Alan Bromborsky wrote:
> Vinzent Steinberg wrote:
>
>> On 1 déc, 21:28, Scott<scotta_2...@yahoo.com> wrote:
>>
>>
>>> Is there a sympy function for making symbolic tensors?
>>> Basically I want to treat 3x3 rotation tensor symbol that is constant
>>> inside of an integral while preserving the tensor algebra (A*B .ne.
>>> B*A). The 3x3 rotain tensor is multiplied by a 3x1 position tenosr
>>> which is integrated.
>>>
>>> Is there a built in function for converting a 3*1 tensor to a skew
>>> antisymetric cross product matrix then back again?
>>> V/R
>>>
>>> Scott
>>>
>>>
>> This is one of our oldest issues:
>> http://code.google.com/p/sympy/issues/detail?id=16
>>
>> You may have a look at the geometric algebra module, but there are no
>> tensors in sympy yet.
>>
>> Vinzent
>>
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>>
>>
>>
> In geometric algebra rotations are done via rotors (spinors) as follows:
>
> For a the rotation of a vector x around an axis u (|u| = 1) through
> an angle theta, the rotor is:
>
> R = cos(theta)-I*u*sin(theta)
>
> where I is the pseudo scalar (I=i*j*k if i, j, and k are orthogonal
> basis vectors and * the geometric product).
> The components of I*u form a rank 2 antisymmetric tensor. The
> reverse of R, rev(R) is:
>
> rev(R) = cos(theta)+I*u*sin(theta)
>
> The rotated x, x' is then:
>
> x' = R*x*rev(R)
>
> where again * is the geometric product.
>
> What domain (path, area, or volume) do you want to integrate
> R*x*rev(R) over?
>
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Here is your example:
First the code -
#!/usr/bin/python
import sys
import os,sympy,time
from sympy.galgebra.GA import *
from sympy import cos,sin
set_main(sys.modules[__name__])
if __name__ == '__main__':
make_symbols('a x y z')
MV.setup('e',metric='[1,1,1]',coords=[x,y,z])
MV.set_str_format(1)
u = MV('u','vector')
x = MV('x','vector')
print u
print x
print MV.I*u
R = cos(a)-MV.I*u*sin(a/2)
Rrev = R.rev()
print R
print Rrev
MV.set_str_format(2)
rot_x = R*x*Rrev
rot_x.simplify()
print rot_x
Second the result -
u__x*e_x+u__y*e_y+u__z*e_z
x__x*e_x+x__y*e_y+x__z*e_z
u__z*e_xe_y-u__y*e_xe_z+u__x*e_ye_z
cos(a/2)-u__z*sin(a/2)*e_x^e_y+u__y*sin(a/2)*e_x^e_z-u__x*sin(a/2)*e_y^e_z
cos(a/2)+u__z*sin(a/2)*e_x^e_y-u__y*sin(a/2)*e_x^e_z+u__x*sin(a/2)*e_y^e_z
(-2*u__z*x__y*cos(a/2)*sin(a/2)+2*u__y*x__z*cos(a/2)*sin(a/2)+x__x*cos(a/2)**2+x__x*u__x**2*sin(a/2)**2-x__x*u__y**2*sin(a/2)**2-x__x*u__z**2*sin(a/2)**2+2*u__x*u__y*x__y*sin(a/2)**2+2*u__x*u__z*x__z*sin(a/2)**2)*e_x
+(-2*u__x*x__z*cos(a/2)*sin(a/2)+2*u__z*x__x*cos(a/2)*sin(a/2)+x__y*cos(a/2)**2+x__y*u__y**2*sin(a/2)**2-x__y*u__x**2*sin(a/2)**2-x__y*u__z**2*sin(a/2)**2+2*u__x*u__y*x__x*sin(a/2)**2+2*u__y*u__z*x__z*sin(a/2)**2)*e_y
+(-2*u__y*x__x*cos(a/2)*sin(a/2)+2*u__x*x__y*cos(a/2)*sin(a/2)+x__z*cos(a/2)**2+x__z*u__z**2*sin(a/2)**2-x__z*u__x**2*sin(a/2)**2-x__z*u__y**2*sin(a/2)**2+2*u__x*u__z*x__x*sin(a/2)**2+2*u__y*u__z*x__y*sin(a/2)**2)*e_z
In the previous post I forgot you had to use the half-angle!
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