Johnny5

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".

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 So Matt, why did you say that they arent a group under multiplication? Wolfram seems to imply that multiplication is the group operation.

It's not deliberate i assure you.

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.

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.

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

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... 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.

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?

Well i saw the potential for the problem there yesterday.

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)]

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

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.

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.

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.

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

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:

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: 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 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.

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.

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 eiq Let me see if thats true. eiq = cosq+i sinq Hence: Reiq = 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: Reiq = 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 eiq hence Z* = R e-iq The previous statement is true by definition. Hence: ZZ* = (R eiq )(R e-iq) Adding exponents, we have: ZZ* = R2eiq-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: 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 ei (B-A) Which we can write as: Z`=R ei B ei (-A) Let the rectangular form of Z` be x`+iy` Therefore: Z`= R ei B ei (-A)= x`+iy` Now, we just want to write x`,y` in terms of x,y, and we are done. Recall that: Z= R ei B Thus, we have: Z`= Z ei (-A)= x`+iy` Thus, we have: (x+iy) ei (-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: 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?