## The problem The classical projectile treatment of light gives the well-known half-value: $$ \theta_{\text{Newton}} = \frac{2GM}{bc^2}. $$ At the solar limb this is about $$ \theta_{\text{Newton}} \approx 0.875 \text{ arcseconds}. $$ Observation gives twice this: $$ \theta_{\text{obs}} \approx 1.75 \text{ arcseconds}. $$ In a Maxwell universe, where gravity is interpreted as refraction rather than spacetime curvature, the question is direct: **what refractive profile gives the observed value, and why?** ## The half-result comes from a half-medium The standard dielectric-style argument treats the gravitational background as an effective change in permittivity only: $$ \varepsilon_{\text{eff}}(r)=\varepsilon_0\bigl(1+2\eta(r)\bigr),\qquad \mu_{\text{eff}}(r)=\mu_0, $$ with $$ \eta(r)=\frac{GM}{rc^2}\ll 1. $$ The resulting refractive index is $$ n(r)=\sqrt{\frac{\varepsilon_{\text{eff}}\mu_{\text{eff}}} {\varepsilon_0\mu_0}} =\sqrt{1+2\eta} \approx 1+\eta =1+\frac{GM}{rc^2}. $$ Using the weak-deflection ray integral, $$ \theta \approx \int_{-\infty}^{\infty}\nabla_\perp n\,dz, $$ this gives $$ \theta = \frac{2GM}{bc^2}. $$ That is the Newtonian half-value. The result is not wrong. The model is incomplete. It perturbs only one constitutive channel of an electromagnetic medium. ## Gravity acts through a full electromagnetic background In this framework, the background around a massive body is not inert matter. It is organized electromagnetic energy. A passing wave does not encounter an electric-only response. It encounters a full electromagnetic response. That matters because light propagation depends on both constitutive coefficients: $$ c_{\text{local}} = \frac{1}{\sqrt{\varepsilon_{\text{eff}}\mu_{\text{eff}}}}, \qquad n = \sqrt{\frac{\varepsilon_{\text{eff}}\mu_{\text{eff}}} {\varepsilon_0\mu_0}}. $$ If the background is itself electromagnetic, there is no reason to privilege the electric channel and ignore the magnetic one. In the Maxwell closure, $\mathbf{E}$ and $\mathbf{B}$ are complementary aspects of one organized flow, not separate substances. An electric-only perturbation would split what the theory has already identified as inseparable. The weak-field constitutive law must therefore be symmetric: $$ \varepsilon_{\text{eff}}(r)=\varepsilon_0\bigl(1+2\eta(r)\bigr), \qquad \mu_{\text{eff}}(r)=\mu_0\bigl(1+2\eta(r)\bigr), $$ with the same $$ \eta(r)=\frac{GM}{rc^2}. $$ ## Why the symmetry is physically required This symmetric choice does two things at once. First, it reflects the ontology. The background is electromagnetic energy, not a purely electric dielectric. Second, it preserves the local vacuum impedance: $$ Z_{\text{eff}}=\sqrt{\frac{\mu_{\text{eff}}}{\varepsilon_{\text{eff}}}} =\sqrt{\frac{\mu_0}{\varepsilon_0}} = Z_0. $$ So the background changes propagation speed without introducing an arbitrary electric-magnetic mismatch. The medium bends rays by delay, not by inventing a new polarization asymmetry. ## The full weak-field refractive index With both channels modified equally, $$ n(r)=\sqrt{(1+2\eta)(1+2\eta)} = 1+2\eta + O(\eta^2). $$ Therefore $$ n(r)\approx 1+\frac{2GM}{rc^2}. $$ This is the full electromagnetic refractive profile. Compared with the electric-only half-medium, the first-order index shift is doubled. ## Weak-field bending calculation Let a ray pass the gravitating body with impact parameter $b$, and use the straight-line approximation $$ r=\sqrt{b^2+z^2}. $$ Then $$ n(r)=1+\frac{2GM}{c^2\sqrt{b^2+z^2}}. $$ The transverse gradient is $$ \frac{\partial n}{\partial b} = -\frac{2GM}{c^2}\frac{b}{(b^2+z^2)^{3/2}}. $$ The total deflection magnitude is $$ \theta = \int_{-\infty}^{\infty}\left|\frac{\partial n}{\partial b}\right|dz = \frac{2GM}{c^2}\int_{-\infty}^{\infty} \frac{b\,dz}{(b^2+z^2)^{3/2}}. $$ Using $$ \int_{-\infty}^{\infty}\frac{b\,dz}{(b^2+z^2)^{3/2}} = \frac{2}{b}, $$ we obtain $$ \theta = \frac{4GM}{bc^2}. $$ This is exactly the observed weak-field result. ## Solar-limb value For a ray grazing the Sun, $b=R_\odot$, so $$ \theta_\odot = \frac{4GM_\odot}{R_\odot c^2} \approx 8.48\times10^{-6}\ \text{rad} \approx 1.75 \text{ arcseconds}. $$ ## Interpretation The factor of 2 does not require curved spacetime. It requires that the gravitational background be treated as what it is in this framework: - electromagnetic, - source-free, - and therefore symmetric in its electric and magnetic response. The Newtonian half-value is the bending produced by a one-channel constitutive model. The full value comes from the full electromagnetic medium. This is also why the earlier flux-versus-mass intuition was pointing in the right direction. Light is not averaged trapped matter. It is directed electromagnetic transport. But the exact coefficient is not fixed by that intuition alone. It is fixed by the constitutive symmetry forced by the electromagnetic character of the background. ## Conclusion In a Maxwell universe, gravity is refraction produced by organized background electromagnetic energy. If that background is modeled as electric-only, one recovers the Newtonian half-value: $$ \theta = \frac{2GM}{bc^2}. $$ If it is modeled as a full electromagnetic medium, with symmetric weak-field changes in both permittivity and permeability, one obtains $$ \theta = \frac{4GM}{bc^2}. $$ At the solar limb this is 1.75 arcseconds. The factor of 2 is not evidence that space itself is curved. It is evidence that gravity acts through the full electromagnetic constitutive structure of the vacuum.