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Geosciences LibreTexts

2.1: The Diffusion Equation

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  • We will start the discussion of diffusion by considering the familiar process of thermal conduction. 

    Thermal Conduction

    Diffusion of heat through a crystal lattice occurs through two processes. First, heat can be carried by electrons in the form of kinetic energy in metallic rocks. If a region of the rock is heated at one end, the thermal energy excites the electrons, giving them extra energy to move through the rock from regions of high heat to regions of low heat. As the electrons moves, they lose some of this extra energy as heat, thus heating up the region they have moved to. Second, heating a lattice structure adds kinetic energy to the lattice itself, by increasing the vibrational bonds between atoms (think of atoms connected to each other by little springs). The movement of this vibrational energy through the lattice structure is termed a phonon. Both modes of passing energy through the lattice structure act to conduct heat from regions of higher temperature to regions of lower temperature, changing the distribution (and perhaps the total amount) of heat within the rock. 

    2.1 Crystal Lattice.png
    Figure \(\PageIndex{4}\): Heat Diffusion Through a Crystal Lattice

    Heat Flow Equation: A 1-D Rod

    Start with a cylindrical rod, of length \(L\), and cross-sectional area \(A\), aligned with the z axis. Fix the temperature at \(z=0\) to \(T_1\) and the temperature from \(z>0\) to  \(z=L\) to \(T_2\), where \(T_2>T_1\). The heat, \(Q\), will move from the high temperature region to the low temperature region, in the -z direction.

    2.1 Pipe.png
    Figure \(\PageIndex{5}\): Heat Flow in a 1D Rod

    The change in heat within the rod, \(\Delta Q\), is given by:

    \[\Delta Q=kA\left(d\frac{T_2-T_1}{L}\right)\Delta t \nonumber\]

    where \(k\) is the thermal conductivity with units of \(\frac{W}{mK}\)(watts per meter Kelvin). Therefore, \(\Delta Q\) has units of joules (note 1 W=1\(\frac{J}{s}\)). Rearrange the equation above as,

    \[\frac{\Delta Q}{A\Delta t}=k\frac{\Delta T}{L} \nonumber\]

    The left hand side is the change in heat per unit area per unit time, which is a heat flux, given the symbol, \(q\).

    \[q=\frac{1}{A} \frac{dQ}{dt} \nonumber\]

    Substituting \(q\) into the equation above and then rewriting the other side as a full derivative we arrive at the Heat Flow Equation: 

    \[q=-k\frac{dT}{dz} \label{heat}\]

    Note, that the minus sign shows that the heat moves in the -z direction from high temperature to low temperature. 


    Diffusion Equation for a 3-D Volume

    More generally, we should consider a 3-D volume element (of a rock) and examine how the heat content changes within this volume as a function of time. Consider a cubic volume element with sides of length \(dx\), \(dy\), and \(dz\) and volume, \(dV=dxdydz\).

    2.1 Cube.png
    Figure \(\PageIndex{6}\): Heat Flow in 3D

    Consider heat flowing along the z axis. The heat that enters the volume at z+dz is \(Q_z\) and the heat that leaves the volume at z is \(Q_z -dQ_z\). The change in heat within the volume is then:

    \[\Delta Q=Q_z-(Q_z-dQz) \nonumber\]

    We an rerewrite as:

    \[\Delta Q=Q_z-(Q_z-\left(\dfrac{dQz}{dz}\right)(-dz)) \nonumber\]


    \[\Delta Q=-\left(\frac{dQz}{dz}\right)dz \nonumber\]

    From the heat flow equation (Equation \ref{heat}) we also know that

    \[q=\frac{1}{A} \frac{dQ}{dt} \nonumber\]


    \[dQ=qA\,dt \nonumber\]


    \[q=-k\frac{dT}{dz}. \nonumber\]

    Substituting \(q\) into the equation for \(dQ\) and then subsituting \(dQ\) into the equation for \(\Delta q\) we find,

    \[\Delta Q=-\frac{d}{dz}(qAdt)dz=-\frac{d}{dz}(-k\frac{dT}{dz})dV\,dt \nonumber\]

    where \(dV=Adz=dxdydz\). If the conductivity, \(k\), is independent of position (a constant throughout the rock), then this can be written as:

    \[\Delta Q=k(\frac{d^2T}{dz^2})dV\, dt \nonumber\]

    where both negative signs have canceled out so that \(\Delta Q\) is positive. This equation states that the changes the heat flux in the volume element of the rock are proportional to the curvature (second derivative) of the temperature profile. This is interesting, because we usually think of conduction as being faster where the gradient  is largest, but its actually faster where the change in gradient is the largest.

    The total heat within the rock can also be considered from the perspective of the total heat capacity of the rock (how much heat can the rock hold), which depends on the specific heat, \(c\), and the density, \(\rho\), of the rock:

    \[\Delta Q=c\rho dV\Delta T \nonumber\]

    Setting the two expressions for \(\Delta Q\) equal:

    \[c\rho dV\Delta T= k(\frac{d^2T}{dz^2})dV dt \nonumber\]

    which can be rearranged as,

    \[\frac{dT}{dt}=\frac{k}{c\rho}\frac{d^2T}{dz^2} \nonumber\]

    where \(dT\) has been substituted for \(\Delta T\). The combination of physical constants, \(\frac{k}{c\rho}\) is the thermal diffusivity and has units of \((\frac{m^2}{s})\). The thermal diffusivity is given the greek symbol \(\kappa\) (kappa). Substituting in \(\kappa\) we arrive at 1-D Diffusion Equation:

    \[\frac{dT}{dt}=\kappa\frac{d^2T}{dz^2} \nonumber\]

    Note, this equation relates the change in temperature per unit time to the curvature of the temperature profile in the direction of heat flow.

    Full Form of the Diffusion Equation

    In general, heat flow can come from any direction, so the temperature will depend on x, y, z, and t. Because \(T=T(x, y, z, t)\) and is not just dependent on one variable, it is necessary to rewrite the derivatives in the diffusion equation as partial derivatives:

    \[\frac{\partial T}{\partial t}=\kappa \left(\frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}+\frac{\partial^2 T}{\partial z^2}\right) \nonumber \]

    In this case for example, for the partial derivative with respect to time, each of the variables (x, y, z) are treated as constants, and so on for each other variable. Don’t let the partial derivatives confuse you, they are just derivatives. In general, you still want to think of this equation as a bunch of small changes in temperature, \(\Delta T\), occurring over small intervals of time, \(\Delta t\), and over small distances (\(\Delta z\)):

    \[\frac{\Delta T}{\Delta t}=\kappa\frac{\Delta T}{(\Delta z)^2} \nonumber\]

    This is the "Simplified Diffusion Equation"

    Estimating Time/Length-scale for Diffusive Processes

    Length-scale of Diffusion

    The above simplified version of the diffusion equation can be solved for \(\Delta z\) to give an indication of how far heat can move in a given amount of time:

    \[\Delta z\simeq\sqrt{\kappa\Delta t} \nonumber\]

    This is the Diffusion Length-scale

    For the earth, a good average value for the thermal diffusivity is \(\kappa=1x10^{-6}\frac{m^2}{s}\). How far does heat diffuse through a rock in 1 million years?

    \[\Delta z=\sqrt{(1x10^{-6})(1x10^6)(3.15x10^7)} \nonumber\]

    \(\Delta z\simeq 5.6 km\) where there are about \(3.15 \times 10^7\) sec per year

    What does this length-scale tell us? It does not tell us the temperature at any particular distance, but instead tells us that a noticeable amount of heat will travel this distance in the time specified (we will look more at this in the homework).

    Time-scale of Diffusion

    Another way of building your intuition for diffusion is to solve the simplified diffusion equation for \(\Delta t\):

    \[\Delta t\simeq\frac{\Delta z^2}{\kappa} \nonumber\]

    This is the Diffusion Timescale

    How long does it take to cool off a dike intrusion that is 100 meters wide?

    \[\Delta t=\frac{(100^2)}{(1x10^{-6})(3.15x10^7)} \nonumber\]

    \[\Delta t=318\;years.\]

    What does this timescale tell us? It does not tell us the temperature at the distance specified, but instead tells us that a noticeable amount of heat will have reached the distance specified in this time (we will look more at this in the homework).

    2.1 Cooling.png
    Figure \(\PageIndex{7}\): Exponential Cooling

    A rule of thumb is that cooling is often exponential. Most heat is lost or moved within \(t=3\cdot\Delta t\)