Basically, the equivalence principle (EP) states that if someone is in a rocket in empty space with acceleration $g$ equal to that at the surface of the earth, any experiment he does cannot distinguish whether the rocket is accelerating in that manner or the rocket is just sitting on the surface of the earth. So if he were to let go of a ball, it would fall to the floor of the rocket in either case. If he were to throw the ball horizontally, it would follow a parabolic path.

Now let's consider that the observer inside the box shines a beam of light horizontally. Because in the first case the rocket is accelerating, the light will follow a parabolic path and strike the wall at a slightly lower height. EP therefore predicts light would also follow a curved path inside the stationary rocket, i.e., gravity bends light.

But the bending will be based on the acceleration $g$, so isn't EP incorrectly predicting the Newtonian bending of light, which is one half of the value obtained using general relativity?

Let me guess at an answer: EP only holds for a "small" rocket and within that approximation the general relativity and Newtonian predictions match?


4 Answers 4


We can compare the deflection of light in an uniformly accelerated rocket and the same deflection near the surface of a celestial body. The requirement is that the distance travelled by the light ray is small enough to take the body surface as a plane, and the acceleration of gravity be taken as constant.

In this case the Newton theory is a good aproximation, and the ray deflects as any projectile, in a parabolic path.

It is different when trying to measure the deflection of a light ray (coming from a star) passing near the sun, and observed from the earth.

In this case the equivalent principle can not be used because the requirement of a limited spacetime region (which can be approximated as a flat spacetime) is no more fulfilled.

  • $\begingroup$ Are you saying that the GR bending prediction will match the Newtonian parabolic path if the light path is small enough, without having to resort to the Newtonian limit of GR? $\endgroup$ Aug 31, 2020 at 1:13
  • $\begingroup$ Yes. That type of high school problems where $g$ is constant is a newtonian limit for "flat earth" so to speak. It is the same idea for the GR limit for flat spacetime, that is the principle of equivalence. $\endgroup$ Aug 31, 2020 at 2:06

The Newtonian prediction for the bending of a test particle that just grazes the sun, as a function of its speed $v$ as it grazes the sun, is $$\frac{2GM}{R\,v^2}.$$ The GR prediction, for large enough $v$ (or small enough deflection), is roughly $$\frac{2GM}{R}\left(\frac{1}{v^2}+\frac{1}{c^2}\right).$$

In other words, it's not really twice the Newtonian prediction; it's the Newtonian prediction plus another term that's independent of the speed.

The second term can be understood as a consequence of the "angular defect" around the sun, the same thing that's responsible for the anomalous precession of Mercury. Another way it manifests is that if you transport a gyroscope around a circle centered on the sun, it will come back rotated by an angle that depends only on the radius of the circle and the sun's mass and not on the speed at which you transported it.


I added a note, then a second later saw what you were actually asking:

But the bending will be based on the acceleration g, so isn't EP incorrectly predicting the Newtonian bending of light, which is one half of the value obtained using general relativity?

I think you are mixing metaphors, but I'm, not sure so here goes...

general relativity and Newtonian predictions match

We do not expect the general relativity and Newtonian predictions match. We expect the two GR predictions to match each other. That's the basis for the EP - it's illustrating a difference with classical mechanics, not suggesting they have to be the same.

Indeed, in the classical case you would expect no bending in the gravitational examples because under classical EM light has no gravitational mass. One can go back to the corpuscular model (Newton would!) and develop a limiting trajectory based on a test mass' behaviour as its mass goes to zero. And yes, that will result in a different answer too. And, well, of course it would, it's a different theory.

The entire model of GR is totally different than classical, and it predicts that all sorts of things will be different than classical. One of those differences is the EP stating that gravitational and inertial mass are one and the same. Any physics that depends on that is going to have different results.

  • $\begingroup$ Understood - I should not expect the GR bending of light from EP as GR is different than classical. But that leads me to a followup question. GR reduces to Newton's law of gravity in the appropriate limit e.g. weak gravity. So does GR bending of light reduce to what one gets from Newton's corpuscular model in that Newtonian limit or is there still a factor of 2? $\endgroup$ Feb 15, 2019 at 13:44
  • $\begingroup$ @Not_Einstein The factor of 2 remains even in the weak limit. So it's an exception to the rule that GR in the weak limit is Newtonian gravity. But light bending is a weak effect anyway, it's only relevant when the light beam passes close enough to the Schwarzschild radius, otherwise a straight line is a reasonable approximation. $\endgroup$
    – Cuspy Code
    Feb 20, 2019 at 18:11
  • 1
    $\begingroup$ Someone on another forum suggested this article which addresses this question: mathpages.com/rr/s8-09/8-09.htm Skip down to the paragraph that begins with "People sometimes wonder how this doubling of the deflection can be reconciled with the equivalence principle." I can't really say that I understand the explanation but get the general gist of it. $\endgroup$ Feb 23, 2019 at 21:41
  • 3
    $\begingroup$ This just seems wrong. The e.p. says that GR locally equates to SR. $\endgroup$
    – user4552
    Nov 18, 2019 at 3:07

When Einstein was working on the field equations of GR, the initial value that he had obtained for the bending of light was the same as the Newtonian value. He even hired a team of astronomers to measure it, but thy were captured during the World War (which turned out to be good for Einstein, as he had the wrong predictions.) It was only later, when refining his equations to match other observation like the perihelion of Mercury, did he realize that the correct value for the bending of light should be twice his original (and even Newtonian) one.

So, why does the Equivalence Principle predict the bending of light according to $g$? Because it was dreamt of before the realization that gravity is related to the curvature of spacetime. After that realization, bending of light went beyond the value of $g$, it was a question of pure geometry. And why didn't anyone notice this? Because try measuring he bending of light at normal $g$ values, and it is too small to measure, impossible for the 1910's. Even the actual experiment done to measure bending of light by the Sun was too small to be observed. But, in thought experiments, it works just fine.

  • $\begingroup$ So I take it this means that the two situations in the Equivalence Principle are not totally equivalent. $\endgroup$ Jul 29, 2020 at 16:37
  • $\begingroup$ @Not_Einstein they are indeed equivalent. It is only that gravity is governed by different, rules. $\endgroup$
    – PNS
    Jul 30, 2020 at 1:37
  • $\begingroup$ But a constant acceleration elevator gives a different result for the bending of light. $\endgroup$ Jul 30, 2020 at 14:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.