SME
Mineralogy
Igneous, sedimentary and metamorphic petrology
Use this for “A schist formed at 0.6 GPa (6 kbar). How deep was it, and how hot on a 25 °C/km geotherm from 15 °C at the surface?”
Chart
T(z) = 25z + 15
The vertical axis grows downward.
This is an example. Type your own values to start a new problem.
P (GPa)
=
ρ (kg/m³)
×
g
×
z (km)
T (°C)
=
T₀ (°C)
+
G (°C/km)
×
z (km)
Formulas
z
=
P
×
10
6
9.81 ×
ρ
21.8436
=
0.6 ×
10
6
9.81 × 2,800
T
=
T₀
+
G
×
z
561.09 = 15 + 25 × 21.8436
Assumptions
•
Lithostatic pressure: the weight of the rock above, P = ρgz, with one mean density and g = 9.81 m/s².
•
1 GPa is 10 kbar; 10⁹ Pa per GPa over 1,000 m per km leaves the factor 10⁶ in z.
•
A steady, straight geotherm: the rock warms G degrees for every kilometer down.
Step-by-step
WE KNOW
P = 0.6 GPa, ρ = 2,800 kg/m³, G = 25 °C/km, T₀ = 15 °C
FIND
depth (z), temperature at depth (T)
Step 1 · Find depth (z)
Use z = P × 10⁶ ÷ (9.81 × ρ)
The
pressure
is
the
weight
of
the
rock
above,
ρgz,
so
divide
it
by
ρg;
the
10
6
turns
GPa
over
meters
into
km.
z
=
0.6 ×
10
6
9.81 × 2,800
z
=
0.6 ×
10
6
27,468
z
=
21.8436
km
Step 2 · Find temperature at depth (T)
Use T = T₀ + G × z
Start
at
the
surface
temperature
and
add
G
degrees
for
each
kilometer
down.
T
=
15
+
25
×
21.8436
T
=
15
+
546.09
T
=
561.09
°C
Check
Put the numbers back in. Both sides should match.
21.8436
=
0.6 ×
10
6
9.81 × 2,800
✓
561.09 = 15 + 25 × 21.8436 ✓
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Grain size on the φ scale
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