Skip to main content
Sandbox Physics
Exoplanet day and night

AC21 · L2

Follow a gas parcel into the night

Sandbox Physics ·

A parcel stays warm after sunset when it crosses the dark hemisphere faster than radiation removes its stored heat. To calculate that competition, follow the temperature of a parcel as it moves through the changing field.

Choose the moving frame

Let latitude be β\beta and longitude λ\lambda, measured eastward from the point directly under the star. This is the substellar frame. Longitude zero stays under the star; the dayside lies between the evening and morning boundaries at right angles from it. The orbit is circular, its mean angular speed is nn, and the spin and orbit axes are aligned.

The body has inertial spin Ωcore\Omega_{\rm core}. Give the atmosphere an equatorial eastward speed uequ_{\rm eq} relative to the body, with speed at latitude proportional to cos⁡β\cos\beta. Dividing that speed by the latitude circle's radius gives the same angular wind rate at every latitude. Subtract the orbital rate to express the material motion relative to the substellar meridian:

ωeff=Ωcore+ueqRp−n.\omega_{\rm eff}=\Omega_{\rm core}+\frac{u_{\rm eq}}{R_p}-n.
A synchronized body has Ωcore=n\Omega_{\rm core}=n. The atmosphere can still move because its wind is relative to the body.

The model prescribes that flow. It does not calculate pressure forces that accelerate the wind or supply a three-dimensional weather solution. Independent latitude bands carry heat around their own circles; no heat crosses between bands.

Reading: Nicolas B. Cowan and Eric Agol, A Model for Thermal Phase Variations of Circular and Eccentric Exoplanets · Taylor J. Bell and Nicolas B. Cowan, Increased Heat Transport in Ultra-Hot Jupiter Atmospheres Through H2 Dissociation/Recombination

Account for the temperature of moving material

Let CC be heat stored per surface area per degree, in joules per square metre per kelvin. For constant CC, a patch stores CTCT per unit area. A parcel moving through the field changes temperature for two reasons: the field may change in time, and the parcel may enter a different longitude. Their sum is the material derivative:

dTdt=∂T∂t+ωeff∂T∂λ.\frac{dT}{dt}=\frac{\partial T}{\partial t}+\omega_{\rm eff}\frac{\partial T}{\partial\lambda}.
A partial time derivative watches one location. The material derivative follows the gas occupying different locations in succession.

At latitude and longitude, the perpendicular fraction of the incident rays is cos⁡βcos⁡λ\cos\beta\cos\lambda. On the dark side the fraction is zero. Subtract the reflected stellar fraction, add a specified uniform internal heat flux, and subtract thermal emission. The resulting equation is

C(∂T∂t+ωeff∂T∂λ)=(1−AB)F⋆cos⁡βmax⁡(cos⁡λ,0)+Fint−σT4.C\left(\frac{\partial T}{\partial t}+\omega_{\rm eff}\frac{\partial T}{\partial\lambda}\right) =(1-A_B)F_\star\cos\beta\max(\cos\lambda,0)+F_{\rm int}-\sigma T^4.
Every term is a power per surface area. The layer uses unit thermal emissivity and constant configured capacity, with distinct effective incoming and outgoing spectral bands.

This equation combines transport and radiation. Raising a temperature by hand after solving it would require an additional heat source. The internal term is instead specified as σTint4\sigma T_{\rm int}^4, so its energy is visible in the equation.

Reading: Nicolas B. Cowan and Eric Agol, A Model for Thermal Phase Variations of Circular and Eccentric Exoplanets · Nicolas B. Cowan and Eric Agol, The Statistics of Albedo and Heat Recirculation on Hot Exoplanets

Two cooling calculations

For an isolated dark parcel, temporarily set both illumination and internal heating to zero. Separate temperature and time in the differential equation, then integrate from its initial temperature TiT_i. The integral of the inverse fourth power supplies an inverse cube:

CdTdt=−σT4,∫TiT(t)T−4 dT=−σtC,T(t)=(Ti−3+3σtC)−1/3.C\frac{dT}{dt}=-\sigma T^4,\qquad \int_{T_i}^{T(t)}T^{-4}\,dT=-\frac{\sigma t}{C},\qquad T(t)=\left(T_i^{-3}+\frac{3\sigma t}{C}\right)^{-1/3}.
Nonlinear cooling slows as the parcel cools. A single exponential is not this exact dark solution.

For example, start at 1800 K1800\,\mathrm K with C=107 J m−2K−1C=10^7\,\mathrm{J\,m^{-2}K^{-1}}. After 20000 s20000\,\mathrm s, about five and a half hours, the formula gives approximately 1250 K1250\,\mathrm K. An independent adaptive ODE integration agrees within a few billionths of a kelvin in the numerical reference; that agreement checks the equation's implementation, not the accuracy of a one-layer atmosphere.

A different useful calculation asks how a small perturbation relaxes near an equilibrium temperature Tˉ\bar T. Expand the fourth power and keep only the term linear in a small δT\delta T. Its restoring coefficient is 4σTˉ34\sigma\bar T^3, giving

τlin=C4σTˉ3.\tau_{\rm lin}=\frac{C}{4\sigma\bar T^3}.
At 1500 K1500\,\mathrm K and the same capacity, the local linear relaxation time is about 3.63 h3.63\,\mathrm h.

Cowan and Agol define a nonlinear temperature scale whose corresponding time is C/(σT03)C/(\sigma T_0^3). At the same reference temperature that scale is four times the linear relaxation time. Both definitions are useful, but inserting one where the other was intended changes the transport comparison substantially.

Reading: Nicolas B. Cowan and Eric Agol, A Model for Thermal Phase Variations of Circular and Eccentric Exoplanets · Nicolas B. Cowan and Eric Agol, The Statistics of Albedo and Heat Recirculation on Hot Exoplanets

A small sinusoid gives an independent sign check

To check the direction of advection, consider a weak sinusoidal heating perturbation of amplitude Q1Q_1 around a uniform reference temperature. Uniform background heating first balances the reference emission; the negative half of the perturbation means slightly less heating, with the total source still positive. Write k=4σTˉ3k=4\sigma\bar T^3. In the steady linearized problem, try a cosine shifted by an angle δ\delta and match its sine and cosine coefficients:

CωeffdδTdλ=Q1cos⁡λ−kδT,δT=Q1k2+(Cωeff)2cos⁡(λ−δ),δ=arctan⁡Cωeffk.C\omega_{\rm eff}\frac{d\delta T}{d\lambda}=Q_1\cos\lambda-k\delta T, \qquad \delta T=\frac{Q_1}{\sqrt{k^2+(C\omega_{\rm eff})^2}}\cos(\lambda-\delta), \qquad \delta=\arctan\frac{C\omega_{\rm eff}}{k}.
Positive flow shifts the small temperature maximum eastward. The larger denominator reduces its contrast.

With Q1=20000 W m−2Q_1=20000\,\mathrm{W\,m^{-2}}, Tˉ=1500 K\bar T=1500\,\mathrm K, radius seventy million metres and wind two kilometres per second on a synchronous body, the amplitude is about 24.5 K24.5\,\mathrm K and the shift about 20.520.5 degrees. The perturbation is small compared with the reference temperature, as the approximation requires. This is a linear test problem; it is not the default nonlinear temperature map or its projected phase maximum.

Reading: Nicolas B. Cowan and Eric Agol, A Model for Thermal Phase Variations of Circular and Eccentric Exoplanets · Nicolas B. Cowan and Eric Agol, Inverting Phase Functions to Map Exoplanets

Sum the entire sphere

Integrate the thermal equation over the sphere. A periodic longitude integral of a longitude derivative is zero: the heat leaving one end enters the other. Constant capacity and the prescribed longitude-independent angular rate make this cancellation exact in the model. Defining total stored energy UU, we obtain

U=Rp2∫CT dΩ,dUdt=Labs+Lint−Lthermal.U=R_p^2\int CT\,d\Omega,\qquad \frac{dU}{dt}=L_{\rm abs}+L_{\rm int}-L_{\rm thermal}.
Advection rearranges heat. Radiation and the specified sources change the total.

A numerical cell represents an area average. The conservative implementation uses the same heat flux on both sides of a shared edge; opposite contributions cancel when cells are summed. The upwind side is the side the gas comes from, so it reverses when the flow changes sign. An implicit timestep solves storage, transport and emission together. It remains stable at short cooling times, but its spatial and temporal diffusion still require refinement checks.

Changing CC during an experiment changes the relationship between temperature and stored energy. If temperature is retained, storage jumps by the integral of ΔC T\Delta C\,T. That is an external configuration event, saved separately from subsequent heating and cooling. Starting a preset instead creates a new steady record with its own initial energy.

Reading: Nicolas B. Cowan and Eric Agol, A Model for Thermal Phase Variations of Circular and Eccentric Exoplanets · Nicolas B. Cowan and Eric Agol, The Statistics of Albedo and Heat Recirculation on Hot Exoplanets

What a steady map can identify

For steady circular illumination the partial time derivative vanishes. Capacity and relative flow occur only through their product. Doubling capacity and halving the angular rate gives exactly the same steady equation and the same temperature field. Their transient response need not be the same.

The default discrete result gives a net nightside fraction near 0.1390.139; reversing flow mirrors the map, and increasing capacity carries more warmth around each latitude before it radiates. At high transport, each latitude becomes nearly uniform in longitude. Its mean illumination remains latitude dependent, so the model retains a pole-to-equator temperature variation.

A phase curve views the same map through changing projection weights. Separating the two thermal parameters requires additional information. The next chapter derives that projection and shows another source of ambiguity: different brightness maps can give the same total light.

Reading: Nicolas B. Cowan and Eric Agol, A Model for Thermal Phase Variations of Circular and Eccentric Exoplanets · Nicolas B. Cowan and Eric Agol, Inverting Phase Functions to Map Exoplanets

Exercises and explanations

  1. Double heat capacity and halve relative angular flow in a steady circular model. What changes in the temperature map?

    Read the explanation

    Their product stays fixed in the steady equation, so the map is unchanged. Transient response times do change.

  2. The same angular flow carries heat at all latitudes. Does very rapid flow make the sphere globally uniform?

    Read the explanation

    It averages the heating around each latitude, but bands receive different average starlight. Without meridional exchange, latitude differences survive.

Try the prediction

Test the steady capacity–flow ridge

  1. Pause Eastward heat before changing either parameter.
  2. Double capacity and halve the eastward wind while the body stays synchronous.
  3. Compare the unchanged profile, then run time and explain why no new steady field is required.

Sources and calculation scope

  1. Nicolas B. Cowan and Eric Agol, A Model for Thermal Phase Variations of Circular and Eccentric Exoplanets

    arXiv:1011.0428v1, submitted 2010-11-01. Sections 2–3, equations 1–10; section 4, equation 13; section 6; figures 1 and 6.

  2. Nicolas B. Cowan and Eric Agol, Inverting Phase Functions to Map Exoplanets

    arXiv:0803.3622v1, submitted 2008-03-25. Section 2, equations 1–4; section 3.2, equation 7; figure 1.

  3. Nicolas B. Cowan and Eric Agol, The Statistics of Albedo and Heat Recirculation on Hot Exoplanets

    arXiv:1001.0012v2, revised 2010-12-20. Sections 1, 2.1 and 2.2; equation 3.

  4. Taylor J. Bell and Nicolas B. Cowan, Increased Heat Transport in Ultra-Hot Jupiter Atmospheres Through H2 Dissociation/Recombination

    arXiv:1802.07725v2, revised 2018-04-09. Sections 2.1–2.3, equations 1–8; figures 1–4.

FeedbackSomething broken, confusing, or missing on this page?Tell us
What kind of feedback?

Sent with this page’s address, its Lab ID, the site language and your device type. Please leave out personal details. Feedback is kept for up to 12 months.Privacy policy