Mechanics Oscillations, Lecture Notes - Physics, Study notes of Mechanics

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Mechanics Physics 151 Lecture 12 Oscillations (Chapter 6) What We Did Last Time Analyzed the motion of a heavy top Reduced into 1-dimensional problem of θ Qualitative behavior Precession + nutation Initial condition vs. behavior Magnetic dipole moment of spinning charged object M = γL, where γ = q/2m is the gyromagnetic ratio L precesses in magnetic field by ω = –γB γ of elementary particles contains interesting physics Problem Definition Kinetic energy is a 2nd-order homogeneous function of velocities This requires that the transformation functions do not explicitly depend on time, i.e. mij generally depend on {qi} Taylor expansion 1 1 1( , , ) 2 2ij n i j ij i j T m q q q q m ηη= =… ( ) 0 0 ij ij ij k ij k m m m T q η ∂⎛ ⎞ = + + ≈⎜ ⎟∂⎝ ⎠ 1 2 ij i j TT ηη≈ 1( , , , )i i Nq q x x t= … Constant symmetric matrix Lagrangian For small deviation {ηi} from equilibrium The equations of motion are This looks similar to Difficulty: Tij and Vij have off-diagonal components If T and V were diagonal 1 1 1 1 2 2 2 2ij i j ij i j L T V T Vηη ηη= − = − = −ηTη ηVη 0ij j ij jT Vη η+ = (Textbook Eq 6.8 wrong) 0mx kx+ = 0ij j ij j ii i ii iT V T Vη η η η+ → + = no sum over i Can we find a new set of coordinates that diagonalizes T and V? Mechanics Physics 151 Lecture 12 Oscillations (Chapter 6) What We Did Last Time Analyzed the motion of a heavy top Reduced into 1-dimensional problem of θ Qualitative behavior Precession + nutation Initial condition vs. behavior Magnetic dipole moment of spinning charged object M = γL, where γ = q/2m is the gyromagnetic ratio L precesses in magnetic field by ω = –γB γ of elementary particles contains interesting physics Problem Definition Kinetic energy is a 2nd-order homogeneous function of velocities This requires that the transformation functions do not explicitly depend on time, i.e. mij generally depend on {qi} Taylor expansion 1 1 1( , , ) 2 2ij n i j ij i j T m q q q q m ηη= =… ( ) 0 0 ij ij ij k ij k m m m T q η ∂⎛ ⎞ = + + ≈⎜ ⎟∂⎝ ⎠ 1 2 ij i j TT ηη≈ 1( , , , )i i Nq q x x t= … Constant symmetric matrix Lagrangian For small deviation {ηi} from equilibrium The equations of motion are This looks similar to Difficulty: Tij and Vij have off-diagonal components If T and V were diagonal 1 1 1 1 2 2 2 2ij i j ij i j L T V T Vηη ηη= − = − = −ηTη ηVη 0ij j ij jT Vη η+ = (Textbook Eq 6.8 wrong) 0mx kx+ = 0ij j ij j ii i ii iT V T Vη η η η+ → + = no sum over i Can we find a new set of coordinates that diagonalizes T and V? Solving Lagrange’s Equations Assume that solution will be It’s a slightly odd eigenvalue equation Solution can be found by n-th order polynomial of λ Expect n solutions for λ λ must be real and λ = ω2 > 0 0ij j ij jT Vη η+ = i t i iCa e ωη −= 2 0ij j ij jT a V aω− + = 0λ− =Va Taor 0λ− =V T Can be proven by a bit of work 2λ ω= We now have an all-real equation We now have a guarantee that each solution of the eigenvalue equation gives an oscillating solution with a definite frequency Positive Definiteness λ=aVa aTa λ = aVa aTa Already shown to be positive Positive because for any real η if V is minimum at the equilibrium 1 2 0V = ≥ηVη λ = ω2 is positive definite i tC e ω−=η a 2λ ω= Normalization Eigenvector satisfying Va = λTa has arbitrary scale a Ca can absorb such scale as well as imaginary phase We fix the normalization by declaring Just the sign (±) remains ambiguous This turns 1=aTa λ = aVa aTa λ = aVa Principal Axis Transformation There are n eigenvectors Call them aj Take transpose If we stack aj to make j j jλ=Va Ta 1, 2,...,j n= k k kλ=a V a T ( ) 0j k k jλ λ− =a Ta k j jkδ=a Ta Assuming λj ≠ λk for j ≠ k =ATA 1 1 0 0 n λ λ ⎡ ⎤ ⎢ ⎥= = ⎢ ⎥ ⎢ ⎥⎣ ⎦ AVA λ T and V are diagonalized by the principal axis transformation no sum over j or k [ ]1 2 n=A a a a k j j jkλ δ=a Va Linear Triatomic Molecule Consider a molecule like CO2 Consider only motion along the axis We want to solve eigenvalue equation m mm 2 2 2 1 2 3( )2 mT η η η= + + 2 22 1 3 2( ) ( )2 2 k kV η η η η= − + − 0 0 0 0 0 0 m m m ⎡ ⎤ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎣ ⎦ T 0 2 0 k k k k k k k −⎡ ⎤ ⎢ ⎥= − −⎢ ⎥ ⎢ ⎥−⎣ ⎦ V 2( ) 0ω− =V T a Linear Triatomic Molecule Solutions are 2 2 2 2 0 2 0 0 k m k k k m k k k m ω ω ω ω − − − = − − − = − − V T 2 2 2( )(3 ) 0k m k mω ω ω− − = 1 0ω = 2 k m ω = 3 3k m ω = 1 1 1 1 3 1m ⎡ ⎤ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎣ ⎦ a 2 1 1 0 2 1m ⎡ ⎤ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥−⎣ ⎦ a 3 1 1 2 6 1m ⎡ ⎤ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥⎣ ⎦ a Is this OK? Linear Triatomic Molecule First solution is linear movement of the molecule This is not an “oscillation” Consider it as an oscillation with infinitely long period Although V is minimum at the equilibrium, it does not increase when the whole molecule is shifted Position of the CoM is a cyclic coordinate Total momentum is conserved 1 0ω = 1 1 1 1 3 1m ⎡ ⎤ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎣ ⎦ a m mm Degenerate Solutions We assumed λj ≠ λk for j ≠ k What if the eigenvalue equation has multiple roots? Can we still achieve ? Quick answer: Don’t worry Multiple root corresponds to multiple eigenvectors Any linear combination cjaj is also an eigenvector It is always possible to find a set of m orthogonal vectors =ATA 1 ( ) ( ) 0m fλ λ κ λ− = − =V T λ = κ is an m-fold root ( ) 0 ( 1, , )j j mκ− = =V T a … m eigenvectors Recipe For an m-fold root and m eigenvectors First normalize a1 with Next turn a2 into This satisfies Normalize a2’ with Next turn a3 into This satisfies and Continue… It is now guaranteed that we can make 1 1 1=a Ta 2 2 1 2 1( )′ = −a a a Ta a 1 2 0′ =a Ta 2 2 1′ ′ =a Ta 3 3 1 3 1 2 3 2( ) ( )′ ′ ′= − −a a a Ta a a Ta a 1 3 0′ =a Ta 2 3 0′ ′ =a Ta =ATA 1 Summary Studied oscillation Discussed general features of multi-dimensional oscillators Equation of motion Eigenvalue problem Showed that oscillating solutions exist Eigenvalues ω2 are positive definite Provided that V is minimum at the equilibrium Principal axis transformation diagonalizes T and V Normal coordinates behave as independent oscillators Next: forced oscillation λ=Va Ta i t i iCa e ωη −=