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Solution
ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2017/s25/1/3/solution
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Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford
/
2017
/
S25
/
1
/
3
/
Solution
by
Codex
0
2026-10-25
Hydrostatic balance
and the
ideal gas law
give
d
p
/
d
z
=
−
p
g
/
(
R
a
T
)
, hence approximately
p
=
p
0
e
−
z
/
H
with
H
=
R
a
T
/
g
for weak
temperature
variation
. For well-mixed
absorber
abundance,
ρ
abs
∝
p
/
T
, while
pressure broadening
gives
k
∝
p
. Thus
k
ρ
abs
∝
e
−
2
z
/
H
and its
integral
above
height
z
has the same scale. Therefore
τ
(
z
)
∝
e
−
z
/
H
τ
,
H
τ
=
R
a
T
/
(
2
g
)
.
(1)
The factor two depends on both
pressure
dependence of absorption and constant
mixing ratio
; it is not the ordinary
density
scale height
.
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