% Longitudinal Delay of an Isolated Bright Interference Branch % An M. Rodriguez % 2026-04-21 # Longitudinal Delay of an Isolated Bright Interference Branch ## Abstract This proposal tests whether the bright center of a spatially isolated interference stripe propagates with the same delay as an ordinary reference beam, or whether its effective longitudinal advance is reduced by coherent loading. Two equal coherent beams are recombined at a small angle to produce stable straight fringes. Standard interference gives a spatial redistribution of density: bright regions exceed the two-beam mean while neighboring dark regions fall below it, with the full fringe average conserving the two-beam budget. The experiment isolates a narrow bright stripe and measures its modulation delay over a known distance against a reference beam. The standard expectation is that the selected bright stripe and the reference beam have the same propagation speed, up to ordinary geometric and aperture effects. The loaded-branch expectation is that a bright stripe carrying the available two-beam flux on a higher local density must advance more slowly. For equal beams at a bright center, the predicted speed is approximately half the reference speed. The experiment is therefore a direct time-of-flight test between two readings of the same interference pattern: ordinary output transport and loaded raw-overlap transport. --- # 1. Goal The goal is to measure the longitudinal delay of the center of an isolated bright interference stripe. The experimental question is: $$ \boxed{ \text{Does an isolated bright stripe propagate like ordinary output light, or like a loaded raw-overlap branch?} } $$ The standard expectation is $$ v_{\mathrm{stripe}}=v_{\mathrm{ref}} $$ within experimental error. The loaded-branch expectation is $$ v_{\mathrm{stripe}}0$ is the conventional proportionality factor between squared field amplitude and energy density. Let the relative phase be $$ \Delta\phi=\phi_1-\phi_2. $$ The raw coherent overlap field is $$ f_{\mathrm{raw}}=f_1+f_2. $$ The raw density is $$ u_{\mathrm{raw}}=C|f_{\mathrm{raw}}|^2. $$ For one freely propagating beam in vacuum, the flux magnitude is $$ J_0=u_0c. $$ If a beam travels at angle $\theta$ relative to the longitudinal $z$-axis, its longitudinal projected flux is $$ J_{0,z}=u_0c\cos\theta. $$ For two equal symmetric beams, $$ J_{\mathrm{in},z}=2u_0c\cos\theta. $$ --- # 4. Derivation of the Raw Interference Density This section derives the standard raw overlap density $$ u_{\mathrm{raw}}(x) = 4u_0\cos^2\!\left(\frac{qx}{2}\right). $$ No loaded-branch assumption is used here. ## 4.1 Source-Free Monochromatic Wave Component In a source-free region, each Cartesian component of a transverse electromagnetic plane wave satisfies the scalar wave equation $$ \nabla^2 f-\frac{1}{c^2}\partial_t^2 f=0. $$ A monochromatic plane-wave solution is $$ f(\mathbf r,t) = A e^{i(\mathbf k\cdot\mathbf r-\omega t)}, $$ with dispersion relation $$ \omega=c|\mathbf k|. $$ The physical real field is the real part of this complex representation. The complex expression is a compact way to track phase. Quadratic observables are obtained from the squared amplitude with the chosen normalization. Define $$ u_0=C|A|^2. $$ ## 4.2 Two Symmetric Beams Let two equal beams cross symmetrically about the $z$-axis in the $x$-$z$ plane. Let $\theta$ be the half-angle between each beam and the $z$-axis. Then $$ \mathbf k_1 = k\sin\theta\,\hat{\mathbf x} + k\cos\theta\,\hat{\mathbf z}, $$ and $$ \mathbf k_2 = -k\sin\theta\,\hat{\mathbf x} + k\cos\theta\,\hat{\mathbf z}. $$ Both beams have $$ |\mathbf k_1|=|\mathbf k_2|=k, \qquad \omega=ck. $$ The corresponding fields are $$ f_1(x,z,t) = A e^{i(kx\sin\theta+kz\cos\theta-\omega t)}, $$ and $$ f_2(x,z,t) = A e^{i(-kx\sin\theta+kz\cos\theta-\omega t)}. $$ Define the transverse phase-gradient parameter $$ q:=2k\sin\theta. $$ The relative phase is $$ \Delta\phi(x) = (\mathbf k_1-\mathbf k_2)\cdot\mathbf r = 2k\sin\theta\,x = qx. $$ Thus $q$ is the transverse spatial phase gradient. ## 4.3 Linear Superposition Because the wave equation is linear, $$ f_{\mathrm{raw}}=f_1+f_2 $$ is also a solution. Substitute the two fields: $$ f_{\mathrm{raw}}(x,z,t) = A e^{i(kx\sin\theta+kz\cos\theta-\omega t)} + A e^{i(-kx\sin\theta+kz\cos\theta-\omega t)}. $$ Factor the common longitudinal and temporal phase: $$ f_{\mathrm{raw}}(x,z,t) = A e^{i(kz\cos\theta-\omega t)} \left( e^{ikx\sin\theta}+e^{-ikx\sin\theta} \right). $$ Using $$ e^{ia}+e^{-ia}=2\cos a, $$ with $$ a=kx\sin\theta, $$ gives $$ f_{\mathrm{raw}}(x,z,t) = 2A\cos(kx\sin\theta) e^{i(kz\cos\theta-\omega t)}. $$ Since $$ kx\sin\theta=\frac{qx}{2}, $$ we obtain $$ \boxed{ f_{\mathrm{raw}}(x,z,t) = 2A\cos\!\left(\frac{qx}{2}\right) e^{i(kz\cos\theta-\omega t)}. } $$ ## 4.4 Quadratic Density Readout The raw density is $$ u_{\mathrm{raw}}(x) = C|f_{\mathrm{raw}}(x,z,t)|^2. $$ Using the expression above, $$ |f_{\mathrm{raw}}|^2 = \left| 2A\cos\!\left(\frac{qx}{2}\right) e^{i(kz\cos\theta-\omega t)} \right|^2. $$ The phase factor has unit magnitude: $$ \left|e^{i(kz\cos\theta-\omega t)}\right|^2=1. $$ Therefore $$ |f_{\mathrm{raw}}|^2 = 4|A|^2\cos^2\!\left(\frac{qx}{2}\right). $$ Multiplying by $C$ gives $$ u_{\mathrm{raw}}(x) = 4C|A|^2\cos^2\!\left(\frac{qx}{2}\right). $$ Since $$ u_0=C|A|^2, $$ we get $$ \boxed{ u_{\mathrm{raw}}(x) = 4u_0\cos^2\!\left(\frac{qx}{2}\right). } $$ This is the standard raw overlap density used in the proposal. --- # 5. Equivalent Phase-Difference Derivation The same result follows without specifying the crossing geometry. Let $$ f_1=Ae^{i\phi_1}, \qquad f_2=Ae^{i\phi_2}. $$ Then $$ f_{\mathrm{raw}}=f_1+f_2. $$ The raw density is $$ u_{\mathrm{raw}}=C|f_1+f_2|^2. $$ Expand: $$ |f_1+f_2|^2 = (f_1+f_2)(f_1^*+f_2^*). $$ Thus $$ |f_1+f_2|^2 = |f_1|^2+|f_2|^2+f_1f_2^*+f_1^*f_2. $$ Since $$ |f_1|^2=|f_2|^2=|A|^2, $$ and $$ f_1f_2^* = |A|^2e^{i(\phi_1-\phi_2)}, $$ while $$ f_1^*f_2 = |A|^2e^{-i(\phi_1-\phi_2)}, $$ we have $$ |f_1+f_2|^2 = 2|A|^2 + |A|^2 \left( e^{i\Delta\phi} + e^{-i\Delta\phi} \right), $$ where $$ \Delta\phi:=\phi_1-\phi_2. $$ Using $$ e^{i\Delta\phi}+e^{-i\Delta\phi}=2\cos\Delta\phi, $$ we get $$ |f_1+f_2|^2 = 2|A|^2(1+\cos\Delta\phi). $$ Using $$ 1+\cos\Delta\phi = 2\cos^2\!\left(\frac{\Delta\phi}{2}\right), $$ this becomes $$ |f_1+f_2|^2 = 4|A|^2 \cos^2\!\left(\frac{\Delta\phi}{2}\right). $$ Multiplying by $C$ gives $$ u_{\mathrm{raw}} = 4u_0 \cos^2\!\left(\frac{\Delta\phi}{2}\right). $$ For the symmetric crossing geometry, $$ \Delta\phi(x)=qx. $$ Therefore $$ \boxed{ u_{\mathrm{raw}}(x) = 4u_0 \cos^2\!\left(\frac{qx}{2}\right). } $$ --- # 6. Bright Centers, Nodes, Bounds, and Fringe Period Because $$ 0\le \cos^2\!\left(\frac{qx}{2}\right)\le 1, $$ the raw density satisfies $$ 0\le u_{\mathrm{raw}}(x)\le 4u_0. $$ Bright centers occur where $$ \cos^2\!\left(\frac{qx}{2}\right)=1. $$ This requires $$ \frac{qx}{2}=n\pi, \qquad n\in\mathbb Z. $$ Therefore $$ x_n=\frac{2\pi n}{q}. $$ At those points, $$ u_{\mathrm{raw}}(x_n)=4u_0. $$ Nodes occur where $$ \cos^2\!\left(\frac{qx}{2}\right)=0. $$ This requires $$ \frac{qx}{2} = \frac{(2n+1)\pi}{2}. $$ Thus $$ x_n^{\mathrm{node}} = \frac{(2n+1)\pi}{q}. $$ At those points, $$ u_{\mathrm{raw}}=0. $$ The fringe period $\Lambda$ is the smallest positive shift satisfying $$ q(x+\Lambda)=qx+2\pi. $$ Thus $$ \Lambda=\frac{2\pi}{q}. $$ Since $$ q=2k\sin\theta, \qquad k=\frac{2\pi}{\lambda}, $$ we get $$ \boxed{ \Lambda=\frac{\lambda}{2\sin\theta}. } $$ For small $\theta$, $$ \Lambda\approx\frac{\lambda}{2\theta}. $$ --- # 7. Full-Period Mean Density The full-period average is $$ \langle u_{\mathrm{raw}}\rangle = \frac{1}{\Lambda} \int_0^\Lambda 4u_0\cos^2\!\left(\frac{qx}{2}\right)dx. $$ Let $$ y=\frac{qx}{2}, \qquad dx=\frac{2}{q}dy. $$ When $x=0$, $$ y=0. $$ When $x=\Lambda=2\pi/q$, $$ y=\pi. $$ Therefore $$ \langle u_{\mathrm{raw}}\rangle = \frac{4u_0}{\Lambda} \frac{2}{q} \int_0^\pi\cos^2y\,dy. $$ Since $$ \int_0^\pi\cos^2y\,dy=\frac{\pi}{2}, $$ we obtain $$ \langle u_{\mathrm{raw}}\rangle = \frac{4u_0}{\Lambda} \frac{2}{q} \frac{\pi}{2} = \frac{4\pi u_0}{q\Lambda}. $$ Using $$ \Lambda=\frac{2\pi}{q}, $$ so that $$ q\Lambda=2\pi, $$ we find $$ \boxed{ \langle u_{\mathrm{raw}}\rangle=2u_0. } $$ Thus the raw interference pattern conserves the two-beam density budget over a complete fringe period: $$ u_0+u_0=2u_0. $$ The density is redistributed. It is not created. --- # 8. Forward Flux Budget Each incoming beam has flux magnitude $$ J_0=u_0c. $$ Since each beam makes angle $\theta$ with the $z$-axis, its longitudinal projected flux is $$ J_{0,z}=u_0c\cos\theta. $$ For two equal symmetric beams, $$ \boxed{ J_{\mathrm{in},z}=2u_0c\cos\theta. } $$ Dividing by the full-period mean density gives the mean longitudinal advance rate of the full pattern: $$ \frac{J_{\mathrm{in},z}}{\langle u_{\mathrm{raw}}\rangle} = \frac{2u_0c\cos\theta}{2u_0} = c\cos\theta. $$ Therefore the full fringe-period average has the ordinary longitudinal projected speed: $$ \boxed{ v_{\mathrm{mean}}=c\cos\theta. } $$ No anomalous delay follows from standard interference alone. The anomalous prediction enters only when a selected bright branch is assigned the raw local density while carrying the available two-beam flux. --- # 9. Ordinary Output Reading At a lossless 50/50 recombiner, the output modes are $$ f_+ = \frac{f_1+f_2}{\sqrt2}, \qquad f_- = \frac{f_1-f_2}{\sqrt2}. $$ The factor $1/\sqrt2$ is the usual lossless beam-splitter amplitude factor. It ensures that the total output energy equals the total input energy. The corresponding densities are $$ u_+ = C|f_+|^2 = \frac{C}{2}|f_1+f_2|^2, $$ and $$ u_- = C|f_-|^2 = \frac{C}{2}|f_1-f_2|^2. $$ Using the phase-difference result, $$ C|f_1+f_2|^2 = 4u_0\cos^2\!\left(\frac{\Delta\phi}{2}\right), $$ so $$ u_+ = 2u_0\cos^2\!\left(\frac{\Delta\phi}{2}\right). $$ Similarly, $$ u_- = 2u_0\sin^2\!\left(\frac{\Delta\phi}{2}\right). $$ Therefore $$ \boxed{ u_+(x)+u_-(x)=2u_0. } $$ At a bright center of the $+$ output, $$ u_+=2u_0, \qquad u_-=0. $$ So ordinary output transport assigns the bright output density $2u_0$, not $4u_0$. It can therefore carry the two-beam budget at the usual projected speed $$ c\cos\theta $$ without a conservation conflict. This is the standard null prediction. --- # 10. Loaded Raw-Overlap Branch Reading The tested alternative is: $$ \boxed{ \text{An isolated bright stripe propagates as a raw loaded branch with density }u_{\mathrm{raw}}. } $$ At a bright center, $$ u_{\mathrm{raw,peak}}=4u_0. $$ But the available two-beam longitudinal flux is still $$ J_{\mathrm{in},z}=2u_0c\cos\theta. $$ If the bright branch were assigned speed $c\cos\theta$, its local throughput would be $$ J_{\mathrm{bright,naive}} = u_{\mathrm{raw,peak}}c\cos\theta = 4u_0c\cos\theta. $$ This is twice the available two-beam longitudinal flux: $$ 4u_0c\cos\theta = 2J_{\mathrm{in},z}. $$ Therefore, under the loaded raw-overlap reading, conservation requires a reduced effective advance: $$ J_{\mathrm{in},z} = u_{\mathrm{raw,peak}}v_{\mathrm{peak}}. $$ Solving, $$ v_{\mathrm{peak}} = \frac{J_{\mathrm{in},z}}{u_{\mathrm{raw,peak}}} = \frac{2u_0c\cos\theta}{4u_0} = \frac{c\cos\theta}{2}. $$ Thus $$ \boxed{ v_{\mathrm{peak}}=\frac{c\cos\theta}{2}. } $$ For $\theta\ll1$, $$ \boxed{ v_{\mathrm{peak}}\approx\frac c2. } $$ This is the central experimental prediction. --- # 11. Dielectric-Style Derivation of the Loaded-Branch Law Matter is standing electromagnetic organization. Dielectric slowing is therefore field-field interaction in a stable organized background. The fringe-delay test asks whether the same loading rule appears in the simplest unbound case: two coherent light fields overlapping in phase. ## 11.1 Ordinary Dielectric Form In a dielectric, $$ \mathbf D=\varepsilon_0\mathbf E+\mathbf P. $$ For a linear response, $$ \mathbf P=\varepsilon_0\chi_e\mathbf E. $$ Therefore $$ \mathbf D = \varepsilon_0(1+\chi_e)\mathbf E = \varepsilon_{\mathrm{eff}}\mathbf E, $$ where $$ \varepsilon_{\mathrm{eff}}=\varepsilon_0(1+\chi_e). $$ If the magnetic sector is similarly loaded, $$ \mu_{\mathrm{eff}}=\mu_0(1+\chi_m), $$ then the propagation speed is $$ v = \frac{1}{\sqrt{\varepsilon_{\mathrm{eff}}\mu_{\mathrm{eff}}}}. $$ For symmetric electric and magnetic loading, $$ \chi_e=\chi_m=k, $$ so $$ \varepsilon_{\mathrm{eff}}=\varepsilon_0(1+k), \qquad \mu_{\mathrm{eff}}=\mu_0(1+k). $$ Then $$ v = \frac{1} {\sqrt{\varepsilon_0\mu_0(1+k)^2}} = \frac{c}{1+k}. $$ Thus $$ \boxed{ v_{\mathrm{eff}}=\frac{c}{1+k}. } $$ ## 11.2 Coherent Response Field Let a primary branch have electric field amplitude $$ E_1. $$ Let the coherent response field be proportional: $$ E_2=kE_1, \qquad k\ge0. $$ Then $$ E_{\mathrm{tot}} = E_1+E_2 = (1+k)E_1. $$ Because the density readout is quadratic, $$ u\propto E^2. $$ Therefore $$ u_{\mathrm{tot}} \propto E_{\mathrm{tot}}^2 = (1+k)^2E_1^2. $$ If $$ u_1\propto E_1^2, $$ then $$ \boxed{ u_{\mathrm{tot}}=(1+k)^2u_1. } $$ For equal coherent contribution, $$ k=1, $$ so $$ E_{\mathrm{tot}}=2E_1, $$ and $$ u_{\mathrm{tot}}=4u_1. $$ The corresponding amplitude-loading factor is $$ n_{\mathrm{eff}}=1+k. $$ Therefore $$ u_{\mathrm{tot}}=n_{\mathrm{eff}}^2u_1, $$ and $$ v_{\mathrm{eff}}=\frac{c}{n_{\mathrm{eff}}}. $$ For $k=1$, $$ n_{\mathrm{eff}}=2, $$ so $$ \boxed{ v_{\mathrm{eff}}=\frac c2. } $$ ## 11.3 Flux Form of the Same Result The same law can be written directly as $$ v_{\mathrm{eff}}=\frac{J}{u}. $$ For one branch, $$ J_1=u_1c. $$ For two equal branches in phase, the available total flux is $$ J_{\mathrm{available}}=2u_1c. $$ The joined raw density is $$ u_{\mathrm{joined}}=4u_1. $$ Therefore $$ v_{\mathrm{eff}} = \frac{J_{\mathrm{available}}}{u_{\mathrm{joined}}} = \frac{2u_1c}{4u_1} = \frac c2. $$ This is the same result as the symmetric dielectric loading derivation. The general proportional-response result is $$ f_2=kf_1, $$ so $$ f_{\mathrm{tot}}=(1+k)f_1, $$ $$ u_{\mathrm{tot}}=(1+k)^2u_1, $$ and $$ \boxed{ v_{\mathrm{eff}}=\frac{c}{1+k}. } $$ --- # 12. Spatially Varying Effective Advance Across a Fringe For the raw overlap fringe, $$ u_{\mathrm{raw}}(x) = 4u_0\cos^2\!\left(\frac{qx}{2}\right). $$ The available two-beam longitudinal flux is $$ J_{\mathrm{in},z}=2u_0c\cos\theta. $$ Applying the loaded-branch law locally gives $$ v_{\mathrm{eff}}(x) = \frac{J_{\mathrm{in},z}}{u_{\mathrm{raw}}(x)}. $$ Thus $$ v_{\mathrm{eff}}(x) = \frac{2u_0c\cos\theta} {4u_0\cos^2(qx/2)} = \frac{c\cos\theta} {2\cos^2(qx/2)}. $$ Therefore $$ \boxed{ v_{\mathrm{eff}}(x) = \frac{c\cos\theta} {2\cos^2(qx/2)} } $$ on open bright intervals where the raw branch survives. At a bright center, $$ \cos^2(qx/2)=1, $$ so $$ v_{\mathrm{eff,peak}}=\frac{c\cos\theta}{2}. $$ At points where $$ \cos^2(qx/2)=\frac12, $$ the density equals the mean, $$ u_{\mathrm{raw}}=2u_0, $$ and $$ v_{\mathrm{eff}}=c\cos\theta. $$ At nodes, $$ u_{\mathrm{raw}}=0. $$ The quotient is not interpreted as an infinite speed. A node is not a surviving forward loaded branch. --- # 13. Exact Finite-Band Prediction A real slit samples a finite band around a bright center. Let the selected band be $$ |x-x_n|\le \frac a2. $$ Set $$ \xi=x-x_n. $$ Because $x_n$ is a bright center, $$ u_{\mathrm{raw}}(\xi) = 4u_0\cos^2\!\left(\frac{q\xi}{2}\right). $$ The band-averaged density is $$ \bar u_a = \frac{1}{a}\int_{-a/2}^{a/2} 4u_0\cos^2\!\left(\frac{q\xi}{2}\right)d\xi. $$ Use $$ \cos^2 y=\frac{1+\cos 2y}{2}. $$ Here $$ 2y=q\xi. $$ Thus $$ \bar u_a = \frac{4u_0}{a} \int_{-a/2}^{a/2} \frac{1+\cos(q\xi)}{2}d\xi. $$ So $$ \bar u_a = \frac{2u_0}{a} \left[ \int_{-a/2}^{a/2}1\,d\xi + \int_{-a/2}^{a/2}\cos(q\xi)\,d\xi \right]. $$ The first integral is $$ \int_{-a/2}^{a/2}1\,d\xi=a. $$ The second integral is $$ \int_{-a/2}^{a/2}\cos(q\xi)d\xi = \frac{2}{q}\sin\!\left(\frac{qa}{2}\right). $$ Therefore $$ \boxed{ \bar u_a = 2u_0 \left[ 1+ \frac{2}{qa}\sin\!\left(\frac{qa}{2}\right) \right]. } $$ The predicted band velocity is $$ v_a = \frac{J_{\mathrm{in},z}}{\bar u_a} = \frac{2u_0c\cos\theta} { 2u_0 \left[ 1+ \frac{2}{qa}\sin(qa/2) \right] }. $$ Thus $$ \boxed{ v_a = \frac{c\cos\theta} { 1+ \frac{2}{qa}\sin(qa/2) } }. $$ In the narrow-band limit, $$ qa\ll1, $$ we have $$ \sin(qa/2)\approx qa/2. $$ So $$ \frac{2}{qa}\sin(qa/2)\approx1, $$ and $$ v_a\approx\frac{c\cos\theta}{2}. $$ As the band widens, the sampled density falls toward the two-beam mean and the predicted speed rises toward the ordinary mean value. --- # 14. Practical Fringe Scale The fringe period is $$ \Lambda=\frac{\lambda}{2\sin\theta}. $$ For a HeNe laser, $$ \lambda=632.8\ \mathrm{nm}. $$ If the half-angle is $$ \theta=0.1\ \mathrm{mrad}, $$ then the total crossing angle is $0.2\ \mathrm{mrad}$ and $$ \Lambda \approx \frac{632.8\times10^{-9}} {2\times10^{-4}} = 3.164\times10^{-3}\ \mathrm{m}. $$ So $$ \Lambda\approx3.16\ \mathrm{mm}. $$ The region satisfying $$ u_{\mathrm{raw}}>3u_0 $$ obeys $$ 4u_0\cos^2\!\left(\frac{q\xi}{2}\right)>3u_0. $$ Canceling $u_0$ gives $$ \cos^2\!\left(\frac{q\xi}{2}\right)>\frac34. $$ Thus $$ \left|\frac{q\xi}{2}\right|<\frac{\pi}{6}. $$ So $$ |\xi|<\frac{\pi}{3q}. $$ The full above-$3u_0$ width is $$ w_{>3u_0} = \frac{2\pi}{3q} = \frac{\Lambda}{3}. $$ For $$ \Lambda\approx3.16\ \mathrm{mm}, $$ we get $$ w_{>3u_0}\approx1.05\ \mathrm{mm}. $$ A practical slit width is therefore $$ a=0.25\ \mathrm{mm} \quad\text{to}\quad a=0.50\ \mathrm{mm}, $$ centered on a bright maximum. --- # 15. Experimental Design ## 15.1 Apparatus Minimum requirements: - stable coherent laser source, - amplitude modulation source, - Mach-Zehnder or equivalent two-beam interferometer, - controlled small recombination angle, - stable straight fringes, - spatial filter or slit selecting one bright ridge, - reference beam path, - equal downstream propagation length $L$ for reference and selected stripe, - fast photodiodes, - oscilloscope, lock-in amplifier, or phase-delay measurement system, - variable path length or multiple known path lengths. ## 15.2 Procedure 1. Generate a coherent laser beam. 2. Apply sinusoidal amplitude modulation at angular frequency $\Omega$. 3. Split the beam into two equal arms. 4. Recombine the arms at small symmetric angle $\theta$. 5. Produce stable straight fringes. 6. Select the center of one bright fringe using a slit or spatial filter. 7. Let the selected bright stripe propagate over known distance $L$. 8. In parallel, propagate a reference beam over the same distance. 9. Detect both signals using matched photodiodes. 10. Measure the modulation phase delay of each channel relative to the common modulation source. 11. Repeat for several propagation lengths $L_i$. 12. Fit delay versus distance. For each channel, $$ \tau(L)=mL+b. $$ Here $b$ is a fixed electronic and geometric offset. The slope is $$ m=\frac{d\tau}{dL}. $$ The measured speed is $$ v=\frac{1}{m}. $$ Thus $$ v_{\mathrm{ref}}=\frac{1}{m_{\mathrm{ref}}}, \qquad v_{\mathrm{stripe}}=\frac{1}{m_{\mathrm{stripe}}}. $$ ## 15.3 Predictions Standard prediction: $$ m_{\mathrm{stripe}}\approx m_{\mathrm{ref}}, $$ so $$ v_{\mathrm{stripe}}\approx v_{\mathrm{ref}}. $$ Loaded-branch prediction at a narrow bright center: $$ v_{\mathrm{stripe}}\approx\frac{c\cos\theta}{2}. $$ For small $\theta$, $$ v_{\mathrm{stripe}}\approx\frac c2. $$ Therefore $$ m_{\mathrm{stripe}}\approx\frac{2}{c}, $$ while $$ m_{\mathrm{ref}}\approx\frac{1}{c}. $$ So $$ \boxed{ m_{\mathrm{stripe}}\approx2m_{\mathrm{ref}}. } $$ Equivalently, for equal propagation distance $L$, $$ \boxed{ \tau_{\mathrm{stripe}}-\tau_{\mathrm{ref}}\approx\frac{L}{c}. } $$ At $$ L=1\ \mathrm{m}, $$ the predicted extra delay is $$ \Delta\tau\approx3.34\ \mathrm{ns}. $$ At $$ L=10\ \mathrm{m}, $$ the predicted extra delay is $$ \Delta\tau\approx33.4\ \mathrm{ns}. $$ This is large enough to test with ordinary fast photodiodes and phase-sensitive electronics. --- # 16. Controls ## 16.1 Reference Beam The reference beam should have: - the same carrier frequency, - the same modulation frequency, - the same optical components where possible, - the same downstream length $L$, - similar detector electronics. ## 16.2 Single-Beam Slit Control Send only one beam through the same slit geometry, with no interference. The selected beam should propagate at the ordinary reference speed. This checks that the slit itself is not producing the predicted delay. ## 16.3 Off-Bright Sampling Move the slit away from the bright center. The loaded-branch prediction varies with sampled density: $$ v_a = \frac{J}{\bar u_a}. $$ Thus lower-density bands should show smaller delay. The standard prediction remains no density-dependent delay. ## 16.4 Fringe-Period Averaging Open the aperture to collect an entire fringe period. The average density returns to $$ \langle u_{\mathrm{raw}}\rangle=2u_0. $$ The loaded-branch anomaly should disappear under full-period collection. This distinguishes local branch loading from ordinary total-power conservation. ## 16.5 Power Scaling The prediction depends on coherent loading ratio, not absolute power, in the proportional regime. Reducing both beam powers equally should reduce signal amplitude but not change the predicted velocity ratio. If the delay depends strongly on absolute optical power, thermal, detector, or nonlinear-medium effects must be suspected. --- # 17. Interpretation of Outcomes ## 17.1 Null Result If $$ v_{\mathrm{stripe}}=v_{\mathrm{ref}} $$ within experimental error, then the isolated stripe behaves as ordinary propagated output light. The raw local $4u_0$ density does not act as a loaded branch with reduced effective advance. ## 17.2 Positive Result If $$ v_{\mathrm{stripe}}\approx\frac{v_{\mathrm{ref}}}{2} $$ for a narrow bright-center stripe, and if the effect tracks the sampled density as predicted, then the loaded-branch law is supported. The result would indicate that coherent local density loading changes the effective longitudinal advance of a surviving branch. ## 17.3 Intermediate Result If $$ v_{\mathrm{ref}}>v_{\mathrm{stripe}}>\frac{v_{\mathrm{ref}}}{2}, $$ then the selected stripe may not remain a pure raw bright branch. Partial mixing, diffraction, finite aperture averaging, imperfect visibility, or incomplete branch isolation may contribute. In that case the measured velocity should be compared against the finite-band prediction $$ v_a = \frac{c\cos\theta} { 1+ \frac{2}{qa}\sin(qa/2) }. $$ --- # 18. Theoretical Fork Both readings conserve total energy. The disagreement is not about energy conservation. It is about how a selected bright region transports energy. ## 18.1 Ordinary Output Transport The recombiner output densities satisfy $$ u_+(x)+u_-(x)=2u_0. $$ At a bright center, $$ u_+=2u_0. $$ No speed reduction is required. ## 18.2 Loaded Raw-Overlap Transport The selected bright stripe is identified with the raw overlap density: $$ u_{\mathrm{raw,peak}}=4u_0. $$ The available two-beam flux is $$ J=2u_0c\cos\theta. $$ Therefore $$ v_{\mathrm{peak}} = \frac{J}{u_{\mathrm{raw,peak}}} = \frac{c\cos\theta}{2}. $$ The experiment tests which reading describes the propagation of an isolated bright stripe. --- # 19. Summary This proposal establishes: 1. Two equal coherent beams produce the raw density profile $$ u_{\mathrm{raw}}(x)=4u_0\cos^2\!\left(\frac{qx}{2}\right). $$ 2. The full fringe average is $$ \langle u_{\mathrm{raw}}\rangle=2u_0, $$ so the two-beam budget is conserved over a full period. 3. Bright centers have $$ u_{\mathrm{raw,peak}}=4u_0. $$ 4. The two-beam longitudinal flux budget is $$ J_{\mathrm{in},z}=2u_0c\cos\theta. $$ 5. The loaded-branch law is $$ v_{\mathrm{eff}}=\frac{J}{u}. $$ 6. Therefore the bright-center prediction is $$ v_{\mathrm{peak}} = \frac{2u_0c\cos\theta}{4u_0} = \frac{c\cos\theta}{2}. $$ 7. The dielectric-style proportional response gives the same result: $$ E_2=kE_1 \Rightarrow E_{\mathrm{tot}}=(1+k)E_1 \Rightarrow u_{\mathrm{tot}}=(1+k)^2u_1 \Rightarrow v_{\mathrm{eff}}=\frac{c}{1+k}. $$ For $k=1$, $$ v_{\mathrm{eff}}=\frac c2. $$ 8. A longitudinal delay measurement of an isolated bright stripe directly distinguishes the two models. The decisive experimental signature is $$ \boxed{ m_{\mathrm{stripe}}\approx2m_{\mathrm{ref}} } $$ for a narrow bright-center stripe, where $m=d\tau/dL$ is the measured delay slope. --- # Appendix A — Compact Derivation of the Bright-Core Width The above-$3u_0$ condition is $$ 4u_0\cos^2\!\left(\frac{q\xi}{2}\right)>3u_0. $$ Canceling $u_0$, $$ \cos^2\!\left(\frac{q\xi}{2}\right)>\frac34. $$ Thus $$ \left|\frac{q\xi}{2}\right|<\frac{\pi}{6}. $$ Therefore $$ |\xi|<\frac{\pi}{3q}. $$ The full width is $$ w_{>3u_0}=\frac{2\pi}{3q}. $$ Since $$ \Lambda=\frac{2\pi}{q}, $$ we get $$ \boxed{ w_{>3u_0}=\frac{\Lambda}{3}. } $$ For $\Lambda\approx3.16\ \mathrm{mm}$, $$ w_{>3u_0}\approx1.05\ \mathrm{mm}. $$ A slit width of $$ 0.25\ \mathrm{mm}\le a\le0.50\ \mathrm{mm} $$ samples the high-density bright core while avoiding neighboring low-density regions. --- # Appendix B — Finite-Band Velocity Formula For a centered slit of width $a$ around a bright maximum, $$ \bar u_a = 2u_0 \left[ 1+ \frac{2}{qa}\sin\!\left(\frac{qa}{2}\right) \right]. $$ Therefore $$ v_a = \frac{2u_0c\cos\theta}{\bar u_a}. $$ So $$ \boxed{ v_a = \frac{c\cos\theta} { 1+ \frac{2}{qa}\sin(qa/2) } }. $$ Limits: For $qa\ll1$, $$ v_a\to\frac{c\cos\theta}{2}. $$ For wider averaging, the sampled density decreases and $v_a$ increases toward the mean-pattern value. --- # Appendix C — Status of the Claim The derivations of the interference pattern, fringe average, output densities, and longitudinal flux budget are standard wave calculations. They prove: $$ u_{\mathrm{raw}}(x) = 4u_0\cos^2\!\left(\frac{qx}{2}\right), $$ $$ \langle u_{\mathrm{raw}}\rangle=2u_0, $$ and $$ J_{\mathrm{in},z}=2u_0c\cos\theta. $$ They do not by themselves prove reduced speed. The reduced speed follows from the tested loaded-branch transport law $$ v_{\mathrm{eff}}=\frac{J}{u}. $$ Thus the proposal is not merely a reinterpretation of a screen fringe. It is a direct time-of-flight test between two alternatives: $$ \boxed{ \text{ordinary output transport} } $$ and $$ \boxed{ \text{loaded raw-overlap transport with }v_{\mathrm{eff}}=J/u. } $$