The Basic Physics of Geomorphology

At its root, process geomorphology is simply applied physics: the physics tend to be rather simple and the problems to which it is applied tend to be exceedingly complex.

Forces

Consider a 1 m x 1m x 1m cube of rock at the top of a mountain that is z meters above sea level, the bottom of which is a fracture plane cutting through the bedrock. What are the forces acting on this block (which we will call element A):

gravity ( $F_{g}=m\cdot g$ )
friction ( $F_{f}$ , which depends on the normal force)
shear forces (due to flowing water, air)
tensile & elastic forces (such as cohesion, electrostatic bonds)
pressure force (if the fracture plane is below the groundwater table)

where m is the mass of the block, equal to the product of its volume (V) and its density ( $\rho _{s}$ ), g is the acceleration due to gravitational attraction

Example 1: What is the gravity force acting on our block if g=9.8 m³/s and  $\rho _{s}$  = 2650 kg/m³?

Gravity force components

When thinking about the potential for something like a landslide to occur, the gravity force is typically divided into a component acting normal to the potential plane-of-failure and a component acting parallel to it. In this example, the potential failure plane is the fracture at the base of the block. If the fracture is horizontal, then the entire gravity force is acting normal to the plane. If the fracture is inclined at some angle to the horizontal (which we will call $\beta$ ), then by drawing a simple force triangle we can divide the following components:

$\tau =\rho _{s}\cdot g\cdot \sin {\beta }=\gamma _{s}\cdot \sin {\beta }$
$\sigma =\rho _{s}\cdot g\cdot \cos {\beta }=\gamma _{s}\cdot \cos {\beta }$

In our example, the volume of A was 1 m³ and the area of the failure plane upon which F_g was acting was 1 m², so the force of gravity could be expressed per-unit-area (forces per-unit-area are called stresses), so $\tau$ is the shear stress acting on the plane due to gravity acting on the mass of the block, $\sigma$ is the normal stress acting on the plane, and $\rho _{s}$ replaces m because V= 1 m^{^3}. The term $\gamma _{s}$ is the unit weight of element A, given by the product $\gamma _{s}=\rho _{s}\cdot g$ .

Example 2: Assuming that the failure plane is inclined at 15^o to the horizontal, what are the shear and normal 
stresses acting on element A? If the element A does not move, then there must be forces that are acting in the  
opposite direction that are of equal or greater magnitude ... what are the opposing forces?

Pressure force

At the bottom of a swimming pool, a swimmer experiences a pressure that is proportional to the overlying water. The pressure, per unit of surface area is:

$u=\gamma \cdot d_{w}$

where $\gamma$ is the unit weight of water and d_w is the depth of the swimmer below the water surface.

Similarly, for a well having a solid, waterproof casing which is open only at the bottom that is installed a some depth below the water table, the pressure is related to the weight of the water in the well. Sometimes the water level in the well rises above the water table, indicating that groundwater is flowing upwards towards the surface, and sometimes it is below the water table, indicating downward flow. The depth of water in the well is referred to as the pressure head, h, and the fluid pressure acting at the end of the well is given by:

$u=\gamma \cdot h$

A complete understanding of this phenomenon is beyond the scope of this article, but is addressed in courses on groundwater hydrology. The fluid pressure acts on the water in the pores of the soil, exerting a force in all directions. The effect of this force in the context of slope stability is to reduce the normal force and thus reduce the friction acting upon a potential failure plane. That is, fluid pressure acts to reduce slope stability and promote slope failures.

Example 3: What is the pressure, u, acting at the bottom of a well in which the water depth reaches 12 m? How thick 
a slab of rock ( $\rho _{s}$  = 2700 kg/m³) could be lifted by this pressure? Why don't rocks at the bottom of the ocean float?

Friction Force

The frictional force mobilized upon a given surface depends on:

the roughness of the two surfaces making up a potential failure plane
the normal force pushing those two surfaces together

The effect of surface roughness is typically paramterized using a static coefficient of friction ( $\alpha$ ) which, for cohesionless soils, is related to the angle of repose, $\phi$ , for that soil ( $\alpha =\tan {\phi }$ ).

Energy

Energy is related to the capacity to do work (i.e. to change landscapes). Generally, the amount of energy a body has is given by:

$E=\int _{o}^{x}{(m\cdot a)}dx$

That is, the energy is equal to the integration of a force ( $F=m\cdot a$ ) over the distance (x) that it displaces the body. If a force is not sufficient to overcome the resisting forces, then there will be no acceleration of the body and it will gain no energy.

Potential Energy

The most important source of energy in geomorphology is potential energy. A body at rest z meters above some reference level (like mean sea level) has a potential energy, E_p (relative to that reference level) equal to:

$E_{p}=m\cdot g\cdot z$

Since the acceleration of gravity is effectively constant at the surface of the Earth, the integral is replaced by a simple product.

Kinetic Energy

Another type of energy that is important in geomorphology is kinetic energy. To figure out the equation for this, we again integrate. However, this is a bit messy, and the same intuition can be gained by considering the equations-of-motion:

$v=v_{o}+a\cdot t$
$x=x_{o}+v_{o}\cdot t+{\frac {a\cdot t^{2}}{2}}$
$v^{2}=v_{o}^{2}+2a\cdot (x-x_{o})$

The first equation-of-motion states that the velocity at time t depends on the initial velocity v_o, the rate-of-change of velocity dv/dt = a and, of course, the time for which that (constant) acceleration was applied. The second equation states that the position of a body depends on where the body started out (x_o), the initial velocity of the body (v_o) and the rate of change of velocity (a): it comes from integrating the first equation. The last equation comes from re-arranging the first equation ( $t=(v-v_{o})/a$ ) and substituting for t in the second equation, and can be written:

${\frac {v^{2}-v_{o}^{2}}{2a}}=x-x_{o}$

It basically relates changes in position $(x-x_{o})$ to changes in velocity $(v^{2}-v_{o}^{2})$ . The kinetic energy that a body at a constant velocity has, is proportional to the distance required to decelerate the body to a velocity of zero, given a constant deceleration. Thus, we set v = 0, assume that we started decelerating the body at x_o=0, and thus the equation above becomes $x=v_{o}^{2}/2a$ , and so:

$E_{k}=m\cdot a\cdot x=m\cdot a\cdot {\frac {v_{o}^{2}}{2a}}={\frac {m\cdot v_{o}^{2}}{2}}$

In this equation, v_o refers to the velocity for which E_k was evaluated. As expected, the deceleration distance x depends on the acceleration a, but E_k does not.