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inverting the inertia tensor

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Handwritten-style working in a Word file dated 3.27.17 by Phil, part of his rigid body appendix. He writes the dumbbell inertia matrix in spherical coordinates, notes that Maple gives a zero determinant, and proves det(I)=r^6-r^6=0 using the epsilon-tensor expansion of I_ij = r^2 delta_ij - x_i x_j. He also checks the trace (2r^2) and the det(e^I) identity, which says nothing about det(I).

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Determinant of an inertia matrix PhL 3.27.17 From (I.1.6), Li = Iij (ω)j Iij = Σαmα [ rα2δij - (rα)i(rα)j ] // rα2 = (rα)i(rα)i (I.1.6) Since we have done no examples of computing inertia matrices in the space frame, we do it here. This is a special case I where in effect there is only one term in the Σα sum. This is for a dumbbell of two equal masses and equal radii. This is based on x = rsinθcosφ y = rsinθsinφ z = rcosθ I11 = 2m(r2 - x2) = 2m(r2 - r2sin2θcos2φ) = 2mr2 (1 - sin2θcos2φ) I22 = 2m(r2 - y2)= 2mr2 (1 - sin2θsin2φ) I33 = 2m(r2 - z2) = 2m(r2 - r2cos2θ ) = 2mr2sin2θ I12 = 2m(-xy) = -2mr2sin2θcosφsinφ I13 = 2m(-xz) = -2mr2sinθcosθcosφ I23 = 2m(-yz) = -2mr2sinθcosθsinφ Then I = 2mr2 Maple shows this matrix has zero determinant so cannot be inverted!!!! I was not expecting that. I may have suggested using that method somewhere!!! Why is the determinant 0? Look at the general single-term expression for Iij Iij = r2δij - xixj det(I) = εijkI1iI2jI3k = εijk(r2δ1i - x1xi)(r2δ2j - x2xj)(r2δ3k - x3xk) There are many terms. Here is the first term εijkr2δ1ir2δ2jr2δ3k = r6 εijkδ1iδ2jδ3k = r6 ε123 = r6 Then three terms like this εijk(r2δ1i)(r2δ2j)(-x3xk) = r4 ε12k(-x3xk) = r4 ε123(-x3x3) = - r4x32 So I think the sum of those three terms will be -r4 r2 = - r6 Then three terms like this εijkr2δ1i(-x2xj)(-x3xk) = r2 x2x3 ε1jk xjxk = 0 by symmetry The last term is - εijk x1xi x2xjx3xk = 0 by symmetry Therefore the determinant is det(I) = r6 - r6 = 0 so can NEVER invert it!!! But if diagonal, you CAN invert it, so what gives?? People on the web do invert some inertia tensors, so I think it is my special case that is not invertible where there is only a single particle. What is the trace of I? For single particle, I = r2δij - xixj tr(I) = tr( r2δij - xixj) = Σi ( r2δii - xixi) = 3r2- Σi xi2 = 2r2 ?? The diagonal elements are: r2 - x2, r2-y2, r2- z2 Their sum is definitely 2r2. Also, det(eI) = etr(I) = exp(2r2) This says nothing about det(I).