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If an array is antisymmetric in every orthonormal basis and its contractions with every antisymmetric second-rank tensor are invariant scalars, it obeys the Cartesian second-rank tensor transformation law. The difference $D=R^{T}A\prime R-A$ is antisymmetric and orthogonal to every antisymmetric test matrix. Choosing the test matrix equal to $D$ gives $D:D=0$, hence $D=0$. The Frobenius inner product is nondegenerate on the antisymmetric subspace.

A Cartesian second-rank tensor invariant under every proper rotation preserving an unoriented axis $n$ has this form. Rotations about the axis give a planar block $aI+bJ$ and eliminate mixed axial-planar entries. A half-turn about a transverse axis reverses $J$, forcing $b=0$. If only rotations about the oriented axis are imposed, this planar antisymmetric term may survive.

The Frobenius inner product of a symmetric matrix and an antisymmetric matrix is zero. Thus invariant contractions of an array with all symmetric second-rank tensors test only its symmetric part; an arbitrary basis-dependent antisymmetric part is invisible. A nonzero antisymmetric matrix in one basis and the zero matrix in another passes all such tests with scalar zero but is not a Cartesian second-rank tensor.

A Cartesian tensor has components in each orthonormal basis of a Euclidean space, related by the displayed rule under an orthogonal component change $R$. Each index contributes one transformation matrix. The Euclidean inner product identifies upper and lower indices in orthonormal coordinates, so the distinction between covariant and contravariant components does not change this rule. An isotropic tensor has unchanged components under all proper rotations; requiring invariance under reflections as well excludes the ordinary odd-parity cases. A Cartesian second-rank tensor is the matrix case, with transformation $T^{\prime }=RT{R}^T$.

A Cartesian second-rank tensor invariant under half-turns about all three coordinate axes is diagonal. Each half-turn has diagonal signs, and each off-diagonal entry changes sign under at least one of them. Invariance therefore kills all off-diagonal entries. It does not make the diagonal entries equal: that requires additional rotational invariance, as for an isotropic second-rank tensor.

A Cartesian second-rank tensor in three dimensions invariant under quarter-turn rotations about all three coordinate axes is a scalar multiple of the identity matrix. Squaring the quarter-turns gives coordinate half-turns; invariance under these forces every off-diagonal component to vanish. The quarter-turns then equate the three diagonal components. This conclusion does not require the original tensor to be symmetric.

For a constant orthogonal Cartesian coordinate change $x^{\prime }=Rx$, a Cartesian second-rank tensor obeys $T_{ij}^{\prime }=R_{ik}{R}_{jl}{T}_{kl}$. The chain rule gives ${\partial }_j^{\prime }=R_{jm}{\partial }_m$. Consequently ${\partial }_j^{\prime }{T}_{ij}^{\prime }=R_{ik}{R}_{jl}{R}_{jm}{\partial }_m{T}_{kl}=R_{ik}{\partial }_l{T}_{kl}$, the transformation rule for a vector. Under non-Cartesian position-dependent coordinate changes, partial derivatives must be replaced by the appropriate covariant derivative.

A Cartesian second-rank tensor invariant under all rotations about an oriented unit axis $n$ has the displayed form. A half-turn about the axis eliminates the mixed axial-transverse entries, and a quarter-turn makes the transverse block a scalar matrix plus an antisymmetric planar rotation generator. Conversely both transverse terms commute with every rotation about the axis. The antisymmetric term is allowed unless symmetry or an additional axis-reversing rotation is imposed. This differs from an axially invariant second-rank tensor for an unoriented axis.

A Cartesian Cartesian second-rank tensor has components transforming under a constant orthogonal change of axes asHere ${\mathbf{X}}^{\prime }=P\mathbf{X}$ and ${\mathbf{u}}^{\prime }=P\mathbf{u}$. By the chain rule and orthogonality,Thus the displacement gradient is a Cartesian second-rank tensor. Its symmetric part transforms in the same way:Hence the infinitesimal strain tensor is also second-rank. This is a passive coordinate rotation, under which both reference and spatial coordinates rotate; it does not assert finite-deformation frame indifference of the linearized strain formula.

For a Cartesian second-rank tensor, write $S_{ij}=(A_{ij}+A_{ji})/2$ and $W_{ij}=(A_{ij}-A_{ji})/2$. Under an orthogonal change of frame,where the last two equalities follow by interchanging the dummy indices in $A_{ji}^{\prime }$. Thus the symmetric and antisymmetric parts obey the same tensor transformation law. If $A=S+W$ with $S^T=S$, $W^T=-W$, adding or subtracting its transpose forces $S=(A+A^T)/2$ and $W=(A-A^T)/2$. This also proves uniqueness.

For the given matrix, the scalar part is $a=\mathrm{Tr}(A)/3=3$. The symmetric part is $S=\left(\begin{array}{ccc}1& 3& 2\\ 3& 5& 4\\ 2& 4& 3\end{array}\right)$, so the traceless second-rank tensor is $B=S-3I$. The skew part is $W=\left(\begin{array}{ccc}0& -1& 1\\ 1& 0& 2\\ -1& -2& 0\end{array}\right)$. Comparing with the cross product matrix $[p{]}_{\times }=\left(\begin{array}{ccc}0& -p_3& p_2\\ p_3& 0& -p_1\\ -p_2& p_1& 0\end{array}\right)$ givesThen $A=aI+[p{]}_{\times }+B$ proves the decomposition for every vector. The vector dual to the antisymmetric tensor is an axial vector under reflections, and an ordinary vector under proper rotations.

Under a Cartesian frame change $B_i^{\prime }=R_{ik}{B}_k$, where $R$ is an orthogonal matrix. Orthogonality gives $B_k^{\prime }{B}_k^{\prime }=B_k{B}_k$ and $R_{ik}{R}_{j\ell }{\delta }_{k\ell }={\delta }_{ij}$. Thereforewhich is the tensor component transformation law. Also $T_{ij}=T_{ji}$ directly, proving that it is a symmetric Cartesian second-rank tensor for every choice of the two scalars.

For $\mathbf{B}\ne 0$, the direction of $\mathbf{B}$ is an eigenvector with eigenvalue $(\alpha +\beta )|\mathbf{B}{|}^2$. Every vector perpendicular to $\mathbf{B}$ is an eigenvector with eigenvalue $\beta |\mathbf{B}{|}^2$, a two-dimensional eigenspace. If $\alpha =0$, these eigenvalues coincide and every nonzero vector is an eigenvector. If $\mathbf{B}=0$, the tensor is zero and again every nonzero vector is an eigenvector, now with eigenvalue zero.

For the position-dependent field with zero divergence, the product rule givesContraction of the two Levi-Civita symbols in the curl and cross product givesConsequentlyThese constants give the magnetic part of the Maxwell stress tensor, with its physical prefactor suppressed. They are forced if the identity is to hold for every divergence-free field: $\mathbf{B}=(y,0,0)$ forces $2\beta =-1$; at the origin $\mathbf{B}=(y,1,0)$ then forces $\alpha =1$.

Apply the divergence theorem separately to each component:Since the whole vector field vanishes on $S$, all components of $T$ vanish there, and hence the vector volume integral is zero. The argument assumes a differentiable field and a boundary for which the divergence theorem applies.

Let $R_{ij}$ be the components of an orthogonal matrix relating two Cartesian frames by $v_i^{\prime }=R_{ij}{v}_j$. Repeated indices are summed from $1$ to $3$. The tensor component transformation law for a Cartesian second-rank tensor isFor a proper rotation, $\det R=1$; the transformation law itself also holds for orthogonal changes of Cartesian frame with determinant $-1$.

To find a cubically invariant second-rank tensor, first square each of the prescribed quarter-turns. For example the half-turn about the first axis is $S_x=\mathrm{diag}(1,-1,-1)$. Invariance under $S_{x}A{S}_x^{\mathsf{T}}$ forces $A_{12},A_{13},A_{21},A_{31}$ to vanish. The corresponding half-turns about the other axes eliminate the remaining off-diagonal components, even though symmetry of $A$ was never assumed.

Now write $A=\mathrm{diag}(a,b,c)$. A quarter-turn about the third axis interchanges $a,b$, so $a=b$; a quarter-turn about the first axis gives $b=c$. Conversely a scalar multiple of the identity matrix is invariant under every rotation. Thus the most general answer is

For a Cartesian second-rank tensor, writeThe first term is a symmetric second-rank tensor and the second is an antisymmetric second-rank tensor. Taking the trace of the symmetric term and subtracting it gives the scalar, symmetric-traceless and axial tensor decomposition:Here $S_{ij}=S_{ji}$ and $S_{ii}=0$. To check reconstruction of the antisymmetric part, contract the Levi-Civita symbols:Thus $P_{ij}=P{\delta }_{ij}+S_{ij}+{ϵ}_{ijk}{A}_k$, and the displayed contractions also prove uniqueness. The vector associated with the antisymmetric part is an axial vector under reflections, and an ordinary vector under proper rotations.

For the specified isotropic tensor law, contract the Kronecker deltas first:Write $T_{ij}=T{\delta }_{ij}+W_{ij}+{ϵ}_{ijk}{V}_k$, where $T=T_{kk}/3$ and $W$ is symmetric and traceless. Transposition preserves its scalar and symmetric-traceless parts and reverses its antisymmetric part. ThereforeUniqueness of the decomposition gives the isotropic elasticity on tensor components relations. Provided their respective coefficients are nonzero, the requested inverses areIf one coefficient vanishes, that strain component has zero stress response: the corresponding stress component must be zero for a solution to exist, and that strain component is not uniquely recoverable. Isotropy by itself does not guarantee invertibility.

If $T_{ij}$ is symmetric then $V=0$, hence $A=0$ and the stress is symmetric. Equivalently, direct transposition gives $P_{ij}-P_{ji}=(\beta -\gamma )(T_{ij}-T_{ji})=0$. Restricting to this symmetric-strain case givesThis final two-parameter expression uses the symmetry assumption from the preceding request. For a general nonsymmetric array, the transpose term cannot be absorbed into the same coefficient: it acts with $\beta +\gamma$ on the symmetric-traceless part and with $\beta -\gamma$ on the antisymmetric part. The stated coefficient $\mu$ is the coefficient of $T_{ij}$ itself; if one instead writes the conventional elastic law with $2\mu$ multiplying the strain, that shear-modulus convention is half this coefficient.

Use the right-handed rotation matrix which sends the positive $x$ direction to the positive $y$ direction:The Cartesian second-rank tensor transformation law givesHere isotropic tensor means invariant under every proper rotation, as is needed for the rank-three assertion involving the Levi-Civita symbol. A passive coordinate rotation through the same angle uses $R^{\mathsf{T}}$ instead; the invariance conclusions are identical.

Setting $S^{\prime }=S$ forces $S_{11}=S_{22}$, $S_{12}=-S_{21}$, and $S_{13}=S_{23}=S_{31}=S_{32}=0$. No symmetry assumption on $S$ has been made: it now has the formInvariance under a right-handed quarter-turn about the $x$ axis next gives $S_{22}=S_{33}$ and $S_{12}=-S_{13}$, hence $a=c$ and $b=0$. Conversely $R(\lambda I)R^{\mathsf{T}}=\lambda I$ for every orthogonal matrix. The general isotropic second-rank tensor is therefore $S_{ij}=\lambda {\delta }_{ij}$.

An isotropic vector must be fixed by the half-turn about the $z$ axis, so its first two components vanish. The half-turn about the $x$ axis then forces its third component to vanish. Thus the only isotropic vector is zero. The general isotropic third-rank tensor under proper rotations is $A_{ijk}=\lambda {ϵ}_{ijk}$. If invariance under the whole orthogonal group were required of an ordinary rank-three tensor, inversion $R=-I$ would send $A$ to $-A$, forcing $A=0$ instead.

For a constant orthogonal Cartesian coordinate change $x_i^{\prime }=R_{ij}{x}_j$, a Cartesian second-rank tensor has components satisfyingBoth indices transform by the coordinate-change matrix; summation over repeated indices is understood by the Einstein summation convention. A vector would have the one-index rule $U_i^{\prime }=R_{ik}{U}_k$.

Define $U_i=\partial T_{ij}/\partial x_j$. Since $x_m=R_{jm}{x}_j^{\prime }$, the chain rule gives $\partial /\partial x_j^{\prime }=R_{jm}\partial /\partial x_m$. The rotation coefficients are constant, soThe identity $R_{jm}{R}_{jl}={\delta }_{ml}$ follows from orthogonality. Hence the contracted derivative transforms as a vector. This proves the divergence of a Cartesian second-rank tensor rule for global Cartesian rotations. In curvilinear coordinates the corresponding construction uses a covariant derivative; ordinary partial derivatives alone would not have the stated tensorial transformation law.

A Cartesian second-rank tensor has components in every orthonormal Cartesian frame that transform according towhen vector components transform as $x^{\prime }=Qx$ for an orthogonal matrix $Q$. Repeated indices are summed. If $A=B$ in one frame, their transformed difference is $Q(A-B)Q^T=0$, so their equality holds in every such frame, at the same physical point.

Now impose the scalar contraction hypothesis on $C$. For every tensor $A$, invariance in the two frames saysAt a fixed point, choose the components of $A$ to be each matrix unit in turn; each is an admissible tensor once its components in other frames are transformed. Thus every coefficient of $A_{pq}$ agrees, yielding $Q^T{C}^{\prime }Q=C$, orTherefore $C$ is necessarily a Cartesian second-rank tensor. This is the scalar contraction test for a Cartesian tensor. Testing all tensors, rather than only symmetric tensors, is essential to the proof.

A Cartesian second-rank tensor is an isotropic tensor if its components are unchanged under every proper rotation matrix $R$:The Kronecker delta is isotropic because $R_{ik}{R}_{jk}={\delta }_{ij}$, the orthogonality relation $R{R}^T=I$.

No symmetry of $T$ needs to be assumed. A half-turn about the first axis has matrix $\mathrm{diag}(1,-1,-1)$; invariance under it forces $T_{12},T_{13},T_{21},T_{31}$ to vanish. A half-turn about the second axis additionally kills $T_{23}$ and $T_{32}$. Thus $T=\mathrm{diag}(d_1,d_2,d_3)$. Each half-turn is the square of an allowed quarter-turn. A quarter-turn about the third axis interchanges $d_1,d_2$, so they are equal. A quarter-turn about the second axis interchanges $d_1,d_3$, so all three are equal. Consequently the complete family of isotropic second-rank tensors isConversely, every member of this family is invariant because $R(\lambda I)R^T=\lambda I$.

For a Cartesian change of basis represented by an orthogonal matrix $Q$, vectors transform as $E_i^{\prime }=Q_{ia}{E}_a$ and $B_i^{\prime }=Q_{ia}{B}_a$. Their squared norms are invariant and $Q_{ia}{Q}_{jb}{\delta }_{ab}={\delta }_{ij}$. Substitution into the given expression therefore givesthe transformation law of a Cartesian second-rank tensor. The magnetic field's extra axial sign under an improper physical reflection, if included, appears twice and cancels in its quadratic contribution.

Write the Maxwell stress tensor as $T=\mathbf{EE}+\mathbf{BB}-\frac{1}{2}(E^2+B^2)I$. Its divergence isThe identity $(\mathbf{A}\cdot \nabla )\mathbf{A}-\frac{1}{2}\nabla A^2=-\mathbf{A}\times (\nabla \times \mathbf{A})$ follows by contracting two Levi-Civita symbols, or directly by differentiating components. Using Maxwell's equations consequently yieldsHence the local conservation of electromagnetic momentum isThe two subtracted terms are the Lorentz force density, while $\mathbf{E}\times \mathbf{B}$ is the electromagnetic momentum density in the units used here.

Under a proper rotation matrix $R$, invariance of a Cartesian second-rank tensor reads $T=RT{R}^T$, equivalently $TR=RT$. The half-turn $R_z(\pi )=\mathrm{diag}(-1,-1,1)$ forces all entries mixing the $z$ direction with the $xy$ plane to vanish. Thus $T$ consists of a planar $2\times 2$ block $B$ and the entry $T_{33}=c$.

Commutation with the planar quarter-turn $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ gives $B=aI+bJ$. This block already commutes with every planar rotation. Now use the half-turn $R_x(\pi )=\mathrm{diag}(1,-1,-1)$: on the planar block its conjugation fixes $I$ and sends $J$ to $-J$, so invariance forces $b=0$. ThereforeThe formula uses the Kronecker delta and is sufficient as well as necessary: rotations preserving the unoriented $z$-axis fix both $I$ and $e_3{e}_3^T$. This is an axially invariant second-rank tensor with the planar antisymmetric part removed by the horizontal half-turn.

Use the Cartesian second-rank tensor transformation convention $T^{\prime }=RT{R}^T$, where $R$ is the orthogonal matrix converting components to the new orthonormal basis. The coordinate half-turns areFor a diagonal rotation $R=\mathrm{diag}(s_1,s_2,s_3)$, invariance gives $t_{ij}=s_i{s}_j{t}_{ij}$, with no summation in this equation. For each $i\ne j$, one of the listed half-turns makes $s_i{s}_j=-1$, so $t_{ij}=-t_{ij}=0$. ThusThis is coordinate half-turn invariance of a second-rank tensor. It does not force equal diagonal entries; invariance under all rotations would be the stronger isotropic second-rank tensor condition.