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Solutions_HW10

# Solutions_HW10 - Theoretical Dynamics Homework 10...

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Theoretical Dynamics November 19, 2010 Homework 10 Instructor: Dr. Thomas Cohen Submitted by: Vivek Saxena Note : Goldstein problems 5 . 8 and 5 . 14 are solved at the end. Problem 1 (a) The coefficients of the moment of inertia tensor in the usual 3 × 3 matrix representation are given by I jk = Z V ρ ( r )( r 2 δ jk - x j x k ) dV (1) Therefore, for the given mass density, we have to evaluate the integral, I jk = ρ 0 Z -∞ Z -∞ Z -∞ dx dy dz e - x 2 + y 2 + z 2 + xy 2 l 2 ( r 2 δ jk - x j x k ) (2) The argument of the exponential is a quadratic form which can be written as x 2 + y 2 + z 2 + xy 2 l 2 = 1 2 x T A x (3) where x = x y z , A = 1 l 2 1 2 l 2 0 1 2 l 2 1 l 2 0 0 0 1 l 2 (4) so, A is real and symmetric. Define F jk = Z -∞ Z -∞ Z -∞ dx dy dz e - 1 2 x T A x x j x k (5) and G = Z -∞ Z -∞ Z -∞ dx dy dz e - 1 2 x T A x (6) In terms of F jk , the moment of inertia coefficient I jk is I jk = ρ 0 [( F 11 + F 22 + F 33 ) δ jk - F jk ] = ρ 0 [ tr ( F ) δ jk - F jk ] (7) To find the coefficient, we use Wick’s Theorem, stated below. 10 - 1

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Wick’s Theorem : If A is a real symmetric N × N matrix, and x is an N × 1 vector, then h x i x j · · · x k x l i Δ = Z -∞ Z -∞ · · · Z -∞ | {z } N integrations dx 1 dx 2 . . . dx N e - 1 2 x T A x x i x j · · · x k x l Z -∞ Z -∞ · · · Z -∞ | {z } N integrations dx 1 dx 2 . . . dx N e - 1 2 x T A x = X σ ( A - 1 ) ab · · · ( A - 1 ) cd (8) where the sum runs over all permutations σ ≡ { a, b, · · · , c, d } of the set of indices { i, j, · · · , k, l } . In particular, for N = 3, we have h x j x k i = F jk G = ( A - 1 ) jk (9) To evaluate G , we use the following result on Gaussian integration. Result : If A is a real symmetric N × N matrix, and x is an N × 1 vector, then Z -∞ Z -∞ · · · Z -∞ | {z } N integrations dx 1 dx 2 . . . dx N e - 1 2 x T A x = (2 π ) N det[ A ] 1 / 2 (10) For the matrix A defined in Eqn. (4), det( A ) = 3 4 l 6 . For N = 3, G = s (2 π ) 3 3 / 4 l 6 = 4 r 2 3 π 3 / 2 l 3 (11) From Eqns. (9) and (10), we get F jk = 4 r 2 3 π 3 / 2 l 3 × ( A - 1 ) jk (12) The inverse of A is A - 1 = 4 l 2 3 - 2 l 2 3 0 - 2 l 2 3 4 l 2 3 0 0 0 l 2 (13) So, using Eqn. (12), the matrix F is found to be F = 16 3 q 2 3 π 3 / 2 l 5 - 8 3 q 2 3 π 3 / 2 l 5 0 - 8 3 q 2 3 π 3 / 2 l 5 16 3 q 2 3 π 3 / 2 l 5 0 0 0 4 q 2 3 π 3 / 2 l 5 (14) 10 - 2
and finally, using Eqn. (7), the matrix representation of the moment of inertia tensor is ←→ I = 28 ρ 0 3 q 2 3 π 3 / 2 l 5 8 ρ 0 3 q 2 3 π 3 / 2 l 5 0 8 ρ 0 3 q 2 3 π 3 / 2 l 5 28 ρ 0 3 q 2 3 π 3 / 2 l 5 0 0 0 32 ρ 0 3 q 2 3 π 3 / 2 l 5 (15) (b) The moment of inertia matrix obtained above is of the form I = a b 0 b a 0 0 0 c (16) where a = 28 ρ 0 3 q 2 3 π 3 / 2 l 5 , b = 8 ρ 0 3 q 2 3 π 3 / 2 l 5 and c = 32 ρ 0 3 q 2 3 π 3 / 2 l 5 . If λ is an eigenvalue, then it satisfies the secular equation, | I - λ I 3 | = 0 (17) which is ( c - λ )( λ 2 - 2 + a 2 - b 2 ) = 0 (18) the solutions to which are λ = a - b, a + b, c . The eigenvector matrix V is given by V = - 1 1 0 1 1 0 0 0 1 (19) So, the diagonal form of the moment of inertia tensor, which is the principal moment of inertia tensor, is given by ←→ I D = V ←→ I V - 1 = a - b 0 0 0 a

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