# Bending of light - photon's inertia instead of mass

In Newtonian gravity the acceleration due to gravity is independant of the mass of the object - a falling elephant accelerates downwards at the same speed as an accelerating gnat (ignoring air resistance). That means the orbit of an object does not depend on the object's mass (provided the object is much lighter than the star).

The deflection of an object by a massive body is just a hyperbolic orbit, and like any other orbit the trajectory doesn't depend on the mass of the orbiting object, just its velocity and initial position. This means that if we calculate the angular deflection of an object travelling at $c$ in a hyperbolic orbit the result turns out to be independant of the mass of the object:

$$\theta \approx \frac{2GM}{r_0 c^2}$$

where $M$ is the mass of the star/planet/whatever and $r_0$ is the distance of closest approach.

The point is that assuming some hypothetical value for the photon mass doesn't affect the Newtonian prediction because the Newtonian prediction doesn't depend on mass. So the answer to your question is that no, using an effective photon mass of $m = h/\lambda c$ does not give the correct result.

The GR calculation gives the deflection of light as:

$$\theta \approx \frac{4GM}{r_0 c^2}$$

which is twice the Newtonian result. But this isn't some special case that applies only to light because GR gives different results to Newtonian gravity for all objects regardless of mass - the deflection of light is just a limiting case.

1. Since gravity couples to energy rather than rest mass it is natural to speculate that the two masses in Newton's law of gravitation should be replaced with the relativistic masses in a post-Newtonian approximation?

2. The above proposal fails already for the bending/deflection of a massless or massive point particle of rest mass $$m$$ around a mass $$M$$. In a coordinate system where $$M$$ is at rest, a naive relativistic Newton's law would then read $$\frac{d(\gamma m {\bf v})}{dt}~ \stackrel{?}{=}~-G\gamma mM\frac{\bf r}{r^3}.\tag{1}$$ Since the velocity $$|{\bf v}|$$ is approximately constant, we can effectively remove the $$\gamma$$ factors on both sides, and we are back where we started, cf. the equivalence principle :(

3. On the other hand, we know from the correct general relativistic formula that we are missing a factor $$\color{red}{1+\frac{v_0^2}{c^2}~=~ 2-\gamma_0^{-2}}\tag{2}$$ on the rhs. of eq. (1). The general relativistic bending/deflection formula is a factor $$\color{red}{2-\gamma_0^{-2}}$$ times the Newtonian result.

4. The factor $$\color{red}{2-\gamma_0^{-2}}$$ can be understood via the ray equation $$\frac{d}{ds}\left(n\sqrt{\frac{E^2}{c^2}-\frac{(mc)^2}{n}}\frac{dr_i}{ds} \right)~=~\frac{\frac{E^2}{c^2}-\frac{(mc)^2}{2n}}{\sqrt{\frac{E^2}{c^2}-\frac{(mc)^2}{n}}}\frac{\partial n}{\partial r^i} , \tag{3}$$ with effective index of refraction $$n({\bf r})~=~1-\frac{2\phi({\bf r})}{c^2} ,\tag{4}$$ and specific gravitational potential $$\phi({\bf r})~=~-GM/|{\bf r}|.\tag{5}$$ Here the COM $$E~=~\gamma_0 mc^2,\qquad \gamma_0~=~\gamma(v_0), \tag{6}$$ is determined asymptotically at spatial infinity $$|{\bf r}|=\infty$$.

The leading approximation of the ray equation (3) yields a specific Newton's 2nd law $$\frac{d^2r_i}{dt^2}~\approx~v_0^2\frac{d^2r_i}{ds^2}~~\stackrel{(3)+(6)}{\approx}~ v_0^2\frac{\gamma_0^2-\frac{1}{2}}{\gamma_0^2-1}\frac{\partial n}{\partial r^i}~\stackrel{(4)}{=}~-(\color{red}{2-\gamma_0^{-2}})\frac{\partial \phi}{\partial r^i} \tag{7}$$ up to the sought-for factor $$\color{red}{2-\gamma_0^{-2}}$$.