# 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. ## Coherent overlap and quadratic loading Because the loading is positive, it can be written as the square of an amplitude-like quantity: $$ u = |f|^2. $$ Let two coherent contributions of the same transporting flow be $$ f_1, \qquad f_2. $$ When they join, the amplitudes add first: $$ f_{\mathrm{join}} = f_1 + f_2. $$ So the joined local loading 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 $$ u_{\mathrm{join}} = u_1 + u_2 + 2\sqrt{u_1u_2}\cos\delta. $$ For equal contributions, $$ u_1 = u_2 = u_0, $$ the joined local readout 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 loading 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 square-law readout 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. ## 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 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 readout reaches $$ u_{\mathrm{join}} = 4u_0. $$ Now suppose the overlap region recovers the two inflows on one loaded branch over the same interval. Then 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 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 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 readout 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 readout. The correct conclusion is not that a sink has appeared. It is that the joined forward branch is not the channel on which the incoming content is recovered. For a finite overlap region, continuity leaves only two possibilities: - the incoming content reappears through other parts of the boundary, - or the flows separate again after the overlap and recover the original two-branch distribution. This is also the place to distinguish the amplitude-like quantity from the recovered density. The sign or phase can reverse in $f$, but the recovered 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 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 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 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 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 and does not prove that no real overlap region is possible. It proves only that an everywhere-dark joined branch cannot itself be the full recovery geometry. If a finite overlap region produced zero joined readout everywhere and no reappearance elsewhere on the outgoing pattern, that region would act as a sink and would therefore be forbidden. To obtain a fresh `4u_0` readout after destructive overlap, the transport must undergo a new constructive recombination into one branch. ## What the square law does and does not do The square form $$ u = |f|^2 $$ does not create or remove energy. It does two precise jobs: - it gives the local readout of the branch being sampled, - it makes the phase-sensitive cross term explicit. The continuity argument is separate. It tells us how the transported content is recovered across the whole overlap region. ## 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 must recover 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 density is the local compression statement. 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, one returns instead to the finite-region continuity statement and the redirected or re-separated recovery geometry. ## 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. ## Proportional background response There is also a direct local route to the same effective advance on the reinforcing branch. Treat the second contribution not as a second independent substance but as the background response induced by the first. In the same way one writes $$ P = \chi E, $$ one may write the reinforcing response of the same flow as $$ f_2 = k f_1, \qquad k \ge 0. $$ This shortcut concerns only the reinforcing branch. When the local sum vanishes, one does not assign a negative effective advance to a non-existent forward branch; one returns instead to the finite-region continuity statement above. Then the joined flow is $$ f_{\mathrm{tot}} = f_1 + f_2 = (1+k)f_1. $$ So the joined local loading is $$ u_{\mathrm{tot}} = |f_{\mathrm{tot}}|^2 = |1+k|^2 |f_1|^2 = (1+k)^2 u_1. $$ In the reinforcing case it is therefore natural to write $$ n_{\mathrm{eff}} := 1+k. $$ Then $$ u_{\mathrm{tot}} = n_{\mathrm{eff}}^2 u_1. $$ Continuity then gives the matching lag statement: if one surviving outflow must recover what is now loaded by the factor $n_{\mathrm{eff}}$, the effective advance is $$ c_{\mathrm{eff}} = \frac{c}{n_{\mathrm{eff}}}. $$ So the proportional-response shortcut closes the same circle: $$ f_2 = kf_1 \quad\Longrightarrow\quad f_{\mathrm{tot}} = (1+k)f_1 \quad\Longrightarrow\quad u_{\mathrm{tot}} = (1+k)^2u_1 \quad\Longrightarrow\quad c_{\mathrm{eff}} = \frac{c}{1+k}. $$ For the equal-contribution case, $$ k = 1, $$ so $$ f_{\mathrm{tot}} = 2f_1, \qquad u_{\mathrm{tot}} = 4u_1, \qquad c_{\mathrm{eff}} = \frac{c}{2}. $$ This is the easiest local route from superposition to the effective-medium summary. ## Local bending If different regions of a wavefront have different loading, then they do not advance equally. - higher density implies lower $c_{\mathrm{eff}}$, - lower 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. ## 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 produces quadratic local loading, - the local joined readout can range from $0$ to $4u_0$, - the conserved two-stream budget remains $2u_0$, - the square law counts the joined state but does not by itself explain it, - continuity forbids assigning the difference to primitive sinks or sources, - where two inflows are locally recovered on one outflow, continuity forces compression of the same total content, - that constructive one-branch recovery doubles the mean density, - spatially varying phase produces nodes where the density vanishes and antinodes where it reappears more densely, - an everywhere-dark joined branch cannot be the full recovery geometry for a nonzero incoming budget, - reduced effective advance is the coarse-grained summary of that loaded-branch lag, - the proportional-response form $f_2 = kf_1$ gives the direct local route to $c_{\mathrm{eff}}$ and closes onto the dielectric effective-medium writing, - differential loading bends transport, - 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?