Ablative Equilibrium Condition

Ablative Equilibrium Condition
πœŒπ‘Ž , π‘π‘Ž , π‘£π‘Ž
Ablation
area
shell
g
z
We define an Ablative RT equilibration problem by defining 𝜌, 𝑝, 𝑣 at the ablation front (dotdash line in Fig. 1) and the gravity 𝑔. The hydro profiles are then calculated with different
approximations in two regions below and above the ablation front.
(1) In the shell above the ablation front, density is high, temperature is low, so the thermal
conduction is neglected here. The shell is performing like a solid piston. So the
hydrodynamic equations are approximated as:
𝑣 = π‘£π‘Ž
𝑝 = 𝑠(𝑧) πœŒπ›Ύ
�𝑑
(𝑝 + πœŒπ‘£ 2 ) = βˆ’πœŒπ‘”
𝑑𝑧
where 𝛾 = 5/3,π‘£π‘Ž is the fluid velocity at the ablation front and should have a negative
value. 𝑠(𝑧) is a prescribed entropy function which is used to adjust the shape of the
shell.
Combining these 3 equations above we obtain:
π‘‘πœŒ
𝑠 β€² (𝑧)πœŒπ›Ύ + 𝜌 𝑔
=βˆ’
𝑑𝑧
𝑠(𝑧) 𝛾 πœŒπ›Ύβˆ’1 + π‘£π‘Ž2
𝜌(𝑧 = π‘§π‘Žπ‘π‘™π‘Žπ‘‘π‘–π‘œπ‘› ) = πœŒπ‘Ž
This ODE can be readily solved to get the density profile and each quantity is then
determined for the shell.
(2) In the ablation area below the ablation front, density is low, temperature is very high.
The profile in this region is in equilibrium, that is d/dt=0. Simply drop the d/dt terms in
hydrodynamic equations:
𝜌 𝑣 = πœŒπ‘Ž π‘£π‘Ž
⎧
𝑑
(𝑝 + πœŒπ‘£ 2 ) = βˆ’πœŒπ‘”
βŽͺ
βŽͺ
𝑑𝑧
2
⎨ 𝑑 �𝑣 οΏ½ 𝛾 𝑝 + 𝜌 𝑣 οΏ½ βˆ’ ΞΊ 𝑑𝑇� = βˆ’πœŒπ‘”π‘£
π›Ύβˆ’1
2
𝑑𝑧
βŽͺ
βŽͺ𝑑𝑧
⎩
𝑝 = πœŒπ‘‡/𝐴
𝐴 is the average particle mass 𝐴 = 𝑀𝑖 /(𝑍 + 1). For notational convenience, define a
constant 𝜎 ≑ πœŒπ‘Ž π‘£π‘Ž . Combine the 4 equations above to 2 equations:
𝑑 𝜎 2 πœŒπ‘‡
⎧
οΏ½ + οΏ½ = βˆ’πœŒπ‘”
βŽͺ
𝐴
𝑑𝑧 𝜌
π›ΎπœŽπ‘‡
𝜎3
𝑑𝑇
βŽ¨π‘‘
οΏ½
+
βˆ’
πœ…
οΏ½ = βˆ’π‘”πœŽ
βŽͺ
2
𝑑𝑧
βŽ©π‘‘π‘§ (𝛾 βˆ’ 1)𝐴 2𝜌
The second equation can be integrated analytically and yield:
𝜎3
𝑑𝑇
π›ΎπœŽπ‘‡
+ 2 βˆ’ πœ… οΏ½ = βˆ’π‘”πœŽπ‘§ + π‘π‘œπ‘›π‘ π‘‘
οΏ½
(𝛾 βˆ’ 1)𝐴 2𝜌
𝑑𝑧
The integration constant is chosen as π‘π‘œπ‘›π‘ π‘‘ = π‘”πœŽπ‘§π‘Žπ‘™π‘π‘Žπ‘‘π‘–π‘œπ‘› so the RHS terms cancel out
at the ablation front.
Now we need to solve the ODE system simultaneously:
𝑑𝑇
π›ΎπœŽπ‘‡
𝜎3
⎧ =�
+ 2 + π‘”πœŽπ‘§ βˆ’ π‘π‘œπ‘›π‘ π‘‘οΏ½ /πœ…
(𝛾 βˆ’ 1)𝐴 2𝜌
𝑑𝑧
βŽͺ
βŽͺ
𝑑𝑄
= βˆ’πœŒπ‘”
𝑑𝑧
⎨
βŽͺ
𝜎 2 πœŒπ‘‡
βŽͺ
𝑄=
+
⎩
𝜌
𝐴
This system need to be solved numerically. Also note that πœ… is a function of T in Spitzer’s
model. After 𝑇 and 𝑄 are advanced in each step, 𝜌 can be found via
𝜌=
𝑄 + �𝑄 2 βˆ’ 4𝜎 2 𝑇/𝐴
2𝑇/𝐴