Johnny5

You are confusing the Quarternionic group with the Quarternionic algebra (strictly speaking there are an infinitenumber of quarternionic algebras' date=' this being the one we commonly talk about), this isn't that surprising, since they are both commonly called just quarternions. They are different. Only the second is a division algebra. The first is not an algebra. Why do I suspect you do not know what an algebra is?[/quote']

 

Well i don't want to confuse one with the other, but that cannot happen since i dont know what either of them are.

 

Quaternionic group

Quaternionic algebra

 

What's the difference between them?

 

What kind of algebra are you thinking of? There are so many "things" called algebras, there is no one definition that would fit them all.

 

Regards

This is my first post using the new server.

[math]

\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} +\frac{\partial^2}{\partial z^2} - \frac{1}{c^2} \frac{\partial^2}{\partial t^2}

[/math]

Great you have latex back.

Also, i think the new colors are great, they give the forum an appealing look, easy to read.

Ok so as for quaternions.

It doesn't make sense that Hamilton would invent quaternions before inventing vectors. In fact, it doesnt make sense to say that he invented vectors at all, since he didn't, Newton was using them, perhaps Newton didn't call them vectors, but certainly they were being used prior to Hamilton's birth.

So it does indeed make sense that Hamilton developed quaternions in order to divide one vector by another. In fact, it makes perfect sense. Anything else, doesn't make sense.

Ok, I managed to find an 1899 reprint of Hamilton's lectures on quaternions.

On page 60 he says:

"But, there was still another view of the whole subject, sketched long afterwards in another communication to the R.I. Academy, on which it is unecessary to say more than a few words in this place, because it is, in substance, the view adopted in the following Lectures, and developed with some fullness in them: namely, that view according to which a QUATERNION is considered as the QUOTIENT of two directed lines in tridimensional space."

So the ratio of two vectors is a quaternion.

In the case where the magnitudes of the two vectors are equal, the ratio is called a versor, which is another name for a unit vector.

Then on the very next page, he begins do describe what he calls, "The method of Argand".

A group with respect to what operation Johnny. Trivially they are a group under addition since they are a vector space. They are not a group under multiplication.

I know the answer to your question is "multiplication." I recently read that but I don't remember at which site, and ive looked at maybe 15-20 so far trying to find someone who makes sense.

Let me find the site, since you say addition.

At wolfram you will read that they are a group under multiplication.

Wolfram on quaternions

The quaternions ± 1, ±i, ±j, ±k, and form a non-Abelian group of order eight (with multiplication as the group operation).

So Matt, why did you say that they arent a group under multiplication?

Wolfram seems to imply that multiplication is the group operation.

you are being very unclear as to what vectors you mean to divide.

It's not deliberate i assure you.

Do you mean the vectors in H' date=' the quarternions, or not? I suspect you mean, from you original post, that you want to divide two vectors in R^3 in some sense, though you haven't been clear on this.

 

It is you that needs to start explaining yourself clearly.[/quote']

 

 

I mean whats in the manual, there's a link in the first post to it.

 

They speak about

 

Versors

 

And I quote:

 

A versor is a quotient of two non-parallel vectors of equal length.

 

But you are right i mean two arbitrary vectors in R³.

 

 

Suppose that you take the quotient of two non-parallel vectors successfully. Because they were not pointing in the same direction, there is a remainder, which will have some direction, hence the remainder is a vector. Furthermore, since the vectors you took the quotient of had equal magnitudes, the quotient, which is a vector, should be a unit vector?

This thread is supposed to be about the group axioms.

I just read that quaternions are a group, they are denoted by H, after Hamilton.

Also used is the symbol Q₈, but I think thats for Octonians, though Wolfram seems to imply otherwise.

Regardless, quaternions are a group, and specifically as matt says a division algebra.

It is trivial Johnny since in a division algebra one can divide by definition. The quartnions possess a multiplicative norm.

Matt let me ask you straight out, do you understand the things in that manual?

Because here is what I'm getting from you:

1. 'it' is a division algebra.

2. Figure out what your division is going to be of.

I don't quite get your reasoning. Are you saying that time has a beginning simply because you hold it to be axiomatically true, and that's that? You mentioned some contradictions, like heat approaching infinity, but I fail to the reasoning behind this.

Here is my reasoning...

I can develop symbolic logic, design it to terminate at the statement:

There is at least one moment in time A, such that for any moment in time B, if not(A=B) then A before B.

The statement above is almost pure first order logic, and sentential calculus.

But exactly when, in the course of me actually contriving this argument, do I in reality "realize the conclusion, which im busy contriving" is true???

And if it truly is a deductive argument, then the argument form has to be a valid argument form, such as modus ponens. Which means the argument runs somewhat like this:

Suppose I know that the following 5 statements are true

1. if A then B

2. if B then C

3. if C then D

4. if D then E

5. If E then (there is at least one moment in time X, such that for any moment in time Y, if not(X=Y) then X before Y)

So all 5 of these statements I could know, but without also knowing a sixth

6. A

I cannot draw the conclusion that time had a beginning.

So this begs the question what statement 'A' do I already know, which I could use to reach the conclusion that time had a beginning deductively?

These are logistical problems in logic, and so the rationally best thing to do is to design an axiomatic system of time, with the following axiom:

1. There is at least one moment in time A, such that for any moment in time B, if not (A simultaneous B) then A before B.

As long as your other axioms are consistent with this axiom, the system you develop will be free from contradiction, and the very fact that time really had a beginning, will mean that your system will coincide with ultimate temporal reality.

Regards

It's a divisoin algebra' date=' Johnny.....

[/quote']

 

 

You make it sound trivial.

 

I've just read through the quaternion manual, and i saw them referring to

 

tensors, versors, vectors, and scalars.

 

The terminology seems worked out by one person.

 

Is the usage of the term 'tensor', in this manual, that of the tensor calculus?

 

Tensor operator applies tension to a vector.

Versor operator applies version to a vector.

 

It's in the manual.

So what? No disrespect, but I do not trust your judgement on matters mathematical. A binary relation requires sets A, B, AxB, and a subset of AxB. I see nothing about a set of all sets being mentioned (which is Cantor's Paradox - Russell is the set of sets that do not contain themselves)

And you should not trust me on matters mathematical, thats what binary logic is for.

so anyhow...

ok set of all sets Cantor's paradox

Also, Russell's paradox is the set of sets which are not elements of themselves. I knew he found that one.

But the two are related, i discovered that some time ago.

Moving on

Let me have a look at what you say here...

A binary relation requires sets A, B, AxB, and a subset of AxB.

Let me see if i can develop the problem...

Definition:

For any set A, and any set B: AXB ={(x,y)| x in A AND y in B}

The LHS is called the cartesian product of set A with set B.

e.g.

Permit A={1,2,3,4}

Permit B={2,3,8}

AxB={(1,2),(1,3),(1,8),(2,2),(2,3),(2,8),(3,2),(3,3),(3,8),(4,2),(4,3),(4,8)}

Count and you shall see there are 12 elements in set AXB.

Definition: For any set A, and any set B:

R is a binary relation from set A to set B if and only if R is a subset of set AXB.

Definition: For any set A:

R is a binary relation on set A if and only if R is a subset of AXA.

By one of the definitions above, the following is true:

For any set A: AXA ={(x,y)| x in A AND y in A}

The LHS is called the cartesian product of set A with itself.

Now, we need the definition of subset of a set.

Suppose that A denotes a set and B denotes a set.

Further suppose that the following statement is true:

If x is an element of A then x is an element of B, for any x.

Then, by definition, A is a subset of B.

If the converse is false, then A is a proper subset of B, and if the converse is true then A=B.

Definition: A Í B iff

1. A denotes a set.

2. B denotes a set.

3. If x is an element of A then x is an element of B, for any x.

I am going to write 3 above symbolically.

3. "x [xÎAÞxÎB].

A Í B is read "A is a subset of B"

Now, from here we can branch to the meaning of set theoretic equality, as well as the concept of a proper subset.

Suppose we are in a situation where we know that A denotes a set, and that B denotes a set, and that A is a subset of B.

There are now two remaining possibilities. Either Set A=Set B, or set A is a proper subset of Set B. And this just an issue of whether or not the converse follows. If the converse does follow, then A=B, and if the converse doesn't follow then A is a proper subset of B. In other words, if the converse is a true statement, then A=B, and if the converse is a false statement then A proper subset B.

Here is the statement that is the converse:

3. "x [xÎBÞxÎA].

So we have two more definitions:

Definition: A Ì B iff

1. A denotes a set.

2. B denotes a set.

3. "x [xÎAÞxÎB].

4. Ø"x [xÎBÞxÎA].

A Ì B is read "A is a proper subset of B"

Definition: Let A,B denote sets.

A=B if and only if

1. "x [xÎAÞxÎB].

2. "x [xÎBÞxÎA].

A=B is read "set A is equivalent to set B"

We can combine the two first order statements above into one, if we so choose:

Definition: Let A,B denote sets.

A=B if and only if "x [xÎAÛxÎB].

So now, I think theres enough above to show that defining a binary relation on a set A, as a subset of AXA leads to the concept of a set of all sets.

We have above:

Definition: For any set A:

R is a binary relation on set A if and only if R is a subset of AXA.

X is a binary relation hence:

Definition: For any set A:

X is a binary relation on set A if and only if X Í AXA.

We have above:

Definition: A Í B iff

1. A denotes a set.

2. B denotes a set.

3. "x [xÎAÞxÎB].

A Í B is read "A is a subset of B"

So, using the definitions above, it is necessarily true that:

For any set A:

X is a binary relation on set A if and only if

1. X denotes a set.

2. AXA denotes a set.

3. "x [xÎXÞxÎAXA].

Now, given that A is a set, if follows that AXA must denote a set, so we can shorten the above to:

For any set A:

X is a binary relation on set A if and only if

1. X denotes a set.

2. "x [xÎXÞxÎAXA].

Yesterday, I was concerned with whether or not the set A above, now denotes the set of sets.

Number 2 above seems funny. But again, its not my fault, i gave the "accepted" definitions.

Well I don't see a way to get to A={x|x is a set} from the above, and its bored me to tears so i'll leave it here.

Well i could look at one more thing. Write two above as:

2. "(x,y) [(x,y)ÎXÞ(x,y)ÎAXA].

Now, from a definition above, AXA is the set of ordered pairs (x,y) such that X in A, and Y in A.

That means that (x,y) in AXA if and only if (x in A AND y in A). So they are tautologically equivalent. So we can rewrite 2 above as:

2. "(x,y) [(x,y)ÎXÞ(xÎA AND yÎA)]

I think i should have used prenex normal form for this.

I'm going to leave things here, its going to give me a headache.

This is exactly why I couldn't stand axiomatic set theory.

"4 element divison algebra" would indicate that it had 4 elements. There are an infinite number of quarternions (as there must be in any non-zero real vector space), so I really advise you to stop inventing notation that makes no sense.

You should be telling that to the quaterniation folk, not me.

On a serious note, can you perform vector division with quaternions?

You have two 3D vectors in three dimensional Euclidean space, and you take their quotient, and you get a 4D quaternion in a ? space?

How? I'd put money on it not doing.

Well i saw the potential for the problem there yesterday.

Right' date=' let's clear up a few things straight away.

 

1. S is not a rotation, but it can be *interpreted* that way if you so choose I imagine. They are *just* a 4-dimensional real division algebra. Saying the first is an angle is like saying the real numbers are lengths. If it helps to visualize them in a certain way feel free to do so but don't confuse the visuallization with them as abstract objects. The unit quartrnions aer iso to the 4-sphere, and thus SU(2), and various things like the hopf fibration can be described using them.

 

2. The quarternions (denoted H for hamilton) *are* real 4-dimensional vector space, they are also a 2-dimensional complex vector space. I could have used the word algebra too instead of vector space.

 

3. The elements i,j,k *are not* a priori orthogonal, since there is no such thing as a canonincal inner product, though there is an obvious one with respect to which we can delcare them orthogonal.

 

 

4. Hamilton was attempting to define a 3-d space over the reals that was a field, liek C is a 2-d real vector space that is also a field. He realized that in fact it needed to be 4-d, and that it couldn't be commutative, hence only a division algebra. we also have the 8-d octonions.[/quote']

 

Ok thanks for the answers, its a lot to figure out all at once, but you are helping.

 

Thanks again.

 

4 dimensional real division algebra

 

4 element real division algebra sounds better.

 

Unit quaternions are isomorphic to the 4-sphere, hence to SU(2).

 

SU(2) group is the set of unitary 2x2 hermitian matrices

The terminology just keeps getting more bizarre by the minute

 

Non-commutative instantons on the 4-sphere from quantum groups

 

So what the hell is an instanton?

 

 

(I am thinking do these people know what they are talking about)

 

Are you going to post "that matrix"? I looked at the page (light font dark background) and didn't see anything like that.

Sure' date=' its on the link but here it is:

F =

(1+a²+b²+c²)^-1 *

[(1+a²-b²-c²) (2ab-2c) (2ac+2b)]

[(2ab+2c) (1-a²+b²-c²) (2bc-2a)]

[(2ac-2b) (2bc+2a) (1-a²-b²+c²)]

One of my algebra texts defines a binary relation like this, as a subset of the Cartesian product. It seems perfectly logical to me now but when I first read it I was like "Woooooaaaaaaah!"

A minor technical point.

I think defining a binary relation as a subset of the Cartesian product, leads to the Russell paradox of set theory, or possibly to the set of all sets, which has the Russell problem.

I haven't worked it out yet, I'm just gonna save it for later.

Regards

SO, if you "divide" two vectors, presumably in R^3, you get something that is in a 4-dimensional real space, or a 2-d complex space, or the quarternions. Hmm. And you think that's a reasonable notion of division of two vectors, which Tom, as most poeple would, took to mean as some operation on R^3 that 'undoes' some multiplicative operation. Of course, since we can embed R^3 into the quarternions in any number of ways we can take inverses there - again the operation will not be closed with respect to the embedding.

Well I don't know that much about quaternions yet, but if i find a good article then i will.

I do know that a quaternion is composed of four elements:

[s,i,j,k]

I know that S represents an amount of rotation, an angle.

And that i,j,k represent unit vectors, an orthogonal basis.

Lets see what else do i know...Oh, I know the thing Hamilton supposedly scribbled down, on that bridge in Ireland...

i^2=j^2=k^2=ijk=-1

so clearly, i^2 is an imaginary unit vector, as with j, and k.

So the 'space' as it were is complex, rather than strictly real.

In fact, the three dimensions are actually in Imaginary 3D space, yet to visualize that one is going to think 3 dimensionally anyways, so I'm not sure about all this yet.

As to your question Matt, I don't know enough about them yet, to know what set they are closed on.

The fourth element of a quaternion is an angle, which doesnt constitute an extra spatial dimension, it constitutes a rotation.

I'm only just beginning to even wonder what Hamilton was trying to do, to say much about quaternions. Or Gibbs, see I'm not even sure who did what, why, and when.

While trying to learn about rotation matrices I came across this:

Quaternion Tutorial

If you go to page 15, you will see this:

Definition of a Quaternion

A quaternion is the geometrical quotient of two vectors.

Let A denote a vector, and let B denote a vector.

Q = quaternion = A/B

Tom Mattson said that vector division isn't defined. Apparently it is.

Right here:

 

 

 

You forgot to put -A into the sine function.

 

 

 

So the second factor on the RHS should be:

 

( cos A - isinA)

Ok thanks, i'll go back and fix it, also I'd like you to have a look at something, I'm going to put in the general mathematics category. It's about vector division... again.

Take it to its (il)logical conclusion: if < denotes less than on the real numbers is duck<lion true? No, but neither is its negation: it is meaningless, since < isn't a relation on the animal kingdom.

Exactly. And thinking in this fashion, is highly advanced to say the least.

Kind regards Matt

PS: Not too many people realize this nuance here, or if they do they certainly don't appreciate its importance.

Yes, but Hamilton is usually credited with creating the Quarternions. I've no idea who Gibbs is.

I presume the same Gibbs that Gibbs free energy is named after.

Diana posted that stuff as Gibbs work, although elsewhere I saw Hamilton credited with it, but I don't buy that "etched it in a bridge" story. She was trying to understand quaternions, which is what led her to that Matrix. And that matrix has something to do with rotations. And apparently, something about quaternions avoids gimbal lock, which afflicts Euler angle approach to rotations.

Regards

its a standard forumla in any physics text

What does A represent?

Electromotive force = E = NBAwsin(wt)

i know what sinwt does, so ignoring that, N is number of turns?

B is magnetic field i know, and A?

EMF has units of volts, B has units of teslas

in SI F=qvB

so B has units of Newton second/coulomb meter

W=work=qV

volt has units of joule/coulomb

So the LHS has units of joule/coulomb

RHS has units of

N times (newton second)/(coulomb meter) times A times (1/seconds)

So we have to have:

N times (newton second)/(meter) times A times (1/seconds) come out in Joules.

A newton meter is a joule, hence

N times (second)/(meter^2) times A times (1/seconds) must be dimensionless

so

N times 1/(meter^2) times A must be dimensionless

If N is number of turns then it is dimensionless, which means that A stands for area.

Yes?

You called that a generator. Do you mean an electric motor, like in an electric fan? With a stator and rotor?

I'm not sure the best category to post this in, so I'm putting it here.

In the the link I am about to post, there is a 3D game programmer Diana Gruber discussing quaternions. During the thread, you can see her progressing towards understanding them. In trying to figure them out, she was led all the way to their inventor Gibbs. I don't know where she finally found Gibbs' work, but she did, and she pasted his equations.

Gibbs equations for quaternions

My question is about the matrix which Gibbs developed.

Is anyone here comfortable enough with quaternions to discuss them?

I want to understand that matrix.

PS:

A few posts later, Ms. Gruber writes:

> Then what the $%^?@! are we arguing about?

 

Imaginary numbers. Don't need 'em. Gibbs got all the way there without ever having to take the square root of -1.

 

Diana

If x and y are not in T then xBy makes absolutely no sense.

Part of the definition, if you make it so. But I understand what you said perfectly, if x,y are not in T, then xBy makes absolutely no sense.

Exactly.

Treating 'before' in the customary way, to say house before tree is totally meaningless, its not even false, it is meaningless.

Just a logical point, which you appear to understand too.

Regards Matt

Ahh, i read a little further, and I quote you:

Moreover if x and y are not both in T then xBy is neither true nor false - it is out of the domain of B and is meaningless.

I will brush up on the definition of relation.

Let A denote a set.

Then we can talk about the set A X A, where X is the cartesian product.

Now, we can define a binary relation on A, using the cartesian product somehow, i just forget the formalism.

Definition- Binary relation

Definition (binary relation):

A binary relation from a set A to a set B is a set of ordered pairs <a, b> where a is an element of A and b is an element of B.

When an ordered pair <a, b> is in a relation R, we write a R b, or <a, b> R. It means that element a is related to element b in relation R.

When A = B, we call a relation from A to B a (binary) relation on A .

Ok, so a binary relation on A, is a subset of A X A, where X denotes the binary relation "cartesian product" defined on the set of sets.

Odd, but there you go.

Tom said binary *operation* not binary *relation*, so what was the point of that I mean in what sense can a binary relation even be thought of as producing another element of the set?

I was thinking of something else, i said strike it. I had something else in mind.

Here is what I was thinking of anyways:

Suppose you define the binary relation -B- on the set of moments in time T.

suppose that x,y are elements of T.

Then if the statement

x-B-y is true, then since B is a relation on T, it must follow that x,y are elements of T. So that by assigning the binary relation x-B-y the value true, it must be the case that x,y are elements of the set the binary relation was defined on. I just got slightly confused, but this is what i was thinking of.

To say what i was thinking of one more time, if you know that R is a relation on set A, and in your work you are busy treating f-R-g as true, it is inferrable just by looking at the expression f-R-g, that f,g are elements of the set A, which R was defined on.

Like i said, my mistake. I was thinking of something else.

Anyway' date=' it depends on how they define binary operation. Some leave the codomain implicit some explicit. Often they are are defined to satisfy closure since * will be referred to as a binary operation from GxG to G.

 

Though quite why some 'fault' of binary relations means we ought to state something about a completely unrelated thing is a mystery. (nb no one should ever define functions as relations)[/quote']

I said I was going to run through Tom's argument, and that's what I intend to do here. Let me see if I can reproduce it, without looking.

Let Z denote an element of the complex numbers.

Hence, for some real numbers x,y we have:

Z = x+iy

Now, this is the rectangular form of the complex number Z.

But, we can also write Z in "polar form."

The complex number Z, can be represented as a vector in the Argand plane.

Let the magnitude of Z be represented by R. And let the angle from the real axis to the 'vector' be denoted by q.

Feel free to correct my terminology.

Ok so...

For some reason I just remember the answer is:

Z = R e^iq

Let me see if thats true.

e^iq = cosq+i sinq

Hence:

Re^iq = Rcosq+Ri sinq

R cos theta is the component of the vector in the X direction, and Rsin theta is the component of the vector in the Y direction hence:

Re^iq = x+iy

So it is indeed true.

The magnitude of the complex number is ZZ* i think.

In other words, it should be the case that ZZ*=R²

Z* is the complex conjugate of Z.

Check:

Let Z = R e^iq

hence

Z* = R e^-iq

The previous statement is true by definition.

Hence:

ZZ* = (R e^iq )(R e^-iq)

Adding exponents, we have:

ZZ* = R²e^iq-iq

The exponent is zero, hence:

ZZ* = R²

QED

Ok so now we have to figure out the formula for a rotation.

Ok, I don't feel like playing around, I just want to run through the argument, so here it is:

Neither classical trigonometry nor quaternions is the way to go. The quickest' date=' most direct route to the rotation matrices is to use complex exponentials. This is the case for 2 reasons:

 

1. Complex numbers in C1 can be represented as vectors that obey the same algebra as real vectors in R2.

2. Exponentials are easier to work with than trigonometric functions by several orders of magnitude.

 

That said let's develop our formula.

 

Let z=x+iy be a complex number with the usual vector representation. Its polar form is z=r exp(iB), where B is the angle the vector makes with the polar axis. Now rotate the vector clockwise (remember this is the same as rotating the axes counterclockwise) by an angle A. The new complex number is z'=r exp[i(B-A)']=r exp(iB)exp(-iA)=x'+iy'.

 

Now the question is: What are x' and y'? Noting that r exp(B)=x+iy, we have:

 

x'+iy'=(x+iy)[cos(A)-i sin(A)]

x'+iy'=[x cos(A)+y sin(A)]+i[-x sin(A)+y cos(A)]

 

or...

 

x'=x cos(A)+y sin(A)

y'=-x sin(A)+y cos(A)

Tom, you said that B is the angle that the vector makes with the polar axis. Is that what the real axis is called?

You say rotate the vector clockwise... by angle A.

So we have a new complex number Z`, and the angle is A+B.

Wait i did that wrong, you said to rotate the vector clockwise.

Ok, so then the new angle is B-A, like you have.

Therefore:

Z`=R e^{i (B-A)}

Which we can write as:

Z`=R e^{i B} e^{i (-A)}

Let the rectangular form of Z` be x`+iy`

Therefore:

Z`= R e<sup>i B</sup> e<sup>i (-A)</sup>= x`+iy`

Now, we just want to write x`,y` in terms of x,y, and we are done.

Recall that:

Z= R e^{i B}

Thus, we have:

Z`= Z e<sup>i (-A)</sup>= x`+iy`

Thus, we have:

(x+iy) e^{i (-A)}= x`+iy`

Thus, we have:

(x+iy) ( cos(-A) + isinA) = x`+iy`

Hence we have:

(x+iy) ( cos A + isinA) = x`+iy`

From which it follows that:

xcosA+xisinA+iycosA+iyisinA=x`+iy`

Hence;

xcosA+xisinA+iycosA-ysinA=x`+iy`

Hence:

(xcosA-ysinA) + i(xsinA+ycosA) = x`+iy`

Hence it follows that:

xcosA - ysinA = x`

and that

xsinA+ycosA = y`

QED

Now, here is what caused this slight digression:

Start by looking at a vector v=<x' date='y> in the xy-plane. Then rotate the axes by an angle A and see what the new components are. You should get:

 

[b']v[/b]'=<xcos(A)+ysin(A),-xsin(A)+ycos(A)>

 

You should be able to write the RHS of the above equation as a matrix times the original vector v. That matrix is:

 

[cos(A) sin(A)]

[-sin(A) cos(A)]

So now for the matrix...

We started off with

V=<x,y> in the complex plane.

Then were told to rotate that vector clockwise to obtain

V` = <x`,y`>

and discovered that:

V` = (xcosA - ysinA, xsinA+ycosA)

Tom there's a disagreement in our minus sign, how come?