An Introduction to Tensors for Students of Physics and Engineering (2002) [pdf] | Hacker News Reader

An Introduction to Tensors for Students of Physics and Engineering (2002) [pdf]

grc.nasa.gov

88 pointsbrudgers21 comments

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...

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

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!)

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.)

Items 1-3 in the "Specific Statements for Tensors per se" (pqge 6) seems to allude to your point.

"1. All scalars are not tensors, although all tensors of rank 0 are scalars (see below). 2. All vectors are not tensors, although all tensors of rank 1 are vectors (see below). 3. All dyads or matrices are not tensors, although all tensors of rank 2 are dyads or matrices."

It's not perfectly clear whether "dyads or matrices" means "dyads, which we also call matrices" (wrong!) or "either dyads or matrices" (right, kinda!), which is why I picked on a slightly later bit of the document to support my accusation.

It's also not perfectly clear what the author really means by saying that not all matrices "are" tensors but all tensors "are" matrices; this is part of what he says is summarizing previous remarks but I don't see any previous remarks addressing the point.

The usual answer is to say that a tensor is a matrix that "transforms correctly"; in something closer to puremathematicianese, what this really means is that we're looking not exactly at matrices but at functions from coordinate systems to matrices (i.e., you plug in the coordinate system you want to use, and out come the components of the thing-that-might-be-a-tensor as a 3x3 array), and that what you've got is a tensor of a particular kind if and only if the way in which these numbers change as the coordinate system does is right. But at this point in the document he hasn't said anything remotely like that. He does eventually mention it but not until the very end, where he seems to think he's getting it as a consequence of some other definition -- but unless that definition is the (definitely wrong) one that says the only tensors are tensor products of vectors, it's not clear what it is.

(It is in fact possible to define tensors in another way that's neater and coordinate-free, and derive the transformation rules as theorems. One way to do this is closely related to what this document does, but instead of (wrongly) saying that all tensors are tensor products of vectors ("dyads", etc.) what you (rightly) say is that all tensors are linear combinations of tensor products of vectors. Er, and their duals. And you actually have to say what all that means, and it turns out to be a rather elegant bit of theory. Perhaps that's what this document is gesturing towards, but it really doesn't do more than gesture.)

I have exactly this experience whenever I read something by a physicist! It is another universe where everything seems vague, but they somehow manage to get the correct answer anyway, and usually very quickly too.

> and it turns out to be a rather elegant bit of theory

Do you have a recommendation on what to read for that?

> 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?)

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.

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?

You can say that R^2 has "more" points than R by inclusion: there is a natural injection of R into R^2 as a proper subspace. Think of the odd numbers in the integers, for example.

"More" is a fairly loose concept in that there is no strict definition. You can use it to refer to cardinality, or some other sense of order.

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.

Well, it's really not necessarily a "lie" as much as it is a "mistake." Not everything that is "flatly untrue" is a "lie" as there is no apparent intent to deceive.

Basically, when you say that "this document tells an important lie" you are likely lying. At least you are accusing somebody of being a liar.

> 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.

Then again, that whole section is in the context of material stresses and permeability, which are described as symmetrical tensors, for which the claim is true.

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?

Last night I came across this tutorial on Tensor diagrams:

http://research.microsoft.com/pubs/79791/usingtensordiagrams...

I love the way Jim Blinn writes.

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

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

I was inspired by 'Interstellar'.

> inner product of a matrix and a vector

how does that work?

I think he just means the normal matrix product Av (where A is a n x m dimensional matrix and v a n dimensional vector). Wikipedia covers matrix multiplication in depth:

http://en.wikipedia.org/wiki/Matrix_multiplication

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

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