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Mechanics Physics 151 Lecture 18 Hamiltonian Equations of Motion (Chapter 8) What’s Ahead We are starting Hamiltonian formalism May not cover Hamilton-Jacobi theory Hamiltonian equation – Today and 11/26 Canonical transformation – 12/3, 12/5, 12/10 Close link to non-relativistic QM Cute but not very relevant What shall we do in the last 2 lectures? Classical chaos? Perturbation theory? Classical field theory? Send me e-mail if you have preference! Hamiltonian Formalism Newtonian Æ Lagrangian Æ Hamiltonian Describe same physics and produce same results Difference is in the viewpoints Symmetries and invariance more apparent Flexibility of coordinate transformation Hamiltonian formalism linked to the development of Hamilton-Jacobi theory Classical perturbation theory Quantum mechanics Statistical mechanics Lagrange Æ Hamilton Lagrange’s equations for n coordinates d ⎛ ∂L ⎞ ∂L = 0 i = 1,… , n ⎜ ⎟− dt ⎝ ∂qi ⎠ ∂qi 2nd-order differential equation of n variables n equations Æ 2n initial conditions qi (t = 0) qi (t = 0) Can we do with 1st-order differential equations? Yes, but you’ll need 2n equations We keep qi and replace qi with something similar ∂L(q j , q j , t ) We take the conjugate momenta pi ≡ ∂qi Configuration Space We considered (q1 ,… , qn ) as a point in an n-dim. space Called configuration space Motion of the system Æ A curve in the config space qi = qi (t ) When we take variations, we consider qi and qi as independent variables i.e., we have 2n independent variables in n-dim. space Isn’t it more natural to consider the motion in 2n-dim space? Phase Space Consider coordinates and momenta as independent State of the system is given by (q1 ,… , qn , p1 ,… , pn ) Consider it a point in the 2n-dimensional phase space We are switching the independent variables (qi , qi , t ) → (qi , pi , t ) A bit of mathematical trick is needed to do this qi = qi (t ) pi = pi (t ) Legendre Transformation Start from a function of two variables f ( x, y ) Total derivative is ∂f ∂f df = dx + dy ≡ udx + vdy ∂x ∂y Define g ≡ f − ux and consider its total derivative dg = df − d (ux) = udx + vdy − udx − xdu = vdy − xdu i.e. g is a function of u and y ∂g If f = L and ( x, y ) = ( q, q ) ∂g = −x =v ∂u ∂y L(q, q ) → g ( p, q ) = L − pq This is what we need Hamiltonian Opposite sign from Legendre transf. Define Hamiltonian: H (q, p, t ) = qi pi − L(q, q, t ) Total derivative is ∂L ∂L ∂L dH = pi dqi + qi dpi − dqi − dqi − dt ∂qi ∂qi ∂t ∂L Lagrange’s equations say = pi ∂qi ∂L dH = qi dpi − pi dqi − dt ∂t This must be equivalent to ∂H ∂H ∂H dH = dpi + dqi + dt ∂pi ∂qi ∂t Putting them together gives… Hamilton’s Equations ∂H ∂H ∂H ∂L =− = qi = − pi and We find ∂t ∂t ∂pi ∂qi 2n equations replacing the n Lagrange’s equations 1st-order differential instead of 2nd-order “Symmetry” between q and p is apparent There is nothing new – We just rearranged equations First equation links momentum to velocity This relation is “given” in Newtonian formalism Second equation is equivalent to Newton’s/Lagrange’s equations of motion Quick Example Particle under Hooke’s law force F = –kx L= m 2 k 2 x − x 2 2 p= ∂L = mx ∂x m 2 k 2 H = xp − L = x + x 2 2 p2 k 2 = + x 2m 2 Hamilton’s equations ∂H ∂H p = − kx p=− x= = ∂x ∂p m Replace x with p m Usual harmonic oscillator Energy Function Definition of Hamiltonian is identical to the energy function h(q, q, t ) = qi ∂L − L(q, q, t ) ∂qi Difference is subtle: H is a function of (q, p, t) This equals to the total energy if Lagrangian is L = L0 (q, t ) + L1 (q, t )qi + L2 (q, t )q j qk Constraints are time-independent This makes T = L2 ( q, t ) q j qk See Lecture 4, or Forces are conservative Goldstein Section 2.7 This makes V = − L0 ( q ) Hamiltonian and Total Energy If the conditions make h to be total energy, we can skip calculating L and go directly to H For the particle under Hooke’s law force p2 k 2 H = E = T +V = + x 2m 2 This works often, but not always when the coordinate system is time-dependent e.g., rotating (non-inertial) coordinate system when the potential is velocity-dependent e.g., particle in an EM field Let’s look at this Particle in EM Field For a particle in an EM field m 2 L = xi − qφ + qAi xi 2 pi = mxi + qAi We can’t jump on H = E because of the last term, but mxi2 H = (mxi + qAi ) xi − L = + qφ This is in fact E 2 We’d be done if we were calculating h For H, we must rewrite it using pi = mxi + qAi ( pi − qAi ) 2 H ( xi , pi ) = + qφ 2m Particle in EM Field Hamilton’s equations are ∂H pi − qAi xi = = ∂pi m ( pi − qAi ) 2 H ( xi , pi ) = + qφ 2m p j − qAj ∂Aj ∂H ∂φ pi = − =q −q ∂xi ∂xi ∂xi m Are they equivalent to the usual Lorentz force? Check this by eliminating pi ∂Aj d ∂φ (mxi + qAi ) = qxi −q ∂xi ∂xi dt A bit of work d (mvi ) = qEi + q ( v × B)i dt Conservation of Hamiltonian Consider time-derivative of Hamiltonian dH (q, p, t ) ∂H ∂H ∂H q+ p+ = ∂q ∂p ∂t dt ∂H = − pq + qp + ∂t Hamiltonian is conserved if it does not depend explicitly on t H may or may not be total energy If it is, this means energy conservation Even if it isn’t, H is still a constant of motion Cyclic Coordinates A cyclic coordinate does not appear in L By construction, it does not appear in H either H ( q , p, t ) = qi pi − L( q , q, t ) Hamilton’s equation says ∂H Conjugate momentum of a =0 p=− cyclic coordinate is conserved ∂q Exactly the same as in the Lagrangian formalism Cyclic Example Central force problem in 2 dimensions m 2 2 2 L = (r + r θ ) − V (r ) 2 pr = mr pθ = mr 2θ 1 ⎛ 2 pθ2 ⎞ H= ⎜ pr + 2 ⎟ + V (r ) r ⎠ 2m ⎝ 1 ⎛ 2 l2 ⎞ = ⎜ pr + 2 ⎟ + V (r ) r ⎠ 2m ⎝ θ is cyclic pθ = const = l Hamilton’s equations pr l2 ∂V (r ) r= pr = 3 − m mr ∂r Cyclic variable drops off by itself Going Relativistic Practical approach Purist approach Find a Hamiltonian that “works” Does it represent the total energy? Construct covariant Hamiltonian formalism For one particle in an EM field Don’t expect miracles Fundamental difficulties remain the same Practical Approach Start from the relativistic Lagrangian that “works” L = −mc 2 1 − β 2 − V (x) mvi ∂L = pi = ∂vi 1− β 2 H =h= Did this last time p 2 c 2 + m 2 c 4 + V ( x) It does equal to the total energy Hamilton’s equations ∂H ∂V pi c 2 pi ∂H pi = − =− = Fi = = xi = ∂xi ∂xi ∂pi p 2 c 2 + m2 c 4 mγ Practical Approach w/ EM Field Consider a particle in an EM field L = −mc 2 1 − β 2 − qφ (x) + q ( v ⋅ A) Hamiltonian is still total energy H = mγ c 2 + qφ Can be easily checked = m 2γ 2 v 2 c 2 + m2 c 4 + qφ Difference is in the momentum pi = mγ vi + qAi H = (p − qA) 2 c 2 + m 2 c 4 + qφ Not the usual linear momentum! Practical Approach w/ EM Field H = (p − qA) 2 c 2 + m 2 c 4 + qφ Consider H – qφ ( H − qφ ) 2 − (p − qA)2 c 2 = m2 c 4 constant It means that ( H − qφ , pc − qAc) is a 4-vector, and so is ( H , pc) Similar to 4-momentum (E/c, p) of a relativistic particle Remember p here is not the linear momentum! This particular Hamiltonian + canonical momentum transforms as a 4-vector True only for well-defined 4-potential such as EM field Purist Approach Covariant Lagrangian for a free particle Λ = 12 muν uν ∂Λ µ = mu µ p = ∂uµ H= We know that p0 is E/c We also know that x0 is ct… pµ p µ 2m Energy is the conjugate “momentum” of time Generally true for any covariant Lagrangian You know the corresponding relationship in QM Purist Approach Value of Hamiltonian is pµ p µ mc 2 = H= This is constant! 2m 2 What is important is H’s dependence on pµ Hamilton’s equations µ µ dx ∂H p = = dτ ∂pµ m µ dp ∂H =− =0 dτ ∂xµ Time components are d (ct ) E d ( E c) = =γc =0 dτ mc dτ 4-momentum conservation Energy definition and conservation Purist Approach w/ EM Field With EM field, Lagrangian becomes Λ( x µ , u µ ) = 12 muµ u µ + qu µ Aµ H= muµ u µ 2 = p µ = mu µ + qAµ ( pµ − qAµ )( p µ − qAµ ) 2m Hamilton’s equations are ( pν − qAν ) ∂Aν ∂H dx µ ∂H p µ − qAµ dp µ =− =− = = dτ ∂xµ m ∂xµ dτ ∂pµ m A bit of work can turn them into ⎛ ∂Aν ∂Aµ ⎞ du µ = q⎜ − m uν = K µ ⎟ 4-force ⎜ ∂x ⎟ ∂ dτ x ν ⎠ ⎝ µ EM Field and Hamiltonian In Hamiltonian formalism, EM field always modify the canonical momentum as p Aµ = p0µ + qAµ With EM field Without EM field A handy rule: Hamiltonian with EM field is given by replacing pµ in the field-free Hamiltonian with pµ – qAµ Often used in relativistic QM to introduce EM interaction Summary Constructed Hamiltonian formalism Equivalent to Lagrangian formalism Simpler, but twice as many, equations Hamiltonian is conserved (unless explicitly t-dependent) Equals to total energy (unless it isn’t) (duh) Cyclic coordinates drops out quite easily A few new insights from relativistic Hamiltonians Conjugate of time = energy pµ – qAµ rule for introducing EM interaction