Exempel variationsräkning 2, SI1142 Fysikens matematiska metoder, vt08.

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1 Fysik KTH Exempel variationsräkning 2, SI1142 Fysikens matematiska metoder, vt8. These problems demonstrate various applications of variational calculus as used in physics. Problems (1) (3) illustrate an efficient method to derive differential equations in general curved coordinate systems. Problems (4) (6) are standard examples of dynamical which can be treated efficiently with variational calculus. In problems 7 and 8 we apply a powerful variational approximation method to some simple problems in quantum mechanics. 1. Particle moving in 3 in a potential U: Use the Hamilton principle to write the corresponding Newton s equations mẍ = U(x) in cylinder coordinates. 2. Use variational calculus to write the Helmholtz equation u + k 2 u = in 3 in cylinder coordinates. 3. We now consider the so-called parabolic coordinates (u, v, φ) on 3 defined as follows: x = uv cos(φ) y = uv sin(φ) z = 1 2 (u2 v 2 ). (i) Use variational calculus to derive Newton s equations mẍ = U(x) in this coordinate system. (ii) Use variational calculus to write the Helmholtz equation U + k 2 U = in this coordinate system. 4. A pearl of mass m moves frictionless on a rotating ring with radius which is rotating with constant angular frequency ω about an axis parallel to the earth gravitation force: if r, θ, φ are spherical coordinates then the pearl is constrained by r = and φ = ωt, and its only degree of freedom is θ. (i) Use the Hamilton principle to derive the equations of motion for the pearl. (ii) Give a qualitative discussion of the possible motions of the pearl. You can treat the pearl as a mass point. 5. (i) Derive the equation of motion for two coupled pendulums in the earth gravitational field using Hamilton s principle: a string 1 of length l 1 is fixed at the ceiling with a point mass with mass m 1 at its other end. Another string of length l 2 is fixed at this mass point, with another point mass m 2 at its other end. You can assume that the strings are rigid and ignore their masses, and that the motion of the point masses is restricted to the xz-plane where the z-axis is parallel to the gravitation force: the only degrees of freedom are the angles θ 1 and θ 2 between the z-axis and the lines parallel to the two strings 1 and 2. (ii) Find the solution of this systems assuming that θ 1,2 always remain small (harmonic approximation) and that the velocities of the mass points are zero and θ 1 =, θ 2 = a > at time t =. 1

2 6. (i) Use the Hamilton principle to derive the equations of motion of a particle with mass m and charge q in a electric and magnetic field E and B. (ii) Find the solution of these equations for the case when E = and B is constant. Hints: (i) The Lagrangian is L = m 2 ẋ2 + qẋ A qφ where A = A(x, t) and φ = φ(x, t) are scalar- and vector potential of the electro magnetic field: E = φ Ȧ and B = A. (ii) To determine the solution it is helpful to use the complex variable z = x + iy. 7. Betrakta den endimensionella Schrödingerekvationen där ψ xx (x) + V (x)ψ(x) = Eψ(x), V (x) = 9x 4. x Det är omöjligt att beräkna grundtillståndent (= lösningen ψ = ψ till Schrödinger ekvationen med minsta egenvärd E = E ) för denna potential V analytisk exakt, men man kan få en bra approximation genom att använda följande varationsprincip: Grundtillståndet till Schrödingerekvationen ovan är den reella funktion ψ (x) som minimerar funktionalen K[ψ] = dx (ψ x(x) 2 + V (x)ψ(x) 2 ) dxψ(x)2. (a) Bevisa variationsprincipen ovan. (b) Använda variationsprincipen för att beräkna konstanten a > så att funktionen φ(x) = e ax2 /2 ger den bästa approximationen till grundtillståndet ψ, dvs., så att K[φ] är minst. Beräkna också motsvarande approximation till minsta egenvärdet E. Ledning: dxx 2k e ax2 /2 (2k 1)!! 2π = a k a, a >, k =, 1, 2,..., och (2k 1)!! = (2k 1). 8. Same as in the previous problem, but for the 2D Schrödinger equation with the potential V (x, y) = g(x 2 + y 2 ) 2, g >, (x, y) In your mechanics course you discussed Newton s equations in rotating coordinate systems. The problem below derives these equations of motion using Variational calculus. Use Hamilton s principle to derive Newton s equations of motion for a particle moving in a rotation invariant potential, V = V ( r ), and in a frame of reference rotating about a fixed axes going through r = with constant angular frequency ω. 2

3 Exempel: outlines of solutions. NOTE: The solutions are in part rather sketchy! 1. Use method explained in the solution of problem 3 below. 2. Use method explained in the solution of problem 3 below. 3. (i) We know that the equations of motion are the Euler-Lagrange equations for the functional dt L(x,ẋ) with L = m 2 ẋ2 U(x). We thus only need to express L as functions of (u, v, φ). We compute L ẋ 2 = ( x u u + x v v + x φ φ) 2 + (x y) + (x z) where (x y) means the same expression as before but with x replaced by y; u = u t etc., of course. This gives ẋ 2 = (u 2 + v 2 )( u 2 + v 2 ) + (uv) 2 φ2, and the Lagrangian above becomes L = m ( ) (u 2 + v 2 )( u 2 + v 2 ) + (uv) 2 φ2 Ũ(u, v, φ) 2 where U(u, v, φ) = U(x(u, v, φ)). It is now easy to write down the Euler-Lagrange equations L u = d dtl u etc. and get the equations of motion. Note that the key computation is ẋ 2, which proves that the coordinate system is orthogonal, from this result we also can read of the scale factors h u, h v, h φ : ds 2 = ẋ 2 dt 2 = h 2 u du2 + h 2 v dv2 + h 2 φ dφ2, implies h u = u 2 + v 2, h φ = uv. (1) (ii) We know that the Helmholtz equation U + k 2 U = is identical with the Euler-Lagrange equation for the functional S = d 3 x 1 ( ( U) 2 k 2 U 2). 2 We thus only need to express S in terms of (u, v, φ). The fastest way to compute U is to remember that U = 1 h u U u e u + 1 h v U u e v + 1 h φ U φ e φ 3

4 where (e u,e v,e φ ) are a system of orthonormal vectors (which we need not to compute here!), and thus ( U) 2 = 1 (U h u) (U 2 u h v) (U 2 v h φ) 2 2 φ with the scale factors in (??). (Of course one could also find U 2 by computing (u, v, φ) as functions of (x, y, z) and using the chain rule. This method would work even for coordinate systems which are not orthonormal.) We also need to change the integration measure. We could do that by computing the Jacobian: d 3 x = dxdydz = (x, y, z) (u, v, w) dudvdw =, but we can also remember that dxdydz = h u h v h φ dudvdw with the scale factors in (??). Anyway, we get S = dudvdw (u 2 + v 2 )uv 1 2 [ 1 u 2 + v 2((U u )2 + (U v )2 ) + 1 (uv) 2(U φ )2 + k 2 U] }{{} :=F where U = U(u, v, φ), and from that it is easy to work out the Euler-Lagrange equation: d F + d F + d F = F du U u dv dφ U U v U φ identical with the Helmholtz equation in our coordinates. emark: Note that it is important to distinguisg the notation d/du from / u: df du = F u + F U U u + F U uu + F U vu + F U φu U u U v U φ 4. ead my notes on variational calculus for background. (i) We use spherical coordinates. Then the constraints give: z = cos(θ), x = sin(θ) cos(ωt), y = sin(θ) sin(ωt), and the Lagrangian becomes L = m 2 ẋ2 mgz = m2 2 ( θ 2 + sin(θ) 2 ω 2 ) mg cosθ. The Newton equations d (L θ) = L dt θ are now easily computed. Note that here we should allow for θ (i.e. not restrict θ to the intervall [, π]): for large energies the particle will rotate about the ring, i.e., the motion θ(t) will be a strictly increasing and unbounded function. It is somewhat nicer to use u = π θ as coordinate (since then u = is the minimum of potential energy for small enough values of ω). I do that in the following. 4

5 (ii) Compute the energy of the systems (first integral of motion), which looks like H = p2 u 2m 2 + V (u), p u = L u for some function V. Interpreting this as the energy of a particle on the circle moving in a potential V is is easy to understand qualitatively what different motions one can have: for small energies one has oscillation about the minimum of V, for large energies unbounded motion where the peal rotates around the ring so that u(t) is a monotonic, unbounded function. It is interesting to note that, for small ω, V has a minimum at u =, but there is a critical value ω = ω c beyond which one gets two new minima u = ±u away from u =, and u = turns into a local maximum. Thus we have a nice example of a system with symmetry breaking: the symmetry u u can be broken in the ground state (= minimum of V = stable solution of Newton s equation where u = ). 5. (i) For the positions of the mass points we use spherical coordinates. For the first mass point: x 1 = l 1 sin(θ 1 ), y 1 =, z 1 = l 1 cos(θ 1 ), and for the second: x 2 = x 1 + l 2 sin(θ 2 ), y 2 =, z 2 = z 1 + l 2 cos(θ 2 ). The rest is straightforward: compute etc., as in the previous example. L = 1 2m 1 ẋ m 2 ẋ 2 2 m 1gz 1 m 2 gz 2 (ii) Introduce u 1 = π θ and similarly for u 2 and Taylor expand L in u 1 and u 2 so that there are at most quadratic terms in the variables and the Euler-Lagrange equations become linear. These linearized equations of motion can be easily solved. 6. Answer: 7. (a) Vi skall visa att mẍ = q(e + x B). δk[ψ ] := d dε K[ψ + εη] =. ε= Vi skriver K = S 1 /S 2 där S 1 [ψ] = dx ( ψ x (x) 2 + V (x)ψ(x) 2), S 2 [ψ] = dxψ(x) 2 och ber knar δk = S 2δS 1 S 1 δs 2 S = 1 S 2 (δs 1 KδS 2 )

6 där δs 1 [ψ ] = dx (2η (x)ψ (x) + 2V (x)η(x)ψ (x)) = 2 dxη(x) ( ψ (x) + V (x)ψ (x)) (vi använde partiell integration) och δs 1 [ψ ] = 2 Detta ger δk[ψ ] = dxη(x)ψ (x). 1 S 2 (ψ ) 2 dxη(x) ( ψ (x) + (V (x) K[ψ ])ψ (x), ) som visar att δk[ψ ] = för alla funktioner η om och bara om ψ är en lösning till Schrödingerekvationen ovan med E = K[ψ ]. Lösningen ψ = ψ till Schrödinger ekvationen med minsta möjliga värdet E = E därföre maåste vara strikt minimum till funktionalen K, och E = K[ψ ]. (b) Vi beräknar k(a) := K[φ] = dx (φ (x) 2 + V (x)φ(x) 2 ) dxφ(x)2 och får k(a) = dx ((ax) 2 + 9x 4 ) e ax2 dx = 1 2 a a 2. e ax2 Bästa approximatioen kan beräknas genom att minimera k: k (a) = , a och det finns bara en lösning > till k (a) = : a = 3. lim a k() = lim a + k( ) = + visar att det detta är ett strikt minimum. Svar: Bäste approximationen till grundtillståndet är ψ (x) = e 3x2 /2, och motsvarende approximationen till grundtillståndsenergin är E k(3) = 9/4 = emark 1: It is important to note that this computation gives an exact upper bound to E : the computation proves that E emark 2: A more general ansatz: f = exp( (c x ) b /2) 6

7 with two variational parameters b, c >, gives and K[f] = f = (bc/2)(c x ) b 1 ( dx e (cx)b (bc/2) 2 (cx) 2b 2 + 9(cx) 4 /c 4) = dx e (cx)b ( du e ub (bc/2) 2 u 2b 2 + 9y 4 /c 4) du e ub and with u = y 1/b, du = (1/b)y 1/b 1 dy, dy e 1 y ( 1/b 1 (bc/2) 2 y 2 2/b + c 4 u 4/b) K[f] = = (bc/2)2 Γ(2 1/b) + 9c 4 Γ(5/b) dy e 1 y 1/b 1 Γ(1/b) where Γ(s) = duu s 1 exp( u) is the Gamma function. For fixed b the minimum of this is ( ) 243 b 4 Γ(5/b)Γ(2 1/b) 2 1/6, 64 Γ(1/b) 3 and the minimal value for this is for b = We thus improve the exact bound to E 2.21, which is surprisingly close to what we got for b = The Schrödinger eq. now is ψ(x) + V (x)ψ = Eψ(x), x 2, and the functional K is K[ψ] = 2 d 2 x [( ψ(x)) 2 + V (x)ψ(x) 2 ] 2 d 2 xψ(x) 2. Introducing polar coordinate the angular integrals are trivial, one one is left with radial integrals. The computation which one is left to do is very similar as in 7, only easier: now one get integrals drr r 2n e ar2 which by a change of variables, u = ar 2, are transformed to duu n e n = n! for n =, 1, 2,... (up to some factor, of course). 9. We derive the Euler-Lagrange equations from skratch, to show that in this way one also gets the correct boundary conditions: Compute δu[u] = ε U[u + εη] ε= 7

8 with the variational function η(x) such that η() = η () =, whereas η(l) and η (l) are unconstrained. Introducing a = EJ, b = ρga this gives l δu[u] = l dx ( a ) ε 2 (u (x) + εη (x)) 2 b(u(x) + εη(x)) ε= = l dx (au (x)η(x) bη) = (au (x)η (x) au (x)η(x) l x= + dxη(x)[au (x) b] = au (l)η (l) au (l)η(l) + l dxη(x)[au (x) b], which is zero for all allowed variational functions η(x) (note that η(l) and η (l) can be arbitrary!) if au (x) b =, u (l) = u (l) =. This is a 4th order ODE with two boundary conditions. To have a unique solution we need four boundary conditions, but the missing ones we had already before: u() = u (). The solution of this I leave to the reader. 1. (a) Choose a karthesian coordinate system (x, y, z) that the rotation axis coincides with the z-axis. The Lagrangian for this system is In cylinder coordinates L = m 2 (ẋ2 + ẏ 2 + ż 2 ) U( x 2 + y 2 + z 2 ). x = ρ cos(φ), y = ρ sin(φ), z = z this yields L = m 2 (ṙ2 + ρ 2 φ2 + ż 2 ) U (ρ 2 + z 2 ). Polar coordinates in the rotating frame of reference are ρ, θ, z where This yields φ 2 = (ω + θ) 2 and φ = ωt + θ. L = m 2 (ṙ2 + ρ 2 θ2 + 2ω θ + ż 2 + ω 2 ρ 2 ) U (ρ 2 + z 2 ) and the equations of motion are obtained as Euler-Lagrange equations from this. Alternative solution: If we rotate about the z-axis by an angle θ we get new coordinates x, y, z such that x = cos(θ)x + sin(θ)y, y = sin(θ)x + cos(θ)y, z = z. 8

9 In the rotating coordinate system we have θ = ωt, i.e. the coordinates in the rotating coordinate system are (u, v, z) such that x = cos(ωt)u + sin(ωt)v, y = sin(ωt)u + cos(ωt)v. (2) We now compute x y 2 + z 2 = u 2 v 2 + z 2 and ẋ 2 + ẏ 2 = ( ) = u 2 + v 2 + ω 2 (u 2 + v 2 ) + 2ω( u v + v u). This gives the Lagrangian in the rotating frame of reference: L = m 2 ( u2 + v 2 + ω 2 (u 2 + v 2 ) + 2ω( u v + v u)) V ( u 2 + v 2 + z 2 ) from which the equations of motion can be obtained as usual. 9