Hyper Dimensional Philosophizing

[Fredrik]

E1=(1,0,0)
E2=(1,1,0)
E3=(1,1,1)

These three vectors do not lie in three independent dimensions. They are described in a three dimensional vector space. E1 lies in only 1 dimension of the vector space. E2 lies in 2 dimensions of the vector space. E3 lies in all three dimensions of the vector space.

To show that the three dimensions of the vector space are independent, you need three representative vectors, each lying only in one of the three dimensions. They don’t even have to have the same scalar value in each dimension, but they must lie exclusively in one and only one of the three dimensions of the three dimensional vector space.

V1=(a,0,0)
V2=(0,b,0)
V3=(0,0,c)

Each of these three vectors are perpendicular to each other.

You couldn't be more wrong about these things. Have you even read #160 where this is explained? E1, E2 and E3 are linearly independent, i.e. they define three independent directions, and they do it just as well as your D1, D2 and D3.

I'm not going to try to explain that to you again unless I see some evidence that you have tried to understand the previous explanation.

There is no chance whatsoever that you will ever understand dimensions unless you learn about linear independence first. If you need more details than you can find in #160, then check out Wikipedia or a textbook on linear algebra.

Here's a good excercise for you: Suppose that v₁, v₂,...,v_n are vectors in some vector space V. Prove that one of them can be expressed as a linear combination of the others (i.e. as a sum of multiples of the others) if and only if {v₁,v₂,...,v_n} is linearly dependent (i.e. if this set of vectors is not linearly independent).

 

[Oppressed]

Fredrik,

I have stated very clearly the type of dimension whose meaning I am referring to and it is not based on vector space, it is based on the real observable physical space around us. We might use a vector space to provide an example of vectors lying in the three (type a) dimensions of that space, but that three dimensional vector space would be orthogonal.

Take out a real 8” by 11” sheet of paper and place it on a flat table top.

Draw a point 1” in from the left hand side and 5.5” down from the top of the paper and label it “D1a”.

Draw a point 1” in from the right hand side and 5.5” down from the top of the paper and label it “D1b”.

Using a straight edge, draw the shortest distance between D1a and D1b. You now have a line segment that lies in one of the three (type a) dimensions of space.

To at least temporarily remove the argument about polar coordinates being used, the second (type a) dimension we are looking at is also a length.

Draw a point halfway between D1a and D1b, label it O for origin. Draw a circle of a 3” radius around the point O. This circle should pass through points D1a and D1b.

Draw a point on the circle towards the top half of the paper such that the point is furthest away from the line segment D1a to D1b and label it D2a.

The shortest distance between D2a and the line segment D1a to D1b is the point O.

Draw a point on the line segment 0.5” to the right of point O on the line segment D1a to D1b and label it D1c.

Draw a point on the circle that is above point D1c along a perpendicular path to the line segment D1a to D1b and label this point D3a. The line segment D1c to D3a should be (3/4)^0.5 inches.


The line segment O to D1a lies in one (type a) dimension of length.
The line segment O to D2a lies in one (type a) dimension of length.
The line segment O to D3a lies in one (type a) dimension of length.

But, movement along the (type a) dimensional length that the line segment O to D2a lies in is accomplished with NO movement along the (type a) dimensional length that the line segment O to D1a lies in.

This is not true of the (type a) dimensional length that the line segment O to D3a lies in. Movement along the (type a) dimensional length that the line segment O to D3a lies in is accomplished with a 0.5” movement along the (type a) dimensional length that the line segment O to D1a lies in.

 

[Oppressed]

Continuing from the points and line segments I defined in the previous post.

(Type a) dimension “D1” is the dimension the line segment O to D1b lies in.
(Type a) dimension “D2” is the dimension the line segment O to D2a lies in.
(Type a) dimension “D3” is the dimension the line segment O to D3a lies in.

Measure out a 1” sided square in two (type a) dimensions.

If the two (type a) dimensions being used are D1 and D2, such the a point is referenced by (value for D1, value for D2), then use the four points consisting of the square (0,0), (1,0), (1,1) and (0,1). Calculate the measured area.

Now, perform the same measurement using the two (type a) dimensions D1 and D3, such the a point is referenced by (value for D1, value for D3), then use the four points consisting of the square (0,0), (1,0), (1,1) and (0,1). Calculate the measured area.

Notice that the area is not the same.

In the two (type a) dimensions of this real plane space, if you measure out a distance along only one of those (type a) dimensions and then measure out a distance the other (type a) dimension and repeat so that you measure out the four points in the two dimensional space (0,0), (1,0), (1,1) and (0,1) and then calculate the area you get 1 units of measurement squared which is in this case 1 inch squared.

Whatever you want to call area, it is still a measurement that is made within the two (type a) dimensions of this real plane space. Since the area measured along the two (type a) dimensions of D1 and D3 fail to accomplish this, it is another example of why the two (type a) dimensions must be perpendicular to each other.

 

[Oppressed]

Fredrik,

I was looking at your example again.

These three basis vectors will do just as well:
E1=(1,0,0)
E2=(1,1,0)
E3=(1,1,1)

I think you are confusing a “basis” for a vector space with a “dimension” for a vector space.

For example, consider a vector in the x-y plane. Lets define three vectors:
Vector = ( x, y)
R1 = (-2,+3)
R2 = (+3,+1)
R3 = (+3,-1)

R1 and R2 could serve as the basis for the vector space.
R1 and R3 could serve as the basis for the vector space.
R2 and R3 could serve as the basis for the vector space.

For a two dimensional vector the notation is:
“Vector” = ( “value of dimension 1” , “value of dimension 2” )

If we were to define a vector in two dimensions (length r and angle θ):
Vector = ( r, θ )
V1 = (+2,-1.4)
V2 = (-1,+4.1)

That is why I used the example of two unit vectors, each lying in a different dimension.

Vector = (D1,D2)
R1 = (+1,+0)
R2 = (+0,+1)
R3 = (+a,+0) such that “a” does not equal zero
R4 = (+0,+b) such that “b” does not equal zero

R1 and R3 lie in the D1 dimension only.
R2 and R4 lie in the D2 dimension only.

R1 and R2 are perpendicular.
R3 and R4 are perpendicular.

 

[Fredrik]

Oppressed, there's a reason why mathematicians abandoned that kind of arguments and that kind of language hundreds of years ago. This is not how to do mathematics. This is not how to do physics.

I don't know what else to say to you. If you would just bother to learn about vectors, and specifically the concept of linear (in)dependence, you would immediately see why it's wrong (or worse) to say that the dimensions of space must be perpendicular.

 

[Oppressed]

GodMark2,

You did agree that the three dimensions of length would have to be mutually perpendicular.

Fredrik disagrees.

Would you care to expound upon the subject?

 

[Fredrik]

I think you are confusing a “basis” for a vector space with a “dimension” for a vector space.

I'm not confusing anything with anything, but yes, I am talking about basis vectors. The dimension of a vector space is defined as the maximum number of linearly independent vectors. That number happens to be the same as the number of vectors in a basis.

That is why I used the example of two unit vectors, each lying in a different dimension.

But you haven't defined what a dimension is yet (or acknowledged that I have). This is something that can only be done within the framework of mathematics, and in the context of a specific kind of mathematical structure.

 

[Fredrik]

Fredrik disagrees.

Would you care to expound upon the subject?

Am I supposed to teach a class on linear algebra in this thread? This stuff is trivial to anyone who understands vector spaces, so I can only suggest that you try to learn something about them.

 

[Skepticybe]

Disclaimer: I've only read about a third of the posts in this thread.

First, regarding the ultra-liberal use of the word "dimension". The loose definition of a measurable attribute that is (more or less) independent from other attribute measurements isn't useful. Taken to its extreme, in precisely one gram of water we would have the following dimensions:
- the number of hydrogen atoms
- the number of oxygen atoms
- the number of protons
- the number of electrons
- the number of neutrons
- the amount of radioactive emissions
- the mass (one gram)
- the volume
- etc. etc. etc.

You can see that by calling non-spatial measurements "dimensions", the definition is so watered down that the number of "dimensions", under the loose definition, is infinite.

For a discussion of dimensions to be meaningful, I think the definition must be restricted to spatial dimensions.

Second, my understanding from Astronomy 101 is that the universe is expanding, and this is generally accepted within the scientific community. Meaning that the distance between the galaxies is increasing, that each individual galaxy is a galaxy moving away from the other galaxies, which in turn are each moving away from the other galaxies.

My understanding is that this is only possible with the existence of a fourth spatial dimension. So isn't a fourth (as yet unobserved) spatial dimension assumed in the current model of an expanding universe?

To visualize this, spread a bunch of quarters on a table. Select one quarter from the middle and try to move it so that it is farther from all of the other quarters. The only way to do this is to lift the quarter vertically off of the table. To move two quarters from the middle so that they move farther away from all other quarters would require both of those quarters to be lifted vertically off of the table.

I tend to think of it more like a 4d (ignoring time) bubble with a 3d surface. As the bubble increases in size, you have a proportional increase in the distance between all 3d points.

 

[GodMark2]

GodMark2,

You did agree that the three dimensions of length would have to be mutually perpendicular.

Fredrik disagrees.

Would you care to expound upon the subject?

I thought that you and Fredrik had decided to limit yourselves to Eucledian space (a subset of general vector spaces). Fredrik's arguments (and his problems with my arguments) are quite correct if you are talking about the generalized concepts of vector spaces.

Are you talking about a three dimensional Eucledian space, or are you talking about any three dimensional space?

 

[GodMark2]

E1=(1,0,0)
E2=(1,1,0)
E3=(1,1,1)

These three vectors do not lie in three independent dimensions. They are described in a three dimensional vector space. E1 lies in only 1 dimension of the vector space. E2 lies in 2 dimensions of the vector space. E3 lies in all three dimensions of the vector space.

To show that the three dimensions of the vector space are independent, you need three representative vectors, each lying only in one of the three dimensions. They don’t even have to have the same scalar value in each dimension, but they must lie exclusively in one and only one of the three dimensions of the three dimensional vector space.

V1=(a,0,0)
V2=(0,b,0)
V3=(0,0,c)

Each of these three vectors are perpendicular to each other.

You couldn't be more wrong about these things. Have you even read #160 where this is explained? E1, E2 and E3 are linearly independent, i.e. they define three independent directions, and they do it just as well as your D1, D2 and D3.

As an aside, {V1,V2,V3} is a vector representation of Cartesian coordinates in a Euclidean space, and {E1,E2,E3} is a vector representation of spherical coordinates. ("a" meaning "one possible", to avoid further confusion)

 

[Fredrik]

OK, there's one thing that I would like to explain further. Here it is:

There is no reason whatsoever to prefer one basis over another.

We're both using a notation that at first glance makes one particular basis seem "special", but there really isn't anything special about it. It's an illusion caused by the notation.

Let V=(a,b,c) be an arbitrary vector. Note that V can be expressed either as a*D1+b*D2+c*D3 or as (a-b)*E1+(b-c)*E2+c*E3. When a vector is expanded in the "D" basis the coefficients match the numbers (a,b,c). When the same vector is expanded in the "E" basis, the coefficients don't match those numbers.

There's nothing that prevents us from simply writing V=(a-b,b-c,c) instead of V=(a,b,c). It's merely a convention to let the coefficients of the vector expanded in the "D" basis appear in the "(number,number,number)" notation. We could have picked any other basis, and it doesn't matter if the basis vectors are perpendicular or not.

 

[Fredrik]

About the angles...

First of all, the angle between vectors isn't even well-defined until we have defined something called a scalar product (also called an inner product) on the vector space. We can for example define the scalar product <u,v> of two arbitrary vectors u=(u1,u2,u3) and v=(v1,v2,v3) by <u,v>=u1*v1+u2*v2+u3*v3. Then we can define the angle between u and v to be cos^-1(<u,v>/sqrt(<u,u><v,v>)).

Without definitions like these it makes no sense to say that two vectors are perpendicular and that two others are not. It's actually possible to make the basis vectors of any basis perpendicular to each other, simply by defining the scalar product in a certain way. So it would make just as much sense to say that the vectors in the "E" basis are perpendicular to each other and the ones in the "D" basis are not, as to say what you've been saying (which is the exact opposite).

The reason I haven't mentioned this (that a scalar product is needed for the concept of "angles" to make sense) until now is that the scalar product I defined above is almost always used when the vector space is R³. It's used because it agrees with our pre-existing ideas about what an angle should be. So even though I know that the phrase "these vectors are perpendicular" doesn't really make sense and should be replaced with "these vectors are perpendicular with respect to this scalar product", I've been letting this sort of thing slide because one particular choice of scalar product is almost always used.

This is another way of looking at it: Define a function f:R³-->R³ by f(a,b,c)=(a-b,b-c,c). This function maps R³ onto itself. It is continuous, bijective, linear, and differentiable infinitely many times, so it preserves all the relevant structures of R³ (the topology, the vector space structure and the manifold structure). However, it doesn't preserve angles.

You say that e.g. (1,0,0) and (0,1,0) are perpendicular. That may be true before we apply the function f, but f transforms them to (1,0,0) and (-1,1,0) which are not perpendicular with respect to the standard scalar product.

You say that e.g. (1,0,0) and (1,1,0) are not perpendicular. That may be true before we apply the function f, but f transforms them to (1,0,0) and (0,1,0) which are perpendicular with respect to the standard scalar product.

Since the function f preserves all the relevant structures, it would make no sense to say that it maps "space" onto something that isn't "space". If R³ is space, then so is f(R³), but vectors that are perpendicular in R³ aren't in f(R³) and vice versa. (I am of course aware that f(R³) isn't just isomorphic to R³ here, it is R³, but it's the fact that f is an isomorphism that's important).

Just to avoid one potential misunderstanding: It's not what I've said in this post that proves that dimensions don't have to be perpendicular. The stuff I've said in other posts before this one is more than sufficient. This post is "just" an explanation of why the concept of "angles" isn't well defined from the start. My previous posts explain why it wouldn't be relevant even if it had been well defined from the start.

 

[Fredrik]

As an aside, {V1,V2,V3} is a vector representation of Cartesian coordinates in a Euclidean space, and {E1,E2,E3} is a vector representation of spherical coordinates. ("a" meaning "one possible", to avoid further confusion)

You may want to replace the word "spherical" with "some other".

 

[Oppressed]

I thought that you and Fredrik had decided to limit yourselves to Eucledian space (a subset of general vector spaces). Fredrik's arguments (and his problems with my arguments) are quite correct if you are talking about the generalized concepts of vector spaces.

Are you talking about a three dimensional Eucledian space, or are you talking about any three dimensional space?

As an aside, {V1,V2,V3} is a vector representation of Cartesian coordinates in a Euclidean space, and {E1,E2,E3} is a vector representation of spherical coordinates. ("a" meaning "one possible", to avoid further confusion)

GodMark2,

I too believe that I have tried to limit the type of dimension I am talking about to that of Euclidian three dimensional space or a subset of that space. This space can be discussed in a subset of general vector spaces.

In this we are in agreement.

Not getting into the discussion yet of angle, but restricting ourselves for the moment to that three dimensional space as being three dimensions of length, you do agree that these lengths must be perpendicular, don’t you?

If you do, can you explain this to Fredrik?

 

[GodMark2]

You may want to replace the word "spherical" with "some other".

Yes, that would be more accurate. I picked a single example that we'd already been discussing, but any of a number of other coordinate systems could be represented. And upon further inspection, the second "is" really should be "could be". That's what I get for posting at work with the boss roaming the lab floor.

 

[Fredrik]

First, regarding the ultra-liberal use of the word "dimension". The loose definition of a measurable attribute that is (more or less) independent from other attribute measurements isn't useful. Taken to its extreme, in precisely one gram of water we would have the following dimensions:
- the number of hydrogen atoms
- the number of oxygen atoms
- the number of protons
- the number of electrons
- the number of neutrons
- the amount of radioactive emissions
- the mass (one gram)
- the volume
- etc. etc. etc.

You can see that by calling non-spatial measurements "dimensions", the definition is so watered down that the number of "dimensions", under the loose definition, is infinite.

These are all dimensions according to definition #3 on my list (post #161), but #3 isn't used much in physics.

For a discussion of dimensions to be meaningful, I think the definition must be restricted to spatial dimensions.

We're trying to do that, but Oppressed is refusing to see that according to that same definition (#3), which is a definition that he posted earlier in this thread, space also contains an infinite number of "dimensions". But #3 is of course not the definition that should be used when talking about space. Either #1 or #2 is required.

Second, my understanding from Astronomy 101 is that the universe is expanding,...
...
My understanding is that this is only possible with the existence of a fourth spatial dimension.

This is incorrect. No fourth spatial dimension is needed to describe an expanding universe. What you need is differential geometry.

 

[Oppressed]

GodMark2,

I too believe that I have tried to limit the type of dimension I am talking about to that of Euclidian three dimensional space or a subset of that space. This space can be discussed in a subset of general vector spaces.

In this we are in agreement.

Not getting into the discussion yet of angle, but restricting ourselves for the moment to that three dimensional space as being three dimensions of length, you do agree that these lengths must be perpendicular, don’t you?

If you do, can you explain this to Fredrik?

GodMark2,

I believe that we are in agreement that the Euclidian model for space as I have been talking about can be defined as having three (type a) dimensions of length which are mutually perpendicular to each other. We can express this space using vectors and when we do, the vector space is orthogonal.

Fredrik is saying this is false.

Why don’t you explain why you agree that it is true?

 

[Jimbo07]

With all due respect to Tom Cochrane:

"Round and round and round we go
Where it stops don't no one know"

 

[Oppressed]

Jimbo07,

By narrowing things down hopefully we can limit the subject to one thing we disagree on at a time and work that out. Id does seem hard to narrow things down to a single thing, but I think we have such a single thing now.

Give we are talking about the real physical space we exist in and are using the most common model for this space which is a flat uniform Euclidean space with 3 (type a) dimensions of length.

I think we can all agree on what space we are talking about and that this space can be defined as having 3 (type a) dimensions of length.

Do you agree with this?

Not yet going into whether or not this space can be defined in other than 3 (type a) dimensions of length, we have a single point of disagreement.

I state the 3 (type a) dimensions of length must be mutually perpendicular for this space.

Fredrik states these 3 (type a) dimensions of length do not have to be mutually perpendicular to each other.

GodMark2 states 3 (type a) dimensions of length must be mutually perpendicular for this space, but GodMark2 does not seem to be arguing this with Fredrik.

What do you think, do these (type a) dimensions of length need to be mutually perpendicular for this space or not?