<p>The Eddington limit represents the maximum luminosity that can be achieved by a body (such as the star) when there is hydrostatic equilibrium (<a href="http://astronomy.swin.edu.au/cosmos/H/Hydrostatic+Equilibrium" rel="noreferrer">http://astronomy.swin.edu.au/cosmos/H/Hydrostatic+Equilibrium</a>). For luminosity greater than Eddington limit, the radiative force of the luminosity on matter exceeds the gravitational force on the matter. If the luminosity radiated by an accretion disk exceeds the Eddington limit, the matter falling towards the supermassive black hole can be blown away. </p>

CC BY-SA 3.0
2017-04-17T11:16:54.447Z
Welcome to Astronomy SE. Thanks for this useful answer. The question asks about black holes; does that have any bearing the answer?
James K
CC BY-SA 3.0
2017-04-17T12:58:25.103Z
Note that the Eddington luminosity is just a type of luminosity unit. It does not represent an absolute maximum because Super-Eddington accretion rates are possible - see https://physics.stackexchange.com/questions/269393/the-difference-between-sub-eddington-and-super-eddington-accretion-disk-models
ProfRob
CC BY-SA 3.0
2017-04-17T06:38:41.580Z
2017-04-17T06:38:41.580Z
Ruchi Pandey
<p>We know Eddington Limit (Eddington Luminosity) is calculated with this formula:</p>

<p>$$L_{Ed} \approx 3 * 10^4 * L_{sun} * \frac{M}{M_{sun}}$$</p>

<p>Example: For a black hole with $10^4 M_{sun}$, we get
$$L_{Ed} = 1.16 * 10^{35} W$$</p>

<p>My question is what does it really show? What do we get by calculating this Number? Is this formula really used to calculate the luminosity or it is used for calculating mass?</p>

<astrophysics><luminosity><radiation><eddington-luminosity>
What does Eddington Limit (Eddington Luminosity) Show?
titansarus
# What does Eddington Limit (Eddington Luminosity) Show?

(Asked by: username_0 on 2017-04-17)

We know Eddington Limit (Eddington Luminosity) is calculated with this formula:

$$L\_{Ed} \approx 3 \* 10^4 \* L\_{sun} \* \frac{M}{M\_{sun}}$$

Example: For a black hole with $10^4 M\_{sun}$, we get
$$L\_{Ed} = 1.16 \* 10^{35} W$$

My question is what does it really show? What do we get by calculating this Number? Is this formula really used to calculate the luminosity or it is used for calculating mass?

**Tags:** astrophysics, luminosity, radiation, eddington-luminosity


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## 1 answers

### Answer by username_1 on 2017-04-17. Score: 6

The Eddington limit represents the maximum luminosity that can be achieved by a body (such as the star) when there is hydrostatic equilibrium (<http://astronomy.swin.edu.au/cosmos/H/Hydrostatic+Equilibrium>). For luminosity greater than Eddington limit, the radiative force of the luminosity on matter exceeds the gravitational force on the matter. If the luminosity radiated by an accretion disk exceeds the Eddington limit, the matter falling towards the supermassive black hole can be blown away.

> **Comments:**
> **username_2**: Welcome to Astronomy SE. Thanks for this useful answer. The question asks about black holes; does that have any bearing the answer?  
> **username_3**: Note that the Eddington luminosity is just a type of luminosity unit. It does not represent an absolute maximum because Super-Eddington accretion rates are possible - see https://physics.stackexchange.com/questions/269393/the-difference-between-sub-eddington-and-super-eddington-accretion-disk-models  

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