The Form of the Inertia Tensor for a Rigid Body: The angular momentum is to be represented as the matrix product of the moment of inertia and the angular velocity G G 3 vector. That is: L = ω or, in terms of the elements and components. LIi = ∑ iAAω . A=1 With the target form identified, one begins with the general definition of the angular G GG GG G momentum of a particle: L =×rmvmr =⎡ ×ω × r⎤ . Each particle in a rigid body ⎣⎢ ( )⎦⎥ G that has an angular velocity ω directed along a fixed axis is moving along a circular G G G path about that axis. The velocity relative to a point on that axis is ω × r where r is the position of the particle relative to that axis point. The identification of the velocity as G G ω × r is valid for rigid body rotation, and the results that follow are valid for the limited case of rigid bodies. The component form of the angular momentum equation is: ⎡⎤GGG Li =××mr()ω r . ⎣⎦⎢⎥i G ˆˆˆ ˆ ˆ ˆ Reviewing some notation, r=+ xiyjzkxixjxk + =12 + + 3, and the inner product 3 3 GG 2 GG 2 rr⋅= r =∑ xxii or rr⋅= r =∑ xxijijδ . The cross products can be represented using the i=1 ij,1= G permutation symbol and the components xi of r . In general, the x component of a vector is to be identified as the 1 component, the y component as the 2 component and the z component as the 3 component. For example, G G 3 ωε×=rxω ()k ∑ knA A n An=1 Applying this notation to the angular momentum, GGGG33G ⎛⎞3 L =××=mr⎡⎤ωεωεε r m x ×= r m x⎜⎟ω x. ij⎣⎦() ∑∑∑ijk ()k ijk j⎜⎟kA n A n i jk,1==jk,1 ⎝⎠A,1n= Two interchanges convert εijk to ε kij with no net sign change. 3 Lij= mx∑ εεkij kA n ωA xn jkA n=1 This form is dramatically simplified by using the identities: 3 33 GG 2 ∑εkijεδδδδ kA n =−ijAAjn in and ∑∑δstx sxxxrr t= s s =⋅=r k=1 st,1==s 1 Note that the first identity is easily established by computing the 81 possible outcomes for each form and demonstrating that they agree. The second identity is established in standard homework problems. 33⎛⎞33 Lmik==∑∑εεijkAAn xjnk ω x m ⎜⎟∑∑εεijkAAnxjni ω x =− m () δδδδAjninjj AAx ω xn jkAA n==11j n⎝⎠ k=1j A n = 1 33 Lmij=−∑∑()δijAAAAAδδδninxj ω xnn =− m () δδδδij jninxxj ωA jn,,AA== 1 jn,, 1 3 =−mx∑ ()δδijAAjnnnxx j δδ in xj ωA jn,,A = 1 The ordering of the components of the vector was shuffled in the last step. They are scalars so their multiplication is commutative. Next, the sum over j is executed. The Kronecker delta picks out the one term with j = n in the first part and the term with j = A in the second part. 3 Lmii=−∑ ()δi AA xxnnδωn xx n A A,1n= Finally, the sum over the index n is executed. In the first part, the sum of the squares of the components of r is just r2, and the Kronecker delta picks out the one term with n = i in the second part. 33 2 Lmii=−=∑∑()( rδiiAAA xxωω I A)A AA==11 Comparison with the target form identifies the elements of the inertia tensor for a particle as: 2 IiA = m (r δiA – xi xA) leading to the form for a collection of particles: 2 IiiAA=−∑ mrαα()δ xxαi αA particles α Matrix Form: ⎡⎤rxxxyxz22−− −⎡ yz +−−2 xyxz⎤ ⎢⎥22⎢2⎥ =−myxryyyzmyxxz⎢⎥ − − =−⎢ + − yz⎥ ⎢⎥22⎢2⎥ ⎣⎦−−zx zy r − zz⎣ −− zx zy x + y ⎦ This truly ugly relation becomes more transparent in matrix form. 222 ⎡⎤Lxx⎡⎤ r−− xx xy − xz ⎡⎤ω ⎡y+−− z xy xz ⎤⎡⎤ωx ⎢⎥ ⎢⎥22⎢⎥ ⎢2⎥⎢⎥ L=− m yx r − yy − yzω =− m yx x + z − yz ω ⎢⎥yy⎢⎥⎢⎥ ⎢⎥⎢⎥y ⎢⎥ ⎢⎥22⎢⎥ ⎢2⎥⎢⎥ ⎣⎦Lzxzyrzzzxzyxyzz⎣⎦−− −⎣⎦ω ⎣ −− +⎦⎣⎦ωz .