# Calculating Proportional Luminosity of Fictitious Star

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• Chem10
In summary: Thank you all for your input.In summary, the equation L ∝ 4π · (r)^2 · (T)^4 states that the proportional increase in luminosity will be 10 times greater than that of the Sun.
Chem10
TL;DR Summary
I am attempting to calculate the proportional luminosity of a fictitious star using surface area and kelvin
To what level of accuracy can I expect from the formula L ∝ AT^4?
Where L = Luminosity, A = surface area and T = Temperature

L ∝ AT^4
L ∝ 4π · (r)^2 · (T)^4

If I replace r and T with the proportional increase/decrease in comparison to the Sun I should get the proportional increase in luminosity.

I would be very grateful if someone would be able to confirm my calculations.
I am attempting to calculate the proportional luminosity of a fictitious star using surface area and temperature in kelvin.
To what level of accuracy can I expect from the formula L ∝ AT^4?
Where L = Luminosity, A = surface area and T = Temperature

L ∝ AT^4
L ∝ 4π · (r)^2 · (T)^4

If I replace r and T with the proportional increase/decrease in radius and temperature in comparison to the Sun, I should get the proportional increase/ decrease in proportional luminosity.

If the value is less than 1* then it is this value times less luminous than the Sun and if the value is more than 1* then it is this value times more luminous than the Sun.

I would be very grateful if someone would be able to confirm my calculations and above statements to ensure I've understood correctly.

Thank you again for your time and help.

*(edited from "0"s to "1"s) highlight above by *)

Last edited:
Stefan-Boltzmann law says $$Radiant Emittance =\sigma T^4 ~~~~~where ~\sigma=5.7x10^{-8} ~W/m^2K$$ This number multiplied by the surface area of the star gives the total output power radiated by star, so yes you are correct.
This is not the same as optical brightness because of the color sensitivity of your eye and the T is the surface Temperature of the photosphere

Chem10
hutchphd said:
Stefan-Boltzmann law says $$Radiant Emittance =\sigma T^4 ~~~~~where ~\sigma=5.7x10^{-8} ~W/m^2K$$ This number multiplied by the surface area of the star gives the total output power radiated by star, so yes you are correct.
This is not the same as optical brightness because of the color sensitivity of your eye and the T is the surface Temperature of the photosphere
Thank you very much for confirming. I've edited a sentence slightly. Could you also confirm this statement?

If the value is less than 1 then it is this value times less luminous than the Sun and if the value is more than 1 then it is this value times more luminous than the Sun.

Thank you again.

Chem10 said:
the value
What value?
Are you asking me how to do algebra? Sorry but I do not understand your question.

Chem10
Sorry no, I mean, if the answer to this equation L ∝ 4π · (r)^2 · (T)^4 is 10 for example, would this mean that the proportional increase in luminosity would be 10 times greater than that of the Sun?

I think I am confusing myself with the fact that the Sun has a luminosity figure of 1.

Chem10 said:
I think I am confusing myself with the fact that the Sun has a luminosity figure of 1.
Not in "normal" units.
For the sun luminosity in S.I. is 3.846 × 1026Watts.

If the answer to this equation L ∝ 4π · (r)^2 · (T)^4 would be 10 for example, would this mean that the proportional increase in luminosity would be 10 times greater than that of the Sun?

But is this correct?

Chem10 said:
But is this correct?
No, that;s not an equation. An equation has an equals sign.

malawi_glenn and hutchphd
Chem10 said:
If the answer to this equation L ∝ 4π · (r)^2 · (T)^4 would be 10 for example, would this mean that the proportional increase in luminosity would be 10 times greater than that of the Sun?

But is this correct?
It cannot be 10, that doesn't have units.

If you express everything relative to the Sun then yes, 10 is 10 times the luminosity of the Sun, by construction.
The relation is very accurate because the surfaces of stars are very good approximations to ideal black bodies. Stars can have slightly different temperatures in different places, so T is an average temperature.

Chem10 said:
TL;DR Summary: I am attempting to calculate the proportional luminosity of a fictitious star using surface area and kelvin
To what level of accuracy can I expect from the formula L ∝ AT^4?
Where L = Luminosity, A = surface area and T = Temperature

L ∝ AT^4
L ∝ 4π · (r)^2 · (T)^4

If I replace r and T with the proportional increase/decrease in comparison to the Sun I should get the proportional increase in luminosity.

I would be very grateful if someone would be able to confirm my calculations.

I am attempting to calculate the proportional luminosity of a fictitious star using surface area and temperature in kelvin.
The OP is specifically asking for proportional increase in luminosity, as such, no units are needed.

The half dozen Google responses I checked (https://www.google.com/search?q=luminosity+based+on+size+and+temperature) agree that the answer to the OPs question is "Yes."

Cheers,
Tom

I also agree that, if you plug in temperature and radius as multiples of the values for the Sun, the result will be luminosity in multiples of the Sun's luminosity.

There are probably some traps around making sure you use the right temperature for the Sun and the star and whether the luminosity is actually the metric of interest. And, as mfb says, the temperature varies across the surface so it's all an average.

Indeed, one normally defines a concept of "effective temperature" such that the equation you are using becomes exact. So if someone tells you the factor that r changes and the factor that T changes, you can assume they mean a T that is defined to make the equation true, so your answer is not approximate at all, it is exactly true. If one tries instead to use a "real surface temperature", one is bringing in a whole bunch of problems. First of all, the surface is not at a single temperature, especially if the star is a rapid rotator. Second, the radius is not a single number, because stars are oblate from their rotation. (Again that problem is quite severe for rapid rotators.) Third, there is not even a "surface" to a star, it is a ball of gas with T that varies with depth. So it's not the equation that is the approximate thing here, it is the concepts of r and T that go into the equation. By using carefully chosen effective values of T and r, the equation is made exact, so the answer you have given above is exactly correct if the inputs are given in terms of these effective values.

(There is even a worse problem-- if the star is a rapid rotator, it looks less bright when seen from its equatorial plane than when seen from the polar direction. So even the concept of luminosity L has to be clarified-- are we talking about the luminosity we would infer if we took the brightness we see and assume it was the same in all directions, or are we talking about the star's true luminosity? Those differences wouldn't amount to anything important for the Sun, but can be quite different for rapidly rotating stars, which some stars are. No doubt the question you are given is not intended to delve into any of those issues, however.)

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