Radiative Hydrodynamics

Here is how the internal energy in the grid can change due to radiation:

\frac{\partial (\rho e)}{\partial t} = - \kappa_R \rho (4\pi B-cE)

where B=B(T)=\sigma T^4 is the bolometric Planck function for a blackbody (BB), and 4 \pi B=a T^4 is the energy output due to BB radiation. cE is the radiative energy density in the grid. This equation shows that when 4 \pi B > cE,

\frac{\partial \rho e}{\partial t}<0

which is interpreted as the matter losing energy via BB radiation. That is, the internal energy of that zone will decrease, having been transferred into radiative energy. This equation also tells us that when 4 \pi B<cE, there is more energy in the radiation field than in the BB, and so it gets absorbed by the matter. This causes the internal energy to increase,

\frac{\partial \rho e}{\partial t}>0

Changes to the internal energy lead to changes in temperature, and thus the next time step the amount of radiative energy from the BB will change (recall B=B(T)).

Now, since the matter and the radiation are coupled in this way, the equation that governs the radiative energy in the grid is the inverse of the internal energy. Additionally, the radiative energy can diffuse through the grid, so there is an extra term for diffusion:

\frac{\partial E}{\partial t}= \nabla \cdot (\frac{c \lambda}{\kappa_R \rho}) \nabla E + \kappa_p \rho(4\pi B-cE)

Note the coupling term comes in with a '+' sign now (rightmost term on the RHS), and the diffusion term (left term). Note also that how strongly the matter and radiation couple depends on the opacity.

Uniform ambient evolution

The rad. energy in the grid at t=0, assuming no 'sources' is given by the temperature field through the term:

E=4\pi B(T) = aT^4

In the code the constant out in front is given by "scalerad". Checking a uniform ambient medium with this expression for Erad, shows that Erad at t=0 is given by scalerad*T⁴. Thus, any gas with finite temperature is producing radiative energy in the grid through black body radiation. Since

\frac{\partial E}{\partial t} \propto \nabla \cdot \nabla E + (4 \pi B - cE),

cE=4\pi B ~(at ~t=0), and

\nabla E =0, we have

\frac{\partial E}{\partial t}=0

There is no change in the radiative energy over time. Now, other dynamics in the simulation, e.g. gravity, could change the temperature distribution in the grid and thus B(T). Only after this would happen, would we begin to see changes in E. Thus, while 4 \pi B = cE, the system is in radiative equilibrium (right?).

Jeans unstable gas

Imagine starting with a uniform, Jeans unstable gas mass. Initially it is in radiative equilibrium, but on a freefall timescale, the gas will begin to collapse — becoming denser and hotter as it does (recall equation for internal energy has a gravitational energy term). This will lead to regions in the grid where 4piB>cE. This will will increase the radiative energy field, and thus act to 'cool' the gas (as Erad increases, e decreases). Is this right? Thinking of radiation as a source of cooling for optically thin gas…?

Eventually the material should become dense enough that is becomes optically thick to the radiation. Thus, despite the gas heating up through infall, it should no longer be adding these photons to the radiation field (for if it does, it will continue to cool given the equations are inverses of each other)…? We can imagine the

the material shouldn't be able to cool through radiation.

Radiation with a source

Now we add a radiating source to the grid.