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Next: Curl Up: Vectors and Vector Fields Previous: Divergence

Laplacian Operator

So far we have encountered

$\displaystyle \nabla\phi = \left(\frac{\partial \phi}{\partial x},\, \frac{\partial \phi} {\partial y},\, \frac{\partial \phi}{\partial z}\right),$ (A.139)

which is a vector field formed from a scalar field, and

$\displaystyle \nabla\cdot{\bf A} = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} +\frac{\partial A_z}{\partial z},$ (A.140)

which is a scalar field formed from a vector field. There are two ways in which we can combine gradient and divergence. We can either form the vector field $ \nabla(\nabla\cdot{\bf A})$ or the scalar field $ \nabla\cdot(\nabla\phi)$ . The former is not particularly interesting, but the scalar field $ \nabla\cdot(\nabla\phi)$ turns up in a great many physical problems, and is, therefore, worthy of discussion.

Let us introduce the heat flow vector $ {\bf h}$ , which is the rate of flow of heat energy per unit area across a surface perpendicular to the direction of $ {\bf h}$ . In many substances, heat flows directly down the temperature gradient, so that we can write

$\displaystyle {\bf h} = - \kappa \,\,\nabla T,$ (A.141)

where $ \kappa$ is the thermal conductivity. The net rate of heat flow $ \oint_S {\bf h}\cdot d{\bf S}$ out of some closed surface $ S$ must be equal to the rate of decrease of heat energy in the volume $ V$ enclosed by $ S$ . Thus, we have

$\displaystyle \oint_S {\bf h}\cdot d{\bf S} = - \frac{\partial}{\partial t}\left( \int c\, T\,dV\right),$ (A.142)

where $ c$ is the specific heat. It follows from the divergence theorem that

$\displaystyle \nabla\cdot{\bf h} = -c\,\frac{\partial T}{\partial t}.$ (A.143)

Taking the divergence of both sides of Equation (A.141), and making use of Equation (A.143), we obtain

$\displaystyle \nabla\cdot\left(\kappa \,\nabla T\right) = c\,\frac{\partial T}{\partial t}.$ (A.144)

If $ \kappa$ is constant then the previous equation can be written

$\displaystyle \nabla\cdot (\nabla T) = \frac{c}{\kappa} \frac{ \partial T}{\partial t}.$ (A.145)

The scalar field $ \nabla\cdot (\nabla T)$ takes the form

$\displaystyle \nabla\cdot(\nabla T)$ $\displaystyle = \frac{\partial}{\partial x}\!\left(\frac{\partial T}{\partial x...
...right)+ \frac{\partial}{\partial z}\!\left(\frac{\partial T}{\partial z}\right)$    
  $\displaystyle =\frac{\partial^{\,2} T}{\partial x^{\,2}}+\frac{\partial^{\,2} T...
...rtial y^{\,2}} + \frac{\partial^{\,2} T}{\partial z^{\,2}} \equiv \nabla^{2} T.$ (A.146)

Here, the scalar differential operator

$\displaystyle \nabla^{\,2} \equiv \frac{\partial^{\,2}}{\partial x^{\,2}}+ \frac{\partial^{\,2}}{\partial y^{\,2}}+ \frac{\partial^{\,2}}{\partial z^{\,2}}$ (A.147)

is called the Laplacian. The Laplacian is a good scalar operator (i.e., it is coordinate independent) because it is formed from a combination of divergence (another good scalar operator) and gradient (a good vector operator).

What is the physical significance of the Laplacian? In one dimension, $ \nabla^{\,2} T$ reduces to $ \partial^{\,2} T/\partial x^{\,2}$ . Now, $ \partial^{\,2} T/\partial x^{\,2}$ is positive if $ T(x)$ is concave (from above), and negative if it is convex. So, if $ T$ is less than the average of $ T$ in its surroundings then $ \nabla^{\,2} T$ is positive, and vice versa.

In two dimensions,

$\displaystyle \nabla^{\,2} T = \frac{\partial^{\,2} T}{\partial x^{\,2}}+ \frac{\partial^{\,2} T}{\partial y^{\,2}}.$ (A.148)

Consider a local minimum of the temperature. At the minimum, the slope of $ T$ increases in all directions, so $ \nabla^{\,2} T$ is positive. Likewise, $ \nabla^{\,2} T$ is negative at a local maximum. Consider, now, a steep-sided valley in $ T$ . Suppose that the bottom of the valley runs parallel to the $ x$ -axis. At the bottom of the valley $ \partial^{\,2} T/\partial y^{\,2}$ is large and positive, whereas $ \partial^{\,2} T/\partial x^{\,2}$ is small and may even be negative. Thus, $ \nabla^{\,2} T$ is positive, and this is associated with $ T$ being less than the average local value.

Let us now return to the heat conduction problem:

$\displaystyle \nabla^{\,2} T = \frac{c}{\kappa} \frac{\partial T}{\partial t}.$ (A.149)

It is clear that if $ \nabla^{\,2} T$ is positive in some small region then the value of $ T$ there is less than the local average value, so $ \partial T/\partial t>0$ : that is, the region heats up. Likewise, if $ \nabla^{\,2} T$ is negative then the value of $ T$ is greater than the local average value, and heat flows out of the region: that is, $ \partial T/\partial t<0$ . Thus, the previous heat conduction equation makes physical sense.

next up previous
Next: Curl Up: Vectors and Vector Fields Previous: Divergence
Richard Fitzpatrick 2016-03-31