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Phys.org's TESS delivers new insights into an ultrahot world links to KELT-9 b's Asymmetric TESS Transit Caused by Rapid Stellar Rotation and Spin–Orbit Misalignment (readable in arXiv)

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) – change the total irradiance on KELT-9 b (Ahlers 2016).

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

Wikipedia's 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, 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

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.

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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 .

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

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  • $\begingroup$ 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? $\endgroup$ – uhoh Jul 2 at 8:27
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    $\begingroup$ @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). $\endgroup$ – Rob Jeffries Jul 2 at 8:36
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    $\begingroup$ In fact, that is the simple argument as to why $f(\phi)$ can just be replaced with $f(\phi_0)$ at the photosphere. $\endgroup$ – Rob Jeffries Jul 2 at 8:41
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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.

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  • $\begingroup$ 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. $\endgroup$ – uhoh Jul 1 at 10:42
  • $\begingroup$ 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. $\endgroup$ – sunra Jul 1 at 11:22
  • $\begingroup$ 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? and Photosphere is relatively transparent. Is that right? It is the temperature of the photosphere that determines the brightness of a given area. $\endgroup$ – uhoh Jul 1 at 12:07
  • $\begingroup$ 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. $\endgroup$ – ELNJ Jul 1 at 13:00
  • $\begingroup$ 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 cambridge.org/core/books/astrophysics-for-physicists/… $\endgroup$ – sunra Jul 1 at 13:41

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