# How to calculate a star's luminosity/radius

• B
Bipo10
TL;DR Summary
Difficulties with some equations and how to use them.
Hello there.
I'm new to this forum and astrophysics in general. I'm also in early high school, so please be patient if I can't grasp the math at first.

Anyway, I've been looking for a way to calculate a star's radius without depending on luminosity, and/or a way to calculate luminosity without depending on the radius of the star. I have found this equation (from http://cas.sdss.org/dr4/en/proj/advanced/hr/radius1.asp): R/Rs = (Ts/T)^2(L/Ls)^1/2. The problem, though, is that although the numbers are quite accurate for stars such as Sirius the calculations start getting off for stars such as Vega or Polaris. Is this equation correct? Or am I lacking some variable I don't know?

Another problem I've encountered is with L = 4πR^2σT^4. No matter what numbers I plug into it, it never gives the correct luminosity or radius. What am I missing here? I'm pretty sure I did my calculations correct, but the results never match... Also, what are the all units I have to use in it? All numbers in solar units? Or in kilometers and watts? (and by extention, the luminosity I get will be in which unit?)

There's this equation too (from http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit1/bright.html): L = 4πd^2B. Again, no explanations on which units I should use and/or if it works only for a set of stars, etc.

I've been searching for 2 days now and all the sites I find are either ambiguous in some way or they use some crazy math I've never encountered (solid angles, spherical coordinates, integral calculus and so on)

If anyone can give me some insight i'd gladly appreciate it!

• Mlesnita Daniel

Let's start with the equation L = 4πR^2σT^4, and why you can't get it to give the correct results. This starts with the Stefan-Boltzmann law, which says that the total radiated power per unit area from a black body is given by P = σT^4, where σ is the Stefan-Boltzmann constant, which in SI units has the value of 5.67×10−8 W⋅m−2⋅K−4. To get the total radiated luminosity (say in Watts), you multiply this P by the surface area of the star, which is A = 4πR^2. Try calculating this for the sun, which has a surface temperature of about 5600 K. This should give you the luminosity of the sun in Watts. If it doesn't work out, post your calculations and we'll see if we can figure out what you're doing wrong.

Note that once you have understood this, your other equation follows from L = 4πR^2σT^4. If I take two stars, L1 = 4πR1^2σT1^4, and L2 = 4πR2^2σT2^4, then divide these two equations, I get that $\frac{L_1}{L_2} = \frac{R_1^2}{R_2^2} \frac{T_1^4}{T_2^4}$. If you solve for R, you get $\frac{R_1}{R_2} = \frac{L_1^{1/2}}{L_2^{1/2}} \frac{T_2^2}{T_1^2}$, which is what you had if you take "2" to be the sun. Then you can use solar radius and solar luminosity in this equation.

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Bipo10

Ok so using the numbers you provided we have L=4π 695,700000^2σ 5600^4 =3.3914×10^26. According to the IAU, This should've been 3.828×10^26. Close but not quite! When I try it with only Google numbers I get L=4π695,700000^2σ5778^4= 3.843×10^26. I used the sun's radius in metres, and temperature in kelvin.
The results were much better! The latter equation is numerically closer then any previous try but both still are a bit off compared to the estimated value from IAU (although it could be this is as best as I can get without any outside equipment... )
Is there any way for me to use solar units in this equation though? Working with numbers as big as 695700000 is a bit jarring for me...

As for R/Rs = (Ts/T)^2(L/Ls)^1/2 the results weren't so good though... I tried calculating the radius of Barnard's star and it resulted in R=695,700000×((5778/3100)^2(2.51^-8.46)^1/2)= 50574000 metres or about 50574 kilometres when in actuality it has a radius of 136.360 km...
(The site from which I got this equation says that L/Ls=2.51^Δm where Δm=ms-m (Sun's absolute magnetude minus thes star's absolute magnetude ))
What can I do about that?

Thanks again for the help!

Staff Emeritus
Anyway, I've been looking for a way to calculate a star's radius without depending on luminosity, and/or a way to calculate luminosity without depending on the radius of the star.

Let's go back to the beginning. What are your inputs to these calculations? Clearly if you knew the diameter, you could do a pretty good job on calculating the radius. So what are known quantities and what are you trying to derive?

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Bipo10
Ok, so basically what I'm trying to find is a way for me to calculate luminosity without depending solely on this equation: L = 4πR^2σT^4. What I thought about it at first was to use an equation were I could find luminosity (and after that use the aforementioned equation to get the radius) with only data such as temperature, magnetude (either apparent or absolute) and/or distance. The problem is that even after finding one (" R/Rs = (Ts/T)^2(L/Ls)^1/2, using 2.51^Δm where Δm=msolar-m to replace L/Ls ") it still didn't work for most stars... Then I looked for an equation where I could find the radius without needing the luminosity, but I couldn't find any...

For "known quantities" I'd like to depend on temperature, magnetude and distance only, so I could use Hipparcus data for experimental calculations, for example. What I'm trying to derive is an equation that works for all stars, or atleast a set of equations I can use for each type of star so I get accurate results regardless of spectral type. I don't really know to to derive an correct equation out of these values since I've recently taught myself derivatives in the first place (so I don't have that deep of an understanding on what I can do with it) and even if i could, I don't know how all the variables relate to each other very well.

If you could help me sort this out, explaining to me the "whys and why nots" of these variables and how they correlate would give me a huge insight. Also avoiding complicated calculus in general as much as possible would be highly appreciated (Thanks for the reply btw!)

OK, a good start, but I have several comments:

(1) Yes, you can use the equation L = 4πR^2σT^4 in solar units. To do this you need to convert σ from Watts/(m^2 K^4) to Solar luminosities/(Solar radii^2 T^4). Try doing this and tell us what the value of σ is in these units.

(2) You need to understand that this is just a model of the luminosity of the star and will never be perfectly accurate for several reasons. Getting within 10-20% is probably as good as you will ever do. First, the model is for a black body, and stars are not perfect black bodies. They have absorption lines, emission lines, and all sorts of features. This graph shows the solar spectrum as compared to a black body. It's close, but not exactly a black body. Second, what exactly is the radius of the sun? The density of the sun drops off from very high values in the interior to essentially zero at some large radius. It has no solid surface like the Earth. So where is the edge? It is somewhat arbitrary where the density has dropped low enough to call it the edge of the sun. For the same reason, defining the surface temperature is somewhat arbitrary, since the temperature drops off as you move out from the center. Usually, people choose values of the radius and surface temperature so that the equation L = 4πR^2σT^4 works.

(3) As for Barnard's star, there is another thing you need to understand. The equation L = 4πR^2σT^4 holds for the bolometric luminosity, which is the total energy emitted at all wavelengths. For Barnard's star, you are probably using the visual magnitude, which only includes the light emitted in the visual part of the spectrum. A red star like Barnard's star emits most of its light in the infrared. Try using the bolometric magnitude of the sun and Barnard's star, and see if that gives a closer result.

For "known quantities" I'd like to depend on temperature, magnetude and distance only ...

If this is your goal, you don't even need the temperature. The absolute magnitude and bolometric luminosity of an object are related by: $$M - M_{sun} = -2.5 log_{10}(\frac{L}{L_{sun}})$$
and the absolute bolometric magnitude of the sun is about 4.75. Do you know how to convert from apparent magnitude to absolute magnitude?

Bipo10
Let's start then : σ in Solar units is (3.828×10^26)/(695700000^2× 5778^4)= 7.096×10^-7 (I used watts, metres and kelvins for this, since these are the units the normal constant is in...)
Now I don't know if that's right because when I try calculating Barnard's star (lets focus on this one for the time being) luminosity it gives me 4π(0.196^2)(7.096×10^-7)(3141/5778)^4 = 2.99*10^-8 , quite far from the bolometric luminosity of 0.0035 Ls.

I do know how to find absolute magnitude using m-M=5log10(d)-5, but i don't know if there's a way to do it without the distance (Maybe an apparent to absolute magnitude without depending on distance would be more efficient)

When calculating the bolometric magnitude for the same star I get X-4.75=-2.5log100.0035 where X equals 13.24, almost exactly the same absolute magnitude from the distance equation above. And when using bolometric magnitudes for Δm it still gives me an wrong result for the radius (perhaps since Δm goes negative?)
By the way, can I convert a star's abolute magnetude into absolute bolometric magnitude? maybe that's what is failing the equation.

Also, thanks for the insight on the accuracy and for the absolute magnitude and bolometric luminosity equation. That helps me alot! (I've seen some sites saying that the magnitudes for that formula are the absolute bolometric magnitudes, so is "absolute magnitude" always the bolometric one unless specified?)

Let's start then : σ in Solar units is (3.828×10^26)/(695700000^2× 5778^4)= 7.096×10^-7 (I used watts, metres and kelvins for this, since these are the units the normal constant is in...)

This isn't right. The best way to convert something into different units is to use the trick of "multiplying by one". You can always multiply an equation by one without changing it. Then you cancel the units as you go along. So for example, suppose you want to change 50 miles/hour into meters/second. You do the following:
$$v = 50 \frac{mile}{hour}\times \frac{5280 \, foot}{1 \, mile}\times \frac{12 inches}{1 \, foot}\times \frac{1 \, meter}{39.37 \, inches}\times \frac{1 \, hour}{3600 \, seconds} = 22.35 \frac{meters}{second}$$

So for σ, we have:
$$\sigma = 5.67\times 10^{-8} \frac{Watt}{m^2 K^4}\times \frac{L_{sun}}{3.83\times 10^{26} \, Watt}\times (\frac{6.96 \times 10^8 m}{R_{sun}})^2 = \\ 7.17 \times 10^{-17} \frac{L_{sun}}{R_{sun}^2 K^4}$$

So then if you take L = 4πR^2σT^4, for the sun you get
$$L = 4 \pi * (1.0 \, R_{sun})^2 * 7.17 \times 10^{-17} \frac{L_{sun}}{R_{sun}^2 \, K^4} * (5778 K)^4 = 1.00 L_{sun}$$

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I do know how to find absolute magnitude using m-M=5log10(d)-5, but i don't know if there's a way to do it without the distance (Maybe an apparent to absolute magnitude without depending on distance would be more efficient)

There is no way to convert between apparent and absolute magnitude without knowing the distance.

(I've seen some sites saying that the magnitudes for that formula are the absolute bolometric magnitudes, so is "absolute magnitude" always the bolometric one unless specified?)
Magnitudes (both apparent and absolute) can be of many different types, bolometric, visual, etc. There is no way of knowing unless it is specified.

Bipo10
Thanks for all the clarifications! Your explanations have helped me a lot!
But I still do have a question...

When using the converted σ (I'll call it σSun from now on) I do get the results accurate for Sirius

L=4π*(1,7144)^2*σSun*9845^4=24.87Lsun (Super precise!)

But when I try Acrux (α1) I get L=4π*(7.8)^2*σSun*24000^4=18184.4 Lsun instead of the estimated 25000 Lsun.
When I try Bellatrix I get L=4π(5.75)^2*σSun*22000^4=6977.36 Lsun out of the 9211 Lsun estimated total
And for stars colder then the Sun I tried Aldebaran which resulted in L=4π(44.13)^2*σSun*3900^4=427.1 Lsun (Actually precise!) and Barnard's star which resulted in L=4π(0.196)^2*σSun*3141^4=0.00336 Lsun (small difference of 0.00014, so its pretty precise)
For the sake of confirmation I also tried Canopus and got L=4π(71)^2*σSun*6998^4=10891Lsun (191 more than the estimates)

It seems that the equation starts to get more and more imprecise the higher the temperature. Could you please shed some light on that?

Thanks for all the clarifications! Your explanations have helped me a lot!
But I still do have a question...

When using the converted σ (I'll call it σSun from now on) I do get the results accurate for Sirius

L=4π*(1,7144)^2*σSun*9845^4=24.87Lsun (Super precise!)

But when I try Acrux (α1) I get L=4π*(7.8)^2*σSun*24000^4=18184.4 Lsun instead of the estimated 25000 Lsun.
When I try Bellatrix I get L=4π(5.75)^2*σSun*22000^4=6977.36 Lsun out of the 9211 Lsun estimated total
And for stars colder then the Sun I tried Aldebaran which resulted in L=4π(44.13)^2*σSun*3900^4=427.1 Lsun (Actually precise!) and Barnard's star which resulted in L=4π(0.196)^2*σSun*3141^4=0.00336 Lsun (small difference of 0.00014, so its pretty precise)
For the sake of confirmation I also tried Canopus and got L=4π(71)^2*σSun*6998^4=10891Lsun (191 more than the estimates)

It seems that the equation starts to get more and more imprecise the higher the temperature. Could you please shed some light on that?
Where are you getting your information on radius, luminosity and temperature for these stars? Are they bolometric luminosities?

If they're not bolometric luminosities (and I suspect they are not), you could try applying a Bolometric correction to the visual magnitudes. The Wikipedia article explains how this works and why it is necessary.

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Bipo10
(I've been getting my data from wikipedia, since its basically the only site I found that had all the values "ready-to-use")

Ok so I retried the calculations with bolometric corrections
Based on the values from https://www.astro.princeton.edu/~gk/A403/constants.pdf I got for Acrux a correction of -2.384 (Since there isn't a correction value for 24000k specifically, I've calculated the proportion of correction from 15500k / 28000k (it goes from -1.5 to -2.8) to 15500k / 24000k (Proportionally it would go from -1.5 to -2.384) (My logic is that since the difference of 28000-15500=125000k yields an increase of -1.3 of BC, then the 24000-15500=8500 should yield an correction of -2.384). In this calculation I got an absolute bolometric magnitude of -2.384 = Mbol - (-3.77) -> Mbol=-6.154.
Using -6.154−4.75=−2.5log10(L/Lsun) I got an luminosity of 22993 Lsun , which probably would be even closer to the real value if I actually had an accurate BC value.
Unfortunately, when trying the same method, I get a BC of -2.176 for Bellatrix, which ends up giving me -2.176 = Mbol - 1.64 -> Mbol = -0.535. ->
-0.535-4.75 = -2.5log10(L/Lsun) = 102.144 = 130 Lsun. That's just bollocks. I got such a good answer for one star and something completely off for the other! When using the values from http://www.isthe.com/chongo/tech/astro/HR-temp-mass-table-byhrclass.html I got 244 for Lsun for Acrux (B1V) and 356.45 for Bellatrix (B3III).
But I did understand the concept of BC correction. I guess the only thing I'm lacking now is a good and consistent source for data, which seems pretty rare as I'm barely able to find any.

Ps: Can I calculate bolometric correction based on some variable? Or is there a list that has a wider range of values for more precise correction ? (besides the one I used?). Do you know any good sources for data?

Thanks again for everything, Phyzguy.

I think you just made a sign error in your correction for Bellatrix. Bellatrix has an absolute magnitude of -2.78, and if I use your BC of -2.18, this gives Mb = -2.78+(-2.18) = -4.96. Then -4.96 - 4.75 = -2.5 log10(L/Lsun), which gives L = 7655 Lsun, which is a little low. Looking at your table, Bellatrix is a Type B2-III giant star, so it looks to me like the BC should be closer to -2.4. Using this give Mb = -5.18, which gives L = 9375 Lsun, which is closer.

Keep at it, you're making progress. These bolometric corrections are approximate, and I think this is as accurate as they get To do it right, you need to actually measure the luminosity of the star across the whole spectral range so you actually know the bolometric luminosity rather than just estimate it.

One other thing you should know. Most stars are too far away and too small to actually measure the radius. So the radius values you are finding were calculated from the L = 4πR^2σT^4 equation, using some value for the temperature and bolometric luminosity. If you can find the source of these radius values, you can find what L and T values were actually used. Since the T appears to the 4th power, only a small change in T makes big changes in L and R.

Bipo10
I see now. Thanks a lot for all the help Phyzguy! I'm sure I'll be able to sort things out now!