Take The Big Plunge OK. So I've already introduced the general quaternion, a number type composed of four other numbers, as follows :
q = q0 + q1i + q2j + q3k
..... where q0, q1, q2 & q3 are real numbers, q0 is the real part of q and and q1, q2, q3 are the imaginary parts. Here, as with complex numbers, the + sign means grouping the parts together and not an arithmetic operation. But again, as with complex numbers, we will sometimes behave as if the + is arithmetic while we understand what we are really doing. As with complex numbers we could have written :
q = (q0, q1, q2, q3 )
provided one understood that quaternions we being referred to and not, for example, just an ordered quadruple from some other number system eg. vectors in R4. So the first point to note is that quaternions are an extension of the complex numbers, just with more imaginary parts. Hypercomplex is the word used to describe such extensions and will include the case of octonions later on. BTW You can go to say, 16 component hypercomplex numbers called sedenions. There are others too ....
The properties of i, j & k are best defined by Hamilton's Rules, which I write out in groups to hopefully clarify things :
(1) i2 = - 1, j2 = - 1, k2 = - 1
(2) (a) ij = k, jk = i, ki = j (b) ji = -k, kj = -i, ik = -j
(3) ijk = -1
The first are easiest to understand. It says that we have three different imaginary axes each with the same property as for the imaginary axis of the complex numbers. These are distinct axes ie. we are talking about a 4D space. So if I were to start at (1, 0, 0, 0) ie. one on the real axis, then two 900 rotations through the any of the imaginary axes gets me to minus one.
The second rules are cyclicand have two versions depending on which way you traverse this diagram :
.... if you go with the arrows the product is positive, but against the arrows the product is negative. So products of the imaginary units don't commute and neither does multiplication for entire quaternions ie. generally p * q <> q * p.
The last rule and others like it can be deduced eg :
ijk = (ij) * k = k * k = -1 OR ijk = i * (jk ) = i * i = -1
.. so associativity holds for these units and, perchance, the same holds for entire quaternions ;
pqr = p * (qr) = (pq) * r
But I haven't defined what the operations on whole quaternions look like. Addition and subtraction are simple and are like complex numbers :
(a, b, c, d) + (e, f, g, h) = (a + e, b + f, c + g, d + h)
(a, b, c, d) - (e, f, g, h) = (a - e, b - f, c - g, d - h)
Multiplication is a real doozy, that is if you express it component-wise and multiply/add terms as if in ordinary algebra :
(a, b, c, d) * (e, f, g, h) = (a + bi + cj + dk) * (e + fi + gj + hk)
= a * (e + fi + gj + hk) + bi * (e + fi + gj + hk) + cj * (e + fi + gj + hk) + dk * (e + fi + gj + hk)
..... and so on with very many boring steps .... 16 terms, gather them up and simplify ... yada yada ....
= a pretty horrible expression I won't bore you with ( but you could do it at home if you like )
I say again : it doesn't commute. What we will do now is an important detour to two important 3D vector operations called the dot product and cross product. Then we will come back to quaternion multiplication as we can express quaternion products in terms of those vector operations, and that's up next .... WHAT FUN WE'RE HAVING !! ;-)
Cheers, Mike.
I have made this letter longer than usual because I lack the time to make it shorter ...
... and my other CPU is a Ryzen 5950X :-) Blaise Pascal
3D Vector Products in R3: The Dot Product Also called the scalar product in that the result is a scalar, or pure number without direction. It is defined as follows for any a = (a0, a1, a2) & b = (b0, b1, b2) :
.... that is the sum of the component-wise products. This has the advantage that the scalar product of a vector with itself :
a.a = a02 + a12 + a22 = ||a||2
... is the square of the norm ( or length or magnitude or absolute value or size ) annotated with double vertical lines either side || .... ||. In other words, how long is the vector ?
... this being Pythagorus' Theorem in disguise, which gives a non-negative answer for the norm, and a zero norm if and only if a0 = a1 = a2 =0 ie. a = 0.
But if the vectors are different then some information may still be gained about their relationship as per the following ( that I won't prove ) :
a.b = ||a||*||b||*cos(theta)
.... theta being the angle included b/w a and b. If theta is zero then a and b are parallel ie. one is multiple/fraction of the other and they both point the same way. If theta is 1800 ( or PI ) then a and b are anti-parallel ie. one is a multiple/fraction of the other but pointing the opposite way. If theta = 900 ( or PI/2 ) then cos(theta) = 0 and then a and b are orthogonal ( at right angles to each other ) and a.b = 0 in that case. The dot product gives an idea of the how well aligned vectors are. For intermediate values of theta then you can tell roughly whether they are pointing toward (a.b > 0) or away from (a.b < 0) the same half of the total space.
The scalar product commutes ie.
a.b = b.a
The scalar product can apply to spaces of any dimension and not just three. However ...
3D Vector Products in R3: The Cross Product Only makes sense in 3D I'm afraid. Falls over in other dimensional spaces due to the manner of it's construction. It produces a vector at right angles to the first two. Written as
and likewise with b.(a x b) = 0, confirming that the cross product is indeed orthogonal to the composing vectors. The vectors a and b will span/define a plane and thus the cross product of them will be at right angles to this plane, which gives two choices ie. 'above' or 'below' that plane. The sense is right-handed so using your right hand in a semi closed fist, rotate your fingers from a towards b then the thumb sticking out is the direction of a x b.
You need three directions to satisfy these constructs, and no more than three as there would be ambiguity about the direction(s) of orthogonality in four dimensions and more. The cross product is the zero vector if a & b are parallel, and a maximum length if a and b are orthogonal. In fact :
||a x b|| = ||a||*||b||* sin (theta)
where theta is the angle b/w a and b. This lends itself to many applications from representing areas to angular momentum etc.
The cross product does not commute, in fact
a x b = - b x a
Cheers, Mike.
I have made this letter longer than usual because I lack the time to make it shorter ...
... and my other CPU is a Ryzen 5950X :-) Blaise Pascal
The dot product measures how much two vectors point in the same direction, but the cross product measures how much two vectors point in different directions
The main difference between the dot product and the cross product of two vectors is that the result of the dot product is a scalar quantity, whereas the result of the cross product is a vector quantity.
A dot product of two vectors is also called the scalar product.
It is the product of the magnitude of the two vectors and the cosine of the angle that they form with each other.
A cross product of two vectors is also called the vector product.
It is the product of the magnitude of the two vectors and the sine of the angle that they form with each other.
I was still awake at 5am and started reading the Mike posts but the Sun was coming up and I figured I better try that sleep thing.......only problem was I got up after 2pm
(1) What is the likely form of the quaternion conjugate ( by analogy with the complex conjugate ) ? What would we use it for ?
(2) What is the likely form of the unity quaternion ie. that quaternion which may multiply any other and give the same quaternion back as a result of the product ? Is the unity quaternion unique ?
Cheers, Mike.
I have made this letter longer than usual because I lack the time to make it shorter ...
... and my other CPU is a Ryzen 5950X :-) Blaise Pascal
Well back when I was -50 years old and in the Quaternion Society I always used the four-dimensional algebra of hyperbolic quaternion multiplication for the study of quaternions and devoted to the study of quaternions and other hypercomplex number systems.
Mainly to explain why I was awake at 1:30am reading this and watching Lost in Space at the same time.
(actually because I have to wait until 2am to start up my beta CMS testing multicores)
I'm reading too, but I don't know what I'm reading... ;^) ...Maybe it's because the last time I had any physics which I liked was over 50 years ago, and I just can't remember that far back. Plus, I'm not sure - maybe I didn't realize it - but I didn't take physics to the level where you are using these equations.
.... what's this then ? If I call qR-1 = q+/ ||q||2 then I have q * qR-1 = 1 ie. the inverse of q is the conjugate divided by the norm squared ! Strictly speaking that's only the right inverse, however I could repeat all the above :
q+ * q = (a - bi - cj - dk) * (a + bi + cj + dk)
= .... etc
= ||q||2
(q+/ ||q||2) * q = 1
with qL-1 = q+/ ||q||2 then I have qL-1 * q = 1
so here the left inverse also happens to equal the right inverse. I had to check that because quaternion products are not generally commutative.
(2) No suspense here, the unity quaternion is again the real number one or (1, 0, 0, 0) or 1 + 0i + 0j + 0k and there is only one of them.
Cheers, Mike.
I have made this letter longer than usual because I lack the time to make it shorter ...
... and my other CPU is a Ryzen 5950X :-) Blaise Pascal
When young I did a degree in physics and sums, then went on to do medicine & surgery to be a 'bush doctor' for twenty five years ( alas ... the stories I can never tell of ). In middle age I did a degree in computing out of ennui. I am compelled to learn. So in ..... errr ..... later middle age I still practise medicine part time, teach the young-uns coming up, play Factorio for laughs, listen to Supertramp, Dire Straits & America ( best bands ever ) and doodle with maths in this online forum. ;^]
Next up : I complete the definition of multiplication for quaternions and reveal a surprising use !
Cheers, Mike.
I have made this letter longer than usual because I lack the time to make it shorter ...
... and my other CPU is a Ryzen 5950X :-) Blaise Pascal
Take The Big Plunge OK. So
)
Take The Big Plunge OK. So I've already introduced the general quaternion, a number type composed of four other numbers, as follows :
q = q0 + q1i + q2j + q3k
..... where q0, q1, q2 & q3 are real numbers, q0 is the real part of q and and q1, q2, q3 are the imaginary parts. Here, as with complex numbers, the + sign means grouping the parts together and not an arithmetic operation. But again, as with complex numbers, we will sometimes behave as if the + is arithmetic while we understand what we are really doing. As with complex numbers we could have written :
q = (q0, q1, q2, q3 )
provided one understood that quaternions we being referred to and not, for example, just an ordered quadruple from some other number system eg. vectors in R4. So the first point to note is that quaternions are an extension of the complex numbers, just with more imaginary parts. Hypercomplex is the word used to describe such extensions and will include the case of octonions later on. BTW You can go to say, 16 component hypercomplex numbers called sedenions. There are others too ....
The properties of i, j & k are best defined by Hamilton's Rules, which I write out in groups to hopefully clarify things :
(1) i2 = - 1, j2 = - 1, k2 = - 1
(2) (a) ij = k, jk = i, ki = j (b) ji = -k, kj = -i, ik = -j
(3) ijk = -1
The first are easiest to understand. It says that we have three different imaginary axes each with the same property as for the imaginary axis of the complex numbers. These are distinct axes ie. we are talking about a 4D space. So if I were to start at (1, 0, 0, 0) ie. one on the real axis, then two 900 rotations through the any of the imaginary axes gets me to minus one.
The second rules are cyclic and have two versions depending on which way you traverse this diagram :
.... if you go with the arrows the product is positive, but against the arrows the product is negative. So products of the imaginary units don't commute and neither does multiplication for entire quaternions ie. generally p * q <> q * p.
The last rule and others like it can be deduced eg :
ijk = (ij) * k = k * k = -1 OR ijk = i * (jk ) = i * i = -1
.. so associativity holds for these units and, perchance, the same holds for entire quaternions ;
pqr = p * (qr) = (pq) * r
But I haven't defined what the operations on whole quaternions look like. Addition and subtraction are simple and are like complex numbers :
(a, b, c, d) + (e, f, g, h) = (a + e, b + f, c + g, d + h)
(a, b, c, d) - (e, f, g, h) = (a - e, b - f, c - g, d - h)
Multiplication is a real doozy, that is if you express it component-wise and multiply/add terms as if in ordinary algebra :
(a, b, c, d) * (e, f, g, h) = (a + bi + cj + dk) * (e + fi + gj + hk)
= a * (e + fi + gj + hk) + bi * (e + fi + gj + hk) + cj * (e + fi + gj + hk) + dk * (e + fi + gj + hk)
..... and so on with very many boring steps .... 16 terms, gather them up and simplify ... yada yada ....
= a pretty horrible expression I won't bore you with ( but you could do it at home if you like )
I say again : it doesn't commute. What we will do now is an important detour to two important 3D vector operations called the dot product and cross product. Then we will come back to quaternion multiplication as we can express quaternion products in terms of those vector operations, and that's up next .... WHAT FUN WE'RE HAVING !! ;-)
Cheers, Mike.
I have made this letter longer than usual because I lack the time to make it shorter ...
... and my other CPU is a Ryzen 5950X :-) Blaise Pascal
3D Vector Products in R3: The
)
3D Vector Products in R3: The Dot Product Also called the scalar product in that the result is a scalar, or pure number without direction. It is defined as follows for any a = (a0, a1, a2) & b = (b0, b1, b2) :
a.b = a0 * b0 + a1 * b1 + a2 * b2 = a0b0 + a1b1 + a2b2
.... that is the sum of the component-wise products. This has the advantage that the scalar product of a vector with itself :
a.a = a02 + a12 + a22 = ||a||2
... is the square of the norm ( or length or magnitude or absolute value or size ) annotated with double vertical lines either side || .... ||. In other words, how long is the vector ?
norm = SQRT[||a||2] = SQRT[a.a] = SQRT[a02 + a12 + a22]
... this being Pythagorus' Theorem in disguise, which gives a non-negative answer for the norm, and a zero norm if and only if a0 = a1 = a2 =0 ie. a = 0.
But if the vectors are different then some information may still be gained about their relationship as per the following ( that I won't prove ) :
a.b = ||a||*||b||*cos(theta)
.... theta being the angle included b/w a and b. If theta is zero then a and b are parallel ie. one is multiple/fraction of the other and they both point the same way. If theta is 1800 ( or PI ) then a and b are anti-parallel ie. one is a multiple/fraction of the other but pointing the opposite way. If theta = 900 ( or PI/2 ) then cos(theta) = 0 and then a and b are orthogonal ( at right angles to each other ) and a.b = 0 in that case. The dot product gives an idea of the how well aligned vectors are. For intermediate values of theta then you can tell roughly whether they are pointing toward (a.b > 0) or away from (a.b < 0) the same half of the total space.
The scalar product commutes ie.
a.b = b.a
The scalar product can apply to spaces of any dimension and not just three. However ...
3D Vector Products in R3: The Cross Product Only makes sense in 3D I'm afraid. Falls over in other dimensional spaces due to the manner of it's construction. It produces a vector at right angles to the first two. Written as
a x b = (a1b2 - b1a2, a2b0 - b2a0, a0b1 - a1b0)
and hence for example :
a.(a x b) = a0(a1b2 - b1a2) + a1(a2b0 - b2a0) + a2(a0b1 - a1b0) = a0a1b2 - a0b1a2 + a1a2b0 - a1b2a0 + a2a0b1 - a2a1b0) = (a0a1b2 - a1b2a0) + (a1a2b0 - a2a1b0 ) + (a2a0b1 - a0b1a2 ) = 0 + 0 + 0 = 0 { Phew! }
and likewise with b.(a x b) = 0, confirming that the cross product is indeed orthogonal to the composing vectors. The vectors a and b will span/define a plane and thus the cross product of them will be at right angles to this plane, which gives two choices ie. 'above' or 'below' that plane. The sense is right-handed so using your right hand in a semi closed fist, rotate your fingers from a towards b then the thumb sticking out is the direction of a x b.
You need three directions to satisfy these constructs, and no more than three as there would be ambiguity about the direction(s) of orthogonality in four dimensions and more. The cross product is the zero vector if a & b are parallel, and a maximum length if a and b are orthogonal. In fact :
||a x b|| = ||a||*||b||* sin (theta)
where theta is the angle b/w a and b. This lends itself to many applications from representing areas to angular momentum etc.
The cross product does not commute, in fact
a x b = - b x a
Cheers, Mike.
I have made this letter longer than usual because I lack the time to make it shorter ...
... and my other CPU is a Ryzen 5950X :-) Blaise Pascal
The dot product measures how
)
The dot product measures how much two vectors point in the same direction, but the cross product measures how much two vectors point in different directions
The main difference between the dot product and the cross product of two vectors is that the result of the dot product is a scalar quantity, whereas the result of the cross product is a vector quantity.
A dot product of two vectors is also called the scalar product.
It is the product of the magnitude of the two vectors and the cosine of the angle that they form with each other.
A cross product of two vectors is also called the vector product.
It is the product of the magnitude of the two vectors and the sine of the angle that they form with each other.
I was still awake at 5am and started reading the Mike posts but the Sun was coming up and I figured I better try that sleep thing.......only problem was I got up after 2pm
At least I now someone is
)
At least I know someone is reading ! ;-)
Two quick questions then :
(1) What is the likely form of the quaternion conjugate ( by analogy with the complex conjugate ) ? What would we use it for ?
(2) What is the likely form of the unity quaternion ie. that quaternion which may multiply any other and give the same quaternion back as a result of the product ? Is the unity quaternion unique ?
Cheers, Mike.
I have made this letter longer than usual because I lack the time to make it shorter ...
... and my other CPU is a Ryzen 5950X :-) Blaise Pascal
Well back when I was
)
Well back when I was -50 years old and in the Quaternion Society I always used the four-dimensional algebra of hyperbolic quaternion multiplication for the study of quaternions and devoted to the study of quaternions and other hypercomplex number systems.
Mainly to explain why I was awake at 1:30am reading this and watching Lost in Space at the same time.
(actually because I have to wait until 2am to start up my beta CMS testing multicores)
......almost the speed of light ...goodnight!
Mike Hewson wrote: At least
)
I am reading, but left dot products behind with Maxwell's equations some years ago.
You can't expect more than that. But keep up the good work.
Mike Hewson wrote:At least
)
I'm reading too, but I don't know what I'm reading... ;^) ...Maybe it's because the last time I had any physics which I liked was over 50 years ago, and I just can't remember that far back. Plus, I'm not sure - maybe I didn't realize it - but I didn't take physics to the level where you are using these equations.
And yes, do keep up the good... no, GREAT work !!
Proud member of the Old Farts Association
Thanks for reading guys !
)
Thanks for reading guys ! ;-)
Now for the answers to the mini quiz.
(1) If q = a + bi + cj + dk then the conjugate, call it q+ = a - bi - cj - dk and you would do a product with it just like complex numbers :
q * q+= (a + bi + cj + dk) * (a - bi - cj - dk)
= a(a - bi - cj - dk) + bi(a - bi - cj - dk) + cj(a - bi - cj - dk) + dk(a - bi - cj - dk)
= a2 - abi - acj - adk + bai - b2i2 - bcij - bdik + caj - cbij - c2j2 - cdjk + dak - dbki - dckj - d2k2
= a2 - b2i2 - c2j2 - d2k2 + (- ab + ba)i + (- ac + ca)j + (- ad +da)k - bc(ij + ji) - bd(ik + ki) - cd(jk + kj)
= a2 + b2 + c2 + d2 + 0i + 0j + 0k - bc(0) - bd(0) - cd(0)
= a2 + b2 + c2 + d2
... that's the square of the norm !
So bonus content here :
q * q+= ||q||2
q * (q+/ ||q||2) = 1
.... what's this then ? If I call qR-1 = q+/ ||q||2 then I have q * qR-1 = 1 ie. the inverse of q is the conjugate divided by the norm squared ! Strictly speaking that's only the right inverse, however I could repeat all the above :
q+ * q = (a - bi - cj - dk) * (a + bi + cj + dk)
= .... etc
= ||q||2
(q+/ ||q||2) * q = 1
with qL-1 = q+/ ||q||2 then I have qL-1 * q = 1
so here the left inverse also happens to equal the right inverse. I had to check that because quaternion products are not generally commutative.
(2) No suspense here, the unity quaternion is again the real number one or (1, 0, 0, 0) or 1 + 0i + 0j + 0k and there is only one of them.
Cheers, Mike.
I have made this letter longer than usual because I lack the time to make it shorter ...
... and my other CPU is a Ryzen 5950X :-) Blaise Pascal
This brings back memories. I
)
This brings back memories. I did a degree in 'sums' fifty years ago and had completely forgotten this.
Richard
Fond memories I hope
)
Fond memories I hope !?
When young I did a degree in physics and sums, then went on to do medicine & surgery to be a 'bush doctor' for twenty five years ( alas ... the stories I can never tell of ). In middle age I did a degree in computing out of ennui. I am compelled to learn. So in ..... errr ..... later middle age I still practise medicine part time, teach the young-uns coming up, play Factorio for laughs, listen to Supertramp, Dire Straits & America ( best bands ever ) and doodle with maths in this online forum. ;^]
Next up : I complete the definition of multiplication for quaternions and reveal a surprising use !
Cheers, Mike.
I have made this letter longer than usual because I lack the time to make it shorter ...
... and my other CPU is a Ryzen 5950X :-) Blaise Pascal