<p>Hot, massive stars tend to be (a) asymmetric, due to their rapid rotation (at least on the main sequence); and (b) very luminous, and thus nearer to the Eddington limit, but the two aren’t causally connected to each other.  So perhaps that answer was using correlation in place of causation.</p>
<p>The rapid rotation is due to the fact that hot stars lack the dynamo-driven magnetic fields that let lower-mass stars shed angular momentum and spin down over time.  The high luminosity is just a natural consequence of maintaining the higher temperature and pressure needed to support the higher mass, i.e. to maintain hydrostatic equilibrium.</p>
<p>It is closer to being true the other way around: because of rapid rotation, the effective gravitational field at the star’s equator is lower, and thus that part of the star will be closer to the Eddington limit at a given luminosity than a non-rotating star would be.</p>

CC BY-SA 4.0
2020-07-07T12:47:49.117Z
I see, we're calling oblate stars with cylindrical symmetry "asymmetric" because they are not spherically symmetric? And the "Eddington limit" can really be applied to discrete regions of a star's surface?
uhoh
CC BY-SA 4.0
2020-07-07T12:56:36.263Z
Yes, correct on both counts - I meant not *spherically* symmetric.  And what matters for mass loss is whether the local radiation pressure exceeds the local effective gravitational force; it doesn’t have to be the same everywhere.
Eric Jensen
CC BY-SA 4.0
2020-07-08T08:20:03.713Z
If part of a star is close to the Eddington limit then small variations (analagous to sunspots) might be enough to lift large masses of material significant distances -- it's quite finely balanced between radiation pressure and gravity to start with. So the star is less strongly forced into a spheroidal shape than less bright one.
Steve Linton
CC BY-SA 4.0
2020-07-07T12:36:54.877Z
2020-07-07T12:36:54.877Z
Eric Jensen
<p><a href="https://astronomy.stackexchange.com/a/36808/7982">This answer</a> to <em>Is the gravitational field of the sun uniform?</em> says:</p>
<blockquote>
<p>Very large stars are asymmetric due to being close to the Eddington limit. The Sun is too small of a star to have the kind of gravitational anomalies that very large stars and rocky planets exhibit.</p>
</blockquote>
<p>which is really interesting but leaves me unsatisfied.</p>
<p><strong>Question:</strong> How exactly does the Eddington limit make the gravitational field of stars asymmetric?</p>
<hr />
<ul>
<li><a href="https://astronomy.stackexchange.com/q/20761/7982">What does Eddington Limit (Eddington Luminosity) Show?</a></li>
</ul>
<p>From wikipedia:</p>
<ul>
<li><a href="https://en.wikipedia.org/wiki/Eddington_luminosity" rel="nofollow noreferrer">Eddington luminosity</a></li>
<li><a href="https://en.wikipedia.org/wiki/Eddington_luminosity#Humphreys%E2%80%93Davidson_limit" rel="nofollow noreferrer">Humphreys–Davidson limit</a></li>
</ul>

<stellar-evolution><mass><luminosity><eddington-luminosity>
How does the Eddington limit make the gravitational field of stars asymmetric?
uhoh
# How does the Eddington limit make the gravitational field of stars asymmetric?

(Asked by: username_0 on 2020-07-06)

[This answer](https://astronomy.stackexchange.com/a/36808/7982) to *Is the gravitational field of the sun uniform?* says:

> Very large stars are asymmetric due to being close to the Eddington limit. The Sun is too small of a star to have the kind of gravitational anomalies that very large stars and rocky planets exhibit.

which is really interesting but leaves me unsatisfied.

**Question:** How exactly does the Eddington limit make the gravitational field of stars asymmetric?

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* [What does Eddington Limit (Eddington Luminosity) Show?](https://astronomy.stackexchange.com/q/20761/7982)

From wikipedia:

* [Eddington luminosity](https://en.wikipedia.org/wiki/Eddington_luminosity)
* [Humphreys–Davidson limit](https://en.wikipedia.org/wiki/Eddington_luminosity#Humphreys%E2%80%93Davidson_limit)

**Tags:** stellar-evolution, mass, luminosity, eddington-luminosity


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## 1 answers

### Answer by username_1 on 2020-07-07. Score: 2 | Accepted answer

Hot, massive stars tend to be (a) asymmetric, due to their rapid rotation (at least on the main sequence); and (b) very luminous, and thus nearer to the Eddington limit, but the two aren’t causally connected to each other. So perhaps that answer was using correlation in place of causation.

The rapid rotation is due to the fact that hot stars lack the dynamo-driven magnetic fields that let lower-mass stars shed angular momentum and spin down over time. The high luminosity is just a natural consequence of maintaining the higher temperature and pressure needed to support the higher mass, i.e. to maintain hydrostatic equilibrium.

It is closer to being true the other way around: because of rapid rotation, the effective gravitational field at the star’s equator is lower, and thus that part of the star will be closer to the Eddington limit at a given luminosity than a non-rotating star would be.

> **Comments:**
> **username_0**: I see, we're calling oblate stars with cylindrical symmetry "asymmetric" because they are not spherically symmetric? And the "Eddington limit" can really be applied to discrete regions of a star's surface?  
> **username_1**: Yes, correct on both counts - I meant not *spherically* symmetric.  And what matters for mass loss is whether the local radiation pressure exceeds the local effective gravitational force; it doesn’t have to be the same everywhere.  
> **username_2**: If part of a star is close to the Eddington limit then small variations (analagous to sunspots) might be enough to lift large masses of material significant distances -- it's quite finely balanced between radiation pressure and gravity to start with. So the star is less strongly forced into a spheroidal shape than less bright one.  

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