
Re: Tensor Definition
Posted:
Jul 9, 2013 3:17 PM


On 20130708, Sam Sung <no@mail.invalid> wrote: > J.B. Wood wrote: >> On 07/08/2013 11:03 AM, Sam Sung wrote: >>> J.B. Wood write:
>>>> Hello, all. Just when I think I've got a good handle on tensors (after >>>> painstakingly reading and working problems in Louis Brand's "Vector and >>>> Tensor Analysis"), I come across the following in MerriamWebster's >>>> Collegiate Dictionary:
>>>> "A generalized vector with more than three components each of which is a >>>> function of an arbitrary point in space of an appropriate number of >>>> dimensions"
>>>> That definition seems to exclude wellknown dyadics (stress tensor, >>>> permeability tensor, etc) having 9components that are constants (for >>>> the material involved.) That aside, I'm still having a problem grasping >>>> the foregoing definition. Thanks for your time and any comment. Sincerely,
>>> You might read this: http://en.wikipedia.org/wiki/Tensor
>> Hello, and thanks for responding. I read the Wiki entry prior to my OP. >> I don't see the correlation to the Webster definition.
> I read this as an example for the use of tensors.
>> Given the >> usual 3D orthonormal coordinate systems commonly used in physics and >> engineering (rectangular, cylindrical, spherical) I can view unit dyads >> (having "two directions" just as easily as unit vectors having but one >> direction (x, y, or z) and a dyadic having a total of 9 components vice >> 3. Still, that dictionary definition confounds me. Sincerely,
> Tensors are/canbe used by more than this kind of stuff; see > (the first sentence of) http://www.wolframalpha.com/input/?i=tensor
Still thinking like geometers or physicists. The uses given are essentially of "square" (or cube or higher order) tensors, which are a special case, but only a special case. I have used tensors which do not have the same number of dimensions in each direction these are useful algebraically. An interesting argument in computational algebra is that the number of nonsacalar operations needed to compute the matrix product AB where A and B are arbitrary matrices of the appropriate sizes is the rank of the tensior product tr ABC, where A,B, and C are arbitrary matrices of the appropriate sizes and the product is square. This will not satisfy the cited criteria for tensors.
> Be sure to also know manifolds and dual objects, understand > covectors, Grassmann products, vector fields and their covector > fields, nforms, volume elements and their integrals, understand > that there is math without reference to any coordinate system, > remember that a tensor Q_a...c^f...h with p lower and q upper > indices for any p,q >= 0 is a multilinear function (or mapping) > of p vectors A,...,C and q covectors F,...,H where > Q(A,...,C; F,...,H) = A^a...C^c Q_a...c^f...h F_f...H_h.
 This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)4946054 FAX: (765)4940558

