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For $r=|x|>0$, the gradient satisfies ${\partial }_{i}r=x_i/r$. The product rule and Einstein notation then give the displayed divergence formula in dimension $n$. In dimension three, its derivative matrix is symmetric, so the curl vanishes. Zero divergence forces $f(r)=C{r}^{-n}$ on each positive radial interval. This formula does not assert regularity at the origin.

Away from the origin, differentiate $r^2=x_j{x}_j$ using Einstein notation:HenceThis gradient is the radial unit vector. Division by $r$ is legitimate on the punctured domain; the distance function is not differentiable at the origin.

In Einstein notation, $(\mathbf{n}\cdot \mathbf{x})n_i=n_i{n}_j{x}_j$, so the matrix and, for $\alpha \ne 0$, its matrix inverse are

Using Einstein notation andwe obtainThus

The scalar triple product isPut $\Delta =[a,b,c]$. When $\Delta \ne 0$, the three vectorsform the reciprocal basis to $a,b,c$, so $e_i^{\prime }\cdot e_j={\delta }_{ij}$. Their scalar triple product is $1/\Delta$. Thereforesince the cyclic permutation preserves orientation. The identity also holds when $\Delta =0$, either by continuity or directly because the cross products are then linearly dependent.

For the given basis $e_1,e_2,e_3$, the same identities giveIf ${\sum }_i{\alpha }_i{e}_i^{\prime }=0$, dotting with $e_j$ gives ${\alpha }_j=0$, so the $e_i^{\prime }$ are linearly independent and hence form a basis. Moreover, $e_i\cdot e_j^{\prime }={\delta }_{ij}$, so the original basis is reciprocal to the primed basis. Uniqueness of a reciprocal basis gives

Every vector $K$ has a unique expansion $K={\sum }_i{\kappa }_i{e}_i^{\prime }$, and thenThis is an integer for every integer triple $(n_1,n_2,n_3)$ exactly when every ${\kappa }_i$ is an integer. Hence all such points areThey form the reciprocal lattice in the convention without a factor of $2\pi$.

In Einstein notation,This is

Similarly,after cyclically relabelling the first term and swapping two indices in the second. Hence

Use Einstein notation, with repeated suffixes summed. The matrix multiplication and transpose giveInterchange the free indices, then relabel the dummy indices $k,\ell$:The negative sign uses $B_{k\ell }=-B_{\ell k}$. Therefore $C$ is skew-symmetric. No orthogonality of $A$ is needed for this part: matrix congruence preserves skew-symmetry.

By the definition of matrix multiplication and trace, with all indices summed from $1$ to $D$,The entries are scalars, so their multiplication commutes. Reordering them and renaming the summed indices givesEquivalently, in Einstein notation, $P_{ij}{Q}_{jk}{R}_{ki}=Q_{jk}{R}_{ki}{P}_{ij}$. This proves the cyclic property of the trace.

Using Einstein notation and the Kronecker delta,The unit vector condition is essential in the last step.

For Cartesian tensors, let $R$ be the orthogonal matrix taking old vector components to new ones, $v_i^{\prime }=R_{ij}{v}_j$. A rank-$n$ tensor transforms by one copy of $R$ on each index:Repeated indices are summed according to Einstein notation. In general coordinates a contravariant tensor index instead carries the corresponding Jacobian matrix $\partial x^{\prime i}/\partial x^j$, and a covariant tensor index carries its inverse; in orthonormal bases of Cartesian axes their component laws coincide with the displayed form.

An isotropic tensor has identical components after every proper orthogonal transformation, so $T^{\prime }=T$ for every $R$ with $R^{T}R=I$ and $\det R=1$. Requiring reflection invariance as well is a stronger convention for odd tensor ranks. The Kronecker delta is invariant under all orthogonal transformations, while the Levi-Civita symbol is invariant under proper rotations and acquires a determinant sign under reflections.

Apply the product rule in Einstein notation, using the Levi-Civita symbol to express the cross product:The third line uses a cyclic permutation of indices in the first term and an interchange of two indices in the second. Rearranging proves the divergence of a cross product identity:

An order-$n$ Cartesian tensor has $3^n$ components whose values in two orthonormal Cartesian frames satisfywith repeated indices summed as in Einstein notation. It is an isotropic tensor when its components are unchanged by every proper rotation, $R\in SO(3)$.

For rank two, a half-turn about any coordinate axis changes the sign of each component having exactly one index along that axis. Invariance therefore makes every off-diagonal component zero. Quarter-turns interchanging two coordinate axes then require $T_{11}=T_{22}=T_{33}$. Thus, without assuming symmetry in advance,Conversely, the Kronecker delta is invariant because $R_{ik}{R}_{jk}={\delta }_{ij}$, so every tensor of this form is isotropic.

For rank three the general proper-rotation invariant form iswhere ${ϵ}_{ijk}$ is the Levi-Civita symbol. The convention matters: the isotropic rank-three tensor and reflections distinction says that invariance of a polar tensor under all $R\in O(3)$ would force $b=0$. A nonzero ${ϵ}_{ijk}$ is invariant under proper rotations, and is a pseudotensor under the full orthogonal group. The shell integrals below are invariant under reflections as well, so their odd moments vanish.

The contraction in the PDF is $I_{iijj}$. By Einstein notation, its integrand is $({\sum }_i{x}_i^2)({\sum }_j{x}_j^2)=r^4$. In spherical coordinates,

For a Cartesian coordinate change $x_i^{\prime }=R_{ij}{x}_j$ with $R$ an orthogonal matrix, a rank-$n$ tensor obeyswith repeated indices summed according to Einstein notation. A tensor is isotropic if $T_{i_1\cdots i_n}^{\prime }=T_{i_1\cdots i_n}$ for every proper rotation matrix $R$. Under proper rotations the Levi-Civita symbol is an isotropic third-rank tensor; if reflections are included, it is a pseudotensor. The convention using proper rotations is the one relevant to the rank-three classification here.

Write $u_i=U_i+u_i^{\prime }$ and $p=P+p^{\prime }$, with $\overline{u_i^{\prime }}=\overline{p^{\prime }}=0$. Assume a statistically stationary mean and sufficient regularity for Reynolds averaging to commute with spatial derivatives. Thenprovided the endpoint difference grows sublinearly. Averaging the continuity equation givesIn Einstein notation, the nonlinear term expands asThus the Reynolds-averaged Navier-Stokes equation isThe extra term survives because the average of a product of zero-mean fluctuations need not vanish; it represents turbulent transport of mean linear momentum.

At an oriented internal surface with unit normal vector $\mathbf{n}$, the traction $\mathbf{t}(\mathbf{x},t;\mathbf{n})$ is the limiting contact force per unit surface area exerted by the material on the positive-$\mathbf{n}$ side on the material on the negative-$\mathbf{n}$ side:For a boundary of a chosen material volume, this convention gives the external traction when $\mathbf{n}$ is outward. The Cauchy stress tensor has componentsso the second index specifies the surface orientation and the first specifies the force component. The vectors ${\mathbf{e}}_i$ form the standard Cartesian coordinate system basis of unit vectors.

Cauchy's stress theorem states that the dependence of traction on the normal vector is linear:Repeated indices are summed according to Einstein notation. In particular, reversing the normal vector reverses the traction, and knowing the tractions on three mutually perpendicular coordinate planes determines the traction on every plane. Symmetry of the Cauchy stress tensor, for an ordinary fluid without body couples, follows separately from angular momentum conservation.