Empirical

Stellar inlicnation from empirical relation

Let’s use the relation in Louden et al. (2021) to calculate the stellar rotation speed at the equator, \(v\), and subsequently the stellar inclination, \(i_\star\), using the projected stellar rotation speed, \(v \sin i_\star\), and effective temperature, \(T_{\rm eff}\).

We’ll assume we have measured \(T_{\rm eff}=6450 \pm 100\) K and \(v \sin i_\star=6.1 \pm 0.5\) km/s.

import coPsi
## Instantiate iStar with values for Teff and vsini
incs = coPsi.iStar(Teff=(6450,100,0,7000,'gauss'),vsini=(6.1,0.5,0,20,'gauss'))

Before we do anything, let’s see how these values compare to the relation:

## Compare to relation from Louden
incs.plotLouden(Teff=6450,vsini=6.1,sTeff=100,svsini=0.5)

Given those values, we’ll create some (normal/gaussian) distributions for these variables and then calculate \(i_\star\).

## Create distributions for Teff and vsini
incs.createDistributions()
## Calculate the stellar incliation using the Louden et al. (2021) relation
incs.stellarInclinationLouden()

/Users/emilkn/Library/CloudStorage/OneDrive-Chalmers/Desktop/postdoc/coPsi/src/coPsi/coPsi.py:242: RuntimeWarning: invalid value encountered in arcsin
  incs = np.rad2deg(np.arcsin(si))

Let’s look at the results

## Calculate median and credible interval and plot the resulting KDE
incs.diagnostics('incs')

Median and confidence level (0.68 credibility):
incs=20.772+4.588-6.840

If we have measures for the obliquity, \(\lambda\), and the orbital inclination, \(i_{\rm o}\), we can calculate the obliquity, \(\psi\). Here we’ll create distributions externally (which could have come from an MCMC), but they could also be created like before.

import numpy as np
## Let's say we also have distributions for lambda and the orbital inlination
## of a length similar to the distributionk for the stellar inclination
## (here we'll simulate them)
incs.dist['lam'] = np.random.normal(-192,5,len(incs.dist['incs']))
incs.dist['inco'] = np.random.normal(81.81,0.12,len(incs.dist['incs']))

## Now let's calculate the obliquity psi
incs.coPsi()
incs.diagnostics('psi')

Median and confidence level (0.68 credibility):
psi=102.097+4.307-6.902