vol integral of tensor field

This has been added to tensor doc, delete a few days after 4.12.12 This is the Title PhL 3.26.05 Note that page numbering is turned on in this template. (f) Volume integration of a tensor field under linear transformations In section (d) above it was shown that, with its interpretation as a distribution, dV' = J-1 dV . In the language of Appendix D, this says that the volume element transforms from x-space to x'-space as a scalar density of weight +1. Since g1/2 transforms as a scalar density of weight -1, according to App. D (b) item 3 the object g1/2dV then transforms as an ordinary scalar (weight 0) (see for example Weinberg p 99 (4.4.6) ), Then if one were to define Tijk... ≡ ∫D g1/2dV Aijk..(x) (*) one might expect that, if Aijk..(x) transforms as a tensor field, then Tijk... might transform as tensor. To investigate this conjecture, consider the above integral in x'-space, T'ijk... ≡ ∫D' g'1/2dV' A'ijk..(x') = ∫D g1/2dV A'ijk..(x'(x)) = ∫D g1/2dV Rii'Rjj'..... Ai'j'k'...(x) If the underlying transformation x' = F(x) is linear, then the Rab are independent of position and one has = Rii'Rjj'..... ∫D g1/2dV Ai'j'k'...(x) = Rii'Rjj'..... Ti'j'k'... Therefore, for linear transformations x' = F(x), an integral of the form (*) of a tensor field is itself a tensor of the same type. Examples: If g = 1 and x' = Rx where R is a rotation (independent of position), then g' = 1 as well and one may conclude that the volume integral of a tensor field of any type is a tensor of the same type. Here are two simple examples: Ji = ∫D dV xi Jij = ∫D dV [ r2δij - xixj] Under rotations, xi is a true vector, and r2δij - xixj is a true rank-2 tensor (traceless). It follows that Ji and Jij are also tensors. Under rotations, mass density ρ transforms as a scalar, so the following objects are tensors as well, Ii = ∫D dV ρ(x) xi // vector Iij = ∫D dV ρ(x) [ r2δij - xixj] // rank-2 tensor Vector components Ii are the first moments of a mass distribution, while Iij is the usual inertia tensor. Since Iij is real and symmetric, it can be diagonalized by a certain rotation R. We can think of this as a transformation from x-space to x'-space where the resulting tensor I'ij is diagonal. In x'-space, the diagonal elements of the tensor I'ij are then its eigenvalues (λi = I'ii), while the eigenvectors are the axis-aligned e'i of Section 3. Back in x-space, the eigenvectors of the non-diagonal Iij are then en ≡ Se'n where S = R-1. This can be verified as follows (developmental notation) I' = R I R-1 => I'ij = Rii'Ii'j'(R-1)j'j = Rii'Rjj'Ii'j // contravariant tensor I' e'n = λn e'n // I' is diagonal so I en = [ R-1I' R ] [Se'n] = R-1 I' (RS) e'n = S I' e'n = S λne'n = λn Se'n = λnen .