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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.

Apply the product rule and the preceding gradient formula to the vector field $F_i=f(r)x_i$. Summing over the repeated index givesHere ${\delta }_{ii}=n$ for the Kronecker delta. Thus the divergence and curl of a radial vector field give, in any dimension,

For the positive Euclidean norm $r=(x_1^2+x_2^2+x_3^2{)}^{1/2}$, the chain rule givesRepeated indices use the Einstein summation convention. The origin is excluded because the Euclidean norm is not differentiable there. Another derivative, useful below, is ${\partial }_i{\hat{x}}_j={\delta }_{ij}/r-x_i{x}_j/r^3$, with ${\delta }_{ij}$ the Kronecker delta.

First eliminate the density-dependent definition of the auxiliary $\mathbf{g}$. The Newtonian gravitational stress tensor can be written without derivatives of its normalization:Here ${\delta }_{ij}$ is the Kronecker delta, and repeated indices are summed. Differentiating givesThe first and third terms cancel because mixed partial derivatives commute; the final step uses the Poisson equation for Newtonian gravity. Therefore the gravitational force density is a stress divergence:The identity holds for spatially varying mass density: substituting the definition of $\mathbf{g}$ before differentiating prevents erroneous extra density-gradient terms.

For the gravitational stress contribution to angular-momentum transport, the anisotropic term $\rho g_i{g}_j$ supplies angular momentum transport. In cylindrical components its radial–azimuthal entry isIn the momentum conservation equation, $r{T}_{r\varphi }$ is the radial stress contribution to the angular-momentum flux, whose sign depends on the correlated radial and azimuthal field components. It vanishes for a perfectly axisymmetric potential but can be nonzero for spiral disturbances.

The other term is isotropic and acts as an effective negative pressure, $P_g=-\rho g^2/2=-|\nabla \Phi {|}^2/(8\pi G)$. Its force contribution is $-\nabla P_g=+\nabla (\rho g^2/2)$, modifying normal compression and force balance. It has no off-diagonal shear component and therefore no direct radial angular momentum transport. In an axisymmetric averaged disk it supplies no azimuthal torque. The complete symmetric Newtonian gravitational stress tensor also expresses conservation of angular momentum without an internal couple.

Within each fixed unitary irreducible representation, choose the selected right singular vectors to be orthonormal. Their corresponding left singular vectors are also orthonormal, even when a singular value is repeated: choose an orthonormal basis in each eigenspace of $T_{\rho }^{\ast }{T}_{\rho }$ and put $B_p=T_{\rho }{A}_p/{\lambda }_p$. Therefore, when ${\rho }_p={\rho }_q$,This is (iv), with the unnormalized Hilbert-Schmidt inner product and the Kronecker delta. No orthogonality of differently shaped matrices is being asserted.

For the unheaded continuation, partition $b\in {\mathbb{C}}^m$ into blocks $b_p\in {\mathbb{C}}^{n_p}$. ThenThe Schur averaging of rectangular matrices formula isfor any $n_p\times n_q$ matrix $M$. The average is an intertwiner; the Schur lemma makes it zero for inequivalent irreducible representations, and a scalar multiple of the identity for the same representative. The trace determines that scalar in the latter case.

Apply this with $M=U(p{)}^{\ast }U(q)$. The Hilbert-Schmidt inner product normalization just proved givesExpanding the squared Euclidean norm now yieldsThus the final identity follows from averaging, although $U^{\ast }U$ itself need not be the identity. The whole construction is the spectral inverse theorem for the matrix-valued U2 quantity.

Set $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$, $M=M_1+M_2$ and $\mu =M_1{M}_2/M$, the reduced mass. Newton's law of universal gravitation gives ${\ddot{\mathbf{x}}}_1=-G{M}_2\mathbf{r}/r^3$ and ${\ddot{\mathbf{x}}}_2=G{M}_1\mathbf{r}/r^3$. Subtraction establishesFor fixed masses, the orbital energy and angular momentum areTaking derivatives gives $\dot{E}=\mu \mathbf{v}\cdot (\ddot{\mathbf{r}}+GM\mathbf{r}/r^3)=0$ and $\dot{\mathbf{J}}=\mu \mathbf{r}\times \ddot{\mathbf{r}}=0$. These are conservation of energy and conservation of angular momentum in the unperturbed Kepler orbit. To see directly that the orbital eccentricity is fixed, put $\mathbf{h}=\mathbf{r}\times \mathbf{v}$ and $\mathbf{n}=\mathbf{r}/r$. The eccentricity vector isSince $\dot{\mathbf{h}}=0$, the cross product identity $\mathbf{r}\times (\mathbf{r}\times \mathbf{v})=\mathbf{r}(\mathbf{r}\cdot \mathbf{v})-r^2\mathbf{v}$ gives $\ddot{\mathbf{r}}\times \mathbf{h}=GM\dot{\mathbf{n}}$, so $\dot{\mathbf{e}}=0$. Its magnitude $e$ is the orbital eccentricity. Equivalently,Thus the semi-major axis and orbital eccentricity of a bound noncollision Kepler orbit are constant, including $e=0$.

Take the centre of mass as origin. The two positions are ${\mathbf{x}}_1=(M_2/M)\mathbf{r}$ and ${\mathbf{x}}_2=-(M_1/M)\mathbf{r}$. For $x\gg r$, a Taylor expansion givesThe mass dipole vanishes because ${\sum }_A{M}_A{\mathbf{x}}_A=0$, and the second mass moment tensor is ${\sum }_A{M}_A{x}_{Ai}{x}_{Aj}=\mu r_i{r}_j$. The gravitational quadrupole potential of a point-mass binary is consequentlyBoth printed tensors are trace-free. Contracting them with the Kronecker delta summed over three spatial dimensions givesThis proves the requested expression through second order. In particular the printed convention is $q_{ij}=\frac{3}{2}{Q}_{ij}$, where $Q_{ij}=\mu (r_i{r}_j-r^2{\delta }_{ij}/3)$ is the standard mass quadrupole moment; the radiation coefficient must be adjusted accordingly. No mass dipole term may be retained in centre of mass coordinates.

To evaluate the quadrupole formula without assuming a circular orbit, put $k=GM$, $u=\dot{r}=\mathbf{n}\cdot \mathbf{v}$ and ${\mathbf{v}}_{\perp }=\mathbf{v}-u\mathbf{n}$. Differentiating the gravitational acceleration gives the kinematic jerkFor $S_{ij}=r_i{r}_j$, the product rule then givesInserting these into the printed mass quadrupole moment convention yieldsThe two tensors in brackets are orthogonal in their tensor contraction, because $\mathbf{n}\cdot {\mathbf{v}}_{\perp }=0$. Their squared norms are $6$ and $2v_{\perp }^2$, respectively. HenceThe instantaneous quadrupole luminosity of a Kepler binary is thereforeThe radial derivative $\dot{r}$ is not the vector speed. The bracket equals $v_{\perp }^2+u^2/12$, so the radiated luminosity is nonnegative for every instantaneous velocity.

For a slowly evolving circular orbit, there is no preferred orbital phase at which to excite a persistent eccentricity vector. More precisely, the rotating mass quadrupole moment emits at twice the orbital frequency, with angular harmonic number two; its gravitational-wave energy and angular-momentum balance is $\dot{E}=\Omega \dot{J}$. The circular Kepler orbit sequence has $J=\mu \sqrt{GMa}$ and $dE/dJ=\Omega$, so this loss is tangent to the sequence. At leading adiabatic order it preserves zero secular orbital eccentricity. This is a circular gravitational-wave inspiral, with a small radial drift rather than an exactly fixed-radius Newtonian circle. The energy flux alone would not establish circularity without this symmetry and angular momentum balance.

To leading radiation order use $r=a$, $u=0$ and ${\mathbf{v}}^2=GM/a$ in the loss formula. Combining it with $E=-GM\mu /(2a)$ givesKeeping the fixed masses and leading quadrupole formula, integration givesThis formal coalescence time displays the strong $a_0^4$ dependence: gravitational-wave emission from a binary system matters far more for close binary stars than for wide ones. The orbital period decreases as $a^{3/2}$ and the gravitational-wave frequency increases, producing a chirp. A detached binary can be driven into Roche-lobe overflow; further evolution then depends on mass-transfer stability. A sufficiently close compact pair may merge. The weak-field, slow-motion, point-mass approximation ceases to apply before literal $a=0$; finite stellar radii or strong relativistic effects determine the final interaction.

In suffix notation,where ${\delta }_{ij}$ is the Kronecker delta. Since $r=(x_k{x}_k{)}^{1/2}$, the chain rule gives

In coordinates $x^{\mu }$, write the type-$(1,1)$ tensor asFor a covector $\eta ={\eta }_{\mu }d{x}^{\mu }$ and vector $V=V^{\nu }{\partial }_{\nu }$,Since this must equal $\eta (V)={\eta }_{\nu }{V}^{\nu }$ for every $\eta$ and $V$, the components areThus $T$ is the identity tensor.

The tensor component transformation law givesThe components are therefore the same Kronecker delta in every coordinate system.

The identity matrix has entries given by the Kronecker delta, soEach of the $n$ diagonal entries is one.

Since $M^T=I-\mathbf{a}{\mathbf{b}}^T=I-N$, nilpotence already gives $(I+N)(I-N)=I$. Explicitly, in index notation,Thus $\boxed{L{M}^T=I}$, or equivalently $M^T=L^{-1}$. Here ${\delta }_{ij}$ is the Kronecker delta.

The constant coefficient is $a_0=0$. For every $n\ge 1$ the two mode equations arewhere ${\delta }_{n1}$ is the Kronecker delta. The first equation gives $a_n=-\sqrt{3}n{b}_n$, so the second becomes $(1+3n^2)b_n=4{\delta }_{n1}$. HenceNo other modes survive. This is the single-frequency case of the Fourier solution of a first-order equation with periodic forcing.

Contract the Kronecker deltas in the first term to obtain $a_i{b}_i$. In the second term, ${\delta }_{ij}$ identifies the two $a$ components, while the two $b_k$ components already contract. Therefore

The displayed matrix is not orthogonal: its first column has squared Euclidean norm three, whereas every column of an orthogonal matrix must have squared length one.

For the general assertion, put $C=AB$. In the Einstein summation convention, orthogonality gives $A_{ki}{A}_{kj}={\delta }_{ij}$ and $B_{ki}{B}_{kj}={\delta }_{ij}$. The entries of $C$ satisfyThis is exactly $C^{T}C=I$, proving that the matrix product $AB$ is orthogonal. The Kronecker delta contracts the two middle indices without changing the multiplication order of the matrices.

Contracting the Kronecker delta identifies $i=k$. The remaining two independent sums are dot products, so the Einstein summation convention givesThe index expression here is verified against the PDF because the OCR corrupts it completely.

In three dimensions each contracted Kronecker delta has value three, while $b_k{b}_k=|\mathbf{b}{|}^2$. Thus

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

The Kronecker delta restricts the Einstein summation convention to equal indices, giving the dot product

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.

The Cartesian tensor law is $T_{ik}^{\prime }=Q_{ia}{Q}_{kb}{T}_{ab}$. In every complete tensor contraction, summation over a transformed index contracts a pair of $Q$ factors to a Kronecker delta. In detail,Thus all three are invariant under rotations. They are $\mathrm{tr}T$, $\mathrm{tr}(T^2)$ and $\mathrm{tr}(T^3)$. For a real symmetric tensor, orthogonal diagonalization givesExpanding ${\prod }_i(\lambda -{\lambda }_i)$ and using Newton's identities to recover the characteristic polynomial from traces of powers yieldsComparison identifies $\boxed{c_2=-1,c_1=\frac{1}{2},c_0=-\frac{1}{6}}$.

An isotropic tensor has unchanged components under every rotation of the Cartesian axes. Under proper rotations in three dimensions, rank one has no nonzero isotropic tensor, rank two has $\boxed{T_{ij}=A{\delta }_{ij}}$, and rank three has $\boxed{T_{ijk}=B{ϵ}_{ijk}}$. Here $\delta$ is the Kronecker delta and $ϵ$ the Levi-Civita symbol. If invariance under reflections is also required for a true tensor, the rank-three case vanishes; $ϵ$ is then a pseudotensor.

The velocity gradient tensor for the given field isIts symmetric part, the strain-rate tensor if $\mathbf{u}$ is interpreted as a velocity, isThe antisymmetric part has the specified ordering $D_{ik}-e_{ik}={ϵ}_{ijk}{\omega }_j$, withThe index ordering matters for this sign; it agrees with a local rotation velocity $\mathbf{\omega }\times \mathbf{x}$. This is the axial part of the scalar, symmetric-traceless and axial tensor decomposition, retaining the trace inside the symmetric part here.

At the specified point, $e=\left(\begin{array}{ccc}9& 6& 0\\ 6& 0& 0\\ 0& 0& 4\end{array}\right)$. The upper block has characteristic polynomial ${\lambda }^2-9\lambda -36=(\lambda -12)(\lambda +3)$. Thus the principal values and unit eigenvectors areThese mutually orthogonal directions are the principal strain rates' axes for this local field; reversing an eigenvector does not change its axis.