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Sorry, a very general question, I'm just looking around with sympy.
I was confused by sympy having two implementations of vectors, sympy.vector
an sympy.physics.vector
. They seem to not be sharing code and the documentation for either doesn't mention another one and the reason to have them separate. Why is that? Is there a way to make them interoperate?
Thanks!
=
I'm using sym.Eq(a, b)
sym.solve()
with a list of the expressions
Fr
and Frstar
outputs of Kane's method. Here I've printed them. Fr
, shows Tx
, which is the load/torque. Cool. Frstar
has 3 terms. First one is Ixx ddeta
; inertia x angular acceleration. ok. But the next 2 terms are Ixx Omega^2
; inertia x angular velocity^2. why is it like that? Fr
has the unit Nm
, term 1 of Frstar
has Nm.rad
, others have Nm.rad^2
. Is it because rad
is an SI derived unit?
Hi guys, question about root finding with symbolic coefficients. Is it possible to get symbolic results for something like the following:
import sympy as sp
t = sp.Symbol('t')
P = sp.Poly(t**2 + 2*t + sp.log(2))
P.all_roots()
This does not work, as I get sympy.polys.polyerrors.PolynomialError: only univariate polynomials are allowed
P = sp.Poly(t**2 + 2*t + sp.log(2), t, domain='RR')
works for numerical values. Looking at the docs, I thought something like P = sp.Poly(t**2 + 2*t + sp.log(2), t, domain='QQ<log(2)>')
might work, but this triggers a sympy.polys.polyerrors.NotAlgebraic: log(2) doesn't seem to be an algebraic element
QQ<log(2)>
does not work because log(2)
is not algebraic. One must use something like QQ(log(2))
instead. Or, simply give t
as the sole generator; then log(2)
will automatically become a constant.In [26]: p = Poly(t**2 + 2*t + log(2), t)
In [27]: roots(p)
Out[27]:
⎧ ____________ ____________ ⎫
⎨-1 - ╲╱ 1 - log(2) : 1, -1 + ╲╱ 1 - log(2) : 1⎬
⎩ ⎭
Sum
, replicating in a "single command" things likeGeneralSumOfSquares
class used here. Is there a paper I can reference/explanation, etc?
I'm having trouble learning the diffgeom
module. I'm trying to compute covariant derivatives in spherical coordinates and I feel like I'm missing something.
Here's what I'm working with:
from sympy import *
from sympy.abc import x, y, z, r, theta, phi
from sympy.diffgeom import CoordSystem, Manifold, Patch, TensorProduct as TP
R3 = Manifold("R3", 3)
S = Patch("S", R3)
relations = {
("Car3D", "Sph") : Lambda(
(x, y, z),
Matrix([
sqrt(x**2 + y**2 + z**2),
acos(z / sqrt(x**2 + y**2 + z**2)),
atan2(y, x),
])
),
("Sph", "Car3D"): Lambda(
(r, theta, phi),
Matrix([r * sin(theta) * cos(phi), r * sin(theta) * sin(phi), r * cos(theta)])
)
}
Car3D = CoordSystem("Car3D", S, [x, y, z], relations)
Sph = CoordSystem("Sph", S, [r, theta, phi], relations)
I can compute the metric tensor in spherical coordinates by doing the following:
g = (Car3D.jacobian(Sph) @ Car3D.jacobian(Sph).T).simplify()
However, this computes the metric tensor with respect to $x, y, z$, while I'd like to write the metric tensor with respect to $r, \theta, \phi$:
$\left[\begin{matrix}1 & 0 & 0\\0 & \frac{1}{x^{2} + y^{2} + z^{2}} & 0\\0 & 0 & \frac{1}{x^{2} + y^{2}}\end{matrix}\right]$
I can do a substitution, but it's kind of awkward:
x, y, z = Car3D.symbols
r, theta, phi = Sph.symbols
forward_subs = {
x: r * sin(theta) * cos(phi),
y: r * sin(theta) * sin(phi),
z: r * cos(theta)
}
g_sph = g.subs(forward_subs).simplify()
$\left[\begin{matrix}1 & 0 & 0\\0 & \frac{1}{r^{2}} & 0\\0 & 0 & \frac{1}{r^{2} \sin^{2}{\left(\theta \right)}}\end{matrix}\right]$
This gives me the expected metric tensor as a matrix but there's a lot of extra effort:
forward_subs
since the CoordinateSymbol
"x" does not exist until I create Car3D
, and I have to provide the forward and backward transformations to create Car3D
.I can convert the metric tensor Matrix
to a tensor expression as follows:
g_sph_tens = 0
for i, e_i in enumerate(Car3D.base_oneforms()):
for j, e_j in enumerate(Car3D.base_oneforms()):
g_sph_tens += g_sph[i, j] * TP(e_i, e_j)
However, computing the Christoffel symbol for g_sph_tens
using metric_to_Christoffel_2nd
gives all zeros, which is incorrect.
Can somebody help me out with this? It seems like I'm missing some fundamentals.
Hi, I want to learn SymPy to do some calculations. I am practicing with the electromagnetic Lagrangian term $F_{\mu\nu}F^{\mu\nu}$. I have defined $F$ in the way that is explained in the documentation but cannot figure out how to go from there to $E^2 + B^2$. How am I supposed to do this?
Up to now I have this:
Ex, Ey, Ez, Bx, By, Bz = sp.symbols('E_x E_y E_z B_x B_y B_z')
c = sp.symbols('c', positive=True)
F = spT.TensorHead('F', [Lorentz, Lorentz], spT.TensorSymmetry.fully_symmetric(-2))
repl = {Lorentz: sp.diag(1, -1, -1, -1)}
repl.update({F(-mu,-nu): [
[0, Ex/c, Ey/c, Ez/c],
[-Ex/c, 0, -Bz, By],
[-Ey/c, Bz, 0, -Bx],
[-Ez/c, -By, Bx, 0]]}
)
and I have tried (F(mu,nu)*F(-mu,-nu)).replace_with_arrays(repl, [mu,nu])
and also F(mu,nu).replace_with_arrays(repl, [mu,nu])*F(-mu,-nu).replace_with_arrays(repl, [mu,nu])
.
sympy.preview(expr, output="pdf", viewer="file", filename="DFN_equations.pdf", dvioptions=['-D', '1100'], euler=False)
I wanted to expand on my question from earlier since it may be too specific.
I am working through some equations from the theory of elasticity.
The differential equation I am trying to work with is the equation
for compatibility of small displacements in a strained isotropic elastic body.
This is written as
$(1 - 2\eta) v^i|^j_j + v^j|^i_j = 0$
where
$f^i|_j$
represents the covariant derivative of f with respect to j, and
$f^i|^j$
represents the covariant derivative of f with respect to j
with the j index raised (this convention is from Green and Zerna -
I don't know if this is a standard notation).
We can use solutions of the above differential equation to compute
the stress tensor as follows:
$\frac{\tau^{ij}}{\mu} = g^{js} v^i|_s + g^{ir} v^j|_r + \frac{2 \eta}{1 - 2 \eta} g^{ij} v^r|_r$
One of the solutions to the displacement equation is as follows:
$v_i = F|_i$
where F is a harmonic function.
What's a good way to
\limits
is.
\limits
puts them directly above or below the integral sign. That does seem better for non-inline mode, especially if the limits can be long.