More on thermal instability
Let the cooling be due to
(1) A line that radiates most at a tempertaure T 0:
" µ
¶2 #
T
¤ line = nLlin e » An exp ¡
¡1
T0
(2) Bremmsstrahlung
¤B r = nLB r » Bn2 T 1=2
Thus
" µ
¶2 #
T
L » A exp ¡
¡1
+ BnT 1=2
T0
At constant pressure
n=
So
L»
P
kT
(
" µ
)
¶2 # µ
¶
T
P
A exp ¡
¡1
+
BT 1=2
T0
kT
This function
³ looks like
´
p
2
10 exp ¡ (x ¡ 1) + 1= x
This graph shows P versus T at constant L:
1
There are three solutions at the same pressure and L; but only one is stable.
Herezs another way to look at it. The red curve is greater L: In the central
region, moving to higher temperature increases the cooling, so the temperature
drops again. On the far right, lower temperature gets to a region of more cooling, and the temperature keeps dropping. On the far left, higher temp implies
less cooling, and the temp keeps increasing.
T
1
At constant entropy:
P » ½° or T » ½°¡ 1 » ½2=3
so
n » T 3=2
Thus
" µ
¶2 #
T
L » A exp ¡
¡1
+ BT 5=2
T0
³
´
2
10exp ¡ (x ¡ 1) + x5=2
2
3