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ericagol avatar ericagol commented on August 26, 2024 1

If both stars are isotropic, then by definition they have the same flux for every observer. So, when you integrate the specific intensity over the visible disk (and over wavelength if the bolometric luminosities are equal; otherwise you require the specific luminosities to be equal), then the fluxes must match.

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ericagol avatar ericagol commented on August 26, 2024 1

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rodluger avatar rodluger commented on August 26, 2024

Stellar radius and luminosity are now fixed to unity. I'm going to tackle the normalization issue next.

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rodluger avatar rodluger commented on August 26, 2024

Case in point:

import matplotlib.pyplot as pl
import starry
import numpy as np

star = starry.Star()
planet = starry.Planet(L=1e-2, r=0.1, a=3)
sys = starry.System([star, planet])
time = np.linspace(0.25, 1.25, 10000)
sys.compute(time)
pl.plot(time, sys.flux / np.nanmedian(sys.flux), label="Uniform")

star[1] = 0.4
star[2] = 0.26
sys.compute(time)
pl.plot(time, sys.flux / np.nanmedian(sys.flux), label="Limb-Darkened")

pl.legend()
pl.show()

image

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ericagol avatar ericagol commented on August 26, 2024

The luminosity is equal to the integral over the surface brightness of the l=0 m=0 integral.

To see this, integrate the spherical harmonic over the sphere:

image

Now, if l' = m' = 0, then this just gives the integral of the spherical harmonic over the sphere (times a constant), so the only spherical harmonic with a non-zero mean flux (integrated over the sphere) is l=m=0.

All to say: the total luminosity is just governed by the l=0 spherical harmonic.

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rodluger avatar rodluger commented on August 26, 2024

OK, I think I understand what I'm doing wrong. Currently my Y_{0,0} coefficient depends on the limb darkening coefficients:
image
because that's what I get when I express
image
in spherical harmonics. So I just need to normalize the map so the luminosity doesn't change when the limb darkening coefficients are changed.

But I'm still a bit confused about this, because the stellar brightness isn't really a sum of the Y_{l,0} harmonics -- it just looks that way from our vantage point. If we go around to the backside of the star, we don't see negative flux! So I'm not convinced the limb darkening terms integrate to zero over all 4 * pi steradians. Does that make sense?

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ericagol avatar ericagol commented on August 26, 2024

Right - there is some sleight-of-hand here, and so it depends on what quantity you are interested in.

If you want the apparent luminosity, then you should integrate the observed flux and multiply by 4\pi d^2. If you want the total luminosity (i.e. sum of fluxes of observers at same distance distributed over 4\pi steradians), then it's only the monopole that counts.

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rodluger avatar rodluger commented on August 26, 2024

Ok, let me ask you a specific question then: say I have two solar-type stars with the same total luminosity, but one is strongly limb-darkened and the other isn't. When I integrate the specific intensity of each one over their visible disk, should I get the same value for both? Or should the limb-darkened one be dimmer?

This will help me decide how to normalize the units...

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rodluger avatar rodluger commented on August 26, 2024

Ha, turns out I had done this before! From Equation (67) in our PPO paper, all we need to do is multiply the flux by
image
to ensure the integral over the visible disk is unity.

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rodluger avatar rodluger commented on August 26, 2024

Actually the above expression normalizes the flux to pi. Anyways, this issue is now resolved. The stellar flux is now always unity.

import matplotlib.pyplot as pl
import starry
import numpy as np

star = starry.Star()
planet = starry.Planet(L=1e-2, r=0.1, a=3)
sys = starry.System([star, planet])
time = np.linspace(0.25, 1.25, 10000)
sys.compute(time)
pl.plot(time, sys.flux, label="Uniform")

star[1] = 0.4
star[2] = 0.26
sys.compute(time)
pl.plot(time, sys.flux, label="Limb-Darkened")

pl.legend()
pl.show()

image

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