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I am new to stellar astrophysics and trying to understand the energy transports in the interior of stars.

Can the energy transport by radiation occur in the convection region of a star?

Here are my thoughts:

Since the gas pressure in the convection region dominates, $\nabla_{\rm gas}$ $<$ $\nabla_{\rm radiative}$, so energy transport by radiation will not occur in the region.

Please correct me if I am wrong.

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    $\begingroup$ Welcome to astronomy SE! $\endgroup$ – B--rian Mar 30 at 11:19
  • $\begingroup$ I have always wondered about this as well. Radiation transport must still take place in a region with convective energy transport, but I don't know whether the ratio between convective/radiative fluxes is ~2-3 or rather several orders of magnitude. I suspect the latter, as radiative fluxes go as $F~1/(2/3+\tau)$ for an optical depth $\tau$, and hence can drop to arbitrary low levels. One would have to compare this with convective fluxes obtained from mixing length-theory. $\endgroup$ – AtmosphericPrisonEscape Mar 30 at 12:24
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    $\begingroup$ Best question of the year so far for me. I've never seen this calculation done. $\endgroup$ – ProfRob Mar 30 at 15:45
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Radiative energy transport continues. The point is that the radiative flux, which is proportional to $dT/dr$ can be overtaken when the temperature gradient achieves the adiabatic value and convection starts. Once convection is started, it is very efficient and the majority of energy flux will be transported by convection.

Details

Broadly speaking, radiative flux is ever-present and proportional to $T^3(dT/dr)/\kappa$, where $\kappa$ is the opacity.

Convective flux begins when the temperature gradient exceeds the adiabatic value and is proportional to the difference between these two to the power of $3/2$ and to the temperature, local gravity and other thermodynamic properties of the gas.

In most cases what happens is that convection is very efficient when it turns on. This means that most of the energy flux is transported by convection and the temperature gradient is almost equal to the adiabatic value. Thus you can estimate a fraction of the flux carried by radiation by assuming the adiabatic temperature gradient$^1$.

I think this fraction is small throughout the bulk of most convection zones.

There are however some circumstances where you need to be a bit more careful. These are where the density/temperature/gravity/opacities conspire to stop convection. So at the edges of convection zones there will be more complicated regions where the fraction of radiative flux grows and the temperature gradient can be superadiabatic. An obvious example is the outer parts of cool stars where the convective envelope changes into a photosphere dominated by radiative heat transport.

$^1$ An example:

In the convection zone

In the convection zone in the Sun at $r\sim 0.9R_{\odot}$, with an enclosed mass of $m\sim 1M_{\odot}$, the temperature $T \sim 10^6$ K, density $\rho \sim 20$ kg/m$^3$ and pressure $P \sim 10^{11}$ Pa.

The adiabatic temperature gradient $$ \left(\frac{dT}{dr}\right)_{\rm ad} = \frac{T}{P}\left(\frac{\gamma-1}{\gamma}\right)\frac{Gm\rho}{r^2} \simeq 3\times 10^{-2}\ {\rm Km}^{-1}\ ,$$ where the ratio of specific heats $\gamma=5/3$.

The radiative heat flux is given by $$F_{\rm rad} = \frac{16\sigma_{sb} T^3}{3\kappa \rho}\left(\frac{dT}{dr}\right)\ , $$ where $\kappa$ is the opacity. From a plot in Turck-Chieze & Couvidat (2011), I estimate $\kappa \simeq 30$ m$^2$/kg at this radius.

Plugging the numbers in: $F_{\rm rad} \sim 1.5\times 10^7$ W m$^{-2}$.

This is to be compared with the actual energy flux of $L_\odot/4\pi r^2 = 8\times 10^7$ W m$^{-2}$.

Thus, the convective flux must be about a factor of 5 times larger than the radiative flux.

In the radiative zone

Then to check the methodology:

Consider somewhere well inside the radiative zone at $r=0.2R_\odot$, where $\rho \sim 4\times 10^4$ kg/m$^3$, $T\sim 9\times 10^6$ K, the enclosed mass $m \sim 0.35M_\odot$ and $\kappa \sim 0.2$ m$^2$/kg.

The adiabatic temperature gradient is $\sim 0.1$ K m$^{-1}$ and using this in the radiative flux formula gives $F_{\rm rad} \sim 3\times 10^9$ W m$^{-2}$ compared with the actual flux of $1.6\times 10^9$ W m$^{-2}$. i.e. In this case the temperature gradient must be about half of the adiabatic gradient, such that all of the flux is carried by radiation.

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  • $\begingroup$ This is curious, shouldn't the convective flux be comparable to the radiative flux at the base of the convection zone? $\endgroup$ – AtmosphericPrisonEscape Mar 30 at 17:21
  • $\begingroup$ @AtmosphericPrisonEscape my numbers aren't for the base of the convection zone. But I see what you mean. I need to repeat the calculation actually in the radiative zone and check nothing has gone wrong... $\endgroup$ – ProfRob Mar 30 at 17:24
  • $\begingroup$ @AtmosphericPrisonEscape a problem with the opacities I was using. Kramer's law not accurate enough at all. $\endgroup$ – ProfRob Mar 30 at 18:42

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