Light carries momentum, as you saw in energy and momentum of light: a flux $F$ presses on what absorbs or scatters it with a force $F/c$ per unit area of target. In a very luminous star the outflowing light pushes on the gas hard enough to compete with gravity. That sets the most luminous a star of a given mass can be, and it explains why the mass–luminosity relation cannot climb forever.
💡 Why it works
In ionized hydrogen, light scatters off free electrons, each presenting a target area $\sigma_T = 6.65 \times 10^{-29}$ m². At distance $r$ the flux is $L/(4\pi r^2)$, so the push on one electron is $\sigma_TL/(4\pi r^2c)$. The electrons cannot leave the protons behind (their electric attraction is enormous), so the push on an electron must hold up a proton against gravity, $GMm_p/r^2$. Setting the two equal, $r^2$ cancels: $$L_{\text{Edd}} = \frac{4\pi GMm_pc}{\sigma_T}.$$ Above it, radiation would blow the outer layers away.
🔑 Key idea
The Eddington luminosity is $L_{\text{Edd}} = 4\pi GMm_pc/\sigma_T \approx 1.3 \times 10^{31}\text{ W} \times (M/M_\odot)$, about $3.3 \times 10^{4}\,L_\odot$ per solar mass. It is the same at every radius, and it applies to anything shining by its own light while held by gravity, from stars to the gas falling into black holes.
Radiation pressure against gravity on a parcel of gas. For the most massive stars, the luminosity on the main sequence approaches the Eddington limit.
Worked examples
✏️ Worked example: The Sun's limit
Evaluate $L_{\text{Edd}}$ for $1\,M_\odot$.
$$\frac{4\pi Gm_pc}{\sigma_T} = \frac{4\pi(6.67 \times 10^{-11})(1.67 \times 10^{-27})(3.00 \times 10^{8})}{6.65 \times 10^{-29}}$$ which is $6.3$ W/kg.
Times $1.99 \times 10^{30}$ kg: $1.26 \times 10^{31}$ W.
Answer: About $3.3 \times 10^{4}\,L_\odot$. The Sun shines at only $3 \times 10^{-5}$ of its limit.
✏️ Worked example: A very massive star
A $60\,M_\odot$ star following $L \approx 1.4\,M^{3.5}$ would have $L \approx 2.3 \times 10^{6}\,L_\odot$. Compare with its Eddington limit.
Answer: It would exceed the limit, so real stars this massive cannot follow the steep relation; their luminosity rises only about in proportion to $M$, and they drive powerful winds.
✏️ Worked example: Weighing a quasar
A quasar shines with $1.0 \times 10^{40}$ W. If it is at or below its Eddington limit, what is the least its black hole can weigh?
$M \ge L/(6.3\text{ W/kg}) = 1.6 \times 10^{39}$ kg.
Divide by $1.99 \times 10^{30}$ kg.
Answer: At least about $8 \times 10^{8}\,M_\odot$.
⚠️ Common mistake
Thinking a star can be as luminous as it likes, whatever its mass. Light pushes on matter, and that push grows with luminosity while gravity grows with mass. At $L_{\text{Edd}}$ they balance, so a star of mass $M$ cannot shine much above $3.3 \times 10^{4}\,(M/M_\odot)\,L_\odot$ and stay intact. That is one reason the most massive stars known, in the R136 cluster of the Large Magellanic Cloud, have on the order of 200 solar masses and not thousands. Check: radiation force over gravity is $L/L_{\text{Edd}}$, independent of distance.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.