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In the straight-trajectory dipole model, retain the Coulomb-coupling convention of $V(t)=Z{e}^2(bx+vty)/(b^2+v^2{t}^2{)}^{3/2}$. In SI units the coupling denoted $e^2$ here is $e^2/(4\pi {ϵ}_0)$. First-order time-dependent perturbation theory expresses a transition through the Fourier transform of this pulse at the transition angular frequency.

For positive $\omega$ put $\beta =\omega b/v$. The two Cartesian pulse transforms arewhere $K_{\nu }$ is a Modified Bessel function of the second kind. One may derive these from the Fourier transform of $(1+s^2{)}^{-1/2}$, differentiating in its scale and integrating its spatial derivative by parts. For $\beta \ll 1$, $\beta K_1(\beta )\to 1$ while $\beta K_0(\beta )\to 0$, so the transverse impulse gives the probability in the title. If the $x$ matrix element vanishes, the smaller longitudinal term can be the leading nonzero contribution.

The conditions include $b$ large compared with the atomic extent for the electric-dipole approximation, a small total transition probability for first-order theory, and a collision time $b/v$ short compared with $1/|\omega |$. Large $\beta$ instead gives exponentially suppressed excitation in this model through the large-argument behavior of $K_{\nu }$.

For fixed real $\nu$ as $z\to +\infty$, the Modified Bessel function of the first kind and Modified Bessel function of the second kind have asymptotic expansionsThe expansion and its error bounds establish actual growing and decaying solutions, beyond formal matching of a differential equation. These formulas are documented in NIST DLMF, equations 10.40.1–2.

For the positive quadratic kernel $\tilde{\Delta }(p)={\kappa }^{-1}{p}^2+m^2$, $\kappa ,m>0$, the connected correlation function is its inverse. Fourier inversion gives $G(r)=\kappa (\sqrt{\kappa }m/r{)}^{D/2-1}{K}_{D/2-1}(\sqrt{\kappa }mr)/(2\pi {)}^{D/2}$. The Modified Bessel function of the second kind has an exponentially decaying large-argument tail, so the correlation length is $1/(\sqrt{\kappa }m)$ at fixed positive gradient coefficient. In three dimensions this becomes $\kappa e^{-\sqrt{\kappa }mr}/(4\pi r)$. The statement concerns the continuum long-distance theory; regulator-dependent contact terms do not define the physical correlation length.

Expand the time-dependent quantum state in unperturbed energy eigenstates, factoring out free dynamical phases:The Schrödinger equation gives the exact interaction picture coefficient equationsInitially $c_n(-\infty )={\delta }_{ni}$. To first order in time-dependent perturbation theory, replace the coefficients on the right by these zeroth-order values and integrate. ThusFor a transition $j\ne i$, this is the desired leading probability amplitude, and $P_{ji}=|{\gamma }_{ji}^{(1)}{|}^2$ to leading order. The formula is understood with free asymptotic phases removed. For $j=i$, the full survival amplitude also contains the zeroth-order $1$; the displayed integral alone is its first-order correction, not the whole amplitude. This specifies the usual scattering operator meaning of the paper's notation at infinite times.

For the ion pulse, write $x_{ji}=⟨j|x|i⟩$ and $y_{ji}=⟨j|y|i⟩$. ThenConsider an excitation with ${\omega }_{ji}>0$ and put $\beta ={\omega }_{ji}b/v$, $u={\omega }_{ji}t$. The first real pulse is even in time and the second is odd. Their respective surviving Fourier components are therefore cosine and imaginary sine:Since $\beta \log (1/\beta )\to 0$, the longitudinal term is smaller for comparable position matrix elements. Hence the electric-dipole excitation by a passing ion hasThis leading limit extends to either sign of ${\omega }_{ji}$ using $|{\omega }_{ji}|b/v\ll 1$. If $x_{ji}=0$ by a selection rule, the retained leading term vanishes and the smaller longitudinal term may give the first nonzero probability. The conditions also include a small total transition probability, approximately straight constant-velocity motion, and $b$ large compared with atomic size for the electric-dipole approximation. The paper uses a Coulomb-coupling convention in which the factor is written $Z{e}^2$; in SI units replace $e^2$ throughout by $e^2/(4\pi {ϵ}_0)$.

The collision duration is of order $b/v$. Excitations with angular frequency ${\omega }_{ji}$ receive an unsuppressed impulse when this is short relative to $1/{\omega }_{ji}$. In contrast, for ${\omega }_{ji}b/v\gg 1$, cancellations in the Fourier transform suppress the transition: a slowly varying distant field approximately preserves the atomic energy eigenstate. The exact dipole-pulse transforms make this explicit:where $K_{\nu }$ is a Modified Bessel function of the second kind. Their large-argument exponential decrease explains the slow-passage cutoff. These transforms are optional refinements of the small-$\beta$ derivation above, not a separate assumption about the target.

To connect the microscopic transition probability to energy loss, suppose the atom begins in a low-energy state. Exciting bound electronic states and causing ionization transfer positive energies $E_j-E_i$ from the ion to the material. For target atomic number density $n_{\mathrm{at}}$ and path length $d$, a schematic independent-encounter sum iswith a continuum integral as well as a sum for ionized final states. The short-collision formula gives $P_{ji}\propto Z^2/(b^2{v}^2)$, so the impact-parameter integral has logarithmic dependence on its cutoffs. Its upper effective cutoff is of order $v/{\omega }_{ji}$, where the transition ceases to be sudden; the lower cutoff must exclude regions where the dipole or weak-collision approximation fails. Thus the leading high-speed charge and isotope scaling of electronic stopping isAlthough a faster ion can access more energetic transitions and a wider range of distant encounters, its shorter interaction time reduces each ordinary excitation probability. In the fast-ion regime shown, the $1/v^2$ decrease dominates the slower variation of the excitation factor. The energy loss consequently falls with increasing incident kinetic energy. The original figure has $\delta E$ increasing downward, so the curves' rise toward the top of the page means decreasing loss, not increasing loss. A low-speed extrapolation of the inverse-square law is invalid because excitation thresholds, slow-passage suppression, changing ion charge and nuclear recoil can then matter.

For nonrelativistic ion mass $M$, $E=M{v}^2/2$. At fixed incident energy the leading prediction becomesThe greater atomic number of nitrogen than carbon than boron therefore gives larger loss; at the same velocity the leading species factor is $Z^2$. Within an isotopic family, the heavier ion moves more slowly at the same $E$ and loses more energy. Equivalently, a heavier isotope needs proportionally greater kinetic energy to achieve the same velocity and approximately the same electronic loss. Comparing at fixed energy per nucleon, rather than fixed total $E$, approximately removes this leading isotope mass dependence. This accounts for the separate isotope bands in the original graph without assigning different electronic excitation spectra to isotopes of the same projectile element.

The charge in this scaling is the effective ion charge if screening by bound projectile electrons is important. The interpretation assumes the same traversed material and thickness and uses the paper's small-fractional-loss regime, $\delta E/E\ll 1$; points with loss comparable to incident energy cannot be described quantitatively by treating $v$ as constant. The plotted curves support the qualitative ordering and high-energy decrease, rather than determining a universal normalization from the supplied single-atom model.

The vanishing total pressure must be imposed at the displaced free surface, not at fixed $r=R$. The first-order Taylor expansion isSince $p_0(R)=0$ and $p_0^{\prime }(R)=-2\pi G{\rho }^{2}R$, the Lagrangian pressure perturbation condition isThe supplied gravitational perturbation and the combined potential, evaluated at $R$, then giveHere $K_0$ is the Modified Bessel function of the second kind. Equating mode amplitudes and using part (c) yieldsRearrange to obtain the dispersion relation for the axisymmetric gravitational instability of an incompressible cylinder:Every quantity on the right is dimensionless, consistent with $G\rho$ having units ${\mathrm{s}}^{-2}$. The stabilizing term $1/2$ comes from displacing the background hydrostatic pressure profile; omitting the Lagrangian pressure perturbation would lose it.

The contrast is weak: $(n_1-n_2)/n_2=0.01$. Use a scalar weak-guidance Helmholtz equation for a transverse field component, with longitudinal wave number $b$ and vacuum wave number $k_0=2\pi /{\lambda }_0$. Separating variables as $\psi =F_l(r)e^{il\varphi }{e}^{ibz}$ givesA bound mode requires $n_2{k}_0<b<n_1{k}_0$. DefineThe step-index fiber normalized frequency is $V$; $u$ is a dimensionless core transverse wave number. Regularity at $r=0$ selects the Bessel function of the first kind, while decay at infinity selects the Modified Bessel function of the second kind:In this scalar approximation, continuity of the field and its radial derivative givesAt cutoff the evanescent wave extends arbitrarily far into the cladding, so $b\downarrow n_2{k}_0$, $w\downarrow 0$ and $u\to V$. For the first azimuthally varying family $l=1$, the recurrence relations reduce the last equality toAs $w\to 0$, $K_1(w)\sim 1/w$ and $K_0(w)\sim -\ln w$, making the right side tend to zero. Thus the onset is at the first positive zero of $J_0$:The scalar $l=0$ fundamental family has no positive cutoff. The first additional spatial family is the $l=1$ family just found; the next radial $l=0$ and first $l=2$ families require the larger zero $j_{1,1}\simeq 3.832$. This establishes the first multimode boundary rather than merely finding one possible cutoff. The two polarizations of the fundamental family are not counted as two spatial modes. Exact vector interface conditions split some higher-mode degeneracies at finite contrast; the displayed condition is the weak-guidance result intended by the supplied scalar equations.

Use $\Xi (u)={\pi }^{-u/2}\Gamma (u/2)\zeta (u)$ for the completed Riemann zeta function, without the polynomial factor used to define the entire Riemann xi function. For complex $\nu$ putThis is the Modified Bessel function of the second kind. The substitution $t=e^v$ writes it as $\frac{1}{2}{\int }_{\mathbb{R}}{e}^{-Y\cosh v+\nu v}dv$, proving that it is entire in $\nu$, that $K_{\nu }=K_{-\nu }$, and that it decays exponentially as $Y\to \infty$, uniformly for $\nu$ in a compact set.

For $\Re s>1$ all the lattice sums and the integrals below converge absolutely. The $m=0$ terms in the completed nonholomorphic Eisenstein series contribute $\Xi (2s)y^s$. For $m\ne 0$, use the Gamma integral and then Poisson summation in $n$:Here the Poisson identity is ${\sum }_n{e}^{-\pi u(n+mx{)}^2}=u^{-1/2}{\sum }_h{e}^{-\pi h^2/u}{e}^{2\pi ihmx}$. At $h=0$, evaluating the Gamma integral and summing over $m$ gives $\Xi (2s-1)y^{1-s}$. At $h\ne 0$, the substitution $u=|h|t/(|m|y)$ and the integral representation of the modified Bessel function of the second kind giveCollect the frequency $r=mh\ne 0$. There are two signed choices of $m$ for each positive divisor of $|r|$. Thus, with ${\sigma }_a(n)={\sum }_{d\mid n}{d}^a$, the full Fourier expansion isThe factor $2$ is important: the lattice sum includes both signs, while the defining factor $1/2$ was already used in its Mellin integral. This expansion uses exactly the completion in the question.

We can deduce both of the offered conclusions. First recall, with a short proof, the needed functional equation of the Riemann zeta function. Put $\Theta (t)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}t}$. Poisson summation gives $\Theta (t)=t^{-1/2}\Theta (1/t)$. Splitting the Mellin integral of $\Theta -1$ at one yieldsThe integral is entire and invariant under $u\mapsto 1-u$. Hence $\Xi (u)=\Xi (1-u)$, with only simple poles at $u=0,1$, of residues $-1,1$. This is the pole-bearing completion, not the entire xi function.

For fixed $y>0$, the nonconstant part of the Eisenstein expansion is entire in $s$ and locally normally convergent: the divisor sums grow at most polynomially on a compact set of $s$, whereas $K_{s-1/2}(2\pi |r|y)$ decays exponentially in $|r|$. The same argument gives local uniformity for $z$ in compact subsets of the complex upper half-plane. Thus only the constant terms can have poles. Their apparent poles at $s=1/2$ cancel: $\Xi (2s)$ has principal part $1/(2(s-1/2))$, while $\Xi (2s-1)$ has its negative, and both multiply $y^{1/2}$ there. The remaining poles are simple, at $s=0,1$, with residues $-1/2,1/2$.

Under $s\mapsto 1-s$, the two constant terms interchange by the zeta functional equation. The nonconstant terms are unchanged because $K_{\nu }=K_{-\nu }$ andwhich follows by replacing a divisor $d$ by $|r|/d$. Therefore

For the Kronecker limit formula, evaluate the finite Laurent coefficient at $s=1$. Write ${\gamma }_E$ for the Euler--Mascheroni constant. From $\zeta (u)=1/(u-1)+{\gamma }_E+O(u-1)$ and ${\Gamma }^{\prime }(1/2)/\Gamma (1/2)=-{\gamma }_E-2\log 2$ we obtainThe gamma derivative follows by differentiating the Gamma duplication formula and using ${\Gamma }^{\prime }(1)=-{\gamma }_E$. Multiplication by $y^{1-s}$ adds $-\frac{1}{2}\log y$ to the constant term; the other constant term contributes $\Xi (2)y=\pi y/6$.

The half-order Modified Bessel function of the second kind is $K_{1/2}(Y)=\sqrt{\pi /(2Y)}{e}^{-Y}$. For example, in its even integral representation set $v/2$ as variable and then $u=\sinh (v/2)$: $\cosh v=1+2u^2$ and the $\cosh (v/2)$ factor converts the integral to a Gaussian. At $s=1$ the nonconstant Eisenstein coefficients are consequently ${\sigma }_{-1}(|r|)e^{-2\pi |r|y}$. On the other hand, the absolutely convergent logarithmic series givesAdding this identity to its conjugate identifies the nonconstant sum as $-2\log |{\prod }_{n\ge 1}(1-q^n)|$. With the analytic Dedekind eta function product $\eta (z)=q^{1/24}{\prod }_{n\ge 1}(1-q^n)$, the resulting answer isThis derivation of the Kronecker limit formula uses the analytic eta product only; it does not assume its modular transformation. The lattice series is invariant for $\Re s>1$ by an integral change of the pair $(m,n)$ and $\Im (\gamma z)=y/|cz+d{|}^2$. Analytic continuation preserves this invariance, so its finite Laurent coefficient is invariant too. This supplies the independent weight-two transformation argument used in Question 2.

The connected generating functional is $W_0[J]=\log Z_0[J]$. Its second functional source derivative gives the connected correlation function; the source-independent determinant drops out. Translational invariance therefore givesThis covariance is independent of the source in a purely Gaussian theory; at zero source the mean field vanishes.

Fourier inversion can be evaluated with the Schwinger parameterization. Since $(p^2+{\mu }^2{)}^{-1}={\int }_0^{\infty }ds{e}^{-s(p^2+{\mu }^2)}$, the elementary Fourier transform of a Gaussian giveswhere $K$ is the Modified Bessel function of the second kind. This integral formula also gives both asymptotic regimes directly.

For $\mu r\ll 1$, set $u=r^2/(4s)$. The integral tends to the gamma function integral ${\int }_0^{\infty }{u}^{D/2-2}{e}^{-u}du=\Gamma (D/2-1)$; this is finite for $D>2$. Hence, at distances larger than the microscopic cutoff,For $\mu r\gg 1$, the exponent ${\mu }^{2}s+r^2/(4s)$ has its minimum at $s=r/(2\mu )$, with value $\mu r$. Gaussian expansion about that saddle, or the large-argument asymptotic expansion of a modified Bessel function, gives $K_{\rho }(z)\sim \sqrt{\pi /(2z)}{e}^{-z}$. ThereforeThis is the massive Gaussian field correlation tail with its normalization displayed. For example $D=3$ gives exactly $G_0(r)=\alpha e^{-r/\xi }/(4\pi r)$, verifying both regimes and the exponential length scale.

Put $w=1/z$ and $Y(w)=y(1/w)$. The equation becomesAt $w=0$ the regular-singular criterion requires $w(2/w)$ and $w^2(-w^{-n-4})$ to be analytic. This holds precisely for $n\le -2$. Thus the singularity at infinity is irregular, corresponding to essential exponential behaviour of solutions, exactly for $\boxed{n\ge -1}$. For $n\le -2$ it is regular singular, with powers and possibly logarithms rather than such essential exponentials.

For $y^{\prime \prime }=z^{3}y$, the Liouville–Green approximation uses $Q=z^3$, amplitude $Q^{-1/4}$ and phase integral $\int \sqrt{Q}dz=2z^{5/2}/5$. On a consistently chosen branch, the two leading solutions areThe derivative correction is smaller by order $z^{-5/2}$, and the two exponential signs give independent solutions.

Using the convention that Stokes lines have equal exponential phase, $\mathrm{Im}(z^{5/2})=0$, their rays are $\arg z=2\pi k/5$. The equal-magnitude, oscillatory rays, often called anti-Stokes lines, satisfy $\mathrm{Re}(z^{5/2})=0$, giving $\arg z=(2k+1)\pi /5$. If the alternative naming convention is used, these two names interchange; the stated equations specify the rays unambiguously.

To give the sector precisely, the solution recessive on the positive axis can be represented bywhere $K_{\nu }$ is the Modified Bessel function of the second kind. Substitution verifies the differential equation. Its asymptotic $K_{\nu }(w)\sim \sqrt{\pi /(2w)}{e}^{-w}$ is valid on closed subsectors of $|\arg w|<3\pi /2$ on the continued branch. ConsequentlyActual exponential decay occurs in the narrower sector $|\arg z|<\pi /5$; in the adjacent portions of the larger asymptotic sector the same continued exponential grows. This distinguishes the validity of the approximation to the positive-axis recessive solution from the directions where its modulus decays.

Take a positive gradient coefficient ${\kappa }^{-1}$, a positive squared mass, a finite-volume regulator and an ultraviolet cutoff. Use the dimensionless statistical action convention $Z=\int \mathcal{D}\varphi e^{-H}$: inverse temperature is absorbed into the coefficients. If $H$ denotes dimensional energy instead, the exponent below gains a factor ${\beta }_{\mathrm{th}}$ and the covariance gains ${\beta }_{\mathrm{th}}^{-1}$. These conventions cannot be mixed.

Choose the Fourier transform conventionIntegration of the quadratic gradient and source terms givesComplete the square with ${\tilde{\varphi }}_{\mathrm{cl}}(p)=\tilde{\Delta }(p{)}^{-1}\tilde{J}(p)$. Translation of the regulated real Gaussian variables leaves their measure unchanged, so the Gaussian integral givesHere $Z_G=Z_0[0]$ is the source-free Gaussian partition function. For $n$ independent regulated real coordinates and a positive quadratic matrix $A$, it is $(2\pi {)}^{n/2}(\det A{)}^{-1/2}$ under Lebesgue normalization. The continuum determinant needs the stated regulator and a measure convention. Reality pairs the nonzero Fourier modes, so they must not be counted as two independent real fields.

A connected correlation function subtracts the product of the means. The logarithm $W[J]=\log Z_0[J]$ generates its cumulants: differentiating $W$ twice with respect to the source, using the measure $d^{D}p/(2\pi {)}^D$ in the functional derivative convention, yieldsThe source changes the mean but not the covariance; the stated zero-source limit therefore follows immediately. Momentum conservation uses $p+q$, appropriate for a real field without complex conjugation on either displayed factor.

For the continuum massive Gaussian field correlation tail, set $\mu =\sqrt{\kappa }m>0$ and $r=|x|$. Fourier inversion givesThe large-argument Modified Bessel function of the second kind has $K_{\rho }(z)\sim \sqrt{\pi /(2z)}{e}^{-z}$. Hence $G_0(r)$ is proportional to $r^{-(D-1)/2}{e}^{-\mu r}$ at large distance andAt zero mass the exponential length diverges. For $D>2$ the critical continuum tail is algebraic, $G_0(r)\propto r^{2-D}$; in lower dimensions infrared regulation needs additional care.

For a uniform source $h$, perform a momentum-shell renormalization group step: separate slow modes $|p|<\Lambda /b$ and fast modes $\Lambda /b<|p|<\Lambda$. In a Gaussian field theory they decouple; the uniform source couples only to the retained zero mode. Integrating fast modes contributes only a field-independent determinant to the free energy. Restore the cutoff by $x^{\prime }=x/b$, $p^{\prime }=bp$ andThe factors from the measure, gradient and two fields cancel in the gradient term. The mass term gains $b^2$, and the linear source gains $b^{(D+2)/2}$. ThereforeHere $m$ is the positive mass; for signed thermal coordinates it is the squared-mass relation that applies. The Gaussian fixed point has thermal scaling eigenvalue ${\lambda }_t=2$, source eigenvalue ${\lambda }_h=(D+2)/2$ and zero anomalous dimension.

Under the formal homogeneous Gaussian scaling hypothesis, substitution of those eigenvalues givesThe last two values are independently visible in $\xi \propto t^{-1/2}$ and $\chi ={\tilde{G}}_0(0)=1/m^2\propto t^{-1}$, agreeing with Landau-Ginzburg theory. The specific-heat power follows directly from the determinant: after two thermal derivatives its singular contribution is proportional toThe change of variables $p=\sqrt{\kappa t}q$ proves this power because the rescaled integral converges at large $q$ exactly when $D<4$. The two-dimensional free energy itself has a $t\log t$ resonance, while its second derivative still has the indicated $t^{-1}$ power.

The printed magnetization exponent is a formal Gaussian scaling index, not a proved ordered-phase law of the purely quadratic model. The ordered-phase obstruction in a Gaussian scalar model is explicit: the uniform field has energy density $m^2{M}^2/2-hM$. At $h=0,m^2<0$ it tends to $-\infty$ as $|M|\to \infty$, and the partition function diverges. For $m^2>0$, the mean is $M=h/m^2$ and tends to zero as $h\to 0$, so there is no spontaneously ordered branch. Thus no literal $M\sim |t{|}^{\beta }$ for $t<0$ can be established from this Statistical Hamiltonian alone. The value $(D-2)/4$ follows from the scaling dimension of the field if one assumes such a branch; for $D\le 2$ its nonpositive value further warns against that interpretation. Adding a stabilizing quartic interaction defines an ordered phase, but below four dimensions that interaction is relevant and generally leads to the Wilson-Fisher fixed point, not these Gaussian thermodynamic powers.

For the Gaussian specific-heat infrared threshold, at $D=4$ the same determinant integral grows like $\log ({\Lambda }^2/t)$, so $\alpha =0$ as a power exponent, with a logarithmic divergence. For $D>4$, it has a finite cutoff-dependent limit; the leading heat capacity is a regular constant and has no divergent power, again described by $\alpha =0$ in the requested convention. After subtraction of this background the Gaussian singular term can still have the negative power index $2-D/2$; it is not correct to erase that distinction. In a stable quartic Landau-Ginzburg theory above four dimensions, the quartic coefficient is a dangerously irrelevant coupling: it is needed for the ordered phase and contributes $f_s\propto -t^2/u$, giving the mean-field specific-heat jump and $\alpha =0$ while invalidating naive hyperscaling relation. At four dimensions, marginal interactions can add further logarithmic corrections. The agreement with Landau theory is in the leading heat-capacity power; it does not assert the absence of logarithms or supply a missing stable Gaussian ordered phase.

Change independent variable from proper time to the scale factor. Then $\dot{\delta }=\dot{a}{\delta }_a$ and $\ddot{\delta }={\dot{a}}^2{\delta }_{aa}+\ddot{a}{\delta }_a$. Under radiation domination, $\ddot{a}=-{\dot{a}}^2/a$ and $H^2\simeq 8\pi G{\bar{\rho }}_r/3$. Since ${\bar{\rho }}_m/{\bar{\rho }}_r=a/a_{\mathrm{eq}}$, the linear matter perturbation growth equation becomesPutting $\delta =au$ givesThis keeps the matter self-gravity term in a radiation-dominated background. It is not an exact background equation through radiation-matter equality.

At $a/a_{\mathrm{eq}}\ll 1$, neglecting that small term gives ${\delta }_{aa}+{\delta }_a/a=0$. Integration yieldsThus matter has at most logarithmic growth during this leading radiation approximation, together with a constant independent mode. The constant is non-growing, not a mode that literally falls as $a^{-1}$; that power belongs to $u$ rather than ${\delta }_m$.

For an increasing/decreasing basis of the displayed equation with matter self-gravity retained, put $x=\sqrt{6a/a_{\mathrm{eq}}}$. Its equation becomes $x^2{\delta }_{xx}+x{\delta }_x-x^2\delta =0$. HenceThe Modified Bessel function of the first kind gives $I_0(x)=1+3a/(2a_{\mathrm{eq}})+O((a/a_{\mathrm{eq}}{)}^2)$, which increases slowly. The Modified Bessel function of the second kind gives $K_0(x)=-\frac{1}{2}\ln (a/a_{\mathrm{eq}})+\frac{1}{2}\ln (2/3)-{\gamma }_E+O((a/a_{\mathrm{eq}})|\ln (a/a_{\mathrm{eq}})|)$, where ${\gamma }_E$ is Euler's constant; this mode decreases as $a$ increases. Their leading span is precisely the constant/logarithmic pair above. Mode labels depend on the chosen basis and normalization; no rapid matter-era growth $\delta \propto a$ occurs here. Neglected background corrections can change subleading terms, so the Bessel basis should not be extrapolated through equality.

Fix the convention $\Delta q-m^{2}q=0$, $m>0$, and let $c_m=m^2/4$. With $z=x+iy$, two families of modified Helmholtz adjoint plane waves areBoth satisfy $\Delta E_{\pm }=m^2{E}_{\pm }$, since the products of their $z$ and $\bar{z}$ exponents are $m^2/4$. They obey $E_{-}(z,k)=\overline{E_{+}(z,\bar{k})}$. Green second identity gives the two conjugate global relations for the modified Helmholtz equationFor real boundary traces, ${\mathcal{G}}_{-}(k)=\overline{{\mathcal{G}}_{+}(\bar{k})}$. Thus the second identity is the conjugate spectral companion, rather than an unrelated extra boundary condition. In complex differential form the first relation is equivalentlyIts integrand is a closed one-form: differentiating its coefficients gives $2E_{+}(q_{z\bar{z}}-c_{m}q)=0$. This supplies a direct derivation of the global relation as well as its outward-normal sign convention.