Inertia, gravity and the meaning of mass

Newton's law of universal gravitation is an expression of the force experienced by a particle with established inertia m_i and gravitational mass m_g , due to a gravitating mass M_g located at r = 0,

$$\begin{eqnarray}&&{{\bf{F}}}_{g}={m}_{i}{\bf{g}}=-\displaystyle \frac{{{Gm}}_{g}{M}_{g}}{{r}^{2}}\hat{r},\end{eqnarray} \tag{ 1 }$$

which is understood relativistically in the limit of weak, static fields and low velocities. Of necessity, the latter condition implies that particles coupling to the force must have non-zero inertia, m_i ≠ 0, for the acceleration g would otherwise be infinite. Equation (1) may also be written in terms of the gravitational potential,

$$\begin{eqnarray}&&{\rm{\Phi }}({\bf{r}})\equiv -\displaystyle \frac{{{GM}}_{g}}{r},\end{eqnarray} \tag{ 2 }$$

such that

$$\begin{eqnarray}&&{m}_{i}{\bf{g}}=-{m}_{g}\vec{{\rm{\nabla }}}{\rm{\Phi }}.\end{eqnarray} \tag{ 3 }$$

Equations (2) and (3) actually represent two distinct phenomena. The first accounts for the potential created by the hypothesized gravitating mass M_g , while the second describes the response of a test particle m_g to the presence of this potential. Thus, to ensure consistency between general relativity in the weak-field, static, low-velocity limit and Newton's law of universal gravitation, two separate physical effects must be considered. First, the response of the particle given by equation (3) is most naturally inferred from the geodesic equation,

$$\begin{eqnarray}&&\displaystyle \frac{{d}^{2}{x}^{\mu }}{d{\lambda }^{2}}+{{{\rm{\Gamma }}}^{\mu }}_{\alpha \beta }\,\displaystyle \frac{{{dx}}^{\alpha }}{d\lambda }\displaystyle \frac{{{dx}}^{\beta }}{d\lambda }=0,\end{eqnarray} \tag{ 4 }$$

describing the trajectory of a free particle through the spacetime created by the gravitating mass M_g (see equation (30) below). The affine parameter λ is often chosen to be the proper time τ in the particle's rest frame when m_i ≠ 0, though not for massless particles, such as a photon, for which τ is always zero. In this expression, ${{{\rm{\Gamma }}}^{\mu }}_{\alpha \beta }$ are the Christoffel symbols containing information about the spacetime curvature, most directly calculated from the metric coefficients themselves:

$$\begin{eqnarray}&&{{{\rm{\Gamma }}}^{\mu }}_{\alpha \beta }\equiv \displaystyle \frac{1}{2}{g}^{\mu \delta }\left[{g}_{\beta \delta ,\alpha }+{g}_{\delta \alpha ,\beta }-{g}_{\alpha \beta ,\delta }\right],\end{eqnarray} \tag{ 5 }$$

where the metric is formally written as

$$\begin{eqnarray}&&{{ds}}^{2}={g}_{\alpha \beta }{{dx}}^{\alpha }{{dx}}^{\beta }.\end{eqnarray} \tag{ 6 }$$

The procedure for reducing equation (4) to its Newtonian form is well known, so we won't dwell on the details, but mention only the key points, mostly in preparation for section 2.2 below. For a particle with inertia moving well below the speed of light, the second term is dominated by the α = β = 0 component, and therefore

$$\begin{eqnarray}&&\displaystyle \frac{{d}^{2}{x}^{\mu }}{d{\tau }^{2}}+{{{\rm{\Gamma }}}^{\mu }}_{00}\,\displaystyle \frac{{{dx}}^{0}}{d\tau }\displaystyle \frac{{{dx}}^{0}}{d\tau }=0.\end{eqnarray} \tag{ 7 }$$

Then, to calculate ${{{\rm{\Gamma }}}^{\mu }}_{00}$ for a weak gravitational field, we put

$$\begin{eqnarray}&&{g}_{\alpha \beta }={\eta }_{\alpha \beta }+{h}_{\alpha \beta },\end{eqnarray} \tag{ 8 }$$

where η_{α β} = diag( + 1, − 1, − 1, − 1) and ∣h_{α β} ∣ ≪ 1. The inverse metric tensor is simply g^{α β} = η^{α β} − h^{α β} . If in addition the field is static,

$$\begin{eqnarray}&&{{{\rm{\Gamma }}}^{\mu }}_{00}=-\displaystyle \frac{1}{2}{\eta }^{\mu \nu }{\partial }_{\nu }{g}_{00}=-\displaystyle \frac{1}{2}{\eta }^{\mu \nu }{\partial }_{\nu }{h}_{00}.\end{eqnarray} \tag{ 9 }$$

Substituting equation (9) into (7), we thus find that

$$\begin{eqnarray}&&\displaystyle \frac{{d}^{2}{x}^{\mu }}{d{\tau }^{2}}=\displaystyle \frac{1}{2}{\eta }^{\mu \nu }{\partial }_{\nu }{h}_{00}{\left(\displaystyle \frac{{{dx}}^{0}}{d\tau }\right)}^{2},\end{eqnarray} \tag{ 10 }$$

so that d² x⁰/d τ² = 0, which means that dt/d τ is constant. That is, time progresses forward at a steady rate for a particle in Newtonian gravity, consistent with the prevailing view on the nature of time during Newton's era.

For the spatial coordinates (j = 1, 2, 3), we instead find that

$$\begin{eqnarray}&&\displaystyle \frac{{d}^{2}{x}^{j}}{d{\tau }^{2}}=-\displaystyle \frac{{c}^{2}}{2}{\left(\displaystyle \frac{{dt}}{d\tau }\right)}^{2}{\partial }_{j}{h}_{00}\end{eqnarray} \tag{ 11 }$$

or, using the chain rule of differentiation,

$$\begin{eqnarray}&&\displaystyle \frac{{d}^{2}{x}^{j}}{{{dt}}^{2}}=-\displaystyle \frac{{c}^{2}}{2}{\partial }_{j}{h}_{00}.\end{eqnarray} \tag{ 12 }$$

A comparison of equations (3) and (12) therefore shows that, in order for Einstein's theory to correctly describe the motion of a particle with established inertia in a classical gravitational field, we must have m_g → m_i in the Newtonian limit (see section 3) which is, afterall, the basis for the Equivalence Principle that gave rise to the general relativistic description of particle trajectories in the first place. In addition, h₀₀ ≡ 2Φ/c², so that

$$\begin{eqnarray}&&{g}_{00}=1+\displaystyle \frac{2{\rm{\Phi }}}{{c}^{2}}.\end{eqnarray} \tag{ 13 }$$

In the next section, we shall learn that the gravitational coupling of particles believed to have zero inertia is very similar to this result, but differs from it in at least one very significant aspect. Before we begin to examine that situation, however, we must first understand the second phenomenon associated with Newtonian gravity—that giving rise to the gravitational potential itself (equation (2)). For this, we need to begin with the gravitational field equations in general relativity, which may be written

$$\begin{eqnarray}&&{G}_{\alpha \beta }\equiv {R}_{\alpha \beta }-\displaystyle \frac{1}{2}\,{g}_{\alpha \beta }\,R=-\displaystyle \frac{8\pi G}{{c}^{4}}\,{T}_{\alpha \beta }.\end{eqnarray} \tag{ 14 }$$

G_{α β} is the Einstein tensor, written in terms of the Ricci tensor,

$$\begin{eqnarray}&&{R}_{\beta \delta }\equiv {g}^{\alpha \gamma }\,{R}_{\alpha \beta \gamma \delta }={R}_{\,\ \ \beta \gamma \delta }^{\gamma },\end{eqnarray} \tag{ 15 }$$

the contracted Ricci tensor

$$\begin{eqnarray}&&R\equiv {g}^{\mu \nu }\,{R}_{\mu \nu }\end{eqnarray} \tag{ 16 }$$

(also known as the curvature scalar) and the stress-energy tensor T_{α β} . In equation (15), the quantity

$$\begin{eqnarray}&&{{R}^{\alpha }}_{\delta \beta \gamma }\equiv {{{\rm{\Gamma }}}^{\alpha }}_{\delta \beta ,\gamma }-{{{\rm{\Gamma }}}^{\alpha }}_{\delta \gamma ,\beta }-{{{\rm{\Gamma }}}^{\alpha }}_{\epsilon \beta }\,{{{\rm{\Gamma }}}^{\epsilon }}_{\delta \gamma }+{{{\rm{\Gamma }}}^{\alpha }}_{\epsilon \gamma }\,{{{\rm{\Gamma }}}^{\epsilon }}_{\delta \beta }\ \ \end{eqnarray} \tag{ 17 }$$

is known as the Riemann-Christoffel (or curvature) tensor.

The appearance of the curvature scalar on the left-hand side of equation (14) is sometimes inconvenient. Contracting this equation with α and β reduces it to the form

$$\begin{eqnarray}&&R=\displaystyle \frac{8\pi G}{{c}^{4}}\,{{T}^{\gamma }}_{\gamma }.\end{eqnarray} \tag{ 18 }$$

Thus, an alternative representation of the field equations is

$$\begin{eqnarray}&&{R}_{\alpha \beta }=-\displaystyle \frac{8\pi G}{{c}^{4}}\left({T}_{\alpha \beta }-\displaystyle \frac{1}{2}\,{g}_{\alpha \beta }\,{{T}^{\gamma }}_{\gamma }\right).\end{eqnarray} \tag{ 19 }$$

It is beyond the scope of the present paper to describe how and why these equations are derived, but this topic is well covered in both the primary and secondary literature [5, 13]. We do point out, however, that the coefficient multiplying T_{α β} on the right-hand side of equations (14) and (19) was chosen in order for Einstein's equations to correctly reproduce the Newtonian potential in equation (2) for a gravitating inertial source M_g in the weak-field, static, low-velocity limit, which we now describe.

For simplicity, we adopt the so-called perfect fluid approximation, in which the stress-energy tensor excludes all possible shear forces associated with the transport of momentum components in directions other than those associated with the components themselves. The covariant form of this tensor may be written

$$\begin{eqnarray}&&{T}_{\alpha \beta }=\displaystyle \frac{1}{{c}^{2}}\left(\rho +p\right){u}_{\alpha }{u}_{\beta }-p\,{g}_{\alpha \beta },\end{eqnarray} \tag{ 20 }$$

where u_α is the local value of dx_α /d τ for a comoving fluid element in the source, and p and ρ are the pressure and energy density, respectively, measured by an observer in a locally inertial frame comoving with the fluid at the instant of measurement.

The trace of this stress-energy tensor is simply

$$\begin{eqnarray}&&T\equiv {{T}^{\gamma }}_{\gamma }=\rho -3p,\end{eqnarray} \tag{ 21 }$$

providing a clear, unequivocal affirmation that the source of spacetime curvature in general relativity is all forms of energy and momentum. This aspect of Einstein's theory cannot be overstated because it represents a clear departure from the Newtonian framework, in which gravity is due to an intrinsic 'mass' associated with the source, considered by Newton to be synonymous with inertia, and neither having anything to do with energy. Our continued examination of the meaning of m_g , m_i and M_g below will be heavily based on this crucial distinction between Einstein's and Newton's theories, and their behavior in the classical limit. Equation (19) may thus also be written in the more suggestive form

$$\begin{eqnarray}&&{R}_{\alpha \beta }=-\displaystyle \frac{8\pi G}{{c}^{4}}\left({T}_{\alpha \beta }-\displaystyle \frac{1}{2}\,{g}_{\alpha \beta }\,[\rho -3p]\right).\end{eqnarray} \tag{ 22 }$$

This expression is completely valid for all forms of energy, with or without inertia, and we shall use it also in the following section where we consider the gravitational coupling of particles believed to have no inertia. Here, however, we use the fact that matter (or its more common designation as 'dust') has essentially zero pressure, so T = ρ, and equation (22) therefore implies that, to first order in the weak-field (Newtonian) limit (see equation (8)),

$$\begin{eqnarray}&&{R}_{00}=-\displaystyle \frac{4\pi G}{{c}^{4}}\rho ,\end{eqnarray} \tag{ 23 }$$

which relates derivatives of the metric coefficients (specifically h₀₀ and Φ) to the energy density, providing a direct link to the Poisson equation for Φ, from which equation (2) is derived.

From equation (15), we see that

$$\begin{eqnarray}&&{R}_{00}={{R}^{\gamma }}_{0\gamma 0},\end{eqnarray} \tag{ 24 }$$

and ${{R}^{0}}_{000}=0$, so

$$\begin{eqnarray}&&{R}_{00}={{R}^{i}}_{0i0}={\partial }_{i}{{{\rm{\Gamma }}}^{i}}_{00}-{\partial }_{0}{{{\rm{\Gamma }}}^{i}}_{i0}+{{{\rm{\Gamma }}}^{i}}_{i\gamma }{{{\rm{\Gamma }}}^{\gamma }}_{00}-{{{\rm{\Gamma }}}^{i}}_{0\gamma }{{{\rm{\Gamma }}}^{\gamma }}_{i0}.\ \ \end{eqnarray} \tag{ 25 }$$

The second term on the right-hand side is zero for a static field, while the third and fourth terms are second order in h₀₀. This leaves

$$\begin{eqnarray}&&{R}_{00}={\partial }_{i}{{{\rm{\Gamma }}}^{i}}_{00}=-\displaystyle \frac{1}{2}{\eta }^{{ij}}{\partial }_{i}{\partial }_{j}\,{h}_{00}\end{eqnarray} \tag{ 26 }$$

and, combining equation (23) with (26), gives

$$\begin{eqnarray}&&{\vec{{\rm{\nabla }}}}^{2}{\rm{\Phi }}=4\pi G\left(\displaystyle \frac{\rho }{{c}^{2}}\right),\end{eqnarray} \tag{ 27 }$$

which is the relativistically derived Poisson's equation for the gravitational potential in the presence of an energy density ρ in the weak-field, static limit. It must be emphasized that this equation excludes any momentum (and therefore the pressure this would create) of the gravitating source specifically because we attributed inertia to the medium, allowing it to reside near the origin with a very low (or even zero) velocity. It is for this reason that T = ρ − 3p simply reduces to ρ in equation (22), and we acknowledge the fact that the coefficient 4π G in equation (27) was chosen to ensure that Einstein's and Newton's theories yield the same potential, Φ, when the gravitational mass, M_g , is calculated solely from the energy density in the system, and nothing else.

But there is another subtle, yet crucial, feature of this equation that we must fully understand, particularly when we begin to compare this result with its counterpart in the following section, addressing gravity in the presence of energy with what we believe to be zero inertia, i.e. in cases where T = ρ − 3p is not merely ρ. Equation (27) is fully consistent with equation (2) only if we follow Newton in attributing the gravitational influence to a hypothesized 'gravitational mass,' M_g , which necessarily would have to correspond to a gravitational mass density ρ_g ≡ ρ/c² in equation (27).

Those of us accustomed to the language of general relativity would be tempted to consider this as being self-evident. After all, isn't rest-mass energy simply given by this relation? Well yes, but not completely, as we shall soon find out. Later in this paper we shall better understand the distinction between gravitational and inertial mass, and realize that the interpretation of M_g in equation (2) as the 'rest-mass'—from the conversion of the energy density ρ to ρ_g ≡ ρ/c² in equation (27)—is valid only because Newton's law of universal gravitation was specifically formulated for particles with non-zero inertia in the low-velocity limit, where the gravitational and inertial masses are proportional to each other—or even equal, with an appropriate choice of units for the gravitational constant G. The situation is very different for photons, because m_g for them is not zero.

Whether light has inertia and/or gravitational mass, and whether these two are equal, was something that could not easily be discussed or explored prior to the advent of relativity theory. But this did not prevent classical physicists from entertaining the idea that large heavenly bodies, such as the Sun, could in principle accelerate rays of light and cause them to deviate in discernible ways from straight-line trajectories. Very famously, Einstein himself used his Principle of Equivalence couched in Newtonian theory, essentially equation (1) with m_i and m_g cancelled from both sides, to predict how much starlight would be bent upon grazing the surface of the Sun on its way toward Earth [14].

Without the full theory of general relativity to support this calculation, he partially relied on intuition, arguing that the rate of proper time (as seen by an observer fixed with respect to the source of gravity) varies with distance from the center of the gravitating body, thereby creating a speed of light varying with radius if one ignores the spatial variations. The latter assumption is key to understanding why he had to correct his prediction once general relativity was completed five years later. Applying Huygenss principle to a wave front passing through such a region, he could then calculate the degree of bending based on the time dilation produced by the central mass. This prediction amounted to about 0.875 seconds of arc, which turns out to be wrong by a factor 2. We shall see below why ignoring the spatial variations results in this not insignificant mistake. One may still predict the correct deflection angle using Newtonian theory, however, but only by using an alternative form of equation (1) appropriate for light (see equation (52)), which was not available to him at that time.

This type of speculation had already been carried out by others before him, even by Newton who, in his treatise on Opticks[15] published in 1704, asked the question: "Do not Bodies act upon Light at a distance, and by their action bend its Rays, and is not this action strongest at the least distance?" A century later, Johann Georg von Soldner had used Newton's Law of universal gravitation to calculate the deflection of starlight by the Sun treating 'a light ray as a heavy body' and predicted a deflection angle of 0.84 arcseconds, virtually identical to Einstein's estimate based on his first attempt [16].

Einstein redid his calculation five years later once General Relativity was completed, taking into account all spacetime curvature effects and corrected his mistake, predicting a deflection angle of 1.745 arcseconds. As is well known by now, Sir Arthur Eddington subsequently led an expedition to an island off the coast of Africa, with a second group in Brazil, to measure the deflection of starlight grazing the edge of the Sun during the total eclipse of May 29, 1919. Their measurement provided a spectacular (if controversial) confirmation that General Relativity is the correct theory of gravity [17].

Questions have been raised about Eddington's analysis of their data because turbulence in Earth's atmosphere causes deflections of starlight comparable to those predicted by Einstein's theory. One must rely on the assumption that these are random in nature, so that they can be averaged away with the use of many images, leaving only the relativistic effect. By the end of their observations, Eddington and his coworkers had only two reliable images (with about five stars) at one site, and eight usable plates (with at least seven stars) at the Sobral location. Nineteen other plates taken with a second telescope had to be abandoned. Given this paucity of measurements, did Eddington really have the evidence to support Einstein's theory [18]? Some have argued that Eddington's enthusiasm for general relativity biased his approach. But many re-analyses between 1923 and 1956 of the plates from those expeditions yielded similar results within ten percent. A reanalysis in 1979 using the Zeiss Ascorecord and its data reduction software [19] yielded the same deflection as that calculated by Eddington, though with even smaller errors. The Sobral plates gave similar results, all consistent with general relativity. A modern assessment of Eddington's work has thus tended to show no credible evidence of bias in his conclusions [20].

Having said this, we now know, that photons do not have rest mass in the conventional form seen in the standard model of particle physics. So why are we justified in using General Relativity to calculate the gravitational acceleration of a photon when Einstein's theory is based on the equality of m_i and m_g (i.e. the Principle of Equivalence)? This question will become even more acute later in this paper, when we learn that Newton's approach of identifying gravitational mass suggests that m_g strictly cannot be zero for light. To place this in context, we should ask ourselves whether we are missing a law of universal of gravitation, analogous to equation (1), representing the weak-field, static limit of General Relativity for photons and other particles that do not have inertia in the conventional sense.

Such particles do not follow trajectories described by equation (7). The magnitude of their velocity is always c, so one cannot adopt an asymptotically small speed to go along with the assumption of weak, static fields. As was the case in section 2.1, there are two effects we need to consider: the first arises from the impact of non-zero momentum on the source of gravity, modifying Newton's definition of M_g , and the second provides the gravitational coupling of a particle we believe to have zero rest mass to the 'force' created by the former.

The first of these phenomena is quite straightforward to understand. Whereas the trace of the stress-energy tensor, T = ρ − 3p, reduces to ρ for 'dust,' the pressure cannot be ignored when the source is radiation since its momentum makes a significant contribution to the overall energy budget. For example, isotropic radiation exerts a pressure p = ρ/3, so T = 0. Thus, instead of equation (23), we now get

$$\begin{eqnarray}&&{R}_{00}=-\displaystyle \frac{8\pi G}{{c}^{4}}\rho ,\end{eqnarray} \tag{ 28 }$$

resulting in the modified Poisson's equation

$$\begin{eqnarray}&&{\vec{{\rm{\nabla }}}}^{2}{\rm{\Phi }}=4\pi G\left(\displaystyle \frac{2\rho }{{c}^{2}}\right).\end{eqnarray} \tag{ 29 }$$

The factor 2 multiplying the density is directly due to the momentum within the source, which cannot be ignored when the gravitating particles cannot come to rest. When the source of gravity is radiation, Newton's 'gravitational mass' M_g producing the potential in equation (2) must therefore be twice the value one would naively have calculated from the energy density alone. (But beware that this is not the factor 2 associated with the deflection of starlight by the Sun. A correction such as this, in cases where the source of gravity is not completely inertial, is typically absorbed into the empirically determined value of M_g ).

Once the potential Φ and the Newtonian gravitational mass M_g have been properly identified in equation (2), we may then proceed with equation (4) to derive the correct equations of motion for a photon in the vicinity of a spherically symmetric object, where the appropriate spacetime is described by the Schwarzschild metric:

$$\begin{eqnarray}&&{{ds}}^{2}=\left(1-\displaystyle \frac{{r}_{{\rm{S}}}}{r}\right){c}^{2}{\left({dt}\right)}^{2}-{\left(1-\displaystyle \frac{{r}_{S}}{r}\right)}^{-1}{\left({dr}\right)}^{2}-{r}^{2}\left[{\left(d\theta \right)}^{2}+{\sin }^{2}\theta {\left(d\phi \right)}^{2}\right].\end{eqnarray} \tag{ 30 }$$

In this expression,

$$\begin{eqnarray}&&{r}_{{\rm{S}}}\equiv \displaystyle \frac{2{{GM}}_{g}}{{c}^{2}}\end{eqnarray} \tag{ 31 }$$

is the Schwarzschild radius for an object with gravitational mass M_g . Our goal is to uncover the radial acceleration experienced by the photon at a radius r ≫ r_S, i.e. in the weak-field limit and, to facilitate the calculation, we shall assume that the photon's velocity is perpendicular to ${\bf{r}}=r\hat{r}$ at that instant.

The non-zero Christoffel symbols for this metric are simply

$$\begin{eqnarray}\begin{array}{ccl}{{{\rm{\Gamma }}}^{r}}_{{tt}} & = & \frac{1}{2}\frac{{r}_{{\rm{S}}}}{{r}^{2}}\left(1-\frac{{r}_{{\rm{S}}}}{r}\right)\\ {{{\rm{\Gamma }}}^{r}}_{{rr}} & = & -\frac{1}{2}\frac{{r}_{{\rm{S}}}}{{r}^{2}}{\left(1-\frac{{r}_{{\rm{S}}}}{r}\right)}^{-1}\\ {{{\rm{\Gamma }}}^{t}}_{{tr}} & = & \frac{1}{2}\frac{{r}_{{\rm{S}}}}{{r}^{2}}{\left(1-\frac{{r}_{{\rm{S}}}}{r}\right)}^{-1}={{{\rm{\Gamma }}}^{t}}_{{rt}}\\ {{{\rm{\Gamma }}}^{\theta }}_{r\theta } & = & \frac{1}{r}={{{\rm{\Gamma }}}^{\theta }}_{\theta r}\\ {{{\rm{\Gamma }}}^{r}}_{\theta \theta } & = & -r\left(1-\frac{{r}_{{\rm{S}}}}{r}\right)\\ {{{\rm{\Gamma }}}^{\phi }}_{r\phi } & = & \frac{1}{r}={{{\rm{\Gamma }}}^{\phi }}_{\phi r}\\ {{{\rm{\Gamma }}}^{r}}_{\phi \phi } & = & -r{\sin }^{2}\theta \left(1-\frac{{r}_{{\rm{S}}}}{r}\right)\\ {{{\rm{\Gamma }}}^{\theta }}_{\phi \phi } & = & -\sin \theta \cos \theta \\ {{{\rm{\Gamma }}}^{\phi }}_{\theta \phi } & = & \cot \theta ={{{\rm{\Gamma }}}^{\phi }}_{\phi \theta }.\end{array}\end{eqnarray} \tag{ 32 }$$

Thus, letting overdot denote differentiation with respect to λ, we obtain the following expression from the μ = 0 component of equation (4):

$$\begin{eqnarray}&&\ddot{t}+\dot{t}\dot{r}\displaystyle \frac{{r}_{{\rm{S}}}}{{r}^{2}}{\left(1-\displaystyle \frac{{r}_{{\rm{S}}}}{r}\right)}^{-1}=0.\end{eqnarray} \tag{ 33 }$$

Integrating this equation once gives

$$\begin{eqnarray}&&\dot{t}\left(1-\frac{{r}_{{\rm{S}}}}{r}\right)={K}_{1},\end{eqnarray} \tag{ 34 }$$

where K₁ is a constant of integration.

The μ = 2 component gives

$$\begin{eqnarray}&&\ddot{\theta }+\displaystyle \frac{2}{r}\dot{r}\dot{\theta }-{\dot{\phi }}^{2}\sin \theta \cos \theta =0,\end{eqnarray} \tag{ 35 }$$

while μ = 3 results in

$$\begin{eqnarray}&&\ddot{\phi }+\displaystyle \frac{2}{r}\dot{r}\dot{\phi }+2\dot{\theta }\dot{\phi }\cot \theta =0.\end{eqnarray} \tag{ 36 }$$

We align our coordinate system so that the photon's trajectory is restricted to the equatorial plane. Then, θ(λ) = π/2 and $\dot{\theta }(\lambda )=0$, and equation (35) becomes irrelevant. Equation (36) reduces to

$$\begin{eqnarray}&&\ddot{\phi }+\displaystyle \frac{2}{r}\dot{r}\dot{\phi }=0,\end{eqnarray} \tag{ 37 }$$

whose solution is simply

$$\begin{eqnarray}&&{r}^{2}\dot{\phi }={K}_{2},\end{eqnarray} \tag{ 38 }$$

where K₂ is a second constant of integration.

Finally, each tangent vector along a null geodesic is light-like, so

$$\begin{eqnarray}&&{g}_{\alpha \beta }\displaystyle \frac{{{dx}}^{\alpha }}{d\lambda }\displaystyle \frac{{{dx}}^{\beta }}{d\lambda }=0,\end{eqnarray} \tag{ 39 }$$

which provides us with the last equation we need:

$$\begin{eqnarray}&&\left(1-\displaystyle \frac{{r}_{{\rm{S}}}}{r}\right){c}^{2}{\dot{t}}^{2}-{\left(1-\displaystyle \frac{{r}_{{\rm{S}}}}{r}\right)}^{-1}{\dot{r}}^{2}-{r}^{2}{\dot{\phi }}^{2}=0.\end{eqnarray} \tag{ 40 }$$

Now using equations (34) and (38) with a change in variable to u ≡ 1/r, together with

$$\begin{eqnarray}&&\displaystyle \frac{{dr}}{d\lambda }=\displaystyle \frac{{dr}}{d\phi }\displaystyle \frac{d\phi }{d\lambda }\end{eqnarray} \tag{ 41 }$$

(which is valid because the motion has no θ dependence), we can modify equation (40) to read as follows:

$$\begin{eqnarray}&&{c}^{2}{K}_{1}^{2}-{K}_{2}^{2}{\left(\displaystyle \frac{{du}}{d\phi }\right)}^{2}-{K}_{2}^{2}{u}^{2}\left(1-{r}_{{\rm{S}}}u\right)=0.\end{eqnarray} \tag{ 42 }$$

Differentiating this equation once more with respect to ϕ gives us the equation of motion for a photon,

$$\begin{eqnarray}&&\displaystyle \frac{{d}^{2}u}{d{\phi }^{2}}+u=\displaystyle \frac{3}{2}{r}_{{\rm{S}}}{u}^{2},\end{eqnarray} \tag{ 43 }$$

which may be thought of as a relativistic version of the classical Binet orbit equation [21], best known for its use in calculating the deflection angle of starlight grazing the surface of the Sun on its way toward Earth.

In the absence of gravity (r_S → 0), the solution to this equation would be a straight line perpendicular to r, given as

$$\begin{eqnarray}&&u=\displaystyle \frac{1}{{r}_{0}}\cos \phi ,\end{eqnarray} \tag{ 44 }$$

where r₀ is the radius of closest approach to the origin, corresponding to the value of r at ϕ = 0. If we now make a weak-field approximation, consistent with r ≫ r_S , the corresponding null trajectory will be a perturbation of this straight line, allowing us to write

$$\begin{eqnarray}&&\displaystyle \frac{{d}^{2}u}{d{\phi }^{2}}+u=\displaystyle \frac{3}{2}\displaystyle \frac{{r}_{{\rm{S}}}}{{r}_{0}^{2}}{\cos }^{2}\phi ,\end{eqnarray} \tag{ 45 }$$

whose solution is

$$\begin{eqnarray}&&u=\displaystyle \frac{{r}_{{\rm{S}}}}{{r}_{0}^{2}}+\displaystyle \frac{1}{{r}_{0}}\cos \phi -\displaystyle \frac{1}{2}\displaystyle \frac{{r}_{{\rm{S}}}}{{r}_{0}^{2}}{\cos }^{2}\phi .\end{eqnarray} \tag{ 46 }$$

Our interest here is not so much how the null geodesic is modified due to the accumulated effect of gravity over the ray's transit to Earth but, rather, the instantaneous acceleration experienced by a photon in the vicinity of ϕ = 0. The acceleration is obtained by differentiating this expression twice with respect to the observer's time, t, starting with

$$\begin{eqnarray}&&-\frac{1}{{r}^{2}}\frac{{dr}}{{dt}}=-\frac{\sin \phi }{{r}_{0}}\frac{d\phi }{{dt}}+\frac{{r}_{{\rm{S}}}}{{r}_{0}^{2}}\cos \phi \sin \phi \frac{d\phi }{{dt}}.\end{eqnarray} \tag{ 47 }$$

We also have

$$\begin{eqnarray}&&r\displaystyle \frac{d\phi }{{dt}}=c\cos \phi ,\end{eqnarray} \tag{ 48 }$$

and so

$$\begin{eqnarray}&&\displaystyle \frac{{dr}}{{dt}}=c\cos \phi -\displaystyle \frac{{{cr}}_{{\rm{S}}}}{r}\sin \phi ,\end{eqnarray} \tag{ 49 }$$

using also the relation in equation (44), since gravity perturbs this trajectory only slightly in the weak-field limit.

A second differentiation results in the expression

$$\begin{eqnarray}&&\displaystyle \frac{{d}^{2}r}{{{dt}}^{2}}=\displaystyle \frac{{c}^{2}}{r}{\cos }^{2}\phi +\displaystyle \frac{{{cr}}_{{\rm{S}}}}{{r}^{2}}\sin \phi \displaystyle \frac{{dr}}{{dt}}-\displaystyle \frac{{c}^{2}{r}_{{\rm{S}}}}{{r}^{2}}{\cos }^{2}\phi .\end{eqnarray} \tag{ 50 }$$

In this equation, however, the first term is simply the effect of seeing the straight line trajectory (equation (44)) in spherical coordinates. The radial acceleration due to gravity in equation (50) is the rest of the righthand side. Thus, subtracting the first (geometric) term, substituting for dr/dt from equation (49), and noting that r_S ≪ r in the weak-field limit, we arrive at the Newtonian acceleration for a photon

$$\begin{eqnarray}&&\displaystyle \frac{{d}^{2}r}{{{dt}}^{2}}=-\displaystyle \frac{2{{GM}}_{g}}{{r}^{2}}\left(1-2{\sin }^{2}\phi \right).\end{eqnarray} \tag{ 51 }$$

The dependence of this expression on the angle ϕ is simply due to the fact that the photon experiences zero acceleration in its longitudinal direction, so equation (51) gives solely the component of acceleration in the radial direction. Be aware, however, that the approximations we have made in reaching this result are valid only near ϕ = 0, so this expression is not valid when $\sin \phi \to 1$. On the other hand, given that ϕ → 0 when r → r₀, this simply reduces to our final result,

$$\begin{eqnarray}&&\displaystyle \frac{{d}^{2}r}{{{dt}}^{2}}=-\displaystyle \frac{2{{GM}}_{g}}{{r}^{2}},\end{eqnarray} \tag{ 52 }$$

which is the maximal radial acceleration experienced by a photon transverse to its direction of motion. Equation (52) is the weak-field (r ≫ r_S), radial acceleration experienced by a photon moving perpendicular to $\hat{r}$ in the vicinity of a gravitational mass M_g . It represents the Newtonian, static limit of General Relativity for particles we believe to have zero rest mass, the analog of equation (1). Aside from their evident similarity, the other feature that stands out is the additional factor 2 emerging in the latter, which owes its appearance to the same physics responsible for the famous factor 2 in the calculation of the deflection angle of starlight grazing the surface of the Sun. This factor 2 appears as long as we use the same value of the gravitational constant, G, in both equations (1) and (52). It is not due to a doubling of the source in Poisson's equation (29), which arises from the contribution of momentum to the active gravitational mass, and would be absorbed into the overall 'measured' value of M_g , independently of G.

It is instead due to the different geometries of particles with and without inertia along their longitudinal direction of motion. Simply put, particles with inertia satisfy equation (1) in the low-velocity limit because, for them, the only metric coefficient in the Schwarzschild spacetime (equation (30)) that matters is g_tt , i.e. the time dilation. As noted earlier in this section, Einstein himself used solely the effects of g_tt to estimate the bending of light passing near the Sun, even though a lightwave was thought to be massless. These particles are moving too slowly compared to the speed of light for the distance covered during their acceleration to contribute significantly to ds². For particles propagating at lightspeed, however, the impact of g_rr cannot be ignored compared to g_tt . In essence, the effects of spacetime curvature are doubled for photons compared to particles with established non-zero inertia. The inclusion of spatial variations resulting from g_rr , once the full theory of general relativity was available to him, is the reason Einstein's recalculation of the deflection angle of starlight passing near the Sun doubled the effect he had anticipated in 1911.