0) q˙kδt (5.9) − ' − dt0 ' − where we have used a Taylor expansion in δt to linear order since δt is small. We also have used dqk/dt0 = dqk/dt since this term already multiplies a power of δt: to linear order in δt, δtdq /dt = δtdq /dt δtq˙ . We then see that shifting time k 0 k ⌘ k results in a shift in the degrees of freedom δq = q (t0) q (t)=q ˙δt(t, q) . (5.10) k k − k Figure 5.2(b) shows how you can think of this e↵ect graphically. 5.1. INFINITESIMAL TRANSFORMATIONS 193 5.1.3 Combined transformations In general, we want to consider a transformation that may include both direct and indirect pieces. We would write δq = q0 (t0) q (t)=q0 (t0)+[ q (t0)+q (t0)] q (t) k k − k k − k k − k =[q0 (t0) q (t0)] + [q (t0) q (t)] = ∆q (t0,q)+q ˙δt(t, q) k − k k − k k =∆qk(t, q)+q ˙δt(t, q) . (5.11) where in the last line we have equated t and t0 since the first term is already linear in the small parameters. To specify a particular transformation, we would then need to provide a set of functions δt(t, q) and δqk(t, q) . (5.12) Equation (5.11) then determines ∆qk(t, q). For N degrees of freedom, that’s N +1 functions of time and the qk’s. Let us look at a few examples. EXAMPLE 5-2: Translations Consider a single particle in three dimensions, described by the three Cartesian coordinates x1 = x, x2 = y, and x3 = z. We also have the time coordinate x0 = ct. An infinitesimal spatial translation can be realized by δxi(t, x)=✏i ,δt(t, x)=0 ∆xi(t, x)=✏i (5.13) ) where i =1, 2, 3, and the ✏i’s are three small constants. A translation in space is then defined by δt(t, x)=0,δxi(t, x)=✏i (5.14) { } A translation in time on the other hand would be given by δxi(t, x)=0 ,δt(t, x)=" ∆xi(t, x)= x˙ i" (5.15) ) − for constant ". Notice that we require that the total shifts in the xi’s — the δxi(t, x)’s — vanish. This then generates direct shifts, the ∆xi’s, to compensate for the indirect e↵ect on the spatial coordinates from the shifting of the time. A translation in time is then defined by δt(t, x)=✏,δxi(t, x)=0 (5.16) { } 194 CHAPTER 5. SYMMETRIES AND CONSERVATION LAWS EXAMPLE 5-3: Rotations To describe rotations, let us consider a particle in two dimensions for simplicity. We use the coordinates x1 = x and x2 = y. We start by specifying δt(t, x)=0 δxi(t, x)=∆xi(t, x) . (5.17) ) Next, we look at an arbitrary rotation angle ✓ using (2.21) x10 cos ✓ sin ✓ x1 = . (5.18) x20 sin ✓ cos ✓ x2 ✓ ◆ ✓ − ◆✓ ◆ We however need to focus on an infinitesimal version of this transformation: i.e. we need to consider small angle ✓. Using cos ✓ 1 and sin ✓ ✓ to second order ⇠ ⇠ in ✓, we then write x10 1 ✓ x1 = . (5.19) x20 ✓ 1 x2 ✓ ◆ ✓ − ◆✓ ◆ This gives δx1(t, x)=x10 (t) x1(t)=✓x2(t)=∆x1(t, x) − δx2(t, x)=x20 (t) x2(t)= ✓x1(t)=∆x2(t, x) . (5.20) − − We see we have a more non-trivial transformation. Rotations can then be defined through δt(t, x)=0,δxi(t, x)=✓"ijxj(t) (5.21) { } where j is summed over 1 and 2. We have also introduced a useful shorthand: "ij. It is called the totally antisymmetric matrix in two dimensions and defined as: "11 = "22 =0 ,"12 = "21 =1. (5.22) − It allows us to write the transformation in a more compact notation. EXAMPLE 5-4: Lorentz transformations To find the infinitesimal form of Lorentz transformations, we can start with the general transformation equations (2.15) and take β small. We need to be careful however to keep the leading order terms in β in all expansions. Given our previous example, it is easier to map the problem onto a rotation with hyperbolic trig functions using (2.27).