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Classical Mechanics Symmetry and Conservation Laws

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Classical Mechanics Symmetry and Conservation Laws Classical Mechanics Symmetry and Conservation Laws Dipan Kumar Ghosh UM-DAE Centre for Excellence in Basic Sciences Kalina, Mumbai 400085 September 7, 2016 1 Concept of Symmetry If the property of a system does not change under some defined operations, the system is said to have a symmetry with respect to so defined operations. Thought he universe is both homogeneous and isotropic, there are local inhomogeneity and lack of isotropy. Nevertheless, a closed system (i.e. a system which does not interact with particles or fields outside of the system) may exhibit space homogeneity and/or isotropy. 1.1 Homogeneity of Space: Homogeneity of space implies that the physical laws are invariant under space translation. This means that if we change the position of every particle (and field) by the same constant vector ~, the Lagrangian of our system would remain the same. Mathematically, it implies 0 that when ~ri ! ~ri = ~ri + ~, the change in the Lagrangian of our system, X @L X @L δL = · ~ = ~ · (1) @~r @~r i i i i Here the summation index i is over all the particles of the system. Symmetry under arbitrary translation implies δL = 0, which results in X @L = 0 (2) @~r i i Using Euler-Lagrange equation, this implies X d @L = 0 dt @~v i i 1 c D. K. Ghosh, IIT Bombay 2 For a closed system, V depends only on positions of the particles, 1 X L = m v2 − V (f~r g 2 i i i i We then get ! ! d X d X m ~v = ~p = 0 (3) dt i i dt i i i i.e. the total momentum of the system is constant. Thus homogeneity of space leads to conservation of linear momentum of the system. 1.2 Isotropy of Space: If L is independent of the orientation of the system, it results in the conservation of angular momentum of the system. Consider (see figure) the rotation of the system about the z-axis by an angle θ so that OA~ ! OA~ 0. z C δϕ r r’ θ O It follows from the figure that OA~ 0 = OA~ + AA~ 0 where the magnitude of AA~ 0 is given by j AA~ 0 j= r sin θδ'. Defining δ'~ to be of magnitude δ' and directed along the axis of rotation, we can write, AA~ 0 = δ'~ × ~r Thus as ~r ! ~r0 = ~r + δ~r, δ~r = δ'~ × ~r. We then have ~v ! ~v0 = ~v + δ~v where δ~v = δ'~ × ~v. c D. K. Ghosh, IIT Bombay 3 If the Lagrangian is invariant under rotation described agove, we have X @L X @L δL = · δ~r + · δ~r_ @~r i ~ i i i i @r_i X d @L X @L = · δ~r + · δ~r_ dt ~ i ~ i i @r_i i @r_i X d @L X @L = − · (~r × δ'~ ) − · (~v × δ'~ ) (4) dt ~ i ~ i i @r_i i @r_i Once again, if we consider a closed system, 1 X L = m v2 − V (f~r g 2 i i i i we have @L = mi~r_i = ~pi @~r_i (4) can be written as X h ~ ~ i δL = − p~_i · (~ri × δ') + ~pi · (~r_i × δ') (5) i Writing δ'~ =nδ' ^ , we can express (5) as X d δL = δ'n^ · (~r × ~p ) (6) dt i i i Thus δL = 0 implies d X (~r × ~p ) = 0 dt i i i ~ P so that the angular momentum of the system J = i(~ri × ~pi) is constant. 1.3 Homogeneity of Time and Energy Conservation: Closed system also exhibit symmetry with respect to time translation. What this implies is that repeating an experiment under identical conditions after a lapse of some time will yield identical result. Time homogeneity is a consequence of the fact that the Lagrangian @L of the system has no explicit time dependence: = 0, so that L = Lq; q_ only. @t The total time derivative of the Lagrangian, in such a case is given by dL X @L X @L = q¨ + q_ dt @q_ i @q i i i i i X @L X d @L = q¨ + q_ @q_ i dt @q_ i i i i i X d @L = q_ dt @q_ i i i c D. K. Ghosh, IIT Bombay 4 We define the Hamiltonian of the system as X @L H = q_ − L (7) @q_ i i i dH so that the preceding equation gives us = 0. For a closed system, the Hamiltonian dt also gives the total energy of the system (see below), so that the time homogeneity results in conservation of energy. Let us consider a closed system for which the kinetic energy is expressible as a bilinear product of generalized velocities and the potential energy is a function of the generalized coordinates alone, 1 X T = C q_ q_ 2 i;j i j i;j where Ci;j = Cj;i could be functions of generalized coordinates but not of velocities or of time. Thus we have L = T (fq_ig) − +V (fqig) We then have ! X @L X @ 1 X q_k = q_k Ci;jq_iq_j @q_k @q_k 2 k k i;j " # X 1 X 1 X = q_ C δ q_ + C q_ δ k 2 i;j i;k j 2 i;j i j;k k i;j i;j ! 1 X X X = C q_ + C q_ q_ 2 k;j j i;k i k k j i X = Cj;kq_jq_k j;k = 2T where in the last but one step, we have used Ci;k = Ck;i and have changed the dummy summation index in the second term from i to j. Using (7) for the definition of the Hamiltonian, we then have, X @L H = q_ − L @q_ i i i = 2T − L = 2T − (T − V ) = T + V which is the total energy. Thus for a closed system, homogeneity of time leads to a conservation of the total energy. c D. K. Ghosh, IIT Bombay 5 2 Invariance under Galilean Transformation Let S and S0 be two inertial frames. Though the values of kinematical quantities will be measured differently in two frames, by definition of inertial frames, Newton's laws will be valid in both the frames, i.e. d2~r d2~r 0 m i = m i (= F ) i dt2 i dt2 i which gives d2(~r − ~r 0) i i = 0 dt2 which can be integrated to yield 0 ~ri − ~ri = ~u0(t − t0) ≡ ~u0t , where u0 is a constant velocity, the last relation follows if the two frames are made to coincide at t = t0. The two frames are thus related by Galilean transformation: 0 ~ri = ~ri − ~u0t How does the Lagrangian transform? If the potential energy depends only on the relative coordinates of two particles then 0 0 ~r2 − ~r1 = ~r2 − ~r1 so that in S0, 1 L0 = mv02 − V 2 1 = m(~v − ~u )2 − V 2 0 1 1 = mv2 − V − m~v · ~u + mu2 2 0 2 0 d 1 = L + mu2t − m~u · ~r (8) dt 2 0 0 dF ≡ L + (9) dt where the function F is defined through (8) and (9). Note that both L and L0 satisfy Euler-Lagrange equation. This is an example of gauge symmetry of the Lagrangian. If L satisfies the Euler-Lagrange equation and F is any differentiable function of (q; t), then dF L + also satisfies the Euler-Lagrange equation. dt This is readily proved by observing that the extra term in the Euler-Lagrange equation c D. K. Ghosh, IIT Bombay 6 dF due to is dt d @ dF @ dF − = dt @q_i dt @qi dt " ! !# d @ X @F @F @ X @F @F q_ + − q_ + dt @q_ @q j @t @q @q j @t i j j i j j (10) Note that d @ @F @ d @F @ @F = = dt @q_i @t @t dt @q_i @t @qi since the explicit differentiation can always be carried out at the end. Further, " !# ! d @ X @F @ X @F X @ @F q_ = q_ = q_ dt @q_ @q j @q @q j @q @q j i j j i j j j j i sinceq _j is independent of the coordinates and the independent coordinates qi and qj are interchangeable. Using the above two relations, the right hand side of (10) becomes X @2F @2F X @2F @2F q_ + − q_ − = 0 @q @q j @t@q @q @q j @q @t j j i i j i j i which proves the assertion. The reason for the existence of the gauge symmetry can be traced back to the minimisation of the action integral. Let, dF (q; t) L0(q; q;_ t) = L(q; q;_ t) + dt The corresponding action integral would then be given by Z tf S0[q(t)] = L0dt ti Z tf dF = L + dt ti dt tf = S + F (q; t) jti = S + constant since δq(ti) = δq(tf ) = 0 for the variational function, adding a constant to the action integral is inconsequential to the determination of t. Example: Consider the Lagrangian for a charged particle in electromagnetic field. We know that c D. K. Ghosh, IIT Bombay 7 the following gauge transformations of the scalar and vector potentials leave the fields unchanged: A~0 = A~ + rF (~r; t) @ '0 = ' − F @t We have seen earlier that the electromagnetic forces can be included in the Lagrangian formulation if we define a velocity dependent potential, e' − e~v · A~, where e is the charge of the particle. Thus 1 L = mv2 − e' + e~v · A~ 2 dF (~r; t) Suppose we add to the Lagrangian above a term equal to e The new Lagrangian dt is then given by 1 dF L0 = mv2 − e' + e~v · A~ + e 2 dt 1 @F h i = mv2 − e ' − + ~v · e(A~ + erF ) 2 @t where we have used dF @F @F @F @F = v + v + v + dt @x x @y y @z z @t @F = rF · ~v + @t which gives 1 L0 = mv2 − e'0 + e~v · A~0 2 3 Symmetry Operations and Noether's Theorem 3.1 Coordinate Transformation vrs.
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