RECALL: $\grad{f} = {\partial f\over\partial x}\,\ii + {\partial f\over\partial y}\,\jj + {\partial f\over\partial z}\,\kk$

It is convenient to think of $\grad$ as a differential operator (called “del”), written as $$\grad = {\partial\over\partial x}\,\ii + {\partial\over\partial y}\,\jj + {\partial\over\partial z}\,\kk$$ Thus, $\grad{f}$ (“del f”) is obtained by “multiplying” $\grad$ by $f$, being careful of course to interpret “multiplication” as partial differentiation.

The operator $\grad$ can also be used to express divergence and curl. We have seen that if $$\FF = F_x\,\ii + F_y\,\jj + F_z\,\kk$$ then $$\Curl\FF = \left( \Partial{F_z}{y } - \Partial{F_y}{z } \right) \,\ii + \left( \Partial{F_x}{z } - \Partial{F_z}{x } \right) \,\jj + \left( \Partial{F_y}{x } - \Partial{F_x}{y } \right) \,\kk $$ But this looks like a cross product! Rather than memorizing this formula, it is much easier to compute the curl directly using the determinant rule for cross product $$\Curl\FF = \grad\times\FF = \left| \matrix{\ii& \jj& \kk\cr \noalign{\smallskip} {\partial\over\partial x}& {\partial\over\partial y}& {\partial\over\partial z}\cr \noalign{\smallskip} F_x& F_y& F_z\cr} \right| $$ Similarly, the divergence of $\FF$ looks like a dot product, and can be written $$\hbox{div}\,\FF = \grad\cdot\FF = \Partial{F_x}{x } + \Partial{F_y}{y } + \Partial{F_z}{z }$$ In both cases, care must be taken to correctly interpret “multiplication” as differentiation, as above.

This notation also helps you remember that the curl of a vector field is again a vector field, but the divergence of a vector field is a scalar field (i.e. a function), and of course the gradient of a function is a vector field.

There are several useful identities involving these operators. First of all, like all derivatives, they are linear. Second, there are the product rules \begin{eqnarray*} \grad\times(f\GG) &=& \grad{f}\times\GG + f\,(\grad\times\GG) \\ \grad\cdot(f\GG) &=& \grad{f}\cdot\GG + f\,(\grad\cdot\GG) \end{eqnarray*} which are both of the form:

• The derivative of a product is the derivative of the first factor times the second factor, plus the first factor times the derivative of the second factor.

where of course “derivative” and “times” must be interpreted appropriately. Be careful not to switch the order of the factors in the first identity!

Two further identities are \begin{eqnarray*} \grad\times\grad f &=& 0 \\ \grad\cdot(\grad\times\FF) &=& 0 \end{eqnarray*} The first identity follows from Stokes' Theorem, which we now write in the form $$ \Oint \FF\cdot d\rr = \Sint (\grad\times\FF) \cdot d\SS $$ If $\FF=\grad f$, it is conservative, so the left-hand side must be zero by path independence for any closed curve $C$, which forces the integrand on the right-hand side to vanish identically, which proves the identity.

To prove the second identity, suppose that both $S_1$ and $S_2$ meet the requirements of Stokes' Theorem for the same curve $C$. We can think of $S_1$ and $S_2$ as being two different butterfly nets (or soap bubbles) on the same wire frame $C$. Supposing that $S_1$ is above $S_2$, the outward unit normal to the closed surface $S=S_1+S_2$ is the upward unit normal to $S_1$, but the downward unit normal to $S_2$. Thus, \begin{eqnarray*} \Int_V \grad\cdot(\grad\times\FF) \, dV &=& \Sint (\grad\times\FF)\cdot\nn_{\rm out} \, \dS \\ &=& \Int_{S_1} (\grad\times\FF)\cdot\nn_{\rm up} \, \dS - \Int_{S_2} (\grad\times\FF)\cdot\nn_{\rm up} \, \dS \\ &=& \Oint\FF\cdot d\rr - \Oint\FF\cdot d\rr = 0 \end{eqnarray*} where the Divergence Theorem was used in the first step. This shows that the above integrals are zero over any volume $V$, forcing the integrand to vanish, which is the second identity.

The new notation for the divergence also allows us to write the Divergence Theorem as $$ \Sint\FF\cdot d\SS = \Int_V\grad\cdot\FF\>dV $$ There is also a 2-dimensional version of the Divergence Theorem, which is really a corollary of Green's Theorem, and which says $$\Oint \FF\cdot\nn \, ds = \Int_R \grad\cdot\FF \> dA$$ where $C$ is a closed curve in the plane, $\nn$ is its outward pointing unit normal vector, and $R$ is the region inside the curve. The integral on the left defines flux in two dimensions.

Finally, a further useful combination is given by $$\triangle{f} = \nabla^2{f} = \grad\cdot(\grad{f})$$ and is called the Laplacian of $f$.

GOALS