The big idea: Two stars can look equally bright in the sky yet pour out wildly different amounts of power — one might just be much closer.
So we keep two separate ideas:
- Luminosity (L) = the total power the star radiates (its own property, in watts). - Apparent brightness (b) = the power we receive per square metre here on Earth (it depends on distance).
Luminosity L
- The total power the star pours out, in all directions
- A property of the star itself — it does not depend on how far away you are
- Measured in watts (W), just like a light-bulb's power rating
Apparent brightness b
- The power you actually receive, spread over each square metre
- Depends on how far away you are — the same star looks dimmer farther off
- Measured in watts per square metre (W m⁻²)
New words, plainly: Luminosity (L) = how powerful the star is in total — like a bulb's wattage. It never changes with where you stand.
Apparent brightness (b) = how bright it looks to us — the power landing on each m² of our detector. Move farther away and b drops, even though L is unchanged.
Inverse-square law = brightness falls off as 1 ÷ (distance)².
| As the light travels outward… | What happens to the brightness |
|---|---|
| The star sends out a fixed total power L every second | L never changes — it's the star's own property |
| That power spreads over a sphere of area 4π d² | The sphere gets bigger as d grows |
| Each square metre therefore catches a smaller share | The apparent brightness b falls |
| Double the distance → 4× the area → ¼ the brightness | This is the inverse-square law |
Why distance changes b but not L: The star always emits the same total power L.
But by the time that light reaches us it has spread over a huge sphere of area 4π d². The farther away we are, the bigger that sphere, so each square metre gets a smaller slice — and b is smaller.
The light isn't destroyed; it's just spread thinner.
Because the star's power L spreads evenly over a sphere of area 4π d², the brightness we measure is simply L divided by that area. That single formula links luminosity, brightness and distance.
- apparent brightness — power received per unit area at Earth (W m⁻²)
- luminosity — total power the star radiates in all directions (W)
- distance from the star to the observer (m)
- surface area of the sphere the light has spread over by distance d (m²)
The shortcut you'll use most: Because b depends on 1/d², distance changes brightness fast:
- 2× farther → ¼ as bright - 3× farther → 1/9 as bright - 10× farther → 1/100 as bright
Watch your powers of ten — luminosities and distances are huge numbers in standard form.
IB-style question — apparent brightness of a star
A star has a luminosity of L = 2.5 × 10²⁸ W. It is at a distance of d = 4.0 × 10¹⁸ m from Earth. Calculate the apparent brightness of the star as seen from Earth.
Solution
- Write the given data-booklet formula first — the brightness is the luminosity spread over the sphere of area 4π d²:
- Substitute L = 2.5 × 10²⁸ W and d = 4.0 × 10¹⁸ m straight into it:
- Work out the bottom (the area of the sphere) first:
- Now divide — keep the unit:
Final answer
b = L / (4π d²) = 1.2 × 10⁻¹⁰ W m⁻². A tiny number — the star's huge power is spread over an enormous sphere.
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How we know how far a star is: As Earth orbits the Sun, a nearby star seems to shift slightly back and forth against the distant background stars over a year. That shift is called parallax.
The closer the star, the bigger the shift. Measure the shift angle and you get the distance — no need to travel there.
| Parallax — the idea | What it tells you |
|---|---|
| Earth moves to the other side of its orbit over six months | We view the star from two ends of a baseline |
| A nearby star appears to shift against the distant background | The shift angle is bigger for closer stars |
| The parallax angle p = half that total yearly shift, in arc-seconds | Smaller p → the star is farther away |
| Distance in parsecs = 1 ÷ p | 1 arc-second of parallax = 1 parsec away, by definition |
- distance to the star, in parsecs (pc)
- parallax angle — half the star's apparent shift over a year, in arc-seconds (″)
Units matter here: This formula only works when p is in arc-seconds and you read d straight out in parsecs.
1 parsec (pc) ≈ 3.26 light-years ≈ 3.1 × 10¹⁶ m.
1 arc-second (″) = 1/3600 of a degree — far too small to see by eye, which is why parallax needs telescopes.
IB-style question — distance from a parallax angle
A star has a measured parallax angle of p = 0.025 arc-seconds. Calculate its distance from Earth, in parsecs.
Solution
- Write the given data-booklet formula first — distance in parsecs is just one over the parallax angle in arc-seconds:
- Substitute the parallax angle p = 0.025 arc-seconds:
- Work it out — the answer comes straight out in parsecs:
Final answer
d = 1 / p = 1 / 0.025 = 40 parsec. The smaller the parallax angle, the farther the star.
How this is tested: This topic shows up most often as quick Paper 1A multiple-choice, and occasionally inside a longer Paper 2 star question.
- Paper 1A — calculate: turn a parallax angle into a distance with d = 1/p, or find a brightness with b = L/(4π d²). - Paper 1A — determine: compare two stars. The classic version: two stars look equally bright, but one is N× more luminous — how much farther away is it? (Equal b ⇒ d ∝ √L.) - Paper 2: brightness/luminosity/distance can feed into a bigger black-body question (linking to L = σAT⁴ in 5.5.4).
Classic trap: mixing up luminosity (total power out — fixed) with apparent brightness (received per m² — depends on distance). Read which one the question gives you.
IB-style question — same brightness, different luminosity
Two stars, A and B, appear equally bright from Earth (the same apparent brightness b). Star B is 9 times more luminous than star A. Determine how many times farther away star B is than star A.
Solution
- Write the given data-booklet formula for each star — they have the same apparent brightness b:
- The 4π cancels. Cross-multiply to compare the two — equal brightness means L is proportional to d²:
- Substitute LB = 9 LA, so the luminosity ratio is 9:
- Take the square root to get the distance ratio:
Final answer
Star B is 3 times farther away. Equal brightness ⇒ d ∝ √L, and √9 = 3 — the more luminous star must be 3× farther to look just as bright.