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Light doesn’t accelerate in a gravitational field, which things with mass would do, because light has a universally constant velocity. Why is that exception?

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    $\begingroup$ Light is affected by gravity in general, that's why there's gravitational lensing. Why would gravity from black holes be any different? $\endgroup$
    – Barmar
    Jan 11, 2019 at 16:12
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    $\begingroup$ You might be interested in How can gravity affect light?, Why doesn't gravity speed up light? and Speed of light in a gravitational field? on the Physics SE. $\endgroup$ Jan 11, 2019 at 16:28
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    $\begingroup$ Title was changed from "Black Holes" to "Gravity", which made the existing comments and answers confusing. $\endgroup$ Jan 11, 2019 at 19:58
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    $\begingroup$ Great, now it's just a bunch of confusing answers. Gravity bends space-time, the medium through which light travels. $\endgroup$
    – Mazura
    Jan 13, 2019 at 23:57
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    $\begingroup$ Photons have no rest mass -- they never rest, do they. But each existing photon has the mass which is equivalent to its energy ($m = \frac{E}{c^2}$, or $\frac{\hbar*\nu}{c^2}$). $\endgroup$ Jan 14, 2019 at 10:15

6 Answers 6

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Another way to answer this question is to apply the Equivalence Principle, which Einstein called his "happiest thought" (so you know it has to be good). The equivalence principle says that if you are in an enclosed box in the presence of what Newton would call a gravitational field, then everything that happens in that box must be the same as if the box was not in a gravitational field, but accelerating upward instead. So when you release a ball, you can imagine the ball is accelerated downward by gravity, or you can imagine everything but the ball is accelerated upward, and the ball is simply being left behind (which, ironically, checks better with the stresses you can easily detect on every object around you that are not present on the ball, including the feeling you are receiving from your bottom right now).

Given that rule, it is easy to see how light would be affected by gravity-- simply imagine shining a laser horizontally. In the "left behind" reference frame, we see what would happen-- the beam would start from a sequentially higher and higher point, and that raising effect is accelerating. So given the finite speed of light, the shape of the beam would appear to curve downward, and the beam would not strike the point on the wall of the box directly opposite the laser. Therefore, this must also be what is perceived from inside the box-- the beam does not strike the point directly across from the laser (as that point is getting higher then the point across from it where the light was emitted), and its path appears to curve downward. Ergo, light "falls."

Indeed, this is the crucial simplification of the Equivalence Principle-- you never need to know what the substance is, all substances "fall the same" because it's nothing happening to the substance, it is just the consequences of being "left behind" by whatever actually does have forces on it and is actually accelerating.

Incidentally, it is interesting to note that even in Newtonian gravity, massless objects would "fall the same" as those with mass, but to see it requires taking a limit. Simply drop a ball in a vacuum, then a lower mass ball, then a lower still mass. All objects fall the same under Newtonian gravity. So simply proceed to the limit of zero mass, you will not see any difference along the path of that limit. Nevertheless, Newtonian gravity doesn't get the answer quite right for the trajectory of light in gravity, because Newtonian physics doesn't treat the speed of light correctly.

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    $\begingroup$ Well, Newtonian gravity, $ F = \frac{Gm_1m_2}{r^2}$ certainly goes to zero force when the mass is zero. $\endgroup$ Jan 11, 2019 at 17:25
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    $\begingroup$ @CarlWitthoft, F = ma also goes to zero when the mass is zero. Applying zero force to zero mass gives you any acceleration you wish. $\endgroup$
    – Mark
    Jan 12, 2019 at 0:38
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    $\begingroup$ I'm torn with this answer... it just explains that light is affected by gravity (by way of a thought experiment, no less). The OP knew that already - he wanted to know why (which is a can of worms, sure...). The equivalence principle is not the reason for the light behaving its way, it is just a didactic helper... $\endgroup$
    – AnoE
    Jan 12, 2019 at 17:12
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    $\begingroup$ @JohnDuffield, ok, my comment was a bit flippant maybe. But in an accepted answer, I would at least like to see curved spacetime mentioned. Or a mention that all massless things always move at the speed of light. Again, the E.P. is a "principle" - it is a thought experiment, not the physical reason for light behaving as it does. $\endgroup$
    – AnoE
    Jan 12, 2019 at 21:04
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    $\begingroup$ @AnoE - Newtonian physics had no answer to why gravity apparently acts instantaneously, with the force proportional to the product of the masses divided by the square of the distance. It's an ad hoc explanation. General relativity explains that (and also explains where that Newtonian model falls short), but it does not explain why the speed of light is (locally) constant, nor does it explain what makes space-time curve. Quantum mechanics has similar issues with "why" questions. Physics answers "what" questions. "Why" questions are the purview of metaphysics, philosophy, and religion. $\endgroup$ Jan 12, 2019 at 23:01
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There are a couple of ways one could approach your question:

Black holes are regions of space that have been deformed by a sufficiently concentrated mass. Light waves/particles always travel in a straight line at a constant velocity ($c$). Although a photon approaching a black hole will continue traveling in a straight line through space, space itself has curved so the photon's path will curve.

While photons don't speed up in the presence of a gravity well, they are affected by it in other ways. In specific, photons entering a gravity well are blue-shifted while photons leaving one are red-shifted. This red/blue-shifting happens because time passes slower within a gravity well than without. In all frames of reference, though, the speed of light remains constant. There's more info on this on the wiki.


Note: The question originally referred specifically to black holes. The above hold for any concentration of matter (of which black holes are an extreme example).

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  • $\begingroup$ Alex - I removed "black hole" from the title -- just in case some other commentor complains about that. $\endgroup$ Jan 11, 2019 at 17:23
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    $\begingroup$ "In all frames of reference, though, the speed of light remains constant." Well, all inertial frames of reference anyway, locally anyway. $\endgroup$ Jan 11, 2019 at 18:46
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    $\begingroup$ Adding to Shufflepants comment, the speed of light is constant locally. If a beam of light approaching a black hole was observed from a sufficiently far away distance, the beam would appear to be slowing down, since time would be passing slower in the gravity well around the black hole. $\endgroup$
    – rcgldr
    Jan 13, 2019 at 17:08
  • $\begingroup$ @rcgldr, Shufflepants Quite true; I simplified it a bit. Things tend to get a bit messy in the presence of deep gravity wells. $\endgroup$ Jan 14, 2019 at 5:50
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    $\begingroup$ Photons travel in straight lines through curved space-time, not space. $\endgroup$
    – ProfRob
    Apr 18, 2019 at 16:51
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TL;DR Light is affected by gravity because it travels along the space-time grid and its curvature which IS gravity. This gets very visible in black holes. also: Einstein > Newton

Black holes are black because no light that crosses the "Event Horizon" can escape ever again. Mass bends the "grid" of space-time. Light - 2-dimensionally speaking - travels along the floor of the space-time grid and follows its curvature i.e. it goes down a cone created by a presence of mass, and moves along the shortest path outwards again. This makes the journey of the light take longer. Now for a black hole things are more extreme: A black hole forms, when a lot of matter is crammed into a space that is at or smaller than the Schwarzschild Radius. The Schwarzschild Radius of any stellar object is determined solely by its mass. Any mass with a high enough denisty turns into a black hole:

rs = 2 * G / 2 c
Schwarzschild Radius = 2* the gravitational constant / 2 * the speed of light.
Multiply that with M, the mass of an object in kg and you got the rs for that mass.

To understand however how black holes curve space so much that they let no light escape, we must look at only a small part of Schwarzschilds equation.

To paint an image for understanding black holes, we only need this middle section:
1)Schwarzschild Radius equation rs/r 2)Schwarzschild Radius equation rs/rs 3)Schwarzschild Radius equation 1/1 4) Schwarzschild Radius equation 0,singularity,doom
We've already established rs as being the Schwarzschild Radius of a particular object, r is the radius of the stellar object. When r becomes as small as rs you get a singularity1 and weird stuff starts happening, most importantly to OPs question, the space-time curvature at the black hole becomes infinite(!), this means that any light that intersects the event horizon at any point will take an infinite amount of time to travel across the black holes funnel. Even at a very flat angle relative to the event horizon, where its just ever so slightly poking it, it is lost because set theory teaches us: any subset of infinity is also infinite.

Here are some extra visalisiations:
Gravity space-time cone of earth:

Gravity space-time funnel of a black hole:

(source: interestingengineering.com)


  1. Singularity: A singularity is basically, in calculus/algebra terms, just when you divide by zero. A 2D singularity might just look like this: f(x) = 1/x (the singularity is there in the middle at x=0).

    A 3D singularity can look like this /, singularity at x=1 (this is Riemanns zeta function).
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    $\begingroup$ There is no infinite curvature at the event horizon, only a coordinate singularity. $\endgroup$
    – ProfRob
    May 2 at 19:32
  • $\begingroup$ What do you mean with "there is no infinite curvature", the funnel just stops right after the event horizon? As far as I understand (and I don't), it just gets too funnely for c at the event horizon $\endgroup$
    – Penguin9
    19 hours ago
  • $\begingroup$ BTW, you can write equations using MathJax on this site. See physics.stackexchange.com/help/notation & math.stackexchange.com/help/notation $\endgroup$
    – PM 2Ring
    18 hours ago
  • $\begingroup$ There's some good info in this answer, but also some stuff that's not quite right. Eg, "set theory teaches us: any subset of infinity is also infinite". That's not true: infinite sets have both finite and infinite subsets. As ProfRob says, there's a coordinate singularity at the EH (event horizon) if you use Schwarzschild coordinates, which means that to the Schwarzschild observer it takes infinite time for anything (including light) just to cross the EH, so you can't really use Schwarzschild t to talk about what happens after something crosses the EH. $\endgroup$
    – PM 2Ring
    17 hours ago
  • $\begingroup$ BTW, a photon going towards a BH is doomed once it crosses the photon sphere at $\frac32r_s$, as discussed in this recent question on our sister site: physics.stackexchange.com/q/715922/123208 $\endgroup$
    – PM 2Ring
    17 hours ago
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Acceleration is not relevant here. Any given gravity well has a definable escape velocity. Particles faster then that velocity escape the well, particles slower do not. The very definition of a black hole is a gravity well (hole) where the escape velocity exceeds 'c' the speed of light particles, so by definition, light can not escape from the hole, making it 'black'.

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    $\begingroup$ I don't think that's what the OP was asking - more along the lines of "why gravitational lensing?" $\endgroup$ Jan 11, 2019 at 17:25
  • $\begingroup$ The current question no longer resembles the one originally asked. $\endgroup$
    – Paul Smith
    Jan 12, 2019 at 19:24
  • $\begingroup$ That isn't the definition of a black hole. $\endgroup$
    – ProfRob
    May 2 at 19:33
  • $\begingroup$ @ProfRob - perhaps you would care to offer a more succinct definition of a black hole? $\endgroup$
    – Paul Smith
    May 3 at 22:37
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Interestingly light or photons do not (according to theory) have “rest mass” ie no mass at rest. But when it is moving (photons are never at rest) they do have mass given by

E = mc2 = hν ⇒ m = E/c2 = hν/c2=h/cλ

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    May 2 at 20:41
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If light has no mass, why is it affected by gravity?

Because light has a wave nature, and a gravitational field is a place where the speed of light varies. So light curves downwards. It's rather like the way sonar waves tend to curve downwards in the sea:

enter image description here

Image from FAS and the US Navy, see course ES310 chapter 20

Light doesn’t speed up its acceleration, which things with mass would do, because light has a universally constant velocity. Why is that exception?

That's not right I'm afraid. See what Einstein said:

1912: “On the other hand I am of the view that the principle of the constancy of the velocity of light can be maintained only insofar as one restricts oneself to spatio-temporal regions of constant gravitational potential”.

1913: “I arrived at the result that the velocity of light is not to be regarded as independent of the gravitational potential. Thus the principle of the constancy of the velocity of light is incompatible with the equivalence hypothesis”.

1914: “In the case where we drop the postulate of the constancy of the velocity of light, there exists, a priori, no privileged coordinate systems.”

1915: “the writer of these lines is of the opinion that the theory of relativity is still in need of generalization, in the sense that the principle of the constancy of the velocity of light is to be abandoned”.

1916: “In the second place our result shows that, according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity and to which we have already frequently referred, cannot claim any unlimited validity”.

1920: “Second, this consequence shows that the law of the constancy of the speed of light no longer holds, according to the general theory of relativity, in spaces that have gravitational fields. As a simple geometric consideration shows, the curvature of light rays occurs only in spaces where the speed of light is spatially variable”.

Einstein also spoke of “the refraction of light rays by the gravitational field”. So did Newton, see Opticks query 20: “Doth not this aethereal medium in passing out of water, glass, crystal, and other compact and dense bodies in empty spaces, grow denser and denser by degrees, and by that means refract the rays of light not in a point, but by bending them gradually in curve lines?” It really is a refraction, and gravitational lensing is an appropriate phrase. Also see the GR section of Is The Speed of Light Everywhere the Same? It’s a PhysicsFAQ article by editor Don Koks. He talks about Einstein and refraction, and says this: "light speeds up as it ascends from floor to ceiling, and it slows down as it descends from ceiling to floor; it's not like a ball that slows on the way up and goes faster on the way down". Isn't that interesting?

A lot of people will tell you that light curves "because it follows the curvature of spacetime", but that isn't right either. Spacetime curvature is associated with the tidal force, not the force of gravity. See my "physics detective" articles on the speed of light and how gravity works for details and references.

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  • $\begingroup$ Would the downvoters like to explain why they think Einstein, Newton, and Koks were all wrong? $\endgroup$ Jan 13, 2019 at 16:46
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    $\begingroup$ More importantly, would the upvoter care to explain that up-vote? This answer is at best fringy. $\endgroup$ Jan 14, 2019 at 3:03
  • $\begingroup$ @David Hammen : It isn't fringe. Einstein wasn't wrong, and nor is this answer. Also see Professor Ned Wright's deflection and delay of light: "In a very real sense, the delay experienced by light passing a massive object is responsible for the deflection of the light". $\endgroup$ Jan 14, 2019 at 8:03
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    $\begingroup$ You tend to confuse the coordinate speed of light, which is just a computational tool but is often regarded as depending on location in a gravitational field, with what any observer would locally measure for the speed of light using their own local ruler and clock. For example, in your link you say " There is no time flowing inside an optical clock, so the height at some location on the plot depicts the real speed of light at that location", but the only thing a physicist can mean by "the real speed of light" is a measurement. Nonunique coordinate systems involve nonunique quantities. $\endgroup$
    – Ken G
    Jan 14, 2019 at 13:15
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    $\begingroup$ You can use whatever quotes you like, the fact is, all local observers will always measure the same c. Any other version of "the speed of light" is a ramification of the chosen coordinates. Coordinate language is not unusual-- we say "time slows down in a gravitational potential" etc.-- but that's also just a coordinate comparison. No observer experiences slowed time-- ever. It's not the invariant. $\endgroup$
    – Ken G
    Jan 15, 2019 at 0:06

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