Spatial displacement action

Summary: Full spatial deformation gives the exact constant-gradient action and a common leading interaction between two polarized displacement waves, with the complementary field retained explicitly.

Retain both incoming spatial polarizations

Use the declared unit-vector field action of Null spatial pulses:

$$S=\frac Z2\int dt\,dy\,d^2x\, \partial_\mu n\cdot\partial^\mu n,\qquad n^2=1. \tag{1}$$

Its static profile is

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

The coordinates are ordinary spatial directions in one containing manifold; $n$ has three constrained field components. Equation (1) remains a conditional component action to be derived from the master field before any physical photon identification.

Each one-way pulse $n_0(x-X(t\mp y))$ solves the complete field equation. To examine an interaction, retain both real vector profiles:

$$X(t,y)=A(t-y)+B(t+y),\qquad a=A'(t-y),\quad b=B'(t+y). \tag{3}$$

Both $a$ and $b$ have their own two-dimensional spatial polarization. A relative angle is physical data; replacing either vector by its norm before varying the field would discard it.

Compute the omitted spatial deformation

Let $n_i=\partial_i n_0$ and define the tangent second derivative

$$\mathcal H_{ij}=\partial_i\partial_jn_0 +(n_i\cdot n_j)n_0. \tag{4}$$

The static field equation gives $\mathcal H_{11}+\mathcal H_{22}=0$. Substituting the rigidly shifted profile into the complete field equation leaves

$$\mathcal R= (\dot X_i\dot X_j-X_{y,i}X_{y,j})\mathcal H_{ij} =S_{ij}\mathcal H_{ij}, \tag{5}$$

because $\Box_{t,y}X=0$, with

$$\boxed{ S=2\left(ab^T+ba^T-(a\cdot b)I_2\right).} \tag{6}$$

The trace has dropped out using (4). The residual is tangent to the field-value constraint and represents an actual change of the spatial profile.

In two bound directions,

$$\boxed{S^2=4|a|^2|b|^2I_2,\qquad \operatorname{tr}S^2=8|a|^2|b|^2.} \tag{7}$$

Thus a nonzero pair of opposite pulses always has a nonzero quadrupole source. Perpendicular polarizations have $a\cdot b=0$ but still $S\ne0$. For example, $a=e_1,b=e_2$ gives $S=2\begin{pmatrix}0&1\\1&0\end{pmatrix}$; $a=b=e_1$ gives $S=2\operatorname{diag}(1,-1)$. The source orientation changes while its norm stays the same. This is a full-field example of the distinction between a vanishing overlap and a vanishing reciprocal response in Bilateral polarization coupling.

For (2), direct differentiation sharpens (5):

$$\mathcal R =-\frac4{|x|^2+\lambda^2}(Sx)_i n_i, \qquad \boxed{\int|\mathcal R|^2d^2x =\frac{128\pi}{3\lambda^2}|a|^2|b|^2.} \tag{8}$$

Its projection onto each translation tangent vanishes by spatial parity: $\int n_i\cdot\mathcal R\,d^2x=0$. A calculation retaining only $X$ would therefore miss this entire complementary source, just as a projected polarization texture can miss its full field residual.

Complete a uniformly moving and tilted plane exactly

For constant velocity $v$ and slope $w$, put

$$C=I_2+ww^T-vv^T,\qquad C>0,\qquad L=C^{-1/2}. \tag{9}$$

The full profile

$$\boxed{n(t,x,y)=n_0\!\left(L(x-vt-wy)\right)} \tag{10}$$

solves the nonlinear field equation. In fact, $LCL^T=I_2$, so the derivatives of its two spatial arguments have exactly the static orthonormal contraction. The positive-definiteness condition states that their normal plane is spacelike.

This construction includes the required rotation and contraction of the spatial profile. It does not hold that profile rigid while changing its motion.

Integrating the original action over the actual $x$ plane gives

$$\boxed{\mathscr L(v,w) =-\mathcal T\sqrt{\det(I_2+ww^T-vv^T)}.} \tag{11}$$

The factor is the Jacobian $1/\det L$. Equivalently,

$$\det C=(1-|v|^2)(1+|w|^2)+(v\cdot w)^2. \tag{12}$$

Equation (11) is the induced-area action, often called the Nambu–Goto form, here obtained by integrating the explicit full spatial solution. No independent interaction coefficient is assigned to each polarization.

Derive the common quartic term in two ways

Expanding (11) about a straight resting background gives

$$\begin{aligned} \mathscr L ={}&-\mathcal T+\frac{\mathcal T}{2}(|v|^2-|w|^2)\\ &+\frac{\mathcal T}{8}(|v|^2-|w|^2)^2 +\frac{\mathcal T}{2} \left[|v|^2|w|^2-(v\cdot w)^2\right] +O((v,w)^6). \end{aligned} \tag{13}$$

Using $v=a+b,w=-a+b$, its quartic part is

$$\boxed{\mathscr L_4=2\mathcal T|a|^2|b|^2 =\frac2{\mathcal T} |\partial_{u_-}\varphi|^2|\partial_{u_+}\varphi|^2,\qquad \varphi=\sqrt{\mathcal T}\,X,} \tag{14}$$

where $u_-=t-y$ and $u_+=t+y$. Here $\varphi$ is the canonically normalized displacement field. The same $\mathcal T$ fixes propagation normalization and the interaction coefficient.

There is an independent spatial-response derivation of (14). Let $\mathcal J$ be the constrained static energy Hessian divided by $Z$, on tangents perpendicular to $n_0$. Its quadratic form is

$$\langle\eta,\mathcal J\eta\rangle =\int\left[|\nabla\eta|^2-|\nabla n_0|^2|\eta|^2\right]d^2x. \tag{15}$$

The trace-free part of the completed profile's second-order change is

$$\eta=\frac12(Sx)_i n_i,\qquad \mathcal J\eta=-\mathcal R. \tag{16}$$

The trace part is a static size variation and has zero Hessian at this order. Direct integration gives

$$\boxed{ Z\langle\eta,\mathcal R\rangle=-4\mathcal T|a|^2|b|^2,\qquad Z\langle\eta,\mathcal J\eta\rangle=4\mathcal T|a|^2|b|^2.} \tag{17}$$

Substituting this response into $-Z\langle\eta,\mathcal R\rangle -\tfrac Z2\langle\eta,\mathcal J\eta\rangle$ gives $2\mathcal T|a|^2|b|^2$, precisely (14). The complementary spatial field supplies the interaction that the rigid translation projection misses.

Domain of the interaction result

Equation (11) is exact for constant gradients. Equation (14) is its leading local term with four first derivatives, with its coefficient also fixed by the static response calculation. It is not the complete scattering law of arbitrary finite pulses.

The correction (16) has a $1/r$ tail. Its static quadratic-form energy is finite, but its $L^2$ norm grows logarithmically with transverse cutoff. The original sigma model also has a gapless transverse continuum. An arbitrary time-dependent elimination consequently needs boundary conditions and a retarded spatial response; a gapped-mode approximation cannot simply be assumed. A finite collision must retain the additional profile deformation and any emitted field.

The equal-strength result in (7) and (14) is specific to this two-dimensional profile and this component action. It establishes a coherent geometrical interaction calculation with both wave polarizations, but neither a measured electromagnetic coefficient nor an interaction with the occupied electron or neutrino geometry has been derived.

Retarded displacement response constructs that causal spatial inverse and derives its low-frequency logarithm, positive spectral normalization and outgoing radiation.

Polarized pulse collisions solves the local induced-area equations with both full vector waveforms and derives their reciprocal energy-dependent delays.

Verification

null_spatial_wave_checks.py checks the complete finite pulses and moving planes using independent spacetime differences. Physical-space quadrature verifies (11); separate derivative, source and Hessian integrals verify (8) and (15)–(17). Both parallel and perpendicular polarizations are retained. The executable also checks the constant-gradient expansion and its canonical coefficient. Reports have timestamped names and all checks remain active under Python optimization.