We know that gravity speeds up a body; for instance, a meteor which enters the earth gets constantly accelerated up by earth's gravity. And from relativity we know that light bends near a massive body, because Newton's law of gravitation is just an approximation and actually gravity depends on energy and momentum. So my question is: If a ray of light is aimed exactly at the center of a body, then will it get accelerated like a meteor? And if does get accelerated, then won't it surpass the universal speed limit of 3,00,000 km/s (approx.)?

  • $\begingroup$ Among other considerations, "no" because $c$ is invariant according to the theory(ies) of relativity. $\endgroup$ Feb 11, 2014 at 18:07
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    $\begingroup$ To expand slightly: the easiest way to see this in the formulation of the theory is that gravity is a force that couples to (acts on) mass, and light (photons) have zero mass. Gravity does not exert a force on light. In terms of light curving, it's more accurate to think of bending light as light travelling in a "straight path" within a curved spacetime than light travelling on a curved path within a flat spacetime. $\endgroup$
    – Kyle Oman
    Feb 11, 2014 at 18:13
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    $\begingroup$ @Kyle I'd actually say that's quite misleading because gravity couples to all forms of energy, not only mass. (Otherwise it would have no effect on light at all!) But its effect on light is not something that we would recognize as a force. $\endgroup$
    – David Z
    Feb 11, 2014 at 18:30
  • $\begingroup$ @DavidZ agreed, which is what I tried to mitigate with the second part of the comment. Probably reads better if you ignore the words "couples to". Photons don't experience gravitational force but do experience gravitationally curved spacetime. $\endgroup$
    – Kyle Oman
    Feb 11, 2014 at 18:41
  • $\begingroup$ As David alludes to, gravity isn't considered a force in GR. Instead, free-falling is defined as being free of any external forces, feeling only the effects of the local gravitational field. This is backed by the (Einstein) equivalence principle and means we have to view spacetime as being curved due to mass (and energy), which changes the free-falling paths. Light always follows these paths, or (null) geodesics, and therefore bends in the presence of mass/energy, but there is no force accelerating it because gravity isn't actually a force. $\endgroup$
    – Wouter
    Feb 11, 2014 at 18:46

3 Answers 3


If a ray of light is aimed exactly at the center of a body, then will it get accelerated like a meteor?

Short answer: no. However, when falling in a gravity field, the momentum of light increases.

Some background...

In Newtonian mechanics, the rate of change of momentum of a (massive) particle is proportional to the acceleration:

$$\frac{d\vec p}{dt} = m \vec a $$

In Relativistic mechanics, these quantities are not proportional. In fact, an accelerating massive particle can never reach speed $c$ but the momentum can reach arbitrarily large values.

This is because relativistic momentum is a non-linear function of velocity

$$\vec p = \frac{m \vec v}{\sqrt{1 - \frac{v^2}{c^2}}} $$

which diverges as $v \rightarrow c$.

In the special case of a massless particle, which must travel at speed $c$ in all frames, the numerator and denominator in the above are zero so, by this formula, the momentum of a massless particle is indeterminate.

However, the relativistic energy-momentum relation

$$E^2 = (pc)^2 + (mc^2)^2 $$

gives the momentum of a massless particle:

$$p = \frac{E}{c} $$

Thus, the momentum can change even though the speed does not. In falling from a higher potential to a lower potential, the massless particle gains energy and thus momentum but not additional speed.

For light, the momentum and frequency are proportional:

$$p = \frac{h\nu}{c} $$

so, while the speed of light does not increase as it falls, the frequency of light increases. From the Wikipedia article "Blueshift":

Photons climbing out of a gravitating object become less energetic. This loss of energy is known as a "redshifting", as photons in the visible spectrum would appear more red. Similarly, photons falling into a gravitational field become more energetic and exhibit a blueshifting

  • $\begingroup$ The formula to calculate speed of a wave is speed = wavelengthfrequency. So, if frequency increases during blueshift and wavelength remains constant then won't the speed increase? And secondly, momentum = massvelocity. So, no massless object will have momentum. Is that right? $\endgroup$
    – Yashbhatt
    Feb 18, 2014 at 0:30
  • $\begingroup$ @user40382, why do you assume wavelength remains constant? And, as I wrote, relativistic momentum is not $mv$. $\endgroup$ Feb 18, 2014 at 2:06
  • $\begingroup$ So how can we say that wavelength varies? Because the velocity is constant? $\endgroup$
    – Yashbhatt
    Feb 22, 2014 at 18:21
  • $\begingroup$ @user40382, $c = \lambda \nu$ and $c$ is constant thus, if frequency increases, wavelength decreases. $\endgroup$ Feb 22, 2014 at 19:40
  • $\begingroup$ Sorry made a mistake.... You are right. $\endgroup$
    – Yashbhatt
    Feb 23, 2014 at 5:10

The short answer has been given a few times in the comments: gravity only bends light, it doesn't speed it up. Of course this statement in and of itself is useless if you want to understand nature. So let's dig a bit deeper. To understand why gravity acts this way, the first step is the equivalence principle. Now, there are several versions of this but we'll look at the Weak Equivalence Principle (WEP) and the Einstein Equivalence Principle (EEP) in particular.

Inertial mass and gravitational mass

Every particle has a mass, which makes it necessary to apply a force on it in order to change its velocity (to make it move). In other words, there is a relation between the applied force and the change in velocity, and the proportionality constant is what we call the inertial mass (it determines exactly how difficult it is to change the particle's velocity). This statement is summarized in Newton's second law:

$$\vec{F} = m_i\vec{a}$$

Now, in a (static) gravitational field $\Phi(\vec{r})$, we have Newton's law of gravitation, which basically states that the force felt by a particle in this field is proportional to the gradient of the field. In this case we call the proportionality constant the gravitational mass $m_g$:

$$\vec{F} = m_g\vec{\nabla}\Phi$$

A priori, there is no need for these two proportionality constants to have any relation to each other whatsoever. However, experiments have shown that they are in fact the same.

The equivalence principle

The Weak Equivalence Principle (WEP) comes down to the statement $m_g = m_i$ (gravitational mass is numerically equal to inertial mass).$^1$ As mentioned before, this is a well-known experimental observation. It was most famously made by Galilei and by the Apollo 15 crew on the moon but most accurately by Eötvös' torsion experiments and subsequent similar experiments. The conclusion is this: all (point) particles fall at the same rate in a gravitational field, regardless of their mass. Or: the "gravitational charge" is universal, i.e. the same for all particles.

This suggests that we may view gravity as something special. Since the "gravitational charge" is the same for all particles, nothing is unaffected by gravity. Therefore, there is no such thing as a "gravitationally neutral" object with respect to which we could reliably measure the acceleration due to gravity. We are led to the conclusion that we must drop the notion of an acceleration due to gravity and define "unaccelerated" as "freely falling", i.e. as being subject only to gravity. This means gravity does not cause acceleration in the usual sense and by our definition of the notion it is not a force.

The Einstein Equivalence Principle simply follows from extending the argument from the motion of freely-falling particles to the whole of physics. The EEP is then a natural extrapolation of the WEP:

In sufficiently small regions of spacetime, the laws of physics reduce to those of special relativity; it is impossible to detect the existence of a gravitational field by means of local experiments.

The EEP is what GR is based on and as mentioned before (everything that was a consequence of the WEP is also a consequence of the extension that is the EEP), it means that we no longer think of gravity as a force and that we are led to define "unaccelerated" as "freely falling". This has some profound consequences to how we do physics and in particular to how we measure distances, velocities, ... But those aren't what I'll focus on here.

Curved spacetime

To incorporate the conclusions of the EEP into our physical theory (most relevantly to us the fact that gravity is not a force field propagating through spacetime, but seems to be an inherent property of nature itself) we have to adjust our model of the physical universe. Thus, we imagine that spacetime is curved. We assume it has some curved geometry to it and gravity is a manifestation of that curvature.

The mathematical description for this curved spacetime will be what is known as a differentiable manifold. There is no need to go into the details here, but for our understanding a (differentiable) manifold is simply a space that looks locally like the real plane $\mathbb{R}^2$ but might look very different on a global scale. This corresponds well to what the EEP requires for our physical model: a general, curved global geometry that locally looks like flat space.

Now, in flat space we know what people mean when they say "a straight line": a straight line is the shortest distance between two points. However, things become a bit more complicated in general geometries. Again, no need to go into the details, but we can generalize the idea of a straight line to curved spacetime and we call this generalization the geodesic. Geodesics in flat space are straight lines, in curved spacetime they are less straightforward but can be determined from the metric on the spacetime in question.

The idea is that freely-falling particles follow these geodesics. More precisely: there are three types of geodesics: timelike, lightlike (or null) and spacelike. Again I don't want to go into the details, but physical particles are restricted to move on timelike paths and therefore will follow timelike geodesics when no other forces are acting on them. Light always follows lightlike geodesics. Spacelike geodesics are unreachable for physical particles since they require faster-than-light travel.

Bending of light and gravitational frequency-shift

Now, suppose a light ray approaches a massive object, like our beloved sun. In the vicinity of the sun the spacetime is strongly curved and therefore so are the geodesics. Consequently the light ray will curve around the sun, something you would never expect from a massless particle in Newtonian mechanics. In fact, this experiment was one of the first confirmations of GR.

Another consequence of the EEP is fairly separate from the notion of geodesics. Imagine an experiment in which you have two identically uniformly accelerating rockets at a constant (small) distance from each other. If the trailing rocket sends out a photon which is received by the leading rocket some time later, the latter will have picked up some speed in the mean time and a Doppler shift will occur.

However, since the EEP states that we shouldn't be able to tell the difference between uniform acceleration and gravitation, the same effect should arise if we perform the following experiment. Imagine two physicists are at rest in the earth's (static) gravitational field: one on surface, the other at a (small) height above it (in a tower). If the first physicist sends out a photon up to the tower, we expect the second one to measure a redshift as well. Indeed this effect of redshift as a photon climbs out of a gravitational potential well has been measured, as well as the converse in which a photon falls into the well (Alfred explains this nicely).


So in conclusion: the reason why light rays bend in the presence of a large mass (energy) is that spacetime is curved strongly there and this causes the geodesics to be bendy instead of straighty. Since gravity shouldn't be seen as a force, this is not due to an acceleration of the photons in the usual sense. Similarly, we shouldn't think of a photon falling into a gravitational potential well as being accelerated in the usual sense, rather its frequency changes - as Alfred illustrates in his answer.

$^1$ An alternative, more precise, statement of the WEP is:

In sufficiently small regions of spacetime, the motion of freely falling particles is the same in a gravitational field and in a uniformly accelerated frame.

Note that I added the qualifier in sufficiently small regions of spacetime, which is needed since in general a gravitational field varies spatially and so the difference with a uniformly accelerated frame could be detected in extended regions of spacetime.


This question has a twist to it! As others have answered, gravity does not change the local speed of light as it falls past you like a meteor.

However, gravity does change the apparent speed of light you see if it passes through a gravitational potential that is different than your own. For example, we see radar pulses take longer to bounce of Venus and return to Earth when the Sun is near the path of the radar photons (Shapiro Delay). We interpret this as the light slowing down as it passes deep thru the sun's gravitational potential and can even calculate the bending of the light by the part of the wave farther from the Sun going a little faster than the wave nearer to the sun (like refraction at an interface). Likewise, if we were deep in the sun's gravitational potential timing the passage of far away light, we would infer that it had sped up to greater than our standard local c.


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