Vector fields and their derivatives

Write $L_{ijlm} = \varepsilon_{ijk}\varepsilon_{klm}$ for the left side and $R_{ijlm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}$ for the right side.

1. Repeated indices in a pair. If $i = j$ then $\varepsilon_{ijk} = 0$ for every $k$, so $L_{ijlm} = 0$; and $R_{iilm} = \delta_{il}\delta_{im} - \delta_{im}\delta_{il} = 0$. The same argument with the antisymmetry in the last two slots of $\varepsilon_{klm}$ disposes of $l = m$. So assume from now on that $i \ne j$ and $l \ne m$.

2. At most one term survives. Since $i \ne j$, the factor $\varepsilon_{ijk}$ is non-zero for exactly one value of $k$, namely the single element $k_0$ of $\{1,2,3\}\setminus\{i,j\}$; for the other two values of $k$ it has a repeated index. Likewise $\varepsilon_{klm}$ is non-zero only for $k = k_1$, the element of $\{1,2,3\}\setminus\{l,m\}$. Hence $L_{ijlm} = 0$ unless $k_0 = k_1$, which — the complements being singletons — happens exactly when $\{i,j\} = \{l,m\}$.

3. The case $i = l$, $j = m$. Then, using the cyclic invariance of $\varepsilon$ in the second factor, $\varepsilon_{ijk}\varepsilon_{kij} = \varepsilon_{ijk}\varepsilon_{ijk}$ (no summation yet: only $k = k_0$ contributes), and a symbol with no repeated index squares to $1$, so $L_{ijij} = 1$. On the right, $\delta_{il}\delta_{jm} = 1$, while $\delta_{im}\delta_{jl} = \delta_{ij}\delta_{ji} = 0$ because $i \ne j$. So $R_{ijij} = 1 = L_{ijij}$.

4. The case $i = m$, $j = l$. Now the second factor is $\varepsilon_{kji}$, which is $-\varepsilon_{kij}$ by antisymmetry in its last two indices, so by step 3 the surviving term is $L_{ijji} = -1$. On the right, $\delta_{il}\delta_{jm} = \delta_{ij}\delta_{ji} = 0$ and $\delta_{im}\delta_{jl} = 1$, so $R_{ijji} = -1 = L_{ijji}$.

5. The remaining cases. If $i \ne j$, $l \ne m$ and $\{i,j\} \ne \{l,m\}$, then $L_{ijlm} = 0$ by step 2. For the right side, $\delta_{il}\delta_{jm} \ne 0$ forces $i = l$ and $j = m$, hence $\{i,j\} = \{l,m\}$, which is excluded; and $\delta_{im}\delta_{jl} \ne 0$ forces $i = m$, $j = l$, excluded for the same reason. So $R_{ijlm} = 0$ as well. Steps 1–5 exhaust all index values, so $(\varepsilon\delta)$ holds.

6. The contractions. Set $l = i$ in $(\varepsilon\delta)$ and sum over $i$ as well as over $k$. On the right, $\delta_{ii}\delta_{jm} - \delta_{im}\delta_{ji} = 3\delta_{jm} - \delta_{jm} = 2\delta_{jm}$, by $\delta_{ii} = 3$ and the renaming rule for $\delta$. On the left we have $\varepsilon_{ijk}\varepsilon_{kim}$, which we bring to a symmetric shape in two steps: a cyclic shift gives $\varepsilon_{kim} = \varepsilon_{imk}$, and then interchanging the last two indices of each factor, $\varepsilon_{ijk} = -\varepsilon_{ikj}$ and $\varepsilon_{imk} = -\varepsilon_{ikm}$, costs two signs that cancel. So

   $$\varepsilon_{ijk}\varepsilon_{kim} = \varepsilon_{ijk}\varepsilon_{imk} = \varepsilon_{ikj}\varepsilon_{ikm} = 2\delta_{jm},$$

   in which the summed pair is now $(i,k)$ and the free pair is $(j,m)$. Renaming the dummies to $i,j$ and the free indices to $k,m$ gives $\varepsilon_{ijk}\varepsilon_{ijm} = 2\delta_{km}$. Setting $m = k$ in that and summing gives $\varepsilon_{ijk}\varepsilon_{ijk} = 2\delta_{kk} = 6$. $\square$

1. Take the $i$th component of $\mathbf{a}\times(\mathbf{b}\times\mathbf{c})$ and expand the inner product in components, choosing dummy indices that are not already in use:

The two symbols share the dummy index $k$ in the position required by $(\varepsilon\delta)$, so

each $\delta$ having renamed one index. The first term is $(\mathbf{a}\cdot\mathbf{c})b_i$ and the second is $(\mathbf{a}\cdot\mathbf{b})c_i$, which is the first identity since $i$ was arbitrary. The second bracketing follows from the first: by anticommutativity $(\mathbf{a}\times\mathbf{b})\times\mathbf{c} = -\,\mathbf{c}\times(\mathbf{a}\times\mathbf{b})$, and the first identity applied to the right-hand side gives $-\,(\mathbf{c}\cdot\mathbf{b})\mathbf{a} + (\mathbf{c}\cdot\mathbf{a})\mathbf{b}$, as claimed.

2. For Lagrange's identity, expand both cross products and contract:

A cyclic shift turns $\varepsilon_{kij}$ into $\varepsilon_{ijk}$, so $(\varepsilon\delta)$ applies and gives

which is $(\mathbf{a}\cdot\mathbf{c})(\mathbf{b}\cdot\mathbf{d}) - (\mathbf{a}\cdot\mathbf{d})(\mathbf{b}\cdot\mathbf{c})$. $\square$

Part 1. Differentiability at $\mathbf{a}$ means $\phi(\mathbf{a}+\mathbf{h}) - \phi(\mathbf{a}) = L\mathbf{h} + o(|\mathbf{h}|)$ for a linear form $L$ on $\mathbb{R}^3$ (Partial derivatives and the chain rule). Taking $\mathbf{h} = t\mathbf{e}_i$ and dividing by $t$ shows $L\mathbf{e}_i = \partial_i\phi(\mathbf{a})$, so $L\mathbf{h} = \partial_i\phi(\mathbf{a})h_i = \nabla\phi(\mathbf{a})\cdot\mathbf{h}$ by the index form of the dot product. This is part 1, and it is the precise content of the physicists' $d\phi = \nabla\phi\cdot d\mathbf{r}$.

Part 2. Fix $\mathbf{u}$ and put $\psi(t) = \phi(\mathbf{a}+t\mathbf{u})$ for $|t|$ small enough that $\mathbf{a}+t\mathbf{u} \in U$. By the chain rule for a differentiable map composed with a differentiable curve, $\psi$ is differentiable at $0$ with $\psi'(0) = \nabla\phi(\mathbf{a})\cdot\mathbf{u}$; the hypothesis used is differentiability of $\phi$ at $\mathbf{a}$, not merely the existence of its partial derivatives. Replacing the line by a differentiable curve $\mathbf{r}(u)$ and applying the same chain rule at each parameter value gives the curve formula, and the arc-length case is that formula with $|d\mathbf{r}/ds| = 1$, so that $d\mathbf{r}/ds = \hat{\mathbf{t}}$.

Part 3. Let $\mathbf{g} = \nabla\phi(\mathbf{a}) \ne \mathbf{0}$ and let $\hat{\mathbf{a}}$ be a unit vector. By part 2, $d\phi/ds = \mathbf{g}\cdot\hat{\mathbf{a}}$. By the Cauchy–Schwarz inequality on $\mathbb{R}^3$ (Inner products and norms),

with equality on either side only when $\hat{\mathbf{a}}$ and $\mathbf{g}$ are linearly dependent, that is $\hat{\mathbf{a}} = \pm\,\mathbf{g}/|\mathbf{g}|$ since both are then unit vectors; the sign of $\mathbf{g}\cdot\hat{\mathbf{a}}$ decides which. Both bounds are attained, so the maximum is $|\mathbf{g}|$ at $\hat{\mathbf{a}} = \mathbf{g}/|\mathbf{g}|$ and the minimum is $-|\mathbf{g}|$ at $-\mathbf{g}/|\mathbf{g}|$. If instead $\mathbf{g} = \mathbf{0}$, then $\mathbf{g}\cdot\hat{\mathbf{a}} = 0$ for all $\hat{\mathbf{a}}$.

Normality, part 1 of the second statement. Let $\boldsymbol{\gamma}$ be a differentiable curve with $\boldsymbol{\gamma}(0) = \mathbf{a}$ lying in $\phi^{-1}(c)$, so $\phi(\boldsymbol{\gamma}(t)) = c$ for all small $t$. Differentiating the constant function $t \mapsto \phi(\boldsymbol{\gamma}(t))$ at $t = 0$ and using the curve form of part 2,

so $\nabla\phi(\mathbf{a})$ is orthogonal to every such tangent vector. Note that this direction of the argument needs no regularity assumption at all: it holds even where $\nabla\phi$ vanishes, where it says nothing.

Normality, part 2. Since $\mathbf{a}$ is a regular point, some partial derivative is non-zero there; renumber the coordinates so that $\partial_3\phi(\mathbf{a}) \ne 0$. Then the $C^1$ function $\Phi(x_1,x_2;x_3) = \phi(x_1,x_2,x_3)$ has invertible partial derivative in its last variable at $\mathbf{a}$, so by the implicit function theorem (Implicit and inverse function theorems) there are a neighbourhood $W$ of $\mathbf{a}$ and a $C^1$ function $g$ with $\phi^{-1}(c)\cap W = \{(x_1,x_2,g(x_1,x_2))\}$. For each $\mathbf{v}$ in the plane $P = \{\mathbf{v} : \nabla\phi(\mathbf{a})\cdot\mathbf{v} = 0\}$ — a plane, because $\nabla\phi(\mathbf{a}) \ne \mathbf{0}$ makes the condition one non-trivial linear equation — the curve $t \mapsto (a_1 + tv_1,\, a_2 + tv_2,\, g(a_1+tv_1, a_2+tv_2))$ lies in $\phi^{-1}(c)$, passes through $\mathbf{a}$ at $t=0$, and is differentiable; its velocity there is a vector of $P$ with first two components $v_1, v_2$, and since a vector of $P$ is determined by its first two components (the coefficient of $x_3$ in the equation of $P$ being $\partial_3\phi(\mathbf{a}) \ne 0$), that velocity is $\mathbf{v}$ itself. So every vector of $P$ is tangent to the level surface, and by the previous paragraph no vector outside $P$ is. Hence the tangent vectors form exactly $P$, whose orthogonal complement is the line spanned by $\nabla\phi(\mathbf{a})$.

Normality, part 3. A point $\mathbf{r}$ lies in the plane through $\mathbf{r}_0$ with normal $\mathbf{n}_0$ precisely when $(\mathbf{r}-\mathbf{r}_0)\cdot\mathbf{n}_0 = 0$, and on the line through $\mathbf{r}_0$ in the direction $\mathbf{n}_0$ precisely when $(\mathbf{r}-\mathbf{r}_0)\times\mathbf{n}_0 = \mathbf{0}$, since a cross product vanishes exactly for parallel vectors (Vector algebra and index notation). Finally $\nabla\phi(\mathbf{a})$ is a positive multiple of $\hat{\mathbf{n}}$ by construction, and part 3 of the first statement identifies that multiple as $|\nabla\phi(\mathbf{a})| = \partial\phi/\partial n$. $\square$