Exemplo n.º 1
0
def test():
    """
    We are considering electron muon scattering, following the feinmann diagram below.
    Assuming all particles have helicity one, and the CM-fram, motion in z-dir, meaning 
    line 1 and 4 have spin up, line 3 and 4 have spin down. This gives the spinnors below.
    p2\     /p4
       \   ^
        \ / μ
       γ §
         §
        / \e
       ^   \.
    p1/     \p3
    """

    # This is to make sure sympy knows these are real numbers
    me, Ee, Em, mm = sp.symbols("me, Ee, mm, Em", positive=True)
    ap = sp.symbols("ap", positive=True)  # sqrt(Ee + me)
    am = sp.symbols("am", positive=True)  # sqrt(Ee - me)
    bp = sp.symbols("bp", positive=True)  # sqrt(Em + mm)
    bm = sp.symbols("bm", positive=True)  # sqrt(Em - mm)

    u1 = np.array([ap, 0, am, 0], dtype=type(I))
    u2 = np.array([0, bp, 0, bm], dtype=type(I))
    u3 = np.array([0, ap, 0, am], dtype=type(I))
    u4 = np.array([bp, 0, bm, 0], dtype=type(I))

    M = 0
    for i in range(dim):
        M += g[i, i] * ((adj(u3) @ γ[i] @ u1) * (adj(u4) @ γ[i] @ u2))
        
    M *= g ** 2 / (4 * am * bp * bm * ap)

    pprint(M)
Exemplo n.º 2
0
    def deriveMetricFromConnection(self, nameOfTargetCoordinateSystem):
        # check if the patch stores a coordsystem with the right name
        matches = [
            cs for cs in self.coord_systems
            if cs.name == nameOfTargetCoordinateSystem
        ]
        if len(matches) < 1:
            raise "no coordinate system of the specified name found"
        elif len(matches) > 1:
            raise "more than one coord system with the specified name found"
        else:
            targetCoordinateSystem = matches[0]
        connections = targetCoordinateSystem.transforms
        #connections is a dictionary whose keys are CoordSystems
        # check if there is at least one metric representation in a connected coord system
        connectionsWithMetricRepresentation = {
            key: val
            for key, val in connections.items()
            if key in self.metricRepresentations
        }
        # in most cases there will be only one match (e.g. the cartesian cs) but it is possible to have more
        # so we chose the first that comes our way

        firstCoordSystem = list(connectionsWithMetricRepresentation.keys())[0]
        # now the question is to which of the metric representations of the patch it refers

        symbols, equations = connectionsWithMetricRepresentation[
            firstCoordSystem]
        #compute the jacobian
        J = targetCoordinateSystem.jacobian(firstCoordSystem, symbols)
        pprint(J)
        # the first column of the jacobian contains the components of the
        # new base vectors in the old base
        # therefor the components of the new metric tensor
        # can be derived from the components of the old
        # by J.transposeLU()*g_old*J
        g_old = self.metricRepresentations[firstCoordSystem]
        oldBases = g_old.bases
        oB0 = oldBases[0]  # first base
        print("oldBases")
        print(oldBases)
        if isinstance(oB0, VectorFieldBase):
            # the second base must be of the same type
            # in this case also
            newBase = VectorFieldBase("newBase", targetCoordinateSystem)
            newBase.connect(oB0, J)
        i, j = map(Idx, ['i', 'j'])
        g_new = TensorIndexSet("g_new")
        g_new[i, j] = g_old[i, j].rebase2([newBase, newBase])
        return (g_new)
Exemplo n.º 3
0
    def _interactive_traversal(expr, stage):
        if stage > 0:
            print

        cprint("Current expression (stage ", BYELLOW, stage, END, "):")
        print BCYAN
        pprint(expr)
        print END

        if isinstance(expr, Basic):
            if expr.is_Add:
                args = expr.as_ordered_terms()
            elif expr.is_Mul:
                args = expr.as_ordered_factors()
            else:
                args = expr.args
        elif hasattr(expr, "__iter__"):
            args = list(expr)
        else:
            return expr

        n_args = len(args)

        if not n_args:
            return expr

        for i, arg in enumerate(args):
            cprint(GREEN, "[", BGREEN, i, GREEN, "] ", BLUE, type(arg), END)
            pprint(arg)
            print

        if n_args == 1:
            choices = '0'
        else:
            choices = '0-%d' % (n_args - 1)

        try:
            choice = raw_input("Your choice [%s,f,l,r,d,?]: " % choices)
        except EOFError:
            result = expr
            print
        else:
            if choice == '?':
                cprint(
                    RED,
                    "%s - select subexpression with the given index" % choices)
                cprint(RED, "f - select the first subexpression")
                cprint(RED, "l - select the last subexpression")
                cprint(RED, "r - select a random subexpression")
                cprint(RED, "d - done\n")

                result = _interactive_traversal(expr, stage)
            elif choice in ['d', '']:
                result = expr
            elif choice == 'f':
                result = _interactive_traversal(args[0], stage + 1)
            elif choice == 'l':
                result = _interactive_traversal(args[-1], stage + 1)
            elif choice == 'r':
                result = _interactive_traversal(random.choice(args), stage + 1)
            else:
                try:
                    choice = int(choice)
                except ValueError:
                    cprint(BRED,
                           "Choice must be a number in %s range\n" % choices)
                    result = _interactive_traversal(expr, stage)
                else:
                    if choice < 0 or choice >= n_args:
                        cprint(BRED, "Choice must be in %s range\n" % choices)
                        result = _interactive_traversal(expr, stage)
                    else:
                        result = _interactive_traversal(
                            args[choice], stage + 1)

        return result
Exemplo n.º 4
0
    def _interactive_traversal(expr, stage):
        if stage > 0:
            print

        cprint("Current expression (stage ", BYELLOW, stage, END, "):")
        print BCYAN
        pprint(expr)
        print END

        if isinstance(expr, Basic):
            if expr.is_Add:
                args = expr.as_ordered_terms()
            elif expr.is_Mul:
                args = expr.as_ordered_factors()
            else:
                args = expr.args
        elif hasattr(expr, "__iter__"):
            args = list(expr)
        else:
            return expr

        n_args = len(args)

        if not n_args:
            return expr

        for i, arg in enumerate(args):
            cprint(GREEN, "[", BGREEN, i, GREEN, "] ", BLUE, type(arg), END)
            pprint(arg)
            print

        if n_args == 1:
            choices = '0'
        else:
            choices = '0-%d' % (n_args-1)

        try:
            choice = raw_input("Your choice [%s,f,l,r,d,?]: " % choices)
        except EOFError:
            result = expr
            print
        else:
            if choice == '?':
                cprint(RED, "%s - select subexpression with the given index" % choices)
                cprint(RED, "f - select the first subexpression")
                cprint(RED, "l - select the last subexpression")
                cprint(RED, "r - select a random subexpression")
                cprint(RED, "d - done\n")

                result = _interactive_traversal(expr, stage)
            elif choice in ['d', '']:
                result = expr
            elif choice == 'f':
                result = _interactive_traversal(args[0], stage+1)
            elif choice == 'l':
                result = _interactive_traversal(args[-1], stage+1)
            elif choice == 'r':
                result = _interactive_traversal(random.choice(args), stage+1)
            else:
                try:
                    choice = int(choice)
                except ValueError:
                    cprint(BRED, "Choice must be a number in %s range\n" % choices)
                    result = _interactive_traversal(expr, stage)
                else:
                    if choice < 0 or choice >= n_args:
                        cprint(BRED, "Choice must be in %s range\n" % choices)
                        result = _interactive_traversal(expr, stage)
                    else:
                        result = _interactive_traversal(args[choice], stage+1)

        return result
Exemplo n.º 5
0
    
The cnot performs the following transformation:
    
|00> -> |00>
|01> -> |01>
|10> -> |11>
|11> -> |10>
"""

theta = Symbol('theta')

X = np.array([[0, 1j], [-1j, 0]])

CNOT = Matrix(
    np.array([[1, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0], [0, 1, 0, 0]]))

I = Matrix(np.eye(2))
sigmax = Matrix(1j * X * theta)
sigmax_inv = Matrix(-1j * X * theta)

R = simplify(exp(sigmax))
R_ = simplify(exp(sigmax_inv))
IX = simplify(TensorProduct(I, R))
IX_ = simplify(TensorProduct(I, R_))

U = IX_ * CNOT * IX

M_ij = Matrix([[U[0, 0], U[0, 2]], [U[2, 0], U[2, 2]]])

pprint(simplify(M_ij))
Exemplo n.º 6
0
Algumas funções são conhecidas:
 - exponenciação: use ** em vez de ^
 - multiplicação: sempre escreva explicitamente 2*a, nunca 2a
 - algumas funções suportadas: cos(x), sin(x), ln(x), log(x), etc.
 
exemplo: 2*(ρ - ρl)*g / (9*a)

""")

    ok = False
    while not ok:
        print("escreva a fórmula:\n")
        try:
            form = parse(input())
            print("\nSua expressão:")
            pprint(form)
            if input("\nParece OK? (s/n)\n") == 's':
                ok = True
        except SyntaxError:
            print("Há um problema com a fórmula... tente de novo")

    print()
    print(f"""Como você quer a formatação da incerteza?
    (r): relativa (padrão) - a resposta sai na forma '{simb} * raiz de alguma coisa'
    (a): absoluta - a resposta sai na forma 'raiz de alguma coisa'""")
    if input() == 'a':
        inc = incerteza_absoluta(form)
    else:
        inc = incerteza_relativa(form, simb)

    print()
Exemplo n.º 7
0
#from LAPM.linear_autonomous_pool_model import LinearAutonomousPoolModel
from sympy.printing import pprint
from sympy import var 
var("lambda_1 lambda_2 t")
M=Matrix([[lambda_1,0,0],[1,lambda_1,0],[0,0,lambda_2]])
pprint(M.eigenvects())
var("t")
pprint(M.exp())
M.jordan_form()
pprint((M*t).exp())
T=Matrix([[1,2,0],[2,1,0],[1,1,1]])
T**-1
A=T*M*T**-1

Exemplo n.º 8
0
from sympy import sqrt,exp,log,Symbol,oo,Rational,sin,cos,limit,I,pi,Mul
from sympy.core import basic
from sympy.printing import print_pygame, pprint

x=Symbol("x") 
y=Symbol("y") 

expressions = (
    x**x,
    x+y+x,
    sin(x)**x,
    sin(x)**cos(x),
    sin(x)/(cos(x)**2 * x**x +(2*y)),

    sin(x**2+exp(x)),
    sqrt(exp(x)),

    #print (1/cos(x)).series(x,10)
)

print "sympy print:"
for expr in expressions:
    print expr
print ''

print "pygame print:"
for expr in expressions:
    pprint(expr)
    print_pygame(expr)
Exemplo n.º 9
0
# Parameters
params.update(eq_params())
params.update({**vy(), **vq()})

# Forward operator
forward = forward_op(config.misc['symbolic'])

# Mapped backward operator
factor_x, factor_y, factor = factors(config.misc['symbolic'], config.num['λ'])
backward = eq.map_operator(forward, f, factor)

forward, backward = evaluate([forward, backward])
factor_x, factor_y, factor = evaluate([factor_x, factor_y, factor])

if config.misc['verbose']:
    pprint(forward)
    pprint(backward)

for key in params:
    params[key].value = params[key].eval()

degree = config.num['degree']
n_points_num = config.num['n_points_num']

# }}}
# DEFINE QUADRATURES {{{
vprint("Defining quadratures")


def compute_quads():
    μx = params['μx'].value
import numpy
import sympy
from sympy.printing import pprint
from sympy import init_printing
from matplotlib import pyplot
init_printing(use_latex=True)

x, nu, t = sympy.symbols(
    'x nu t')  # Setting symbols for x, nu, and t from symbol library.
# An example equation to show the look of output.
phi = (sympy.exp(-(x - 4 * t)**2 / (4 * nu * (t + 1))) +
       sympy.exp(-(x - 4 * t - 2 * numpy.pi)**2 / (4 * nu * (t + 1))))
pprint("Phi")  # Labeling the following equation.
pprint(
    phi)  # Prints Phi using pretty print, outputs the result in equation form.

pprint("Phi prime")  # Labeling the following equation.
phiprime = phi.diff(x)  # Solves for the derivative of phi.
pprint(
    phiprime
)  # Prints Phi' using pretty print, outputs the result in equation form.

from sympy.utilities.lambdify import lambdify
u = -2 * nu * (phiprime / phi) + 4
pprint(u)

ufunc = lambdify((t, x, nu), u)
pprint(ufunc(1, 4, 3))

# Burger's Equation
nx = 101  # Set number of grid points in x dir