The last column introduced the exterior derivative operator $$d$$ and, using this operator and the language of differential forms, developed analogous expressions for the classical vector concepts of gradient, divergence, and curl, each being given by
\[ \nabla f \Leftrightarrow d f \; ,\]
\[ \nabla \cdot \vec A \Leftrightarrow *d* \phi_A \; , \]
and
\[ \nabla \times \vec A \Leftrightarrow * d \phi_A \; , \]
respectively. These relations, collectively, assume the name of the primitive correspondences.
This installment tackles the extension of these ideas to the classical set ‘vector derivative identities’ that tend to crop up whenever vector calculus is employed (usually in electrodynamics or fluid mechanics) and which are often unclear or unmotivated. One of the main thrusts behind the work of Schleifer to publish the correspondences between vector calculus and the language of differential forms was the hope that a deeper understanding of the underlying, unified structure would become apparent. It was this aim that originally attracted my attention to his paper.
The first set of identities to be discussed are the curl of the gradient is zero,
\[ \nabla \times \nabla f = 0 \; , \]
and the divergence of a curl is zero,
\[ \nabla \cdot \nabla \times \vec A = 0 \; .\]
From the point-of-view of vector calculus, these two identities, while looking similar, don’t flow from any common source. But each of these is an expression of the much simpler identity $$d^2 = 0$$. Indeed, it was the desire to unite these dispparate relations that led, in great part, to the invention of the language of differential forms.
To prove that the curl of a gradient is zero explicitly use the first and third primitive correspondences
\[ \nabla \times (\nabla f) \Leftrightarrow *d(d f) = * d^2 f = 0 \; .\]
Likewise, the demonstration that the curl of a vector field is divergence-free involves the second and the third of the primitive correspondences
\[ \nabla \cdot \nabla \times \vec A \Leftrightarrow (*d*)(*d\phi_A) = * d ** d \phi_A = * d^2 \phi_A = 0 \; .\]
Most of the more exotic vector identities require a bit more work to demonstrate them explicitly and are not done with abstract manipulation but, instead, with coordinates. This is not especially concerning, since the aim here is to bridge the gap between the two approaches and not to demonstrate internal consistency or proficiency in either one.
Curl-of-the-curl
One such vector identity is the curl-of-the-curl, $$\nabla \times (\nabla \times \vec A)$$, which arises often in the study of Maxwell’s equations. The traditional way to perform the identity is by index-gymnastics involving the Levi-Cevita tensor density and the Kronecker delta. Appling this technology yields
\[\nabla \times (\nabla \times \vec A) = \epsilon_{ijk} \partial_j ( \epsilon_{k \ell m} \partial_\ell A_m ) = (\delta_{i\ell}\delta_{jm} – \delta_{im}\delta_{j\ell} ) \partial_j \partial_\ell A_m \; .\]
Resolving the Kronecker deltas yields the index equation
\[\nabla \times (\nabla \times \vec A) = \partial_j \partial_i A_j – \partial_j \partial_j A_i \; ,\]
which has the immediate translation into ‘vector language’ of
\[ \nabla \times (\nabla \times \vec A) = \nabla (\nabla \cdot \vec A) – \nabla^2 \vec A \;. \]
The differential forms way is to start with the standard correspondence
\[\phi_A = A_x dx + A_y dy + A_z dz \]
and then to apply the third primitive correspondence
\[ *d \phi_A = (A_{z,y} – A_{y,z}) dx + (A_{x,z} – A_{z,x}) dy + (A_{y,x} – A_{x,y}) dz \; . \]
Applying the exterior derivative an additional time and accounting for the properties of the wedge product yields the two-form
\[ d*d \phi_A =( A_{x,zx} – A_{z,xx} – A_{z,yy} + A_{y,zy} ) dx \wedge dy \\ + ( A_{y,xy} – A_{x,yy} – A_{x,zz} + A_{z,xz} ) dy \wedge dz \\ + ( A_{z,yz} – A_{y,zz} – A_{y,xx} + A_{x,yx} ) dz \wedge dx \; .\]
Applying the Hodge star operator to get the dual gives
\[ *d*d \phi_A = (\partial_j \partial_i A_j – \partial_j \partial_j A_i) dq^i \; ,\]
where $$dq^i = dx, dy, dz$$ for $$ i = i, 2, 3$$.
Employing the vector-to-form correspondence, one sees that these are the same components as in the vector form, once the common, self-cancelling terms associated with $$i=j$$ are taken into account.
Gradient-of-the-dot
The next identity is the gradient of a dot-product. Based on the primitive correspondences, one can start with
\[ \nabla (\vec A \cdot \vec B) \Leftrightarrow d(*(\phi_A \wedge *\phi_B)) \; .\]
To see how this comes out, begin with the recognition that the dot-product is a scalar function so that it is particularly easy to move between vectors and forms. The form-equivalent of the gradient of this scalar function is
\[ d (A_x B_x + A_y B_y + A_z B_z) = \partial_i (A_x B_x + A_y B_y + A_z B_z ) dq^i \; ,\]
where $$dq^i$$ is as defined above.
What remains is to expand the action of the partial derivative and show that the result forms translate into the desired vector identity.
\[ \partial_i (A_j B_j) = (A_{x,i} B_x + A_{y,i} B_y + A_{z,i} B_z) + ( A_x B_{x,i} + A_y B_{y,i} + A_z B_{z,i} ) \; .\]
For simplicity, focus on the $$i=x$$ and strategically add and subtract two groups of terms
\[ A_{x,x} B_x + A_{y,x} B_y + A_{z,x} B_z \\ + (A_{x,y} B_y – A_{x,y} B_y) + (A_{x,z} B_z – A_{x,z} B_z) + B \leftrightarrow A \; .\]
Next group the terms into the suggestive form
\[ B_x A_{x,x} + B_y A_{x,y} + B_z A_{x,z} \\ + B_y (A_{y,x} – A_{x,y}) + B_z(A_{z,x} – A_{x,z} ) + B \leftrightarrow A \; .\]
The next step requires a peek back into the language of vectors. Since the curl is given by
\[ \nabla \times \vec A = (A_{z,y} – A_{y,z} ) \hat e_x + (A_{z,y} – A_{y,z} ) \hat e_y + (A_{z,y} – A_{y,z} ) \hat e_x \; ,\]
the re-arranged term can be re-written as
\[ B_x A_{x,x} + B_y A_{x,y} + B_z A_{x,z} + B_y (A_{y,x} – A_{x,y}) + B_z(A_{z,x} – A_{x,z} ) \\ + B \leftrightarrow A = (\vec B \cdot \nabla) A_x + B_y [\nabla \times \vec A]_z – B_z [\nabla \times \vec A]_y + B \leftrightarrow A\; .\]
The final two terms on the right can be further re-written as
\[ B_x A_{x,x} + B_y A_{x,y} + B_z A_{x,z} + B_y (A_{y,x} – A_{x,y}) + B_z(A_{z,x} – A_{x,z} ) \\ + B \leftrightarrow A = (\vec B \cdot \nabla) A_x + [\vec B \times (\nabla \times \vec A) ]_x + B \leftrightarrow A \; .\]
And so we arrive at the final vector calculus identity for the gradient of the dot-product, without a really meaningful differential form analog.
\[ \nabla ( \vec A \cdot \vec B ) = (\vec B \cdot \nabla) \vec A + \vec B \times (\nabla \times \vec A) + (\vec A \cdot \nabla) \vec B + \vec A \times (\nabla \times \vec B) \]
This was to be expected since the motivation for the addition and subtraction of those additional terms was only to manipulate the partial derivative expression into something familiar from the vector calculus.
Divergence of a cross
The final identity to be explored in this column is the divergence of a cross. In vector calculus, the divergence of a cross expands to
\[ \nabla \cdot (\vec A \times \vec B) = (\nabla \times \vec A) \cdot \vec B – (\nabla \times \vec B) \cdot \vec A \]
Using the second and third primitive correspondences, the proposed analog in the language of differential forms is
\[ \nabla \cdot (\vec A \times \vec B) \Leftrightarrow *d*( *(\phi_A \wedge \phi_B) ) \; .\]
Tackling the right-hand side is easier by ommitting the leading Hodge star operator to give
\[ d*( *(\phi_A \wedge \phi_B) ) = d\left[ (A_y B_z – A_z B_y) dy \wedge dz \\ + (A_z B_x – A_x B_z) dz \wedge dx + (A_x B_y – A_y B_x) dx \wedge dy \right] \; .\]
Re-applying this operator yields
\[ *d*( *(\phi_A \wedge \phi_B) ) = (A_{y,x} B_z – A_{z,x} B_y ) + (A_{z,y} B_x – A_{x,y} B_z ) \\ + ( A_{x,z} B_y – A_{y,z} B_x ) – B \leftrightarrow A \; .\]
Grouping terms suggestively gives
\[ *d*( *(\phi_A \wedge \phi_B) ) = (A_{z,y} – A_{y,z}) B_x + (A_{x,z} – A_{z,x}) B_y \\ + (A_{y,x} – A_{x,y}) B_z – B \leftrightarrow A \; ,\]
from which it is obvious that
\[ *d*( *(\phi_A \wedge \phi_B) ) = (\nabla \times \vec A) \cdot \vec B – (\nabla \times \vec B) \cdot \vec A \; .\]
Next week, I’ll close out this subject with a look at the equivalent operator to the Laplacian and with a comment on the identities that Schleifer claims he couldn’t prove.