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.
Every decent man is ashamed of the government he lives under - Mencken
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.
"When the King is far the people are happy." Chinese proverb
For Alexander Zinoviev and the free market there is a shared delight:
"Where there are problems there is life."
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.
Scalars (rank-0 tensors), vectors (rank-1 tensors), and tensors (0 or more ranks) are more than just piles of numbers.
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
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)