I am trying to understand the issue of light bending in relativity vs classical physics. I will clearly describe my understanding/assumptions in case I am wrong on those.

  • In relativity, unlike classical physics, gravitation is not a force that act between two bodies, rather it is the effect of the space-time distortion caused by a large mass
  • The "curving" is independent of the mass of the second body (to an approximation, because the second body will also distort space-time, but the geodesics in the "curved" space by the first object are the same for any other object)
  • In classical physics, since photons have no mass, their trajectory will not be bent by the large body
  • In classical physics, if we add to the 0-inertial mass the relativistic mass $E=m*c^2$, then, even in classical physics the bending of the trajectory for a photon should be apparent
  • However, the bending due to the relativistic mass in classical physics is different than the bending predicted by the theory of relativity
  • This link describes the bending using relativistic mass and Newtonian physics to be a deflection of 0.87, while the deflection due to Relativity should be 1.75 in the famous 1919 experiment

Is this difference due to the fact that in Newtonian physics the deflection depends on the energy of the light (using the formula $E=m*c^2$) and, in the experiment, it was calculated to be about 0.87, while, in Relativity, such deflection is for the most part independent of the energy of the photon?

More in general, how were those values calculated? Did they know the energy for the photons for the particular star, and calculated the photons relativistic mass using $E=m*c^2$ in the first case, while they simply calculated the space-time distortion caused by the mass of the Sun in the second case, calculated the geodesics, and ignored the mass of the photons and assumed the photons would just move along those geodesics to calculate the deflection? Or am I totally off in my assumptions and overly simplifying a much more complex theory?


2 Answers 2


The value of 0.87 arcseconds is $2GM/Rc^2$, where $G$ is Newton's gravitational constant, $M$ the mass of the Sun, and $R$ the radius of the Sun. This is derivable by considering the photon to have any nonzero Newtonian mass $m$, and considering the hyperbolic trajectory under Newtonian gravity of a particle of mass $m$ moving with velocity $c$ and having impact parameter $R$ (because the light ray just grazes the Sun).

For a quickie derivation, you can look up the deflection angle for Rutherford scattering and simply substitute $\frac{Z_1 Z_2 e^2}{4\pi\epsilon_0}\rightarrow GMm$ to change from electrostatics to Newtonian gravity. They're both inverse square forces, so the hyperbolic trajectories are the same. Take the velocity $v$ to be $c$ and the impact parameter $b$ to be $R$. Using the small-angle approximation for the arctangent, you get $2GM/Rc^2$.

Another derivation is here.

The angle is independent of whatever nonzero Newtonian mass the photon is assumed to have, for the same reason that a cannonball and a golf ball fall at the same rate in vacuum. The gravitational force is proportional to $m$, but the gravitational acceleration is independent of $m$ because in


the $m$ simply cancels out. If you want to consider the photon's Newtonian mass to be the relativistic mass of its energy, as in $m=E/c^2=hf/c^2$, it doesn't matter what its energy is or its frequency is.

The observed, Einsteinian, twice-as-large value, $4GM/Rc^2$, comes from computing the deflection of a Sun-grazing null geodesic in the Schwarzschild metric using General Relativity. All photons have zero invariant mass, regardless of their energy, and thus move on null geodesics.

So you can see that the two theories both have the feature that the deflection is independent of the photon energy, but they achieve it in different ways.

  • $\begingroup$ Why is there a difference from the deflection for a photon, but not, say, for any object with non-zero mass? $\endgroup$
    – user
    Commented May 12, 2019 at 20:21
  • 1
    $\begingroup$ @user A difference between what and what? Between the Newtonian value and the Einsteinian value? I’m fairly sure that there is a difference for massive particles as well! Elliptical planetary orbits are definitely slightly different in GR than in Newtonian gravity, as the precession of the perihelion of Mercury shows. Hyperbolic orbits of, say, an asteroid or a cosmic proton passing through the solar system are surely slightly different in the two theories. $\endgroup$
    – G. Smith
    Commented May 12, 2019 at 20:52
  • $\begingroup$ @user For highly relativistic massive particles, I would expect the deflections to approach those for a photon. For non-relativistic massive particles, I would expect the difference in deflections to be small. $\endgroup$
    – G. Smith
    Commented May 12, 2019 at 21:00
  • $\begingroup$ So basically, for massive slow moving objects, the deflection predicted using Newtonian physics just happens to match that calculated using the space-time geodesics. Is this because, if an object move slowly, the geodesics in the space-time look close to those in the space-only coordinates? $\endgroup$
    – user
    Commented May 12, 2019 at 21:25
  • $\begingroup$ @user I didn’t say they match. For example, elliptical orbits have zero precession in Newtonian gravity and nonzero precession in GR. I haven’t looked at how the GR precession behaves in the limit of low velocity, so I don’t know whether it goes to zero but it probably does. I assume this is because GR has Newtonian gravity as its weak-field limit. $\endgroup$
    – G. Smith
    Commented May 12, 2019 at 21:56

In the paradigm of Newtonian gravity one must have a mass (inertial = gravitational mass) in order to couple to gravity. However, the acceleration (the effect of gravity on a mass) is independent of the mass. A truly massless particle would not be acted upon in Newtonian theory though back then they probably did not have a clear view of what light was back then. They assumed a small mass and argued that regardless of the value the acceleration would be present. The previous answer did a good job of showing this by comparison to other processes.

In GR the curvature of space-time is responsible for this. The geodesics of the space-time manifold are the paths traveled by test particles. However, in contrast to Newton's law of gravity, every form of energy and momentum can generate curvature (not just mass). This mean that even force fields and other types of energy are capable of curving space-time. This can be interpreted as saying everything couples to everything else in GR. As for the bending of light observation and other such phenomenon, a very large mass is considered as the source of curvature and all other particles (smaller in mass or energy than the source) zip around on the geodesics created by the "source". Massive particles will move on time-like geodesics while photons, as well as other massless particles, will move on Null geodesics (curves of zero length in 4-dim space-time). Using GR the angle calculation is done by comparing the null geodesic path to that of the straight line path one would have with no massive source present.

One thing I want to point out in my answer is that, regardless of the way the deflection is calculated, the two theories differ not only in how gravity is generated (force field versus space-time curvature), but also in what can generate gravitational effects (all sources of energy-momentum-stress, etc). Everything contributes not just mass. This is a key point in evaluating the theoretical validity of previous estimates.


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