<p>Stars are far from perfect blackbodies due to scattering/reflection.  This is especially true for hotter stars, because of all the free electrons, but even cooler stars can reflect a significant amount.  For example, in aanda.org/articles/aa/pdf/2001/19/aa1009.pdf you will see that they use a reflection albedo of 0.30 for the K star and 1.00 for the F star, but the latter number is not meant to be taken seriously, they simply don't care if the light is reflected or absorbed and re-emitted because it isn't an important term. But the value of 0.30 for the K star might be meant more seriously, though it is still not regarded as a critical parameter because it only affects the color of the light that is reflected, not the total amount of light (given that stars are in radiative equilibrium, so must ultimately return all the incident light, whether it happens by reflection or heating).</p>

<p>Indeed, stellar emissions are often characterized by "effective temperature," to connect the surface flux of a star to the Stefan-Boltzmann formula for the emission of a blackbody by using a T parameter that is not necessarily the actual temperature.  When using this notion, as is quite common for dealing with stars, there is no essential difference between heating of and reflection from the surface of the star in question. The details of the difference have to do with the shape of the spectrum, but that shape is generally not a Planck function anyway, so as soon as one is using the "effective temperature" concept one has already parted company from a detailed understanding of the shape of the spectrum.  (When you do want the details of the spectrum, you will have to model the situation with some care.)</p>

<p>You are right that the albedo depends on the type of star (hotter stars having high scattering albedo due to all those free electrons), and many stars do show a lot of scattering from their surfaces.  Scattering/reflection elevates the effective temperature above the actual temperature of the layers being looked at, but again, if you stick to effective temperature, it makes no difference if there is scattering or heating of a blackbody, because effective temperature explicitly refers to the outgoing  flux with no claims on the actual temperature.  It is quite common for this distinction to be swept under the rug, and most stellar temperatures are effective temperatures, not actual temperatures.</p>

<p>Another consequence of radiative equilibrium is that the "reflection effect" cannot alter the total luminosity of the binary system seen from all angles, so if the effective temperature of the surface of the stars is elevated by reflection and/or heating, as seen from some directions, this must be exactly compensated by the reduced brightness due to eclipses seen from other directions.  So the reflection effect is part of the study of eclipse light curves, which are quite useful for understanding things like the sizes of the stars.</p>

<p>Added:  a nice paper on the polarization of reflected light was provided in the comments by uhoh, it is  nature.com/articles/s41550-019-0738-7 .  The idea is that if the stars are not too close to each other, there's not a lot of reflected light, but it is still noticeable by virtue of the fact that it is highly linearly polarized in the direction perpendicular to the line between the stars.  So it shows up much better in linearly polarized light, a potentially important new diagnostic of binary systems.</p>

CC BY-SA 4.0
2018-05-04T19:51:51.743Z
I don't see anything here that mentions a value for a reflectivity yet. 1%? 99% 1E-7? This is more of an essay stating "it depends", than answer to the question as asked. Can you fortify it with at least a rough value?
uhoh
CC BY-SA 4.0
2018-05-05T04:37:53.580Z
It depends on the degree of ionization, but a ballpark is at least 50% reflective for hot stars and considerably less for cool stars.  It is also wavelength dependent.  But my point was, the difference may not be nearly as important as you might think.
Ken G
CC BY-SA 4.0
2018-05-04T19:15:51.440Z
2019-05-01T04:22:07.453Z
2019-05-01T04:22:07.453Z
Ken G
<p>This is a long-standing (and complicated!) problem in modeling eclipsing binary light curves. <a href="https://ui.adsabs.harvard.edu/abs/1985Ap%26SS.113..349V/abstract" rel="nofollow noreferrer">Here’s a full review</a> from 1985 (with later citations <a href="https://ui.adsabs.harvard.edu/abs/1985Ap%26SS.113..349V/citations" rel="nofollow noreferrer">here</a>), which suggests (from a very quick skim) that an albedo of about 0.5 has been found for some systems, but the details depend also on temperature and wavelength.</p>
<p>For the current state of the art in modeling this, you could look at papers and documentation for the <a href="https://ui.adsabs.harvard.edu/abs/2016ApJS..227...29P/abstract" rel="nofollow noreferrer">PHOEBE eclipsing binary code</a>.</p>

CC BY-SA 4.0
2020-07-09T21:27:15.707Z
This is very helpful, *thank you!* From your [list of citations](https://ui.adsabs.harvard.edu/abs/1985Ap%26SS.113..349V/citations) of [Vaz 1985](https://ui.adsabs.harvard.edu/abs/1985Ap%26SS.113..349V/abstract) I found Heber et al. 2004 [HS 2333+3927: A new sdB+dM binary with a large reflection effect](https://ui.adsabs.harvard.edu/abs/2004A%26A...420..251H/abstract). Table 2 gives two solutions with $T_1$ = 36,000 K and an albedo of 1.0 and $T_2$ = 3,800 K and albedo of 0.39 or  $T_2$ = 3,400 K and albedo of 1.0  so it seems that average reflectance can be very high!
uhoh
CC BY-SA 4.0
2020-07-09T17:57:37.587Z
2020-07-09T17:57:37.587Z
Eric Jensen
<p>Complementing @KenG's <a href="https://astronomy.stackexchange.com/a/26173/7982">answer</a>, Here's an actual datapoint.</p>
<p>The new paper in Nature <a href="https://www.nature.com/articles/s41550-019-0738-7" rel="nofollow noreferrer">Polarized reflected light from the Spica binary system</a> (downloadable <a href="https://www.researchgate.net/publication/332133160_Polarized_reflected_light_from_the_Spica_binary_system" rel="nofollow noreferrer">here</a>) is notable in that the measurement of the <em>polarized component of the reflected light speaks to it being an actual reflection</em>, rather than one star heating the other producing a more brightly radiating area.</p>
<p>Their model of the Spica system uses mostly adopted parameters, but of the derived parameters, the <em>geometric albedo</em> of the A and B components are about 3.6% and 1.4% respectively.</p>
<p>You can read further about Bond albedo and geometric albedo in @zephyr's <a href="https://astronomy.stackexchange.com/a/20803">excellent answer</a> to the question <em>Why is Enceladus's albedo greater than 1?</em></p>

CC BY-SA 4.0
2019-05-01T10:23:35.223Z
2020-07-09T08:41:10.030Z
2020-07-09T08:41:10.030Z
uhoh
<p>Ignore the Blender Monkey! (from <a href="https://blender.stackexchange.com/q/51585/5334">here</a>)</p>
<p><a href="https://i.sstatic.net/0SYjI.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/0SYjI.gif" alt="from https://blender.stackexchange.com/q/51585/5334" /></a></p>
<p>In a close-orbiting binary pair, a small but significant fraction of the light from each star falls upon the other, and the result has to be carefully modeled to simulate the observed spectrum of the system as it rotates.</p>
<p>One effect is that each star's radiation will non-uniformly heat the surface of the other star and alter its thermal radiation, but here I'm asking about light from one star being scattered or diffusely reflected by the other.</p>
<p>How is this scattered/reflected light calculated, and roughly what is the spectral reflectance?</p>
<p>Of course each star will be different, so one example or maybe two would be sufficient.</p>
<p>We often treat stars as blackbodies as an approximation, suggesting that stars have low albedo, but at some depth the electron density will be high enough so that the plasma frequency reaches visible wavelengths, increasing reflectivity, though that may be too deep to matter.</p>

CC BY-SA 3.0
2018-04-06T14:57:20.040Z
My suspicion is that stars really are incredibly close to black bodies, so the fraction of reflected photons is so small as to be lost in the noise.
Carl Witthoft
CC BY-SA 4.0
2018-05-04T09:08:37.253Z
@CarlWitthoft one does not necessarily follow the other. Consider for example hot glass, while it radiates a blackbody-like spectrum, it is simultaneously nearly transparent! https://physics.stackexchange.com/q/254099/83380
uhoh
CC BY-SA 4.0
2018-05-04T09:24:14.987Z
See @RobJeffries answer there in particular.
uhoh
CC BY-SA 4.0
2018-05-04T10:08:14.697Z
@CarlWitthoft They can still be pseudo-blackbodies but that doesn't mean that every part you can see is at the same temperature. The reflection effect is most definitely a "thing" that is considered when modelling the spectra and light curves of close binaries.
ProfRob
<star><binary-star><albedo><eclipsing-star-systems>
What is the spectral reflectance of starlight in a close binary?
uhoh
# What is the spectral reflectance of starlight in a close binary?

(Asked by: username_0 on 2018-04-05)

Ignore the Blender Monkey! (from [here](https://blender.stackexchange.com/q/51585/5334))

[![from https://blender.stackexchange.com/q/51585/5334](https://i.sstatic.net/0SYjI.gif)](https://i.sstatic.net/0SYjI.gif)

In a close-orbiting binary pair, a small but significant fraction of the light from each star falls upon the other, and the result has to be carefully modeled to simulate the observed spectrum of the system as it rotates.

One effect is that each star's radiation will non-uniformly heat the surface of the other star and alter its thermal radiation, but here I'm asking about light from one star being scattered or diffusely reflected by the other.

How is this scattered/reflected light calculated, and roughly what is the spectral reflectance?

Of course each star will be different, so one example or maybe two would be sufficient.

We often treat stars as blackbodies as an approximation, suggesting that stars have low albedo, but at some depth the electron density will be high enough so that the plasma frequency reaches visible wavelengths, increasing reflectivity, though that may be too deep to matter.

**Tags:** star, binary-star, albedo, eclipsing-star-systems
> **Comments:**
> **username_1**: My suspicion is that stars really are incredibly close to black bodies, so the fraction of reflected photons is so small as to be lost in the noise.  
> **username_0**: @CarlWitthoft one does not necessarily follow the other. Consider for example hot glass, while it radiates a blackbody-like spectrum, it is simultaneously nearly transparent! https://physics.stackexchange.com/q/254099/83380  
> **username_0**: See @RobJeffries answer there in particular.  
> **username_2**: @CarlWitthoft They can still be pseudo-blackbodies but that doesn't mean that every part you can see is at the same temperature. The reflection effect is most definitely a "thing" that is considered when modelling the spectra and light curves of close binaries.  

---

## 3 answers

### Answer by username_3 on 2018-05-04. Score: 4

Stars are far from perfect blackbodies due to scattering/reflection. This is especially true for hotter stars, because of all the free electrons, but even cooler stars can reflect a significant amount. For example, in aanda.org/articles/aa/pdf/2001/19/aa1009.pdf you will see that they use a reflection albedo of 0.30 for the K star and 1.00 for the F star, but the latter number is not meant to be taken seriously, they simply don't care if the light is reflected or absorbed and re-emitted because it isn't an important term. But the value of 0.30 for the K star might be meant more seriously, though it is still not regarded as a critical parameter because it only affects the color of the light that is reflected, not the total amount of light (given that stars are in radiative equilibrium, so must ultimately return all the incident light, whether it happens by reflection or heating).

Indeed, stellar emissions are often characterized by "effective temperature," to connect the surface flux of a star to the Stefan-Boltzmann formula for the emission of a blackbody by using a T parameter that is not necessarily the actual temperature. When using this notion, as is quite common for dealing with stars, there is no essential difference between heating of and reflection from the surface of the star in question. The details of the difference have to do with the shape of the spectrum, but that shape is generally not a Planck function anyway, so as soon as one is using the "effective temperature" concept one has already parted company from a detailed understanding of the shape of the spectrum. (When you do want the details of the spectrum, you will have to model the situation with some care.)

You are right that the albedo depends on the type of star (hotter stars having high scattering albedo due to all those free electrons), and many stars do show a lot of scattering from their surfaces. Scattering/reflection elevates the effective temperature above the actual temperature of the layers being looked at, but again, if you stick to effective temperature, it makes no difference if there is scattering or heating of a blackbody, because effective temperature explicitly refers to the outgoing flux with no claims on the actual temperature. It is quite common for this distinction to be swept under the rug, and most stellar temperatures are effective temperatures, not actual temperatures.

Another consequence of radiative equilibrium is that the "reflection effect" cannot alter the total luminosity of the binary system seen from all angles, so if the effective temperature of the surface of the stars is elevated by reflection and/or heating, as seen from some directions, this must be exactly compensated by the reduced brightness due to eclipses seen from other directions. So the reflection effect is part of the study of eclipse light curves, which are quite useful for understanding things like the sizes of the stars.

Added: a nice paper on the polarization of reflected light was provided in the comments by uhoh, it is nature.com/articles/s41550-019-0738-7 . The idea is that if the stars are not too close to each other, there's not a lot of reflected light, but it is still noticeable by virtue of the fact that it is highly linearly polarized in the direction perpendicular to the line between the stars. So it shows up much better in linearly polarized light, a potentially important new diagnostic of binary systems.

> **Comments:**
> **username_0**: I don't see anything here that mentions a value for a reflectivity yet. 1%? 99% 1E-7? This is more of an essay stating "it depends", than answer to the question as asked. Can you fortify it with at least a rough value?  
> **username_3**: It depends on the degree of ionization, but a ballpark is at least 50% reflective for hot stars and considerably less for cool stars.  It is also wavelength dependent.  But my point was, the difference may not be nearly as important as you might think.  

---

### Answer by username_4 on 2020-07-09. Score: 1

This is a long-standing (and complicated!) problem in modeling eclipsing binary light curves. [Here’s a full review](https://ui.adsabs.harvard.edu/abs/1985Ap%26SS.113..349V/abstract) from 1985 (with later citations [here](https://ui.adsabs.harvard.edu/abs/1985Ap%26SS.113..349V/citations)), which suggests (from a very quick skim) that an albedo of about 0.5 has been found for some systems, but the details depend also on temperature and wavelength.

For the current state of the art in modeling this, you could look at papers and documentation for the [PHOEBE eclipsing binary code](https://ui.adsabs.harvard.edu/abs/2016ApJS..227...29P/abstract).

> **Comments:**
> **username_0**: This is very helpful, *thank you!* From your [list of citations](https://ui.adsabs.harvard.edu/abs/1985Ap%26SS.113..349V/citations) of [Vaz 1985](https://ui.adsabs.harvard.edu/abs/1985Ap%26SS.113..349V/abstract) I found Heber et al. 2004 [HS 2333+3927: A new sdB+dM binary with a large reflection effect](https://ui.adsabs.harvard.edu/abs/2004A%26A...420..251H/abstract). Table 2 gives two solutions with $T_1$ = 36,000 K and an albedo of 1.0 and $T_2$ = 3,800 K and albedo of 0.39 or  $T_2$ = 3,400 K and albedo of 1.0  so it seems that average reflectance can be very high!  

---

### Answer by username_0 on 2019-05-01. Score: 1

Complementing @KenG's [answer](https://astronomy.stackexchange.com/a/26173/7982), Here's an actual datapoint.

The new paper in Nature [Polarized reflected light from the Spica binary system](https://www.nature.com/articles/s41550-019-0738-7) (downloadable [here](https://www.researchgate.net/publication/332133160_Polarized_reflected_light_from_the_Spica_binary_system)) is notable in that the measurement of the *polarized component of the reflected light speaks to it being an actual reflection*, rather than one star heating the other producing a more brightly radiating area.

Their model of the Spica system uses mostly adopted parameters, but of the derived parameters, the *geometric albedo* of the A and B components are about 3.6% and 1.4% respectively.

You can read further about Bond albedo and geometric albedo in @zephyr's [excellent answer](https://astronomy.stackexchange.com/a/20803) to the question *Why is Enceladus's albedo greater than 1?*


---

