By Eka


2012-01-31 08:02:20 8 Comments

Photons travel at the fastest speed in our universe, the speed of light.

Do photons have acceleration?

6 comments

@David Z 2012-02-01 00:33:11

By definition, acceleration is a change in velocity. Velocity is a vector, so it has a magnitude (also known as speed) and a direction. For light, the magnitude does not change, but the direction might.

For example, consider a light ray bouncing off a mirror. The incoming light is going in one direction, the outgoing light is going in another direction, so presumably there is some acceleration involved. However, actually calculating the acceleration gets a little tricky because $\vec a_\text{avg} = \frac{\Delta \vec v}{\Delta t}$, and it's hard to precisely define $\Delta t$, the time that it takes the light to bounce off the mirror. Once you look closely enough, the action of reflection starts to seem more like absorption and re-emission.

You might also ask the same question about light traveling in empty space, free of anything to reflect off of. Here it gets a little tricky though, because light traveling in empty space might or might not accelerate depending on how you define "constant velocity."

In Newtonian physics, we tend to define "constant velocity" relative to the Earth, or the sun, or some other convenient object. In this view, the direction of a light ray (and thus its velocity) will change as it passes by some massive object and is attracted by gravity. So when you look at it that way, light does accelerate in empty space.

But in general relativity, "constant velocity" is defined in terms of parallel transport - not relative to a convenient object, but relative to the way the object's path "naturally" extends itself through spacetime. In this sense, the "natural" extension of a path is a geodesic, precisely the kind of trajectory that an object (or a light ray) follows under the influence of only gravity. So in the general relativistic sense, light rays do not accelerate in the absence of materials.

@anna v 2012-01-31 09:56:09

Gravitational lensing has been observed which means that the photons bend. An acceleration can be defined in its change of direction, angular acceleration in radians/second^2, so the answer is positive, yes, light can be accelerated, but its speed will still be c locally.

@Timtam 2012-02-04 09:58:47

this is incredibly innacurate, gravitational lensing is a GR effect whereas a photon exists in the quantum regime, GR doesn't apply to it on those length scales. You would have to make some argument of effective curvature but this is way beyond anything possible to cover here

@anna v 2012-02-04 11:46:05

@Timtam these photns that take part in the gravitational lensing have been measured, it is and experimental fact. The curvature is known and thus an angular velocity can be estimated.I do not see why you need to complicate the problem.Think of it as a black box, not knowing how the photons bend.

@P O'Conbhui 2012-03-31 01:14:01

This is not exactly acceleration of a photon. The photon is still following a null geodesic, so in the frame of the light, there is no acceleration. There might only appear to be some acceleration to another observer.

@Nikolaj-K 2012-08-08 11:34:57

I think that if you "define acceleration as change in direction" and count gravitational lensing as change in direction, then you should be able to give a procedure to compute that direction in which the light beam is traveling if it does not change direction. The problem is that you can reduce your answer to "Due to gravitation, the notion of 'motion in an unchanging direction' is obsolete." If you want to trick the system by saying the observed path is curved (although "observed bath" is quite abstract here), then the answer becomes trivial again: "The acceleration depends on the observer."

@Eduardo Guerras Valera 2013-01-12 19:51:33

@Timtam, it is not innacurate. Photons, or plane electromagnetic waves, or whatever model you want to use for light, move (or propagate or however you want to call it) along null geodesics, which result not only in curved light paths from a distant observer, but also in time propagation delays. This is bread and butter for extragalactic astronomers and cosmologists... It is there and it is measured and used every day, and cannot be denied or called innacurate.

@Eduardo Guerras Valera 2013-01-12 19:57:23

@Nick Kidman, the procedure to compute that direction change was stablish first by Laplace, although the numerical result was half the correct one (he was using newtonian gravity) and later correctly by Einstein. Google for "Gravitational Lens Equation"

@Richardbernstein 2013-01-13 04:09:05

Under Anna's analysis, light paths "Bend" under the influence of gravity. Since the universe is filled with gravitating objects which in principle all have gravitational fields pervading all of space, wouldnt this mean that photons are never going "straight" because they are always moving through space that is "bent"by gravity?

@anna v 2013-01-13 04:43:08

@Richardbernstein the "bent" is determined by the geodesic in GR framework. en.wikipedia.org/wiki/Geodesics_in_general_relativity

@anna v 2013-01-13 04:53:52

@Richardbernstein see also David's reply which clarifies the matter

@user4552 2013-07-06 00:35:27

@PO'Conbhui: Your objection is correct, but: ...in the frame of the light... There is no frame of reference corresponding to lightlike motion.

@user4552 2013-07-06 00:37:27

@EduardoGuerrasValera: null geodesics, which result not only in curved light paths from a distant observer, but also in time propagation delays. This doesn't make sense. Curvature is a local notion, and for a world-line curvature is measured relative to the curvature of a geodesic, which is zero by definition. What's curved is not the geodesic followed by a light ray. What's curved is the spacetime through which the geodesic is drawn.

@anna v 2013-07-06 04:29:17

@BenCrowell by this logic even a falling apple which is following a space curvature, is not accelerating. In the framework of GR space distortions acceleration has no meaning , imo. It is all space geometry.

@user4552 2013-07-06 04:37:20

@annav: Right, the free-falling apple has no acceleration. Acceleration does have meaning, however. The same apple sitting on the ground has an acceleration of 9.8 m/s2.

@Eduardo Guerras Valera 2013-07-06 22:37:24

@BenCrowell, please note the words from a distant observer. The calculation of the path in space that results from propagation along a null geodesic is by all means, as seen from a distant observer, curved. Yes, curvature is a local concept unless you have a trivial, constant metric tensor. But the resulting light path is, as seen by a distant observer, curved, unless you have a better explanation for the known couple hundreds or so multiple quasars.

@Eduardo Guerras Valera 2013-07-06 22:44:26

@BenCrowell, in quasars, time delays and multiple imaging due to ligth paths bending as seen from our point of view (=distant observers, lens galaxies are tipically at z~0.5 or more, not to mention the quasar sources!) are well modeled by assuming lens potential models and then seeing what results from the propagation along null geodesics (from which you get an equivalent reffractive index). What part doesn't make sense?

@P O'Conbhui 2013-07-07 06:11:14

@BenCrowell: Of course. But it made for a shorter response and was only a little wrong.

@Kilo 2012-08-08 11:34:02

If the redshift factor differs from the change in the rate of change of affine parameter under parallel transport, yes.

@Schroeder 2012-08-08 05:00:32

The speed of light is a constant, but it can accelerate by changing direction. For instance, light accelerates under the influence of gravity like every other falling object. Reflection is not a very meaningful form of acceleration since the $\Delta t \approx 0$, but it certainly a change in velocity.

Light can also gain and loose energy like other accelerating objects. It does this by changing wavelength (red/blue shifting).

@dmckee --- ex-moderator kitten 2012-08-08 11:42:38

As others have discussed, these paths are along null geodesics...what you would call free fallpaths in Newtonian physics and can not properly be said to be accelerating. Sorry, but that is just the way General relativity works.

@sciencefan 2012-01-31 14:48:51

In the vacuum, the light follow the minimun distance of the space-time even if the is the massive object (like gravitational lense).Light alway goes right away according to the space-time metric. Photon doesn't loss energy => it not accelrate/decelerate

@anna v 2012-01-31 15:31:32

It is not losing energy in a gravitational lens case. Light loses energy with the expansion though, the frequency changes. The speed is still c.

@Siyuan Ren 2012-02-01 02:20:09

Light-like geodesic is not "the minimum distance", because all light-like paths are of zero length.

@Timtam 2012-01-31 09:01:27

I think you mean to ask whether a photon can accelerate. The answer is yes. For instance when travelling from one medium to another light changes speeds. In fact, the actual photon is abosrbed and a new photon is reradiated during this process. However, you could call this effectively light acceleration

@Eduardo Guerras Valera 2013-01-12 20:00:59

In contrast with that, a gravitationally deflected photon in vacuum is not absorbed or redadiated, but simply changes direction and speed as seen from a distant observer. So I think the answer by anna v is more accurate. A photon approaching a massive object in vacuum will decelerate as it approaches the deflecting mass, without being absorbed or reemited (for that reason, I think the answer given by anna v is by far the right one)

@P O'Conbhui 2013-07-07 06:14:47

This doesn't really describe acceleration. Is there a point at which the velocity of the photon is half way between the velocity it travels through the two media? If not, the jump in velocity is discontinuous, is not differentiable, and so, cannot be described as having an acceleration.

@Jitter 2015-03-14 20:40:11

You need to study up on group velocity, phase velocity and set velocity;-)

@Alfred 2019-10-15 21:47:10

I don't think this answers the OP's question. Contrary to your argument, anna v's and Schroeder's answers are more to the point.

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