{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solutions_HW10 - Theoretical Dynamics Homework 10...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Background image of page 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
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}