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Unifying Gravity and EM

by sweetser
Tags: gravity, unifying
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sweetser
#613
Nov4-08, 12:10 PM
sweetser's Avatar
P: 361
Hello:

Clifford algebra = Geometric Algebra.

David Hestenes has promoted GA as a label over Clifford algebra because Clifford was one guy who worked on some aspects of this approach, while Geometric Algebra suggests what the math does. I only figured that out at the end of the 8th International Conference on Clifford Algebras and their applications to Mathematical Physics. David was a nice guy, and was fun to talk to so long as you didn't disagree with him on a technical point. As a visionary, he is quick to take out his blade.

According to standard GA,

quaternions = Cl(0,2)

meaning it has a scalar - the 0 - and a bivector - the 2. A bivector is a generalization of a cross product to other dimensions. Look in a mirror, and left hand becomes right hand. It was clear from a brief discussion with David that this bit of dogma was not worth discussing because there were people who were confused on the topic. Anyone who didn't understand that quaternions were Cl(0, 2) needed education.

It is one of these size arguments: GA is bigger than quaternions, ergo study GA. The size of your algebra matters. NOT.

What I think matters most in our efforts to describe Nature is completeness. Quaternions certainly have the property that under a mirror reflection M:

[tex]M: M(t, x, y, z) -> (t, -x, -y, -z)\quad eq~1[/tex]

The problem with this approach is that we need to also consider a time reversal operation (TR = -M):

[tex]TR: TR(t, x, y, z) -> (-t, x, y, z)\quad eq~2[/tex]

The handedness stays the same, which indicates a quaternion should not be viewed as a scalar and a bivector. The artificial constraint that the behavior of quaternions should fit within the GA model doesn't fit.

It was Grassman and Clifford who people credit with the great discovery of the geometric product:

[tex]a b = a \dot b + a \wedge b\quad eq~3[/tex]

Having my groundings with quaternions, this result is incomplete. For quaternions, the dot product will have 4 pairs of number, while the cross product has 6, making 10 out of the possible 16 combinations. Grassman and Clifford came after Hamilton, so they partially recreated quaternions. Arg! Let me write the quaternion a as (a, A) and b as (b, B) where the lowercase letter inside a () is one number, while a capital letter inside is a 3-vector. Compare the products of Hamilton to Clifford:

[tex](a, A)(b, -B) = (ab + A \dot B, Ab - aB - A \times B) \quad eq~4[/tex]
[tex](a, A)(b, B) = (ab + A \dot B, A \times B) \quad eq~5[/tex]

Aside from silly sign issues that arose from the "pro-positive sign" position of Grassman, I won't work with geometric products because there is no accounting for Ab and aB. It is completely legal to adjust the scale of a 3 vector. That completely reasonable operation is missing from the geometric product. It is such a glaring omission that I don't put an effort in bridging the gap between the GA school and my own work. Bad accounting is indefensible.

The even representation of quaternions - or hypercomplex numbers - also doesn't work within the Clifford algebra constraint.

So it goes for Independent Research.

Doug
sweetser
#614
Nov14-08, 08:59 AM
sweetser's Avatar
P: 361
Hello:

On occasion I pitch my quaternion animations to people who have written something on quaternion. The usual reply is no reply. Sometime we exchange a few emails, then go our separate ways.

This time I found out the fellow was visually impared. My animations are meaningless. What I could do is provide a description. Here is what a 4D cube looks like:

Quote Quote by 4D cube
We are all familiar with a 3D cube. Mathematically that is the eight spatial vertices formed from all the combinations of plus or minus one for x, y, and z. In 4D, there are sixteen vertices, the additional ones being plus or minus time. So when time is minus one, we have a 3D cube. At plus one for time we have a 3D cube. In between, when say time is minus one third or plus two sevenths, all we have are the eight vertices, none of the connecting points between the vertices. We see the vertices travel through time from one 3D cube to another.
This approach can be generalized:

Quote Quote by general approach
For fun, you might want to imagine what a polynomial looks like, or a sine function. Take your mind's image of these, and run a pencil along the image to create the animation. Note you are allowed to hold the pencil at any angle, which can create very different looking animations.
Let us think about the sine function, an endless range of mole hills. If the pencil is vertical, we will see one dot oscillating forever. If the pencil is horizontal, then we see nothing until we meet the bottom of the sine functions at -1. Then we see single dots that split in two, run apart quickly at 0, and slow down and join another, then disappear.

I think the single oscillating dot is how classical physics uses a sine function. We can watch the time-ordered oscillation of one particle for as long as we want. The array of creation, then destruction has the look of quantum field theory. To make it a complete field theory, then we would have to pile one sine atop another, making a graph that looks something like chainmail, so we see the creation and annihilation forever.

Doug


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