
An Introduction to Tensors for Students of Physics and Engineering (2002) [pdf] - brudgers
http://www.grc.nasa.gov/WWW/k-12/Numbers/Math/documents/Tensors_TM2002211716.pdf
======
BaryonBundle
There is also a similarly titled, much longer (92 pages vs. 29 pages)
introduction to tensors by the same author, also hosted by NASA, and released
~3 years later: _Foundations of Tensor Analysis for Students of Physics and
Engineering With an Introduction to the Theory of Relativity_

[http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/2005017...](http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20050175884.pdf)

The pdf.js in my Iceweasel 31.2 seems to be choking on the pdf document,
though it looks fine in other readers.

~~~
gjm11
Unreadable in pdf.js in my FF 33.1.1, too.

Interestingly, in this longer document the author gives (more or less) an
actual definition of "dyad" which isn't the same as "tensor product of two
vectors", and with that definition the statement that "every tensor is a dyad"
is (more or less) correct.

I need to qualify everything here with "more or less" because the author is
trying to do mathematics but fairly clearly isn't actually a mathematician,
and so a lot of what he says doesn't really quite make sense, but one can see
what he's trying to say. (E.g., he says "a dyad is any quantity that operates
on a vector through the inner product to produce a new vector with a different
magnitude and direction from the original". Except that actually I'm sure he
wants ii+jj+kk to count as a dyad even though it doesn't change either
magnitude or direction, and except that he hasn't actually said explicitly
what the "inner product" is except for the special case where the dyad is a
tensor product of two vectors, and in order to say what the inner product is
for a vector and a dyad one already has to know (at least kinda) what a dyad
is...)

On the other hand, if you ask a pure mathematician to explain this stuff
you'll generally get something nice and elegant but incomprehensible to most
physicists and engineers, and much less suitable for symbolic calculations
than the physicists' coordinate-based approach. (A tensor field is a section
of a product of the tensor product of some tensor power of the tangent bundle
of your underlying manifold, and some tensor power of the cotangent bundle.
What do you mean you don't know what a section of a tensor power of the
cotangent bundle of a smooth Riemannian manifold means? _Bozhe moi_!)

------
gjm11
This document tells an important lie: it says (e.g., at the top of page 8)
that every rank-2 tensor is an outer product of two vectors. This is flatly
untrue and is an easy misconception for a student to acquire even if the thing
they're learning from doesn't explicitly say it.

I skimmed through looking to see if at some point the author says "oh, by the
way, I told you a lie earlier", but it doesn't look like he does.

(If anyone reading this has difficulty believing that there are more rank-2
tensors than outer products ("dyads") of two vectors, note that in 3
dimensions you can specify two vectors by giving 6 numbers, but it takes 9 to
specify a rank-2 tensor because it can be represented by an arbitrary 3x3
matrix.)

~~~
thomasahle
> If anyone reading this has difficulty believing that there are more rank-2
> tensors than outer products

Small nitpick: Unless you have tensors over a finite universe, there aren't
really "more" of the large ones that the small ones..

So we have to argue in some other way. Anybody's got a counter example?
(matrix that is probably not a dyad?)

~~~
gjm11
Yeah, there really are more, provided you define "more" right. They form a
topological space, and the rank-2 tensors are a 9-dimensional space versus a
6-dimensional space for the outer products.

Alternatively, any matrix whose determinant isn't zero is a non-dyad.

~~~
sampo
But the cardinality of the set of all points in a 9-dimensional space is not
larger than the cardinality of the set of points in a 6-dimensional space, or
even between 0 and 1 in the 1-dimensional real line.

And the set theoretic cardinality is the normal way to define what "more"
means, or to compare sets of an infinite amount of objects.

How would you define "more", to get e.g. R² have "more" points than R?

~~~
gjm11
There are lots of ways to define "more" (or "bigger" or whatever term one
chooses to use). Which one is most appropriate depends on what kind of things
you're working with.

Working with vector spaces over a given field: "bigger" means "higher
dimension". Working with manifolds: "bigger" means "higher dimension". Working
with sets: "bigger" usually means "larger cardinality" but could sometimes
mean "proper superset".

In general, you can take "X is bigger than Y" to mean "Y is isomorphic to some
substructure of X, but not vice versa". If "isomorphic" means "in bijection
with" and "substructure" means "subset" then this is cardinality. If they mean
"isomorphic as vector spaces over F" and "vector subspace" then it's
dimension. If they mean "homeomorphic" and "topological subspace" then
actually it isn't true that lower-dimensional manifolds are always "smaller
than" higher-dimensional ones (so maybe we actually want a different
definition) but in this particular case the rank-1 tensors are a lower-
dimensional proper subspace of all the tensors so the definition gives the
"right" answer.

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mvaliente2001
Excellent, thanks for sharing! I found particularly useful the explanation
that dF = SdA implies that the tensor S must convert the area vector dA in a
force vector dF and that vector can have any direction.

But something that I miss is a visual representation of what kind of change of
directions tensors do. I can see in my mind how a vector multiplication works,
but I haven't the same intuition for tensors. For example, if I have a cube of
marble and I put a heavy weight over it, qualitatively how the tensorial space
in the cube? What kind of values I would see near the top? Which ones near the
bottom? What's the direction of dF if I chose a dA parallel to each side of
the cube? which one if I chose an arbitray vector, say one at 45° of one side?

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lebek
Last night I came across this tutorial on Tensor diagrams:

[http://research.microsoft.com/pubs/79791/usingtensordiagrams...](http://research.microsoft.com/pubs/79791/usingtensordiagrams.pdf)

I love the way Jim Blinn writes.

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krsree
Clear discussion of the basic concepts. Coincidentally, I was reading this two
days back and before this link was posted.

~~~
vimalvnair
Me too, i was reading it a few days back. Trying to understand the math behind
relativity.

I was inspired by 'Interstellar'.

------
return0
> inner product of a matrix and a vector

how does that work?

~~~
RaptorJ
The inner product in the context of tensors is a rank-contraction
operation[0]. The IP you know and love contracts over the covariant and
contravariant indicies of two rank-1 tensors (vectors). You need a metric to
do this so you can raise one of the contravariant indicies to be covariant. If
your metric is the normal one (all 1's) then you just add the product of all
coefficients. Though if you have the Minkowski metric (like in special
relativty) then you negate the time coordinate (or all the space coordinates)
and add. Here the matrix might be regarded as a rank 1,1 tensor (one
covariant, one contravariant index) that you contract with the vector.

[0]
[http://en.wikipedia.org/wiki/Tensor_contraction](http://en.wikipedia.org/wiki/Tensor_contraction)

