November 12, 2003 15:20:52 P. Gutierrez Physics 5153 Classical Mechanics The Virial Theorem and The Poisson Bracket 1 Introduction In this lecture we will consider two applications of the Hamiltonian. The ¯rst, the Virial Theorem, applies to systems that are bounded. It allows us to determine statistical properties of systems. The other is the Poisson bracket, which has allows us to determine properties of systems without solving the equations of motion, and is connected to the commutator in quantum mechanics. 1.1 The Virial Theorem Let's consider the variable Q = piqi, which remains bounded if both pi and qi are ¯nite. From Hamilton's equations of motion we have for the total derivative of Q @H @H Q_ = piq_i + qip_i = pi ¡ qi (1) @pi @qi so that Q(t) ¡ Q(0) 1 T 1 T @H @H = Qdt_ = dt p ¡ q (2) T T T i @p i @q Z0 Z0 µ i i ¶ If the system is periodic and we integrate over one period, then left hand side is zero. If the system is bounded, then as T ! 1 the left hand side goes to zero. In these case, we have @H @H p = q (3) i @p i @q ¿ i À ¿ i À where the brackets denote the time average values of the enclosed quantities. The quantity @H V = ¡qi (4) @qi is a generalization of the virial function introduced into the kinetic theory of gases by Clausius1. Thus the time average virial function is @H hVi = ¡ p (5) i @p ¿ i À For a system of N particles, the right-hand side becomes n n ¡ piq_i = ¡2 Ti = ¡2NT¹ (6) * i=1 + * i=1 + X X where the bar denotes an average over time and kinetic energy of each particle. The virial function for this system can be also expressed as follows n hVi = p_iqi = I + P~r ¢ n^dS (7) *i=1 + X I 1This is just the total work done on the gas, including external and internal forces. The Virial Theorem and The Poisson Bracket-1 lect 14.tex November 12, 2003 15:20:52 P. Gutierrez where the left had side is the work done by the intermolecular forces, and the second term is the work done in containing the gas with in the container. Using the divergence theorem, we arrive at hVi = I ¡ V P r~ ¢ ~r = I ¡ 3P V (8) where V is the total volume. Combining the two expression for the average virial function we arrive at 2 I I P V = NT¹ + = Nkθ + (9) 3 3 3 ¹ 3 where θ is the absolute temperature from the expression T = 2 kθ. If I = 0, this is the ideal gas law. Another application of the virial theorem is the case of a particle in a central potential V (r). In this case the average virial function is given by hVi = ¡ ~p ¢ ~r_ = ~p_ ¢ ~r (10) D E D E which is reduced to 2 hT i = h~r ¢ rV (r)i (11) For a potential of the form V (r) = Kr¡n, this expression becomes 1 hT i = ¡ n hV (r)i (12) 2 For the gravitational or Coulomb forces n = 1. This means that the average kinetic energy is ¡1=2 the average potential energy. 1.2 The Poisson Bracket The Poisson bracket appears often in classical mechanics and translates itself to quantum mechanics as the commutator. Let's consider the total time derivative of a function u that depends on the canonical variables q and p, and can also depend on the time t n du @u @u @u = q_ + p_ + (13) dt @q i @p i @t i=1 i i X µ ¶ Using the canonical equations of motion, the total time derivative becomes n du @u @H @u @H @u = ¡ + (14) dt @q @p @p @q @t i=1 i i i i X µ ¶ which can be written using the following shorthand notation du @u = [u; H] + (15) dt qi;pi @t where we de¯ne the Poisson bracket as n @u @H @u @H [u; H] ´ ¡ (16) qi;pi @q @p @p @q i=1 i i i i X µ ¶ The Virial Theorem and The Poisson Bracket-2 lect 14.tex November 12, 2003 15:20:52 P. Gutierrez Let's consider the case of u not depending explicitly on time, and the Poisson bracket being zero. In this case the quantity u(q; p) is a constant of the motion. This is the same as we have in quantum mechanics. If the commutator of an operator with the Hamiltonian is zero and does not depend explicitly on time, then the operator corresponds to a variable that is a constant of the motion. As an example of how this works, let's consider a simple harmonic oscillator in two dimensions. Using the Poisson bracket, we will show that the angular momentum is conserved (a constant of the motion). The Hamiltonian for this system is 1 1 H = (p2 + p2) + (q2 + q2) (17) 2 1 2 2 1 2 where for simplicity we take m = k = 1. The canonical equations of motion are @H @H p_i = ¡ = ¡qi q_i = = pi with i = 1, 2 (18) @qi @pi The angular momentum associated with this system is J = q1p2 ¡ q2p1 (19) The Poisson bracket is dJ @J @H @J @H = ¡ = (p2p1 ¡ p1p2) + (q2q1 ¡ q1q2) = 0 (20) dt @qi @pi @pi @qi therefore the angular momentum is a constant of the motion. Based on the de¯nition of the Poisson bracket, the canonical equations of motion can be ex- pressed in a compact form. Let's start by calculating q_i using the Poisson bracket @qi @H @H @qi @H q_i = [qi; H] = ¡ ) q_i = (21) @qi @pi @qi @pi @pi Similarly the canonical equation of motion for p_i can be derived @pi @H @pi @H @H p_i = [pi; H] = ¡ ) p_i = ¡ (22) @qi @pi @pi @qi @qi At this point we de¯ne the Poisson bracket of any two functions that depend on the canonical variables n @u @v @v @u [u; v] = ¡ (23) @q @p @q @p i=1 i i i i X µ ¶ If the functions we use are the canonical variables themselves, we get the following \fundamental" brackets [qi; qj] = 0 [qi; pj] = ±ij [pi; pj] = 0 (24) Notice how closely these resemble the commutators in quantum mechanics. Just replace ±ij with i¹h±ij. Before going further, let's consider a few properties of the Poisson bracket. Assume that we have the dynamical variables u(qi; pi), v(qi; pi), w(qi; pi). The Poisson bracket has the following properties The Virial Theorem and The Poisson Bracket-3 lect 14.tex November 12, 2003 15:20:52 P. Gutierrez 1. Linearity: [®u + ¯w; v] = ® [u; v] + ¯ [w; v] (25) 2. Antisymmetric: [u; w] = ¡ [w; u] (26) 3. The product rule: [u; wv] = [u; w] v + w [u; v] (27) 4. The Jacobi identity: [w; [u; v]] + [u; [v; w]] + [v; [w; u]] = 0 (28) 5. The chain rule: if the variables are expressed in terms of a new set Pi and Qi that are functions of the old variables pi and qi, then the Poisson bracket of the dynamical variables is @w @u [w; u] = [´i; ´j] (29) @´i @´j where ´i = Qi and ´i+n = Pi for i = 1 ¢ ¢ ¢ n and we have a total of 2n variables. 6. The fundamental brackets: [qi; qj] = [pi; pj] = 0 [qi; pj] = ±ij (30) 1.3 Poisson's Theorem Assume that we have a set of dynamical variables and we would like to know whether their time evolution is generated by a Hamiltonian or not. Poisson's theorem states that for the time evolution of a set of variables to be generated by a Hamiltonian, the variables must satisfy d [R; S] = R_ ; S + R; S_ (31) dt h i h i Assume that the time evolution is generated by H(pi; qi; t), then according to Eq. 15 we have d @ [R; S] = [[R; S] ; H] + [R; S] (32) dt @t The ¯rst term can be reduce using the Jacobi identity [[R; S] ; H] = [[R; H] ; S] + [R; [S; H]] (33) The second term can be expanded as follows @ @ @R @S @R @S [R; S] = ¡ @t @t @q @p @p @q µ i i i i ¶ @2R @S @2R @S @R @2S @R @2R = ¡ + ¡ @q @t @p @p @t @q @q @p @t @p @q @t µ i i i i ¶ µ i i i i ¶ @R @S = ; S + R; (34) @t @t · ¸ · ¸ The Virial Theorem and The Poisson Bracket-4 lect 14.tex November 12, 2003 15:20:52 P. Gutierrez Next we combine the two sets d @R @S [R; S] = [R; H] + ; S + R; [S; H] + = R_ ; S + R; S_ (35) dt @t @t · ¸ · ¸ h i h i which is the statement of the theorem Notice that this theorem also implies that if R and S are constants of the motion, then so is [R; S]. Let's consider two examples: The ¯rst we take the one-dimensional simple harmonic oscillator. The two dynamical variables are q and p. These have the following equations of motion p_ = ¡q q_ = p (36) where we take m = k = 1. Applying Poisson's theorem we get d [q; p] = 0 (37) dt [p_; q] + [p; q_] = ¡ [q; q] + [p; p] = 0 (38) which implies that the time evolution of the dynamical variables is generated by a Hamiltonian as expected. As a second example, consider the same two variables but with the following equations of motion q_ = pq p_ = ¡pq (39) Again we apply Poisson's theorem d [q; p] = 0 (40) dt [q_; p] + [q; p_] = [pq; p] ¡ [q; pq] = [q; p] p + q [p; p] ¡ [q; p] q ¡ [q; q] p = p ¡ q (41) which does not satisfy Poisson's theorem, therefore the time evolution of these variables is not generated by a Hamiltonian.