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