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Download Sect. 2.7 - TTU Physics
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Astronomy before computers! Sect. 2.7: Energy Function & Energy Conservation • One more conservation theorem which we would expect to get from the Lagrange formalism is: CONSERVATION OF ENERGY. • Consider a general Lagrangian L, a function of the coords qj, velocities qj, & time t: L = L(qj,qj,t) (j = 1,…n) • The total time derivative of L (chain rule): (dL/dt) = ∑j(∂L/∂qj)(dqj/dt) + ∑j(∂L/∂qj)(dqj/dt) + (∂L/∂t) Or: (dL/dt) = ∑j(∂L/∂qj)qj + ∑j(∂L/∂qj)qj + (∂L/∂t) • Total time derivative of L: (dL/dt) = ∑j(∂L/∂qj)qj + ∑j(∂L/∂qj)qj + (∂L/∂t) (1) • Lagrange’s Eqtns: (d/dt)[(∂L/∂qj)] - (∂L/∂qj) = 0 Put into (1) (dL/dt) = ∑j(d/dt)[(∂L/∂qj)]qj + ∑j(∂L/∂qj)qj + (∂L/∂t) Identity: 1st 2 terms combine (dL/dt) = ∑j(d/dt)[qj(∂L/∂qj)] + (∂L/∂t) Or: (d/dt)[∑jqj(∂L/∂qj) - L] + (∂L/∂t) = 0 (2) (d/dt)[∑jqj(∂L/∂qj) - L] + (∂L/∂t) = 0 (2) • Define the Energy Function h: h ∑jqj(∂L/∂qj) - L = h(q1,..qn;q1,..qn,t) • (2) (dh/dt) = - (∂L/∂t) For a Lagrangian L which is not an explicit function of time (so that (∂L/∂t) = 0) (dh/dt) = 0 & h = constant (conserved) • Energy Function h = h(q1,..qn;q1,..qn,t) – Identical Physically to what we later will call the Hamiltonian H. However, here, h is a function of n indep coords qj & velocities qj. The Hamiltonian H is ALWAYS considered a function of 2n indep coords qj & momenta pj • Energy Function h ∑jqj(∂L/∂qj) - L • We had (dh/dt) = - (∂L/∂t) For a Lagrangian for which (∂L/∂t) = 0 (dh/dt) = 0 & h = constant (conserved) • For this to be useful, we need a Physical Interpretation of h. – Will now show that, under certain circumstances, h = total mechanical energy of the system. Physical Interpretation of h • Energy Function h ∑jqj(∂L/∂qj) - L • Recall (Sect. 1.6) that we can always write KE as: T = M0 + ∑jMjqj + ∑jMjkqjqk M0 (½)∑imi(∂ri/∂t)2 , Mj ∑imi(∂ri/∂t)(∂ri/∂qj) Mjk ∑i mi(∂ri/∂qj)(∂ri/∂qk) Or (schematically) T = T0(q) + T1(q,q) + T2(q,q) – T0 M0 independent of generalized velocities – T1 ∑jMjqj linear in generalized velocities – T2 ∑jMjkqjqk quadratic in generalized velocities • With almost complete generality, we can write (schematically) the Lagrangian for most problems of interest in mechanics as: L = L0(q,t) + L1(q,q,t) + L2(q,q,t) L0 independent of the generalized velocities L1 linear in generalized the velocities L2 quadratic in generalized the velocities – For conservative forces, L has this form. Also does for some velocity dependent potentials, such as for EM fields. L = L0(q,t) + L1(q,q,t) + L2(q,q,t) (1) • Euler’s Theorem from mathematics: If f = f(x1,x2,.. xN) = a homogeneous function of degree n of the variables xi, then ∑ixi(∂f/∂xi) = n f (2) • Energy Function h ∑jqj(∂L/∂qj) - L (3) • For L of form (1): (2) h = 0L0 + 1L1+2L2 - [L0 + L1 + L2] or h = L2 - L0 L = L0(q,t) + L1(q,q,t) + L2(q,q,t) Energy function h ∑jqj(∂L/∂qj) - L = L2 - L0 • Special case (both conditions!): a.) The transformation eqtns from Cartesian to Generalized Coords are time indep. In the KE, T0 = T1 = 0 T = T2 b.) V is velocity indep. L2 = T = T2 & L0 = -V h = T + V = E Total Mechanical Energy • Under these conditions, if V does not depend on t, neither does L & thus (∂L/∂t) = 0 = (dh/dt) so h = E = constant (conserved) Energy Conservation • Summary: Different Conditions: Energy Function h ∑jqj(∂L/∂qj) - L ALWAYS: (dh/dt) = - (∂L/∂t) SOMETIMES: L does not depend on t (∂L/∂t) = 0, (dh/dt) = 0 & h = const. (conserved) USUALLY: L = L0(q,t) + L1(q,q,t) + L2(q,q,t) h ∑jqj(∂L/∂qj) - L = L2 - L0 SOMETIMES: T = T2 = L2 AND L0 = -V h = T + V = E Total Mechanical Energy Conservation Theorem for Mechanical Energy: If h = E AND L does not depend on t, E is conserved! • Clearly, the conditions for conservation of energy function h are DISTINCT from those which make it the total mechanical energy E. Can have conditions in which: 1. h is conserved & = E 2. h is not conserved & = E 3. h is conserved & E 4. h is not conserved & E Most common case in classical (& quantum) mech. is case 1. • Stated another way: Two questions: 1. Does the energy function h = E for the system? 2. Is the mechanical energy E conserved for the system? • Two aspects of the problem! DIFFERENT questions! – May have cases where h E, but E is conserved. – For example: A conservative system, using generalized coords in motion with respect to fixed rectangular axes: Transformation eqtns will contain the time T will NOT be a homogeneous, quadratic function of the generalized velocities! h E, However, because the system is conservative, E is conserved! (This is a physical fact about the system, independent of coordinate choices!). • It is also worth noting: The Lagrangian L = T - U is independent of the choice of generalized coordinates. The Energy function h ∑jqj(∂L/∂qj) - L depends on the choice of generalized coordinates. • The most common case in classical (& quantum) mechanics is h = E and E is conserved. Non-Conservative Forces • Consider a non-conservative system: Frictional forces obtained from the dissipation function ₣ . Derivation with the energy function h becomes: (dh/dt) + (∂L/∂t) = - ∑jqj(∂₣ /∂qj) • Ch. 1: The formulation of ₣ shows it is a homogeneous, quadratic function of the q’s. Use Euler’s theorem again: ∑jqj(∂₣ /∂qj) = 2₣ (dh/dt) = - (∂L/∂t) - 2₣ • If L is not an explicit function of time (∂L/∂t =0 ) AND h = E: (dE/dt) = - 2₣ – That is, under these conditions, 2₣ = Energy dissipation rate. Symmetry Properties & Conservation Laws (From Marion’s Book!) • In general, in physical systems: A Symmetry Property of the System Conservation of Some Physical Quantity Also: Conservation of Some Physical Quantity A Symmetry Property of the System • Not just valid in classical mechanics! Valid in quantum mechanics also! Forms the foundation of modern field theories (Quantum Field Theory, Elementary Particles,…) We’ve seen in general that: Conservation Theorem: If the Generalized Coord qj is cyclic or ignorable, the corresponding Generalized (or Conjugate) Momentum, pj (∂L/∂qj) is conserved. • An underlying symmetry property of the system: If qj is cyclic, the system is unchanged (invariant) under a translation (or rotation) in the “qj direction”. pj is conserved Linear Momentum Conservation • Conservation of linear momentum: If a component of the total force vanishes, nF = 0, the corresponding component of total linear momentum np = const (is conserved) • Underlying symmetry property of the system: The system is unchanged (invariant) under a translation in the “n direction”. np is conserved Angular Momentum Conservation • Conservation of angular momentum: If a component of total the torque vanishes, nN = 0, the corresponding component of total angular momentum nL = const (conserved) • Underlying symmetry property of the system: The system is unchanged (invariant) under a rotation about the “n direction”. nL is conserved Energy Conservation • Conservation of mechanical energy: If all forces in the system are conservative, the total mechanical energy E = const (conserved) • Underlying symmetry property of the system: (More subtle than the others!) The system is unchanged (invariant) under a time reversal. (Changing t to -t in all eqtns of motion) E is conserved Summary: Conservation Laws • Under the proper conditions, there can be up to 7 “Constants of the Motion” “1st Integrals of the Motion” Quantities which are Conserved (const in time): Total Mechanical Energy (E) 3 vector components of Linear Momentum (p) 3 vector components of Angular Momentum (L)