Determining blackbody temperatures in Binary Systems (help!)

[Malenfant]

Another somewhat technical/specific question from me that I'm hoping a nice astrophysicist can answer for me... (paging Selden/Fridger/Grant?!)

I've recently been trying to figure out temperature variations in binary systems. So far I've figured out that the blackbody temperature at a given distance from the centre of mass (COM) of the system can be calculated by adding the [L/16(pi)(sigma)(D^2)] values for each star and then taking the fourth root of the sum (L = luminosity of given star in W, D = distance of planet from COM in m). (For solo stars you just take the fourth root of the equation and you have the blackbody temperature at that distance)

I'm trying to figure out the distance to the "frost line" though and I'm stumped - this is the distance at which the temperature of a blackbody would be 175K. I know that for a solo star you can just 175K into the blackbody equation and make D the subject to find the distance... but how do you do that for binary stars?

What I'm after is an equation that tells me the distance from the COM at which the blackbody temperature is 175K. I suspect that I don't just add the luminosities of the stars together linearly and use that total as the L value to figure that out though, I have a feeling I'll have to merge them some other way... anyone got any ideas?

[selden]

I'm not an astrophysicist and my math skills aren't the best, but...

To first approximation you'd just add them, but that assumes both stars stay at the same distance from the planet and always present the same optical profile. However, don't forget that the distances change between the individual stars and a planet in a binary system and the profile of close binaries changes, so it's not really going to be that simple a relationship.

[Malenfant]

Yeah, so far I've got:

T = fourth root of [(L1/16pi.sigma.(R1^2)) + (L2/16pi.sigma.(R2^2))]

where R1 and R2 are the distance between the planet and each star, and L1 and L2 are the luminosities of each star, and T should be set to 175 K.

Problem is, R1 and R2 are calculated using nasty trigonometry:

R1 = SQRT[(R+D1sin(90-th2)^2)+(D1cos(90-th2))^2]
R2= SQRT[(R-D2cos(th2))^2+(D2sin(th2))^2]

where R = distance between planet and COM, D1 = distance of Star1 from COM, D2 = distance of Star2 from COM, th2 = angle between Star2 and planet (0 means Star2 is directly between planet and COM, 180 means Star2 is directly opposite the COM relative to the planet, angle is measure clockwise)

It's beyond my poor equation rearranging ability to substitute those into the top equation and get things solely in terms of R (which is what I want). If I have time later tonight I'll see if I can put this all on a diagram to make it clearer.

[PlutonianEmpire]

Question: will this be implemented in Celestia? I ask because I have a system with a pair that orbits each other tightly, making all the planets in the system orbit the barycenter. First thing i noticed was that all the worlds' temps were zero (because they were orbiting the barycenter instead of a star)

[Malenfant]

Here's an easier way to see things... This image is four snapshots of the same system taken at different times, centred on the planet:



Star1 is the larger white star, Star2 is the smaller yellow one. The planet is at the bottom of the image. All are orbiting a common barycentre (the white dot). D1 and D2 are the distances of Star1 and Star2 from the Barycentre (their orbits are assumed to be circular). R is the distance of the planet from the Barycentre, and R1 and R2 are the distances from the planet to Star1 and Star2 respectively. The angle (theta) is the angle between Star2 and the planet (measured clockwise) - in case(4) theta = 0 degrees, in case(2) theta = 270 degrees, in case(3) theta = 180 degrees, in case(1) theta = 240 degrees.

The general case is (1) (on the left of the image), and you can calculate the distance between the planet and each star (R1 and R2 respectively) by:
(1)

From this, it should be fairly straightforward to see how to calculate R1 and R2 for cases (2), (3), and (4):
(2)

(3)

(4)


Finally, this is the equation to calculate the blackbody temperature at the planet's orbital distance from the Barycentre.



What I need to know is how to calculate the distance from the Barycentre at which I can find a specific blackbody temperature (175K). In other words, I need to substitute the R1 and R2 equations from (1) into the Temperature equation, and then make R the subject of the Temperature equation.

I have a feeling this is going to be horrendously complicated...

[bdm]

I have a feeling this is going to be horrendously complicated...

It is.

I have an idea that may help you to calculate a good approximation. If I get stuck with mathematical complexities like these, I usually get good results by using brute force to get a solution.

If you have access to a spreadsheet then try the following:

Create a new blank spreadsheet. Place a starting value in A1 and an increment in B1. Place distances in one cell (A2), and black body temperatures in another (B2), computed off the first cell (A2). Make A2 equal to the value in A1. Now duplicate B2 into B3. For A3, make it equal to A1 + $B$1 (absolute cell reference for B1 cell). Then duplicate the second row down the page so you have several of them. About 10 to 20 will work well.

Now you can brute force a solution. Enter in suitable starting values into A1 and B1, and you'll find that the spreadsheet will calculate the blackbody temperatures at each distance. If you choose good values you will find that there is a point where one line has a value higher than the desired temperature and one has a value that is lower. You now choose new values for A1 and B1 and repeat. Each iteration should improve your accuracy by about 1 decimal place.

If your spreadsheet has a Goal Seek function (most do), then you can simply Goal Seek to a good approximation. After using Goal Seek, you can then fine tune the values as desired.

[Malenfant]

Just had a wacky thought... take a look at these equations:



The top equation here is basically the one to calculate the temperature at a given distance (R) from a single star (Lc would be the luminosity of the single star).

So... could I just calculate the combined luminosity (Lc) of the two stars by using the second equation, and then use the first equation to find the distance from the Barycentre at which that temperature occurs?

If I use that second equation then if the luminosities are similar the combined luminosity is noticeably larger than that of each individual star - e.g. if both stars are luminosity of 1 Sol then the combined luminosity would be 1.19 Sols. If one star is significantly brighter than the other then the combined luminosity is much closer to that of the brighter star - eg if one star is 5 Sols and the other is 1 Sol, then the combined luminosity would be 5.002 Sols.

Would this make any kind of sense as an approximation? Or is it totally incorrect? It seems to make more sense to me than just linearly adding the two luminosities.

[t00fri]

Malenfant,

I (slightly) corrected, solved and plotted what you were interested in with the help of Maple. The first thing to do is to introduce suitable scaled quantities which reduce the parameters to few and the variables to three.

The two parameters are:

Code: Select all

lambda = sqrt(L2/L1)
w = 4*Pi*r^2*sigma*T^4/sqrt(L1*L2)


The reduced variables are

Code: Select all

rho1 = r/d1
rho2 = r/d2
theta


w should be close to w ~ 1, since w=1 corresponds to the Stephan Boltzman law for an effective star sitting in the center of gravity of the binary system. That effective star has a luminosity sqrt(L1*L2) at a distance r from the center of gravity, i.e. the geometrical mean of the two luminosities!

Here is my Maple worksheet with comments and two representative plots. Note, in these plots I did precisely what you wanted: I prescribed w, i.e. the temperature and looked for patches of solutions in the space of variables ( rho1, rho2, theta)

Worksheet:
http://www.celestiaproject.net/~t00fri/images/malenfant.pdf





Enjoy,
Bye Fridger

PS: if you want to see more configurations, I can easily plot them for you.

[Malenfant]

Awesome, thanks Fridger! I did try plugging it into Maple myself but couldn't get any kind of answer that made sense (got lots of "RootOf"), but then I wasn't reducing the parameters...

I'll take a proper look at it later on this evening after work, but a quick question now: do your comments about "w" mean that I could actually just calculate the distance at which the temperature is 175K by assuming that I have a single star with a luminosity equal to the geometric mean of the two luminosities at the centre of the system? (instead of the fourth root of the sum of the luminosities to the power of four as I guessed earlier?)

PS. what do the colours on the graph indicate?

[t00fri]

Awesome, thanks Fridger! I did try plugging it into Maple myself but couldn't get any kind of answer that made sense (got lots of "RootOf"), but then I wasn't reducing the parameters...

I'll take a proper look at it later on this evening after work, but a quick question now: do your comments about "w" mean that I could actually just calculate the distance at which the temperature is 175K by assuming that I have a single star with a luminosity equal to the geometric mean of the two luminosities at the centre of the system? (instead of the fourth root of the sum of the luminosities to the power of four as I guessed earlier?)

PS. what do the colours on the graph indicate?

Ah, you got maple!? Then you can directly download my worksheet HERE:

http://www.celestiaproject.net/~t00fri/images/malenfant.mw

But this is the latest Maple 10, so you might get some (unimportant warnings) with older versions.

I actually made one or two trivial modifications (eliminating typos...).
The above version is the latest one.

The colors in the plots are just to make them better looking . A standard Maple feature.

Actually you got it quite right: by defining w (~ 1) means perturbing the problem relative to a good 1st approximation which is always a good idea...

You also see that the asymptotic solution for rho1 -> infinity and rho2 -> infinity is w = sqrt(lambda) + 1/sqrt(lambda) > 2 with the minimum w=2 attained at lambda=1 or L1=L2.
Actually there are NO solutions for w<1 for any values of rho1,rho2, theta

This problem has a rich variety of quite different solutions in fact....

Bye Fridger

[Malenfant]

Great, thanks again Fridger! I have Maple 8, but I'll try to plug your worksheet into it and see what happens and play around with it tonight.

[Malenfant]

Ooops. Fridger, did you catch that I'd got the Temperature equation wrong? It should be R^2 at the bottoms of those fractions, not R...

[t00fri]

Ooops. Fridger, did you catch that I'd got the Temperature equation wrong? It should be R^2 at the bottoms of those fractions, not R...

Malenfant,

like I always do, I (re)derive such formulae myself before using them.

I wrote above that I corrected your formulae. That mainly referred to the factors r1^2, r2^2 instead of r1, r2. Also you had a factor 16 instead of 4.

4*Pi*r1^2 is just the irradiated area by which the luminosity has to be divided to arrive at the radiation energy FLUX.

Bye Fridger

[ajtribick]

One thing I'm confused about here, in the simple basic model where the two stars are treated as separate sources both located at the same distance (R) away, would the formula be

T^4 = (L1 + L2) / (16*pi*sigma*R^2)

...there seem to be several conflicting messages on how to combine the luminosities here.

[t00fri]

One thing I'm confused about here, in the simple basic model where the two stars are treated as separate sources both located at the same distance (R) away, would the formula be

T^4 = (L1 + L2) / (16*pi*sigma*R^2)

...there seem to be several conflicting messages on how to combine the luminosities here.

But that cannot be a general ansatz. You must be able to (theoretically) move one of the two stars to infinity while the other one stays at a finite distance from the planet. With your ansatz that's hardly possible.
It does not posses enough degrees of freedom.

Hence the only sensible solution that includes the requested degrees of freedom, is to add the two fluxes of radiation energy, infalling onto the planet.

Bye Fridger

[ajtribick]

So would

T^4 = L1/(16*pi*sigma*R1^2) + L2/(16*pi*sigma*R2^2)

be a reasonable approximation then?

[t00fri]

So would

T^4 = L1/(16*pi*sigma*R1^2) + L2/(16*pi*sigma*R2^2)

be a reasonable approximation then?

That's what I used, yes. Except your factor 1/16. It should be 1/4.

It's more instructive to arrange the formula in a physically intuitive fashion:

4*Pi * r^2 is the circular area with which the FLUX F= rad. energy/area is formed:

F=L/(4*Pi*r^2)

It's F that satisfies the Stefan Boltzmann law:

++++++++++++++
F = sigma * T^4
++++++++++++++

Hence for a binary system, write

F1 + F2 =sigma* T^4

or

L1/(4*Pi*r1^2) + L2/(4*Pi*r2^2) =sigma*T^4

A generalization one may derive is to allow for another factor e besides the Boltzmann constant sigma:

F1 + F2 = e* sigma*T^4

Bye Fridger

[Malenfant]

So would

T^4 = L1/(16*pi*sigma*R1^2) + L2/(16*pi*sigma*R2^2)

be a reasonable approximation then?

I'll check when I get home, but IIRC the "16" comes from a "flux factor" term that's incorporated into the equation, which is determined by the rotation of the blackbody. If it's a rapid rotator then the energy from the star can be evenly spread over the whole body - in this case the Flux Factor = 4. This is multiplied into the bottom part of the formula, so you get that 4 multiplied by the "4" from the area to get 4x4=16 there. Slow rotators have a flux factor of 2 instead of 4, so for them the multiplier would 8 instead of 16.

I probably could have explained that better, but that's the general gist of it as far as I understand it.