Proofs Chapter 4: Curl Formulas

Proof

Use the linearity of partial derivatives componentwise.

The $i$-th component of the left-hand side reads

\begin{equation}[\nabla \times (\mathbf{A} + \mathbf{B})]_i = \varepsilon_{ijk}\,\dfrac{\partial (A_k + B_k)}{\partial x_j} \label{eq:4-29-1}\end{equation}

By linearity of partial differentiation, $\partial(A_k + B_k)/\partial x_j = \partial A_k/\partial x_j + \partial B_k/\partial x_j$, hence

\begin{equation}[\nabla \times (\mathbf{A} + \mathbf{B})]_i = \varepsilon_{ijk}\,\dfrac{\partial A_k}{\partial x_j} + \varepsilon_{ijk}\,\dfrac{\partial B_k}{\partial x_j} \label{eq:4-29-2}\end{equation}

Each term in $\eqref{eq:4-29-2}$ is precisely the $i$-th component of the curl of $\mathbf{A}$ or $\mathbf{B}$.

\begin{equation}[\nabla \times (\mathbf{A} + \mathbf{B})]_i = [\nabla \times \mathbf{A}]_i + [\nabla \times \mathbf{B}]_i \label{eq:4-29-3}\end{equation}

Since the equality holds for $i = 1, 2, 3$, the vector identity follows.

Proof

Pull the constant out of each partial derivative: $\partial(cA_k)/\partial x_j = c\,\partial A_k/\partial x_j$.

The $i$-th component of the left-hand side is

\begin{equation}[\nabla \times (c\,\mathbf{A})]_i = \varepsilon_{ijk}\,\dfrac{\partial (c A_k)}{\partial x_j} = \varepsilon_{ijk}\,c\,\dfrac{\partial A_k}{\partial x_j} \label{eq:4-30-1}\end{equation}

The constant $c$ factors out of the contracted sum.

\begin{equation}[\nabla \times (c\,\mathbf{A})]_i = c\,\varepsilon_{ijk}\,\dfrac{\partial A_k}{\partial x_j} = c\,[\nabla \times \mathbf{A}]_i \label{eq:4-30-2}\end{equation}

The result holds componentwise, so the vector identity follows.

Proof

Apply the Leibniz rule for partial derivatives within the Levi-Civita representation.

The $i$-th component of the left-hand side is

\begin{equation}[\nabla \times (f\mathbf{A})]_i = \varepsilon_{ijk}\,\dfrac{\partial (f A_k)}{\partial x_j} \label{eq:4-31-1}\end{equation}

By the Leibniz rule, $\partial(f A_k)/\partial x_j = f\,\partial A_k/\partial x_j + A_k\,\partial f/\partial x_j$. Substituting into $\eqref{eq:4-31-1}$ gives

\begin{equation}[\nabla \times (f\mathbf{A})]_i = \varepsilon_{ijk}\left( f\,\dfrac{\partial A_k}{\partial x_j} + A_k\,\dfrac{\partial f}{\partial x_j} \right) \label{eq:4-31-2}\end{equation}

Split the expression into two terms.

\begin{equation}[\nabla \times (f\mathbf{A})]_i = f\,\varepsilon_{ijk}\,\dfrac{\partial A_k}{\partial x_j} + \varepsilon_{ijk}\,\dfrac{\partial f}{\partial x_j}\,A_k \label{eq:4-31-3}\end{equation}

The first term equals $f\,[\nabla \times \mathbf{A}]_i$. For the second term, write $\partial f/\partial x_j = (\nabla f)_j$:

\begin{equation}\varepsilon_{ijk}\,(\nabla f)_j\,A_k = [(\nabla f) \times \mathbf{A}]_i \label{eq:4-31-4}\end{equation}

This is the standard Levi-Civita representation $[\mathbf{u} \times \mathbf{v}]_i = \varepsilon_{ijk} u_j v_k$ of the cross product. Combining $\eqref{eq:4-31-3}$ with $\eqref{eq:4-31-4}$:

\begin{equation}[\nabla \times (f\mathbf{A})]_i = f\,[\nabla \times \mathbf{A}]_i + [(\nabla f) \times \mathbf{A}]_i \label{eq:4-31-5}\end{equation}

The vector identity follows since this holds for every component.

Proof

Combine the epsilon-delta identity $\varepsilon_{ijk}\varepsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}$ with the Leibniz rule.

Write the $i$-th component of the left-hand side using Levi-Civita. The $k$-th component of $\mathbf{A} \times \mathbf{B}$ is $\varepsilon_{klm} A_l B_m$:

\begin{equation}[\nabla \times (\mathbf{A} \times \mathbf{B})]_i = \varepsilon_{ijk}\,\dfrac{\partial}{\partial x_j} (\varepsilon_{klm} A_l B_m) \label{eq:4-32-1}\end{equation}

Step 1: apply the epsilon-delta identity

The constant $\varepsilon_{klm}$ pulls outside the derivative. Apply the epsilon-delta identity to $\varepsilon_{ijk}\varepsilon_{klm}$:

\begin{equation}\varepsilon_{ijk}\varepsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl} \label{eq:4-32-2}\end{equation}

Substituting $\eqref{eq:4-32-2}$ into $\eqref{eq:4-32-1}$ gives

\begin{equation}[\nabla \times (\mathbf{A} \times \mathbf{B})]_i = (\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl})\,\dfrac{\partial (A_l B_m)}{\partial x_j} \label{eq:4-32-3}\end{equation}

Step 2: contract with the Kronecker deltas

The product $\delta_{il}\delta_{jm}$ contracts to $l = i$, $m = j$, while $\delta_{im}\delta_{jl}$ contracts to $m = i$, $l = j$:

\begin{equation}[\nabla \times (\mathbf{A} \times \mathbf{B})]_i = \dfrac{\partial (A_i B_j)}{\partial x_j} - \dfrac{\partial (A_j B_i)}{\partial x_j} \label{eq:4-32-4}\end{equation}

Step 3: Leibniz rule and interpretation

Apply the Leibniz rule to each term.

\begin{equation}\dfrac{\partial (A_i B_j)}{\partial x_j} = A_i\,\dfrac{\partial B_j}{\partial x_j} + B_j\,\dfrac{\partial A_i}{\partial x_j} \label{eq:4-32-5}\end{equation}

\begin{equation}\dfrac{\partial (A_j B_i)}{\partial x_j} = A_j\,\dfrac{\partial B_i}{\partial x_j} + B_i\,\dfrac{\partial A_j}{\partial x_j} \label{eq:4-32-6}\end{equation}

Substituting $\eqref{eq:4-32-5}$ and $\eqref{eq:4-32-6}$ into $\eqref{eq:4-32-4}$:

\begin{equation}[\nabla \times (\mathbf{A} \times \mathbf{B})]_i = A_i\,\dfrac{\partial B_j}{\partial x_j} + B_j\,\dfrac{\partial A_i}{\partial x_j} - A_j\,\dfrac{\partial B_i}{\partial x_j} - B_i\,\dfrac{\partial A_j}{\partial x_j} \label{eq:4-32-7}\end{equation}

Identify each term: $\partial B_j/\partial x_j = \nabla \cdot \mathbf{B}$ and $\partial A_j/\partial x_j = \nabla \cdot \mathbf{A}$. The directional derivative operators $\mathbf{B} \cdot \nabla = B_j\,\partial/\partial x_j$ and $\mathbf{A} \cdot \nabla = A_j\,\partial/\partial x_j$ give

\begin{equation}B_j\,\dfrac{\partial A_i}{\partial x_j} = [(\mathbf{B} \cdot \nabla) \mathbf{A}]_i, \quad A_j\,\dfrac{\partial B_i}{\partial x_j} = [(\mathbf{A} \cdot \nabla) \mathbf{B}]_i \label{eq:4-32-8}\end{equation}

Substituting these into $\eqref{eq:4-32-7}$:

\begin{equation}[\nabla \times (\mathbf{A} \times \mathbf{B})]_i = A_i\,(\nabla \cdot \mathbf{B}) - B_i\,(\nabla \cdot \mathbf{A}) + [(\mathbf{B} \cdot \nabla) \mathbf{A}]_i - [(\mathbf{A} \cdot \nabla) \mathbf{B}]_i \label{eq:4-32-9}\end{equation}

The vector identity follows since this holds for every component:

\begin{equation}\nabla \times (\mathbf{A} \times \mathbf{B}) = \mathbf{A}(\nabla \cdot \mathbf{B}) - \mathbf{B}(\nabla \cdot \mathbf{A}) + (\mathbf{B} \cdot \nabla)\mathbf{A} - (\mathbf{A} \cdot \nabla)\mathbf{B} \label{eq:4-32-10}\end{equation}

Proof

Apply (32) with $\mathbf{A} = \nabla f$ and $\mathbf{B} = \nabla g$.

The right-hand side of (32) reads

\begin{equation}\nabla \times (\mathbf{A} \times \mathbf{B}) = \mathbf{A}(\nabla \cdot \mathbf{B}) - \mathbf{B}(\nabla \cdot \mathbf{A}) + (\mathbf{B} \cdot \nabla)\mathbf{A} - (\mathbf{A} \cdot \nabla)\mathbf{B} \label{eq:4-33-1}\end{equation}

Substituting $\mathbf{A} = \nabla f$ and $\mathbf{B} = \nabla g$ yields

\begin{equation}\nabla \times (\nabla f \times \nabla g) = \nabla f\,(\nabla \cdot \nabla g) - \nabla g\,(\nabla \cdot \nabla f) + (\nabla g \cdot \nabla)\nabla f - (\nabla f \cdot \nabla)\nabla g \label{eq:4-33-2}\end{equation}

By the definition of the Laplacian, $\nabla^2 = \nabla \cdot \nabla$, so $\nabla \cdot \nabla f = \nabla^2 f$ and $\nabla \cdot \nabla g = \nabla^2 g$. Substituting into $\eqref{eq:4-33-2}$:

\begin{equation}\nabla \times (\nabla f \times \nabla g) = \nabla f\,\nabla^2 g - \nabla g\,\nabla^2 f + (\nabla g \cdot \nabla)\nabla f - (\nabla f \cdot \nabla)\nabla g \label{eq:4-33-3}\end{equation}

Proof

Apply (31) with $\mathbf{A} = \nabla g$ and use the vanishing curl of a gradient (Identity (38): $\nabla \times \nabla g = \mathbf{0}$, proved in Chapter 5).

From (31):

\begin{equation}\nabla \times (f\,\nabla g) = f\,(\nabla \times \nabla g) + (\nabla f) \times \nabla g \label{eq:4-34-1}\end{equation}

The curl of a gradient is identically zero:

\begin{equation}\nabla \times \nabla g = \mathbf{0} \label{eq:4-34-2}\end{equation}

Substituting $\eqref{eq:4-34-2}$ into $\eqref{eq:4-34-1}$ eliminates the first term:

\begin{equation}\nabla \times (f\,\nabla g) = \nabla f \times \nabla g \label{eq:4-34-3}\end{equation}

Proof

Combine (34) with the antisymmetry of the cross product.

By (34), the left-hand side is

\begin{equation}\nabla \times (f\,\nabla g) = \nabla f \times \nabla g \label{eq:4-35-1}\end{equation}

Apply the same identity (34) with $f$ and $g$ swapped:

\begin{equation}\nabla \times (g\,\nabla f) = \nabla g \times \nabla f \label{eq:4-35-2}\end{equation}

By the antisymmetry of the cross product (Identity (2): $\mathbf{u} \times \mathbf{v} = -\mathbf{v} \times \mathbf{u}$),

\begin{equation}\nabla g \times \nabla f = -(\nabla f \times \nabla g) \label{eq:4-35-3}\end{equation}

Substituting $\eqref{eq:4-35-3}$ into $\eqref{eq:4-35-2}$:

\begin{equation}\nabla \times (g\,\nabla f) = -(\nabla f \times \nabla g) \label{eq:4-35-4}\end{equation}

Comparing $\eqref{eq:4-35-1}$ with $\eqref{eq:4-35-4}$ gives

\begin{equation}\nabla \times (f\,\nabla g) = -\nabla \times (g\,\nabla f) \label{eq:4-35-5}\end{equation}