Talk:Intermediate treatment of tensors

Sorry to be annoying, but this article have several problems:

1. the sentence: "A tensor is an invariant multi-dimensional transformation" is very unclear:

the entry for invariant say: "An invariant is something that does not change under a set of transformations". If we expand this would give something like "A tensor is a multi-dimensional transformation which doesn't change under a set of transformation" !@?

1. It is not true that tensor are always transformation (i.e. classical mass: tensor of rank(0/0); how this can be considered as a transformation?)

What I think this article should be:

1. Start by explaining that a tensor is a generalization of the concept of vector and matrices.
2. Then explain that tensors allow to express physical laws in a form that apply to any coordinate systems.
3. Say that tensors are heavily used in Continuum mechanics and Theory of relativity (beacause of the previous point)
4. Introduce the two species: contravariant/covariant, introduce notation and ranks (~= number of indices).
5. Define the contravariant/covariant component by showing how they transform under a change of coodinate system.
6. Special cases:
   1. tensors of rank(0/0) => scalars,
   2. rank(1/0) => vectors in differential geometry or contravariant vectors in tensor analysis,
   3. rank(0/1) => one-forms in differential geometry or covariant vectors in tensor analysis.
7. Give some example: Curvature tensor, Metric tensor, Stress-energy tensor

P.S. I'm not a tensor specialist myself, so ...

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This article is confusing. First statement: tensors generalise vectors and matrices. Fine: a vector is a 1-dimensional array of numbers, a matrix is a 2-dimensional array of numbers. The only obvious generalisation is to an n-dimensional array of numbers, n integral, possibly > 2. Is that what a tensor is? No.

Second statement: "A tensor is an invariant multi-dimensional transformation". So, it's a transformation? Well, as far as I know, a vector and a matrix can represent a transformation, for example, of coordinates. I'm not sure that representation is identity; however, perhaps this is just quibbling. Provided the representation is unique, that may do for practical purposes.

Definitions: To define <math>{T}^i</math>, we get an equation relating <math>\bar{T}^i</math> and <math>\bar{T}_i</math>. Where is the definiend? Supposing that was meant to be <math>\bar{T}^i</math>, the term <math>\bar{T}_i</math> on the right is undefined. And why must a transformation necessarily take the form of a partial derivative? Similar problems afflict the second definition.

Yahya Abdal-Aziz - 2003/04/29.

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I agree with the two sentiments expressed above, this article is very unclear. I think it is best to merge it with tensor, and redirect. AxelBoldt 17:18 May 1, 2003 (UTC)

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The reason it was split from 'tensor', is that it is a different treatment. Merging the two would be like merging 'Marxism' with 'Plato's Republic'. It would be very confusing, and also would make for an excessively long article. If this article is unclear, then the last thing it needs is to be entangled in a mass of ideas related to another procedure altogether, and especially one which is currently far more unclear.

Instead of trying to push a political agenda, maybe we should attempt to clearly present the information, with a focus on developing concepts( as opposed to constructing a rigourous self-referential mathematical soup).Kevin Baas

I don't understand where your reference to a political agenda comes from. The modern and the classical approach are really talking about the same thing, but from different angles. They need to be explained in the same article so that the reader sees how and why they are about the same thing. By the way, the coordinate approach is already mentioned in tensor, just not very prominently. AxelBoldt 23:20 May 1, 2003 (UTC)

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The political agenda reference was not pointed at you. Since the classical method is being deprecated, there are some who think that it should be quit altogether. It is very difficult, however, to learn the modern approach without a developed geometric intuition, which the classical approach does a good job of establishing. For this, and other reasons, and esp. that this is an Encyclopedia, information on the classical approach should not be withheld or repressed, even if it's usage is 'defered'. There used to be a classical approach presented in the tensor section, but it was 'phased out'; replaced.

You obviously are in agreement with me that both methods must be presented. We are, then, only in disagreement on a more subtle point: whether the two treatments( which are, ofcourse, about the same thing) should be presented simultaneously or in parrallel. I would image that an encyclopedia in book form would present one in whole, followed by the other in whole, and would not entangle them. If the reader cannot clearly identify their equavalency, which is explicitly stated, clearly marked, and geometrically neccessary, then it is clear that they have not geometrically comprehended the material, and thus the manner of presentation is failing. It is of little practicality for one to recognize the equivalence of two things that they cannot understand. However, once they understand them, the equavalence is obvious and trivial.

--Kevin Baas 2003.05.02

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This article text has mostly been replaced by material adapted from the PlanetMath GFDL article on tensors.

Credit: An earlier version of this article was adapted from the GFDL article on tensors at http://planetmath.org/encyclopedia/Tensor from PlanetMath, written by Robert Milson and others

Here are some PlanetMath to Wikipedia TeX transformations:

\cU should be <math> \mathcal{U} </math>

\bv should be <math> \mathbf{v} </math>

\halpha should be <math> \hat{\alpha} </math>

\ve should be <math> \varepsilon </math>

\hve should be <math> \hat{\varepsilon} </math>

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This equation does not work at the moment, so I will paraphrase it for now. Moving it here for future reference when \overbrace is fixed:

<math>\mathcal{U}^{p,q} = \overbrace{\mathcal{U}\otimes\ldots\otimes\mathcal{U}}^{p\mbox{

times}} \otimes \overbrace{\mathcal{U}^*\otimes\ldots\otimes\mathcal{U}^*}^{q\mbox{ times}}.</math>

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Note: bold alphas do not seem to render properly. The Anome 12:13 2 Jul 2003 (UTC)

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ACK! What happened!?! This is unreadable now! The page used to be very clean. Now not only is it bloated, and poorly formatted, but the modern treatment is mixed up in it. No!!! This is completely inaccessible! Who did this!? It must be undone. I'm sure there may be some good pieces in this new part, but that doesn't justify all the other crap. Whoever it was, please revert, or someone else will have to go thru the trouble of doing so. And be a little more considerate and less self-righteous next time. Pay more attention to presention, rather than brute information. Kevin Baas

Also, keep in mind that this is the starting page for the classical treatment, not the final page. You write as if they already knew everything about tensors, in which case, why the hell would they be reading this? Kevin Baas

This page is now renamed Intermediate treatment of tensors, and your article is back at Classical treatment of tensors -- please hack away. -- Anon.