Hamilton’s Bubbles

I like bubbles, but they relate to Hamilton's quaternion reference frame so intuitively that I have to learn to lurve them!

I once called the quaternion Hamilton's umbrella, but my more recent insights has inevitably drawn me into the spheres of influence of the quaternions, and as a basic frame for straightening out math before we get into Grassmann, who will also need a bit of setting straight before I get to Clifford algebras.

So I think it was Ridley Scott who promoted the alien quadrilogy with the by line" in space nobody can hear you scream!" and this is the case too when we look at the complexity of the sign1 quaternion.

I had identified the notation – as a diminutive vector notation, do let me just clarify that . If I have 2 vectors v and w and v is opposite to v I can write sign1v = w, and sign1w = v. Thus I have replaced a – by a quaternion action on a vector. The quaternion rule for the unit quaternions is q1q1 = q2q2 = q3q3 =q1q2q3 = -1 , and it is essentially the definition of sign1

The quaternion rule for the unit quaternions is q1q1 = q2q2 = q3q3 =q1q2q3 = -1 , and it is essentially the definition of sign1

So the associative rule in quaternions actually tells us how sign1 has to be used.
Can I now establish the rule for the tripled? First I need to establish the rule in the couples. To do this I need De Moivre's theorem about the roots of unity, the sons and daughters of Shunya. If I pick the even roots of unity, 2,4 ,6 etc I know I can
Find 3 roots that will act to give me -1 = c1c2c3.

So now I can define a rule for couples Ii = c1c2c3 = -1

This rule now enables me to factories the couples into a closed set .

So now with this model I can tackle Hamilton's triples. I have to find a set of quaternions or triples so that

Ii = JJ = t1t2t3 = -1

Now to solve thes equations we require the De Moivre Cotes Euler formula

e = cos ø +i*sinø with e = -1

We can see the true nature of sign1, It is in fact e.

We have a version for the quaternions and the triples

For example, exp (i + j) = cos sqrt 2 + (i + j)(sin sqrt 2)/sqrt 2,
while exp i * exp j = (cos 1)^2 + (i + j)(sin 2)/2 + k(sin 1)^2, and
exp j * exp i = (cos 1)^2 + (i + j)(sin 2)/2 – k(sin 1)^2. However,
whenever ab = ba (for example, whenever a is real), then
exp (a + b) = exp a * exp b does hold in the quaternions

That fact is the secret to calculating exp (a + ib + jc + kd). Since
a commutes with ib + jc + kd (when a, b, c, and d are real),
exp (a + ib + jc + kd) = exp a * exp (ib + jc + kd). Now, you can't
break exp (ib + jc + kd) into exp ib * exp jc * exp kd, but you don't
need to. Simply divide this vector by its magnitude. The quantity
L = (ib + jc + kd)/sqrt (b^2 + c^2 + d^2) satisfies L^2 = -1 (check it
for yourself). And you can go back to the infinite series to see that
exp Lx = cos x + L sin x, whenever L^2 = -1 and x is real. So let
M = sqrt (b^2 + c^2 + d^2), so ib + jc + kd = LM. Then

exp (a + ib + jc + kd) = exp(a)*(cos M + L sin M)=exp(a)*(cos M + {ib+jc+kd}*sin M}/M}=exp(a)*(cos M + {i*bsin M + j*csin M + k*dsin M } /M)


Now Betchen sent me this where n is the number of roots, and j (unfortunately) the index of the nth root, and
which gives the solutions for the i and j planes only, while the Quaternion ones work for space. So if I wanted to find the second (j=2) sixth root(n=6) I would plug those values into the formula.

[

Thus I can find any combination of roots which multiply to give -1, but they will be polynomials of I or j. For triples we need a polynomial of i+j. These triple combinations are a subgroup of the quaternions and may have to be written as such, due to the calculation axis.

One thing must not escape attention: all these relationships are predicated on the old convention of – , thus we are in a tautology. I have to remember that it is not the function of – that is being expurgated here, it is the archaic and misleading notation. For example it is already clear that – cannot be brought through with the identity, as this is procedurally different and produces different results. Also, in general, commutativity is not common, and so moving terms around leads to different results.

The simplest solution is algebraic, by simply evaluating ij as a lhat a fixed triple in space. But as soon as that is done it is apparent that one has created a new axis. Now everything proceeds exactly as the quaternions, and you realise that this is what Hamilton was experiencing before he accepted that his triples required 4 axes. Thus the quaternions are a 4 axis system for 3d space.

The other powerful thing is that sign becomes another quaternion, but one that does commute with the quaternions which in general do not commute, or rather were not allowed to commute to avoid inconsistencies. sign1 however does commute, but as a property of a 2nd root of unity in a 4th root of unity substrate, or environment.

Thus sign1 is not unique, any second root of unity will commute with the quaternions. To eliminate sign1 then we simply rewrite the unit quaternions squared as some vector marker qpi˙ say, and the identity quaternion as having qi as its vector marker. The vector markers are just labels of unique quaternion positions in the unit sphere surface. The action of the quaternion units is the rotation power of quaternions, the vectors are just markers and should be associated with the quaternion units eg q1i˚ is a specific marker so when a quaternion combination occurs a new marker has to be designated. They will all be fixed in the unit sphere and unique and innumerable. Labeling them by the quaternion product is the only way to keep track, but it is also a recipe for confusing the marker with the quaternion unit.

Why not just allow this confusion? Because the quaternions over a field commute with the elements of that field of numbers and they form a perfect scalar field for the quaternions. Thus the unit quaternions with this scalar field form a vector field. We can say that all vector fields are homeomorphic to the quaternion vector field. Clearly the quaternion vector field has an infinite basis, but only needs 8 to span 3d space. By keeping the vector markers clean from the quaternions they ark, this remarkable find becomes clearly visible, and the basis of Hamiltons life long commitment to them.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s