Electric and magnetic constants are tensors

[mathwonk]

i.e. briefly,

V*tensorV* is the space of non abelian second order polynomials on V,

i.e. V*tensorV* = bilinear functions on VxV.VtensorV is the dual space of V*tensorV*.thus by mere definition,

V*tensorV* is the space of linear functions on VtensorV.thats all there is to it.

 

[mathwonk]

but anything can be made to look different. take a bilinear map from VxV to W.

by fixing one entry, we get a linear map from V to W.

i.e. if <x,y> is bilinear in both entries, then fixing x, we get a map
y --> <x,y> which is linear in y.

and the map from x to linear maps in y, is itslef lienar in x!

thus we can regard Bil(VxV-->R) = Lin(V-->V*),

thus also Bil(VxV*) = Lin(V-->V).

this says that V*tensorV and Hom(V,V) are essentially the same thing!

i.e. certain (1,1) tensors are equivalent to matrices. people can discuss at length whether these are or are not "the same", but this is largely just language.

 

[mathwonk]

let me go out on a limb here and guess that anyhitng that depends multiplicatively on tangent vectors is a tenmsor.

e.g. whatever, is it called momentum? i.e. mv^2, depends multiplicatively and quadratively on velocity, so it should be expressible as a second order tensor.tensors are nothing but an algebraic way of expressing multiplication of thigns that originally only belonged to a vector space and hence could not be multiplied.suppose v and w arte elements of a vector space V, then we write vtensor w to be their product in VtensorV.

we see that vtensorw is determined by the pair <v,w> and yet is different from that pair, because in the space VxW we add <v,w> to <v',w> and get

<v+v', w+w>, but in VtensorV we add vtensorw and v'tensorw and get (v+v')tensorw.

see the difference? this is what changes bilinear functions on VxV into linear functions on VtensorV.

 

[mathwonk]

i went back and read hurkyls explanation and it sounded so much clearer and easier than mine. he said a tensor is just a multilinear function on a sequence of vectors and covectors. that's right.

here let me relate one of my statements to that:

I said: in trying to define second order contravriant tensors,

"one unnatural way to do this is to simply define VtensorV =
{Bil(VxV,R)}*, i.e. the dual of the bilinear maps."

Hurkyls version would define VtensorV instead as {Bil(V*xV*,R)} i.e. as bilinear maps on pairs of covectors. this is the same thing essentially, i.e. there is an isomorphism of vector spaces between

{Bil(VxV,R)}* and {Bil(V*xV*,R)}, that takes let's see,...,ok i think i got it, remember auslanders dictum, basically anything you can think of is correct.

so first note there is a simple map from pairs of loinear functions to bilinear ones, namely multiply. i.e. if f,g are elements of V*, then ftensorg, which takes <v,w> to f(v)g(w), is an element of Bil(VxV,R).

so here we go:

we want to define a map {Bil(VxV,R)}* --> {Bil(V*xV*,R)}

so let H be an elt of {Bil(VxV,R)}*, i.e. if H sees a bilinear map on VxV, it spits out a number. now we want that to give us an elt of
{Bil(V*xV*,R)}, which is a gadget that spits numbers when it sees two linear maps on V. well easy. let f,g, be two linear maps on V, and apply H to ftensorg.

i.e. the compositon of a linear and a bilinear map is bilinear, so
H(ftensorg) is bilinear in f and g, hence gives an element of

{Bil(V*xV*,R)}.

since this map from {Bil(VxV,R)}* to {Bil(V*xV*,R)} is the only one i can think of, by auslanders dictum, it is an isomorphism. you can vheck this yourself by fidning an inverse.i.e. suppoose K is an elt of {Bil(V*xV*,R)}, i.e. something that spits out a number when it sees a pair of linear functions f,g. now we want it to define an elt of {Bil(VxV,R)}*, i,.e. to spit out a number whe it sees a bilinear map.

so let m be a bilinear map on VxV. Uh oh, i have to produce a pair of linear maps on V, but there is no nice way to do this. i.e.w e are using the finite dimensionality here, and as remarked above that is why the isomrphism depends on a certain amp being in jective hence an isomotrihsm, whereas in infinite dimensions that ca fail. so i suspect there is no natural definition of the inverse here independent of coordinates, since if there were it would work in infinite dimensions too where the reult is false. pooh.

well i over extended auslanders dictum, i.e. since there is nothign i can think of in the reverse direction, it may not always be an isomorphism. he really said the only thing you can think of is the right thing, not that is is an isomorphism, since his dictum requiers you tot hink of two inverse thigns for that to hold. my apologies to his memory. he was always very clear that an isomorphism is a map with an inverse.

 

[mathwonk]

nonetheles, here there is an isomorphism since the two spaces are the same dimension, so all we have to check is injectivity, which iw will do in case it happens to b false, which it won't be.

so let H be any linear map on bilinear maps, and let f,g be any two linear maps. then i claim, if H(ftensorg) is zero for all f,g, then H is zero. I.e. I have to show that bilinear maps of form ftensorg span the space of all bilinear maps on VxV.

Hmmmm, this is the same problem as before, but now I am allowed to use coordiantes, to check a coordinate free statement. I.e. I claim every bilinear map on VxV is a linear combination of ones of form ftensorg, where f,g, are linear.

this is easy (i hope, i always say that as cover), choose a basis v1,...vn, and then a bilinear map is determined by tis values on pairs like <vi,vj>.

but i can get any numbers i want from such a pair using the dual basis f1,...fn, where fi(vj) = kronecker delta (i,j), timesa constant.

i.e. (fi)tensor(fj) has value 1 on <vi,vj>.

so the special bilinear maps (fi)tensor(fj), give basis for all of them.

so any bilinear map can be expressed as a linear combination of the special oens of form (fi)tensor(fj). and since H kills this basis, it is zero.

whew!i am getting some idea of why this is so hard for learners, who do not have a good grasp of all these natural isomorphisms between different ways of saying the same thing that I take pretty much for granted.

 

[mathwonk]

but to rephrase hurkyls very clear explanation: a tensor is just a way of assigning a number to a sequence of vectors and covectors, that is linear in each variable separatley, i.e. it is some way of multiplying them.

so anytime you encounter a quantity that depends on several tangent vectors and cotangent vectors, and is linear in each one separately, i.e. somehow is a product of them of some kind, maybe yielding a number or another vector or covector, or even yielding another linear or multilinear map, it seems to be [representable as] a tensor.

 

[mathwonk]

it would be very illuminating now to go back and take the various ophysical exampels athat are said to be tenosrs, and analyze exactly how they fit into this paradigm.

i.e. how does stress appear as a multilinear function on some sequence of vectors and covectors? etc...

then we will actually be talking to and communicating with each other.

 

[mathwonk]

let me indulge in a few more points tht cause confusin and controversy. people here have debated whether a certain thing IS or IS NOT a tensor. e.g. is a vector a tensor? is a matrix a tensor?

well if you accept hurkyl's description of a tensor as a multilinear real valued function on sequences of vectors and covectors and covectors, then no, neither a vector nor a matrix is strictly such an object.But as recently debated in the media in another context, it does all depend on your definition of the word "is".

I.e. there is a natural map from a vector space V to V** = linear functions on linear functions. if v is a vector, then it defiensa linear function on linear functions by evaluation, i.e. v takes f to f(v), which is being thought of as v(f).

this map is very natural and injective, hence in finite dimensions is an isomorphism, although not in infinite dimensions, hence has no such natural inverse.

but this allows one by means of this uniquely natural isomorophism to say that a vector "is" a linear function on covectors and hence a tensor.

similarly there is a natural map from VtensorV* to Hom(V,V), taking a basic tensor of form vtensorf to the linear map sending w to f(w).v.

This too is an isomorophism in finite dimensions and hence permits a matrix, or linear endomorphism in Hom(V,V) to be thiught of as a tensor belonging to VtensorV*.

Since this map is completely natural and unique, there is really only one way to represent a matrix this was a a tensor, so there can be "no confusion" (haha) in saying a matrix is in this sense a tensor of type (1,1)?anyway mathematicians tend to asume that all statements are made up to natural isomorphism, unless they are trying to win an argument.

so anytine somebody says such and such IS something else, ask yourself if there is any natural way to intepret that as true.

 

[mathwonk]

if i have the indices right, by the conventions above a vector "is" a tensor of type (1,0) anda matrix "is" a tensor of type (1,1).

however because there is no natural isomorphism betwen V and V*, even in finite dimensions, although many unnaturl ones, we cannot naturally identify tensors of types (1,0) with those of type (0,1).

now i am geting beyond what i have thought about thoroughly here, but a riemannian metric does allow vectors in V to be identified with vector in V*. nonetheless, the transormation laws get screwed up i believe, so although a field of elements of various V's can thus be changed into a field of elements of V*, they will not transform correctly if one uses the other transofrmation rules.

thus I guess even in the presence of a metric one must distinguish types of vector fields, but i have made mistakes on this before by claiming otherwise.

the problem sems to be that when replaces a linear function by dotting with a vector, one still hS TO TRANSFORM BY THE transpose of the matrix for transforming vectors, not the matrix itself, and in the opposite direction.

so if you replace a linear function by a vector, such as replacing a diferential by a gradient vector, I think now that transforming that vector by the vector transformation laws, will not transform the linear function correctly.

i apolopgize if i hVE MISLED PEOPLE ON THIS POINT BEFORE.

i.e. keep in mind that one does not really replace the differential by the gradient, but by the operation of dotting with the gradient.

 

[mathwonk]

a bit more on when thigns may be considred the same.

a natural construction is also called a"functor" to frighten children.

thus "chANGING" V to V is a functor, the identity functor, "changing V to V* is a functor the dual functor, V to Hom(V,V) is a functor, V to VtensorV*is a functor, and so on...some con structions, i.e.some functors are essentilly equivaklent or at elast related to otger functors, such relationships are expresed by families of maps called natural transformations.

there is a natural transformation from the identity to the double dual functor sending V to V** for every V, and its is an equivalence for finite dimensional V.

there is a naturl transformation from VtensorV* to Hom(V,V) and it is an equivalnece for finite dimensional V.

there is no natural transformation from V to V*. in any case, the right definition of the word "is" above might be "is naturally equivalent to" or "define naturally equivalent functors", to frighten children and adults.

 

[HeilPhysicsPhysics]

Tensor is a general vector.

 

[mathwonk]

i highly recommnend the following text, on tensors, by a math prof and a prof of mechanical engineerinbg: book #60 by biowen and wang on the free site:
http://www.math.gatech.edu/~cain/tex...linebooks.html .in particular their introduction includes this:

In preparing this two volume work our intention is to present to Engineering and Science
students a modern introduction to vectors and tensors. Traditional courses on applied mathematics
have emphasized problem solving techniques rather than the systematic development of concepts.
As a result, it is possible for such courses to become terminal mathematics courses rather than
courses which equip the student to develop his or her understanding further.

As Engineering students our courses on vectors and tensors were taught in the traditional
way. We learned to identify vectors and tensors by formal transformation rules rather than by their
common mathematical structure. The subject seemed to consist of nothing but a collection of
mathematical manipulations of long equations decorated by a multitude of subscripts and
superscripts. Prior to our applying vector and tensor analysis to our research area of modern
continuum mechanics, we almost had to relearn the subject. Therefore, one of our objectives in
writing this book is to make available a modern introductory textbook suitable for the first in-depth
exposure to vectors and tensors. Because of our interest in applications, it is our hope that this
book will aid students in their efforts to use vectors and tensors in applied areas. in particular they explain such things as the natural isomorphisms of V and V**, and of Hom(V,V) with tensors of type (1,1) on V.

 

[mathwonk]

chapter 11.5 of dummitt and foote 2nd edition discusses tensors, symmetric and alternating, and the isomorphism betwen the symmetric algebra of an dimensional vector space and the ring of polynomials in n variables.

 

[JohanL]

"In classical physics it is customary to define a tensor T_ijk... by generalizing

V_i= \sum_{j} R_{ij} Vj

(which is his definition of a vector, a quantity which components transforms like that) as follows

<br /> T_{ijk...}= \sum_{i&#039;}\sum_{j&#039;}\sum_{k&#039;} ... R_{ii&#039;} R_{jj&#039;} R_{kk&#039;} ... T_{i&#039;j&#039;k&#039;...} <br />
under a rotation specified by the 3*3 orthogonal matrix R"

/Modern quantum mechanics by J.J. Sakurai

I understand that a tensor takes N vectors to a scalar, or N components of N vectors to a scalar. And this can be generalized to include upper indices (covectors, which is nothing more than column vectors).

And the tensor above is a cartesian tensor. Now I am trying to learn about spherical tensors. Can someone tell me about them? And is what i wrote above about cartesian tensors correct?

 

[mathwonk]

arggggh! actually there is no thing such as a tensor. the terminology is a joke on the community. try to calm down and forget about wanting to know what a tensor is, babble babble babble...

 

[JohanL]

arggggh! actually there is no thing such as a tensor. the terminology is a joke on the community. try to caln down and forget about wanting to know what a tensor is, babble babble babble...

Are you suggesting that Sakurai is wrong, me is wrong or us both? :)

 

[pmb_phy]

No, that is not right. The dot product is not a tensor, nor is the result of a dot product a (0,2) tensor-it is a (0,0) tensor a.k.a. scalar.

Of course it is. The dot product maps two vectors into a scalar. I.e. g(A,B) -> real number. By the very definition of tensors this is truly a tensor (of second rank).

Pete

 

[George Jones]

now i am geting beyond what i have thought about thoroughly here, but a riemannian metric does allow vectors in V to be identified with vector in V*. nonetheless, the transormation laws get screwed up i believe, so although a field of elements of various V's can thus be changed into a field of elements of V*, they will not transform correctly if one uses the other transofrmation rules.

In a slight abuse of notation, let g be the metric-induced, natural isomorphism from V to V*, denote the image under g of \tilde{w} by \tilde{w}. Let L be an isometry, with w&#039; = Lw. Let \tilde{L} = g^{-1}Lg be the isomorphism on V* that makes

[insert appropriate diagram that I can't seem to copy and paste from my latex editor]

commute for all v.

Introducie a basis for V, and the matrix representation of L gives the transformation law for vectors. The matrix representation of \tilde{L} with respect to the corresponding dual basis of V* is the "correct" transformation law for covectors.

 

[mathwonk]

lets keep this simple: tensor = multiplication.

hence pete is certainly right, the dot product is one our favorite tensors.

 

[ObsessiveMathsFreak]

So are tensors operators or operands?

 

[mathwonk]

or operatoids? or operatives?

is that a diamonoid or a diamonelle? to paraphrase cleo on the cosby show.