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An isotropic tensor has unchanged components under every rotation: for rank four, $A_{ijkl}=R_{ia}{R}_{jb}{R}_{kc}{R}_{ld}{A}_{abcd}$. Orthogonality gives $R_{ia}{R}_{ja}={\delta }_{ij}$, so each product of paired Kronecker deltas is invariant. Every linear combination of ${\delta }_{ij}{\delta }_{kl}$, ${\delta }_{ik}{\delta }_{jl}$ and ${\delta }_{il}{\delta }_{jk}$ is therefore isotropic, for arbitrary coefficients.

For the integral, differentiation away from the origin givesAfter multiplication by $x_i{x}_j$, this behaves as $O(r^{-1})$ and is locally integrable in three dimensions. Interpret the integral as the limit with a small central ball removed. A rotation preserves the integration ball and the radial kernel, so the resulting $B$ is an isotropic tensor integral. The symmetries $i\leftrightarrow j$ and $k\leftrightarrow l$ force the two cross-pair coefficients equal:Taking the trace in $k,l$ uses ${\nabla }^2(r^{-1})=0$ off the origin and gives $3A+2B=0$. For the other contraction, the displayed Hessian gives ${\sum }_{ij}{x}_i{x}_j{\partial }_i{\partial }_j(r^{-1})=2/r$. ThusSolving gives $B=2\pi a^2/5$ and $A=-4\pi a^2/15$. Therefore the requested tensor isThis is the quadratically weighted inverse-radius Hessian integral. The origin creates no extra contribution to this weighted integral; even the distributional point term in the unweighted Hessian is annihilated by $x_i{x}_j$.

A fully ionized helium nucleus supplies two electrons for approximately $4m_p$ of mass, so $n_e=\rho /(2m_p)$. Filling all electron momentum states up to the Fermi momentum givesFor a nonrelativistic electron, the speed is $p/m_e$. The three mean squared Cartesian components of momentum are equal by rotational symmetry and sum to $p^2$, so each is $p^2/3$. This is the isotropic tensor integral. Thus the momentum flux, or electron degeneracy pressure, isSubstituting the expression for $p_0$ and simplifying the powers of two givesHere $h$ is the Planck constant. This is the pure-helium specialization of nonrelativistic electron-degeneracy pressure in a hydrogen-helium mixture.

For a cold helium white dwarf, neglect ion thermal pressure and use this electron degeneracy pressure as the support against Newtonian gravity. Combining hydrostatic equilibrium with mass conservation givesSet $\rho ={\rho }_c{\theta }^{3/2}$ and $r=\alpha \xi$, where ${\alpha }^2=5K{\rho }_c^{-1/3}/(8\pi G)$. Direct substitution gives the Lane-Emden equationIts dimensionless solution is independent of ${\rho }_c$. Let ${\xi }_1$ be its first zero and ${\omega }_1=-{\xi }_1^2{\theta }^{\prime }({\xi }_1)={\int }_0^{{\xi }_1}{\xi }^2{\theta }^{3/2}d\xi$. Integrating mass conservation givesEliminating ${\rho }_c$ yields the polytropic mass-radius relationThus a more massive nonrelativistic white dwarf is smaller at fixed stellar composition. The assumptions require $p_0\ll m_{e}c$ and $k_{B}T\ll p_0^2/(2m_e)$ in the region providing most support; the law is not extended into the relativistic or thermally supported regimes.

Here isotropic tensors are invariant under proper rotation matrices, $R^{T}R=I$ and $\det R=1$. The isotropic second-rank tensor and isotropic third-rank tensor forms in three dimensions arewith scalar coefficients $A,B$. One can see completeness directly. A half-turn about each coordinate axis kills every off-diagonal component of a rank-two tensor, and quarter-turns make its three diagonal components equal. For rank three, the half-turns require that each coordinate label occur an odd number of times in a nonzero component. Thus only permutations of $(1,2,3)$ survive. Quarter-turns interchange two labels while reversing one sign, forcing the surviving components to alternate with the parity of a permutation. This gives a multiple of the Levi-Civita symbol.

The tensor transformation law verifies invariance of these forms:The first equality uses orthogonality and the second uses the determinant formula. If reflections were also required while using the ordinary tensor transformation law, the rank-three coefficient would have to vanish. The proper-rotation convention is the one under which $B{ϵ}_{ijk}$ is an isotropic tensor.

The ball is unchanged by any rotation matrix, and rotating coordinates has Jacobian determinant one. Thus changing variables $\mathbf{y}=R\mathbf{x}$ givesThese moments are therefore isotropic tensor integrals. For the Cartesian moments of a three-dimensional ball, the second moment is $\alpha {\delta }_{ij}$. Taking its tensor contraction givesThe third moment vanishes: $\mathbf{x}\mapsto -\mathbf{x}$ preserves the ball and changes the sign of its integrand. Equivalently, its symmetry under index interchange excludes a nonzero multiple of the antisymmetric Levi-Civita symbol.

The fourth moment is symmetric under every permutation of its four indices. The general rank-four isotropic tensor form consequently has equal coefficients, yieldingFor example, comparing $T_{1122},T_{1212},T_{1221}$ shows that the three coefficients agree. Contracting $i=j$ and $k=l$ givesFinally the vector triple product identity gives $\mathbf{x}\times (\mathbf{\Omega }\times \mathbf{x})=r^2\mathbf{\Omega }-(\mathbf{\Omega }\cdot \mathbf{x})\mathbf{x}$. The first term integrates to $3\alpha \mathbf{\Omega }$ and the second to $\alpha \mathbf{\Omega }$, using the second moment. Hence

Remove the constant trace from the second mass moment tensor to obtain the mass quadrupole moment $Q_{ij}=I_{ij}-{\delta }_{ij}{I}_{kk}/3$. Since $I_{kk}$ is constant, ${\stackrel{...}{Q}}_{ij}={\stackrel{...}{I}}_{ij}$. In units $G=c=1$, the quadrupole formula givesIts normalization can also be seen from the gravitational-wave energy flux: $h_{ij}^{\mathrm{TT}}=2{\Lambda }_{ij,kl}{\ddot{Q}}_{kl}/r$, where $\Lambda$ is the transverse-traceless projector. With $S_{ij}={\stackrel{...}{Q}}_{ij}$ and $P_{ij}={\delta }_{ij}-n_i{n}_j$, the projection is $S_{ij}^{\mathrm{TT}}=P_{ik}{P}_{jl}{S}_{kl}-P_{ij}{P}_{kl}{S}_{kl}/2$. For a symmetric trace-free $S$,The isotropic tensor integrals $\int n_i{n}_{j}d\Omega =4\pi {\delta }_{ij}/3$ and $\int n_i{n}_j{n}_k{n}_{l}d\Omega =4\pi ({\delta }_{ij}{\delta }_{kl}+{\delta }_{ik}{\delta }_{jl}+{\delta }_{il}{\delta }_{jk})/15$ therefore give $\int |S^{\mathrm{TT}}{|}^{2}d\Omega =8\pi S_{ij}{S}_{ij}/5$. Integrating the flux yields $⟨P⟩=(8\pi {)}^{-1}\int ⟨|S^{\mathrm{TT}}{|}^2⟩d\Omega =⟨S_{ij}{S}_{ij}⟩/5$, as above.

Now differentiate the components from part (i). With $\Delta =I_1-I_2$,The off-diagonal component occurs twice in the contraction. ThusIt is already time independent, so averaging givesThis is the leading gravitational radiation from a rotating triaxial body; restoring units multiplies it by $G/c^5$. The source is treated as rotating uniformly over an averaging interval, with radiation reaction negligible at this order.

The outgoing solution obtained from the retarded fundamental solution of the wave equation isIn the radiation zone, put $r=|\mathbf{x}|$, $\mathbf{n}=\mathbf{x}/r$, and retain the leading $1/r$ term at retarded time $u=t-r$. The trace term disappears after applying the transverse-traceless projectorsoTwice using stress-energy conservation, ${\partial }_a{T}^{ab}=0$, and integrating by parts givesThe projector removes the trace, so in terms of the mass quadrupole momentthe far field isThe gravitational-wave energy flux isFor a trace-free symmetric tensor $A_{ij}$, the isotropic tensor integral over the observation direction givesConsequently the standard quadrupole formula, in the units $G=c=1$ used by the paper, isThus the printed coefficient $4\pi /5$ is inconsistent with the stated definition of $Q_{ij}$ and the standard wave-energy normalization; it appears to be a typographical error.

For the planet, choose its circular orbit in the $xy$-plane and write $\omega =2\pi /\tau$. With the star treated as fixed,and henceDirect differentiation givessoRestoring units multiplies this by $G/c^5$. If the printed $4\pi /5$ coefficient is followed literally, the answer is instead $4\pi$ times larger,For two bodies of comparable mass, $M$ is replaced by the reduced mass and $R$ by their separation.

Write $m_{ij}=m_{ij}^{(\pm 2)}$ and $h=h^{(\pm 2)}$. Since $m_{ij}{e}^i{e}^j=\frac{1}{2}(1-{\mu }^2)e^{\pm 2i\phi }$, the line-of-sight solution isThe angular pattern $m_{ij}{e}^i{e}^j$ is proportional to $Y_{2,\pm 2}$. Projecting onto this spherical harmonic multiplies its coefficient bywhere $x=k(\eta -{\eta }^{\prime })$ and $j_2$ is a Spherical Bessel function. Thus the quadrupole isUse the isotropic tensor integralBecause $m$ is symmetric and trace-free, contracting this with $m_{ab}$ gives $2m^{ij}/15$. Substitution into the paper's signed neutrino tensor anisotropic stress therefore givesThe neutrino tensor free-streaming kernel is finite at the origin, with $K(0)=1/15$.

The measure, Gaussian factor, $r^{-1}$, coordinates, and derivatives all transform covariantly under rotations, so $T_{ijkl}$ is an isotropic tensor integral. Symmetry gives $\beta =\gamma$.

Contracting $i=j$ and using ${\nabla }^2(1/r)=0$ away from the origin givesNext contract $i=k$ and $j=l$. Sinceone obtainsSolving yields