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## Re: [Axiom-developer] Clifford Algebra (Physics, Geometry, Algebra)

 From: Martin Baker Subject: Re: [Axiom-developer] Clifford Algebra (Physics, Geometry, Algebra) Date: Fri, 9 Dec 2016 09:15:23 +0000 User-agent: Mozilla/5.0 (X11; Linux x86_64; rv:45.0) Gecko/20100101 Thunderbird/45.5.1

```On 08/12/16 18:03, Tim Daly wrote:
```
```So the real question might be "Can I formulate my physics in
matrix and tensors and shift between them". Which leads to the
question "Can I easily convert between matrices and tensors?"
From what I've seen so far this should be possible, provided it is
implemented properly. The CA wedge product is just the basis
vectors times the determinant so you might have to specify the
map between the matrix basis and the tensor basis.
```
```
```
Although matrix and Clifford algebras are apples and oranges I think that, in the context of a given geometry and physics it may be theoretically possible to have an equivalence between them but in practice I suspect this may be too messy?
```
```
As an example lets look at 3D rotations as either 3x3 Matrix or quaternions (taken to be Clifford algebra with 3D bivector + scalar) in the context of Euclidean space and classical physics (rotation by 360 degrees taken to be identity).
```
```
I think the matrix to Clifford conversion would be something like this untested code:
```
matrix2Quat(m:SquareMatrix(3,DoubleFloat)) == Quaternion(DoubleFloat)
tr:DoubleFloat = elt(m,0,0) + elt(m,1,1) + elt(m,2,2)
if (tr > 0) then
S:DoubleFloat = sqrt(tr+1.0) * 2
qw = 0.25 * S
qx = (elt(m,2,1) - elt(m,1,2)) / S
qy = (elt(m,0,2) - elt(m,2,0)) / S
qz = (elt(m,1,0) - elt(m,0,1)) / S
else if ((elt(m,0,0) > elt(m,1,1)) and (elt(m,0,0) > elt(m,2,2))) then
S:DoubleFloat = sqrt(1.0 + elt(m,0,0) - elt(m,1,1) - elt(m,2,2)) * 2
qw = (elt(m,2,1) - elt(m,1,2)) / S
qx = 0.25 * S;
qy = (elt(m,0,1) + elt(m,1,0)) / S
qz = (elt(m,0,2) + elt(m,2,0)) / S
else if (elt(m,1,1) > elt(m,2,2)) then
S:DoubleFloat = sqrt(1.0 + elt(m,1,1) - elt(m,0,0) - elt(m,2,2)) * 2
qw = (elt(m,0,2) - elt(m,2,0)) / S
qx = (elt(m,0,1) + elt(m,1,0)) / S
qy = 0.25 * S
qz = (elt(m,1,2) + elt(m,2,1)) / S
else
S:DoubleFloat = sqrt(1.0 + elt(m,2,2) - elt(m,0,0) - elt(m,1,1)) * 2
qw = (elt(m,1,0) - elt(m,0,1)) / S
qx = (elt(m,0,2) + elt(m,2,0)) / S
qy = (elt(m,1,2) + elt(m,2,1)) / S
qz = 0.25 * S
quatern(qx,qy,qz,qw)

```
The 'if' statements are required to do the translation in a part of the geometry away from singularities and to avoid root of negative numbers.
```
```
This seems quite complicated to me and I suspect (but I don't know) that any attempt to generalise it may make it more complicated?
```
```
Therefore I was thinking more of being able to plug in either matrix or Clifford but not both.
```
```
So I have a physics problem that needs to work with 3D rotations, I could either plug in the matrix code to do the problem or the Clifford code. They should give the same answer but the will differ in rounding errors, ability to scale etc.
```
Martin

```