Null spatial pulses

Summary: A specified stationary spatial field supports two normalizable displacement modes, exact finite pulses at the underlying wave speed, and circular polarization with a massless-vector angular-momentum ratio.

A constructive propagation test

The scalar Dirac translation branch has two displacement polarizations at bound depth two, but its finite carrier frequency changes the relation between inertia and normal stiffness. A stationary background with identical time and propagation normalizations provides another mechanism to examine.

Specify the following component action as a mathematical benchmark:

$$S[n]=\frac Z2\int dt\,dy\,d^2x\, \left[ |\partial_t n|^2-|\partial_y n|^2 -|\partial_1 n|^2-|\partial_2 n|^2 \right],\qquad n\cdot n=1,\quad Z>0. \tag{1}$$

The two $x$ coordinates are ordinary bound-plane spatial directions; $y$ is an existing perpendicular propagation direction in the same containing manifold. The three real components of $n$ are field values. Their unit-vector constraint adds no physical spatial dimensions.

Equation (1) is not already derived from the master action for $\Psi_\infty$. It tests whether a stationary retained field pattern could supply the missing propagation structure. Any physical use would have to derive this component action, its background and its couplings as parts of that one field. No mediator particle is introduced.

Its complete constrained field equation is

$$\Box n+(\partial_\mu n\cdot\partial^\mu n)n=0, \qquad \Box=\partial_t^2-\partial_y^2-\partial_1^2-\partial_2^2. \tag{2}$$

An explicit spatially localized background

For any $\lambda>0$, take

$$n_0(x)= \frac{(2\lambda x_1,\ 2\lambda x_2,\ |x|^2-\lambda^2)} {|x|^2+\lambda^2}. \tag{3}$$

Direct differentiation gives

$$n_0^2=1,\qquad \partial_i n_0\cdot\partial_j n_0 =\frac{4\lambda^2}{(|x|^2+\lambda^2)^2}\delta_{ij}, \tag{4}$$

and

$$\Delta_x n_0+|\nabla_x n_0|^2n_0=0. \tag{5}$$

Thus (3) solves the full static equation, with finite energy per unit propagation length

$$\mathcal T=\frac Z2\int|\nabla_x n_0|^2d^2x=4\pi Z. \tag{6}$$

This is the familiar unit-winding two-dimensional sigma-model profile. Its mathematical context is the finite-energy spatial construction of Belavin and Polyakov. Equations (3)–(6) give the profile and verification here without requiring that reference.

The background approaches a constant field value at spatial infinity. Its gradient energy is localized; the background itself is not a normalized one-particle wavefunction. The displacement perturbations below do have finite canonical norm.

Exactly two finite-norm position modes in this family

Translations produce tangents $-\partial_i n_0$. Their kinetic Gram matrix is

$$\boxed{Z\int\partial_i n_0\cdot\partial_j n_0\,d^2x =\mathcal T\delta_{ij}.} \tag{7}$$

Both time and $y$ derivatives in (1) use this same matrix. The two linear displacement modes therefore obey

$$\partial_t^2X_i-\partial_y^2X_i=0,\qquad \omega^2=k^2. \tag{8}$$

The size and axial field-phase parameters of (3) are also static moduli, but their infinite-plane kinetic norms diverge logarithmically. For a disk of radius $R$, the size norm divided by $Z$ is

$$\int_{|x|<R}|\partial_\lambda n_0|^2d^2x =4\pi\left[ \log\left(1+\frac{R^2}{\lambda^2}\right) -\frac{R^2}{R^2+\lambda^2} \right]. \tag{9}$$

The axial-phase norm is $\lambda^2$ times (9). By comparison, each translation norm divided by $Z$ is $4\pi R^2/(R^2+\lambda^2)$ and tends to a finite limit. Thus only the two position moduli in this four-parameter, fixed-boundary family give normalizable waves on the infinite transverse plane. This statement does not classify every fluctuation of the sigma model.

Retain the complete finite-amplitude field

Let $u=t-y$ and choose any twice differentiable real two-vector $X(u)$. Then

$$\boxed{n(t,x,y)=n_0(x-X(t-y))} \tag{10}$$

is an exact solution of (2), at arbitrary displacement amplitude. Indeed,

$$\partial_t n=-X_i'\partial_i n_0,\qquad \partial_y n=+X_i'\partial_i n_0, \tag{11}$$

so $\partial_t^2n=\partial_y^2n$ and $|\partial_t n|^2=|\partial_y n|^2$ pointwise. Every acceleration and nonlinear derivative term cancels between $t$ and $y$, leaving exactly (5).

Replacing $t-y$ by $t+y$ gives the opposite direction. More existing normal coordinates may be retained: use $u=t-\widehat k\cdot y$ with $|\widehat k|=1$. The two bound displacement directions remain perpendicular to propagation.

This is the same general null-coordinate mechanism behind arbitrary traveling field deformations discussed by Vachaspati and Vachaspati. The explicit proof above concerns only the stated action.

For $X'$ square-integrable, the excess energy and longitudinal momentum over the stationary background are finite:

$$\boxed{ E_{\rm pulse}=\mathcal T\int|X'(u)|^2du,\qquad P_y=E_{\rm pulse}} \tag{12}$$

for propagation toward $+y$, and $P_y=-E_{\rm pulse}$ in the opposite direction. The transverse momentum must also be retained. When the endpoint limits exist and the momentum integral converges, it is $P_i=\mathcal T[X_i(+\infty)-X_i(-\infty)]$. For a localized pulse returning to the same center, this vanishes and (12) is a null relation for the complete excess energy–momentum. Square-integrability of $X'$ alone does not establish that endpoint condition. The underlying extended background still has positive energy.

The later linked geometric stress calculation obtains an analogous null disturbance and positive supporting inertia from the geometric action used by the current normal carrier, retaining its temporal connection and complete stress.

Linear, elliptical and circular polarization

A monochromatic displacement has

$$X(u)=a\cos(\omega u)+b\sin(\omega u),\qquad a,b\in\mathbb R^2. \tag{13}$$

Parallel $a,b$ give a line; independent $a,b$ give an ellipse; orthogonal equal-length vectors give a circle. Both real amplitudes, their relative phase and the actual spatial orientation are retained.

The orbital angular momentum about the propagation axis follows directly from the spatial stress tensor. The centered profile's angular integral vanishes, leaving

$$L_{12}=\mathcal T\int (X_1\dot X_2-X_2\dot X_1)\,dy. \tag{14}$$

For a circular mode $X=A(\cos\omega u,\pm\sin\omega u)$, averaging per unit propagation length gives

$$\langle E\rangle=\mathcal T A^2\omega^2,\qquad \langle L_{12}\rangle=\pm\mathcal T A^2\omega,\qquad \frac{\langle L_{12}\rangle}{\langle E\rangle} =\pm\frac1\omega. \tag{15}$$

This is the angular-momentum ratio also associated with circular massless vector modes. Here it comes from actual circular displacement of the spatial profile. The ratio alone does not identify the photon's helicity representation or a single-quantum spin. Those require the physical asymptotic state, rotation action and quantization.

The profile is also equivariant under simultaneous coordinate rotation and rotation of its first two field components. That combined symmetry acts pointwise as rotation of $X$. Its component contribution integrates to zero for these translated profiles, consistent with the orbital calculation above.

Coupling still needs the full complementary field

The two normalizable modes, exact speed and handedness are useful propagation data. They do not by themselves give a Maxwell current, electric charges, a Coulomb law, or neutrino selectivity.

Two opposite pulses generally require a further spatial deformation. Spatial displacement action derives that residual and the exact action of uniformly tilted and moving profiles. As with general polarization textures, an equation for selected coordinates alone must not replace the full spatial field equation.

Verification

null_spatial_wave_checks.py checks the complete static field, canonical norms, winding integral and finite-radius tails. Independent spacetime differences verify finite linear, elliptical and circular pulses in both propagation directions. The combined coordinate-plus-component symmetry is checked pointwise, with energy and angular momentum integrated separately. The same diagnostic tests the complementary response needed for opposite pulses.