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July 8

math questions regarding 1 per hundred, 1 per thousand, 1 per ten and other numbers

is there a website that can help you with questions like what is 1 per ten of 300, 1 per thousand of 3000, 1 per hundred of 300 and 2 per hundred of 300 and etc? I mean that help you practice answering the questions. — Preceding unsigned comment added by 70.31.17.66 (talk) 02:28, 8 July 2012 (UTC)[reply]

Google works "1 per [ten, thousand, cent(hundred)] of [enter number here]", I also like Wolfram Alpha, same words. Bluefist ^talk 02:38, 8 July 2012 (UTC)[reply]

The names for such problems are ratios, proportions, and unit fractions. A quick way to solve them, if they start with only 1's and 0's, is by subtracting the number of zeroes in the ratio from the main number. So, 1 per 100 of 3000 is 30 (you just take off two zeroes). Then, if you wanted 2 per 100 of 3000, you would multiply that 30 by 2, to get 60.

If using a calculator, try 3000 * 2 / 100. StuRat (talk) 03:16, 8 July 2012 (UTC)[reply]

Rigid Body Rotation

Hey I was trying to solve the Feynmann plate problem concerning the ratio of the rates of 'wobble' and 'spin' of a free disk. Starting from the Lagrangian $L={\frac {1}{2}}{\boldsymbol {\omega }}^{T}I{\boldsymbol {\omega }}$ with ${\boldsymbol {\omega }}={\bigl (}{\begin{smallmatrix}{\dot {\phi }}\sin {\theta }\sin {\psi }+{\dot {\theta }}\cos {\psi }\\{\dot {\phi }}\sin {\theta }\cos {\psi }-{\dot {\theta }}\sin {\psi }\\{\dot {\phi }}\cos {\theta }+{\dot {\psi }}\end{smallmatrix}}{\bigr )}$ in terms of the usual Eulerian angles and the coordinates chosen so that $I$ is diagonal. The E-L equation wrt $\psi$ gives one, and hence by analogy all, of the Euler equations $I_{1}{\dot {\omega }}_{1}=(I_{2}-I_{3})\omega _{2}\omega _{3}$ (and other cyclic perms).

The case of the plate is considered by setting $I_{1}=I_{2}={\frac {1}{2}}I_{3}$ . Thus the Euler equations predict that ${\dot {\omega }}_{3}=0$ and $\omega _{1}=\omega _{0}\cos {\omega _{3}t}$ and $\omega _{2}=\omega _{0}\sin {\omega _{3}t}$ .

The Lagrangian is then $L={\frac {1}{2}}I_{1}\left({\dot {\theta }}^{2}+{\dot {\phi }}^{2}\sin ^{2}{\theta }+2({\dot {\psi }}+{\dot {\phi }}\cos {\theta })^{2}\right)$ , which as far as I can tell yields the dynamical equations:

${\operatorname {d} \! \over \operatorname {d} \!t}\left(4{\dot {\psi }}\cos {\theta }+{\dot {\phi }}(3+\cos {2\theta })\right)=0$
${\operatorname {d} \! \over \operatorname {d} \!t}\left({\dot {\psi }}+{\dot {\phi }}\cos {\theta }\right)={\operatorname {d} \!\omega _{3} \over \operatorname {d} \!t}=0$
${\ddot {\theta }}=-{\dot {\phi }}\sin {\theta }(2{\dot {\psi }}+{\dot {\phi }}\cos {\theta }$ )

What I am trying to find is the ratio ${\dot {\phi }}:{\dot {\psi }}$ . The literature on the internet seems to suggest that it is achieved by identifying the frequency of the precession of ${\boldsymbol {\omega }}$ given by $\omega _{3}$ with $-{\dot {\psi }}$ , (giving $2{\dot {\psi }}=-{\dot {\phi }}\cos {\theta }$ ) but I cant see why these should equate? Nor can I see a way of extracting this information from the EL equations. Can anyone help at all? Thanks. — Preceding unsigned comment added by 92.11.64.80 (talk) 15:43, 8 July 2012 (UTC)[reply]

matrix vs. tensor

Is there a simple example of a matrix that is not a tensor or a tensor that is not (or cannot be represented as) a matrix?
A similar question was asked on the tensor talk page, but I don't see a clear example in the answers. Searching for tensor matrix in the reference desk archives yielded these [1] [2] [3], but I don't see an example of something that is one but not the other.--Wikimedes (talk) 19:49, 8 July 2012 (UTC)[reply]

Any tensor can be represented as a matrix, once you choose a coordinate system. (What it's a matrix of is a slightly subtler question.) And any matrix can be interpreted as a tensor. So I don't think you're going to get a clear answer of something that "is" a matrix but "is not" a tensor, or vice versa.

It's a question of intent and interpretation. A tensor is supposed to be the underlying abstraction that's independent of the coordinate system, whereas a matrix, if it's representing a tensor, can do so only in one particular coordinate system. Hope this helps. Others may have subtly different takes on the question, but I think by and large they'll agree with the direction I'm pointing. --Trovatore (talk) 20:31, 8 July 2012 (UTC)[reply]

Any tensor can be a matrix? Aren't matrices specifically 2nd order tensors, whereas generally a tensor can have any order? — Preceding unsigned comment added by 92.14.28.221 (talk) 23:17, 8 July 2012 (UTC)[reply]

OK, point taken. I was thinking of rank-2 tensors. But then again you could have multi-dimensional matricies. --Trovatore (talk) 02:14, 9 July 2012 (UTC)[reply]

Here is a concrete example of a tensor that can't be represented as a matrix (excluding multi-dimensional matrices). Let V = Rⁿ, and let T:(V×V×V) → R be the linear map which gives the product of the first coordinates of the 3 vectors. In other words, for v, w, u in V with v = (v₁,...,v_n) etc, let T(v,w,u) = v₁w₁u₁. Tensors encompass any linear map that takes any number of vectors and spits out a scalar. Matrices can only be used to represent the ones that take two vectors as their input. Rckrone (talk) 03:53, 9 July 2012 (UTC)[reply]

The result that you get from applying T appears to depend on the co-ordinate system, so is it actually a tensor ? Gandalf61 (talk) 08:15, 9 July 2012 (UTC)[reply]

If you want to explicitly define a specific tensor, then of course you have to fix a basis. What other choice is there? Rckrone (talk) 05:22, 10 July 2012 (UTC)[reply]

I think Gandalf61 is pointing out one of those subtleties: If your description is intended to apply to every basis (as for example with the Kronecker delta or the Levi-Civita symbol), it is not a tensor. If it is intended to define the tensor only for some preferred basis, it does. This is significant enough that the use of a preferred basis should be highlighted in the definition, since in the absence of such a statement the converse would generally be implied.

Indeed. My point is that if T(v,w,u) = v₁w₁u₁ in any co-ordinate system then T is not a tensor because T(v,w,u) depends on the co-ordinate system. This is what I assumed the defitnition of T meant. I suppose you could say that T(v,w,u) = v₁w₁u₁ in some specific preferred co-ordinate system, and that T transforms as a tensor when you change co-ordinate systems, in which case, yes, T is a tensor by definition - but this does not seem like a natural way to define a tensor. Gandalf61 (talk) 06:37, 10 July 2012 (UTC)[reply]

I guess I will ask again, what other choice is there? How would you more naturally define this tensor T? Rckrone (talk) 14:28, 10 July 2012 (UTC)[reply]

Sorry, let me rephrase that. I should have said that T is not a natural tensor to want to define or use, precisely because its definition involves a preferred co-ordinate system. It is more natural to illustrate the concept with a tensor such as the metric tensor which can be defined in a coordinate-free way in terms of the geometric properties of tangent vectors etc. The representation of the tensor relative to any given co-ordinate system is then a secondary consideration. I think this is the same point that Trovatore makes below. An analogy - Fibonacci numbers are a more natural sequence than emirps because emirps can only be defined in a base-dependent way. Gandalf61 (talk) 15:01, 10 July 2012 (UTC)[reply]

You can only really get by without coordinates at a certain level of abstraction. For instance the claim that a metric tensor can be explicitly defined without a choice of coordinates is just not accurate. You can say the words "the metric tensor" without coordinates, but if I hand you an explicit atlas for some smooth manifold and tell you "define a smooth metric on this manifold" you're going to have to get your hands dirty. Think carefully about the claim you're making about which tensors are "natural". No one would ever say that the only natural linear transformations are the ones that can be defined without matrices. The only linear transformations that are truly "coordinate-free" are the scalar multiples of the identity which are boring. Rckrone (talk) 15:32, 10 July 2012 (UTC)[reply]

Well, I guess it all depends on what you mean there by "explicitly defined" - if you only believe that a tensor is "explicitly defined" when its components are listed in a specific co-ordinate system then we have a circular argument. As a final point, I would refer you to Tensor (intrinsic definition) which says "In differential geometry an intrinsic geometric statement may be described by a tensor field on a manifold, and then doesn't need to make reference to coordinates at all". I am not going to waste any more time on this. Gandalf61 (talk) 16:32, 10 July 2012 (UTC)[reply]

Yes, now you are getting at the problem that I'm trying to describe. By "explicitly defined" I mean give a description that uniquely picks out a specific element of the set without making any arbitrary choices. For example if the set were the integers I could say "4" and you would know which integer I was talking about. Unfortunately, as you were alluding to, this can't be done for vector spaces, or spaces of tensors, unless we make a choice of basis (with some small classes of exceptions). This concept of one tensor being "natural" and another not doesn't exist. That's too bad in some ways, because when we teach about vector spaces we want to instill the idea that these objects exist before we choose a basis, but bases have to be tied up in any explicit descriptions of the elements. You objected to my example above for not being natural, but unfortunately that's the best we can do. Rckrone (talk) 17:31, 10 July 2012 (UTC)[reply]

The question (or answer) should be separated ab initio into the question of the order of the tensor (which equals to the "dimensionality" of the matrix), and the other subtler aspects. The first question of order should be straightforward, so let's assume that is not being asked about (OP please confirm). Adding to Travatore's response, there are also matrices (or matrix-like sets of components) that are not considered to be tensors, because to be a tensor, an additional requirement must be satisfied, essentially that the components must transform appropriately with a change of basis. An example of a non-tensor that still can be represented as a matrix is the Levi-Civita symbol, which in two dimensions is a square matrix. In the converse case, an order-k tensor can always be represented as a matrix of dimensionality k once a basis has been chosen. — Quondum ^☏ 08:08, 9 July 2012 (UTC)[reply]

Yes, I was not asking about dimensionality or order.

Thanks for the Levi-Civita example. I had always heard that tensors had additional properties and figured a matrix existed which didn't have those properties. I still have some work ahead of me to understand why it acquires a minus sign under an improper rotation and why this means it is not a tensor, but I'll give it a try before asking for an explanation.

Is a matrix representation of a tensor considered to be a tensor, or do you also need to define how it operates? Looking at Rckrone's example, I can imagine representing it as an array of 3 ones and a bunch of zeroes, but it seems you'd still have to define what those ones and zeroes do when operating on 3 vectors.--Wikimedes (talk) 09:12, 9 July 2012 (UTC)[reply]

Let's go down one rank; nothing fundamental changes. That is, is an ordered triple of real numbers the same as a (three-dimensional) real vector, or not?

Well, any 3-d real vector can be represented as an ordered triple, and any such ordered triple can be interpreted as a vector. But are they the same thing?

I would say they are not. Imagine an arrow, floating in space. That's a vector. Which ordered triple you associate with it, depends on the coordinate system you assign. Conversely, given an ordered triple, which arrow you associate to it again depends on the coordinate system. --Trovatore (talk) 09:42, 9 July 2012 (UTC)[reply]

I think the key difference is that a matrix is an array of scalars, and a tensor is what you get when you sum the products of such a matrix's entries with a basis (of a –normally abstract– vector space). A specified basis is essential for the matrix to represent a tensor, and changing the basis changes the matrix components in a specific way to keep the resultant tensor unchanged. If the matrix elements are specified to change with a change in basis in a way that conflicts with this "specific way", then they clearly do not represent a well-defined tensor. It is this overspecification with conflict that can prevent a matrix from being a representation of a tensor. (Oh, and you do not have to define how it operates; it is sufficient for a tensor to be a member of a linear space.) I hope this encapsulates the nub of the issue. — Quondum ^☏ 13:20, 9 July 2012 (UTC)[reply]

A "normally abstract" vector space??? That might depend on your field. But I didn't think algebraists used the "tensor" language so much (outside of locutions like tensor product); why wouldn't you just say linear transformation?

My default context for talking about "tensors" specifically is the tangent bundle of a manifold, or something similar to it. --Trovatore (talk) 18:07, 9 July 2012 (UTC)[reply]

My terminology is probably unconventional. I'm trying to force the idea of the vector space being unspecified other than for the number of dimensions and the base field (and certainly don't mean it in the sense of abstract algebra); so many people are taught that a vector is an n-tuple of values. The properties of a tensor as a linear transformation are actually not entirely relevant to its representation via a matrix. Its membership of a vector space (or a tensor power thereof) is sufficient. — Quondum ^☏ 21:36, 9 July 2012 (UTC)[reply]