Solar Luminosity Comparison

Use L/L☉ = (R/R☉)² (T/T☉)⁴ to estimate relative and absolute stellar luminosity.

Starting values — Solar Luminosity Comparison

Radius (R☉)	1
Temperature (K)	5772

Change any control above and the results recompute from these.

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Lesson

The theory — Solar Luminosity Comparison

Luminosity is total radiant power — every watt a star emits in all directions at once. It is not how bright the star looks: apparent brightness falls off with distance, luminosity does not.

What each symbol means

L: the star’s luminosity, given here both as a multiple of the Sun’s and in watts.
R: the star’s radius. It enters squared, because what radiates is the surface, and surface area goes as R².
T: the effective temperature of the surface, in kelvin. It enters to the fourth power, by the Stefan-Boltzmann law.
L☉: the solar subscript, used as the unit. The values built in here are the IAU nominal ones: L☉ = 3.828 × 10²⁶ W and T☉ = 5772 K.

How to read what you see

Two panels, and their order is the point. The first gives luminosity as a ratio to the Sun, straight from (R/R☉)²(T/T☉)⁴ — at the defaults that is 1 · 1 = 1 L☉. Only the second turns the ratio into watts, by multiplying by L☉ = 3.828 × 10²⁶ W. The ratio needs no physical constants whatsoever; the watts need one.

Assumes: A sphere radiating as an ideal blackbody at a single temperature. Real stars are limb-darkened, flattened by rotation and covered in spots — and effective temperature is defined as the blackbody temperature that would produce the observed power, so the fourth-power law holds here by definition rather than by measurement.
Breaks when: Nothing here accounts for what lies between the star and the telescope: interstellar dust dims and reddens, so a luminosity inferred from observed brightness needs an extinction correction this page does not model. And a stellar radius is not a surface anything could stand on — it is the depth at which the gas stops being opaque.

Comparing two stars never needs the Stefan-Boltzmann constant 🖖

The breakdown prints a ratio first and watts second, and that order is not cosmetic. Write L = 4πR²σT⁴ once for the star and once for the Sun, then divide: 4π and σ are identical in both lines and cancel, leaving (R/R☉)²(T/T☉)⁴. The units cancel with them, so two plain ratios are enough — the tool reports Vega at 42.7 times solar before touching a physical constant. Watts appear only in the second row, where that 42.7 is multiplied by one solar luminosity, 3.828 × 10²⁶ W.

Temperature matters more than size 🖖

A star's total light output depends on two things: how big it is and how hot its surface is. But temperature carries far more weight, because it enters the formula to the fourth power while radius is only squared. That's why Vega — barely 2.4 times the Sun's radius — shines about 40 times brighter: its roughly 9600 K surface does the heavy lifting.

A hot star dimmer than the Sun 🖖

Surface temperature alone doesn't make a star luminous — size gets a vote too. A white dwarf can glow at 25,000 K, over four times hotter than the Sun, yet radiate only a few percent of the Sun's light. The reason is its Earth-sized body: with a radius near 0.01 R☉, the R² term shrinks the output despite the fierce heat. Hot does not always mean bright.

Practice

Check yourself

Predict the answer first, then use the controls above to find out. Reveal only after you have committed to a guess — that is what makes it practice.

Set the temperature to half the Sun's, 2886 K. How large would the star have to be to still emit exactly 1 L☉ — and where on the page is the exchange rate stated outright?

Show answer

R = 4 R☉, and the breakdown shows why: 16 · 0.063 = 1 L☉. The radius term gains a factor 16 exactly as the temperature term loses one. The exchange rate is printed in the Local sensitivity panel — +1% T → +4% L against +1% R → +2% L — so one percent of temperature is worth two of radius, and undoing a halved temperature takes a quadrupled radius, not a doubled one.
Now set the radius to 10 R☉ with the temperature back at the Sun's, giving a ratio of 100. What does the magnitude row read, and why is it so round?

Show answer

Exactly −5.00. The magnitude scale is defined so that five magnitudes are a factor of 100: ΔM = −2.5 log₁₀(L/L☉), and log₁₀(100) = 2. That is why one magnitude is the awkward 100^(1/5) ≈ 2.512 rather than anything tidier — the roundness was put at the factor of 100 and had to come out of somewhere. Note the sign too: brighter is lower, so a star 100 times the Sun sits five magnitudes below it.

Problems solved in full

Luminosity and magnitude offset of Rigel at 78.9 solar radii 6 steps
Rigel: 78.9 solar radii at 12,100 K. Work out its luminosity and its magnitude offset, then say which of the two inputs you would rather measure accurately. This is the Rigel state.
1. The whole tool is one equation, and it is Stefan's law divided by itself. Everything cancels except the two ratios.
  
  $L = 4 π R^{2} σ T^{4} ⟹ \frac{L}{L _{⊙}} = (\frac{R}{R _{⊙}})^{2} (\frac{T}{T _{⊙}})^{4}$
2. Evaluate the factors separately, because the point is how unequal they are. The radius contributes three orders of magnitude, the temperature barely more than one.
  
  $(\frac{R}{R _{⊙}})^{2} = 78. 9^{2} = 6.225 \times 1 0^{3}; (\frac{12100}{5772})^{4} = 2.096 3^{4} = 19.311$
3. Multiply, then scale by the solar luminosity to leave the dimensionless world.
  
  $\frac{L}{L _{⊙}} = 6225 \times 19.311 = 1.202 \times 1 0^{5} ⟹ L = 4.602 \times 1 0^{31} W$
4. Magnitudes are −2.5 log₁₀ of a flux ratio, so a factor of 120,000 becomes 12.7 magnitudes brighter. The minus sign is the astronomers' convention that brighter means smaller.
  
  $Δ M = - 2.5 lo g_{10} (1.202 \times 1 0^{5}) = - 2.5 \times 5.080 = - 12.70$
5. The sensitivity rows are just the exponents, read as elasticities. This is the useful form when you are deciding what to measure.
  
  $\frac{\partial ln L}{\partial ln T} = 4, \frac{\partial ln L}{\partial ln R} = 2 ⟹ + 1% T \to + 4% L$
6. Turn the ratio into a duration and it stops being abstract.
  
  $\frac{1.202 \times 1 0 ^{5} days}{365.25} = 329 years of sunlight, per day$
Answer

The temperature, by a factor of two — but the radius, by a factor of three hundred. Those are not contradictory, they are the difference between a relative error and an absolute one. Stefan's law puts the exponent 4 on temperature and 2 on radius, so a 1% error in T costs twice as much as a 1% error in R, and that is the sensitivity the panel prints. But Rigel is 78.9 solar radii and only 2.1 solar temperatures, so the radius factor contributes 6,225 of the 120,000 and the temperature factor only 19.3. The exponent tells you which measurement to be careful with; the value tells you where the light is actually coming from. Rigel radiates in one day what the Sun manages in 329 years.

$\frac{L}{L _{⊙}} = (\frac{R}{R _{⊙}})^{2} (\frac{T}{T _{⊙}})^{4} = 1.202 \times 1 0^{5}, Δ M = - 12.70$
Luminosity of Proxima Centauri at 0.1542 solar radii and 3042 K 5 steps
Proxima Centauri: 0.1542 solar radii at 3042 K. Get its luminosity, then check the magnitude scale by comparing it with Rigel. This is the Proxima Centauri state.
1. Same two factors, both now below one. Note that the temperature ratio is barely half and still costs a factor of thirteen, because it is raised to the fourth.
  
  $(\frac{R}{R _{⊙}})^{2} = 0.154 2^{2} = 0.02378; (\frac{3042}{5772})^{4} = 0.527 0^{4} = 0.07713$
2. The product is 1.834 × 10⁻³ of the Sun — 1 part in 545 — which is why the nearest star to us is invisible without a telescope.
  
  $\frac{L}{L _{⊙}} = 0.02378 \times 0.07713 = 1.834 \times 1 0^{- 3} ⟹ L = 7.022 \times 1 0^{23} W$
3. A luminosity below one gives a positive magnitude offset — fainter than the Sun by 6.84 magnitudes.
  
  $Δ M = - 2.5 lo g_{10} (1.834 \times 1 0^{- 3}) = + 6.84$
4. Now the cross-check. Take the ratio of the two luminosities directly.
  
  $\frac{L _{Rigel}}{L _{Proxima}} = \frac{1.202 \times 1 0 ^{5}}{1.834 \times 1 0 ^{- 3}} = 6.55 \times 1 0^{7}$
5. Then take the difference of the two magnitudes and undo the logarithm. The two routes have to agree, and agreeing is what tells you the magnitude scale is a logarithm and nothing more.
  
  $Δ M_{Proxima} - Δ M_{Rigel} = 6.84 - (- 12.70) = 19.54; 1 0^{19.54/2.5} = 6.55 \times 1 0^{7}$
Answer

1.834 × 10⁻³ L☉, and Rigel outshines it by 6.55 × 10⁷. The check is the point: the two magnitude offsets are −12.70 and +6.84, a gap of 19.54 magnitudes, and 10^19.54/2.5 is 6.55 × 10⁷ — the same number the luminosities give directly. That is not a coincidence, it is the definition; the magnitude scale was fitted to the eye's response in the second century BC and formalised as exactly five magnitudes per factor of 100 in 1856. Both of Proxima's factors work downward here, radius and temperature together, which is how a star 6.5 times smaller than the Sun in radius ends up 545 times fainter.

$1.834 \times 1 0^{- 3} L_{⊙}, Δ M = + 6.84; 19.54 mag = 1 0^{19.54/2.5} = 6.55 \times 1 0^{7}$

Learning path

How far away is it?

Leads to Stellar magnitude the star’s true output, as a multiple of the Sun’s.

References (1)

Where L☉ = 3.828 × 10²⁶ W and T☉ = 5772 K come from — the IAU 2015 nominal values this page uses: A. Prša et al., “Nominal Values for Selected Solar and Planetary Quantities: IAU 2015 Resolution B3.” The Astronomical Journal 152, 41, 2016.

By the 42Math team · Updated 2026-09-03

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Proxima Centauri - Proxima: tiny radius and cool temperature produce very low luminosity.
Sun - Solar reference case: exactly 1 L☉ by normalized definition.
Vega - Vega: hotter photosphere and larger radius produce much higher luminosity.
Rigel - Rigel-like supergiant: both size and temperature drive enormous luminosity.