# Why the habitable zone of a binary star is a disc and not a sphere?

I understand that the orbit of a planet will always be flat, so there is no point on making a sphere out of the HZ when we consider one star, but what about binary stars? For example, this binary system (taken from here):

The habitable zone of this binary surrounds both stars, but what if the orbit of the planet isn't contained in the binary's orbit plane? For example, orbiting the star on the left with an inclination of, say, 20 degrees? Wouldn't it have to be two spheres and not two disks? Or is the HZ calculated assuming coplanar orbits?

• Personally I consider HZ as a sphere. At least if there is not a planetary system around the star. Of course the only planet will be orbiting on a plane. – Alchimista Oct 15 '20 at 9:54

If you go to the associated paper Müller & Haghighipour (2014), it shows that what they're plotting are the points which satisfy:

$$\frac{L_\odot}{l_{\rm In-Sun}^2} \le \sum_{i=1}^{N} W_i (T_{\rm star}) \frac{L_i / L_\odot}{d_i^2} \le \frac{L_\odot}{l_{\rm Out-Sun}^2}$$

where:

• $$L_\odot$$ is the luminosity of the Sun
• $$L_i$$ is the luminosity of the $$i$$th star (of a total $$N$$ stars)
• $$l_{\rm In-Sun}$$ and $$l_{\rm Out-Sun}$$ are the inner and outer boundaries of the Sun's habitable zone (two sets of distances are used for the conservative and extended habitable zones, plotted as dark and light green respectively)
• $$d_i$$ is the distance from the point to the $$i$$th star
• $$W_i(T_{\rm star})$$ is a weight factor that accounts for the different spectral energy distribution at different temperatures $$T_{\rm star}$$. Details of the calculation of this weight factor are given in the paper.

This formula does not rely on the objects being co-planar. If the binary were inclined relative to the plane of the plot, the $$d_i$$ values would be altered at each plotted point to account for the component of the distance out of the plane, but essentially this would just be a different slice of a 3-dimensional volume.

• I am aware of the formula. If $d_i$ is the distance from the planet to the star, wouldn't that be a sphere? Because the planet, if it's inclined, coud have been found in any point of the space away from the binary plane. Maybe the disk is a projection of the sphere in the binary plane... – Carlos Vázquez Monzón Oct 15 '20 at 15:58
• @CarlosVázquezMonzón - for the zone around a single star, the inner and outer boundaries would be spherical. Add a second star and the region becomes non-spherical (although in the case where the stars are located very close together, the circumbinary region would be very close to spherical, while if the stars are sufficiently far apart, the region around each star would also be very close to spherical). Note that $d_i$ is the distance from the point in space to each of the stars at a given time, it isn't an orbital distance. – user24157 Oct 15 '20 at 16:27
• Yeah I meant 3-dimensional, close to a sphere in the example above. So, why is presented as 2-dimensional? – Carlos Vázquez Monzón Oct 15 '20 at 16:31
• @CarlosVázquezMonzón - probably because it's easier to do the visualisation, particularly since it's being depicted on a 2-dimensional computer screen (or sheet of paper if you decide to print it for some reason, etc.). If the stars are at intermediate distances, you end up with a peanut-shaped blob with voids in the middle, which isn't so easy to depict. – user24157 Oct 15 '20 at 16:35
• But can't that be a little misleading? For example, let say a planet is, being the plane of reference that of the binary plane, inside the disk, but in the z-axis is really really far away from any of the stars, so it's not really in the habitable zone. But, if we represent the HZ as a 3-dimensional object, you would see that it's, indeed, not really inside of it. – Carlos Vázquez Monzón Oct 15 '20 at 17:15