Calculating Spin-Loss of a Particle Using Integral Form

In summary: def f(u): \\defining the function f q = -q v = complex(0, math.sqrt(q))
  • #1
amjad-sh
246
13
TL;DR Summary
I'm trying to integrate an imaginary function by using the package quad.
The function is a complex function and also a very complicated one.
This typing error is showing up to me after running the program: """unsupported operand type(s) for *: 'complex' and 'function' """".
I will show you the details below.
The integral has the form:
$$\frac{s^2\nu^4}{(2\pi)^2}\int_{-1}^1 u(1-u^2)k_f^5[|r_1\chi_1|^2+|r_1\chi_2|^2-|r_1|^2\chi_1^*\chi_2\cos(2k_f\sqrt{u^2-\nu^2}a)-|r_1|^2\chi_2^*\chi_1cos(2k_f\sqrt{u^2-\nu^2}a)]\, du$$
##r_1,\chi_1## and ##\chi_2## are also imaginary functions of u, because the form of these imaginary functions is so complicated, I also represented them interms of another imaginary functions like a,b,c,m1,m2 and f.
All the functions ##r_1,r_2,\chi_1,\chi_2,b,a## and ##c## are functions of u.
So I defined them as functions with one parameter u as below:

import numpy as np
from scipy.integrate import quad
import scipy
import matplotlib.pyplot as plt
import cmath as math

spinloss=[]
lnu=[]
a = 1.6*10**(-6)
nu = -0.01 \\defining some constants
kf = 4.2*1.6*10**(-19)

for i in range(0,101):
spinloss.append([])
for i in range(0,101):
lnu.append([])

def b(u):
q=u**2-nu**2
if q<0: \\defining the function b
q = -q
v = complex(0, math.sqrt(q))
else:
v = math.sqrt(u**2-nu**2)
return(kf*(u+v)*math.exp(-2j*kf*v*a)+kf*(u-v)*math.exp(2j*kf*v*a))def c(u):
q = u ** 2 - nu ** 2
if q < 0:
q = -q
v = complex(0, math.sqrt(q)) \\defining the function c
else:
v = math.sqrt(u ** 2 - nu ** 2)
return (kf**2*(u+v)**2*np.exp(-2j*kf*v*a)-kf**2*(u-v)**2*math.exp(2j*kf*v*a))

def a(u):
q = u ** 2 - nu ** 2
if q < 0:\\defining the function a
q = -q
v = complex(0, math.sqrt(q))
else:
v = math.sqrt(u ** 2 - nu ** 2)
return (((kf*b(u)*(u-v)+c(u))*math.exp(1j*kf*v*a)-2*kf**2*u*(u+v)*math.exp(-1j*kf*v*a))/((kf*b(u)*(u+v)+c(u))*math.exp(-1j*kf*v*a)-2*kf**2*u*(u-v)*math.exp(1j*kf*v*a)))def x1(u):
\\defining x1
return((a(u)*b(u)+2*kf*u)/c(u))

def x2(u):
\\defining x2
return((2*a(u)*kf*u+b(u))/c(u))def m1(u):
q = u ** 2 - nu ** 2
if q < 0:
q = -q
v = complex(0, math.sqrt(q)) \\defining m1
else:
v = math.sqrt(u ** 2 - nu ** 2)
return(kf*v*(1-a(u)))def m2(u):
q = u ** 2 - nu ** 2
if q < 0:
q = -q \\defining m2
v = complex(0, math.sqrt(q))
else:
v = math.sqrt(u ** 2 - nu ** 2)
return((-kf*v*(1-a(u))))def f(u):
q = u ** 2 - nu ** 2
if q < 0:
q = -q
v = complex(0, math.sqrt(q)) \\defining f
else:
v = math.sqrt(u ** 2 - nu ** 2)
return((a(u)+1)*math.exp(-1j*math.sqrt(kf*a))*math.cos(kf*v*a))def r1(u):
q = u ** 2 - nu ** 2
if q < 0:
q = -q \\defining r1
v = complex(0, math.sqrt(q))
else:
v = math.sqrt(u ** 2 - nu ** 2)
return((2*math.exp(-2j*kf*u*a))/(2*f(u)-((1/(kf*u))*(m1(u)*math.exp(1j*(kf*v-kf*u)*a)-m2(u)*math.exp(-1j*(kf*v+kf*u)*a))))) \\the error started from this linedef r0(u):
q = u ** 2 - nu ** 2
if q < 0:
q = -q \\defining r0
v = complex(0, math.sqrt(q))
else:
v = math.sqrt(u ** 2 - nu ** 2)
return(math.exp(-2j*kf*u*a)-(r1(u)/(kf*u))*(2*kf*v*math.exp(-1j*kf*(v+u)*a)-(2*kf*v*math.exp(1j*kf*(v-u)*a))))def t0(u):
q = u ** 2 - nu ** 2
if q < 0:
q = -q \\defining t0
v = complex(0, math.sqrt(q))
else:
v = math.sqrt(u ** 2 - nu ** 2)
return(((r1(u))/(u))*v*a(u)*(math.exp(-1j*kf*(v+u)*a)-math.exp(1j*kf*(v-u)*a)))
def ls(u):
q = u ** 2 - nu ** 2
if q < 0: \\defining ls
q = -q
v = complex(0, math.sqrt(q))
else:
v = math.sqrt(u ** 2 - nu ** 2)
return(u*(1-u**2)*kf**5*((abs(r1(u)*x1(u)))**2+(abs(r1(u)*x2(u)))**2-(abs(r1(u)))**2)*np.conjugate(x1(u))*x2(u)*math.cos(2*kf*v*a)-(abs(r1(u)))**2*np.conjugate(x2(u))*x1(u)*math.cos(2*kf*v*a))\\ it showed also here.(the error)
def jx(u):
return((((-kf)**2)/(8*(math.pi)**2))*u*(abs(r0(u)))**2) \\defining jx
def complex_quadrature(func, a, b):
def real_func(u):
return scipy.real(func(u))
\\for integration
real_integral = quad(real_func, a, b)

return (real_integral[0])

for i in range(0, 101):
nu = nu + 0.01
if nu == 1: \\ nu represents ##\nu## and it must be between zero and 1.
spinloss = 0
lnu = nu

if nu != 1:
spinloss = ((complex_quadrature(ls, -1, 1)) * (nu ** 4) /(2*math.pi)**2) / (complex_quadrature(jx, -1, 1)) \\ this gives us the result of the integration of ls from -1 to 1 divided by the integration of jx from -1 to 1.
lnu = nu
plt.plot(lnu,spinloss,color='b')
plt.show()

If some python expert can point out for me the reason why this error is appearing, I will be so appreciated.
 
Technology news on Phys.org
  • #2
It would help to know which multiplication exactly causes the error. The debugger might show that, alternatively simplify this equation step by step until you find a minimal expression that causes the error:
amjad-sh said:
return((2*math.exp(-2j*kf*u*a))/(2*f(u)-((1/(kf*u))*(m1(u)*math.exp(1j*(kf*v-kf*u)*a)-m2(u)*math.exp(-1j*(kf*v+kf*u)*a))))) \\the error started from this line
 
  • Like
Likes amjad-sh

FAQ: Calculating Spin-Loss of a Particle Using Integral Form

1. What is the purpose of calculating spin-loss of a particle?

The purpose of calculating spin-loss of a particle is to understand the behavior of the particle in a magnetic field and to determine the amount of energy it loses due to its spin. This information is important in various fields of study, such as particle physics and materials science.

2. How is spin-loss of a particle calculated using integral form?

Spin-loss of a particle is calculated using the Larmor formula, which involves the use of an integral form. The integral form takes into account the particle's spin and its trajectory in a magnetic field, and calculates the amount of energy lost due to the spin.

3. What factors affect the spin-loss of a particle?

The spin-loss of a particle is affected by several factors, such as the strength of the magnetic field, the mass and charge of the particle, and the particle's velocity and spin orientation. These factors can influence the trajectory of the particle and the amount of energy it loses due to its spin.

4. How is the spin-loss of a particle experimentally measured?

The spin-loss of a particle can be experimentally measured using various techniques, such as electron paramagnetic resonance (EPR) spectroscopy or nuclear magnetic resonance (NMR) spectroscopy. These techniques involve the use of magnetic fields to measure the energy lost by the particle due to its spin.

5. What are the applications of calculating spin-loss of a particle?

The calculation of spin-loss of a particle has various applications in different fields of study. It is used in particle accelerators to understand the behavior of particles in magnetic fields, in materials science to study the properties of magnetic materials, and in medical imaging techniques such as MRI to produce high-resolution images of the body's tissues.

Similar threads

Replies
8
Views
1K
2
Replies
50
Views
5K
Replies
6
Views
1K
Replies
2
Views
2K
Replies
5
Views
3K
Replies
1
Views
856
Replies
3
Views
3K
Replies
1
Views
1K
Back
Top