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For a bounded sufficiently regular region, the divergence theorem applied to the vector field $x_i{e}_j$ gives the displayed identity. Contracting with constant vectors gives ${\int }_{\partial \Omega }(b\cdot x)(c\cdot n)dS=|\Omega |b\cdot c$. Contracting with the Kronecker delta or the Levi-Civita symbol respectively gives $\int x\cdot ndS=3|\Omega |$ and $\int x\times ndS=0$ in three dimensions.

A Cartesian basis is an ordered basis of unit vectors along the mutually perpendicular axes of a Cartesian coordinate system. It is an orthonormal basis of a real Euclidean space: $e_i\cdot e_j={\delta }_{ij}$, with ${\delta }_{ij}$ the Kronecker delta. In the plane, the familiar standard basis is $e_1=(1,0{)}^T$, $e_2=(0,1{)}^T$; the vector $p=(2,1{)}^T$ is $2e_1+e_2$. A rotated pair of perpendicular unit vectors is also a Cartesian basis, even though its basis vectors have different standard basis entries.

The advantage is that coordinates can be measured separately by dot products, and the squared norm is the sum of their squares:These are the real Euclidean space instances of the orthonormal basis expansion.^[ref]

An orthogonal basis only requires perpendicular basis vectors; their lengths need not be one. For example, $(2,0{)}^T,(0,3{)}^T$ are orthogonal, but $p=(2,1{)}^T$ has coordinates $(1,1/3{)}^T$, while its dot products with them are $4,3$. For an orthogonal basis $(a_i)$, the coordinate along $a_i$ is $(p\cdot a_i)/(a_i\cdot a_i)$. Dividing each nonzero basis vector by its norm makes it a unit vector without changing its direction. Unit length alone does not imply orthogonality: $(1,0{)}^T$ and $(1,1{)}^T/\sqrt{2}$ are both unit vectors but have dot product $1/\sqrt{2}$.

The examples below show why these distinctions matter: rotating a Cartesian basis changes the coordinate vector while keeping lengths easy to compute; using nonperpendicular basis vectors still gives unique coordinates, but requires the dual basis or a matrix inverse to extract them.

A complex visibility is a calibrated complex correlation of the signals received at a particular interferometric baseline. In a narrow field, with scalar radiation, the far-field approximation for an outgoing source, a narrow frequency channel, negligible noncoplanar phase curvature and a steady sky of mutually incoherent emitters,Here $l,m$ are small dimensionless sky-direction offsets, $I$ is brightness in the chosen channel, and $A$ is the primary beam. This is the Fourier transform of beam-weighted brightness. The two-dimensional formula requires those approximations: a wide field generally needs the third baseline coordinate and curved sky geometry.

For an explicit one-dimensional derivation, set the primary beam to one over the small field, and place independent emitters at angular offsets ${\theta }_j$. Choose a wave phase convention in which, after removing the phase centre's delay, their received complex amplitudes areHere ${\delta }_{jk}$ is the Kronecker delta and angle brackets denote an expected value estimated by time averaging a steady signal. The last equality states mutual incoherence: different emitters have zero averaged cross term, although each emitter's two received copies remain correlated. Multiplying and averaging givesFor continuous brightness this becomes $V(u)=\int I(\theta )e^{-2\pi iu\theta }d\theta$. Real brightness implies the Hermitian symmetry $V(-u)=V(u{)}^{\ast }$; it does not require brightness to be even. In particular, an asymmetric real source usually has a nonzero imaginary complex visibility. At zero baseline, $V(0)={\sum }_j{I}_j$ is the total beam-weighted source strength. Ordinary separated-element correlations need not measure that zero-baseline value.

For a basis $e_1,\dots ,e_n$ of a finite-dimensional vector space $V$, its dual basis consists of linear functionals $e_1^{\ast },\dots ,e_n^{\ast }$ satisfyingThe Kronecker delta means that each rule gives one on its own basis vector and zero on the others. These rules extract coordinates:Applying $e_i^{\ast }$ to a linear combination removes every term except the $i$th. Consequently the dual basis is a basis of the dual space, and every linear functional has expansion $\phi ={\sum }_i\phi (e_i)e_i^{\ast }$. ^[ref].

Coordinate extraction works for a nonorthogonal basis too. With $b_1=(1,0)$ and $b_2=(1,1)$, the dual basis isThe checks $b_1^{\ast }(b_1)=1$, $b_1^{\ast }(b_2)=0$, $b_2^{\ast }(b_1)=0$, and $b_2^{\ast }(b_2)=1$ fix the two rules. For $v=(3,1)$ they give coordinates $(2,1)$, so $v=2b_1+b_2$. The usual dot products with $b_1,b_2$ instead give $3,4$ and do not extract these coordinates: simple projection formulas depend on an orthonormal basis.

The same idea explains polynomial interpolation. On polynomials of polynomial degree at most $n$, evaluation at distinct interpolation nodes $x_0,\dots ,x_n$ gives the dual basis to the Lagrange cardinal polynomials, because ${\ell }_i(x_j)={\delta }_{ij}$. Samples are coordinates in this basis, so the reconstruction $p={\sum }_{i}p(x_i){\ell }_i$ is a dual-basis expansion. Differentiation weights from nodal evaluations and factorial monomial basis for polynomial differentiation apply the same principle to different measurements.

The identity tensor is the canonical type-$(1,1)$ tensor that maps each tangent vector to itself. It pairs a covector $\eta$ and vector $V$ as $\eta (V)$, and its components are the Kronecker delta ${\delta }^{\mu }{}_{\nu }$ in every coordinate system.

The impulse response records the response to a unit impulse, with specified initial or boundary conditions. In continuous time the impulse is a Dirac delta function; in discrete time it is the Kronecker delta $\delta [n]$. For a linear time-invariant system with zero initial state, the response to a general input is its convolution with the impulse response. The impulse response of a causal filter is zero before the impulse.

For the five-sample moving average, a discrete impulse produces $h[n]=1/5$ at $n=0,1,2,3,4$ and zero otherwise. The response lasts five samples, so this is a finite impulse response system. Its frequency response is the finite sum of the five delayed phase contributions: one short input experiment determines how every sustained sinusoid is filtered.

The inertia tensor packages the moments of inertia about every axis through a specified point $O$ into one symmetric matrix. An axis still needs a direction: if $\mathbf{n}$ is a unit vector, its scalar moment of inertia is $I_{O,\mathbf{n}}={\mathbf{n}}^T{\mathbf{I}}_O\mathbf{n}$. The same mass layout can therefore be easy to spin about one direction and harder about another. The inertia tensor also determines rotational kinetic energy and relates angular velocity to angular momentum in the appropriate rigid body model.

Take position vectors $\mathbf{r}=(x,y,z)$ relative to $O$, expressed in an orthonormal basis. For point masses $m_a$, or a continuous distribution with mass density $\rho$ and $dm=\rho dV$,Here ${\delta }_{ij}$ is the Kronecker delta. Equivalently, using the identity matrix $1$ and an outer product,Every entry has SI units $\mathrm{kg}{\mathrm{m}}^2$. The diagonal entries are axial moments of inertia; the off-diagonal entries carry a minus sign in this convention. Some texts instead define the integrals without the prefixed minus sign $\int xydm$, $\int xzdm$, $\int yzdm$ as their named off-diagonal quantities and put the minus signs into the displayed matrix. Check that convention before combining formulas. ^{[ref] [ref]}

The symmetric matrix is positive semidefinite, because its quadratic form isThe dot product and cross product identify this as a sum of squared distances when $\mathbf{u}$ is a unit vector. A zero moment of inertia occurs precisely when all the mass lies on that axis, up to sets of zero mass. A genuine three-dimensional body with positive mass density in a volume has strictly positive axial moments of inertia. Do not confuse this inertia tensor with the mass second-moment tensor $\mathbf{S}=\int \mathbf{r}{\mathbf{r}}^{T}dm$: they obey ${\mathbf{I}}_O=\mathrm{tr}(\mathbf{S})1-\mathbf{S}$, where $\mathrm{tr}$ denotes the matrix trace.

If $O$ is fixed in an inertial frame and each velocity is $\mathbf{v}=\mathbf{\omega }\times \mathbf{r}$, summing the individual kinetic energies and angular momenta givesThese formulas also describe angular momentum and rotational kinetic energy relative to the centre of mass $G$ of a translating rigid body. In that case the total kinetic energy in the inertial frame is $K_{\mathrm{total}}=\frac{1}{2}M|{\mathbf{V}}_G{|}^2+\frac{1}{2}{\mathbf{\omega }}^T{\mathbf{I}}_G\mathbf{\omega }$. For a general moving point $O$, the body's actual velocities include ${\mathbf{V}}_O$, so ${\mathbf{I}}_O\mathbf{\omega }$ alone need not be its angular momentum about $O$. ^{[ref] [ref]}

Changing the orthonormal basis changes components, rather than the physical distribution: if the columns of an orthogonal matrix $Q$ are the new basis vectors expressed in the old basis, then ${\mathbf{I}}_O^{\prime }=Q^T{\mathbf{I}}_{O}Q$. Changing the reference point is a different operation, governed by the tensor parallel-axis theorem. Both principal axes and principal moments of inertia must therefore be quoted with their reference point.

Contract an isotropic tensor with the Kronecker delta or the Levi-Civita symbol, using invariance under proper rotations. The three displayed contractions have ranks one, two and three. The only isotropic vector is zero; an isotropic second-rank tensor is a scalar delta, and an isotropic third-rank tensor is a scalar epsilon. For $T_{ijkl}=\alpha {\delta }_{ij}{\delta }_{kl}+\beta {\delta }_{ik}{\delta }_{jl}+\gamma {\delta }_{il}{\delta }_{jk}$, direct contraction gives $\mu =3\alpha +\beta +\gamma$ and $\nu =\beta -\gamma$.

Every three-dimensional fourth-rank Cartesian tensor invariant under all proper rotations has the displayed form, without assuming index-interchange symmetry. Half-turns about the coordinate axes force every nonzero component to contain each coordinate label an even number of times. The surviving patterns have four equal labels or two equal pairs. Axis permutations identify the latter with $c_{1122}=\alpha$, $c_{1212}=\beta$ and $c_{1221}=\gamma$, while all four-equal components share a value $A$.

Evaluate the component with four copies of $(e_1+e_2)/\sqrt{2}$. Multilinearity gives $(A+\alpha +\beta +\gamma )/2$, which must equal $A$ by rotational invariance. Hence $A=\alpha +\beta +\gamma$, proving the formula. Conversely each pairing of Kronecker deltas is invariant under any orthogonal transformation. For symmetric strain only $\alpha$ and $\beta +\gamma$ matter, giving the two Lamé parameters of isotropic elasticity on tensor components.

A second-rank Cartesian tensor invariant under every proper rotation matrix is a scalar multiple of the Kronecker delta. Half-turns about the coordinate axes kill its off-diagonal entries, and quarter-turns make its diagonal entries equal. A tensor of the form $v_i{w}_j$ is an outer product with matrix rank at most one, so it is isotropic in three dimensions only when it vanishes, equivalently when $v=0$ or $w=0$.

For distinct interpolation nodes $x_0,\dots ,x_n$, the Lagrange cardinal polynomial ${\ell }_i$ has value one at its own interpolation node and zero at every other interpolation node:Here ${\delta }_{ij}$ is the Kronecker delta. These Lagrange cardinal polynomials form a basis of the vector space of polynomials of polynomial degree at most $n$: every such polynomial has the unique expansion $p(x)={\sum }_{i}p(x_i){\ell }_i(x)$. The nodal evaluation linear functionals $E_j(p)=p(x_j)$ form the dual basis, since $E_j({\ell }_i)={\delta }_{ij}$.

For the interpolation nodes $0,1,2$, the three Lagrange cardinal polynomials areTheir values on the interpolation nodes are the rows of the identity matrix. In the figure, each colored curve passes through its own unit-height marker and through zero at the other two interpolation nodes. Multiplying the curves by the data $1,2,5$ and adding gives the Lagrange polynomial $x^2+1$.

Since the constant polynomial one interpolates data values all equal to one, uniqueness gives ${\sum }_i{\ell }_i(x)=1$. Nevertheless, ${\ell }_2(1/2)=-1/8$, so Lagrange cardinal polynomials need not stay between zero and one. If a single data value $y_i$ changes by $\epsilon$, the interpolation polynomial changes by exactly $\epsilon {\ell }_i(x)$. Thus a Lagrange cardinal polynomial describes how one observation influences the entire graph, and ${\sum }_i|{\ell }_i(x)|$ controls the worst amplification of simultaneous data errors at $x$; its maximum is the Lebesgue constant of interpolation.

A linear time-invariant system combines two properties. Linearity means that the output from a linear combination of inputs is the same linear combination of their outputs. Time invariance means that shifting an input by an integer delay shifts its output by exactly that delay. The operation has no preference for a particular time origin.

For discrete time, write the unit impulse as $\delta [n]=1$ at $n=0$ and zero otherwise, a Kronecker delta. Its response $h[n]=T[\delta ][n]$ is the impulse response. A finite input can be decomposed into shifted impulses, givingThe superposition principle supplies the sum and time invariance supplies the shifted copies of $h$. Infinite sums need convergence assumptions; for a bounded input and an impulse response $h$ with absolute summability, the discrete convolution converges absolutely. For a physical system with stored initial energy, this construction describes the response with zero initial state; a separate free response accounts for nonzero initial conditions.

The gain is reuse. Measure or calculate one impulse response, and the convolution predicts the response to every allowed signal without solving a new dynamical problem each time. Alternatively, the frequency response predicts how each oscillatory component changes. ^[ref].

A mixed tensor has $k$ defining indices and $l$ dual indices. The SU(3) action applies $U$ to every upper index and $U^{-1}$ to every lower index. Young symmetrizers select permutation types, while invariant Kronecker deltas take traces and invariant alternating tensors convert an antisymmetric pair of defining indices into a dual index, or conversely. Repeated fundamental and dual tensor rules decompose it with multiplicities: $(p,q)\otimes (1,0)=(p+1,q)\oplus (p-1,q+1)\oplus (p,q-1)$ and the conjugate rule, omitting negative labels.

An orthogonal polynomial family chooses polynomials that do not overlap under a specified inner product. For real polynomials on an interval $(a,b)$, use the weighted inner productAssume $a<b$, the weight $w$ is measurable and positive almost everywhere, and its absolute moments ${\int }_a^b|x{|}^{k}w(x)dx$ are finite for every integer $k\ge 0$. These conditions make the integrals finite and give every nonzero polynomial positive squared norm. A sequence $p_0,p_1,\dots$ is an orthogonal polynomial family when $p_n$ has real coefficients and polynomial degree exactly $n$, and $⟨p_n,p_k{⟩}_w=0$ for $n\ne k$.^[ref]

Here $h_n=\Vert p_n{\Vert }_w^2>0$ and ${\delta }_{nk}$ is the Kronecker delta. Orthogonality does not require $h_n=1$: an orthonormal set is obtained only after replacing $p_n$ by $p_n/\sqrt{h_n}$. Multiplying each member by any nonzero real constant preserves orthogonality, but changes its norm and the corresponding expansion coefficient. The monic orthogonal polynomial convention instead fixes each leading coefficient to one.

The first four Legendre polynomials, named for Adrien-Marie Legendre, use $w=1$ on $[-1,1]$ and the normalization $P_n(1)=1$:Their orthogonality and squared norms areThus their Gram matrix has a positive diagonal and zero entries elsewhere, although their function graphs cross and overlap on the page. Orthogonality measures an integral of products, rather than separation of function graphs.^[ref]

For these four polynomials, a short verification expands each product and uses ${\int }_{-1}^1{x}^{2j+1}dx=0$ and ${\int }_{-1}^1{x}^{2j}dx=2/(2j+1)$. For example, $⟨P_0,P_2⟩=(3(2/3)-2)/2=0$ and $⟨P_1,P_3⟩=(5(2/5)-3(2/3))/2=0$. The same calculation gives the diagonal entries $2,2/3,2/5,2/7$. More generally, apply the Gram-Schmidt process to the monomial basis $1,x,x^2,\dots$: at step $n$, subtract projections onto the earlier polynomials. The result retains its leading term, has degree $n$, and has positive norm, proving existence without a differential equation. Different degrees also give linear independence, so $p_0,\dots ,p_m$ form a basis of the polynomials of degree at most $m$.

This basis turns least-squares solutions for functions into independent scalar calculations, as the decoupled coefficients of orthogonal polynomial projections show. In a spectral method, increasing the degree enlarges the approximation space while keeping earlier projection coefficients fixed. Weighted integrals also connect the same orthogonal polynomials to Gaussian quadrature: its nodes are polynomial roots of a member of the family. The projection problem minimizes an integrated squared error; Gaussian quadrature uses the family to choose evaluation points for an integral.

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 $x_i{x}_j$.

Using Einstein summation convention and the contraction of two Levi-Civita symbols,This proves the vector triple product identity directly from the Levi-Civita symbol and Kronecker delta.

Define the scalar triple product by $[a,b,c]=a\cdot (b\times c)$. Its expression ${ϵ}_{ijk}{a}_i{b}_j{c}_k$ is unchanged by cyclically relabeling the three indices: a three-cycle has positive sign. ThusIn particular, let $\Delta =[e_1,e_2,e_3]$. Since the three vectors form a basis, their determinant $\Delta$ is nonzero, so their reciprocal basis vectors are well defined. Cyclic invariance and vanishing of a scalar triple product with repeated vectors giveTo calculate the new scalar triple product explicitly, apply the already proved vector triple product identity with $A=e_3\times e_1$:Consequentlyso the reciprocal vectors also form a basis. Taking the reciprocal basis of that basis now givesThe same cyclic calculation gives ${\hat{\hat{e}}}_2=e_2$ and ${\hat{\hat{e}}}_3=e_3$, proving reciprocal basis involution.

Finally, write $V=V_1{e}_1+V_2{e}_2+V_3{e}_3$. Taking a dot product with ${\hat{e}}_j$ isolates $V_j$. Applying the same argument to the reciprocal basis, whose reciprocal is the original basis, gives the two sets of coordinates:No orthonormality assumption is needed; the reciprocal basis is what makes these coordinate-extraction formulas work.

The Kronecker delta is the identity map on the defining representation. Its transformation is $A^{\alpha }{}_{\gamma }{\delta }^{\gamma }{}_{\delta }(A^{-1}{)}^{\delta }{}_{\beta }={\delta }^{\alpha }{}_{\beta }$. For the alternating Levi-Civita symbol, the determinant identity givesThe lower-index alternating tensor transforms with three inverse matrices and acquires $\det (A^{-1})=1$ instead. Thus ${\delta }^{\alpha }{}_{\beta }$, ${ϵ}^{\alpha \beta \gamma }$ and ${ϵ}_{\alpha \beta \gamma }$ are invariant tensors of SU(3). These identities use unimodularity; unitarity identifies the dual representation with the complex-conjugate defining representation.

For a mixed SU(3) tensor representation $T^a{}_b$, the matrix trace is invariant. HencegivesThe trace is the singlet. The traceless matrix transforms by conjugation, so it is the adjoint representation of SU(3). It is irreducible: an invariant complex subspace is invariant under commutators with the complexified algebra ${\mathfrak{sl}}_3$, and would be an ideal in that simple Lie algebra. Only the zero and full traceless-matrix spaces are possible.

For the cubic tensor, define its symmetric and antisymmetric first-pair parts byand let $S^{abc}=T^{(abc)}$ be its completely symmetric part. This has $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$ independent components and transforms as the symmetric cubic representation of SU(3). Its irreducibility can be seen in the homogeneous cubic-polynomial realization: the simultaneous highest-weight conditions $z_1{\partial }_{z_2}P=0$ and $z_2{\partial }_{z_3}P=0$ leave only $P\propto z_1^3$. A finite-dimensional unitary representation is completely reducible, and each irreducible summand has a highest-weight vector, so this single highest-weight line implies a single irreducible summand. Lowering generates all ten monomials.

Remove this fully symmetric part from $T_{+}$, putting $M^{abc}=T_{+}^{abc}-S^{abc}$. DefineIts trace vanishes because $M$ is symmetric in its first two indices. Invariance of $ϵ$ makes $O$ transform as one upper and one lower index. The inverse map isIndeed contraction with ${ϵ}_{ebc}$ gives $(2O^a{}_e+O^a{}_e-{\delta }^a{}_e\mathrm{tr}O)/3=O^a{}_e$. The inverse has the required first-pair symmetry and zero completely symmetric part. Thus this eight-dimensional remainder is another adjoint representation of SU(3).

For the antisymmetric part, defineNow decompose $N^c{}_d=U^c{}_d+{\delta }^c{}_d\mathrm{tr}N/3$, with $U$ traceless. The first term supplies a second independent octet; the trace supplies a singlet proportional to ${ϵ}^{abc}$. With $q={ϵ}_{abc}{T}^{abc}/6=\mathrm{tr}N/3$, the explicit cubic SU3 tensor projections reconstruct the entire tensor:All projections commute with the SU(3) action because they use permutations and invariant tensors. Their inverse maps and the dimension count $10+8+8+1=27$ establishThe two octets are equivalent as group representations but distinct subspaces in this decomposition; one comes from the first-pair symmetric sector and the other from the first-pair antisymmetric sector. General permutations can mix these equivalent copies.

In the quark model, the up quark, down quark and strange quark form the defining triplet of approximate flavor symmetry. Three-quark baryons therefore have precisely these possible flavor representations. The fully symmetric decuplet corresponds to the observed spin-$3/2$ baryon decuplet: the four Delta baryons, three ${\Sigma }^{\ast }$ states, two ${\Xi }^{\ast }$ states and the Omega baryon. Their strangeness rows have flavor hypercharge and isospin $(Y,I)=(1,3/2),(0,1),(-1,1/2),(-2,0)$; $uuu$ and $sss$ occupy the extreme ${\Delta }^{++}$ and ${\Omega }^{-}$ weights.

The observed spin-$1/2$ baryon octet contains the nucleons, three Sigma baryons, the Lambda baryon and two Xi baryons. In its $Y=0$ row, ${\Sigma }^0$ and $\Lambda$ share the same third-component/hypercharge weight but have different total isospin. Approximate flavor symmetry explains multiplet organization, not exact equality of all their masses: the strange-quark mass and electromagnetic interactions break it.

The two octet copies in the flavor tensor do not require two identical ground-state octets. A three-quark colour singlet is antisymmetric in color; for a symmetric ground-state orbital wavefunction, the Pauli exclusion principle requires a symmetric combined spin-flavor state. The mixed flavor octet tensors combine with the mixed spin-$1/2$ tensors to supply that symmetry. The symmetric flavor decuplet combines with symmetric spin $3/2$. Together they form the symmetric spin-flavour SU6 representation, with $10\times 4+8\times 2=56$ spin-flavor states. A completely antisymmetric flavor singlet cannot appear in this spatial ground state because three spin-$1/2$ indices have no fully antisymmetric spin state: ${\bigwedge }^3{\mathbb{C}}^2=0$. Three-quark flavour singlet configurations instead require a suitable excited spatial/spin symmetry. Flavor SU(3) here is distinct from the exact color gauge symmetry; treating them as the same group action would misidentify the physical multiplets.

For an orthogonal change of Cartesian coordinates $x_i^{\prime }=Q_{ij}{x}_j$, an ordinary rank-$n$ tensor transforms byAn isotropic tensor is invariant under such rotations. Orthogonality gives $Q_{ip}{Q}_{jq}{\delta }_{pq}={\delta }_{ij}$, so transforming any of the three products of two Kronecker deltas leaves it unchanged. Their linear combination with arbitrary scalar coefficients is therefore isotropic.

For completeness, these three products exhaust the rank-four isotropic tensors in three dimensions. Half-turns about coordinate axes force a nonzero component to contain each coordinate label an even number of times. Axis interchanges then leave just the values $X=c_{1111}$, $\alpha =c_{1122}$, $\beta =c_{1212}$ and $\gamma =c_{1221}$. Rotating the first two axes by $\pi /4$ gives $X=\frac{1}{2}X+\frac{1}{2}(\alpha +\beta +\gamma )$, so $X=\alpha +\beta +\gamma$. These component values are exactly those of the three-delta-product expression. Thus isotropy of the elastic medium gives this form for its elasticity tensor.

Contracting with the symmetric strain givesThese are the Lamé parameters. Taking the trace determines the scalar part of the strain:Since ${\sum }_{ij}{e}_{ij}^2=3p^2+{\sum }_{ij}{d}_{ij}^2$, the stored strain energy density isA nonzero traceless test strain proves necessity of $\mu \ge 0$, and a pure scalar strain proves necessity of $3\lambda +2\mu \ge 0$. Conversely these two conditions make both terms nonnegative for every strain. Thus nonnegative isotropic elastic strain energy is equivalent to

The event rate counts unordered pairs of reacting nuclei. For unlike species there are $n_i{n}_j$ pairs per unit-volume normalization; for identical species there are approximately $n_i^2/2$ pairs, since counting ordered pairs counts each collision twice. The Kronecker delta therefore supplies exactly the required divisor $1+{\delta }_{ij}$. It is the event rate, not the destruction rate of identical nuclei: one identical-pair reaction destroys two nuclei.

The temperature dependence combines the high-energy tail of the Maxwell-Boltzmann distribution with quantum tunnelling through the Coulomb barrier. The principal exponential factor in the energy integral is $\exp [-E/(k_{B}T)-b/\sqrt{E}]$. Its stationary point, the Gamow peak, has $E_0\propto T^{2/3}$ and exponent $\eta =3E_0/(k_{B}T)\propto T^{-1/3}$. Integrating around this peak gives a prefactor proportional to $T^{-2/3}$, or ${\eta }^2$, when the nuclear factor varies slowly. Thus the supplied thermonuclear reaction rate is locally proportional to ${\eta }^2{e}^{-\eta }$. The slow weak interaction in the proton-proton reaction also makes its overall coefficient very small; its local thermal slope follows the supplied formula.

At fixed number densities, define the local temperature exponent of a thermonuclear reaction by a logarithmic derivative. Since $d\eta /d\log T=-\eta /3$,For proton-proton reactions, $A=1/2$ and $Z_1=1$. At $T_6=15$, ${\eta }_{11}=13.67$ and $\alpha =3.89$. For two helium-3 nuclei, $A=3/2$ and $Z_3^2{Z}_3^2=16$, giving ${\eta }_{33}=49.68$ and $\beta =15.89$. For helium-3 with helium-4, $A=12/7$ and the same charge product gives ${\eta }_{34}=51.95$ and $\gamma =16.65$. HenceThese are local powers near the specified temperature, not constant exponents over an arbitrarily large range. The greater reduced nuclear mass in the $34$ reaction increases its tunnelling exponent and its thermal sensitivity.

For the effective proton-proton reaction network, denote the event rates byThe short-lived beryllium intermediate can be eliminated, so production through $R_{34}$ feeds $n_7$, representing predominantly lithium-7. This rapid conversion is electron capture. Count the nuclei consumed and produced in each reaction: $11$ consumes two protons and makes one deuterium; $21$ consumes a proton and a deuterium and makes one helium-3; $33$ consumes two helium-3 and makes one helium-4 and two protons; $34$ consumes one of each helium isotope; $17$ consumes a proton and lithium-7 and makes two helium-4. It follows thatFor closure, ${\dot{n}}_7={\lambda }_{34}{n}_3{n}_4-{\lambda }_{17}{n}_1{n}_7$. These nuclear terms conserve $n_1+2n_2+3n_3+4n_4+7n_7$, providing an independent stoichiometric check. They describe a fixed local volume; compression or mixing would add transport terms.

The roughly one-second destruction time of deuterium is negligible compared with the proton-burning and helium-3 evolution times. Its abundance rapidly adjusts to the slowly evolving production rate. Set ${\dot{n}}_2\simeq 0$, giving the deuterium quasi-equilibrium in proton-proton burningSubstitution, without setting ${\dot{n}}_1=0$, givesThe helium-3 relaxation time of about $6\times 10^5$ years is likewise short compared with the central proton-burning time and the age of the Sun. On this relaxation time the proton and helium-4 abundances, and the background temperature, can be treated as fixed. Put $A={\lambda }_{33}$, $B={\lambda }_{34}{n}_4$, $C={\lambda }_{11}{n}_1^2/2$. Then ${\dot{n}}_3=C-A{n}_3^2-B{n}_3$, and its nonnegative equilibrium isThe other root is negative and unphysical. Write $n_3=n_{3e}+x$ and subtract the equilibrium equation. Exactly, for a fixed background,Dropping the quadratic displacement givesThus small displacements decay exponentially. A slowly moving equilibrium introduces an additional $-{\dot{n}}_{3e}$ term, negligible to leading order when $\tau$ is short compared with background evolution.

To estimate the temperature for helium-3 freeze-out, use the cooler pp-I-dominated region and neglect $B$ compared with $2A{n}_{3e}$. Then $n_{3e}\simeq n_1\sqrt{{\lambda }_{11}/(2{\lambda }_{33})}$ and ${\tau }^{-1}\simeq n_1\sqrt{2{\lambda }_{11}{\lambda }_{33}}$. With the local powers just obtained and fixed proton number density, $n_{3e}\propto T^{-6}$ and $\tau \propto T^{-10}$. Adopting $t_{\odot }\simeq 4.6\times 10^9\mathrm{yr}$ givesThis is an order-of-magnitude extrapolation of local exponents. Keeping the supplied exponential factors gives the helium-3 freeze-out estimate with Gamow factorswhich yields about $6.8\times 10^6\mathrm{K}$ with the same normalization and fixed-density approximation. Thus the robust estimate is several million kelvin. The supplied data do not fix the local density, composition or pp-II fraction away from the centre, so they cannot determine an exact transition temperature or radius. In particular, using $\tau \propto T^{-16}$ alone would omit the temperature dependence of the equilibrium helium-3 abundance.