Light is bent near a mass (for example when passing close to the sun as demonstrated in the famous sun eclipse of 1919). I interpret this as an effect of gravity on the light.

However, it seems (to me, at least) that light is not accelerated when it travels directly toward the (bary-)center of the sun. The same gravitational force applies yet the speed of light remains constant (viz. $c$).

What am I missing?

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    $\begingroup$ Do you mean that the light does not change its speed or its momentum? Because the first one is simple: light travels always at the speed of light as it is mass-less (special theory of relativity). The bending due to the mass is a different effect. Do you mean: why does it not increase its momentum? $\endgroup$
    – Mayou36
    Commented Jun 7, 2017 at 17:37
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    $\begingroup$ If it satisfy you, it "kind of" accelerates: light is blueshifted if shot directly into a massive body. Since light speed is fixed, it gains energy by increasing its frequency (or shortening its wavelenght). $\endgroup$
    – lvella
    Commented Jun 7, 2017 at 19:07
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    $\begingroup$ Wait a minute - isn't the OP asking precisely about this: wikipedia.org/wiki/Pound–Rebka_experiment physics.aps.org/story/v16/st1 The Pound–Rebka experiment. $\endgroup$
    – Fattie
    Commented Jun 8, 2017 at 15:02
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    $\begingroup$ I've removed some comments about "slow light" and "stopped light" in materials, which is a QED effect irrelevant to this question about general relativity. $\endgroup$
    – rob
    Commented Jun 8, 2017 at 19:54
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    $\begingroup$ It is light travelling in "curved" 4-dimensional space-time. Light travels straight (without any acceleration) but the space time is curved due to gravity, this makes it look like bending of light. $\endgroup$
    – user92340
    Commented Jun 12, 2017 at 4:13

6 Answers 6


You missed a key aspect of general relativity (GR):

The gravitational 'force' we observe is an illusion according to general relativity. Specifically, the 'gravitational force' and 'gravitational acceleration' observed on an object is an illusion generated by the object moving along a geodesic on four-dimensional space.


In general relativity, the presence of mass and energy warps four-dimensional space time, thereby inducing spatial curvature. The greater the presence of mass and energy in a given location, the greater the induced spatial curvature. When any particle (massless or not) travels into this curved space, the particle will continue to travel in a straight line (absent outside forces); but, since the space it is travelling upon is curved, its global path will be curved.

As an analogy, draw two straight lines on a sphere (a curved surface) travelling in different directions. Locally (at small two-dimensional distances), the lines travel in a straight direction, without veering. Globally (at three dimensions), we see that its path is curved and will inevitably intersect (on the other side of the sphere). We call these paths geodesics. The math concerning geodesics involves differential geometry which makes heavy use of multivariate calculus.

Now back to general relativity. GR predicts that the gravitational forces we observe are the manifestation of four-dimensional spacetime being warped by the presence of mass-energy. A common analogy made is the trampoline-well model shown below. A heavy mass sitting upon a trampoline curves the surface of the trampoline. Any objects then move towards the heavy mass has its path deflected towards it. Now, I must stress an important simplification made in such diagrams: these diagrams reduce four-dimensional spacetime to three spatial dimensions. The XY plane of the diagram represents the XYZ components of spacetime whereas the Z axis of the diagram represents the T component of spacetime. For math lovers, they are reducing $(\vec{X}, \vec{Y}, \vec{Z}, \vec{T})$ to $(\sqrt{\vec{X}^2 + \vec{Y}^2}, \vec{Z}, \vec{T})$

Curvature of Light from a star behind the sun.

Now here's the cool part:

Now instead of its path curving along the $\vec{X}\vec{Y}$ plane as seen in the photo above, its path would curve against $\vec{Z}$ (vertical). In this context though, $\vec{Z}$ does not refer to the Z direction, but to $\vec{T}$. What this means is that, observers would see the particle 'accelerate' in through time and apparently 'slow down'. Namely, they will see gravitational time dilation.

EDIT. I made a mistake: the light does accelerate! It merely does so according to the rules of special relativity.* When objects (massless or not) pass near a gravitational well they pick up gravitational energy and accelerate, thereby gaining kinetic energy. For objects with mass, this means an increased velocity (hence gravitational sling shots). For massless particles (such as photons), this typically means an increased frequency or blue-shifting as Jeremy pointed out in a separate answer. (Thank you Peter, Rob, and Jeremy for pointing out this oversight.)


You may have noticed a contradiction here. According to special relativity and observations, objects in gravitational wells do accelerate. Case in point: gravitational sling-shooting. According to general relativity though, the 'gravitational force' we observe is a manifestation of four-dimensional space-time curvature. So which is it: is there a force or not? Not-really: it's a matter of frames-of-reference. From our reference frame we see acceleration; but, from the four-dimensional space-time reference we see pure geodesic motion.

Hence, gravity isn't a force acting upon an object, but rather the object moving along a geodesic path that manifests the appearance of acceleration.

  • $\begingroup$ For comparison, what happens if you send a massive particle in the same direction at, say, 0.999c? $\endgroup$ Commented Jun 7, 2017 at 18:52
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    $\begingroup$ Generally speaking, it'll draw out a similar geodesic (straight line on a curved surface). $\endgroup$ Commented Jun 7, 2017 at 18:59
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    $\begingroup$ Ok, that's what I thought. So really, the reason, in general relativity, that a photon experiences no acceleration in this situation is not because it is massless, but because no particle experiences true acceleration in this situation in the context of general relativity. $\endgroup$ Commented Jun 7, 2017 at 19:05
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    $\begingroup$ I've updated my answer to address both of your questions generally. @kayleeFrye_onDeck: your memory is referencing the introduction to geodesics. $\endgroup$ Commented Jun 7, 2017 at 19:33
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    $\begingroup$ Since no external force acted upon the photon, it never accelerated - How does this relate to gravitational slingshots of massive objects? It's pretty clear that the object gained speed after performing the maneuver and the large mass lost speed. The well doesn't merely change direction for massive objects $\endgroup$
    – Rob
    Commented Jun 9, 2017 at 5:42

One thing that the previous answers are missing -- the light is accelerated; it just is accelerated according to the rules of special relativity, which says that it cannot pick up speed when already travelling at the speed of light.

Instead, it gains kinetic energy the way a photon gains kinetic energy -- by being blueshifted to a higher frequency, which does translate to more energy, according to the Planck relation $E = h\nu$.

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    $\begingroup$ Exactly. Of course the Energy–momentum relation for massless particles $E=pc$ means that the momentum of the photon also changes (in magnitude, not in direction) because of this acceleration. The photon gains momentum. This relates to the very first comment (to the question, not to the above answer), by Mayou36. $\endgroup$ Commented Jun 11, 2017 at 14:37
  • $\begingroup$ Refering to "(in magnitude, not in direction)". What about famous Einstein's photon from a star behind the sun being deviated but not blue-shifted? This amounts to a gain in momentum? If the photon had mass there would be radial acceleration, thus gain in momentum, for sure. Curvature of space seems to relate. $\endgroup$ Commented Nov 21, 2022 at 13:13
  • $\begingroup$ @PeterBernhard in that case, the photon is interacting with the star, and there is a momentum transfer $\endgroup$ Commented Nov 21, 2022 at 17:54
  • $\begingroup$ I'd also note that in that case, to see the momentum transfer, you have to break the assumption that the mass of the star is much larger than $E_{\rm photon}/c^{2}$, and you're really not talking about a photon anymore $\endgroup$ Commented Nov 21, 2022 at 18:44
  • $\begingroup$ "... photon is interacting with the star ... " By star you mean our wandering star, the sun. So it there a momentum transfer that should shift the wavelength. In fear of being removed to chat that is my last comment. Thank you. $\endgroup$ Commented Nov 22, 2022 at 11:02

Currently, there is no evidence that photons have mass, and it is generally accepted that they are massless particles.

Nonetheless, gravitation does affect the path of photons, because the bending of space-time causes all particles to travel on curved paths, including massless ones. But that does not mean than light will be accelerated. The speed of light (299,792,458 m/s) is an absolute maximum, and it may not decrease from that nor may it increase.

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    $\begingroup$ The speed of light (299,792,458 m/s) is a maximum, and it may decrease than that rather than increase. ... no, no, it may not. Massless particles always travel at c, when they're traveling, not above it, nor below it. The difference between c and the apparent speed at which light traverses different mediums is a result of the time spent by photons interacting with particles in the medium they're travelling through. $\endgroup$ Commented Jun 7, 2017 at 14:39
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    $\begingroup$ @Gnudiff If you were to track the path of an individual photon through a medium, it would travel in a straight line, at c, through the mostly empty space of the medium, until it "impacted" one of the particles composing the medium. At this point, the photon would get absorbed by the particle, exciting it to a higher energy state. After a brief time, the photon would be re-emitted by the particle. This process of absorption and re-emission takes time, and the vector of the re-emitted photon is not necessarily the same, so the path a photon takes is not a straight line. ... $\endgroup$ Commented Jun 7, 2017 at 15:32
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    $\begingroup$ @Gnudiff ... the combination of these two effects (the time the photons spend "inside" the atoms of the medium, rather than travelling) and the fact that the photons are re-emitted at different vectors when they "exit" the atoms of the medium, lengthening their path relative to a straight line, account for the difference between c and the apparent speed of light in a medium. In short, when travelling through a medium, photons spend time "inside" atoms and also zig zag between atoms as they are re-emitted at different vectors, which lengthens the distance they travel. $\endgroup$ Commented Jun 7, 2017 at 15:40
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    $\begingroup$ @industry in short (but I'm oversimplifying here) every interaction of light in a medium results in a fixed phase shift for the reemitted photon and if a plane coherent wave is incident on the medium then averaging over all these interactions will result in a coherent plane reemitted wave. $\endgroup$
    – LLlAMnYP
    Commented Jun 7, 2017 at 21:19
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    $\begingroup$ @industry7 to expand on the other response a bit, it comes down to a matter of probabilities. Any one photon could end up being retransmitted in any direction, but the odds are such that in aggregate, the "average" path will conform to the laws of optics (refractive index, angle of incidence = angle of reflection, etc). It's kind of like flipping a coin or a roll of a die. A coin doesn't "know" it's supposed to come up heads half the time, and you can't predict the result of any one flip, but do enough of them, and the results "average out" to 50/50. A simplification, but an accurate one. $\endgroup$ Commented Jun 8, 2017 at 3:17

Does the speed of light change due to the Sun? Well, yes and no. There are two ways in which one can think about speed in General Relativity. One is the coordinate speed, which means the rate of change of the spatial coordinate with respect to time coordinate of the coordinate system that you can choose at your will. Another is the speed as viewed from a special frame, namely the locally inertial frame in the vicinity of the considered light quanta. The fundamental aspect of General Relativity is that the Physics in a locally inertial frame is exactly the same as the Physics of Special Relativity. But due to gravity, these small-small local inertial frames are so arranged globally that a global inertial frame can't be formed. Now, in a generic frame of reference, i.e., in a generic coordinate system, the speed of light can certainly be different from $c$ and, in fact, it can even change with time.

For example, the speed of a photon moving radially in the vicinity of a spherically symmetric and static object is given by $v=\dfrac{dr}{dt}=c\bigg(1-\dfrac{2GM}{rc^2}\bigg)$ if you choose your spatial coordinates to be the spherical coordinates centered at the massive object with radial coordinate $r$ and keep track of time with a clock situated far away from the spherical object (star). As you can clearly see, the speed can vary with the radius $r$. Although, the speeding up and speeding down occurs in somewhat counter-intuitive fashion. An out-going photon seems to be speeding up and an in-going photon seems to be slowing down. Again, if you go to a local inertial frame, the speed is invariable $c$ but you don't have any frame that is inertial and can describe the motion of light for a finite time or within a finite region of space.


You are missing special relativity and general relativity. In special relativity the speed of light in vacuum is always c, no matter the reference frame of measurement.

Also classical electromagnetism, light, emerges from a confluence of the quantum mechanical constituents which are photons and have zero mass. A photon aiming at the barycenter of the sun is attracted by the gravitational field of the sun but the effect is not a change in velocity , but on its energy which is $E=h\nu$ and therefore the extra energy increases the frequency while the velocity remains at $c$.

  • $\begingroup$ "You are missing special relativity and general relativity. In special relativity the speed of light in vacuum is always c, no matter the reference frame of measurement. " This might sound like the solution to the question is that in general relativity, the speed of light actually is increased by gravity. $\endgroup$
    – JiK
    Commented Jun 7, 2017 at 15:34
  • $\begingroup$ @jik not for local systems, those follow special relativity $\endgroup$
    – anna v
    Commented Jun 7, 2017 at 16:26
  • $\begingroup$ What about the bending? Does it amount to change of momentum? There is no mass thus no impulse. And there is no gain of energy as there is no blue-shift, the energy does not change. $\endgroup$ Commented Nov 21, 2022 at 13:17

In Einstein's 1911 paper "The influence of gravitation on the propagation of light", he pointed out that from his generalizrd theory of relativity, the speed of light as viewed from us is different at different locations in a gravitational field. From the Huygens principle, a change in the speed of light causes the wave front to tilt over toward the sun. The exact amount of the deflected angle of star light passing near to the sun is by his calculation 0.85 arcsecond(which he later corrected it to be 1.75 arcsecond). The bending of light path near to the sun is thus caused by the variable speed of light.

  • $\begingroup$ Doesn't this mean that the curvature is vir $\endgroup$ Commented Nov 21, 2022 at 13:18

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