The flow of heat; a point source near an infinite plane boundary
We have discussed one example earlier (Section 3–4)—the flow of heat. Imagine a block of material, which need not be homogeneous but may consist of different materials at different places, in which the temperature varies from point to point. As a consequence of these temperature variations there is a flow of heat, which can be represented by the vector $\FLPh$. It represents the amount of heat energy which flows per unit time through a unit area perpendicular to the flow. The divergence of $\FLPh$ represents the rate per unit volume at which heat is leaving a region: \begin{equation*} \FLPdiv{\FLPh}=\text{rate of heat out per unit volume}. \end{equation*}  (We could, of course, write the equation in integral form—just as we did in electrostatics with Gauss’ law—which would say that the flux through a surface is equal to the rate of change of heat energy inside the material. We will not bother to translate the equations back and forth between the differential and the integral forms, because it goes exactly the same as in electrostatics.) The rate at which heat is generated or absorbed at various places depends, of course, on the problem. Suppose, for example, that there is a source of heat inside the material (perhaps a radioactive source, or a resistor heated by an electrical current). Let us call $s$ the heat energy produced per unit volume per second by this source. There may also be losses (or gains) of thermal energy to other internal energies in the volume. If $u$ is the internal energy per unit volume, $-du/dt$ will also be a “source” of heat energy. We have, then, \begin{equation} \label{Eq:II:12:5} \FLPdiv{\FLPh}=s-\ddt{u}{t}. \end{equation}  We are not going to discuss just now the complete equation in which things change with time, because we are making an analogy to electrostatics, where nothing depends on the time. We will consider only steady heat-flow problems, in which constant sources have produced an equilibrium state. In these cases, \begin{equation} \label{Eq:II:12:6} \FLPdiv{\FLPh}=s. \end{equation}  It is, of course, necessary to have another equation, which describes how the heat flows at various places. In many materials the heat current is approximately proportional to the rate of change of the temperature with position: the larger the temperature difference, the more the heat current. As we have seen, the vector heat current is proportional to the temperature gradient. The constant of proportionality $K$, a property of the material, is called the thermal conductivity: \begin{equation} \label{Eq:II:12:7} \FLPh=-K\,\FLPgrad{T}. \end{equation}  If the properties of the material vary from place to place, then $K=K(x,y,z)$, a function of position. [Equation (12.7) is not as fundamental as (12.5), which expresses the conservation of heat energy, since the former depends upon a special property of the substance.] If now we substitute Eq. (12.7) into Eq. (12.6) we have \begin{equation} \label{Eq:II:12:8} \FLPdiv{(K\,\FLPgrad{T})}=-s, \end{equation}  which has exactly the same form as (12.4). Steady heat-flow problems and electrostatic problems are the same. The heat flow vector $\FLPh$ corresponds to $\FLPE$, and the temperature $T$ corresponds to $\phi$. We have already noticed that a point heat source produces a temperature field which varies as $1/r$ and a heat flow which varies as $1/r^2$. This is nothing more than a translation of the statements from electrostatics that a point charge generates a potential which varies as $1/r$ and an electric field which varies as $1/r^2$. We can, in general, solve static heat problems as easily as we can solve electrostatic problems. Consider a simple example. Suppose that we have a cylinder of radius $a$ at the temperature $T_1$, maintained by the generation of heat in the cylinder. (It could be, for example, a wire carrying a current, or a pipe with steam condensing inside.) The cylinder is covered with a concentric sheath of insulating material which has a conductivity $K$. Say the outside radius of the insulation is $b$ and the outside is kept at temperature $T_2$ (Fig. 12–1a). We want to find out at what rate heat will be lost by the wire, or steampipe, or whatever it is in the center. Let the total amount of heat lost from a length $L$ of the pipe be called $G$—which is what we are trying to find. How can we solve this problem? We have the differential equations, but since these are the same as those of electrostatics, we have really already solved the mathematical problem. The analogous problem is that of a conductor of radius $a$ at the potential $\phi_1$, separated from another conductor of radius $b$ at the potential $\phi_2$, with a concentric layer of dielectric material in between, as drawn in Fig. 12–1(b). Now since the heat flow $\FLPh$ corresponds to the electric field $\FLPE$, the quantity $G$ that we want to find corresponds to the flux of the electric field, i.e., the electric charge over $\epsO$. We have solved the electrostatic problem by using Gauss’ law. We follow the same procedure for our heat-flow problem. From the symmetry of the situation, we know that $h$ depends only on the distance from the center. So we enclose the pipe in a Gaussian cylinder of length $L$ and radius $r$. From Gauss’ law, we know that the heat flow $h$ multiplied by the area $2\pi rL$ of the surface must be equal to the total amount of heat generated inside, which is what we are calling $G$: \begin{equation} \label{Eq:II:12:9} 2\pi rLh=G\quad\text{or}\quad h=\frac{G}{2\pi rL}. \end{equation}  The heat flow is proportional to the temperature gradient: \begin{equation*} \FLPh=-K\,\FLPgrad{T}, \end{equation*}  or, in this case, the radial component of $\FLPh$ is \begin{equation*} h=-K\,\ddt{T}{r}. \end{equation*}  This, together with (12.9), gives \begin{equation} \label{Eq:II:12:10} \ddt{T}{r}=-\frac{G}{2\pi KLr}. \end{equation}  Integrating from $r=a$ to $r=b$, we get \begin{equation} \label{Eq:II:12:11} T_2-T_1=-\frac{G}{2\pi KL}\ln\frac{b}{a}. \end{equation}  Solving for $G$, we find \begin{equation} \label{Eq:II:12:12} G=\frac{2\pi KL(T_1-T_2)}{\ln(b/a)}. \end{equation}  This result corresponds exactly to the result for the charge on a cylindrical condenser: \begin{equation*} \frac{Q}{\epsO}=\frac{2\pi\kappa L(\phi_1-\phi_2)}{\ln(b/a)}. \end{equation*}  The problems are the same, and they have the same solutions. From our knowledge of electrostatics, we also know how much heat is lost by an insulated pipe. Let’s consider another example of heat flow. Suppose we wish to know the heat flow in the neighborhood of a point source of heat located a little way beneath the surface of the earth, or near the surface of a large metal block. The localized heat source might be an atomic bomb that was set off underground, leaving an intense source of heat, or it might correspond to a small radioactive source inside a block of iron—there are numerous possibilities.  We will treat the idealized problem of a point heat source of strength $G$ at the distance $a$ beneath the surface of an infinite block of uniform material whose thermal conductivity is $K$. And we will neglect the thermal conductivity of the air outside the material. We want to determine the distribution of the temperature on the surface of the block. How hot is it right above the source and at various places on the surface of the block?  How shall we solve it? It is like an electrostatic problem with two materials with different dielectric coefficients $\kappa$ on opposite sides of a plane boundary. Aha! Perhaps it is the analog of a point charge near the boundary between a dielectric and a conductor, or something similar. Let’s see what the situation is near the surface. The physical condition is that the normal component of $\FLPh$ on the surface is zero, since we have assumed there is no heat flow out of the block. We should ask: In what electrostatic problem do we have the condition that the normal component of the electric field $\FLPE$ (which is the analog of $\FLPh$) is zero at a surface? There is none! That is one of the things that we have to watch out for. For physical reasons, there may be certain restrictions in the kinds of mathematical conditions which arise in any one subject. So if we have analyzed the differential equation only for certain limited cases, we may have missed some kinds of solutions that can occur in other physical situations. For example, there is no material with a dielectric constant of zero, whereas a vacuum does have zero thermal conductivity. So there is no electrostatic analogy for a perfect heat insulator. We can, however, still use the same methods. We can try to imagine what would happen if the dielectric constant were zero. (Of course, the dielectric constant is never zero in any real situation. But we might have a case in which there is a material with a very high dielectric constant, so that we could neglect the dielectric constant of the air outside.) How shall we find an electric field that has no component perpendicular to the surface? That is, one which is always tangent at the surface? You will notice that our problem is opposite to the one of a point charge near a plane conductor. There we wanted the field to be perpendicular to the surface, because the conductor was all at the same potential. In the electrical problem, we invented a solution by imagining a point charge behind the conducting plate. We can use the same idea again. We try to pick an “image source” that will automatically make the normal component of the field zero at the surface. The solution is shown in Fig. 12–2. An image source of the same sign and the same strength placed at the distance $a$ above the surface will cause the field to be always horizontal at the surface. The normal components of the two sources cancel out. Thus our heat flow problem is solved. The temperature everywhere is the same, by direct analogy, as the potential due to two equal point charges! The temperature $T$ at the distance $r$ from a single point source $G$ in an infinite medium is \begin{equation} \label{Eq:II:12:13} T=\frac{G}{4\pi Kr}. \end{equation}  (This, of course, is just the analog of $\phi=q/4\pi\epsO R$.) The temperature for a point source, together with its image source, is \begin{equation} \label{Eq:II:12:14} T=\frac{G}{4\pi Kr_1}+\frac{G}{4\pi Kr_2}. \end{equation}  This formula gives us the temperature everywhere in the block. Several isothermal surfaces are shown in Fig. 12–2. Also shown are lines of $\FLPh$, which can be obtained from $\FLPh=-K\,\FLPgrad{T}$. We originally asked for the temperature distribution on the surface. For a point on the surface at the distance $\rho$ from the axis, $r_1=$ $r_2=$ $\sqrt{\rho^2+a^2}$, so \begin{equation} \label{Eq:II:12:15} T(\text{surface})=\frac{1}{4\pi K}\,\frac{2G}{\sqrt{\rho^2+a^2}}. \end{equation}  This function is also shown in the figure. The temperature is, naturally, higher right above the source than it is farther away. This is the kind of problem that geophysicists often need to solve. We now see that it is the same kind of thing we have already been solving for electricity.