如何使用将arcsec转换为Mpc？

Question

我需要将下列数量从弧秒转换为超大比例尺：

a = 737.28 # arcsec
z = 0.3 # redshift
d = ? # Mpc

I am using flat lambdaCDM using parameters
H0 = 67.8
omega_m = 0.308

使用的宇宙学: Ade等2016年https://arxiv.org/pdf/1502.01589.pdf table1 2013F(DS)

我已经试过这个了，

from astropy.cosmology import FlatLambdaCDM
import astropy.units as u

cosmo = FlatLambdaCDM(H0=70, Om0=0.3)
cosmo.luminosity_distance(z=0.3)

# I am not sure how to convert arcsec to Mpc here.

替代品：http://arcsec2parsec.joseonorbe.com/index.html

这工作和提供3.38MPC，但我不能简单地引用一个网站，它是希望使用python复制结果。

Answer 1

要找到距离，需要将角直径的距离乘以角的大小。

L= D_A(z)×θ

参考资料：http://arcsec2parsec.joseonorbe.com/about.html

from astropy.cosmology import FlatLambdaCDM
import numpy as np
cosmo = FlatLambdaCDM(H0=67.8, Om0=0.308)

# angular diameter distance in Mpc
d_A = cosmo.angular_diameter_distance(z=0.3)
theta = 737.28 # arcsec

# pi radian = 180 degree ==> 1deg = pi/180 ==> 1arcsec = pi/180/3600
theta_radian = theta * np.pi / 180 / 3600

# arc length = radius * angle
distance_Mpc = d_A * theta_radian

print(distance_Mpc) # 3.3846475 Mpc

更新

正如评论中所建议的，我们也可以使用不稳定的单位，

from astropy.cosmology import FlatLambdaCDM
import numpy as np
from astropy import units as u
cosmo = FlatLambdaCDM(H0=67.8, Om0=0.308)

d_A = cosmo.angular_diameter_distance(z=0.3)
print(d_A) # 946.9318492873492 Mpc

theta = 737.28*u.arcsec
distance_Mpc = (theta * d_A).to(u.Mpc, u.dimensionless_angles()) # unit is Mpc only now

print(distance_Mpc) # 3.384745689510495 Mpc