Can anyone please explain what the heck this video is talking about?
http://www.youtube.com/watch?v=istE1bpoDPg
Description: Economic and political failure result from using systems of buying and selling as well as systems of voting that cannot do what we expect them to do. This failure stems from shortcomings in the process of counting votes and counting dollars and the nature of human values. An alternative method using tensors and vectors is proposed to resolve this difficulty.
Without having watched it, I would guess it is introducing physics into economics. In other words, more engineering/hard science for a discipline that is nothing like them. If this is indeed what is going on, then it is simply another spin on the Fatal Conceit.
Can’t access the video, but a vector is just a rank 1 tensor, so one is a subcategory of the other. It’s rank 1 because you need one index, e.g. x, y or z to identify a component. A matrix is rank 2 since you need both a row number and a column number to specify a component. Economists I’ve encountered don’t know what tensors are.
abskebabs,
Studying vector calculus and tensors for the last few weeks, as a physics undergraduate, I think most physicists do not know either.
A matrix is rank 2 [tensor] since you need both a row number and a column number to specify a component.
Scalars (rank-0 tensors), vectors (rank-1 tensors), and tensors (0 or more ranks) are more than just piles of numbers. They also must obey certain transformation laws. An example of something that’s not a tensor is angular momentum. It may have three components, but it is not really a vector, but rather a pseudovector http://en.wikipedia.org/wiki/Pseudovector
Dr. Acula,
And there it is. Pseudovectors are vectors indeed because they are elements of vector spaces. The proper terms are covariant and contravariant.
Pseudovectors are vectors
I don’t agree, pseudotensors are not tensors. See http://en.wikipedia.org/wiki/Pseudotensor
"In physics and mathematics, a pseudotensor is usually a quantity that transforms like a tensor under a proper rotation, but gains an additional sign flip under an improper rotation (a transformation that can be expressed as an inversion followed by a proper rotation).
There is a second meaning for pseudotensor, restricted to general relativity; tensors obey strict transformation laws, whilst pseudotensors are not so constrained."
If it doesn’t transform like a tensor, then it’s not a tensor. This applies to both of the meanings of the word pseudotensor above.
The proper terms are covariant and contravariant.
I don’t agree, covariant and contravariant vectors are both tensors and obey tensor transformation rules. In fact, tensors can have mixed variance (both upper and lower indices). Pseudotensors are something different.
Is a pseudovector element of a vector space or is it not?
Yeah, sorry you are right. In mathematics pseudovectors and axial vectors qualify as a vector. In physics, they are not considered to be a vector.
Would you call a two-index pseudotensor a “tensor”? I would not, even when working in pure mathematics.
No I would not. Of course, nowadays I would use the definition as elements of tensor products for tensors and pseudotensors do not qualify. For tensors the term “pseudotensor” actually is meaningful as there are some important objects that are similar to tensor but to not transform like tensors, for example the epsilon-tensor or Levi-Civita symbol in three dimensions.
It is just this sloppy use of mathematical terms that confuses students in physics classes. Of course it makes sense to distuingish polar from axial vectors but it is senseless to call one “vector” and the other “pseudovector” because one is a rank (0,1) tensor and the other is a pseudotensor. They both are vectors, the latter terminology hides this fact and suggests that axial vectors do not represent spatial qualities or something.
I stand corrected. Strictly I should have said a rank 2 tensor’s components can be represented in the form of a matrix, with column and row indices both needed in identifying components. I think you’ve proven Metus’ point about physicists(or ex-physicists) not knowing what they’re talking about when it comes to tensors! 
P.S. Dr Acula, you never answered my query in the previous thread we both posted in, regarding utility transformations and knapsack problems.
Video link is fixed! 
it is senseless to call one “vector” and the other “pseudovector” because one is a rank (0,1) tensor and the other is a pseudotensor.
OK, now I’m confused. If a vector is a kind of tensor, then how is it senseless to call something a pseudovector when its a low-rank pseudotensor? It seems rather sensible to me.
They both are vectors, the latter terminology hides this fact and suggests that axial vectors do not represent spatial qualities or something.
But the mathematical defintion of vector has nothing to do with spatial qualities. (3 oranges, 4 apples, 5 peaches) is no less valid a mathematical vector than angular momentum is.
I think we are facing inconsistencies in terminology. Part of the problem is that a mathematical vector in a mathematical vector space doesn’t have to obey the transformation rules of a physics vector. A pseudovector obeys the rules necessary to be an element of mathematical vector space, making it a mathematical vector. But it can’t be a physics vector since it doesn’t obey the right transformation rules - it’s a pseudovector.
LOL, this article from wikipedia contradicts itself:
http://en.wikipedia.org/wiki/Cross_product
the cross product … is a binary operation on two vectors in three-dimensional space. It results in a vector
the cross-product of two vectors is not a (true) vector, but rather a pseudovector
Either it results in a vector, or it results in a non-vector. Which is it?
(deleted - double post)
So I saw the video and:
-I can’t imagine the preference loops existing in real life
-I don’t see how preference loops can possibly be revealed through praxeology. They sound as fanciful as cardinal values.
i.e. they might exist but there is no way to measure or even detect them.
-I don’t see how “tensor money” can work unless the people are forbidden from exchanging the individual components of the money
Dr. Acula,
Of course you are confused. The terminology is confusing. If a vector is a cind of tensor it is meaningful to call it tensor, if its not but almot it is meaningful to call it pseudotensor. If it is a vector it is meaningful to call it a vector, if is a vector but not a tensor it is senseless to call it pseudovector since it is not only almost but it is in fact a vector. The property of being a tensor is just an extension to the properties of being a vector.
I was talking about the properties of vectors as used in physics. Even if I were not, the R^3 as vector space is a model for Euklid’s geometry, the archetype of space models.
I am arguing that the term “pseudovector” is misleading, not that the difference between axial and polar vectors is irrelevant. Your argument suggests that an angular moment, the archetype of a so called “pseudovector”, is not a physical vector and thus, since it is not a real vector, does not carry spatial properties.
Both. Vector as an element of a vector space, “pseudovector” since its axial.
Though I doubt the last point. Take the apparently polar vectors (1,0,0), (0,1,0), (0,0,1), the standard base of R^3. But (1,0,0) x (0,1,0) = (0,0,1) is axial. This is apparently a contradiction. I will think about it tomorrow.
Metus, you are making a lot of sense to me but I’m struggling with accepting your definitions because they lead to loathesome outcomes.
For example, per your definition, isn’t every tensor a vector? After all, a tensor like Aij where 1<=i<=3 and 1<=j<=3, is just an element in a 9-dimensional vector space. If it’s an element of a vector space, it must be a vector.
If every tensor is a vector, then going out of your way to call only rank-1 tensors “vectors” is kind of dishonest.
"Take the apparently polar vectors (1,0,0), (0,1,0), (0,0,1), the standard base of R^3. But (1,0,0) x (0,1,0) = (0,0,1) is axial. This is apparently a contradiction"
Oh, I see. If it makes you feel better you could write pseudovectors as [x,y,z]: (1,0,0)x(0,1,0) = [0,0,1]
baxter,
Yes, every matrix with n rows and m columns is element of a vector space. Yes, every tensor is element of a vector space. I never said that only rank (1,0) tensors are vectors, I am only arguing that “pseudovector” is misleading. In this context, we can further see how this is misleading since a pseudotensor is still a vector of the vector space of all multilinear transformations.
One has to note that the terms “vector”, “tensor” and “scalar” are not mutually exclusive or a hierarchy, they are merely different ways to look at mathematical objects. It is meaningful to talk about scalars, vectors and tensors as if they were different objects if they are used differently which they are in physics and in some branches of mathematics. It is of course not meaningful to do so from a fundamental point of view. For all practical purposes it is useful to talk about scalars as elements of the body of real numbers, vectors as elements of R^3 or R^4, tensors as part of the specific tensor products. If there is ambiguity, the definitions are to be dropped and more precise declarations to be used.
At the time I started to post there was something like “I do not see the contradiction” but apparently you edited it. Anyway, I showed that it is not the result of a cross product that is a pseudovector but the cross product itself. The cross product would be a good example for a pseudotensor as it is an alternating multilinear transformation but not a tensor since it does not obey the tensor transformation law.
Yes, preference loops make no sense. So his system collapses.
Tensors are elements of the tensor product, so they are vectors. All the index transformation physics bs is just headache.