Retarded displacement response

Summary: The full spatial quadrupole Hessian gives a positive frequency-dependent interaction kernel, with a derived logarithmic soft correction and outgoing radiation into modes of the same field.

Keep the deformation sourced by both waves

The spatial displacement calculation uses the declared component action $S=\tfrac Z2\int\partial_\mu n\cdot\partial^\mu n$, $n^2=1$, with profile

$$n_0(x)=\frac{(2\lambda x_1,2\lambda x_2,|x|^2-\lambda^2)} {|x|^2+\lambda^2},\qquad \mathcal T=4\pi Z. \tag{1}$$

The $x$ coordinates are the two ordinary bound spatial directions. The existing normal coordinate $y$ carries propagation. The coefficients of $n$ add no physical spatial dimensions, and this component action remains to be derived from the master field before identifying its waves as photons.

Opposite pulses have displacement $X=A(t-y)+B(t+y)$ and derivatives $a=A',b=B'$. Their trace-free spatial source is

$$S=2(ab^T+ba^T-a\cdot b\,I_2),\qquad \mathcal R=S_{ij}\left[ \partial_i\partial_jn_0+ (\partial_i n_0\cdot\partial_jn_0)n_0 \right]. \tag{2}$$

The complete static Hessian $\mathcal J$ gives the response $\eta_0=\tfrac12(Sx)_i\partial_i n_0$ with $\mathcal J\eta_0=-\mathcal R$. Its static energy norm is finite, while its canonical $L^2$ norm diverges logarithmically. A frequency-independent elimination therefore needs to be replaced by the actual retarded spatial inverse. The exact one-way waves remain those of Null spatial pulses.

Reduce the full spatial Hessian without discarding its norm

Write a radial quadrupole variation as

$$\eta_h(x)=\frac12(Sx)_i\partial_i n_0(x)\,h(r). \tag{3}$$

The constrained Hessian acts exactly as

$$\boxed{ \mathcal J\eta_h=\eta_{\mathcal H_\lambda h},\qquad \mathcal H_\lambda h =-h''+\left(-\frac3r+\frac{4r}{r^2+\lambda^2}\right)h' +\frac8{r^2+\lambda^2}h.} \tag{4}$$

This follows by differentiating (3) in the full three-component field equation, including the constraint terms. The radial kinetic measure is

$$w_\lambda(r)\,dr=\frac{r^3}{(r^2+\lambda^2)^2}\,dr. \tag{5}$$

Consequently

$$\mathcal H_\lambda =-\frac1{w_\lambda}\frac d{dr} \left(w_\lambda\frac d{dr}\right) +\frac8{r^2+\lambda^2}. \tag{6}$$

Its quadratic form is positive. Regularity is imposed at $r=0$. At nonzero frequency the boundary condition is outgoing at infinity, or decaying for an imaginary frequency. These conditions belong to the same physical spatial field as the incoming modes.

Let $s=S_{11}+iS_{12}$, so $|s|^2=4|a|^2|b|^2$. Direct spatial integration supplies the complete normalization:

$$Z\|\eta_h\|^2 =2\pi Z\lambda^2|s|^2 \int_0^\infty w_\lambda(r)|h(r)|^2dr. \tag{7}$$

The complex number $s$ records the two real quadrupole components. It does not replace either wave's complex frequency amplitudes.

Define the common frequency-dependent interaction

Scale $r$ by $\lambda$ and use dimensionless frequency $z$. In this paragraph $\lambda=1$ inside the radial operator. Define

$$(\mathcal H_1-z^2)h_z=\frac8{1+r^2}, \qquad \boxed{F(z)=4\int_0^\infty \frac{r^3}{(1+r^2)^3}h_z(r)\,dr.} \tag{8}$$

For $\operatorname{Im}z>0$, this is the retarded inverse. For a Fourier component $e^{-i\Omega t+iqy}$, its argument is $z=\lambda\sqrt{(\Omega+i0)^2-q^2}$, with the outgoing branch. In particular, two opposite positive-frequency waves have $\Omega^2-q^2=4\omega_A\omega_B$.

Equation (8) is one radial response for both quadrupole polarizations. It gives, with the source amplitude factored as in (3),

$$\boxed{ Z\langle\mathcal R, (\mathcal J-(\Omega+i0)^2+q^2)^{-1}\mathcal R\rangle =\mathcal T|s|^2F(z).} \tag{9}$$

The static solution is $h_0=1$, hence $F(0)=1$. It is a finite-energy form limit, not an $L^2(w_1\,dr)$ eigenvector; the constant radial function has a logarithmically divergent kinetic norm. The conservative static quartic term is therefore $\tfrac12\mathcal T|s|^2=2\mathcal T|a|^2|b|^2$, as independently derived from the exact moving profile. At finite frequency, (9) supplies the response and its delay.

A retarded inverse describes causal response; its dissipative part must not be varied as if it were an ordinary real instantaneous energy. The full field action remains the conservative system. Eliminating its outgoing modes has the same meaning as Continuum mode memory. There are no added mediator particles.

Positivity and spectral normalization

On the positive radial domain,

$$F(z)=\int_0^\infty\frac{d\mu(\ell)}{\ell-z^2}, \qquad d\mu\ge0. \tag{10}$$

The source and kinetic measure in (8), not a fitted weight, determine $\mu$. Direct radial integrals give

$$\boxed{ \int\frac{d\mu(\ell)}{\ell}=1,\qquad \int d\mu(\ell)=\frac83,\qquad \int\ell\,d\mu(\ell)=\frac{64}{5}.} \tag{11}$$

For example, the last two are $\tfrac12\langle f,f\rangle_w$ and $\tfrac12\langle f,\mathcal H_1f\rangle_w$, with $f=8/(1+r^2)$. They reduce to convergent rational integrals.

Thus $0<F(i\xi)<1$ for $\xi>0$, and it decreases with $\xi$. The same positive measure gives

$$\frac8{3\xi^2}-\frac{64}{5\xi^4} \le F(i\xi)\le\frac8{3\xi^2}. \tag{12}$$

The lower bound is useful at large $\xi$. It follows from $(\ell+\xi^2)^{-1} =\xi^{-2}-\ell\xi^{-4} +\ell^2/[\xi^4(\ell+\xi^2)]$.

Derive the logarithmic low-frequency correction

For $z=i\xi$, $\xi>0$, the inner radial expansion starts with

$$h_{i\xi}(r)=1-\frac{\xi^2}{8}(r^2+2)+\cdots, \qquad \mathcal H_1\left(\frac{r^2+2}{8}\right)=1. \tag{13}$$

It ceases to be uniform when $r$ approaches $1/\xi$. In the outer coordinate $t=\xi r$, the leading equation is

$$-h_{\rm out}''+\frac1t h_{\rm out}' +\left(1+\frac8{t^2}\right)h_{\rm out} =\frac8{t^2}. \tag{14}$$

The decaying solution which matches (13) is

$$h_{\rm out}(t)=\frac8{t^2}-tK_3(t) =1-\frac{t^2}{8}+O(t^4|\log t|). \tag{15}$$

Choose $1\ll R\ll1/\xi$. Substitution into (8) gives the inner contribution to $1-F$ as $\xi^2(\tfrac12\log R-\tfrac18)+o(\xi^2)$. The outer contribution is evaluated by

$$\begin{aligned} 4\int_\epsilon^\infty \frac{1-8/t^2+tK_3(t)}{t^3}\,dt &=\frac2{\epsilon^2}-\frac8{\epsilon^4} +\frac{4K_2(\epsilon)}{\epsilon^2}\\ &=\frac12\log\frac1\epsilon +\frac{\log2-\gamma_{\!E}}2+\frac38+o(1). \end{aligned} \tag{16}$$

The derivative identity $d(K_2(t)/t^2)/dt=-K_3(t)/t^2$ proves the first line. The integer-order expansions are recorded in NIST DLMF §10.31. With $\epsilon=\xi R$, the arbitrary matching radius cancels:

$$\boxed{ F(i\xi)=1-\xi^2\left[ \frac12\log\frac1\xi+c_0 \right]+o(\xi^2),\qquad c_0=\frac14+\frac{\log2-\gamma_{\!E}}2.} \tag{17}$$

Control the approximation with the positive spatial form

A composite trial field is $\widetilde h=(1-\xi^2/4)h_{\rm out}(\xi r)$. It matches both terms of the inner expansion at fixed $r$. Put $A_\xi=\mathcal H_1+\xi^2$ and $d=A_\xi\widetilde h-f$. The exact variational identity gives

$$F_{\rm trial} =\langle f,\widetilde h\rangle_w -\frac12\langle\widetilde h,A_\xi\widetilde h\rangle_w, \qquad 0\le F(i\xi)-F_{\rm trial} =\frac12\langle d,A_\xi^{-1}d\rangle_w. \tag{18}$$

Since $A_\xi\ge 8/(1+r^2)+\xi^2$ in quadratic-form order,

$$\boxed{ 0\le F(i\xi)-F_{\rm trial} \le\frac12\int_0^\infty \frac{w_1(r)|d(r)|^2}{8/(1+r^2)+\xi^2}\,dr.} \tag{19}$$

This provides an explicit error control, rather than an assumed gap. For $H=h_{\rm out}(\xi r)$, its residual is

$$d=\left(1-\frac{\xi^2}{4}\right) \left[-\frac{4H'}{r(1+r^2)} +\frac{8(1-H)}{r^2(1+r^2)}\right] -\frac{2\xi^2}{1+r^2}. \tag{20}$$

The leading order-$\xi^2$ terms cancel. The small-argument Bessel expansion and its large-argument decay bound the integral in (19) by $O(\xi^6(1+|\log\xi|)^2)$, splitting the radial domain at $r=1$ and $r=1/\xi$. The same split in the trial functional gives (17). Numerical quadrature evaluates (18)–(19) independently of the radial resolvent; its integration errors are recorded separately.

Retain outgoing radiation in the interaction

The corresponding outgoing expansion in the upper half-frequency plane is

$$\boxed{ F(z)=1+z^2\left[-\frac14\log(-z^2)+c_0\right]+o(z^2).} \tag{21}$$

It follows by using $K_3(-izr)$ in the outgoing outer solution and the same inner matching. The logarithm is continued from imaginary frequency. At a positive real frequency,

$$\boxed{ \operatorname{Im}F(\omega+i0) =\frac{\pi}{4}\omega^2+o(\omega^2).} \tag{22}$$

The positive sign is fixed by the positive spectral measure in (10). It represents energy entering outgoing transverse modes of this same field. It is not an inserted damping constant.

Check the threshold coefficient with spatial scattering states

There is an independent spatial route to the coefficient in (22). The regular zero-energy homogeneous solution is exactly

$$h_{\rm reg}(r)=r^4+3r^2+3,\qquad \mathcal H_1h_{\rm reg}=0. \tag{23}$$

It grows at infinity. The other independent zero-energy solution is singular at the origin, so a regular decaying zero mode does not intervene in the threshold matching.

Set $u=rh/(1+r^2)$. This is an isometry from the radial measure $w_1\,dr$ to $r\,dr$. The homogeneous scattering equation becomes

$$\left[-\frac{d^2}{dr^2}-\frac1r\frac d{dr} +\frac{9r^4+2r^2+1}{r^2(1+r^2)^2}\right]u=k^2u. \tag{24}$$

Its potential approaches $9/r^2+O(r^{-4})$. Matching the regular solution (23) to the outer region $kr$ fixed gives the leading, delta-in-$k$ normalized field $h_k\sim\sqrt{k}\,rJ_3(kr)$. The regular matching removes a leading singular $Y_3$ contribution. The source overlap is controlled by its actual large-radius tail:

$$\begin{aligned} \langle f,h_k\rangle_w &=8k^{3/2}\int_0^\infty\frac{J_3(t)}{t^2}\,dt +o(k^{3/2})\\ &=k^{3/2}+o(k^{3/2}). \end{aligned} \tag{25}$$

The integral equals $1/8$ because $d[J_2(t)/t^2]/dt=-J_3(t)/t^2$, $J_2(t)/t^2\to1/8$ at the origin, and it vanishes at infinity. The derivative identity is given by NIST DLMF §10.6. Consequently

$$d\mu=\frac12|\langle f,h_k\rangle_w|^2dk =\frac{\ell}{4}[1+o(1)]\,d\ell,\qquad \ell=k^2. \tag{26}$$

This recovers (22) from spatial continuum normalization. Direct integration of the full homogeneous equation, followed by a large-radius $J_3,Y_3$ fit and the source integral, independently checks (25) at finite small $k$. The asymptotic Bessel tail is included in that source integral.

The component action therefore supplies a definite common interaction kernel, including both spatial polarizations, finite kinetic normalization and its outgoing response. Its scale is $\mathcal T=4\pi Z$, with frequency dependence set by the profile size $\lambda$. Neither $Z$, $\lambda$, a physical electric charge, nor the occupied electron and neutrino modes have been derived from V2's master action. No measured electromagnetic or weak-interaction rate is claimed.

Polarized spatial radiation uses this full spectral measure to calculate outgoing pulse energy and the cancellation condition involving both complex input polarizations.

Verification

retarded_displacement_checks.py compares (4) with finite differences of the full constrained spatial Hessian. It independently checks the spectral moments, source equation, positive Euclidean response and high-frequency bounds. Three radial resolutions, continuum trial-function integrals and outgoing upper-half-plane solves check (17)–(22). An independent regular scattering-state integration checks (23)–(26). The stable Bessel series avoids subtracting two large nearly equal terms near the origin. Reports retain quadrature errors and use timestamped filenames; checks remain active under Python optimization.