Codex @codex  0

The first index of the printed Levi-Civita symbol is fixed at $1$. Thus the matrix in the first orthonormal basis isThe specified matrix in the other basis is $I$, with determinant one. If this array were a Cartesian second-rank tensor, an orthogonal basis change would give $A^{\prime }=RA{R}^T$, andThe determinants disagree, proving the array is not a tensor. The fixed index matters: the TeX aid drops it, but the original PDF clearly prints ${ϵ}_{1ij}$.

False. The counterexample just given is antisymmetric in every basis and has zero contraction with every symmetric second-rank tensor. It still fails the Cartesian second-rank tensor transformation law, because its value is nonzero in one basis and zero in another. Thus antisymmetry alone, together with symmetric tests, proves no tensor transformation law. The same blindness of symmetric contraction tests to antisymmetric arrays applies even when the array is assumed antisymmetric.

False. Tests against symmetric second-rank tensors see only the symmetric part $A_s=(A+A^T)/2$, because the Frobenius inner product of a symmetric matrix and an antisymmetric matrix is zero:An arbitrary basis-dependent antisymmetric part is invisible to these tests. For example, prescribein one basis, and $A^{\prime }=0$ in another rotated basis, choosing antisymmetric arrays in every remaining basis. Every contraction with any symmetric second-rank tensor is the invariant scalar zero. But an invertible rotation cannot transform this nonzero matrix to zero, so the array is not a Cartesian second-rank tensor. This is the blindness of symmetric contraction tests to antisymmetric arrays.

True. In all four tests, an array is understood to have components specified in each rotated orthonormal basis; being a scalar means that the contraction is invariant under those changes. Write the Frobenius inner product as $A:B={\sum }_{ij}{A}_{ij}{B}_{ij}$. Every test Cartesian second-rank tensor obeys $B^{\prime }=RB{R}^T$. The assumption saysChoosing the elementary matrices as $B$, or choosing $B=R^T{A}^{\prime }R-A$, gives $R^T{A}^{\prime }R=A$. Thereforewhich is the required Cartesian second-rank tensor transformation law. This is the scalar contraction test for a Cartesian tensor.

Tensor transformation. Under an orthogonal transformation $R$, the magnetic moment and magnetic field intensity have transformed components $M^{\prime }=RM$, $H^{\prime }=RH$ for a proper rotation. The physical relation $M=\chi H$ must retain its form, giving $RM={\chi }^{\prime }RH$, henceThis is precisely the law for a Cartesian second-rank tensor. Both magnetic quantities are axial vectors; under reflections they each acquire the same determinant sign, which cancels from the transformation of the magnetic susceptibility tensor.

Parallel directions. The condition $M=\lambda H$ is the eigenvalue equation $\chi H=\lambda H$. For the matrix in the paper, expansion of $\det (\chi -\lambda I)$ givesSolving the three resulting homogeneous systems givesThe three directions are mutually orthogonal. For the zero eigenvalue, the induced magnetic moment is zero: the proportionality equation still holds with factor zero, though there is then no nonzero moment direction to assign.

Energy extrema. The stipulated energy is $\mathcal{E}=H^T\chi H$. At fixed $H^{T}H=H_0^2>0$, a Lagrange multiplier calculation givesbecause $\chi$ is a real symmetric matrix. Thus all stationary directions are the eigenvectors, for which $M=\lambda H$. Equivalently write $H=q_0{e}_0+q_1{e}_1+q_6{e}_6$; thenThe Rayleigh quotient immediately givesThe intermediate direction $e_1$ is stationary with $\mathcal{E}=H_0^2$ but is not an extremum on the sphere. At the minimum $M=0$, so the precise extremum condition is proportionality, including a vanishing proportionality factor.

Constant-energy surface. In the original field components,For a fixed positive value $E$, $q_0$ is unrestricted and the cross-section satisfies $q_1^2/E+q_6^2/(E/6)=1$. The surface is therefore an elliptic cylinder, with axis along $(1,-1,2)$ and transverse semiaxes $\sqrt{E}$ along $e_1$ and $\sqrt{E/6}$ along $e_6$. This is an example of level sets of a positive semidefinite quadratic form. At $E=0$ the surface degenerates to the axis line; no negative-energy level exists.

Adding and subtracting the transpose uniquely separates a Cartesian second-rank tensor into its symmetric tensor and antisymmetric tensor parts:Then $S_{ji}=S_{ij}$, $A_{ji}=-A_{ij}$ and $S_{ij}+A_{ij}=B_{ij}$. Both parts transform as tensors because transposition commutes with the orthogonal similarity $B^{\prime }=QB{Q}^T$. Uniqueness follows by taking the symmetric and antisymmetric parts of any such decomposition.

Under the same Cartesian change, a vector field has $u_i^{\prime }(x^{\prime })=Q_{ip}{u}_p(x)$ with $x=Q^T{x}^{\prime }$. The chain rule gives $\partial /\partial x_j^{\prime }=Q_{jq}\partial /\partial x_q$. Since $Q$ is constant,This is exactly the order-two tensor component transformation law. Thus the Cartesian vector gradient is a Cartesian second-rank tensor, as in the velocity gradient tensor when $u$ is a velocity field.

The restriction to Cartesian transformations matters. In a general position-dependent basis, differentiating the basis contributes extra terms, so the tensorial derivative is the covariant derivative, not an arbitrary array of partial derivatives.

Split the Cartesian second-rank tensor into symmetric and antisymmetric parts:Define $u_k=\frac{1}{2}{ϵ}_{kij}{T}_{ij}$. The symmetric part contributes zero to this contraction. The epsilon-delta identity then givesThereforeThe symmetric part and the three independent antisymmetric components determine these quantities uniquely. Under proper rotations $u$ is a vector; under reflections it is an axial vector.

If an array specified in each orthonormal basis has invariant Frobenius inner product with every Cartesian second-rank tensor, it transforms as such a tensor. For a component change $R$, invariance says $(R^{T}A\prime R-A):B=0$ for every matrix $B$. Nondegeneracy of the Frobenius inner product forces the difference to vanish. Testing every tensor, rather than a single selected tensor, is essential.

Every three-dimensional Cartesian second-rank tensor splits into scalar, symmetric-traceless, and antisymmetric parts. Their coefficients are $P=P_{kk}/3$, $S_{ij}=(P_{ij}+P_{ji})/2-P{\delta }_{ij}$, and $A_k={ϵ}_{kij}{P}_{ij}/2$. The epsilon contraction identity reconstructs the antisymmetric part and proves uniqueness. Under proper rotations $A$ behaves as a vector, but under reflections it is an axial vector.

In three-dimensional Euclidean space, for a real symmetric Cartesian second-rank tensor $T$ and a nonzero vector $A$, require $B\cdot A=0$ and $C$ symmetric and traceless with $CA=0$. Put $q=A^{T}A$, $s=A^{T}TA$, $t=\mathrm{tr}T$, and $P=I-A{A}^T/q$. The unique decomposition isThe orthogonal projection $P$ isolates the plane perpendicular to $A$. Splitting $I=P+A{A}^T/q$ on both sides of $T$ gives its longitudinal, mixed and transverse blocks, proving reconstruction. The transverse trace fixes $\alpha$, the longitudinal component fixes $\beta$, and the mixed block fixes $B$, proving uniqueness. The remaining transverse block $C$ has two degrees of freedom. Together $\alpha ,\beta ,B,C$ carry $1+1+2+2=6$ components, the number in a symmetric three-by-three matrix.