# How do I derive the formula for lithostatic (overburden) pressure?

The title pretty much says it. I have the formula: $P = \rho g h$ where $\rho$ is the density, $h$ is how deep the pressure is in the Earth and $g$ is the gravitational acceleration(?).

I don't get the units either. If I substitute the units for each term I get this:

$$P = \frac{\mathrm{kg}}{\mathrm{m}^3} \times \frac{\mathrm{m}}{\mathrm{s}^2} \times \mathrm{m} = \frac{\mathrm{kg}}{\mathrm{m}*\mathrm{s}^2}$$

Shouldn't it be something like $P = \dfrac{\mathrm{kg}}{\mathrm{m}^2}$? As far as I'm concerned that's the unit for pressure.

Where do I get this formula from and how do I derive the unit of measurement?

-

This isn't that difficult, but anything is if you start from the wrong place. Let's derive this thing:

$$P = \frac{F}{A}$$

Where $P$ is pressure, and $A$ is the area the force is pushing down on. Let's take a break and derive the units first, just so we know our end derivation is correct; $F$ is in Newtons, which comes out to $\mathrm{kg} \times \frac{\mathrm{m}}{\mathrm{s}^2}$ and $A$ is in $\mathrm{m}^2$. This means that pressure is $\frac{\mathrm{kg}}{\mathrm{m}\times\mathrm{s}^2}$. So your first substitution is correct.

Now that that is established, let's think about lithostatic pressure, and break up the original equation:

$$P = \frac{M \times a}{A}$$

We know that $a = g$, gravitational acceleration, but what we really want is to figure out how to get this equation in terms of density, so

$$\rho = \frac{M}{V}$$

where $\rho$ is density, $M$ is mass, and $V$ is volume. If we substitute this equation in for mass in the pressure equation, we get

$$P = \frac{\rho V \times g}{A}$$

let's now separate $V$ and $A$, $$V = \ell \times w \times h$$ $$A = \ell \times w$$

So canceling out $\ell$ and $w$, our final equation shows

$$P = \rho gh$$

or lithostatic pressure.

-
Wow, thank you. This is very easy to understand now indeed. Btw, do you happen to know any good resources about stages of magma solidification, igneous rock formation? – Tamás Jun 10 '14 at 14:23
Unfortunately I do not know where to learn about that; The stuff I have learned has been just through reading various journal articles; perhaps look for a volcanology review or an igneous petrology text book? I haven't taken much geology. – Neo Jun 10 '14 at 14:48

The SI unit for force is the Newton, not the kilogram. This is defined from Newton's Second Law: $F = ma$. Hence, dimensionally, force (Newton) is 1 kg·m·s-2

Pressure is force per distance squared (kg·m·s-2 / m2).

Hence, dimensionally, pressure is kg / m·s2

As for the formula, the pressure is due to the force (weight) of all the overlying rock. $F=ma$ again. Except we're talking about pressure (ie. force per unit area), so we consider this "per unit area" in the formula. i.e.

$$P = \left(\begin{array}\text{\text{mass of column of 1 m}}^2\\\text{ of rock up to the surface}\end{array}\right) \times \text{acceleration due to gravity} \\ P = (\text{depth} \times \rho) \times g$$

-