The differential of the Lagrangian L(q, $, q, §) is dL = dq + (al/dg) ds (0L/as) d d, whence = (0L/d) d. If we define the Routhian as = pq-L, (41.1) in which the velocity q is expressed in terms of the momentum P by means of the equation P = 0L/dq, then its differential is dR = - ds - (aL/a) (41.2) Hence DRIP, p = OR/dq, (41.3) (41.4) Substituting these equations in the Lagrangian for the co-ordinate $, we have (41.5) Thus the Routhian is a Hamiltonian with respect to the co-ordinate q (equations (41.3)) and a Lagrangian with respect to the co-ordinate $ (equation (41.5)). According to the general definition the energy of the system is E -p-L = In terms of the Routhian it is E=R-R, (41.6) as we find by substituting (41.1) and (41.4). The generalisation of the above formulae to the case of several co-ordinates q and & is evident. The use of the Routhian may be convenient, in particular, when some of the co-ordinates are cyclic. If the co-ordinates q are cyclic, they do not appear in the Lagrangian, nor therefore in the Routhian, so that the latter is a func- tion of P, $ and $. The momenta P corresponding to cyclic co-ordinates are constant, as follows also from the second equation (41.3), which in this sense contains no new information. When the momenta P are replaced by their given constant values, equations (41.5) (d/dt) JR(p, $, 5)108 = JR(P, &, §) 128 become equations containing only the co-ordinates $, so that the cyclic co- ordinates are entirely eliminated. If these equations are solved for the func- tions (t), substitution of the latter on the right-hand sides of the equations q = JR(p, $, E) gives the functions q(t) by direct integration. PROBLEM Find the Routhian for a symmetrical top in an external field U(, 0), eliminating the cyclic co-ordinate 4 (where 4, , 0 are Eulerian angles). §42 Poisson brackets 135 SOLUTION. The Lagrangian is = see §35, Problem 1. The Routhian is R = cos 0); the first term is a constant and may be omitted. 42. Poisson brackets Let f (p, q, t) be some function of co-ordinates, momenta and time. Its total time derivative is df Substitution of the values of and Pk given by Hamilton’s equations (40.4) leads to the expression (42.1) where (42.2) dqk This expression is called the Poisson bracket of the quantities H and f. Those functions of the dynamical variables which remain constant during the motion of the system are, as we know, called integrals of the motion. We see from (42.1) that the condition for the quantity f to be an integral of the motion (df/dt = 0) can be written af(dt+[H,f]=0 (42.3) If the integral of the motion is not explicitly dependent on the time, then [H,f] = 0, (42.4) i.e. the Poisson bracket of the integral and the Hamiltonian must be zero. For any two quantities f and g, the Poisson bracket is defined analogously to (42.2): (42.5) The Poisson bracket has the following properties, which are easily derived from its definition. If the two functions are interchanged, the bracket changes sign; if one of the functions is a constant c, the bracket is zero: (42.6) [f,c]=0. (42.7) Also [f1+f2,g]=[f1,g)+[f2,g] (42.8) [f1f2,g] ]=fi[fa,8]+f2[f1,8] = (42.9) Taking the partial derivative of (42.5) with respect to time, we obtain (42.10) 136 The Canonical Equations §42 If one of the functions f and g is one of the momenta or co-ordinates, the Poisson bracket reduces to a partial derivative: (42.11) (42.12) Formula (42.11), for example, may be obtained by putting g = qk in (42.5); the sum reduces to a single term, since dqk/dqi = 8kl and dqk/dpi = 0. Put- ting in (42.11) and (42.12) the function f equal to qi and Pi we have, in parti- cular, [qi,qk] = [Pi, Pk] =0, [Pi, 9k] = Sik. (42.13) The relation [f,[g,h]]+[g,[h,f]]+[h,[f,g]] = 0, (42.14) known as Jacobi’s identity, holds between the Poisson brackets formed from three functions f, g and h. To prove it, we first note the following result. According to the definition (42.5), the Poisson bracket [f,g] is a bilinear homogeneous function of the first derivatives of f and g. Hence the bracket [h,[f,g]], for example, is a linear homogeneous function of the second derivatives of f and g. The left-hand side of equation (42.14) is therefore a linear homogeneous function of the second derivatives of all three functions f, g and h. Let us collect the terms involving the second derivatives of f. The first bracket contains no such terms, since it involves only the first derivatives of f. The sum of the second and third brackets may be symboli- cally written in terms of the linear differential operators D1 and D2, defined by D1() = [g, ], D2(b) = [h, ]. Then 3,[h,f]]+[h,[f,g]] = [g, [h,f]]-[h,[g,f] = D1[D2(f)]-D2[D1(f)] = (D1D2-D2D1)f. It is easy to see that this combination of linear differential operators cannot involve the second derivatives of f. The general form of the linear differential operators is where & and Nk are arbitrary functions of the variables …. Then and the difference of these, §42 Poisson brackets 137 is again an operator involving only single differentiations. Thus the terms in the second derivatives of f on the left-hand side of equation (42.14) cancel and, since the same is of course true of g and h, the whole expression is identi- cally zero. An important property of the Poisson bracket is that, if f and g are two integrals of the motion, their Poisson bracket is likewise an integral of the motion: [f,g] = constant. = (42.15) This is Poisson’s theorem. The proof is very simple if f and g do not depend explicitly on the time. Putting h = H in Jacobi’s identity, we obtain [H,[f,g]]+[f,[g,H]]+[g,[H,fl]=0. Hence, if [H, g] =0 and [H,f] = 0, then [H,[f,g]] = 0, which is the required result. If the integrals f and g of the motion are explicitly time-dependent, we put, from (42.1), Using formula (42.10) and expressing the bracket [H, [f,g]] in terms of two others by means of Jacobi’s identity, we find d [ (42.16) which evidently proves Poisson’s theorem. Of course, Poisson’s theorem does not always supply further integrals of the motion, since there are only 2s-1 - of these (s being the number of degrees of freedom). In some cases the result is trivial, the Poisson bracket being a constant. In other cases the integral obtained is simply a function of the ori- ginal integrals f and g. If neither of these two possibilities occurs, however, then the Poisson bracket is a further integral of the motion. PROBLEMS PROBLEM 1. Determine the Poisson brackets formed from the Cartesian components of the momentum p and the angular momentum M = rxp of a particle. SOLUTION. Formula (42.12) gives [Mx, Py] = -MM/Dy = -d(yp:-2py)/dy = -Pz, and similarly [Mx, Px] = 0, [Mx, P2] = Py. The remaining brackets are obtained by cyclically permuting the suffixes x, y, Z. 6 138 The Canonical Equations