Before we can proceed to the next round of Legendre transformation, in which we replace ˙q2byp2, we need to evaluate∂R1/∂q˙2. From (1.172), we see that
dR1=dL−p1d ˙q1−q˙1dp1
=
f i=1
∑
∂L
∂qidqi+
f i=1
∑
∂L
∂q˙id ˙qi+∂L
∂tdt−p1d ˙q1−q˙1dp1
=
f i=1
∑
∂L
∂qidqi+
f i=2
∑
∂L
∂q˙id ˙qi+∂L
∂tdt−q˙1dp1, (1.174) where we used (1.103). It follows that
∂R1
∂q˙2 = ∂L
∂q˙2=p2. (1.175)
It is important to note that the partial derivative on the left is evaluated by holding q1, . . . ,qf,p1, ˙q3, . . . , ˙qf, andtconstant. In contrast, that on the right is for constant q1, . . . ,qf, ˙q1, ˙q3, . . . , ˙qf, andt.
The left most expression of (1.175) givesp2as a function ofq1, . . . ,qf,p1, ˙q2, . . . , ˙qf, andt. When this is solved for ˙q2, we obtain
˙
q2=q˙2(q1, . . . ,qf,p1,p2,q˙3, . . . ,q˙f,t). (1.176) Using this equation, we now express
R2:=R1−p2q˙2=L−p1q˙1−p2q˙2 (1.177) as a function ofq1, . . . ,qf,p1,p2, ˙q3, . . . , ˙qf, andt. The resulting function is another Routhian. Continuing in this way, we finally arrive at
Rf:=L−
f
∑
i=1piq˙i=Rf(q1, . . . ,qf,p1, . . . ,pf,t). (1.178) Now that all of ˙q1, . . . , ˙qf are replaced by p1, . . . , pf, we do not refer toRf as a Routhian. Instead, it is the negative of the HamiltonianH.
Exercise 1.15.Derive the following set of equations of motion from the Routhian Rn:
∂Rn
∂qi =p˙i, ∂Rn
∂pi =−q˙i, i=1, . . . ,n, (1.179)
and d
dt ∂Rn
∂q˙i
−∂Rn
∂qi =0, i=n+1, . . . ,f . (1.180) ///
Equations of motion derived from a Routhian find practical applications in an analysis of the stability of a steady motion.
1.11 Poisson Bracket 39
1.11 Poisson Bracket
Quantities such as the energy and the total linear momentum of a system are called dynamical variables. More generally, we define a dynamical variableAas a func- tion of the instantaneous state of the mechanical system, which is specified byqf andpf:
A(qf,pf,t). (1.181)
Here we allowed for an explicit time dependence ofA. We also introduced a new notation in whichqf andpf stand forq1, . . . ,qf andp1, . . . ,pf, respectively. Note carefully the distinction between qf andqf. The former is the collective notation just introduced, while the latter refers to the fth generalized coordinate.
As the coordinates and momenta change in accordance with the laws of mechan- ics, the value ofAwill, in general, change as well. It is straightforward to find the expression for the time rate of change ofA. Differentiating (1.181) with respect to time, we find
dA dt =
f i=1
∑
∂A
∂qiq˙i+∂A
∂pip˙i
+∂A
∂t . (1.182)
Using Hamilton’s equations of motion, we may rewrite the right-hand side as dA
dt =
f i=1
∑
∂A
∂qi
∂H
∂pi−∂H
∂qi
∂A
∂pi
+∂A
∂t . (1.183)
We introduce the Poisson bracketof two dynamical variablesA andB, which is defined by
{A,B}:=
f
∑
i=1∂A
∂qi
∂B
∂pi−∂B
∂qi
∂A
∂pi
, (1.184)
and write (1.183) more compactly as dA
dt ={A,H}+∂A
∂t . (1.185)
By definition,Ais aconstant of motionif dA/dt≡0. It should be noted that the definition doesnotdemand∂A/∂t to be zero. Thus, a dynamical variable can be a constant of motion even if it has an explicit time dependence. The Hamiltonian is an exception to this rule. In fact, if we setA=Hin (1.185),
dH
dt ={H,H}+∂H
∂t =∂H
∂t . (1.186)
So,His a constant of motion if and only ifHdoes not depend explicitly on time.
As another example of dynamical variables, let us considerqk. Using (1.185), we find
˙
qk={qk,H}+∂qk
∂t . (1.187)
As in (1.183), the partial derivative with respect tot is taken while holdingqf and pf constant. Thus,∂qk/∂t=0 and we obtain
˙
qk={qk,H}. (1.188)
Similarly, by settingA=pk, we find
˙
pk={pk,H}. (1.189)
As seen in Exercise1.16, (1.188) and (1.189) are nothing but Hamilton’s equations of motion.
We note that, sinceq’s andp’s are independent variables, we have∂pk/∂qi≡0 and∂qk/∂pi≡0 regardless of the values ofiandk. On the other hand,∂qk/∂qiis unity ifi=kand zero ifi=k, that is,
∂qk
∂qi =δik and ∂pk
∂pi =δik, (1.190)
where δik is theKronecker delta, which takes the value unity ifi=k and zero otherwise.
Exercise 1.16.LetAbe a dynamical variable of a mechanical system with fdegrees of freedom, that is,A=A(qf,pf,t). Evaluate{qj,A}and{pj,A}. Then, show that (1.188) and (1.189) are Hamilton’s equations of motion. ///
Exercise 1.17.Using (1.190), show that
{qi,qj}=0, {qi,pj}=δi j, and {pi,pj}=0. (1.191) ///
The Poisson bracket satisfies a number of identities:
{A,A}=0, (1.192)
{c,A}=0, (1.193)
{A,B+C}={A,B}+{A,C}, (1.194) {A,BC}=B{A,C}+{A,B}C, (1.195) wherecis a real number. Through a straightforward but lengthy computation, one can proveJacobi’s identity:
{A,{B,C}}+{B,{C,A}}+{C,{A,B}}=0. (1.196) With the help of this identity, we can show that, ifAandBare constants of motion, then so is{A,B}. This is known as thePoisson theorem, which, on occasion, allows us to generate a new constant of motion from known constants of motion.
Exercise 1.18.Prove the Poisson theorem.
1.12 Frequently Used Symbols 41 a. For simplicity, first assume that neitherAnorBdepends explicitly on time.
b. Generalize the theorem whenAandBmay depend explicitly on time. ///
Finally, there appears to be no general agreement on the sign of the Poisson bracket. For example, Ref. [3] defines it as
{A,H}=
f
∑
i=1∂A
∂pi
∂H
∂qi−∂H
∂pi
∂A
∂qi
, (1.197)
which has the opposite sign to what we defined.
1.12 Frequently Used Symbols
A:=B, the symbolAis defined by the expressionB.
A=:B, the expressionAdefines the symbolB.
a= (a. x,ay,az), thex-,y-,z-components of the vectoraareax,ay, andaz, respectively.
||a||, length of the vectora.
dφ, the first order term of∆ φ.
∆ φ , change inφincluding the higher order terms.
f , the number of mechanical degrees of freedom.
g, gravitational acceleration.
h, energy function.
h.o. , higher order terms of Taylor series expansion, typically second order and higher.
m, mass of a particle.
pi, generalized momentum conjugate toqi. pf , collective notation forp1, . . . ,pf. pi, linear momentum of theith particle.
qi, theith generalized coordinate.
qf , collective notation forq1, . . . ,qf. ri, position vector of theith particle.
t, time.
vi, velocity vector of theith particle.
E , mechanical energy.
F, force.
H, Hamiltonian.
L, Lagrangian.
M , total mass of a many-particle system.
M, total angular momentum of a many-particle system.
P, total linear momentum of a many-particle system.
R, Routhian.
R, position of the center of mass.
V, velocity of the center of mass.
W , work.
S , action.
δq(t), infinitesimal variation of some functionq(t)at timet.
δi j, Kronecker delta.
µ, reduced mass.
φ , potential energy due to interparticle interactions.
ψ , potential energy due to an external field.