It is shown that the Hamiltonian structures for the acoustic field obtained by means of the canonical ... malisms employed in the study of mechanical systems with.
´ INVESTIGACION
REVISTA MEXICANA DE F´ISICA 49 (4) 344–347
AGOSTO 2003
Hamiltonian structures for the acoustic field G.F. Torres del Castillo Departamento de F´ısica Matem´atica, Instituto de Ciencias, Universidad Aut´onoma de Puebla, 72570 Puebla, Pue., M´exico E. Galindo Linares Facultad de Ciencias F´ısico Matem´aticas, Universidad Aut´onoma de Puebla, Apartado Postal 1152, 72001 Puebla, Pue., M´exico Recibido el 10 de diciembre de 2002; aceptado el 25 de febrero de 2003 It is shown that the Hamiltonian structures for the acoustic field obtained by means of the canonical formalism using as field variables the components of the molecular displacements or the variation of the gas density are different and lead to different Poisson brackets. It is shown that by requiring that the values of the Hamiltonians coincide, the Poisson brackets also coincide. Keywords: Acoustic field; Hamiltonian structures. Se muestra que las estructuras hamiltonianas para el campo ac´ustico que se obtienen a trav´es del formalismo can´onico usando como variables de campo las componentes de los desplazamientos moleculares o la variaci´on de la densidad del gas son diferentes y llevan a par´entesis de Poisson diferentes. Se muestra que si se propone que los valores de las hamiltonianas coincidan, entonces los par´entesis de Poisson tambi´en coinciden. Descriptores: Campo ac´ustico; estructuras hamiltonianas. PACS: 11.10.Ef; 03.50.-z; 43.20.+g
1. Introduction As is well known, the Lagrangian and Hamiltonian formalisms employed in the study of mechanical systems with a finite number of degrees of freedom can be applied in the case of continuous media and fields. The Hamiltonian formulation is usually obtained from the Lagrangian formulation by means of the Legendre transformation, but in the case of fields this canonical procedure presents difficulties since not always the momentum densities are independent of the field variables, which is usually mended by the introduction of constraints. However, it is possible to avoid these complications and give a Hamiltonian formulation for a given continuous system, without making reference to the Lagrangian formulation, if its evolution equations can be written in the form δH φ˙ α = Dαβ , δφβ
(1)
where the field variables φα represent the state of the system, H is some functional of the φα , δH/δφα is the functional derivative of H with respect to φα , and the Dαβ are operators that must satisfy certain conditions that allow the definition of a Poisson bracket between functionals of the φα (see, e.g., Refs. 1–3). Here and in what follows a dot denotes partial differentiation with respect to the time and there is summation over repeated indices. In this paper we consider Hamiltonian structures for the acoustic field in a perfect gas. An interesting feature of this simple example is the fact that one can use as field variables either the components of the vector field representing the displacement of the particles of the gas (three variables per space
point) or the variations of the density of the gas (one variable per space point) [4,5]. In Sec. 2 we compare the Hamiltonian structures for the acoustic field obtained from the corresponding Lagrangians using the displacement vector field or the variation of the gas density as field variables and we find that the Poisson brackets and the Hamiltonians obtained in these two cases do not coincide. In Sec. 3 we show that if the value of the Hamiltonian is the same, regardless of which variables are employed, then the Poisson brackets also coincide.
2. Canonical formalism In the study of the acoustic field in a perfect gas one can make use of the Cartesian components of the vector field, η, that represents the small displacements of the gas particles with respect to their positions in the absence of sound waves. A suitable Lagrangian density is [4] 1 1 Lv = µ0 η˙ 2 + P0 ∇ · η − γP0 (∇ · η)2 , (2) 2 2 where µ0 is the mass density of the gas in equilibrium, P0 is the pressure in equilibrium and γ = Cp /Cv , with Cp and Cv being the heat capacities at constant pressure and at constant volume, respectively (the subscript v is introduced to remark the vector character of the field variables). The Lagrangian density (2) has the usual structure found in the case of mechanical systems with a finite number of degrees of freedom, being the difference between a term corresponding to the kinetic energy density and another term identifiable as the potential energy density. The Euler–Lagrange equations applied to (2) yield ¨ − γP0 ∇(∇ · η) = 0. µ0 η
(3)
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HAMILTONIAN STRUCTURES FOR THE ACOUSTIC FIELD
Following the standard procedure one finds that the conjugate momenta to the field variables ηi are πi = ∂Lv /∂ η˙ i = µ0 η˙ i (i, j, . . . = 1, 2, 3) and the Hamiltonian density is given by Hv = πi η˙ i − Lv =
π2 γP0 − P0 ∇ · η + (∇ · η)2 . (4) 2µ0 2
hence {φα (r0 , t0 ), φβ (r00 , t00 )}v = Dαγ Gβγ (r00 , t00 , r0 , t0 ). The relative variation of the gas density, denoted by σ, is related with η according to [4,5] σ = −∇ · η
Then, the Hamilton equations [4,2] η˙ i =
δHv , δπi
π˙ i = −
δHv , δηi
(5)
R where Hv = Hv dv, reproduce Eq. (3). If F and G are two functionals of ηi and πi , their Poisson bracket is defined as ¶ Z µ δF δG δG δF {F, G}v = − dv. (6) δηi δπi δηi δπi Letting (φ1 , . . . , φ6 ) ≡ (η1 , η2 , η3 , π1 , π2 , π3 ) we have Z φα (r0 , t) = δαβ δ(r0 − r)φβ (r, t)dv
and the divergence of Eq. (3) yields the wave equation µ0 σ ¨ − γP0 ∇2 σ = 0.
As can be readily seen, Eq. (15) follows from the Euler– Lagrange equations with Ls =
(7)
¤ 1£ µ0 σ˙ 2 − γP0 (∇σ)2 , 2
and the corresponding Hamiltonian density is given by Hs = ρσ˙ − Ls =
{φα (r0 , t), φβ (r00 , t)}v = Dαβ δ(r0 − r00 ), ·
0 −I
I 0
ρ2 γP0 + (∇σ)2 . 2µ0 2
(8)
(9)
for any pair of functionals of σ and ρ. Thus, by analogy with Eqs. (10) and (11),
¸
and I is the 3 × 3 identity matrix. Explicitly,
{σ(r0 , t), σ(r00 , t)}s = 0 = {ρ(r0 , t), ρ(r00 , t)}s
{πi (r0 , t), πj (r00 , t)}v = 0 = {ηi (r0 , t), ηj (r00 , t)}v
(10)
and {σ(r0 , t), ρ(r00 , t)}s = δ(r0 − r00 ).
and {ηi (r0 , t), πj (r00 , t)}v = δij δ(r0 − r00 ).
(11)
Making use of the matrix (Dαβ ), Eqs. (5) and (6) can be rewritten as δHv φ˙ α = Dαβ δφβ
(12)
{F, G}v =
δG δF Dαβ dv, δφα δφβ
(13)
respectively. It may be pointed out that, owing to the linearity of the evolution equations considered here, there exists a set of functions Gαβ (r0 , t0 ; r, t) such that Z 0 0 φα (r , t ) = Gαβ (r0 , t0 ; r, t)φβ (r, t)dv,
(19)
On the other hand, since σ = −∇· η and ρ = µ0 σ˙ = −µ0 ∇ · η˙ = −∇ · π , the variables σ and ρ can be regarded as functionals of ηi and πi . Then, making use of the expression (6) we obtain {σ(r0 , t), ρ(r00 , t)}v = −∇02 δ(r0 − r00 ),
Z
(17)
With the variables σ, ρ there is associated a Poisson bracket given by ¶ Z µ δF δG δG δF − dv, (18) {F, G}s = δσ δρ δσ δρ
Hence,
and
(16)
ρ = ∂Ls /∂ σ˙ = µ0 σ˙
δφα (r0 , t) = δαβ δ(r0 − r). δφβ (r, t)
(Dαβ ) =
(15)
considering σ as the field variable. The conjugate momentum to σ is therefore
(α, β, . . . = 1, 2, . . . , 6), and
where
(14)
(20)
which differs from (19); hence, the Poisson brackets (6) and (18) are different and define different Hamiltonian structures for the acoustic field. The linear momentum of the acoustic field (defined as the generator of translations) is given by the general expression [4,2] Z ∂ηk dv. (21) Gi = − πk ∂xi
Rev. Mex. F´ıs. 49 (4) (2003) 344–347
346
G.F. TORRES DEL CASTILLO AND E. GALINDO LINARES
But, when σ is the field variable, the components of the linear momentum are Z ∂σ Gi = − ρ dv (22) ∂xi and, apart from the difference in the dimensions of the functionals (21) and (22), it seems clear that their values are not related (though both are conserved, e.g., if there are no boundaries), which adds another difference between the two Hamiltonian structures. We note, in passing, that if the Lagrangian density Ls is expressed in terms of η, then the Euler–Lagrange equations give
and making use of the fact that πi = µ0 η˙ i we have δHv0 δHv0 + Eij δηj δπj µ ¶ µ ¶ 1 ∂ ∂ ¨ + Eij − = Cij µ0 ∇·η ∇·π ∂xj µ0 ∂xj πi = (25) µ0
η˙ i = Cij
and, similarly, δHv0 δHv0 + Gij δηj δπj ¶ µ ¶ µ 1 ∂ ∂ ¨ + Gij − ∇·η ∇·π = Fij µ0 ∂xj µ0 ∂xj
π˙ i = Fij
¨ ) − γP0 ∇[∇2 (∇ · η)] = 0, µ0 ∇(∇ · η which is the gradient of the divergence of Eq. (3).
3. Alternative definition of the Hamiltonian structure We shall consider again the components ηi as the field variables without making use of a Lagrangian, but demanding that the value of the Hamiltonian density coincides with that obtained in the case of the scalar variables, Hs . The idea is to write the equations of motion in the form (12), looking for the appropriate Dαβ , which may be constants, functions or operators, with the con† † dition that Dαβ =−Dβα , where Dαβ is the adjoint of Dαβ and the bar denotes complex conjugation, in order for the Poisson bracket (13) to be antisymmetric (i.e., {F, G}=−{G, F }) [2,3]. When the Dαβ are constants, the Poisson bracket automatically satisfies the Jacobi identity, but in other cases one has to verify that this identity is satisfied [1]. Thus, assuming that the ηi are the field variables, we take as the Hamiltonian density Hv0 =
(∇ · π )2 γP0 + [∇(∇ · η ]2 , 2µ0 2
= µ0 η¨i .
(26)
These equations are satisfied if we take Cij = 0 = Gij , µ Eij
∂ ∇·π ∂xj
∂ 2 ∂ δHv0 ¨, = γP0 ∇ (∇ · η) = µ0 ∇·η δηi ∂xi ∂xi where we have made use of Eq. (3), and
= −πi
(27)
and Fij = −Eij . Taking the curl on both sides of Eq. (3) we obtain ∇ר η =0, and therefore we can assume that ∇ × π also vanishes (otherwise ∇×η would not be bounded). Thus Eq. (27) can be rewritten as µ ¶ µ ¶ ∂ ∂ ∂ ∂ −πi = Eij πk = Eij πk ∂xj ∂xk ∂xk ∂xj µ ¶ ∂ ∂ = Eij πj = Eij (∇2 πj ), ∂xk ∂xk hence, if there are no boundaries present, the Eij are the integral operators
(23)
which is obtained from Hs given in (17), substituting σ and ρ by the corresponding expressions in terms of ηi and πi (it may be noticed that Hv0 cannot be obtained from a Lagrangian density R in the standard way). The functional derivatives of Hv0 = Hv0 dv are then
¶
1 4π
Eij (Aj ) = Indeed, making use of ∇2
Z
Ai (r0 ) 0 dv . |r − r0 |
1 = −4πδ(r − r0 ), from |r − r0 |
Eq. (28) we have µ Eij
∂ ∇·π ∂xj
¶
1 = Eij (∇ πj ) = 4π
Z
2
(∇02 πi )(r0 ) 0 dv |r − r0 |
and integrating by parts twice
δHv0 1 ∂ =− ∇ · π. δπi µ0 ∂xi
µ Eij
Substituting these derivatives into Eq. (12) with (Dαβ ) written as the block matrix µ ¶ (Cij ) (Eij ) (Dαβ ) = , (24) (Fij ) (Gij )
(28)
∂ ∇·π ∂xj
¶ = Z
=− as required by Eq. (27).
Rev. Mex. F´ıs. 49 (4) (2003) 344–347
1 4π
Z πi (r0 )∇02
1 dv 0 |r − r0 |
πi (r0 )δ(r − r0 )dv 0 = −πi (r),
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HAMILTONIAN STRUCTURES FOR THE ACOUSTIC FIELD
From Eq. (28) it follows that for any pair of vector fields, A and B, ·Z ¸ Z Z 1 Ai (r0 ) 0 Bi (r) dv dv Bi Eij (Aj )dv = 4π |r − r0 | ·Z ¸ Z 1 Bi (r) = Ai (r0 ) dv dv 0 4π |r − r0 | ·Z ¸ Z 1 Bi (r0 ) 0 = Ai (r) dv dv 4π |r − r0 | Z Z = Ai Eij (Bj )dv = Aj Eji (Bi )dv
but ∂ ∂ 1 ∂ ∂ 1 = − 00 00 00 00 − r000 | ∂x00k ∂x000 |r ∂x ∂x |r − r000 | k k k = −∇002
1 = 4πδ(r00 − r000 ) |r00 − r000 |
and therefore {σ(r, t), ρ(r0 , t)}0v = δ(r − r0 ),
(30)
which agrees with Eq. (19). † Eij
therefore, = Eji and, since Eij = −Fij , we have † Eij = −Fji , which means that † Dαβ = −Dβα = −Dβα
4. Conclusions
1 {σ(r, t), ρ(r0 , t)}0v = 4π Z Z ∂ ∂ 1 × δ(r00 − r)δ(r0 − r000 ) 00 000 00 dv 00 dv 000 ∂xk ∂xk |r − r000 |
The example considered here shows that despite the difference in the number of field variables employed to deal with the acoustic field, we can obtain equivalent Hamiltonian structures. This example also illustrates the fact that the Hamiltonian functional and the Poisson brackets can be chosen in many ways (cf. also Ref. 6). The Hamiltonian is in all these cases a constant of the motion, but to the authors’ knowledge, it is not known if there exist additional conditions for a constant of motion to be a Hamiltonian (with the corresponding Poisson bracket satisfying the Jacobi identity). As pointed out at the beginning of Sec. 3, when the Dαβ are constants the Jacobi identity is always satisfied [1,2]. Since the Poisson bracket constructed in Sec. 3 agrees with that defined by Eq. (18), which satisfies the Jacobi identity, it is to be expected that it also satisfies this identity. Finally, it should be remarked that it would be wrong to talk about the Lagrangian or Hamiltonian density of a given system. If these densities are defined by the only condition that they reproduce the equations of motion of the system in question through the Euler–Lagrange equations or the Hamilton equations, there may be many acceptable choices, which may not have a direct physical meaning. Of course, with any appropriate Lagrangian or Hamiltonian density the corresponding formalism must yield valid results (such as conservation laws).
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4. H. Goldstein, Classical Mechanics, 2nd ed. (Addison-Wesley, Reading, Mass., 1980), Chap. 12.
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3. G.F. Torres del Castillo and N. Bagatella Flores, Rev. Mex. F´ıs. 40 (1994) 209.
6. G.F. Torres del Castillo and D. Acosta Avalos, Rev. Mex. F´ıs. 40 (1994) 405.
as required in order for the Poisson bracket to be antisymmetric. We now compute the Poisson bracket between σ and ρ using the definition (13) with (Dαβ ) given by (24), with Cij = 0 = Gij , Fij = −Eij and Eij given by Eq. (28) Z δσ(r, t) δρ(r0 , t) 00 {σ(r, t), ρ(r0 , t)}0v = Ekl dv 00 δηk (r , t) δπl (r00 , t) Z δσ(r, t) δρ(r0 , t) 00 − E dv (29) kl δπk (r0 , t) δηl (r, t) hence
Z 0
{σ(r, t), ρ(r
, t)}0v
=
∂ δ(r00 − r) ∂x00k
1 × 4π
Z
∂ 0 000 δ(r − r ) 00 ∂x000 k dv 000 dv 00 000 |r − r |
and integrating by parts both integrals we find that
Rev. Mex. F´ıs. 49 (4) (2003) 344–347
