# 7b. Self-Refraction Chapter 7 recovered source-free transport, and Chapter 7a resolved that transport into two complementary transverse aspects. What remains is to understand how a single field can bend its own path. No second substrate is required. Distinct portions of the same transporting field interact when they are realized in a common overlap region. The question is how this overlap modifies transport. ## Dielectric loading as the known case The familiar laboratory form of this effect is dielectric slowing. In ordinary electromagnetic notation, a probe field in a dielectric is read together with the in-phase response of the medium. For a proportional response with amplitude ratio $k$ in both transverse aspects, write $$ \mathbf E_2 = k\mathbf E_1, \qquad \mathbf H_2 = k\mathbf H_1. $$ Then $$ \mathbf D = \varepsilon_0(\mathbf E_1+\mathbf E_2) = \varepsilon_0(1+k)\mathbf E_1 := \varepsilon_{\mathrm{eff}}\mathbf E_1, $$ and $$ \mathbf B = \mu_0(\mathbf H_1+\mathbf H_2) = \mu_0(1+k)\mathbf H_1 := \mu_{\mathrm{eff}}\mathbf H_1. $$ The source-free wave speed of the loaded field is therefore $$ c_{\mathrm{eff}} = \frac{1}{\sqrt{\varepsilon_{\mathrm{eff}}\mu_{\mathrm{eff}}}} = \frac{1}{\sqrt{\varepsilon_0\mu_0(1+k)^2}} = \frac{c}{1+k}. $$ This is the conventional two-aspect writing of the same compression rule. The response field is not an extra primitive substrate; it is the in-phase contribution that loads the transport. The coherent-overlap step below writes the same rule for the case in which the in-phase response is supplied by another phase-locked portion of the same transporting flow rather than by organized matter. ## Coherent overlap as field-pattern addition The energy density at each point is expressed as the square of an amplitude-like quantity: $$ u = |f|^2. $$ For a monochromatic ray, the phase is the optical phase along the ray, $$ \phi = ks-\omega t+\phi_0, $$ not the joining angle of two beams. The corresponding field value may be written $$ f = A e^{i\phi}=A(\cos\phi+i\sin\phi), $$ so $$ u=|f|^2=A^2(\cos^2\phi+\sin^2\phi). $$ The split $$ \cos^2\phi+\sin^2\phi=1 $$ is first a trigonometric identity, rooted in geometry. In field language it appears as two quadrature projections of one oscillation. As an energy-density expression it is a geometric decomposition of one transported mode. Let two coherent contributions of the same transporting flow be $$ f_1, \qquad f_2. $$ When they join, the amplitudes add first: $$ f_{\mathrm{join}} = f_1 + f_2. $$ The recovery geometry can selectively make coincident the quadrature projections from each branch. What is recovered is therefore determined by field-value addition before squaring. So the joined local energy density is $$ u_{\mathrm{join}} = |f_{\mathrm{join}}|^2 = |f_1+f_2|^2 = |f_1|^2 + |f_2|^2 + 2|f_1||f_2|\cos\delta, $$ where $\delta$ is the relative phase between the two flows, $f_1$ and $f_2$. Writing $$ u_1 := |f_1|^2, \qquad u_2 := |f_2|^2, $$ gives $$ u_{\mathrm{join}} = u_1 + u_2 + 2\sqrt{u_1u_2}\cos\delta. $$ The final term is the interference term. It is not a correction to arithmetic; it is the pointwise overlap of the two field-value patterns. If the patterns are incoherent or orthogonal, this term averages away. If they are equal and in phase, it contributes positively everywhere on the joined branch. For equal contributions, $$ u_1 = u_2 = u_0, $$ the joined local density becomes $$ u_{\mathrm{join}} = 2u_0(1+\cos\delta). $$ Therefore $$ 0 \le u_{\mathrm{join}} \le 4u_0. $$ At first sight, both endpoints are surprising: coherent overlap seems to produce a fourfold local energy density for free, while destructive overlap seems to erase the joined energy altogether. But in all cases energy is conserved, so these endpoints have to be interpreted through continuity on the full overlap region, not by the local density value alone. For two equal inputs, the incoming two-stream budget is still the original $u_0 + u_0 = 2u_0$ before any particular recovery geometry is considered. ## Collinear flux mismatch and selected-branch compression The cleanest case is collinear recovery. There is no fringe pattern and no transverse $\cos^2+\sin^2$ redistribution to average over. Two equal in-phase channels are brought back onto one line. The density change is ordinary wave-pattern addition. Two patterns of field values are added point by point. In-phase values reinforce, out-of-phase values cancel, and shifted patterns generate alternating dense and sparse regions. Thus two equal in-phase contributions do not merely sit side by side as $u+u$ once they are recovered on one line. Their joined field value is $\sqrt u+\sqrt u$, so its density is $$ (\sqrt u+\sqrt u)^2 = 4u. $$ This is no more mysterious than a moire pattern: two static patterns can be superposed to generate denser and sparser visible regions without creating new ink. Coherent waves do the dynamical version. The physical question is then whether the available flux can support that value at the original advance rate. In the laboratory version, let the original laser channel carry energy density $2u$ and advance at $c$. Its flux is $$ J_{\mathrm{in}} = 2u c. $$ After a 50/50 split, the two arms carry $$ 2u c \longrightarrow u c + u c. $$ The split changes the spatial distribution, not the available forward flux. The total available flux remains $$ J_{\mathrm{available}} = 2u c. $$ At collinear constructive recovery, the two coherent contributions add in amplitude on the same outgoing line. Relative to either arriving contribution, the local recovered density is $$ u_{\mathrm{bright}} = 4u. $$ If that recovered branch were assigned the unchanged advance $c$, it would require the flux $$ J_{\mathrm{bright}} = u_{\mathrm{bright}}c = 4u c. $$ But the laser supplies only $2u c$ into the two-arm recovery. In the collinear case there is only one positive recovery channel and no second place with $\mathbf S>0$ that can provide the missing flux. The two split channels can be re-phased into one recovered branch, but the same input cannot feed a $4u$ branch moving at $c$. So the forbidden combination is precise: $$ u_{\mathrm{bright}}=4u \qquad\text{and}\qquad c_{\mathrm{eff}}=c. $$ At most one of those can survive as the description of the recovered branch. If the branch really recovers the $4u$ energy density as its transported density, its advance must slow. If the branch is forced to advance at $c$, then the $4u$ value cannot be sustained as the transported density of that branch. Continuity therefore fixes the effective advance by $$ J_{\mathrm{available}} = u_{\mathrm{bright}}c_{\mathrm{eff}}, $$ so $$ 2u c = 4u c_{\mathrm{eff}}, \qquad c_{\mathrm{eff}} = \frac{c}{2}. $$ The slowdown is not an added mechanism. It is the forced consequence of maintaining the $4u$ density while only $2u c$ of flux is supplied. The small-angle fringe version is a practical variant, not the conceptual core. When the two outgoing rays cross at a nonzero angle, the same accounting is spread across a spatial pattern. The full $\cos^2+\sin^2$ pattern can still carry the original mean flux at the original speed by redistribution. The loaded-branch question then concerns a selected bright part of that pattern, which is the local version of the collinear recovery branch. ## Finite overlap region The physical explanation starts from continuity on a finite overlap region $\Omega$: $$ \frac{d}{dt}\int_{\Omega} u\,dV = -\oint_{\partial\Omega}\mathbf J\cdot\mathbf n\,dS. $$ No primitive source or sink is allowed. So whatever energy enters the overlap region must either - leave through some part of its boundary, or - remain transiently stored inside the region. Over a fixed interval $\Delta t$, let one thin inflow of cross section $A$ occupy the realized transport volume $$ V_0 := A c\,\Delta t. $$ If two equal inflows enter the overlap region, then before interaction the incoming budget is $$ \mu_{\mathrm{in}} = 2E, \qquad V_{\mathrm{in}} = 2V_0. $$ So the incoming mean energy density is $$ \bar u_{\mathrm{in}} = \frac{\mu_{\mathrm{in}}}{V_{\mathrm{in}}} = \frac{2E}{2V_0} = \frac{E}{V_0}. $$ ## Constructive recombination If the two inflows are in phase, $$ \delta = 0, $$ then the joined local density reaches $$ u_{\mathrm{join}} = 4u_0. $$ In the constructive recovery geometry there is exactly one local positive forward channel: the joined branch. There is no second local place where the same branch has $\mathbf S>0$. The two inflows are therefore recovered on one loaded branch over the same interval. The realized transport volume falls from $2V_0$ to $V_0$ while the carried content remains $2E$: $$ \mu_{\mathrm{out}} = 2E, \qquad V_{\mathrm{out}} = V_0. $$ Since in this case $$ \mu_{\mathrm{out}} = \mu_{\mathrm{in}} = 2E, \qquad V_{\mathrm{out}} = V_0. $$ The mean energy density on that loaded branch is $$ \bar u_{\mathrm{out}} = \frac{\mu_{\mathrm{out}}}{V_{\mathrm{out}}} = \frac{2E}{V_0} = 2\bar u_{\mathrm{in}}. $$ This is the exact *compression* statement: the mean energy density doubles because the same transported content is recovered on half the transport volume. ## Destructive overlap If instead the two inflows are out of phase, $$ \delta = \pi, $$ then the joined local density reaches $$ u_{\mathrm{join}} = 0. $$ This needs interpretation. The continuity equation tells us that the incoming energy cannot simply disappear, but the joined forward branch now contributes no local density. The correct conclusion is not that a sink has appeared. It is that this local joined forward branch is absent. At the dark branch, $\mathbf S=0$ for the surviving forward channel. The incoming content is recovered only where the continuous pattern has positive outgoing flux. In a spatial fringe, that recovery occurs at the neighboring bright parts of the same pattern. If no joined branch is formed, the flows separate again and recover the original two-branch distribution. The dark branch itself is not a recovery channel; it is the local absence of the forward joined branch. This is also the place to distinguish the amplitude-like quantity from the recovered energy density. The sign or phase can reverse in $f$, but the recovered energy density $$ u = |f|^2 $$ does not become negative. It can only vanish at a node and then reappear elsewhere. That spatial reappearance becomes explicit when the relative phase varies across the overlap region. Let two equal contributions have the form $$ f_1(x) = A e^{ikx}, \qquad f_2(x) = A e^{-ikx}. $$ Then $$ f_{\mathrm{join}}(x) = f_1(x)+f_2(x) = 2A\cos(kx), $$ so the recovered energy density is $$ u(x) = |f_{\mathrm{join}}(x)|^2 = 4|A|^2\cos^2(kx). $$ Writing $$ u_0 := |A|^2, $$ this becomes $$ u(x) = 4u_0\cos^2(kx). $$ Therefore $$ u(x)=0 \qquad \text{at} \qquad x=\frac{(2n+1)\pi}{2k}, $$ while $$ u(x)=4u_0 \qquad \text{at} \qquad x=\frac{n\pi}{k}. $$ The amplitude-like quantity $f_{\mathrm{join}}(x)$ changes sign across each node because $\cos(kx)$ changes sign there. The energy density does not invert. It vanishes at the node and reappears at separated antinodes. Over one spatial period $$ \lambda = \frac{2\pi}{k}, $$ the mean recovered energy density is $$ \frac{1}{\lambda}\int_0^\lambda u(x)\,dx = \frac{1}{\lambda}\int_0^\lambda 4u_0\cos^2(kx)\,dx = 2u_0. $$ So the two-stream mean budget is recovered exactly, even though some points are dark. The local zero is therefore not a sink. It is one part of a spatially redistributed energy density pattern. This example does not realize $\delta=\pi$ at every point. It shows the generic case in which local destructive points are paired with local constructive points, and the two-stream budget is recovered by spatial redistribution. In the second case, if no new local recompression occurs, the natural recovery is simply $$ u_0 + u_0 = 2u_0. $$ So destructive overlap does not create a contradiction. It proves only that an everywhere-dark joined branch has no positive recovery channel. If a finite overlap region produced zero joined density everywhere and no positive outgoing flux anywhere in the continuing pattern, that region would act as a sink and would therefore be forbidden. To obtain a fresh `4u_0` density after destructive overlap, the transport must undergo a new constructive recombination into one branch. ## Effective advance Transport along each realized path still proceeds at the local rate $c$. What changes is not the microscopic transport speed but the coarse-grained lag of a branch that recovers two inflows on one realized transport volume over the same interval. In the constructive case above, the loaded branch carries $$ \mu_{\mathrm{out}} = 2E $$ on $$ V_{\mathrm{out}} = V_0, $$ whereas the same total transported content had previously been distributed over $$ V_{\mathrm{in}} = 2V_0. $$ Equivalently, $$ \bar u_{\mathrm{out}} = 2\bar u_{\mathrm{in}}. $$ That doubled mean energy density is the local compression statement. Equivalently, the branch obeys the flux-density relation $$ J = u\,c_{\mathrm{eff}}, $$ so when $J$ is fixed and $u$ rises, $c_{\mathrm{eff}}$ must fall. The effective advance is only a compact summary of the corresponding lag on that loaded branch. If one wants a compact summary of that lag without carrying the full inflow and outflow bookkeeping every time, it is convenient to write an effective advance rate $$ c_{\mathrm{eff}} < c $$ and a corresponding effective index $$ n_{\mathrm{eff}} := \frac{c}{c_{\mathrm{eff}}} > 1. $$ On a locally surviving loaded branch, a convenient summary is $$ n_{\mathrm{eff}} := \frac{\bar u_{\mathrm{out}}}{\bar u_{\mathrm{in}}}, \qquad c_{\mathrm{eff}} = \frac{c}{n_{\mathrm{eff}}}. $$ So in the constructively recombined case, $$ \bar u_{\mathrm{out}} = 2\bar u_{\mathrm{in}} \qquad\Longrightarrow\qquad c_{\mathrm{eff}} = \frac{c}{2}. $$ This summary applies only to a branch that actually survives the overlap. When the forward joined branch vanishes in destructive overlap, there is no speed to assign to that local branch. The positive recovery is read only in the nonzero parts of the continuing pattern, or in the re-separated beams if no joined branch is formed. ## Effective index The effective index is therefore not a second mechanism. It is the compact way to write that, on locally surviving loaded branches, more strongly superposed regions lag more strongly than weakly loaded ones. ## Local bending If different regions of a wavefront have different energy density, then they do not advance equally. - higher energy density implies lower $c_{\mathrm{eff}}$, - lower energy density implies higher $c_{\mathrm{eff}}$. So one side of the wavefront lags behind the other. That lag bends the transport toward the more strongly loaded side. This is refraction. No external medium is required. The field bends because different parts of the same flow are superposed and compressed by different amounts. ## Operational test The laboratory test is direct because ordinary interference already gives the energy density profile. A flux detector placed at a screen measures the flux of the superposed fields arriving at that point. The cleanest delay version is collinear. The two recovered rays are brought back onto the same line and phase, so the whole recovered channel is bright. There is no transverse fringe average to appeal to. The available flux is $2u_0c$, while the coherent density is $4u_0$. The loaded-branch prediction is therefore $$ c_{\mathrm{eff,bright}}=\frac{2u_0c}{4u_0}=\frac{c}{2}. $$ A small nonzero crossing angle is useful when one wants a visible bright stripe to select. For two equal coherent rays leaving the final recombination region and crossing symmetrically about the $z$ direction, let $\theta$ be the angle each ray makes with the $z$ axis. At a screen coordinate $x$, one ray has transverse phase $+kx\sin\theta$ and the other has transverse phase $-kx\sin\theta$. Their longitudinal phase is the same: $$ f_1(x,z,t) = A e^{i(kx\sin\theta+kz\cos\theta-\omega t)}, $$ $$ f_2(x,z,t) = A e^{i(-kx\sin\theta+kz\cos\theta-\omega t)}. $$ The opposite signs are not opposite optical phases. They are only the opposite transverse components of the two angled rays. Both beams have the same longitudinal phase $kz\cos\theta-\omega t$. Their relative phase across the screen is therefore $$ \Delta\phi(x)=2kx\sin\theta. $$ In the collinear limit $\theta=0$, the relative phase does not vary across the screen. The result is one recovered bright beam rather than a spatial fringe pattern. With $$ u_0 := |A|^2, $$ the joined density is $$ u(x) = |f_1+f_2|^2 = 4u_0\cos^2(kx\sin\theta). $$ The bright centers therefore have $$ u_{\mathrm{bright}} = 4u_0, $$ while the period average remains $$ \langle u\rangle = 2u_0. $$ The forward incoming flux density is $$ J_{z,\mathrm{in}} = 2u_0 c\cos\theta. $$ On the full fringe average this gives the ordinary projected advance $c\cos\theta$. But if a narrow bright stripe is isolated and propagated as the surviving positive channel, the loaded-branch law gives $$ c_{\mathrm{eff,bright}} = \frac{J_{z,\mathrm{in}}}{u_{\mathrm{bright}}} = \frac{2u_0 c\cos\theta}{4u_0} = \frac{c\cos\theta}{2}. $$ For nearly collinear beams this is $c/2$. The experiment is therefore not "does a screen show a bright fringe?" That part is ordinary interference. The experiment is: > Does an isolated bright stripe, once selected as the surviving positive > channel, acquire the larger longitudinal delay implied by > $c_{\mathrm{eff}}=J/u$? Equivalently, the experiment asks which part of the forbidden pair fails. Either the isolated bright channel keeps the $4u_0$ energy density and slows, or a channel that advances at the ordinary speed cannot sustain the $4u_0$ energy density as an isolated transported branch. A collinear delay implementation is: 1. split one coherent source into two equal arms, 2. recombine the arms collinearly and in phase, 3. propagate the recovered bright channel over a known length $L$, 4. propagate a matched reference beam over the same length, 5. compare modulation or pulse delay. A small-angle stripe implementation is: 1. form a stable two-beam fringe pattern, 2. isolate a narrow region around a bright center, 3. propagate that selected stripe over a known length $L$, 4. propagate a matched reference beam over the same length, 5. compare modulation or pulse delay. The reference prediction is $$ t_{\mathrm{ref}}=\frac{L}{c}, $$ while the bright-stripe prediction for $\theta\ll1$ is $$ t_{\mathrm{bright}}\approx\frac{2L}{c}. $$ So the extra delay is $$ \Delta t \approx \frac{L}{c}, $$ about $3.34\,\mathrm{ns}$ per meter. A geometric version sends the selected bright stripe to a glass boundary. If the bright branch carries effective index $n_{\mathrm{eff}}=2$, Snell's law at a boundary with glass index $n_g\approx1.5$ gives $$ n_{\mathrm{eff}}\sin\theta_i=n_g\sin\theta_r. $$ The loaded branch then has a critical angle $$ \sin\theta_c=\frac{n_g}{n_{\mathrm{eff}}}\approx0.75, \qquad \theta_c\approx48.6^\circ. $$ Above that incidence angle the selected bright stripe should not transmit into the glass, while an ordinary reference beam at the same angle should still transmit. That is the clean refraction form of the same loaded-branch claim. Opening the aperture to average over a full fringe period should erase the loaded-branch signature, because the measured energy density returns to the two-beam mean $2u_0$. The prediction belongs to the isolated bright channel, not to the unselected fringe pattern as a whole. ## Retarded self-overlap In a self-interacting configuration, the overlap is not produced by two independent laboratory beams. It is produced when a later portion of the same flow enters a region already shaped by an earlier portion. Let $\gamma(s)$ be a transport line. Then a point at $s$ interacts with earlier positions $s_{\mathrm{ret}}$ satisfying $$ |\gamma(s)-\gamma(s_{\mathrm{ret}})| = c\,(t-t_{\mathrm{ret}}). $$ For a harmonic field, $$ E(s,t) = \Re\!\left[\widetilde E(s)e^{-i\omega t}\right], $$ the retarded contribution is $$ E_{\mathrm{ret}}(s,t) = \Re\!\left[\widetilde E(s_{\mathrm{ret}})e^{-i\omega t}e^{i\omega\tau}\right], \qquad \tau = t-t_{\mathrm{ret}}. $$ So self-interaction appears as a phase-delayed contribution carried by the earlier flow. In coarse form, this may be summarized by writing $$ D = \epsilon E_{\mathrm{loc}} + P_{\mathrm{self}}, \qquad H = \frac{1}{\mu}B_{\mathrm{loc}} - M_{\mathrm{self}}, $$ with $$ P_{\mathrm{self}} \approx \epsilon \chi_{e,\mathrm{eff}} E_{\mathrm{loc}}, \qquad M_{\mathrm{self}} \approx \chi_{m,\mathrm{eff}} H. $$ Then $$ c_{\mathrm{eff}} = \frac{1}{\sqrt{\mu_{\mathrm{eff}}\epsilon_{\mathrm{eff}}}}, \qquad n_{\mathrm{eff}} = \sqrt{(1+\chi_{e,\mathrm{eff}})(1+\chi_{m,\mathrm{eff}})}. $$ This dielectric-style writing is the conventional two-aspect version of the same proportional-response idea. The response coefficient $k$ above is the one-field analogue of the effective susceptibilities written here through $\chi_{e,\mathrm{eff}}$ and $\chi_{m,\mathrm{eff}}$. The dielectric form therefore does not introduce a different mechanism. It is the conventional electromagnetic writing of the same retarded compression effect. ## What this gives This chapter establishes: - coherent overlap adds field patterns point by point, - the local joined density can range from $0$ to $4u_0$, - the conserved two-stream budget remains $2u_0$, - the interference term records the pointwise overlap of the added field patterns, - continuity forbids assigning the difference to primitive sinks or sources, - where two constructive inflows recover as one local positive branch, continuity forces compression of the same total content, - that constructive one-branch recovery doubles the mean energy density, - spatially varying phase produces nodes where the energy density vanishes and antinodes where it reappears more densely, - a dark branch has no positive flux channel; recovery is read only where the continuing pattern has $\mathbf S>0$, - reduced effective advance is the coarse-grained summary of that loaded-branch lag, - the dielectric/proportional-response form $f_2=kf_1$ gives the direct local route to $c_{\mathrm{eff}}$, - differential energy density bends transport, - an isolated bright stripe gives a direct delay or refraction test of the loaded-branch law, - self-refraction is retarded overlap of the same flow. No second substrate is required. The next step is global: > when self-refraction is strong enough to close the path, what stable > configurations are allowed?