Economics Questions

Re: Tensors and Vectors?

scineram — Thu, 13 Jan 2011 11:42:47 GMT

I am applied math undergrad.

Tensors were taught to us badly back then. I never knew they were what or of what use. I only understood them later, after reading about the abstract approach on my own.

Re: Tensors and Vectors?

Metus — Wed, 12 Jan 2011 21:44:06 GMT

I meant real as in body of real numbers. There is nothing to talk about them.

A base to a vector is always in the same space as the vector, obviously, so I am not sure I understand where this is going. Ultimately you will have to claim that the cross product maps two polar vectors (or contravariant vectors or elements of the vector space) to an axial vector (or covariant vector or element of the dual space) since the cross product is apparently axial. I see now that we are fudging the mathematics even a bit more.

Edit: I just want to make clear what we are discussing. My original point was that the wording "pseudovector" for an axial vector is misleading. Further I argue that the notion of polar and axial is meaningful since it represents properties of physical objects. But their use is erroneous. Finally I suspect that the notion of axial and polar is not equivalent to contra- and covariant, that is being an element of vector or dual space. It all breaks down to bad mathematics.

Re: Tensors and Vectors?

Dr. Acula — Wed, 12 Jan 2011 21:28:52 GMT

>Do tensors have a transformation law?

Yes http://en.wikipedia.org/wiki/Covariant_transformation

>Is a polar vector a tensor?

Yes, it's a (0,1) tensor.

> I am talking about real vectors... about the mathematical framework.

OK, let's talk about your real, mathematical vectors.

You claim that (1,0,0) is polar.

If I establish a basis in the dual space, I can denote a member of that dual space by (1,0,0). Is that also polar?

Re: Tensors and Vectors?

Metus — Wed, 12 Jan 2011 21:19:55 GMT

Do tensors have a transformation law? Is a polar vector a tensor?

Anyway, now it is just silly. I made it clear that I am talking about real vectors. Talking about velocities, oranges or anything is meaningless when we are talking about the mathematical framework.

Re: Tensors and Vectors?

Dr. Acula — Wed, 12 Jan 2011 21:07:21 GMT

"It is not ambiguous. Somewhere above I said I was talking about real vector spaces. Ultimately, you are avoiding the question. (1,0,0), (0,1,0) and (0,0,1) are by all means polar because they transform like polar vectors"

That's nonsense.

There's no such thing as a transformation rule for "(1,0,0)".

Just like there's no transformation rule for "42". It could be a scalar or a pseudoscalar.

A bunch of numbers cannot be polar or axial. Only things that related to actual physical space can be polar or axial. If I use (1,0,0) to indicate that I have 1 apple, 0 bananas, and 0 oranges, then it is not polar or axial. In fact it does not transform at ALL under physical rotation or reflection.

Re: Tensors and Vectors?

Metus — Wed, 12 Jan 2011 20:54:51 GMT

Dr. Acula:
>Is (0,0,1) polar or axial? Is (1,0,0) x (0,1,0) = (0,0,1) polar or axial?

It's ambiguous, because the notation is deficient.

(1,0,0)x(0,1,0)=(0,0,1) could be Polar x Polar = Axial. Or PxA = P or AxP = P or AxA = A.

If (1,0,0) stands for 1 apple, 0 bananas, and 0 oranges, then it's neither a polar nor an axial vector and the equation is meaningless.

Obviously, the notation doesn't express the full meaning needed in real life problems. It conflates the "isomorphic spaces" you referred to. It lacks units, such as distance or time.

It is not ambiguous. Somewhere above I said I was talking about real vector spaces. Ultimately, you are avoiding the question. (1,0,0), (0,1,0) and (0,0,1) are by all means polar because they transform like polar vectors but (1,0,0) x (0,1,0) = (0,0,1) is allegedly axial.

Dr. Acula:
Edit: I don't think pure mathematicians are blameless when it comes to poor, sloppy notation. There is sloppiness when you write f(x)=x^2. Is f a function or is it a variable being multiplied by x? Is x a free variable or is this stating something regarding the value of f at a particular constant, x? The notation is inadequate so common sense and context have to be used.

In principle there can not be ultimate unambiguity. However we can try not to use confusing terms. That is my whole argument.

Re: Tensors and Vectors?

Dr. Acula — Wed, 12 Jan 2011 20:27:47 GMT

>Is (0,0,1) polar or axial? Is (1,0,0) x (0,1,0) = (0,0,1) polar or axial?

(0,0,1) by itself is a bunch of symbols. It's meaning is ambiguous, because the notation is deficient.

(1,0,0)x(0,1,0)=(0,0,1) could mean Polar x Polar = Axial. Or PxA = P or AxP = P or AxA = A.

If (1,0,0) stands for 1 apple, 0 bananas, and 0 oranges, then it's neither a polar nor an axial vector and the equation is meaningless.

Obviously, the notation doesn't express the full meaning needed in real life problems. It conflates the "isomorphic spaces" you referred to. It lacks units, such as distance or time.

Edit: mathematics is full of sloppy notation. There is sloppiness when you write f(x)=x^2. Is f a function or is it a variable being multiplied by x? Is x a free variable or is this stating something regarding the value of f at a particular constant, x? Does "sin pi x + cos pi x" mean "(sin(pi))x + (cos(pi))x=-x"? The notation is inadequate so common sense (e.g. context and tradition), have to be relied upon.

Re: Tensors and Vectors?

Metus — Wed, 12 Jan 2011 20:07:43 GMT

baxter:
>I showed that it is not the result of a cross product that is a pseudovector but the cross product itself.

This is a bunch of doublespeak.

Is (0,0,1) polar or axial? Is (1,0,0) x (0,1,0) = (0,0,1) polar or axial?

baxter:
Metus, I think part of your confusion comes from focusing too much on mathematical notation. In physics, quantities like angular momentum are considered to be real and not just sets of numbers. A velocioty might be denoted by (1,0,0). An angular momentum might be denoted by (-1,0,0). However, these are fundamentally different objects. You cannot add these to obtain another vector, just as you cannot add the number 3 and an apple to obtain anything. Yes, you can add the numeric components together but the result is meaningless and can't be transformed. It is revealed to be meaningless as soon as one chooses to assign a different set of numbers to spatial locations (erect a coordinate system).

There is no confusion. I am purposefully focusing on the mathematical notation since the mathematical notation stands for certain logical concepts that the physical concepts are believed to have. Funny thing is that you can formalize your argument neatly by introducing different isomorphic vector spaces. Still does not change anything about my argument. Of course it is senseless to add up a velocity and an angular moment.

Re: Tensors and Vectors?

baxter — Wed, 12 Jan 2011 18:25:51 GMT

>I showed that it is not the result of a cross product that is a pseudovector but the cross product itself.

This is a bunch of doublespeak.

Metus, I think part of your confusion comes from focusing too much on mathematical notation. In physics, quantities like angular momentum are considered to be real and not just sets of numbers. A velocioty might be denoted by (1,0,0). An angular momentum might be denoted by (-1,0,0). However, these are fundamentally different objects. You cannot add these to obtain another vector, just as you cannot add the number 3 and an apple to obtain anything. Yes, you can add the numeric components together but the result is meaningless and can't be transformed. It is revealed to be meaningless as soon as one chooses to assign a different set of numbers to spatial locations (erect a coordinate system).

Re: Tensors and Vectors?

Metus — Wed, 12 Jan 2011 16:56:25 GMT

scineram,

May I know your specialty?

Re: Tensors and Vectors?

scineram — Wed, 12 Jan 2011 16:01:27 GMT

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.

Re: Tensors and Vectors?

Metus — Wed, 12 Jan 2011 13:04:40 GMT

baxter,

baxter:
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.

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.

baxter:
"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]

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.

Re: Tensors and Vectors?

baxter — Wed, 12 Jan 2011 04:40:56 GMT

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]

Re: Tensors and Vectors?

Metus — Wed, 12 Jan 2011 00:39:47 GMT

Dr. Acula,

Dr. Acula:
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.

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.

Dr. Acula:
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 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.

Dr. Acula:
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.

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.

Dr. Acula:
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?

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.

Re: Tensors and Vectors?

Dr. Acula — Wed, 12 Jan 2011 00:03:06 GMT

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