# Experimental Proposal -- Confirmation of a Dielectric Longitudinal Delay of a Bright Interference Fringe ## Abstract Light propagates more slowly in a transparent dielectric because the electric and magnetic response of the medium increases the effective permittivity and permeability. In the usual electromagnetic description, the propagation speed therefore becomes $$ c_{\mathrm{eff}}=\frac{1}{\sqrt{\varepsilon_{\mathrm{eff}}\mu_{\mathrm{eff}}}} =\frac{c}{n}. $$ The same kind of slowdown follows when two coherent beams are brought into coherent superposition in the recombination phase of a Mach-Zehnder interferometer. The first beam splitter only prepares two bright arm beams. The relevant step appears later, when those beams recombine. These are not the same spatial operation run backward (see Section 3). If each beam is treated as the coherent electromagnetic response seen by the other, then equal-beam constructive interference doubles the fields, quadruples the instantaneous energy density, and gives the same reduced-speed result $$ c_{\mathrm{eff}}=\frac{c}{2}. $$ The standard output reading instead assigns the bright output density $2u_0$ and therefore gives no anomalous delay. The experiment is a direct time-of-flight test between these two readings. --- ## 1. Classic Dielectric Result In a transparent linear dielectric, $$ \mathbf D=\varepsilon_0\mathbf E+\mathbf P, $$ and $$ \mathbf B=\mu_0(\mathbf H+\mathbf M). $$ For a linear response, $$ \mathbf P=\varepsilon_0\chi_e\mathbf E. $$ and $$ \mathbf M=\chi_m\mathbf H. $$ Therefore $$ \mathbf D=\varepsilon_0(1+\chi_e)\mathbf E =\varepsilon_{\mathrm{eff}}\mathbf E, $$ so $$ \varepsilon_{\mathrm{eff}}=\varepsilon_0(1+\chi_e). $$ If the magnetic sector is likewise loaded, $$ \mu_{\mathrm{eff}}=\mu_0(1+\chi_m). $$ In a source-free region, $$ \nabla\times \mathbf E=-\frac{\partial \mathbf B}{\partial t}, \qquad \nabla\times \mathbf H=\frac{\partial \mathbf D}{\partial t}. $$ Substituting the effective constitutive relations gives $$ \nabla\times \mathbf E=-\mu_{\mathrm{eff}}\frac{\partial \mathbf H}{\partial t}, \qquad \nabla\times \mathbf H=\varepsilon_{\mathrm{eff}}\frac{\partial \mathbf E}{\partial t}. $$ Taking the curl of the first equation, and using the second together with $\nabla\cdot\mathbf E=0$, gives the wave equation $$ \nabla^2\mathbf E-\varepsilon_{\mathrm{eff}}\mu_{\mathrm{eff}} \frac{\partial^2\mathbf E}{\partial t^2}=0. $$ Therefore the propagation speed is $$ c_{\mathrm{eff}}=\frac{1}{\sqrt{\varepsilon_{\mathrm{eff}}\mu_{\mathrm{eff}}}}. $$ For symmetric electric and magnetic loading, $$ \chi_e=\chi_m=k, $$ so $$ \varepsilon_{\mathrm{eff}}=\varepsilon_0(1+k), \qquad \mu_{\mathrm{eff}}=\mu_0(1+k), $$ and therefore $$ c_{\mathrm{eff}} =\frac{1}{\sqrt{\varepsilon_0\mu_0(1+k)^2}} =\frac{c}{1+k}. $$ This is the standard reduced-speed result. --- ## 2. Same Structure at Coherent Superposition The same electromagnetic loading structure appears when two coherent beams are coherently superposed in vacuum during recombination. Let one beam have fields $$ \mathbf E_1,\qquad \mathbf H_1, $$ and let the second beam act as a coherent response field proportional to the first: $$ \mathbf E_2=k\mathbf E_1, \qquad \mathbf H_2=k\mathbf H_1. $$ Here the coherent superposition is described in the same form as the dielectric case, but with the second coherent beam supplying the response sector. Define, abbreviating *coherent* by the subscript $\mathrm{coh}$, $$ \mathbf D_{\mathrm{coh}} =\varepsilon_0(\mathbf E_1+\mathbf E_2), \qquad \mathbf B_{\mathrm{coh}} =\mu_0(\mathbf H_1+\mathbf H_2). $$ Then $$ \mathbf D_{\mathrm{coh}} =\varepsilon_0(1+k)\mathbf E_1 =\varepsilon_{\mathrm{coh}}\mathbf E_1, $$ with $$ \varepsilon_{\mathrm{coh}}=\varepsilon_0(1+k), $$ and likewise $$ \mathbf B_{\mathrm{coh}} =\mu_0(1+k)\mathbf H_1 =\mu_{\mathrm{coh}}\mathbf H_1, $$ with $$ \mu_{\mathrm{coh}}=\mu_0(1+k). $$ The same source-free Maxwell form then becomes $$ \nabla\times \mathbf E_1 =-\frac{\partial \mathbf B_{\mathrm{coh}}}{\partial t}, \qquad \nabla\times \mathbf H_1 =\frac{\partial \mathbf D_{\mathrm{coh}}}{\partial t}. $$ So $$ \nabla\times \mathbf E_1 =-\mu_{\mathrm{coh}}\frac{\partial \mathbf H_1}{\partial t}, \qquad \nabla\times \mathbf H_1 =\varepsilon_{\mathrm{coh}}\frac{\partial \mathbf E_1}{\partial t}. $$ Taking the curl again gives $$ \nabla^2\mathbf E_1-\varepsilon_{\mathrm{coh}}\mu_{\mathrm{coh}} \frac{\partial^2\mathbf E_1}{\partial t^2}=0, $$ therefore $$ c_{\mathrm{eff}} =\frac{1}{\sqrt{\varepsilon_{\mathrm{coh}}\mu_{\mathrm{coh}}}} =\frac{1}{\sqrt{\varepsilon_0\mu_0(1+k)^2}} =\frac{c}{1+k}. $$ For equal beams at constructive interference, $$ k=1, $$ so $$ c_{\mathrm{eff}}=\frac{c}{2}. $$ At the bright center of the fringe, where the interference is constructive, the total fields are $$ \mathbf E_{\mathrm{tot}}=2\mathbf E_1, \qquad \mathbf H_{\mathrm{tot}}=2\mathbf H_1, $$ and the instantaneous energy density there is $$ u_{\mathrm{tot}}=4u_1. $$ More generally, across the fringe profile one may write $$ u(x)=4u_1\cos^2\!\left(\frac{\Delta\phi(x)}{2}\right), $$ so the factor $4$ is the bright-center value of the full spatial dependence. That is the tested coherent-overlap result. --- ## 3. Why the Split Phase Is Different A Mach-Zehnder has two distinct phases. At the first beam splitter: - one bright incident beam becomes two bright arm beams - the purpose is to prepare coherent arm beams - no loaded constructive-interference fringe is being claimed At recombination: - two arm beams meet coherently - the fields add or subtract before the ordinary output decomposition is written - the loaded-fringe question is whether the bright raw overlap itself propagates with the reduced speed above Mathematically, the same 50/50 beam-splitter matrix can describe both interfaces, but the occupied input states are different. If $$ \binom{b_1}{b_2} = \frac{1}{\sqrt 2} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \binom{a_1}{a_2}, $$ then the split phase starts from one populated input port, $$ \binom{a_1}{a_2}=\binom{A}{0}, $$ so $$ \binom{b_1}{b_2} = \binom{A/\sqrt 2}{A/\sqrt 2}, $$ which gives two bright arm beams. Recombination starts instead from two populated arm inputs, $$ \binom{a_1}{a_2}=\binom{A}{A}, $$ so $$ \binom{b_1}{b_2} = \binom{\sqrt 2\,A}{0}, $$ which gives one bright output and one dark output. The optical element is the same, but the populated states are different, so the split and recombination phases are not the same spatial decomposition. This also explains why recombination still produces two spatially distributed output arms rather than a single merged beam. The recombiner is a two-port linear map: it sends the two incoming arm fields into the sum and difference output channels. Those channels emerge in different spatial directions, so the recovered field is written across two output arms even when one of them is dark at a bright point. So the slowdown question belongs to recombination, not to the initial split. --- ## 4. Experimental Fork At a "lossless" 50/50 recombiner, the ordinary output fields are $$ \mathbf E_+=\frac{\mathbf E_1+\mathbf E_2}{\sqrt 2}, \qquad \mathbf E_-=\frac{\mathbf E_1-\mathbf E_2}{\sqrt 2}, $$ and likewise for the magnetic fields. Here "lossless" means the splitter is modeled as conserving total optical energy, so all energy is accounted for across the two output arms. Small real losses do not change the core result being tested, and in a well-aligned setup they are negligible compared with the claimed factor-of-two speed difference. The factor $1/\sqrt 2$ appears because the splitter divides the incoming energy into two equal beams. Energy density is a positive quadratic quantity in the field *amplitudes*, so halving the energy means scaling the field amplitudes by $1/\sqrt 2$, not by $1/2$. The same factor is written on both the electric and magnetic fields because they are modeled as vector fields and enter the electromagnetic energy density quadratically. For equal beams at a bright output point, the standard output reading gives $$ u_+=2u_0, \qquad u_-=0. $$ So the experimental fork is: - ordinary output reading: $$ v_{\mathrm{bright}}=v_{\mathrm{ref}} $$ - loaded raw-overlap reading: $$ v_{\mathrm{bright}}=\frac{v_{\mathrm{ref}}}{2} $$ in the equal-beam limit --- ## 5. Experimental Idea Use a coherent source, modulate it, split it into two arms, and recombine the arms so they form stable fringes. Then: 1. isolate one bright interference fringe with an aperture 2. propagate that selected fringe over distance $L$ 3. propagate a matched reference beam over the same distance 4. compare delay slopes $d\tau/dL$ The standard output reading predicts equal slopes. The loaded-fringe reading predicts a larger slope for the bright fringe. For the equal-beam limit, $$ \tau_{\mathrm{bright}}-\tau_{\mathrm{ref}}\approx \frac{L}{c}. $$ So at $1\,\mathrm{m}$ the extra delay is about $3.34\,\mathrm{ns}$, and at $10\,\mathrm{m}$ it is about $33.4\,\mathrm{ns}$. --- ## 6. Detailed Derivation ### 6.1 Dielectric Derivation Start from $$ \mathbf D=\varepsilon_0\mathbf E+\mathbf P, \qquad \mathbf P=\varepsilon_0\chi_e\mathbf E. $$ Then $$ \varepsilon_{\mathrm{eff}}=\varepsilon_0(1+\chi_e). $$ If the magnetic sector is likewise loaded, $$ \mu_{\mathrm{eff}}=\mu_0(1+\chi_m). $$ So $$ c_{\mathrm{eff}}=\frac{1}{\sqrt{\varepsilon_{\mathrm{eff}}\mu_{\mathrm{eff}}}}. $$ For symmetric loading $\chi_e=\chi_m=k$, $$ c_{\mathrm{eff}}=\frac{c}{1+k}. $$ ### 6.2 Instantaneous Constructive-Interference Derivation No time averaging is needed for the basic bright-point argument. At a full constructive-interference point for equal beams, $$ \mathbf E_1(t)=\mathbf E_0(t), \qquad \mathbf E_2(t)=\mathbf E_0(t), $$ and $$ \mathbf B_1(t)=\mathbf B_0(t), \qquad \mathbf B_2(t)=\mathbf B_0(t). $$ Then $$ \mathbf E_{\mathrm{tot}}(t)=2\mathbf E_0(t), \qquad \mathbf B_{\mathrm{tot}}(t)=2\mathbf B_0(t). $$ The instantaneous electromagnetic energy density is $$ u_{\mathrm{tot}}(t) = \frac{\varepsilon_0}{2}\lvert \mathbf E_{\mathrm{tot}}(t)\rvert^2 + \frac{1}{2\mu_0}\lvert \mathbf B_{\mathrm{tot}}(t)\rvert^2. $$ Substituting the doubled fields gives $$ u_{\mathrm{tot}}(t)=4u_0(t). $$ So the bright-point factor of four is already an instantaneous result. ### 6.3 Coherent-Response Form Write the second beam as a coherent response field proportional to the first: $$ \mathbf E_2=k\mathbf E_1, \qquad \mathbf B_2=k\mathbf B_1. $$ Then $$ \mathbf E_{\mathrm{tot}}=(1+k)\mathbf E_1, \qquad \mathbf B_{\mathrm{tot}}=(1+k)\mathbf B_1, $$ so $$ u_{\mathrm{tot}}=(1+k)^2u_1. $$ Applying the same reduced-speed structure as in the dielectric case gives $$ c_{\mathrm{eff}}=\frac{c}{1+k}. $$ For equal beams at constructive interference, $k=1$, hence $$ c_{\mathrm{eff}}=\frac{c}{2}. $$ --- ## 7. What the Experiment Decides This experiment tests which object should be treated as the propagating fringe after recombination: - the ordinary normalized output mode - or the isolated raw constructive-interference fringe If the measured delay matches the reference, the ordinary reading wins. If the measured delay approaches the $c/2$ prediction in the equal-beam limit, the loaded-fringe reading is supported.