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PHY411 Lecture Notes Part 8

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PHY411 Lecture Notes Part 8 PHY411 Lecture notes Part 8 Alice Quillen November 26, 2018 Contents 1 Continuum Mechanics 3 1.1 Strings . 3 2 Infinite dimensional Classical Hamiltonian systems 5 2.1 The functional derivative . 5 3 Examples with canonical coordinates 8 3.1 Wave equation . 8 3.2 Two-dimensional incompressible inviscid fluid dynamics . 12 3.2.1 Dynamics of vortices in 2D . 13 3.2.2 Blinking Vortex model and time dependent models . 14 3.3 Conformal mapping . 14 4 Inviscid Barotropic Compressible Fluids 15 4.1 Eulerian and Lagrangian coordinates . 15 4.2 Inviscid Burger's equation . 16 4.3 Working with the Determinant of the deformation matrix . 17 4.3.1 Functional derivatives involving the Jacobian matrix . 19 4.4 For Barotropic Compressible Fluids . 20 4.5 Lagrangian for Compressible Fluids . 21 4.6 Hamiltonian for Compressible Barotropic Fluids . 22 4.7 Changing coordinate fields and constructing a new Poisson bracket . 23 4.8 Another way to write Inviscid irrotational fluid dynamics . 26 5 Poisson manifolds 27 5.1 Example: Inviscid Burgers' equation . 29 5.1.1 The symplectic two form . 31 5.2 Conserved Quantities and Continuity Equations . 32 1 6 The KdV equation 33 6.1 Rescaling . 35 6.2 Soliton Solutions . 35 6.3 Infinite number of conserved quantities . 36 6.3.1 Integrability . 39 6.4 Lax pair . 39 6.5 Connection to Schr¨odingerequation . 41 6.6 Connection to Inverse Scattering method . 41 6.7 Zero Curvature Condition . 42 6.8 Two different Poisson brackets . 45 7 Symplectic Reduction 48 Figure 1: A string! 2 1 Continuum Mechanics 1.1 Strings Consider a string of linear mass µ(x) and under tension τ(x) where x is horizontal distance along the string when it is at rest. Its vertical displacement we describe with a function y(x; t) that is also a function of time. When the string is at rest y(x; t) = 0. A length of the string due to displacement is p ds = dx2 + dy2 s @y 2 = 1 + dx @x ! 1 @y 2 ∼ dx 1 + (1) 2 @x The change in length locally is dx @y 2 dl = ds − dx = 2 @x The differential potential energy (force times distance) 1 @y 2 dU = τ(x)dl = τ(x) dx 2 @x The differential kinetic energy 1 @y 2 dT = µ(x) dx 2 @t We find the total Lagrangian by integrating over x, Z L = dx L(y; y;_ y0; x; t) and L we call the Lagrangian density 1 @y 2 1 @y 2 L(y; y;_ y0; x; t) = µ(x) − τ(x) (2) 2 @t 2 @x The action is a double integral Z tb Z xb Z tb Z xb 1 S = dt dxL(y; y;_ y0; x; t) = dt dx µy_2 − τy02 ta xa ta xa 2 3 When we minimize the action we can consider minimizing it for paths y(x; t) ! y(x; t)+ h(x; t). Z tb Z xb h i @S = dt dx L(y + h; y_ + h;_ y0 + h0; x; t) − L(y; y;_ y0; x; t) ta xa Z tb Z xb @L @L @L _ 0 = dt dx h + h + 0 h (3) ta xa @y @y_ @y We integrate by parts " # Z xb Z tb Z xb Z tb tb @L d @L @L dx dt h_ = dx − h + h @y_ dt @y_ @y_ xa ta xa ta ta " # Z xb Z tb Z tb Z xb xb @L 0 d @L @L dx dt h = dt − h + h @y0 dx @y0 @y0 xa ta ta xa xa Inserting these into equation 3 Z tb Z xb Z xb tb Z tb xb @L d @L d @L @L @L @S = dt dx − − h + dx h + dt h @y dt @y_ dx @y0 @y_ @y0 ta xa xa ta ta xa With suitable boundary conditions @L d @L d @L − − = 0 @y dt @y_ dx @y0 Evaluating the derivatives with τ; µ independent of x and t for the Lagrangian in equation 2 @L @L = µy_ = −τy0 @y_ @y0 giving @2y @2y µ − τ = 0 @t2 @x2 which is the wave equation. We can perform a Legendre transformation (at each value of x) @L π(x) = = µy_(x) @y_(x) Inverting this y_(π(x)) = π(x)/µ and we find a Hamiltonian or energy density π(x)2 y0(x)2 H(π(x); y(x); y0(x)) = π(x)_y(π) − L(y; y_(π); y0; x) = + τ 2µ 2 4 The total energy is integrated over x Z E = dxH(π(x); y(x); y0(x)) and because the Hamiltonian does not depend on time, the total integrated energy should be a constant. It is confusing to see our functions depend on y0 here. It is helpful to think in the limit of a discrete system as the number of discrete items becomes large. The string can be considered as a large N limit of many coupled bodies, where the coupling depends on the differences between nearby bodies that is here represented by the spatial derivative. Each discrete body is at a different x, so we can think of N different bodies at N different x positions. In the large N limit the number of bodies becomes infinite. In this sense π(x) is an infinite dimensional vector of momenta conjugate to y(x), an infinite dimensional vector of coordinates. Our Hamiltonian is really depend on momenta π(x) and coordinates y(x) and y0(x) is a function of nearby coordinates. 2 Infinite dimensional Classical Hamiltonian systems 2.1 The functional derivative Instead of just conjugate vectors p; q we can consider conjugate fields ρ(x) and φ(x). In the continuum limit the Hamiltonian is an integral over all space. Z H[ρ, φ] = dxf(ρ(x); φ(x)) and f(ρ, φ) a function that depends on the fields ρ(x); φ(x). The value of the fields at each position x serve as coordinates. These are functions of a continuous variable, effectively like an infinite number of coordinates. How do we take derivatives with respect to fields instead of with respect to coordinates? We use a functional derivative δF [u(x)] 1 ≡ lim (F [u(x) + δ(x − y)] − F [u(x)]) δu(y) !0 Here, as is common in physics, the delta function is used. Above F is a functional (operating on the function or field u) and u is a function. More rigorously the functional derivative can be defined Z δF 1 (x)g(x)dx = lim [F [ρ + g] − F [ρ]] δρ !0 d = [F [ρ + g]] (4) d =0 5 where F is a functional (function of functions) and ρ is a function and g is an arbitrary function. We can think of δF/δρ as the gradient of F at the point ρ. In analogy to X @F dF = dρ @ρ i i i Z δF [ρ] δF = g(x)dx δρ(x) with g(x)dx serving as dρi. A function can be written as a functional using an integral and a delta function Z Fx[φ] = dy δ(y − x)φ(y) = φ(x) For this point case the functional derivative δF [φ] 1 x = lim [φ(x) + δ(x − y) − φ(x)] δφ(y) !0 = δ(x − y) This follows as the value of the derivative can only change at the location of x. Consider a slightly more difficult but point case Fx[φ] = rφ(x) The functional derivative δFx[φ] 1 = lim [rx(φ(x) + δ(x − y)) − rxφ(x)] δφ(y) !0 = rxδ(x − y) The functional derivative satisfies the product rule δ(FG)[ρ] δF [ρ] δG[ρ] = G[ρ] + F [ρ] δρ(x) δρ(x) δρ(x) If g is a function then the chain rule is δF [g(ρ)] δF [g(ρ)] dg(ρ) = δρ(y) δg(ρ(y)) dρ(y) If G is an operator or a functional then the chain rule is Z δ δF [G] δG[ρ] F [G[ρ]] = dx δρ(y) δG(x) G=G[ρ] δρ(y) 6 Suppose we have a function Z F [ρ] = f(x; ρ(x); rρ(x))dx The functional derivative is this δF [ρ] @f @f = − r · : (5) δρ(x) @ρ @rρ We now show this using the definition for the functional derivative in equation 4 Z @F [ρ] d Z g(x)dx = f(x; ρ(x) + g(x); rρ + rg)dx δρ(x) d =0 1 Z @f @f = lim dx f(x) + g(x) + rg(x) − f(x) !0 @ρ @rρ Z @f @f = g + · rg dx @ρ @rρ Z @f @f @f = g + r · g − r · g dx @ρ @rρ @rρ Z @f @f = g − r · g dx @ρ @rρ where we have assumed that φ or f are zero on the boundary of the integral. The last line shows that we have derived equation 5. For a Hamiltonian H[ρ, φ], with ρ acting like a momentum and φ acting like a coordi- nate, we can write Hamilton's equations as @ρ(x) δH[φ, ρ] = − @t δφ(x) @φ(x) δH[φ, ρ] = @t δρ(x) in analogy to Hamilton's equations for discrete or finite systems. What is the Poisson bracket? The equation of motion should be consistent with φ_(x) = fφ(x);H[φ, ρ]g ρ_(x) = fρ(x);H[φ, ρ]g In the finite dimensional setting, the Poisson bracket is summed over derivatives with respect to all pairs of momenta and coordinates.
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