<p>The argument goes something like this.</p>
<p>Hydrostatic equilibrium means that the local pressure gradient is proportional to the local density multiplied by a latitude-dependent local gravity. If the pressure just depends on density and temperature, this means that those quantities will also just depend on latitude and therefore will be constant along an equipotential surface. i.e. pressure, temperature and density are functions of the effective gravitational potential <span class="math-container">$\phi$</span>.</p>
<p>For stars with radiative outer envelopes, the heat flux is proportional to the temperature <em>gradient</em>, multiplied by some stuff (like inverse opacity) that just depends on density and temperature.</p>
<p>But
<span class="math-container">$$\nabla T(\phi) = \frac{dT}{d\phi}\nabla \phi = f(\phi)g_{\rm eff}$$</span></p>
<p>If we now say that at the surface <span class="math-container">$\sigma T_{\rm eff}^4$</span> equals the radiative flux, then we recover the Von Zeipel gravity darkening law that <span class="math-container">$T_{\rm eff}$</span> is proportional to <span class="math-container">$g_{\rm eff}^{1/4}$</span>.</p>
<p>The missing step in this argument is to show the <span class="math-container">$f(\phi)$</span> is constant. Given that the photosphere is defined as where the optical depth is some fixed value (usually 1 or 2/3), and may be assumed to depend only on temperature and density, then this also lies on an equipotential. But <span class="math-container">$dT/d\phi$</span> also only depends on <span class="math-container">$\phi$</span> and so must also be constant along an equipotential.</p>
<p>For more detail, although missing the last paragraph above(!), see <a href="https://www.astro.umd.edu/%7Ejph/Stellar_Rotation.pdf" rel="nofollow noreferrer">https://www.astro.umd.edu/~jph/Stellar_Rotation.pdf</a> .</p>
<p>The situation is much more complex for stars with convective envelopes or differential rotation and I think can only be tackled through detailed modelling.</p>

CC BY-SA 4.0
2020-07-02T08:27:30.500Z
Thank you for the answer! Would it be safe then to say that at the equator the photosphere has a lower temperature *and therefore a higher density* than it does at the poles?
uhoh
CC BY-SA 4.0
2020-07-02T08:36:41.133Z
@uhoh Along an equipotential the pressure would be constant. I don't think that is quite the same thing. Where the photosphere is depends on density and temperature in a complex way. But given that I said all the relevant quantitites are functions of the potential, then I guess where the optical depth is 1 will also be on an equipotential. So I think you are right (if we are just talking about a perfect gas).
ProfRob
CC BY-SA 4.0
2020-07-02T08:41:55.767Z
In fact, that is the simple argument as to why $f(\phi)$ can just be replaced with $f(\phi_0)$ at the photosphere.
ProfRob
CC BY-SA 4.0
2020-07-02T08:15:14.310Z
2020-07-02T08:49:13.413Z
2020-07-02T08:49:13.413Z
ProfRob
<p>From the same Wikipedia page:</p>
<blockquote>
<p>This means that equatorial regions of a star will have a greater centrifugal force when compared to the pole. The centrifugal force pushes mass away from the axis of rotation, and results in less overall pressure on the gas in the equatorial regions of the star. This will cause the gas in this region to become less dense, and cooler.</p>
</blockquote>
<p>So it appears that the equatorial bulge is caused centrifugally via rapid rotation (as expected). This outwards-directed force relieves the pressure acting inwards arising from gravitational contraction, and of course temperature is proportional to pressure. Therefore surface temperature will be higher at the poles than the equator.</p>

CC BY-SA 4.0
2020-07-01T10:42:49.320Z
Hmm... pressure is proportional to temperature *at constant volume* but I think it's necessary to explain further here. The temperature of the radiated light is that of the photosphere, and so we need to understand why the photosphere is cooler near the equator. I'm not sure the Wikipedia article really addresses this.
uhoh
CC BY-SA 4.0
2020-07-01T11:22:03.663Z
I don't know these objects, but I assume the rotation speed is constant.

The equations of hydrostatic EQ say pressure gradient is proportional to density. Density will be higher when surface gravity is stronger. The luminosity at the surface is equal to the energy flux at that distance which is proportional to density.
sunra
CC BY-SA 4.0
2020-07-01T12:07:07.733Z
I don't think stars have proper "surfaces". There is a layer called the photosphere where the material becomes transparent enough to let light escape to infinity, and its distance from the star's center is determined by *both temperature and density*. See [What is the density profile within the Sun's photosphere? Which one of these is wrong?](https://astronomy.stackexchange.com/a/32737/7982) and [Photosphere is relatively transparent. Is that right?](https://astronomy.stackexchange.com/q/18970/7982) It is the temperature of the photosphere that determines the brightness of a given area.
uhoh
CC BY-SA 4.0
2020-07-01T13:00:20.933Z
One way to think about the photosphere location is that there needs to be a certain column of material above it, just enough so that a photon can escape from there.  (It’s a surface at a certain optical depth, and absent big equator-to-pole T variations we could take optical depth and column density as proportional to each other.)  So that fixed column above a spot on the photosphere has a certain mass, but at the poles it will be closer to the star’s center, and thus experience more gravitational force, than at the equator.  Therefore, photospheric P and T will need to be higher at the pole.
Eric Jensen
CC BY-SA 4.0
2020-07-01T13:41:20.233Z
The surface is a mathematical boundary a distance r = R from the centre. I recommend you study hydrostatic equilibrium. "Astrophysics for Physicists" is an excellent postgraduate level book https://www.cambridge.org/core/books/astrophysics-for-physicists/9F6DBBDAAF177504899F5BB40B3BDE33
sunra
CC BY-SA 4.0
2020-07-01T13:48:57.567Z
@sunra  In the rotating reference frame of the star, a gravitational equipotential surface won’t be a sphere, i.e. it won’t all be at a constant distance r from the center.
Eric Jensen
CC BY-SA 4.0
2020-07-01T14:25:38.870Z
1/ Basic hydrostatic equilibrium provides an approximation in spherical symmetry. Like most things in astronomy, it is an estimate and in this case even more so since this object is extremely oblate. But, the principle is the same even if the real homologous equations would be much more complicated. Hydrostatic equilibrium compares very well with observation. If we take averages of internal values we get some nice scaling relationships such as: P ∝ M^2/R^4, e.g. pressure inversely proportional to the 4th power of radius, and T ∝ M/R, e.g. temperature is inversely proportional to radius.
sunra
CC BY-SA 4.0
2020-07-01T14:26:52.823Z
2/ Now these are just average values, but the differential equations I mentioned before will give you results at varying radius. Hydrostatic EQ treats the star as a hydrostatic fluid balancing radiation pressure and gravitational collapse and also applies to rotating stars, but in this case a more accurate model would be needed due to the extreme angular velocity. Either way the OP is complicating the situation somewhat. Reason why the temp. at the equator is cooler is explained in the Wiki page. If he needs more technical description, he should definitely read some peer reviewed articles.
sunra
CC BY-SA 4.0
2020-07-01T18:21:57.787Z
A star's rotation period isn't constant. From https://en.wikipedia.org/wiki/Solar_rotation "The solar rotation period is 24.47 days at the equator and almost 38 days at the poles". (Those are sidereal periods, we get slightly different numbers viewing the Sun from Earth). There's an equation in that article which gives the angular velocity as a function of latitude.
PM 2Ring
CC BY-SA 4.0
2020-07-03T16:41:31.453Z
@PM_2Ring really it is an approximation and it is what is taught in a postgraduate level stellar astrophysics course. looking at the answer that was accepted by the OP, it is exactly the same argument I made in my first comment about the pressure gradient of hydrostatic EQ. literally the same. I guess suggesting to rtfm was a bit harsh xD
sunra
CC BY-SA 4.0
2020-07-01T10:36:15.397Z
2020-07-01T10:36:15.397Z
sunra
<p>Phys.org's <a href="https://phys.org/news/2020-06-tess-insights-ultrahot-world.html" rel="nofollow noreferrer">TESS delivers new insights into an ultrahot world</a> links to <a href="https://iopscience.iop.org/article/10.3847/1538-3881/ab8fa3" rel="nofollow noreferrer">KELT-9 b's Asymmetric TESS Transit Caused by Rapid Stellar Rotation and Spin–Orbit Misalignment</a> (readable in <a href="https://arxiv.org/abs/2004.14812" rel="nofollow noreferrer">arXiv</a>)</p>
<p>The assymetric dip in the light curve comes from a near-polar transit across a rotating, oblate star where the poles are hotter and therefore brighter due to gravity darkening:</p>
<blockquote>
<p>KELT-9’s high internal angular momentum (<span class="math-container">$\nu \sin(i)$</span> = 111.4 ± 1.3 km/s) flattens it into an oblate spheroid, making the equatorial radius of the star larger than the polar radius. Additionally, the star’s abundant centrifugal force near its equator distorts its hydrostatic equilibrium, causing its effective temperature to vary by nearly a thousand Kelvin over the surface of the star. These two effects of stellar oblateness and varying effective temperature – together commonly referred to as gravity darkening (<a href="https://iopscience.iop.org/article/10.1088/0004-637X/705/1/683/meta" rel="nofollow noreferrer">Barnes 2009</a>) – change the total irradiance on KELT-9 b (<a href="https://iopscience.iop.org/article/10.3847/0004-637X/832/1/93/meta" rel="nofollow noreferrer">Ahlers 2016</a>).</p>
</blockquote>
<p>Those links discuss gravity darkening but don't offer a simple explanation.</p>
<p>Wikipedia's <a href="https://en.wikipedia.org/wiki/Gravity_darkening" rel="nofollow noreferrer">Gravity darkening</a> says:</p>
<blockquote>
<p>When a star is oblate, it has a larger radius at its equator than it does at its poles. As a result, the poles have a higher surface gravity, and thus higher temperature and brightness.</p>
</blockquote>
<p><strong>Question:</strong> Why exactly does the increased surface gravity at some places on a given star lead to a higher temperature at those locations? Is it related to the difference in scale heights? The surface brightness relates to the temperature at the <a href="https://en.wikipedia.org/wiki/Photosphere" rel="nofollow noreferrer">photosphere</a>, is the reason simply that a higher pressure therefore higher temperature is needed to support the same density in a higher gravitational field?</p>
<p><a href="https://i.sstatic.net/JuH3v.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/JuH3v.png" alt="Figure 2. (Left) KELT-9 b transit from https://arxiv.org/abs/2004.14812" /></a></p>
<blockquote>
<p>Figure 2. (Left) KELT-9 b begins its transit near the star’s hot pole and moves toward the star’s cooler equator. Our transit analysis directly measures the stellar inclination (<em>i</em><sub>★</sub>), the planet’s projected alignment (<em>λ</em>), and the orbital inclination (i.e., the impact parameter <em>b</em>). We find that KELT-9 varies in effective temperature by ∼ 800 K between its hot poles and cooler equator. (Right) KELT-9 b’s phase-folded primary transit from TESS. The transit depth steadily decreases throughout the eclipse, indicating that KELT-9 b begins its transit near one of the host star’s hotter poles and moves toward the dimmer stellar equator.</p>
</blockquote>

<exoplanet><rotation><stellar-astrophysics><stellar-structure>
How to understand exactly why gravity darkening happens on rotating stars?
uhoh
# How to understand exactly why gravity darkening happens on rotating stars?

(Asked by: username_0 on 2020-06-30)

Phys.org's [TESS delivers new insights into an ultrahot world](https://phys.org/news/2020-06-tess-insights-ultrahot-world.html) links to [KELT-9 b's Asymmetric TESS Transit Caused by Rapid Stellar Rotation and Spin–Orbit Misalignment](https://iopscience.iop.org/article/10.3847/1538-3881/ab8fa3) (readable in [arXiv](https://arxiv.org/abs/2004.14812))

The assymetric dip in the light curve comes from a near-polar transit across a rotating, oblate star where the poles are hotter and therefore brighter due to gravity darkening:

> KELT-9’s high internal angular momentum ($\nu \sin(i)$ = 111.4 ± 1.3 km/s) flattens it into an oblate spheroid, making the equatorial radius of the star larger than the polar radius. Additionally, the star’s abundant centrifugal force near its equator distorts its hydrostatic equilibrium, causing its effective temperature to vary by nearly a thousand Kelvin over the surface of the star. These two effects of stellar oblateness and varying effective temperature – together commonly referred to as gravity darkening ([Barnes 2009](https://iopscience.iop.org/article/10.1088/0004-637X/705/1/683/meta)) – change the total irradiance on KELT-9 b ([Ahlers 2016](https://iopscience.iop.org/article/10.3847/0004-637X/832/1/93/meta)).

Those links discuss gravity darkening but don't offer a simple explanation.

Wikipedia's [Gravity darkening](https://en.wikipedia.org/wiki/Gravity_darkening) says:

> When a star is oblate, it has a larger radius at its equator than it does at its poles. As a result, the poles have a higher surface gravity, and thus higher temperature and brightness.

**Question:** Why exactly does the increased surface gravity at some places on a given star lead to a higher temperature at those locations? Is it related to the difference in scale heights? The surface brightness relates to the temperature at the [photosphere](https://en.wikipedia.org/wiki/Photosphere), is the reason simply that a higher pressure therefore higher temperature is needed to support the same density in a higher gravitational field?

[![Figure 2. (Left) KELT-9 b transit from https://arxiv.org/abs/2004.14812](https://i.sstatic.net/JuH3v.png)](https://i.sstatic.net/JuH3v.png)

> Figure 2. (Left) KELT-9 b begins its transit near the star’s hot pole and moves toward the star’s cooler equator. Our transit analysis directly measures the stellar inclination (*i*★), the planet’s projected alignment (*λ*), and the orbital inclination (i.e., the impact parameter *b*). We find that KELT-9 varies in effective temperature by ∼ 800 K between its hot poles and cooler equator. (Right) KELT-9 b’s phase-folded primary transit from TESS. The transit depth steadily decreases throughout the eclipse, indicating that KELT-9 b begins its transit near one of the host star’s hotter poles and moves toward the dimmer stellar equator.

**Tags:** exoplanet, rotation, stellar-astrophysics, stellar-structure


---

## 2 answers

### Answer by username_1 on 2020-07-02. Score: 3 | Accepted answer

The argument goes something like this.

Hydrostatic equilibrium means that the local pressure gradient is proportional to the local density multiplied by a latitude-dependent local gravity. If the pressure just depends on density and temperature, this means that those quantities will also just depend on latitude and therefore will be constant along an equipotential surface. i.e. pressure, temperature and density are functions of the effective gravitational potential $\phi$.

For stars with radiative outer envelopes, the heat flux is proportional to the temperature *gradient*, multiplied by some stuff (like inverse opacity) that just depends on density and temperature.

But
$$\nabla T(\phi) = \frac{dT}{d\phi}\nabla \phi = f(\phi)g\_{\rm eff}$$

If we now say that at the surface $\sigma T\_{\rm eff}^4$ equals the radiative flux, then we recover the Von Zeipel gravity darkening law that $T\_{\rm eff}$ is proportional to $g\_{\rm eff}^{1/4}$.

The missing step in this argument is to show the $f(\phi)$ is constant. Given that the photosphere is defined as where the optical depth is some fixed value (usually 1 or 2/3), and may be assumed to depend only on temperature and density, then this also lies on an equipotential. But $dT/d\phi$ also only depends on $\phi$ and so must also be constant along an equipotential.

For more detail, although missing the last paragraph above(!), see [https://www.astro.umd.edu/~jph/Stellar\_Rotation.pdf](https://www.astro.umd.edu/%7Ejph/Stellar_Rotation.pdf) .

The situation is much more complex for stars with convective envelopes or differential rotation and I think can only be tackled through detailed modelling.

> **Comments:**
> **username_0**: Thank you for the answer! Would it be safe then to say that at the equator the photosphere has a lower temperature *and therefore a higher density* than it does at the poles?  
> **username_1**: @username_0 Along an equipotential the pressure would be constant. I don't think that is quite the same thing. Where the photosphere is depends on density and temperature in a complex way. But given that I said all the relevant quantitites are functions of the potential, then I guess where the optical depth is 1 will also be on an equipotential. So I think you are right (if we are just talking about a perfect gas).  
> **username_1**: In fact, that is the simple argument as to why $f(\phi)$ can just be replaced with $f(\phi_0)$ at the photosphere.  

---

### Answer by username_2 on 2020-07-01. Score: 2

From the same Wikipedia page:

> This means that equatorial regions of a star will have a greater centrifugal force when compared to the pole. The centrifugal force pushes mass away from the axis of rotation, and results in less overall pressure on the gas in the equatorial regions of the star. This will cause the gas in this region to become less dense, and cooler.

So it appears that the equatorial bulge is caused centrifugally via rapid rotation (as expected). This outwards-directed force relieves the pressure acting inwards arising from gravitational contraction, and of course temperature is proportional to pressure. Therefore surface temperature will be higher at the poles than the equator.

> **Comments:**
> **username_0**: Hmm... pressure is proportional to temperature *at constant volume* but I think it's necessary to explain further here. The temperature of the radiated light is that of the photosphere, and so we need to understand why the photosphere is cooler near the equator. I'm not sure the Wikipedia article really addresses this.  
> **username_2**: I don't know these objects, but I assume the rotation speed is constant.  The equations of hydrostatic EQ say pressure gradient is proportional to density. Density will be higher when surface gravity is stronger. The luminosity at the surface is equal to the energy flux at that distance which is proportional to density.  
> **username_0**: I don't think stars have proper "surfaces". There is a layer called the photosphere where the material becomes transparent enough to let light escape to infinity, and its distance from the star's center is determined by *both temperature and density*. See [What is the density profile within the Sun's photosphere? Which one of these is wrong?](https://astronomy.stackexchange.com/a/32737/7982) and [Photosphere is relatively transparent. Is that right?](https://astronomy.stackexchange.com/q/18970/7982) It is the temperature of the photosphere that determines the brightness of a given area.  
> **username_3**: One way to think about the photosphere location is that there needs to be a certain column of material above it, just enough so that a photon can escape from there.  (It’s a surface at a certain optical depth, and absent big equator-to-pole T variations we could take optical depth and column density as proportional to each other.)  So that fixed column above a spot on the photosphere has a certain mass, but at the poles it will be closer to the star’s center, and thus experience more gravitational force, than at the equator.  Therefore, photospheric P and T will need to be higher at the pole.  
> **username_2**: The surface is a mathematical boundary a distance r = R from the centre. I recommend you study hydrostatic equilibrium. "Astrophysics for Physicists" is an excellent postgraduate level book https://www.cambridge.org/core/books/astrophysics-for-physicists/9F6DBBDAAF177504899F5BB40B3BDE33  
> **username_3**: @username_2  In the rotating reference frame of the star, a gravitational equipotential surface won’t be a sphere, i.e. it won’t all be at a constant distance r from the center.  
> **username_2**: 1/ Basic hydrostatic equilibrium provides an approximation in spherical symmetry. Like most things in astronomy, it is an estimate and in this case even more so since this object is extremely oblate. But, the principle is the same even if the real homologous equations would be much more complicated. Hydrostatic equilibrium compares very well with observation. If we take averages of internal values we get some nice scaling relationships such as: P ∝ M^2/R^4, e.g. pressure inversely proportional to the 4th power of radius, and T ∝ M/R, e.g. temperature is inversely proportional to radius.  
> **username_2**: 2/ Now these are just average values, but the differential equations I mentioned before will give you results at varying radius. Hydrostatic EQ treats the star as a hydrostatic fluid balancing radiation pressure and gravitational collapse and also applies to rotating stars, but in this case a more accurate model would be needed due to the extreme angular velocity. Either way the OP is complicating the situation somewhat. Reason why the temp. at the equator is cooler is explained in the Wiki page. If he needs more technical description, he should definitely read some peer reviewed articles.  
> **username_4**: A star's rotation period isn't constant. From https://en.wikipedia.org/wiki/Solar_rotation "The solar rotation period is 24.47 days at the equator and almost 38 days at the poles". (Those are sidereal periods, we get slightly different numbers viewing the Sun from Earth). There's an equation in that article which gives the angular velocity as a function of latitude.  
> **username_2**: @PM_2Ring really it is an approximation and it is what is taught in a postgraduate level stellar astrophysics course. looking at the answer that was accepted by the OP, it is exactly the same argument I made in my first comment about the pressure gradient of hydrostatic EQ. literally the same. I guess suggesting to rtfm was a bit harsh xD  

---

