Sistema binario: Energía y Momentum angular radiado


In [1]:
from sympy import *
init_printing(use_unicode=True)

Primero definimos los símbolos de las variables a usar: $a,e,\mu,L,\omega_0,\varphi$ y $\varphi_0$:


In [2]:
a = Symbol('a', positive=True)
e = Symbol('e', positive=True)
mu = Symbol('mu', positive=True)
L = Symbol('L',positive=True)
omega0 = Symbol('omega0',positive=True)
phi = Symbol('varphi',positive=True)
phi0 = Symbol('varphi0',real=True)

Momento de inercia: $M_{ij}$

Y ahora definimos los símbolos de las componentes (no nulas) del momento cuadrupolar $M_{ij}$, como funciones de $\varphi$


In [3]:
M11 = Function('M11')(phi)
M12 = Function('M12')(phi)
M22 = Function('M22')(phi)
r = Function('r')(phi)

Ahora ingresamos la forma de la órbita Kepleriana, en términos de $a, e, \varphi$ y $\varphi_0$:


In [4]:
r = a*(1-e**2)/(1+e*cos(phi-phi0))

In [5]:
r


Out[5]:
$$\frac{a \left(- e^{2} + 1\right)}{e \cos{\left (\varphi - \varphi_{0} \right )} + 1}$$

Ahora sustituimos los valores de las componentes del momento cuadrupolar en función de las coordenadas del sistema reducido:


In [6]:
M11 = mu*(r**2)*cos(phi)**2
M12 = mu*(r**2)*sin(phi)*cos(phi)
M22 = mu*(r**2)*sin(phi)**2

In [7]:
M11,M12,M22


Out[7]:
$$\left ( \frac{a^{2} \mu \left(- e^{2} + 1\right)^{2} \cos^{2}{\left (\varphi \right )}}{\left(e \cos{\left (\varphi - \varphi_{0} \right )} + 1\right)^{2}}, \quad \frac{a^{2} \mu \left(- e^{2} + 1\right)^{2} \sin{\left (\varphi \right )} \cos{\left (\varphi \right )}}{\left(e \cos{\left (\varphi - \varphi_{0} \right )} + 1\right)^{2}}, \quad \frac{a^{2} \mu \left(- e^{2} + 1\right)^{2} \sin^{2}{\left (\varphi \right )}}{\left(e \cos{\left (\varphi - \varphi_{0} \right )} + 1\right)^{2}}\right )$$

Calculo de la derivada ${d M_{ij}}/{d\varphi}$

Calculamos las derivadas de las componentes de $M_{ij}$ respecto a $\varphi$, usando la función diff de Sympy


In [8]:
dM11_phi = simplify(diff(M11,phi))
dM12_phi = simplify(diff(M12,phi))
dM22_phi = simplify(diff(M22,phi))

In [9]:
dM11_phi,dM12_phi,dM22_phi


Out[9]:
$$\left ( - \frac{2 a^{2} \mu \left(e^{2} - 1\right)^{2} \cos{\left (\varphi \right )}}{\left(e \cos{\left (\varphi - \varphi_{0} \right )} + 1\right)^{3}} \left(e \sin{\left (\varphi_{0} \right )} + \sin{\left (\varphi \right )}\right), \quad \frac{a^{2} \mu \left(e^{2} - 1\right)^{2}}{\left(e \cos{\left (\varphi - \varphi_{0} \right )} + 1\right)^{3}} \left(\frac{e}{2} \left(\cos{\left (\varphi + \varphi_{0} \right )} - \cos{\left (3 \varphi - \varphi_{0} \right )}\right) + \left(e \cos{\left (\varphi - \varphi_{0} \right )} + 1\right) \cos{\left (2 \varphi \right )}\right), \quad \frac{2 a^{2} \mu \left(e^{2} - 1\right)^{2} \sin{\left (\varphi \right )}}{\left(e \cos{\left (\varphi - \varphi_{0} \right )} + 1\right)^{3}} \left(e \cos{\left (\varphi_{0} \right )} + \cos{\left (\varphi \right )}\right)\right )$$

Calculo de las derivadas $\dot{M}_{ij}$, $\ddot{M}_{ij}$ y $\dddot{M}_{ij}$

Ahora calculamos las derivadas respecto al tiempo, usando la regla de la cadena, es decir, $$\frac{dM_{ij}}{dt}=\frac{dM_{ij}}{d\varphi}\frac{d\varphi}{dt},$$ donde luego sustituiremos $$\frac{d \varphi}{d t}=\frac{L}{\mu r^2},$$ con $$L=\omega_0\mu a^2\sqrt{1-e^2}$$


In [10]:
L = omega0*mu*(a**2)*sqrt(1-e**2)
dphi_t = L/(mu*r**2)

In [11]:
dphi_t


Out[11]:
$$\frac{\omega_{0}}{\left(- e^{2} + 1\right)^{\frac{3}{2}}} \left(e \cos{\left (\varphi - \varphi_{0} \right )} + 1\right)^{2}$$

In [12]:
dM11_t = simplify(dM11_phi*dphi_t)
dM12_t = simplify(dM12_phi*dphi_t)
dM22_t = simplify(dM22_phi*dphi_t)

In [13]:
dM11_t,dM12_t,dM22_t


Out[13]:
$$\left ( - \frac{2 a^{2} \mu \omega_{0} \sqrt{- e^{2} + 1} \cos{\left (\varphi \right )}}{e \cos{\left (\varphi - \varphi_{0} \right )} + 1} \left(e \sin{\left (\varphi_{0} \right )} + \sin{\left (\varphi \right )}\right), \quad \frac{a^{2} \mu \omega_{0} \sqrt{- e^{2} + 1}}{2 e \cos{\left (\varphi - \varphi_{0} \right )} + 2} \left(e \left(\cos{\left (\varphi + \varphi_{0} \right )} - \cos{\left (3 \varphi - \varphi_{0} \right )}\right) + 2 \left(e \cos{\left (\varphi - \varphi_{0} \right )} + 1\right) \cos{\left (2 \varphi \right )}\right), \quad \frac{2 a^{2} \mu \omega_{0} \sqrt{- e^{2} + 1} \sin{\left (\varphi \right )}}{e \cos{\left (\varphi - \varphi_{0} \right )} + 1} \left(e \cos{\left (\varphi_{0} \right )} + \cos{\left (\varphi \right )}\right)\right )$$

Repetimos el procedimiento, para calcular las segundas derivadas:


In [14]:
dM11_tt = simplify(diff(dM11_t,phi)*dphi_t)
dM12_tt = simplify(diff(dM12_t,phi)*dphi_t)
dM22_tt = simplify(diff(dM22_t,phi)*dphi_t)

In [15]:
dM11_tt,dM12_tt,dM22_tt


Out[15]:
$$\left ( \frac{2 a^{2} \mu \omega_{0}^{2}}{e^{2} - 1} \left(e \left(e \sin{\left (\varphi_{0} \right )} + \sin{\left (\varphi \right )}\right) \sin{\left (\varphi - \varphi_{0} \right )} \cos{\left (\varphi \right )} - \left(e \cos{\left (\varphi - \varphi_{0} \right )} + 1\right) \left(e \sin{\left (\varphi \right )} \sin{\left (\varphi_{0} \right )} + 2 \sin^{2}{\left (\varphi \right )} - 1\right)\right), \quad \frac{a^{2} \mu \omega_{0}^{2}}{2 e^{2} - 2} \left(2 e^{2} \sin{\left (2 \varphi_{0} \right )} + 5 e \sin{\left (\varphi + \varphi_{0} \right )} + e \sin{\left (3 \varphi - \varphi_{0} \right )} + 4 \sin{\left (2 \varphi \right )}\right), \quad \frac{2 a^{2} \mu \omega_{0}^{2}}{- e^{2} + 1} \left(e \left(e \cos{\left (\varphi_{0} \right )} + \cos{\left (\varphi \right )}\right) \sin{\left (\varphi \right )} \sin{\left (\varphi - \varphi_{0} \right )} + \left(e \cos{\left (\varphi - \varphi_{0} \right )} + 1\right) \left(e \cos{\left (\varphi \right )} \cos{\left (\varphi_{0} \right )} - 2 \sin^{2}{\left (\varphi \right )} + 1\right)\right)\right )$$

Finalmente, hacemos lo mismo, para calcular la tercera derivada temporal:


In [16]:
dM11_ttt = simplify(diff(dM11_tt,phi)*dphi_t)
dM12_ttt = simplify(diff(dM12_tt,phi)*dphi_t)
dM22_ttt = simplify(diff(dM22_tt,phi)*dphi_t)

In [17]:
dM11_ttt,dM12_ttt,dM22_ttt


Out[17]:
$$\left ( \frac{2 a^{2} \mu \omega_{0}^{3}}{\left(- e^{2} + 1\right)^{\frac{5}{2}}} \left(e \cos{\left (\varphi - \varphi_{0} \right )} + 1\right)^{2} \left(- \frac{e}{2} \sin{\left (\varphi - \varphi_{0} \right )} + \frac{5 e}{4} \sin{\left (\varphi + \varphi_{0} \right )} + \frac{3 e}{4} \sin{\left (3 \varphi - \varphi_{0} \right )} + 2 \sin{\left (2 \varphi \right )}\right), \quad - \frac{a^{2} \mu \omega_{0}^{3}}{2 \left(- e^{2} + 1\right)^{\frac{5}{2}}} \left(e \cos{\left (\varphi - \varphi_{0} \right )} + 1\right)^{2} \left(5 e \cos{\left (\varphi + \varphi_{0} \right )} + 3 e \cos{\left (3 \varphi - \varphi_{0} \right )} + 8 \cos{\left (2 \varphi \right )}\right), \quad - \frac{2 a^{2} \mu \omega_{0}^{3}}{\left(- e^{2} + 1\right)^{\frac{5}{2}}} \left(e \cos{\left (\varphi - \varphi_{0} \right )} + 1\right)^{2} \left(\frac{e}{2} \sin{\left (\varphi - \varphi_{0} \right )} + \frac{5 e}{4} \sin{\left (\varphi + \varphi_{0} \right )} + \frac{3 e}{4} \sin{\left (3 \varphi - \varphi_{0} \right )} + 2 \sin{\left (2 \varphi \right )}\right)\right )$$

Potencia promedio radiada

A continuación, calcularemos la potencia promedio radiada, evaluando la expresión $$ \left\langle P\right\rangle=\frac{G}{5 c^5} \left\langle\dddot{Q}{}^{ij}\dddot{Q}{}^{ij}\right\rangle $$ que, en nuestro caso de movimiento en el plano $xy$, se reduce a $$\left\langle P\right\rangle=\frac{2G}{15c^5}\, \left\langle \left(\dddot{M}_{11}\right)^2+ \left(\dddot{M}_{22}\right)^2+3\left(\dddot{M}_{12}\right)^2 -\dddot{M}_{11}\dddot{M}_{22}\right\rangle . $$

Primero, definiremos los símbolos de $G, c$ y $P$:


In [18]:
G = Symbol('G')
c = Symbol('c')
P = Symbol('P')

Y reemplazamos los valores de las derivadas del momento cuadrupolar en la expresión para la potencia (aún sin promediar!):


In [19]:
P = simplify(((2*G)/(15*c**5))*(dM11_ttt**2 + dM22_ttt**2 + 3*dM12_ttt**2 - dM11_ttt*dM22_ttt))

In [20]:
P


Out[20]:
$$- \frac{G a^{4} \mu^{2} \omega_{0}^{6}}{30 c^{5} \left(e^{2} - 1\right)^{5}} \left(e \cos{\left (\varphi - \varphi_{0} \right )} + 1\right)^{4} \left(3 \left(5 e \cos{\left (\varphi + \varphi_{0} \right )} + 3 e \cos{\left (3 \varphi - \varphi_{0} \right )} + 8 \cos{\left (2 \varphi \right )}\right)^{2} + \left(- 2 e \sin{\left (\varphi - \varphi_{0} \right )} + 5 e \sin{\left (\varphi + \varphi_{0} \right )} + 3 e \sin{\left (3 \varphi - \varphi_{0} \right )} + 8 \sin{\left (2 \varphi \right )}\right)^{2} + \left(- 2 e \sin{\left (\varphi - \varphi_{0} \right )} + 5 e \sin{\left (\varphi + \varphi_{0} \right )} + 3 e \sin{\left (3 \varphi - \varphi_{0} \right )} + 8 \sin{\left (2 \varphi \right )}\right) \left(2 e \sin{\left (\varphi - \varphi_{0} \right )} + 5 e \sin{\left (\varphi + \varphi_{0} \right )} + 3 e \sin{\left (3 \varphi - \varphi_{0} \right )} + 8 \sin{\left (2 \varphi \right )}\right) + \left(2 e \sin{\left (\varphi - \varphi_{0} \right )} + 5 e \sin{\left (\varphi + \varphi_{0} \right )} + 3 e \sin{\left (3 \varphi - \varphi_{0} \right )} + 8 \sin{\left (2 \varphi \right )}\right)^{2}\right)$$

Nuevamente, usando la regla de la cadena, podemos reescribir el promedio requerido en términos de una integral sobre el ángulo $\varphi$: \begin{align} \left\langle P(t)\right\rangle &= \frac{1}{T}\int_0^T P(t)\,dt \\ &= \frac{1}{T}\int_0^{2\pi} P(\varphi)\frac{dt}{d\varphi}\,d\varphi \\ &= \frac{1}{T}\frac{\mu}{L}\int_0^{2\pi} r^2(\varphi)P(\varphi)\,d\varphi \\ &= \frac{\mu a^2(1-e^2)^2}{TL}\int_0^{2\pi} \frac{P(\varphi)}{\left[1+e\cos(\varphi-\varphi_0)\right]^2}\,d\varphi \\ &= \frac{(1-e^2)^{3/2}}{2\pi}\int_0^{2\pi} \frac{P(\varphi)}{\left[1+e\cos(\varphi-\varphi_0)\right]^2}\,d\varphi. \end{align}


In [21]:
integrando = factor(simplify(((1-e**2)**(3/2))/(2*pi))*(P/(1+e*cos(phi-phi0))**2))

In [22]:
integrando


Out[22]:
$$\frac{G a^{4} \mu^{2} \omega_{0}^{6} \left(e \cos{\left (\varphi - \varphi_{0} \right )} + 1\right)^{2}}{60 \pi c^{5} \left(e - 1\right)^{4} \left(e + 1\right)^{4}} \left(4 e^{2} \sin^{2}{\left (\varphi - \varphi_{0} \right )} + 75 e^{2} \sin^{2}{\left (\varphi + \varphi_{0} \right )} + 90 e^{2} \sin{\left (\varphi + \varphi_{0} \right )} \sin{\left (3 \varphi - \varphi_{0} \right )} + 27 e^{2} \sin^{2}{\left (3 \varphi - \varphi_{0} \right )} + 75 e^{2} \cos^{2}{\left (\varphi + \varphi_{0} \right )} + 90 e^{2} \cos{\left (\varphi + \varphi_{0} \right )} \cos{\left (3 \varphi - \varphi_{0} \right )} + 27 e^{2} \cos^{2}{\left (3 \varphi - \varphi_{0} \right )} + 240 e \sin{\left (2 \varphi \right )} \sin{\left (\varphi + \varphi_{0} \right )} + 144 e \sin{\left (2 \varphi \right )} \sin{\left (3 \varphi - \varphi_{0} \right )} + 240 e \cos{\left (2 \varphi \right )} \cos{\left (\varphi + \varphi_{0} \right )} + 144 e \cos{\left (2 \varphi \right )} \cos{\left (3 \varphi - \varphi_{0} \right )} + 192 \sin^{2}{\left (2 \varphi \right )} + 192 \cos^{2}{\left (2 \varphi \right )}\right)$$

In [23]:
promedio_P = Symbol('PP')

In [24]:
import time

In [25]:
# tomamos el tiempo de este cálculo, porque es el que más tarda (~ 2 horas!)
st = time.time()
promedio_P =integrate(integrando,(phi,0,2*pi))
ft = time.time()
print(ft-st)


6605.178473

In [26]:
simplify(promedio_P)


Out[26]:
$$\frac{G a^{4} \mu^{2} \omega_{0}^{6} \left(37 e^{4} + 292 e^{2} + 96\right)}{15 c^{5} \left(e^{8} - 4 e^{6} + 6 e^{4} - 4 e^{2} + 1\right)}$$

Como en el apunte:

$$ g(\varphi) := 2 [1 + e \cos(\varphi − \varphi_{0} )]^4 [ 24 + 13e^2 + 48e \cos(\varphi − \varphi_{0} ) + 11e^2 \cos(2\varphi − 2\varphi_{0} )] $$$$ \left\langle g(\varphi)\right\rangle =\frac{(1 − e^2 )^{3/2}}{2\pi}\int_{0}^{2\pi} \frac{g(\varphi)}{[1 + e cos(\varphi − \varphi_{0})]^2} d\varphi $$

In [27]:
M = Symbol('M')
x = Symbol('x')
x0 = Symbol('x0')
g = Function('g')(x)
promedio_g = Symbol('Pg')

In [28]:
g = 2*((1 + e*cos(phi-phi0))**4)*(24 + 13*e**2 + 48*e*cos(phi-phi0) + 11*e**2*cos(2*phi-2*phi0))

In [29]:
g


Out[29]:
$$2 \left(e \cos{\left (\varphi - \varphi_{0} \right )} + 1\right)^{4} \left(11 e^{2} \cos{\left (2 \varphi - 2 \varphi_{0} \right )} + 13 e^{2} + 48 e \cos{\left (\varphi - \varphi_{0} \right )} + 24\right)$$
$$ \left\langle P\right\rangle=\frac{2 G^4 \mu^2 M^3}{15 c^5 a^5 (1 − e^2 )^5} \left\langle g(\varphi)\right\rangle $$

In [30]:
promedio_g = simplify(((1-e**2)**(3/2))/(2*pi))*integrate(g/(1+e*cos(phi-phi0))**2,(phi,0,2*pi))

In [31]:
trigsimp(promedio_g)


Out[31]:
$$\frac{1}{2} \left(- e^{2} + 1\right) \left(37 e^{4} + 292 e^{2} + 96\right)$$

Momentum angular radiado

Similarmente, para calcular el momento angular radiado usamos $$ \left\langle\dot{L}{}^{i}\right\rangle=\frac{2 G}{5 c^5} \epsilon^{ijk}\left\langle\ddot{M}{}^{ja}\ddot{M}{}^{ka}\right\rangle . $$

En nuestro caso, de movimiento en el plano $xy$, encontramos: \begin{align} \left\langle\dot{L}{}^{3}\right\rangle &= \frac{2 G}{5 c^5} \left\langle\epsilon^{31k}\ddot{M}{}^{1l}\dddot{M}{}^{kl}+\epsilon^{32k}\ddot{M}{}^{2l}\dddot{M}{}^{kl}\right\rangle\\ &= \frac{2 G}{5 c^5} \left\langle\epsilon^{312}\ddot{M}{}^{1l}\dddot{M}{}^{2l}+\epsilon^{321}\ddot{M}{}^{2l}\dddot{M}{}^{1l}\right\rangle \\ &= \frac{2 G}{5 c^5} \left\langle\ddot{M}{}^{11}\dddot{M}{}^{21}+\ddot{M}{}^{12}\dddot{M}{}^{22}-\ddot{M}{}^{21}\dddot{M}{}^{11}-\ddot{M}{}^{22}\dddot{M}{}^{12}\right\rangle \\ &=\frac{4 G}{5 c^5} \left\langle\ddot{M}{}^{12}(\dddot{M}{}^{22}-\dddot{M}{}^{11})\right\rangle . \end{align}


In [32]:
L_t = simplify(((4*G)/(5*c**5))*(dM12_tt*(dM22_ttt-dM11_ttt)))

In [33]:
L_t


Out[33]:
$$\frac{2 G a^{4} \mu^{2} \omega_{0}^{5}}{5 c^{5} \left(- e^{2} + 1\right)^{\frac{7}{2}}} \left(e \cos{\left (\varphi - \varphi_{0} \right )} + 1\right)^{2} \left(5 e \sin{\left (\varphi + \varphi_{0} \right )} + 3 e \sin{\left (3 \varphi - \varphi_{0} \right )} + 8 \sin{\left (2 \varphi \right )}\right) \left(2 e^{2} \sin{\left (2 \varphi_{0} \right )} + 5 e \sin{\left (\varphi + \varphi_{0} \right )} + e \sin{\left (3 \varphi - \varphi_{0} \right )} + 4 \sin{\left (2 \varphi \right )}\right)$$

El promedio se calcula de forma similar a como se hizo con la potencia, realizando un cambio de variable para finalmente integrar sobre el ángulo $\varphi$


In [34]:
integrando = (1-e**2)**(3/2)/(2*pi)*L_t/(1+e*cos(phi-phi0))**2

In [35]:
promedio_L_t = trigsimp(integrate(integrando,(phi,0,2*pi)))

In [36]:
promedio_L_t


Out[36]:
$$\frac{4 G a^{4} \mu^{2} \omega_{0}^{5} \left(7 e^{2} + 8\right)}{5 c^{5} \left(- e + 1\right)^{\frac{5}{2}} \left(e + 1\right)^{\frac{5}{2}}}$$