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python - 如何使用 python astropy 将 arcsec 转换为 Mpc？

问题描述

我需要将以下数量从弧秒转换为兆秒差距：

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 et al 2016 https://arxiv.org/pdf/1502.01589.pdf table1 2013F(DS)

到目前为止我已经尝试过了，

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.

替代方案： http: //arcsec2parsec.joseonorbe.com/index.html

这有效并提供 3.38 Mpc，但我不能简单地引用一个网站，希望使用 python 重现结果。

标签： pythonastropy

解决方案

要找到距离，您需要将角直径距离乘以角大小。

l = D_A(z) × θ

参考：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，我们也可以使用天文单位，

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

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