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

<p>Pressure is force per distance squared (kg·m·s<sup>-2</sup> / m<sup>2</sup>).</p>

<p>Hence, dimensionally, pressure is kg / m·s<sup>2</sup></p>

<p>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>

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

CC BY-SA 3.0
2014-06-10T13:32:13.537Z
2014-06-10T15:35:25.363Z
2014-06-10T15:35:25.363Z
winwaed
<p>This isn't that difficult, but anything is if you start from the wrong place. Let's derive this thing:</p>

<p>$$P = \frac{F}{A}$$</p>

<p>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. </p>

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

<p>$$P = \frac{M \times a}{A} $$</p>

<p>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 </p>

<p>$$\rho = \frac{M}{V}$$</p>

<p>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>

<p>$$P = \frac{\rho V \times g}{A}$$</p>

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

<p>So canceling out $\ell$ and $w$, our final equation shows</p>

<p>$$P = \rho gh$$</p>

<p>or lithostatic pressure.</p>

CC BY-SA 3.0
2014-06-10T14:23:50.997Z
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
CC BY-SA 3.0
2014-06-10T14:48:53.643Z
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
CC BY-SA 3.0
2014-06-10T13:39:43.080Z
2014-06-10T14:47:09.483Z
2014-06-10T14:47:09.483Z
Neo
<p>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(?).</p>

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

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

<p>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. </p>

<p>Where do I get this formula from and how do I derive the unit of measurement?</p>

<geophysics><geology><geodynamics>
How do I derive the formula for lithostatic (overburden) pressure?
Tamás
# How do I derive the formula for lithostatic (overburden) pressure?

(Asked by: username_0 on 2014-06-10)

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?

**Tags:** geophysics, geology, geodynamics


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## 2 answers

### Answer by username_2 on 2014-06-10. Score: 14 | Accepted answer

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.

> **Comments:**
> **username_0**: 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?  
> **username_2**: 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.  

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### Answer by username_1 on 2014-06-10. Score: 5

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.

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


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