Here is commented version of my "square_root_of_expr" function. The
functions determines if the input expression is a perfect square and if
it is returns the positive square root of the perfect square. The other
cases for the input expression are now documented in the code. I
assume the expression is real and if
the expression is a number and negative you want the square root of the
absolute value of the expression. This is done for the case of
normalizing the basis
vectors in a Minkowski space. That is if the square of a basis vector
is negative then the square of the normalized basis vector should be
-1. This only works for
basis vectors whose squares are numbers and not functions of the
coordinates since in general sympy cannot determine if a function is
positive or negative over
a given range.
def square_root_of_expr(expr):
"""
If expression is product of even powers then every power is divided
by two and the product is returned. If some terms in product are
not even powers the sqrt of the absolute value of the expression is
returned. If the expression is a number the sqrt of the absolute
value of the number is returned.
"""
if expr.is_number:
if expr > 0:
return(sqrt(expr))
else:
return(sqrt(-expr))
else:
expr = trigsimp(expr)
(coef, pow_lst) = sqf_list(expr)
if coef != S(1):
if coef.is_number:
coef = square_root_of_expr(coef)
else:
coef = sqrt(abs(coef)) # Product coefficient not a number
for p in pow_lst:
(f, n) = p
if n % 2 != 0:
return(sqrt(abs(expr))) # Product not all even powers
else:
coef *= f ** (n / 2) # Positive sqrt of the square of
an expression
return coef
On 12/06/2014 09:58 AM, Francesco Bonazzi wrote:
I published a gist on how to get the gradient in any coordinate
system, given the backward transformation to rectangular coordinates.
I was also trying to write the code to get the divergence, but I ran
into problems with sympy's simplify( ) function.
https://gist.github.com/Upabjojr/34f75ce115f1cf5a6afc
This gist defines a function *get_grad( ... )* which returns the
gradient, it requires as an input a lambda function (backwards
transformation).
Example:
|
from_spherical
=lambdar,theta,phi:(r*sin(theta)*cos(phi),r*sin(theta)*sin(phi),r*cos(theta))
grad =get_grad(from_spherical)
|
Notice that the function /from_spherical/ is simply the easiest way to
define the spherical coordinates.
I get the following output:
|
⎡∂_θ│sin(θ)│⋅∂ᵩ⎤
⎢∂ᵣ,───,───────────⎥
⎢ r 2⎥
⎣ r⋅sin (θ)⎦
|
SymPy doesn't know that in /theta/'s range of values its /sin(theta)/
is always positive, this is the reason why it does not get simplified.
I have only tried the spherical coordinates, but this should work with
any coordinate transformation in three dimensions. It should not be
too complicated to generalize this to any dimension.
That's why I would like coordinate systems to be instances of a class,
and not classes themselves. It is really uncomfortable to define a new
class for every new coordinate system.
On Thursday, November 27, 2014 7:17:49 AM UTC+1, Aaron Meurer wrote:
On Tue, Nov 25, 2014 at 1:58 AM, Francesco Bonazzi
<franz....@gmail.com <javascript:>> wrote:
>
>
> On Wednesday, November 19, 2014 11:17:24 PM UTC+1, Jason Moore
wrote:
>>
>> Sachin Joglekar developed the vector package this past summer.
I was his
>> GSoC mentor on the project. You can check out this implemented
of the
>> Cartesian coordinate system:
>>
>>
https://github.com/sympy/sympy/blob/master/sympy/vector/coordsysrect.py
<https://github.com/sympy/sympy/blob/master/sympy/vector/coordsysrect.py>
>
>
> In sympy.diffgeom coordinate systems are defined as instances of
the
> CoordSystem class, once more coordinate systems are defined, the
> transformation rules are set using the connect_to method:
>
> In [1]: from sympy.diffgeom import *
>
> In [2]: r, theta, phi = symbols('r theta phi', positive=True)
>
> In [4]: M = Manifold('M', 3)
>
> In [5]: P = Patch('P', M)
>
> In [7]: rect = CoordSystem('rect', P, ['x', 'y', 'z'])
>
> In [8]: spherical = CoordSystem('spherical', P, ['r', 'theta',
'phi'])
>
> In [9]: spherical.connect_to(rect, [r, theta, phi],
[r*sin(phi)*cos(theta),
> r*sin(phi)*sin(theta), r*cos(phi)], inverse=False)
>
>
> I put inverse=False in the last line, because SymPy's solvers
are unable to
> reverse the spherical-to-rect transformation.
>
> Manifold and Patch are some boilerplate, Manifold specifies the
space you're
> considering, while Patch is a connected subspace of the
manifold. Patch is
> used because on some manifolds there does not exist a
homeomorphism to a
> coordinate system (e.g. on the sphere).
>
> I think that some algorithmic code could be shared between the
vector and
> the diffgeom modules.
+1 to that, especially if we can avoid hard-coding things and just
let
solve() compute things (similar to how sympy.stats works).
Aaron Meurer
>
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