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Symmetries and Conservation Laws Chapter 4 Symmetries and Conservation Laws xx Section 1 Hamiltonian ∂L Legendre transform p = μ ∂q· μ A system is described by a function f with x, y as independent variables. Legendre’s transform to go from · to as a independent qμ pμ f = f (x, y) variable. The Hamiltonian ∂f ∂f · μ , df = dx + dy = udx + vdy H = ∑ pμq − L ∂x ∂y Hence Now we want to describe the system in terms of x, v. ∂L ∂L ∂L dH = (p δq· μ + q· μδp − δqμ − δq· μ) − δt Since d(vy) = vdy + ydv, ∑ μ μ ∂qμ ∂q· μ ∂t d(vy − f ) = − udx + ydv ∂L dH = (p δq· μ + q· μδp − p· δqμ − p δq· μ) − δt ∑ μ μ μ μ ∂t The new function is g = g(x, v) = vy − f. · μ · μ ∂L ∂g ∂g dH = q δpμ − pμδq − δt = − u; = y ∂t ∂x ∂v Hence Hamiltonian ∂H ∂H ∂H ∂L Lagrangian L = L(q , q· , t) q· μ = ; p· = − ; = − μ μ μ μ ∂pμ ∂q ∂t ∂t 34 dL ∂L ∂L ∂L ∂L Note: = q· μ + q··μ + δt = p· q· μ + p q··μ + dt ∂qμ ∂q· μ ∂t μ μ ∂t dH ∂L = − dt ∂t Homogeneity of time implies that for an isolated system, the potential function is not an explicit function of time. For a pair charge particles, U depends only the distance between the charged particles. Hence, for an isolated system, ∂L /∂t = 0. Hence dH = 0, dt or the total energy of the system is a constant. This is the statement of conservation of energy, which is derived using the homogeneity property of the space. Exercise 1. Argue that the energy is conserved for spring-mass system and planetary system. 2. An oscillator of mass m and spring constant k is forced with . Write down the Lagrangian for the same. Is the F0 cos(ωf t) energy conserved for this system? 35 Section 2 More Conservation laws Conservation of Linear Momentum ∂L d ∂L d ∂L . ∑ = ∑ = ∑ = 0 Homogeneity of space: If each particle of an isolated system is a ∂ra a dt ∂va dt a ∂va shifted by a constant distance ε, then the system’s Lagrangian ∂L should not change. Again think of a collection of charged Therefore, the total linear momentum of the system is P = ∑ ∂va particle. Interaction potential is function only of the distances a between the particles. Here conserved. It is derived from the homogeneity of space. ∂L On many occasions, a system may be symmetric about some δL = ⋅ ϵ = 0 ∑ translations, but not arbitrary translations. For example for a line a ∂ra charge aligned along the z axis, the system is invariant under any Since ε is arbitrary, translation along z. Therefore only ∂L is conserved for Pz = ∑ a ∂va,z ∂L . ∑ = 0 such system. a ∂ra If the Lagrangian is not an explicit function of q , then the Since i generalized force , and the generalized momenta Fi = ∂L /∂qi = 0 ∂L d ∂L = , ∂L ∂r dt ∂v a a pi = · ∂qi we obtain is conserved. 36 Conservation of Angular Momentum is a constant. This is related to the isotropy of space. Isotropy of space: The system or Lagrangian of an isolated If the system is invariant under rotation about an axis, the angular system is invariant under the rotation of the whole system by an momentum about the axis is conserved. arbitrary angle. The above three symmetries (homogeneity and isotropy of space, Let us rotate the system by angle δφ about an axis. The vector and homogeneity in time) have never been broken. So far, we δφ is a vector aligned along the axis, and its magnitude is δφ. have not observed any violation of conservation laws of energy, Under this operation, each particle is shifted by linear momentum, and angular momentum. and . δra = δϕ × ra δva = δϕ × va Robust conservation Therefore, the change in the Lagrangian under the above Example: operation is Galilean invariance: ∂L ∂L or δL = ∑ ⋅ δra + ⋅ δva = 0 a ∂ra ∂va is the relative velocity between the two inertial frames. For a Vr · δL = ∑ pa ⋅ δϕ × ra + pa ⋅ δϕ × va = 0 set of particles, a 1 1 2 and ! 2 Therefore, L = ∑ mava L′ = ∑ ma(va + Vr) a 2 a 2 d . δϕ ⋅ ∑ ra × pa = 0 Hence dt a Since δφ is arbitrary, the total angular momentum of the system 1 about the origin δL = L′! − L = δ m (v + V )2 = m v ⋅ V = P ⋅ V ∑ [ a a r ] ∑ a a r CM r a 2 [ a ] d L = r × p ∑ a a for small . dt a Vr 37 We do not set δL zero like in earlier cases because it will lead to 3. Construct conserved quantities for the system of Exercise 1 of , a special case. We set Section 3.2 as well as a system of two masses m1 and m2 PCM = 0 coupled by a spring of a spring constant k. dF d δL = = [R ⋅ V ] dt dt CM r 4. What are the conserved quantities for a particle moving on the inner side of the cone? Comparing the above we obtain d R P = CM , CM dt which is an identity. Self similarity Exercise: 1. Construct conserved quantities for two masses coupled by a spring. 2. A particle moves in the field for the following system. Construct the conserved quantities of the charged particle: Infinite charged plate, infinite line charge, two charges of equal magnitude, infinite half-plane charge, charged helix, charged torus. 38 Section 3 Discrete Symmetries Nonrelativistic particle dynamics Time Reversal Symmetry: U(−t) = U(t), hence the Lagrangian is symmetric under time reversal. We cannot distinguish between the forward and time-reversed (b) (a) dynamics. Can’t figure out who is the real gun man! (b) (a) Parity or Mirror Symmetry r → − r = [(x, y, z) → (x, − y, z)] + [(x, − y, z) → (−x, − y, − z)] Parity = mirror (about xz plane) + rotation about the y axis by π. g g symmetry of mirror reflection. Can’t figure out the real projectile motion 39 Since A and B are proper vectors, the rules of mirror reflections of Eq. (8.8.1) yield Proper and pseudo vectors C′! = − C , C′! = C , C′! = − C (8.8.5) When a mirror is placed as an xz plane, the symmetry x x y y z z of mirror reflection is mathematically expressed as A vector that transforms like C is called pseudo vector or axial vector. Thus, z′! = z, y′! = − y, x′! = x; t′! = t Hence, Cross product of two proper vectors is a pseudo vector. ! ! ! Two corollaries of the above statements are vx′ = vx, vy′ = − vy, vz′ = vz A cross product of a proper vector and a pseudo vector is a a′! = a , a′! = − a , a′! = a x x y y z z proper vector. A vector that transforms as above is called proper vector A cross product of two pseudo vectors is a proper vectors. or real vector. Examples of pseudo vectors A cross product of two proper vectors transforms very differently under mirror transformation. Let us consider (1) Angular momentum L = r × p since the linear mo- two proper vectors A and B. A cross product of these two mentum p = mv is a proper vector. Also, spin angular vectors is momentum of a body. C = A × B (2) Angular velocity Ω, which is defined using v = Ω × r. = (AyBz − AzBy) x̂ + (AzBx − AxBz) ŷ + (AxBy − AyBx) ẑ (3) Magnetic field B since (8.8.4) μ Idl × r B = 0 4π ∫ r3 40 Wu and coworkers discovered that that the electrons (beta) are emitted preferentially opposite to the direction of the spin of the Cobalt-60 nucleus (see Fig. (a)). Note that the spin vector perpendicular to the mirror does not change sign under mirror reflection, but the velocity vec- B B tor does. Hence in the mirror image, the electrons would F F move preferentially in the direction of the spin, as shown in Fig. (b). Thus the law that the beta particles have a pref- V (b) (a) V erential velocity opposite to their spin is violated in the mirror reflection. This is how mirror symmetry is violated Can’t figure out real experiment! in a beta decay experiment. Does the mirror symmetry hold for all experiment? Mathematically, the violation of the mirror symmetry Ans: No! In Beta decay is due to the appearance of nonzero value of the pseudo scalar Q = S ⋅ v in the Lagrangian. Here S is the spin of the Cobalt-60 nucleus, and v is the velocity of the elec- (b) (a) tron. Such quantities vanish in experiments respecting e e mirror symmetry (e.g., gravitational and electromagnetic v v interactions). The potential of weak nuclear force how- S S ever is a combination of a proper scalar and a pseudo sca- lar. This kind of potential was first proposed by Marshak and Sudarshan, and Feynman and Gell-mann in 1957. 41 Charge Conjugation q → − q In nuclear physics, C, P, T all are violated, but CPT to- gether is not violated. Exercises 1. Which of the following forces would violate mirror symmetry? (a) F=q v×B (b) F=q v×E (c) F=q v×B+mg (d) F=E+B (e) F=E×B In the above, E is the electric field, B is the magnetic fields, and g is the acceleration due to gravity. 42 Section 4 Noether’s theorem ! ! We can derive conservation laws for continuous transformation. ∂L′! ∂L′! dt′! ∂L′! dq′μ ∂L′! dq· ′μ = + + A general transformation is ∂ϵ ∂t′! dϵ ∂qμ dϵ ∂q· μ dϵ ! t′! = t + ϵτ ∂L′! ∂L′! ∂L′! dq· ′μ = τ + ζμ + ∂t′! ∂qμ ∂q· μ dϵ ! q′μ = qμ + ϵζμ ! dq′μ dqμ + ϵdζμ The new action is Using = , dt′! dt + ϵdτ dt′! ! ! ! ′μ! L′dt′ − Ldt = L′ − L dt d dq · ! ∫ ∫ ∫ [ dt ] we obtain = ζμ − q· ′μτ·.
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