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Quark mixing arises because the up-type and down-type quark Yukawa matrices need not be diagonal in the same weak interaction basis. In the Standard Model the three left-handed quark doublets $Q_L^{\prime }$ transform as $(3,2,1/6)$ under $SU(3{)}_c\times SU(2{)}_L\times U(1{)}_Y$, whereas $u_R^{\prime }$ and $d_R^{\prime }$ transform as $(3,1,2/3)$ and $(3,1,-1/3)$. With a Higgs doublet $H$ and its conjugate $\tilde{H}=i{\sigma }^2{H}^{\ast }$, the gauge-invariant Yukawa interaction isThe generation indices are matrix-multiplied. Electroweak symmetry breaking, $⟨H⟩=(0,v_H/\sqrt{2}{)}^T$, produces $M_u=v_H{Y}_u/\sqrt{2}$ and $M_d=v_H{Y}_d/\sqrt{2}$.

Each complex fermion mass matrix admits a biunitary diagonalization,where the diagonal entries are real nonnegative quark masses. Define $u_L^{\prime }=U_{uL}{u}_L$ and $d_L^{\prime }=U_{dL}{d}_L$, and similarly for right-handed fields. The kinetic terms retain their canonical form because the transformations are unitary matrices. However, the weak charged current becomesThis unitary matrix is the Cabibbo-Kobayashi-Maskawa matrix. Its entries multiply transitions between up-type and down-type mass eigenstates. The minimal Standard Model has no analogous right-handed charged weak current.

Quark mixing does not introduce tree-level flavor-changing neutral currents. Photon, gluon and weak neutral current couplings are proportional to the identity within each fixed-charge quark sector, and $U^{†}IU=I$. The single Higgs doublet also has diagonal neutral scalar field couplings after the same transformations that diagonalize the masses. At higher loop orders, flavor changes are possible. For example, an $s\to d$ amplitude contains ${\sum }_i{V}_{is}^{\ast }{V}_{id}f(m_i^2)$. Unitarity of the CKM matrix sets the sum of coefficients to zero, cancelling any mass-independent part. This is the GIM mechanism; mass splittings leave a suppressed, generally nonzero amplitude.

For $N$ generations a unitary matrix has $N^2$ real parameters: $N(N-1)/2$ mixing angles and $N(N+1)/2$ phases. Rephasing the $N$ up-type and $N$ down-type quark fields removes $2N-1$ phases; a common baryon number phase leaves every matrix entry unchanged. Hence the physical numbers areTwo generations therefore give a real Cabibbo rotation,with strangeness-changing charged-current transitions suppressed by $\sin {\theta }_C$. Three generations have three mixing angles and one irreducible weak CP phase. This counting assumes nondegenerate masses; degeneracies allow additional basis rotations.

Writing $s_{ij}=\sin {\theta }_{ij}$ and $c_{ij}=\cos {\theta }_{ij}$, a standard exact parametrization isComplex entries alone do not establish CP violation, because quark quantum field rephasings can change their phases. For nondegenerate quark masses the weak mixing sector conserves CP precisely when a basis makes $V$ real. A useful invariant test is the Jarlskog invariant,A nonzero $J$ requires all three mixing angles and a nontrivial phase. If two same-charge quark masses coincide, the additional basis freedom removes the corresponding physical CP effect. The possible QCD theta angle is a separate CP parameter, not part of this weak-mixing count.

The Wolfenstein parametrization displays the hierarchy of quark mixing economically. Put ${\lambda }_W=s_{12}$, $s_{23}=A{\lambda }_W^2$, and $s_{13}{e}^{i\delta }=A{\lambda }_W^3(\rho +i\eta )$. Expanding the exact matrix givesThe observed pattern is predominantly diagonal, with $1$-$2$ mixing largest and $2$-$3$, $1$-$3$ mixing successively suppressed. The expansion implies $J\simeq A^2{\lambda }_W^6\eta$. The truncated matrix must not be treated as an exactly unitary matrix.

Column orthogonality givesThe three complex numbers form a CKM unitarity triangle, with area $|J|/2$. Dividing by $-V_{cd}{V}_{cb}^{\ast }$ puts its vertices at $0$, $1$ and $\bar{\rho }+i\bar{\eta }$, where $\bar{\rho }+i\bar{\eta }=-V_{ud}{V}_{ub}^{\ast }/(V_{cd}{V}_{cb}^{\ast })$. The barred parameters include higher-order corrections to the Wolfenstein parameters. Comparing independent measurements of its sides and angles tests both quark mixing and the absence of additional flavor dynamics.

Semileptonic decay rates determine magnitudes of CKM entries through factors $|V_{ij}{|}^2$ once hadronic matrix elements are controlled. Neutral kaon and neutral B-meson mixing constrain products of entries through weak box scattering amplitudes. CP asymmetries probe their invariant relative phases, with kaon mixing providing indirect CP violation and decay amplitudes providing direct CP violation. These processes complement tests of the unitarity constraints on rows and columns. The minimal three-generation Standard Model describes weak quark mixing by a unitary matrix with three angles and one physical CP phase; it does not predict the Yukawa couplings or their hierarchy.

Use radiative homology with proton-proton burning and inverse-fourth-power opacity, retaining the composition factors throughout. To distinguish the stellar radius from the stellar gas constant, write them as $R$ and $\mathcal{R}$. With gas pressure alone, introduce the scale factorsThe ideal gas equation gives $\rho ={\rho }_{\ast }b$, $P=P_{\ast }p$ and $T=T_{\ast }p/b$. Substitution into hydrostatic equilibrium and mass conservation givesso $p^{\prime }=-bq/x^2$ and $q^{\prime }=x^{2}b$.

Set $\zeta =Z(1+X)$. The radiative diffusion in a star equation, with the stipulated opacity, becomesHere ${\rho }^2{T}^{-7}={\rho }_{\ast }^2{T}_{\ast }^{-7}{b}^9{p}^{-7}$. ThusThe luminosity equation follows from the specific heating rate of the Proton–proton chain:ConsequentlyThese give all four dimensionless stellar structure equations with their numerical coefficients.

The stellar homology assumption means the dimensionless profiles are shared across the models, not merely that an individual model can be written in dimensionless coordinates. The displayed equations then require $D$ and $E$ to be fixed. Holding the universal constants and ${\kappa }_0,{ϵ}_0$ fixed givesEquating the two cancels $M^6$ and yields $R^6\propto X^2\zeta {\mu }^{-4}$. Therefore the radius and mass-luminosity relation areThe mass-independent radius is a consequence of the particular heating and opacity exponents in this model; it is not a general radius law for main-sequence stars.

Initially $X=X_0$ is the same for all stars. The adopted mean molecular weight $\mu =4/(5X+3)$ also has the same initial value, neglecting its small direct dependence on $Z$. Thus $R\propto Z^{1/6}$. At fixed luminosity, the Stefan–Boltzmann law $L=4\pi R^2\sigma T_e^4$ givesThe stars compared here can have different masses; the luminosity law adjusts the mass to keep $L$ fixed.

For the evolution, take the stellar mass as constant and maintain homogeneous composition. The available energy in the remaining hydrogen is $M{E}_{H}X$, so conservation of energy in the nuclear-powered model givesThis includes the entire mixed hydrogen reservoir, not just the mass in a central burning region.

To determine the turnoff mass of a homogeneously mixed proton-proton-burning population, write the luminosity law as $L=C{M}^6{Z}^{-7/6}H(X)$, whereSeparating variables, the time to evolve from $X_0$ to $f{X}_0$ isFor a common $X_0$ and $0\le f<1$, this integral is a positive finite constant independent of $M,Z,t$. It follows that the turning-off stars satisfyThe integration retains the evolving luminosity and mean molecular weight; replacing $L$ by its initial value is unnecessary.

For the Schwarzschild criterion, displace a small gas parcel upwards. Assume its composition is unchanged, it remains in pressure balance with its surroundings, and it moves quickly enough that heat exchange is negligible. The parcel obeys the adiabatic equation of state $P{\rho }^{-\gamma }=\mathrm{constant}$. Combining this with the ideal gas equation gives $T\propto P^{(\gamma -1)/\gamma }$, so its adiabatic temperature gradient is ${\nabla }_{\mathrm{ad}}=(\gamma -1)/\gamma =2/5$.

Let the surrounding temperature gradient be $\nabla =d\log T/d\log P$. For an upward displacement $\delta \log P<0$, the parcel's temperature relative to its new surroundings isIf $\nabla <{\nabla }_{\mathrm{ad}}$, this is negative: the parcel is cooler and, at equal pressure and composition, denser than its surroundings. It therefore sinks back; a downward displacement likewise gives a restoring buoyancy force. If the inequality is reversed, the displaced parcel continues to rise. The stability condition is thusEquality is the neutral boundary. There is no composition-gradient term because the ideal-gas parcel and the background have the same mean molecular weight.

In the thin upper stellar atmosphere, take a plane-parallel approximation with constant $g$, optical depth increasing inward, and zero overlying pressure $P(0)=0$. Hydrostatic equilibrium in optical depth gives $dP/d\tau =g/\kappa$. The ideal gas equation converts the stipulated opacity intoDefine $u=1+3\tau /2$, so the given grey atmosphere relation is $T^4=T_e^{4}u/2$. Integrating from the surface givesThis proves the required pressure law. In particular $P^2\propto g{T}_e^{-4\beta }$ times a function only of $u$ and $\beta$.

For the grey-atmosphere convection onset with density-linear opacity, differentiate the two profiles to obtain the radiative temperature gradient:It increases monotonically from zero, since $\beta >1$, and first reaches the adiabatic temperature gradient at $u_b^{\beta -1}=(4\beta +1)/5$. ThereforeBelow this point the radiative stratification is unstable and convection must transport some of the outward energy flux.

At the outer convection boundary, $u_b$ depends only on $\beta$, so $T_b=T_e(u_b/2{)}^{1/4}\propto T_e$ and $P_b\propto g^{1/2}{T}_e^{-2\beta }$ for fixed $\mu ,{\kappa }_0,\beta$. The matching constant in $P=K{T}^{5/2}$ is $K=P_b/T_b^{5/2}$. ConsequentlyThis temperature form of the polytropic relation follows from an ideal monatomic adiabat; its $K$ is not the density-polytrope coefficient in Question 3.

For the deep radiative boundary beneath a power-law-opacity convective envelope, negligible envelope mass and absent local energy generation give $m\simeq M$ and $L_r\simeq L$. Divide the radiative temperature equation by the pressure equation to get the stellar radiative temperature gradientThis is the gradient that would be needed if radiation alone carried $L$, and it is the quantity to compare with $2/5$ even while the actual region is convective. The deeper Kramers' opacity law gives $\kappa ={\kappa }_1\mu P{T}^{-9/2}/\mathcal{R}$. Substituting $P=K{T}^{5/2}$ yieldsAs $T$ rises inward this decreases, so the transition back to a stable radiative interior is atThus $T_{\mathrm{in}}^{7/2}\propto K^{2}L/M$. Using the outer matching and the Stefan–Boltzmann law,The mass and radius cancel. Hence, at fixed opacity coefficients and composition,This uses the same adiabat through the convection zone and distinguishes its inner transition temperature from its outer onset temperature.

For a regular, spherical star, hydrostatic equilibrium gives $dP/dr=-Gm(r)\rho (r)/r^2$. Multiply by $4\pi r^3$ and integrate to the surface. Integration by parts givesThe central boundary term vanishes for finite central pressure. Since $dm=4\pi r^2\rho dr$, the right-hand side of the integrated hydrostatic equation isThe last expression is the gravitational energy: adding a thin mass shell $dm$ to the interior mass $m$ contributes $-Gmdm/r$, counting each interacting pair once. HenceWith the specified zero surface pressure this proves the stellar virial theorem in the required form,

To derive the nonrelativistic electron-degeneracy pressure in a hydrogen-helium mixture, take fully ionized hydrogen and helium with hydrogen mass fraction $X$, the electron number density isHere helium nuclei have mass $4m_p$ in the approximation used in the question. Filling the two spin states of each momentum mode up to the Fermi momentum $p_0$ givesThe isotropic momentum flux gives pressure $\frac{1}{3}\int pvdn$. For nonrelativistic electrons $v=p/m_e$, so the electron degeneracy pressure isSubstitute the composition-dependent $n_e$ and simplify powers of two:Comparing $5/3=1+1/n$ gives the polytropic index $n=3/2$. The result assumes complete electron degeneracy, $p_0\ll m_{e}c$, and negligible nuclear mass corrections.

Now adopt the stated approximate partially degenerate equation of state, keeping composition and hence $K,\mu$ fixed during contraction. Set $q=m/M$ and define the mass-averaged temperature ${\overline{T}}_M={\int }_0^{1}Tdq$. The stellar virial theorem from part (a) givesFor an index-$3/2$ stellar polytrope, the permitted gravitational-energy formula is $\Omega =-6G{M}^2/(7R)$. Also, using $\bar{\rho }=3M/(4\pi R^3)$,Thus, with $A=3K{\int }_0^1[3\rho /(4\pi \bar{\rho }){]}^{2/3}dq$,The normalized density $\rho /\bar{\rho }$ is a fixed function of $q$ for this homologous polytropic structure. Therefore its integral and $K$ fix $A$, which is independent of $M$ and $R$ for a common composition. The mean here is mass-weighted, rather than a volume average.

For the mass-averaged temperature maximum in a partially degenerate polytrope, put $\alpha =2\mu GM/(7\mathcal{R})$ and $\delta =\mu A{M}^{2/3}/(3\mathcal{R})$, both positive. As a function of $s=1/R$ the temperature is the concave quadraticIt reaches its maximum when $R=2\delta /\alpha =7A{M}^{-1/3}/(3G)$, withAt large radius, contraction increases the temperature as in a thermally supported star. Once electron degeneracy provides a sufficiently large fraction of the support, further contraction can lower the thermal temperature. The formal zero-temperature radius is $R_{\mathrm{deg}}=\delta /\alpha =R_{\max }/2$, with $R_{\mathrm{deg}}\propto M^{-1/3}$. The continuation to negative temperatures would be unphysical; the adopted sequence only describes its positive-temperature part.

The contraction-limited central temperature has the same mass scaling. Indeed, if the total polytropic pressure is $P=K_{\mathrm{tot}}{\rho }^{5/3}$, the approximate equation of state gives $T=(\mu /\mathcal{R})(K_{\mathrm{tot}}-K){\rho }^{2/3}$. The fixed density profile therefore fixes $T_c/{\overline{T}}_M$. A sufficiently small mass never reaches the central temperature required for sustained hydrogen burning, producing a hydrogen-burning minimum mass. Such an object becomes a cooling, degeneracy-supported brown dwarf instead of settling onto the main sequence. More massive contracting objects can ignite hydrogen before reaching this degeneracy-limited maximum. This conclusion uses the nonrelativistic, fixed-composition model; it does not require extrapolation into the relativistic regime.

Treat the stars as point masses for the orbital motion, neglecting their spin angular momenta. Let $M=M_1+M_2$. Their distances from the centre of mass are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. The sum of their orbital angular momenta isThus the circular-binary orbital angular momentum is

For conservative binary mass transfer, both $M$ and $J$ are fixed. In terms of the binary mass ratio $q=M_1/M_2$, $M_1=Mq/(1+q)$ and $M_2=M/(1+q)$, henceMass conservation also gives $d{M}_2=-d{M}_1$, soDifferentiating $J\propto M_1{M}_2{a}^{1/2}$ at fixed $M,J$ gives $d\log a/d\log M_1=2(q-1)$. Therefore the stipulated Roche lobe law has the Roche-lobe radius response exponent

Define the overfill by $\Delta =\log (R_1/R_L)$. On a time scale short compared with the slow change of $K$, the donor's response is $d\log R_1=\beta d\log M_1$. ThusDuring mass loss $d\log M_1<0$. If ${\zeta }_L>\beta$, mass loss increases the overfill and therefore drives more mass loss: this is positive feedback, giving more rapid Roche-lobe overflow. If ${\zeta }_L<\beta$, mass loss decreases the overfill and can balance the slow expansion due to stellar evolution. More explicitly, with $k_{\mathrm{ev}}=d\log K/dt>0$, contact requiresA slow negative transfer rate can satisfy this only on the stable side $\beta >{\zeta }_L$. The rapid-transfer condition is thereforeIts time scale depends on the physical donor response represented by $\beta$; it need not be a dynamical time scale if $\beta$ is a thermal-equilibrium mass-radius exponent.

For Roche-lobe overflow feedback with a constant wind torque factor, retain mass conservation to the stated approximation but allow the prescribed angular-momentum loss. Since $d\log J=fd\log M_1$, differentiation givesThe new Roche-lobe radius response exponent isThe same overfill calculation givesA loss of angular momentum has $f>0$ because ${\dot{M}}_1<0$, and it lowers the critical ratio by making the orbit contract more strongly.

For the specified solar-mass donor exponent $\beta =1/2$, the answer isWithout the wind torque this becomes $q>22/23\simeq 0.957$, so donors slightly less massive than their companions can already lie on the rapid-transfer side of this model. For $0<f<11/10$ the lower threshold is $(22-20f)/23$, and for $f\ge 11/10$ every positive $q$ satisfies the strict inequality. There is no finite upper bound on $q$ from this criterion. Since the paper does not specify $f$, the wind case has a conditional range rather than a unique numerical cutoff.

Let $\rho =3m/(4\pi R_0^3)$ be the uniform mass density. In spherical coordinates $(r,\vartheta ,\varphi )$ with the chosen diameter as polar axis, the squared perpendicular distance to that axis is $r^2{\sin }^2\vartheta$. The moment of inertia is thereforeThis derives the moment of inertia of a uniform solid sphere about any central axis by spherical symmetry.

Put $a=R_1-R_0$. The center of mass follows the circle of radius $a$, with position ${\mathbf{r}}_C=a(\sin \theta ,-\cos \theta )$. Let ${\mathbf{e}}_r=(\sin \theta ,-\cos \theta )$ and ${\mathbf{e}}_{\theta }=(\cos \theta ,\sin \theta )$. Its velocity is $a\dot{\theta }{\mathbf{e}}_{\theta }$, and the contact point is displaced from the center of mass by $R_0{\mathbf{e}}_r$. With positive spin angular velocity ${\Omega }_s$ counterclockwise, its contact-point velocity isThe cylinder is fixed, so rolling without slipping requires this velocity to vanish. Consequently ${\Omega }_s=-a\dot{\theta }/R_0$. If instead the spin angle $\psi$ is positive clockwise, opposite to increasing $\theta$, then the printed positive-coefficient convention isThe spin and orbital rotation have opposite directions when both are measured about the same oriented axis; the angular speed is $a|\dot{\theta }|/R_0$. This specifies the sign convention in the printed expression.

The instantaneous contact point is at rest, so the normal reaction and static friction do no work. Conservation of energy equates the loss of gravitational potential energy from release to the total kinetic energy:On the descent from $0<\alpha <\pi /2$, $\theta$ decreases, soIntegrating $dt=-d\theta /|\dot{\theta }|$ from release to the bottom givesThe integral is finite: at the release point its denominator is asymptotic to $\sqrt{\sin \alpha (\alpha -\theta )}$. Contact is maintained, since the inward radial equation of motion gives the positive normal reaction $N=m(g\cos \theta +a{\dot{\theta }}^2)$ throughout this range.

For frictionless sliding, the normal reaction passes through the sphere's centre and produces no torque; Newtonian gravity also exerts no central torque. With initial zero spin, the sphere remains nonrotating. All the lost gravitational potential energy becomes translational kinetic energy, soA hollow spherical shell of the same mass and radius places its material farther from the centre, giving a larger moment of inertia. Under rolling without slipping, a given centre speed therefore requires more rotational kinetic energy; the same gravitational drop yields a smaller speed. Replacing the solid sphere by a hollow shell increases $T_R$ and leaves $T_S$ unchanged. The sliding case remains nonrotating and has the same center of mass path and gravitational drop. This is the general comparison for rolling inside a fixed circular cylinder.

During ejection, the remaining mass is $m(t)=m_o+m_w-Qt$, and the water is exhausted at $t_c=m_w/Q$. Consider the momentum of the octopus and the small portion of water expelled during $dt$. Initially it is $mu$. At the end of this interval, the remaining body has mass $m-Qdt$ and velocity $u+du$, while the expelled mass $Qdt$ has laboratory velocity $u-V$, to first order. Hence the total final momentum isThe external impulse from quadratic drag is $-\alpha u^{2}dt$. Equating this to the change in total momentum, and dividing by $dt$, proves the equation of motionThe outgoing water's momentum flux is essential: applying $d(mu)/dt$ to the remaining body alone would omit it.

For dimensional analysis, the independent physical inputs are $m_o,m_w,Q,V,\alpha$, with dimensionsThe scales $m_o$, $m_o/Q$ and $V{m}_o/Q$ fix units of mass, time and length. In these units, the only remaining independent dimensionless variables areFor example, with $s=Qt/m_o$ and $w=u/V$, the equation of motion becomes $(1+\lambda -s)dw/ds=1-\mu w^2$, with $w(0)=0$ and ejection ending at $s=\lambda$. Thus the dimensionless terminal value depends only on $\lambda ,\mu$, proving $u_c=Vf(\lambda ,\mu )$.

For its explicit value, put $m_0=m_o+m_w$ and $U=\sqrt{QV/\alpha }$. The initially stationary solution stays below $U$, since the acceleration vanishes at $U$ and is positive below it. Separation of variables givesThe two integrals yieldThis is constant mass-loss propulsion with quadratic drag. At $t_c$, the remaining mass is $m_o$, givingThus $f(\lambda ,\mu )={\mu }^{-1/2}\tanh (\sqrt{\mu }\log (1+\lambda ))$, verifying the required dimensional form. The limit as drag tends to zero is $V\log (1+\lambda )$, the ordinary rocket equation, while the positive-drag answer is below the terminal velocity $U$.

A renormalizable quantum field theory requires only finitely many independent field and parameter redefinitions to remove regulator dependence from its perturbative correlation functions, order by order. The allowed local counterterms must close within that finite set; it is not necessary that the unrenormalized integrals be finite.

In four spacetime dimensions the kinetic term fixes the mass dimension of $\varphi$ to one. A coefficient multiplying ${\varphi }^k$ therefore has dimension $4-k$. Power counting excludes couplings of negative mass dimension for perturbative renormalizability. For a single scalar with a conventional nonderivative polynomial potential, the allowed form isThe constant controls vacuum energy and the linear term a possible tadpole. A field-reflection symmetry can remove the odd terms but is not necessary for renormalizability. Terms of degree above four generally require an infinite tower of counterterms when treated as fundamental interactions; they can instead be used in an effective field theory. Stability of the potential is a separate requirement, for example a positive quartic leading term, and should not be confused with the power-counting criterion.

For a graph with $E$ external scalar lines, $I$ internal lines, $L$ loops and $V_k$ vertices of degree $k$, the superficial ultraviolet divergence degree is $\omega =4L-2I$. The identities $L=I-{\sum }_k{V}_k+1$ and $2I+E={\sum }_{k}k{V}_k$ giveWhen every $k\le 4$, only the finitely many low-point local divergences, including the kinetic term, can require subtraction. Subdivergences are removed recursively by the same local counterterms. A vertex with $k>4$ instead permits arbitrarily high divergence degrees as more such vertices are added. This explains the restriction rather than merely asserting it.

The bare Lagrangian density is written in terms of regulator-dependent bare quantities, before taking the regulator away:The coefficients in $V_B$ are related to the renormalized parameters by regulator-dependent redefinitions. The wave-function renormalization supplies a kinetic counterterm. In dimensional regularization $d=4-ϵ$, a quartic coupling has a bare relation of the form ${\lambda }_B={\mu }^{ϵ}[\lambda +\delta \lambda (\lambda ,ϵ)]$, with field factors incorporated according to the chosen definition. The bare parameters and bare correlation functions are held fixed when changing the renormalization scale.

Even without a physical mass, interacting loop subtractions require a reference scale $\mu$ to define the renormalized dimensionless coupling and field normalization. Finite logarithms then involve dimensionless combinations such as $\mu |x_r-x_s|$ or $p^2/{\mu }^2$. This scale is a subtraction convention, not a new physical mass parameter. Its explicit dependence is compensated by the running coupling and field normalization.

Let $G_n=⟨\varphi (x_1)\cdots \varphi (x_n)⟩$ at separated points. Its bare counterpart satisfies $G_{B,n}=Z_{\varphi }^{n/2}{G}_n$. DefineDifferentiating $G_{B,n}$ at fixed bare theory givesThis Callan-Symanzik equation expresses independence of the arbitrary subtraction scale. The renormalization-group beta function describes the compensating coupling change; the field-renormalization anomalous dimension describes the compensating rescaling of each insertion. Coincident composite insertions would require their own additional renormalization, rather than automatically obeying this separated-point equation.