In order to extend the results from last week into the realm of classical vector analysis, we must introduce an additional operator called the exterior derivative and which is denoted by $$d$$. The exterior derivative can be thought of as the classical nabla $$\\vec \\nabla$$ on steroids since it is a bit more versatile and works in spaces of arbitrary dimension.<\/p>\n
The defining property of the exterior derivative is that when it operates on a differential form $$\\phi(x,y,z)$$ it produces a new differential form $$d\\phi$$ of one higher order. So if $$\\phi$$ is an $$n$$-form then $$d\\phi$$ is an $$n+1$$-form. To understand how this operation is carried out, we start with its action on a function (i.e. a $$0$$-form). In this case<\/p>\n
\\[ df = \\frac{\\partial f}{\\partial x} dx + \\frac{\\partial f}{\\partial y} dy + \\frac{\\partial f}{\\partial z} dz \\; ,\\]<\/p>\n
which is just what is expected from basic calculus. The remaining piece of the definition is to see what the action of $$d$$ is on a $$1$$-form. Since it results in a $$2$$-form, we can suspect that may involve the wedge product and we would be right. Specifically, suppose that the one form is given by<\/p>\n
\\[ \\phi = A_x dx \\]<\/p>\n
then<\/p>\n
\\[ d \\phi = d (A_x dx) = \\frac{\\partial A_x}{\\partial x} dx \\wedge dx + \\frac{\\partial A_y}{\\partial y} dy \\wedge dx + \\frac{\\partial A_z}{\\partial z} dz \\wedge dx \\; .\\]<\/p>\n
From the anti-symmetry property of the wedge product, this result simplifies to<\/p>\n
\\[ d \\phi = \\frac{\\partial A_y}{\\partial y} dy \\wedge dx + \\frac{\\partial A_z}{\\partial z} dz \\wedge dx \\; .\\]<\/p>\n
The generalization to more complex $$1$$-forms is obvious.<\/p>\n
As a result of these properties, the exterior derivative satisfies $$d^2 = dd = 0$$ regardless of what it operates on.<\/p>\n
We are now in the position to adapt the language of differential forms to some familiar results of vector calculus.<\/p>\n
The core adaptations are for the divergence and curl. The remaining vector calculus identities more or less flow from these.<\/p>\n
The expression for the divergence start with the expression for the corresponding one form<\/p>\n
\\[ \\phi_A = A_x dx + A_y dy + A_z dz \\; . \\]<\/p>\n
Apply the Hodge star operator first<\/p>\n
\\[ *\\phi_A = A_x dy \\wedge dz + A_y dz \\wedge dx + A_z dx \\wedge dy \\; .\\]<\/p>\n
The application of the exterior derivative only produces derivatives for the excluded variable in the wedge product<\/p>\n
\\[ d*\\phi_A = A_{x,x} dx \\wedge dy \\wedge dz + A_{y,y} dy \\wedge dz \\wedge dx + A_{z,z} dz \\wedge dx \\wedge dy \\; , \\]<\/p>\n
where the term<\/p>\n
\\[ A_{x,x} = \\frac{\\partial A_x}{\\partial x} \\]<\/p>\n
and so on for the other terms.<\/p>\n
All of the wedge products are cyclic permutations of $$dx \\wedge dy \\wedge dz$$ so rearrangement involves no changes of sign. A further application of the Hodge star operator give the zero form<\/p>\n
\\[ *d*\\phi_A = A_{x,x} + A_{y,y} + A_{z,z} \\Leftrightarrow \\vec \\nabla \\cdot \\vec A \\; .\\]<\/p>\n
The curl, which is handled in a similar way, is expressed as<\/p>\n
\\[ \\vec \\nabla \\times \\vec A \\Leftrightarrow *d\\phi_A \\; .\\]<\/p>\n
The proof is reasonably easy. Start again with the expression for the corresponding form<\/p>\n
\\[ \\phi_A = A_x dx + A_y dy + A_z dz \\; \\]<\/p>\n
and then operate on it with the exterior derivative to get<\/p>\n
\\[ d \\phi_A = A_{x,y} dy \\wedge dx + A_{x,z} dz \\wedge dx + A_{y,x} dx \\wedge dy \\\\ + A_{y,z} dz \\wedge dy + A_{z,x} dx \\wedge dz + A_{z,y} dy \\wedge dz \\; \\]<\/p>\n
Organizing the terms in lexicographical order (i.e. $$dx \\wedge dy$$, $$dx \\wedge dz$$, and $$dy \\wedge dz$$), accounting for changes in sign as terms are swapped, gives<\/p>\n
\\[ d\\phi_A = (A_{y,x}-A_{x,y}) dx \\wedge dy \\\\ + (A_{z,y}-A_{y,z}) dy \\wedge dz \\\\ + (A_{x,z}-A_{z,x}) dz \\wedge dx \\; \\]<\/p>\n
The final step is to apply the Hodge star operator to arrive at<\/p>\n
\\[ *d\\phi_A = (A_{y,x}-A_{x,y}) dz + (A_{z,y}-A_{y,z}) dx + (A_{x,z}-A_{z,x}) dy \\; , \\]<\/p>\n
which proves the correspondence<\/p>\n
\\[ *d\\phi_A \\Leftrightarrow \\vec \\nabla \\times \\vec A \\; .\\]<\/p>\n
Next column, I’ll expand on the algebra of differential forms just a bit and then will apply the results introduced here to prove a host of vector calculus identities.<\/p>\n","protected":false},"excerpt":{"rendered":"