Astrophysical gases tend to radiate when heated. This radiation is an energy leak which may carry away substantial amounts of energy. A simulation which does not include the radiative cooling (such as the one in the previous section) is often referred to as an adiabatic simulation.
The file corocool.tab (in anonymous ftp) contains what is called a coronal cooling curve. The file contains two columns, the first one lists the log of the temperature, the second one the log of the cooling in units of erg cm s. This particular cooling curve is taken from Dalgarno & McCray (1972), reproduced here as Fig. 3. It was calculated by determining the equilibrium ionization state at every temperature and from that the amount of emission. Below K the cooling depends strongly on the ionization fraction of the gas, the one in the table is the curve labelled in the paper. The cooling rate of a gas can be approximated by , where is the number density of particles.
Numerically, the cooling is a source term for the energy density and can be included in a similar way as the geometric source terms, i.e. in a separate calculation every time step. However, there is the danger that the cooling will reduce the temperature/internal energy density below zero, which is unphysical. The reason for this is the very non-linear behaviour of the cooling function. There are two ways to control this:
A) Solve the problem from Section 2 with radiative cooling included. Still assume that all material is ionized. What differences are there with the non-cooling solution? Why is the effect larger for the outer shock than for the inner shock? Is it possible to resolve the cooling region? What compression (density jump) do you find for the swept up shell?
If you are unsuccessful in getting the code to run with cooling, you can try lowering the overall density. This will reduce the cooling rate. You could for instance try yr and yr for the fast and slow wind respectively. A 10 times lower density gives a 100 times lower cooling. If you reduce the density too much the result of the calculation will be the similar to the one without cooling.
B) Follow the interaction between a moderately fast wind and an outer slow wind. Use the following parameters: yr, km s for the fast wind, and yr, km s for the slow wind. First do a simulation without cooling and then add the cooling. Compare the two results and also compare them to the differences between the non-cooling and cooling simulation for the 1000 km s wind. Why is the behaviour of the inner shock very different?
C) A weakly cooling bubble is often called energy conserving and a strongly cooling bubble momentum conserving. Lamers & Cassinelli (1999) state in Sect. 12.3 of their book that one can observationally check whether a bubble is momentum or energy conserving by considering the following ratios:
=(total kinetic energy in shell)/(total wind energy provided)
=(momentum in the shell)/(momentum imparted by the wind)
If the shell is energy conserving, whereas when it is momentum conserving.
Check whether the shells in your models are indeed energy conserving and momentum conserving according to these definitions, by calculating and for both cases.