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.