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Gravity causes anything with energy to accelerate toward the source. Black holes, for example, have such strong gravity that they pull in light and don't let any escape. But can acceleration still apply to light? The speed of light is constant, of course, but why are photons affected by gravity yet aren't accelerated by it?

Edit: My main question is why photons aren't affected in the same way as most other particles. I'm perfectly aware that it cannot surpass lightspeed, but I want to know what makes it unaffected by acceleration while other particles are affected.

  • This question seems relevant: http://physics.stackexchange.com/q/33816/. – udrv Nov 12 '15 at 05:10
  • More on faster-than-light in GR: http://physics.stackexchange.com/q/2230/2451 , http://physics.stackexchange.com/q/24319/2451 , http://physics.stackexchange.com/q/36500/2451 and links therein. – Qmechanic Nov 12 '15 at 14:30
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    I've removed some comments; a reminder to everyone that comments are mainly for suggesting improvements and requesting clarification, not for extended discussion. – David Z Nov 12 '15 at 15:21
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  • A gravitational field is a place where the "coordinate" speed of light varies. The ascending light beam speeds up, and light curves like sonar. See this answer. Light changes speed and direction, so gravity does accelerate light. But the downward light beam doesn't go faster, it goes slower. it doesn't go – John Duffield Jan 05 '16 at 13:40
  • "Light changes speed and direction, so gravity does accelerate light" LIght gains energy, not speed as it already and always is "c". See gravitational blue shift (rarely heard). I wonder about the change of direction - is pure change of direction without change of direction some change of energy? It's not. – Peter Bernhard Nov 21 '22 at 15:59

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Photons are blue-shifted when attracted by gravity (I mean - moving towards a mass, not moving at right angles to the gravitational field like in an orbit). They can't go faster, but their energy goes up.

Floris
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    But the "right angle" case also occurs with gravitational lenses, doesn't it? And this is considered acceleration, it just doesn't change the size of the vector... – yo' Nov 12 '15 at 17:26
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    @yo' yes, but I was talking about blue shift, which doesn't happen when the attraction is at right angles to the path of the photon. Note that during gravitational lensing, the photon is only at right angles to the gravitational potential at one specific point; before that, it is moving towards the attracting mass (blue shifting) and after, it is moving away again (red shifting). – Floris Nov 12 '15 at 17:28
  • "And this is considered acceleration, it just doesn't change the size of the vector.." The answer should deal with that statement. Direction seems to be one parameter of a vector, of the impulse. Energy doesn't seem to change, when there is no change of wavelength. – Peter Bernhard Nov 21 '22 at 16:01
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You don't feel acceleration. When onboard the ISS, you are accelerating towards the earth (down) due to gravity: if you didn't, you would just fly away from the planet. Because you and the ISS are accelerating exactly the same way, you don't feel a thing. You don't feel a force if it's accelerating you: you feel pressure caused by opposing forces. Here on the ground, I feel the floor beneath my feet opposing my normal gravitational acceleration.

If you run a thruster on the ISS, then the ISS starts accelerating differently than you do, and eventually, one of the walls is going to collide with you. Then you'll feel that wall interfering with your own gravitation acceleration, and feel something like weight.

Light undergoes acceleration due to gravity: look up 'gravitational lensing' for that. To understand how light can accelerate with a constant speed, you have to understand the difference between speed and velocity, and what acceleration really means.

Speed is a 'scalar', just a number with no direction. If you're travelling 30 KPH, that's your speed.

Velocity is a 'vector', a number with a direction. Driving 30 KPH north is much different than driving 30 KPH south: clearly, you'll end up in different locations regardless of your speed.

Acceleration is not a change in speed, it is a change in velocity. Think of a car. There are usually three ways to accelerate a car. To increase your speed (scalar), step on the accelerator, and you'll feel your seat back pushing into you harder as it accelerates you with the car. To decrease the speed (scalar), hit the brake and you'll feel your safety straps accelerating you with the car.

But what happens when you turn? Your speed stays roughly the same (exactly the same if you're skilled enough), but you are changing your direction. Your 30 KPH north is becoming 30 KPH west, and the change in direction is an acceleration. Depending on whether your car is build to drive on the right or the left, you'll have a tendency to either push against your door or into your passenger's lap. That's still acceleration.

If a photon is passing by something heavy, it will be accelerated towards that object, changing its course but not its speed. If a photon is going towards or away from something heavy, it can't properly accelerate by changing speed. I'm not a physicist, but I believe that it increases or decreases energy by changing its frequency. In other words, things that you would expect to increase its speed will instead increase its frequency ('blue-shifting' it if it's visible light), and what you would expect to decrease its speed will instead decrease its frequency ('red-shifting' if it's visible light).

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    doesn't the photon change its trajectory because the space itself is warped? so it keeps following those straight lines in that warped space? – Mehdi Nov 12 '15 at 14:54
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    A wall hitting you in space seems like it would be a very frightening sensation... – corsiKa Nov 12 '15 at 17:29
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    @Mehdi: Two different ways of modelling the same thing. It's all abstractions when you get right down to it. – Lightness Races in Orbit Nov 13 '15 at 18:09
  • @corsiKa: See: USS Jenolen vs dyson sphere – Lightness Races in Orbit Nov 13 '15 at 18:09
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    Just to clear up any doubt: the statements preceded by "I'm not a physicist" in the final paragraph are entirely correct. (Though the terms "blue shifting" and "red shifting" are used for all forms of electromagnetic radiation as well as for visible light.) – N. Virgo Nov 13 '15 at 18:42
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To talk about acceleration in space is a little bit dangerous without exact definition. One has to separate free fall and acceleration from an impulse.

Imagine, you are inside the ISS during an orbit correction. The impulse from the rocket engine you could feel, you get some weight, and and this is an acceleration.

In all other time you are weightless and does not feel any acceleration. But without doubt there is something what held the spaceship in his orbit. To call it a force is not correct because you has to feel an acceleration (doe to Newtons formula F = m a). That is the reason why Einstein does not more talk about gravitational forces but about curvature of space.

What is about the curvature of space and time? Due any particle or body follow the same path, when starting from the same point in the same direction? This is not the case. The straightest reachable path is with photons, they have the maximum possible velocity, but still are influenced by the curvature of space. For all other body the path is more bended under the influence of gravitational masses.

It is possible to answer a question about does the light travel with different speed in space. 1) No, if answer is in a frame close to the particle. 2) Yes, if answer it from a point in deep space. Near a black hole the photon runs slower from the point of view of a far away from this hole observer.

HolgerFiedler
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    "To call it a force is not correct because you has to feel an acceleration." This statement is wrong. Just because you don't "feel" it doesn't mean that it's not there. The acceleration due to gravity is still there (that's what's holding you in orbit). – Beska Nov 12 '15 at 14:11
  • @Beska I'm not saying that there is no acceleration in orbital move. But do you feel this acceleration. The anser has to be No. On earth or in orbit, if a force accelerate you (what is accompanied by an impulse), did you feel this? Answer is Yes. So there has to be a difference between this two accelerations. – HolgerFiedler Nov 13 '15 at 16:56
  • @trichoplax The speed of light is a local constant. If you are nearby the moving EM radiation (you are in its frame), c is always 300.000 km/sec (at least this is one of our basic constants in physics). But if you in space far away and with different gravitational density (potential=, our watches are going faster or slower. Near a black hole with its high gravitational potential the photon runs slower and on the boundary of the black hole it has the slowest possible velocity. – HolgerFiedler Nov 13 '15 at 17:08
  • @HolgerFiedler thanks for clarifying. Since someone upvoted my misunderstanding comment I've deleted it to avoid leading others astray. – trichoplax is on Codidact now Nov 13 '15 at 19:17
  • "Near a black hole with its high gravitational potential the photon runs slower and on the boundary of the black hole it has the slowest possible velocity." Yet the gravity will influence both the photon and the observer, won't it? In such case the rate of both clocks will be the same. – bright magus Jan 18 '16 at 15:00
  • @bright magus 1. It is common sense that c THE one constant number to have a basis for calculations. 2. A observer far away from the black hole has another time than near the hole. This not contradicts your statement – HolgerFiedler Jan 18 '16 at 15:18
  • An observer far away from the black hole can only see light that is coming to his own eye (measurement device). He cannot measure light that is near the black hole because he cannot see it. Also, one-way lightspeed measurement is not possible. – bright magus Jan 18 '16 at 19:04
  • I expected this objection. The question about c as a local constant number troubles Einstein and he say that c depends from the gravitational potential. As well, as you can measure c only on earth or nearby, the dependence of c from gravitational potential has a theoretical solution only. – HolgerFiedler Jan 18 '16 at 19:13
  • c is always local and therefore it will always be c. – bright magus Jan 20 '16 at 10:09
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According to Einstein and many others, light speed c0 in vacuum is universal and measures about 299,792,458 m/s. So it is never possible to change the light speed in vacuum, which is the absolute upper limit for everything.

Gravity does not affect a light ray, but the space and time through which the ray travels.

In a strong gravitational field, the time passes slower, according to observers far away from the field. For observers affected by the field it is still c0 though.

And the space in this field is expanded, stronger the closer we get to the center of the field. If we imagine space as 2D plane, we could say that it gets a deep dent where the mass is. The light ray that travels through this dent now travels perfectly straight in his view, but observers far away see a curved path because the space is curved.

The noticeable "gravitational redshift" (if light travels from an observer into a gravitational field) or "blueshift" (if light comes out of a gravitational field to an observer) for external observers has its cause because the photon gathers energy while entering the gravity well while it loses energy when it leaves it. This is potential energy, depending on the gravity potential the photon is located at. As masses create gravity wells with low potentials, the "absence" of a (strong) gravity field can be called high potential. This energy difference can't express itself in a speed difference (kinetic energy), because we already said that light speed is constant. Instead, the potential energy converts itself into light/electromagnetic (please correct me if the term is wrong) energy, as it can be described with E=hf. That means the photon's energy is proportional to its frequency. An increase of the electromagnetic energy of the photon, because it enters a lower gravity potential and converts its potential energy, therefore results in a higher frequency, which is visible as a blueshift.

  • Please tell about some lensing effect that only changes the direction and path of the light, not its wavelenght, not its energy. Has the photon been accelerated in that case? – Peter Bernhard Nov 21 '22 at 16:03
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I would suppose the short answer is no, but photons are affected by gravity.

What is happening is naturally quite relativ to the observer. Suppose you are sitting on the photon, travelling past the gravity source with the speed of light. As has been argued above you would experience the force of the gravitational pull as an acceleration toward the source. Also, if you were to observe the passage of time during the journey, you would find that your clock is running slower than the clock of an observer travelling further away from the gravity source. To this "further away traveller" you would therefore appear to be travelling faster, even though your both travelling at the speed of light.

Ultimately, what this means is, while the gravity-source affects the path of the photon, the speed isn't affected due to the relative change in time.

At least thats how I understand it anyway. :)

trugr
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