Chapter 6 Lecture 3 Legendre Transformations Lagrange & Poisson Bracket
Dr. Akhlaq Hussain 6.3 Legendre Transformations
Lagrangian of the system 퐿 = 퐿 푞푖, 푞푖ሶ , 푡 푑 휕퐿 휕퐿 Lagrange's equation − = 0 푑푡 휕푞푖ሶ 휕푞푖
휕퐿 푞푖, 푞푖ሶ ,푡 Where momentum 푝푖 = 휕푞푖ሶ
In Lagrangian to Hamiltonian transition variables changes from 푞, 푞푖ሶ , 푡 to 푞, 푝, 푡 , where p is related to 푞 and 푞푖ሶ by above equation. This transformation process is known as Legendre transformation. Consider a function of only two variable 푓(푥, 푦) 휕푓 휕푓 푑푓 = 푑푥 + 푑푦 휕푥 휕푦 푑푓 = 푢푑푥 + 푣푑푦 2 6.3 Legendre Transformations
If we wish to change the basis of description from 푥, 푦 to a new set of variable 푦,푢 Let 푔 푦, 푢 be a function of 푢 and 푦 defined as 푔 = 푓 − 푢푥 (I) Therefore 푑푔 = 푑푓 − 푥푑푢 − 푢푑푥 since 푑푓 = 푢푑푥 + 푣푑푦 So 푑푔 = 푢푑푥 + 푣푑푦 − 푥푑푢 − 푢푑푥 푑푔 = 푣푑푦 − 푥푑푢 And the quantities 푣 and 푥 are function of 푦 and 푢 respectively. 휕푔 휕푔 And 푑푔 = 푑푦 + 푑푢 휕푦 휕푢 휕푔 휕푔 = 푣 & = −푥 휕푦 휕푢
3 6.3 Legendre Transformations
Equation I represent Legendre transformation. The Legendre transformation is frequently used in thermodynamics as 푑푈 = 푇푑푆 − 푃푑푉 for 푈(푆, 푉) And 퐻 = 푈 + 푃푉 퐻 푆, 푃 푑퐻 = 푇푑푆 + 푉푑푃 And 퐹 = 푈 − 푇푆 퐹(푇, 푉) 퐺 = 퐻 − 푇푆 퐺 푇, 푃
U = Internal energy T = Temperature S = Entropy P = Pressure
V = Volume H = Enthalpy F = Helmholtz Free Energy G=Gibbs free energy 4 6.3 Legendre Transformations
The transformation of 푞, 푞ሶ, 푡 to 푞, 푝, 푡 is however different. 휕퐿 휕퐿 휕퐿 Since 푑퐿 = 푑푞 + 푑푞ሶ + 푑푡 휕푞 휕푞ሶ 휕푡 휕퐿 푑 휕퐿 And = 푝 ⇒ = 푝ሶ 휕푞ሶ 푑푡 휕푞ሶ 푑 휕퐿 휕퐿 푑 휕퐿 Therefore, − = 푝 − = 0 푑푡 휕푞ሶ 휕푞 푑푡 휕푞 휕퐿 ⇒ = 푝ሶ 휕푞 휕퐿 휕퐿 휕퐿 Therefore , 푑퐿 = 푑푞 + 푑푞ሶ + 푑푡 휕푞 휕푞ሶ 휕푡 휕퐿 푑퐿 = 푝ሶ푑푞 + 푝푑푞ሶ + 푑푡 휕푡 And 퐻 = 푞ሶ푝 − 퐿 5 6.4 Lagrange’s Bracket
Lagrange brackets were introduced by Joseph Louis Lagrange in 1808–1810.
➢ Mathematical formulation of Classical Mechanics.
➢ The Lagrange bracket is not use in modern mechanics
➢ The Lagrange's bracket are defined as if 푢 and 푣 are functions depending on 푞푖 Zand 푝푖 then
휕푞 휕푝 휕푞 휕푝 휕 푞 ,푝 푢, 푣 = σ 푖 푖 − 푖 푖 = σ 푖 푖 푞,푝 푖 휕푢 휕푣 휕푣 휕푢 푖 휕(푢,푣) And invariant under canonical transformation.
6 6.4 Lagrange’s Bracket
Some properties i. 푢, 푣 푞,푝 = − 푣, 푢 푞,푝 Lagrange's bracket are anti commutative ii. 푞 , 푞 = 0 = 푝 , 푝 For identical function it is zero 푖 푗 푞,푝 푖 푗 푞,푝
1 푖푓 푖 = 푗 iii. 푞 , 푝 = 훿 ቊ Equal to Kronecker delta function. 푖 푗 푝,푞 푖푗 0 푖푓 푖 ≠ 푗
iv. 푢, 푣 푞,푝 = 푢, 푣 푄,푃 Invariance of Lagrange Bracket
7 6.4 Lagrange’s Bracket
Proof: 푢, 푣 푞,푝 = − 푣, 푢 푞,푝
휕푞 휕푝 휕푞 휕푝 푢, 푣 = σ 푖 푖 − 푖 푖 푞,푝 푖 휕푢 휕푣 휕푣 휕푢
휕푞 휕푝 휕푞 휕푝 Taking negative common 푢, 푣 = − σ − 푖 푖 + 푖 푖 푞,푝 푖 휕푢 휕푣 휕푣 휕푢
휕푞 휕푝 휕푞 휕푝 푢, 푣 = − σ 푖 푖 − 푖 푖 푞,푝 푖 휕푣 휕푢 휕푢 휕푣
푢, 푣 푞,푝 = − 푣, 푢 푞,푝 풑풓풐풗풆풅
8 6.4 Lagrange’s Bracket
Proof: 푞 , 푞 = 0 = 푝 , 푝 For identical 푖 푗 푝,푞 푖 푗 푞,푝 function it is zero
휕푞푘 휕푝푘 휕푞푘 휕푝푘 푞푖, 푞푗 = σ푘 − 푞,푝 휕푞푖 휕푞푗 휕푞푖 휕푞푗 Here 푝 and 푞 are treated as independent co-ordinates in phase space. So
휕푝 휕푝 푘 = 0 = 푘 휕푞푗 휕푞푖 Therefore 푞 , 푞 = 0 푖 푗 푞,푝 similarly , we can show 푝 , 푝 = 0 푖 푗 9 6.4 Lagrange’s Bracket
1 푖푓 푖 = 푗 푞 , 푝 = 훿 ቊ Equal to Kronecker delta function. 푖 푗 푖푗 0 푖푓 푖 ≠ 푗 ퟎ
휕푞푘 휕푝푘 휕푞푘 휕푝푘 푞푖, 푝푗 = σ푘 − 푞,푝 휕푞푖 휕푝푗 휕푝푗 휕푞푖 휕푞 휕푝 Since 푝, 푞 are independent, we get 푘 = 푘 = 0 휕푝푗 휕푞푖
휕푞푘 휕푝푘 Therefore 푞푖, 푝푗 = σ푘 푞,푝 휕푞푖 휕푝푗 ⇒ 푞 , 푝 = σ 훿 훿 = 훿 푖 푗 푞,푝 푘 푘푖 푘푗 푖푗
Where 훿푖푗 is a Kronecker delta function 1 푖푓 푖 = 푗 Hence 푞 , 푝 = 훿 ቊ Equal to Kronecker delta function. 푖 푗 푖푗 0 푖푓 푖 ≠ 푗 10 6.4 Lagrange’s Bracket
푢, 푣 푞,푝 = 푢, 푣 푄,푃 Invariance of Lagrange Bracket
휕 푞 ,푝 휕 푄 ,푃 0r σ 푖 푖 = σ 푗 푗 푖 휕(푢,푣) 푗 휕(푢,푣)
Proof: Let 푞푖 and 푝푖 are function of 푄푗 and 푃푗 휕푞 휕푝 휕푞 휕푝 푢, 푣 = σ 푖 푖 − 푖 푖 (I) 푞,푝 푖 휕푢 휕푣 휕푣 휕푢
휕푝푖 휕푝푖 휕푄푗 휕푝푖 휕푃푗 And for 푝푖 = 푝푖 푄푗, 푃푗 ⇒ = σ푗 + 휕푣 휕푄푗 휕푣 휕푃푗 휕푣 Using Maxwell’s equations 휕푝 휕푄 휕푝 휕푃 푖 = 푗 & 푖 = − 푗 putting in above equation 휕푃푗 휕푞푖 휕푄푗 휕푞푖
휕푝푖 휕푃푗 휕푄푗 휕푄푗 휕푃푗 휕푄푗 휕푃푗 휕푃푗 휕푄푗 = σ푗 − + = σ푗 − = 푞푖, 푣 푄,푃 (II) 휕푣 휕푞푖 휕푣 휕푞푖 휕푣 휕푞푖 휕푣 휕푞푖 휕푣 11 6.4 Lagrange’s Bracket
휕푞푖 휕푞푖 휕푄푗 휕푞푖 휕푃푗 And 푞푖 = 푞푖 푄푗, 푃푗 ⇒ = σ푗 + 휕푣 휕푄푗 휕푣 휕푃푗 휕푣 Using Maxwell’s equations 휕푞 휕푃 휕푞 휕푄 푖 = 푗 & 푖 = − 푗 putting in above equation 휕푄푗 휕푝푖 휕푃푗 휕푝푖
휕푞푖 휕푃푗 휕푄푗 휕푄푗 휕푃푗 = σ푗 − = 푝푖, 푣 푄,푃 (III) 휕푣 휕푝푖 휕푣 휕푝푖 휕푣 휕푞 휕푝 휕푞 휕푝 Putting Eqs (II) &(III) in Eq (I) 푢, 푣 = σ 푖 푖 − 푖 푖 푞,푝 푖 휕푢 휕푣 휕푣 휕푢
휕푞푖 휕푄푗 휕푃푗 휕푃푗 휕푄푗 휕푃푗 휕푄푗 휕푄푗 휕푃푗 휕푝푖 푢, 푣 푞,푝 = σ푖푗 − − − 휕푢 휕푞푖 휕푣 휕푞푖 휕푣 휕푝푖 휕푣 휕푝푖 휕푣 휕푢
휕푃푗 휕푄푗 휕푞푖 휕푄푗 휕푝푖 휕푄푗 휕푃푗 휕푞푖 휕푃푗 휕푝푖 푢, 푣 푞,푝 = σ푖푗 + − + 휕푣 휕푞푖 휕푢 휕푝푖 휕푢 휕푣 휕푞푖 휕푢 휕푝푖 휕푢 12 6.4 Lagrange’s Bracket
휕푃푗 휕푄푗 휕푞푖 휕푄푗 휕푝푖 휕푄푗 휕푃푗 휕푞푖 휕푃푗 휕푝푖 σ σ푖 + − σ푖 + 푢, 푣 푞,푝 = 푗 휕푣 휕푞푖 휕푢 휕푝푖 휕푢 휕푣 휕푞푖 휕푢 휕푝푖 휕푢
휕푃 휕푄 휕푄 휕푃 푢, 푣 = σ 푗 푗 − 푗 푗 푞,푝 푗 휕푣 휕푢 휕푣 휕푢
휕푄 휕푃 휕푃 휕푄 푢, 푣 = σ 푗 푗 − 푗 푗 푞,푝 푗 휕푢 휕푣 휕푢 휕푣
푢, 푣 푞,푝 = 푢, 푣 푄,푃
13 6.5 Poisson’s Brackets
Poisson’s bracket: ➢ An important binary operation in Hamiltonian mechanics ➢ Play a central role in Hamilton's equations of motion. ➢ Distinguishes a class of coordinate transformations (canonical transformations) ➢ Very useful tool in quantum mechanics and field theory. The Poisson bracket are defined as
휕푢 휕푣 휕푢 휕푣 푢, 푣 푞,푝= σ푖 − 휕푞푖 휕푝푖 휕푝푖 휕푞푖 휕푢 휕푣 휕푢 휕푣 Or 푢, 푣 푄,푃= σ푖 − 휕푄푖 휕푃푖 휕푃푖 휕푄푖 Where 푢, 푣 are two functions, 푤. 푟. 푡 the canonical variable 푞, 푝 or 푄, 푃 . 14 6.5 Poisson’s Brackets
Consider 푓 is a function such that
푓 = 푓 푞푖, 푝푖
푑푓 휕푓 푑푞푖 휕푓 푑푝푖 = σ푖 + σ푖 푑푡 휕푞푖 푑푡 휕푝푖 푑푡 푑푓 휕푓 휕푓 = σ푖 푞푖ሶ + σ푖 푝푖ሶ 푑푡 휕푞푖 휕푝푖 휕퐻 휕퐻 As we know that 푞ሶ = and 푝ሶ = − 푖 휕푝 푖 휕q
푑푓 휕푓 휕퐻 휕푓 휕퐻 Eq 1 can be written as = σ푖 − 푑푡 휕푞푖 휕푝푖 휕푝푖 휕푞푖 푑푓 = 푓, 퐻 푑푡 푞,푝
Where 푓, 퐻 is called Poisson's Brackets. 15 6.5 Poisson’s Brackets
Generally u and v are two function their Poisson’s brackt is
푛 휕푢 휕푣 휕푢 휕푣 푢, 푣 = σ푖=1 − 휕푞푖 휕푝푖 휕푝푖 휕푞푖 Properties It is obvious from the definition of the Poisson brackets that i. 푢, 푣 = − 푣, 푢 Poisson bracket are anti-commutative ii. 푢, 푢 = 0 = 푣, 푣 Poisson bracket of identical functions are zero. iii. 푢, 푐 = 0 = 푢, 푐 Where c is independent of 푝 or 푞. iv. 푢 + 푣, 푤 = 푢, 푤 + 푣, 푤 Poisson brackets obeys the distributive law v. 푢, 푣푤 = 푢, 푣 푤 + 푣 푢, 푤
1 푖푓 푖 = 푗 vi. 푞 , 푝 = 훿 ቊ Where 훿 is the Kronecker delta function. 푖 푗 푖푗 0 푖푓 푖 ≠ 푗 푖푓 16 6.5 Poisson’s Brackets
1)Skew symmetric OR anti commutative (OR show that Poisson brackets do not obey commutative law)
푛 휕푢 휕푣 휕푢 휕푣 As we know that 푢, 푣 푞,푝 = σ푖=1 − 휕푞푖 휕푝푖 휕푝푖 휕푞푖 Taking –ve sign out of the bracket
푛 휕푢 휕푣 휕푢 휕푣 푢, 푣 푞,푝 = − σ푖=1 − + 휕푞푖 휕푝푖 휕푝푖 휕푞푖
푛 휕푢 휕푣 휕푢 휕푣 푢, 푣 푞,푝 = − σ푖=1 − 휕푝푖 휕푞푖 휕푞푖 휕푝푖 휕푣 휕푢 휕푣 휕푢 푢, 푣 푞,푝 = − − 휕푞푖 휕푝푖 휕푝푖 휕푞푖
푢, 푣 푞,푝 = − 푣, 푢 푞,푝 proved 17 6.5 Poisson’s Brackets
For identical function
푢, 푢 푞,푝 = − 푣, 푣 푞,푝 = 0
푛 휕푢 휕푢 휕푢 휕푢 푢, 푢 푞,푝 = σ푖=1 − 휕푞푖 휕푝푖 휕푝푖 휕푞푖
Similarly 푣, 푣 푞,푝 = 0 Poisson brackets with a-constant
Let 퐹 be function of generalized 푞푖 and 푝푖 and 퐶 is any constant quality.
푛 휕퐹 휕퐶 휕퐹 휕퐶 Then 퐹, 퐶 푞,푝 = σ푖=1 − 휕푞푖 휕푝푖 휕푝푖 휕푞푖 휕퐶 휕퐶 Where = = 0 where 푐 does not depend 표푛 푞푖 and 푝푖. 휕푞푖 휕푝푖
18 퐹, 퐶 푞,푝 = 0 6.5 Poisson’s Brackets
Poisson Brackets obey the distributive law
Consider 퐹1, 퐹2 and G are three functions dependently upon 푞푖 and푝푖.
푛 휕 퐹1+퐹2 휕퐺 휕 퐹1+퐹2 휕퐺 Then 퐹1 + 퐹2 , 퐺 = σ푖=1 − 휕푞푖 휕푝푖 휕푝푖 휕푞
푛 휕퐹1 휕퐹2 휕퐺 휕퐹1 휕퐹2 휕퐺 퐹1 + 퐹2 , 퐺 = σ푖=1 + − + . 휕푞푖 휕푞푖 휕푝푖 휕푝푖 휕푝푖 휕푞푖
푛 휕퐹1 휕퐺 휕퐹2 휕퐺 휕퐹1 휕퐺 휕퐹2 휕퐺 퐹1 + 퐹2 , 퐺 = σ푖=1 + − + 휕푞푖 휕푝푖 휕푞푖 휕푝푖 휕푝푖 휕푞푖 휕푝푖 휕푞푖
푛 휕퐹1 휕퐺 휕퐹1 휕퐺 휕퐹2 휕퐺 휕퐹2 휕퐺 퐹1 + 퐹2 , 퐺 = σ푖=1 − + − 휕푞푖 휕푝푖 휕푝푖 휕푞푖 휕푞푖 휕푝푖 휕푝푖 휕푞푖
푛 휕퐹1 휕퐺 휕퐹1 휕퐺 푛 휕퐹2 휕퐺 휕퐹2 휕퐺 퐹1 + 퐹2 , 퐺 = σ푖=1 − . + σ푖=1 − 휕푞푖 휕푝푖 휕푝푖 휕푞푖 휕푞푖 휕푝푖 휕푝푖 휕푞푖
퐹1 + 퐹2, 퐺 푞,푝 = 퐹1, 퐺 푞,푝 + 퐹2, 퐺 푞,푝 19 Proved it obeys. 6.5 Poisson’s Brackets
Poisson's Brackets are linear
Consider if 퐹 푞푖,푝푖 , 퐺 푞푖,푝푖 and 푊 푞푖,푝푖 are three functions and 푎, 푏 are constant, then we here to show that 푎퐹 + 푏퐺, 푊 = 푎퐹, 푊 + 푏퐺, 푊 푎퐹 + 푏퐺, 푊 = 푎 퐹, 푊 + 푏 퐺, 푊 Let 푎퐹 = 퐹′ and 푏퐺 = 퐺′ Then by distributive property we know that ′ ′ 푎퐹 + 푏퐺, 푊 = 퐹 + 퐺 , 푊 푞,푝 퐹′ + 퐺′, 푊 = 퐹′, 푊 + 퐺′, 푊
푎퐹 + 푏퐺, 푊 = 푎퐹, 푊 + 푏퐺, 푊 20 6.5 Poisson’s Brackets
Now consider
푛 휕 푎퐹 휕푊 휕 푎퐹 휕푊 푎퐹, 푊 = σ푖=1 . − 휕푞푖 휕푝푖 휕푝푖 휕푞푖
푛 휕퐹 휕푊 휕퐹 휕푊 푎퐹, 푊 = 푎 σ푖=1 − 휕푞푖 휕푝푖 휕푝푖 휕푞푖 푎퐹, 푊 = 푎 퐹, 푊 Similarly 푏퐺, 푊 = 푏 퐺, 푊 Hence 푎퐹 + 푏퐺, 푊 = 푎 퐹, 푊 + 푏 퐺, 푊
21 6.5 Poisson’s Brackets
Show that if u, v and w are functions dependent on 퐪퐢, 퐩퐢 , then show that 푢, 푣푤 = 푢, 푣 푤 + 푣 푢, 푤
푛 휕푢 휕 푣푤 휕푢 휕 푣푤 As 푢, 푣푤 푞,푝 = σ푖=1 − 휕푞푖 휕푝푖 휕푝푖 휕푞푖
푛 휕푢 휕w 휕푣 휕푢 휕푤 휕푣 푢, 푣푤 푞,푝 = σ푖=1 푣 + 푤 − 푣 + 푤 휕푞푖 휕푝푖 휕푝푖 휕푝푖 휕푞푖 휕푞푖
푛 휕푢 휕푣 휕푢 휕푣 휕 푢 휕 푤 휕푢 휕푤 푢, 푣푤 푞,푝 = σ푖=1 푤 − 푤 + 푣 − 푣 휕푞푖 휕푝푖 휕푝푖 휕푞푖 휕푞푖 휕푝푖 휕푝푖 휕푞푖
푛 휕푢 휕푣 휕푢 휕푣 휕 푢 휕 푤 휕푢 휕푤 푢, 푣푤 푞,푝 = σ푖=1 푤 − + 푣 − 휕푞푖 휕푝푖 휕푝푖 휕푞푖 휕푞푖 휕푝푖 휕푝푖 휕푞푖
푢, 푣푤 푝,푞 = 푤 푢, 푣 푞,푝 + 푣 푢, 푤 푞,푝
22 6.5 Poisson’s Brackets
휕 휕퐹 푑퐺 Proved that 퐹, 퐺 = , 퐺 + 퐹, 휕푡 휕푡 푑푡
푛 휕퐹 휕퐺 휕퐹 휕퐺 Sol: 퐹, 퐺 = σ푖=1 − 휕푞푖 휕푝푖 휕푝푖 휕푞푖
휕 휕 푛 휕퐹 휕퐺 휕퐹 휕퐺 퐹, 퐺 = σ푖=1 − 휕푡 휕푡 휕푞푖 휕푝푖 휕푝푖 휕푞푖
휕 푛 휕 휕퐹 휕퐺 휕 휕퐹 휕퐺 퐹, 퐺 = σ푖=1 − 휕푡 휕푡 휕푞푖 휕푝푖 휕푡 휕푝푖 휕푞푖
휕 푛 휕퐹 휕 휕퐺 휕 휕퐹 휕퐺 휕퐹 휕 휕퐺 휕 휕퐹 휕퐺 퐹, 퐺 = σ푖=1 + − − 휕푡 휕푞푖 휕푡 휕푝푖 휕푡 휕푞푖 휕푝푖 휕푝푖 휕푡 휕푞푖 휕푡 휕푝푖 휕푞푖
휕 푛 휕퐹 휕 휕퐺 휕퐹 휕 휕퐺 푛 휕 휕퐹 휕퐺 휕 휕퐹 휕퐺 퐹, 퐺 = σ푖=1 − + σ푖=1 − 휕푡 휕푞푖 휕푡 휕푝푖 휕푝푖 휕푡 휕푞푖 휕푡 휕푞푖 휕푝푖 휕푡 휕푝푖 휕푞푖
23 6.5 Poisson’s Brackets
휕 푛 휕퐹 휕 휕퐺 휕퐹 휕 휕퐺 푛 휕 휕퐹 휕퐺 휕 휕퐹 휕퐺 퐹, 퐺 == σ푖=1 − + σ푖=1 − 휕푡 휕푞푖 휕푝푖 휕푡 휕푝푖 휕푞푖 휕푡 휕푞푖 휕푡 휕푝푖 휕푝푖 휕푡 휕푞푖 휕 휕퐺 휕퐹 퐹, 퐺 = 퐹, + , 퐺 proved 휕푡 휕푡 휕푡
Assignment Show that i. 푞푖, 푞푗 = 푝푖, 푝푗 = 0
1 푖푓 푖 = 푗 ii. 푞 , 푝 = 훿 ቊ 푖 푗 푖푗 0 푖푓 푖 ≠ 푗
휕푝 휕푞 푤ℎ푒푟푒 = 0 = 0 휕푞 휕푝
24 6.5 Poisson’s Brackets
Show that Poisson’s brackets is invariant under canonical transformation
퐹, 퐺 푞,푝 = 퐹, 퐺 푄,푃 Proof: let 퐹 and 퐺 be two arbitrary function of 푞 and 푝.Then 휕퐹 휕퐺 휕퐹 휕퐺 퐹, 퐺 푞,푝 = σ푖 − 휕푞푖 휕푝푖 휕푝푖 휕푞푖 Let the Transformation 푄 = 푄 푞, 푝 and 푃 = 푃 푞, 푝 Inverse transformation 푞 = 푞 푄, 푃 and 푝 = 푝(푄, 푃) Therefore 퐺 = 퐺(푄, 푃) 휕퐺 휕퐺 휕퐺 = σ푘 휕푄퐾 + σ푘 휕푃푘 휕푄푘 휕푃푘
휕퐺 휕퐺 휕푄푘 휕퐺 휕푃푘 휕퐺 휕퐺 휕푄퐾 휕퐺 휕푃푘 Then = σ푘 + & = σ푘 + σ푘 휕푞푖 휕푄푘 휕푞푖 휕푃푘 휕푞푖 휕푝푖 휕푄푘 휕푝푖 휕푃퐾 휕푃푖 25 6.5 Poisson’s Brackets
휕퐹 휕퐺 휕푄푘 휕퐺 휕푃푘 휕퐹 휕퐺 휕푄푘 휕퐺 휕푃푘 ⇒ 퐹, 퐺 푞,푝 = σ푖,푘 + − + 휕푞푖 휕푄푘 휕푝푖 휕푃푘 휕푝푖 휕푝푖 휕푄푘 휕푞푖 휕푃푘 휕푞푖
휕퐹 휕퐺 휕푄푘 휕퐹 휕퐺 휕푄푘 휕퐹 휕퐺 휕푃푘 휕퐹 휕퐺 휕푃푘 ⇒ 퐹, 퐺 푞,푝 = σ푖,푘 − + − 휕푞푖 휕푄푘 휕푝푖 휕푝푖 휕푄푘 휕푞푖 휕푞푖 휕푃푘 휕푝푖 휕푝푖 휕푃푘 휕푞푖
휕퐺 휕퐹 휕푄푘 휕퐹 휕푄푘 휕퐺 휕퐹 휕푃푘 휕퐹 휕푃푘 ⇒ 퐹, 퐺 푞,푝 = σ푘 σ푖 − + σ푖 − 휕푄푘 휕푞푖 휕푝푖 휕푝푖 휕푞푖 휕푃푘 휕푞푖 휕푝푖 휕푝푖 휕푞푖
휕퐺 휕퐺 ⇒ 퐹, 퐺 푞,푝 = σ푘 퐹, 푄푘 푞,푝 + 퐹, 푃푘 푞,푝 (I) 휕푄푘 휕푃푘
Replacing F by 푄푖 ퟎ
휕퐺 휕퐺 ⇒ 푄푖, 퐺 푞,푝 = σ푘 푄푖, 푄푘 푞,푝 + 푄푖, 푃푘 푞,푝 휕푄푘 휕푃푘
휕퐺 휕퐺 26 ⇒ 푄푖, 퐺 푞,푝 = σ푘 훿푖푘 = 휕푃푘 휕푃푖 6.5 Poisson’s Brackets
Now replacing G by F in previous equation
휕퐹 ⇒ 푄푖, 퐹 = 휕푃푖
휕퐹 Or ⇒ 퐹, 푄푘 푞,푝 = − (II) 휕푃푘
Similarly, replacing 퐹 by 푃푖 in 푒푞 (I) ퟎ
휕퐺 휕퐺 ⇒ 푃푖, 퐺 푞,푝 = σ푘 푃푖, 푄푘 푞,푝 + 푃푖, 푃푘 푞,푝 휕푄푘 휕푃푘 휕퐺 ⇒ 푃푖, 퐺 푞,푝 = σ푘 푃푖, 푄푘 휕푄푘 휕퐺 ⇒ 푃푖, 퐺 푞,푝 = − σ푘 푄푘, 푃푖 휕푄푘 27 6.5 Poisson’s Brackets
휕퐺 휕퐺 ⇒ 푃푖, 퐺 푞,푝 = − σ푘 훿푖푘 = − for 푖 = 푘 휕푄푘 휕푄푘 휕퐺 And ⇒ 퐺, 푃푘 푞,푝 = 휕푄푘 휕퐹 Now replacing G by F, we get 퐹, 푃푘 = (III) 휕푄푘 Putting 푎 and 푏 in 푒푞 1
휕퐺 휕퐺 휕퐺 휕퐹 휕퐺 휕퐹 ⇒ 퐹, 퐺 푞,푝 = σ푘 퐹, 푄푘 푞,푝 + 퐹, 푃푘 푞,푝 = σ푘 − + 휕푄푘 휕푃푘 휕푄푘 휕푃푘 휕푃푘 휕푄푘 휕퐹 휕퐺 휕퐹 휕퐺 ⇒ 퐹, 퐺 푞,푝 = σ푘 − = 퐹, 퐺 푃,푄 휕푄푘 휕푃푘 휕푃푘 휕푄푘
⇒ 퐹, 퐺 푞,푝 = 퐹, 퐺 푃,푄
28 6.5 Jacobi Identity
푢, 푣, 푤 + 푣, 푤, 푢 + 푤, 푢, 푣 = 0
푛 휕푢 휕푣 휕푢 휕푣 Solution Since 푢, 푣 = σ푘=1 − 휕푞푘 휕푝푘 휕푝푘 휕푞푘
푛 휕푢 휕 휕푢 휕 푢, 푣 = σ푘=1 − 푣 휕푞푘 휕푝푘 휕푝푘 휕푞푘
푢, 푣 = 퐷푢푣
2푛 휕 Note Where 퐷푢 = σ푖 훼푖 휕휉푖 휕푢 휕 휕 For 1 to n 훼푖 = & = 휕푞푘 휕휉푖 휕푝푘 휕 휕푢 휕 휕 σ2푛 For n+1 to 2n 훼 = & = Similarly 퐷푣 = 푗 훽푗 푖 휕푝 휕휉 휕푞 휕휉푗 푘 푖 푘 Now considering first two terms
푢, 푣, 푤 + 푣, 푤, 푢 = 푢, 푣, 푤 − 푣, 푢, 푤 = 푢, 퐷푣푤 − 푣, 퐷푢 푤 29 6.5 Jacobi Identity
⇒ 푢, 푣, 푤 + 푣, 푤, 푢 = 푢, 퐷푣푤 − 푣, 퐷푢 푤 = 퐷푢 퐷푣푤 − 퐷푣 퐷푢푤
2푛 휕 2푛 휕 2푛 휕 2푛 휕 ⇒ 푢, 푣, 푤 + 푣, 푤, 푢 = σ푖 훼푖 σ푗 훽푗 푤 − σ푗 훽푗 σ푖 훼푖 푤 휕휉푖 휕휉푗 휕휉푗 휕휉푖
2 2 2푛 휕 푤 2푛 휕훽푗 휕푤 2푛 휕 푤 2푛 휕훼푖 휕푤 ⇒ 푢, 푣, 푤 + 푣, 푤, 푢 = σ푖,푗 훼푖 훽푗 + σ푖,푗 훼푖 − σ푖,푗 훼푖 훽푗 − σ푖,푗 훽푗 휕휉푖휕휉푗 휕휉푖 휕휉푗 휕휉푗휕휉푖 휕휉푗 휕휉푖
2푛 휕훽푗 휕푤 2푛 휕훼푖 휕푤 ⇒ 푢, 푣, 푤 + 푣, 푤, 푢 = σ푖,푗 훼푖 − σ푖,푗 훽푗 휕휉푖 휕휉푗 휕휉푗 휕휉푖
2푛 휕훽푗 휕훼푗 휕푤 ⇒ 푢, 푣, 푤 + 푣, 푤, 푢 = σ푖,푗 훼푖 − 훽푖 휕휉푖 휕휉푖 휕휉푗 By using the property that sum is not effected if the indices are interchanged (dummy indices)
30 6.5 Jacobi Identity
2푛 휕훽푗 휕훼푗 휕푤 푛 휕훽푗 휕훼푗 휕푤 2푛 휕훽푗 휕훼푗 휕푤 ⇒ σ푖,푗 훼푖 − 훽푖 = σ푖,푗 훼푖 − 훽푖 + σ푖,푗=푛+1 훼푖 − 훽푖 휕휉푖 휕휉푖 휕휉푗 휕휉푖 휕휉푖 휕휉푗 휕휉푖 휕휉푖 휕휉푗
2푛 휕훽푗 휕훼푗 휕푤 휕푤 휕푤 ⇒ σ푖,푗 훼푖 − 훽푖 = σ푖 퐴푗 + 퐵푗 휕휉푖 휕휉푖 휕휉푗 휕푝푖 휕푞푖
휕푝푗 휕푝푗 휕푝푗 Replacing w by 푝푗 ⇒ σ푖 퐴푗 + 퐵푗 = σ푖 퐴푗 = σ푖 퐴푖훿푖푗 = 퐴푗 휕푝푖 휕푞푖 휕푝푖
⇒ 푢, 푣, 푝푗 − 푣, 푢, 푝푗 = 퐴푗
휕푣 휕푢 ⇒ 퐴푗 = 푢, − 푣, 휕푞푗 휕푞푗
휕푣 휕푢 휕 ⇒ 퐴푗 = 푢, + , 푣 = 푢, 푣 휕푞푗 휕푞푗 휕푞푗 31 6.5 Jacobi Identity
Replacing w by 푞푗
푢, 푣, 푞푗 − 푣, 푢, 푞푗 = 퐵푗
휕푣 휕푢 ⇒ 퐵푗 = − 푢, + 푣, 휕푝푗 휕푝푗
휕 ⇒ 퐵푗 = − 푢, 푣 휕푝푗
휕푤 휕 휕푤 휕 ⇒ 푢, 푣, 푤 + 푣, 푤, 푢 = σ푖 푢, 푣 − 푢, 푣 휕푝푖 휕푞푗 휕푞푗 휕푝푗 ⇒ 푢, 푣, 푤 + 푣, 푤, 푢 = − 푤, 푢, 푣 ⇒ 푢, 푣, 푤 + 푣, 푤, 푢 + 푤, 푢, 푣 = 0 32 6.5 Poisson’s Brackets
33 34 35 36