Does Newtonian mechanics predict the bending of the course of light by objects with mass?

$F=Gm_1m_2/d^2$

$F=ma$

$a_g=Gm_{other}/d^2$

In Newtonian mechanics, the acceleration of object A toward object B is not dependent on the mass of object A but on the mass of object B and the distance between objects A and B. Because the mass of object A does not affect its acceleration due to the gravitational influence of object B, does Newtonian mechanics predict that a massless particle (e.g. a photon) would be gravitationally affected by an object with mass?

Also, would it predict that an object with mass would not be gravitationally influence by an object without mass. Would it also predict that two massless particles would have no gravitational influence on each other?

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Related - Newtonian Bending of Light? –  voix Jun 9 '12 at 23:17
Laplace calculated how massive a star would have to be for no light to escape. This calculation can be found in Laplace's essay in the Appendix to Hawking/Ellis Large Scale Structure of Spacetime. –  MadScientist Jul 13 '12 at 15:47

Newton obviously knew that the mass of an object falling under the influence of Earth's gravity has no effect on its acceleration, i.e., all objects accelerate toward Earth at 32 ft/sec/sec regardless of their mass ("weight"). Therefore it follows that an object with no mass, such as a photon, would follow the same rule: it accelerates to earth at 32 ft/sec/sec. (The reason we don't notice this is that photons spend so little time between the object we see and our eyes due to their extreme speed.)

Newton obviously knew this, and logically concluded that photons from distant stars grazing the Sun's limb (edge) would "fall" just a bit towards the Sun as they passed by, resulting in a slightly curved trajectory.

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Good insight. I have never thought of this way.+1 –  Aftnix Jul 11 '12 at 18:49
I edited your question out because answers are not the place to ask questions - but that would be a great one to post as an independent question. –  David Z Jul 12 '12 at 1:55
The answer to the deleted question is because the space-space part of the Schwartzschild metric tensor is of the same form as the time-time part, and the photon has a trajectory where both contribute equally, so you get a factor of 2. It's an accident of the Schwartzschild solution. As for this, Newton thought light was made of massive particles. –  Ron Maimon Jul 12 '12 at 9:53
+1. In the 18th century, objects somewhat similar to black holes were hypothesised for exactly this reason - if a star were heavy enough then light could not escape from it, no matter the "mass" of light. See en.wikipedia.org/wiki/Dark_star_(Newtonian_mechanics) –  Nathaniel Jul 13 '12 at 15:42
@Peter I don't understand why a particle without mass, in Newton theory where the interaction is supposed to be proportional to both the masses of the particles considered, should feel a gravitational field. Could you pleas clarify this point? –  usumdelphini Jul 14 '12 at 4:18
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What theory of light do you want to use? There doesn't seem to be any reasonable way to discuss Newtonian mechanics and photons; photons are innately quantum. We can't very reasonably discuss Newtonian gravity and classical electromagnetism, either, since Newton's gravitational law is not Lorentz-invariant, while classical electromagnetism manifestly is.

There was a time when people did calculate the effects of Newtonian gravity on light. Here is a paper from 1804 that does it. One approach is take an object of mass $m$ and initial velocity $\mathbf{v}$ moving past a star with some specified impact parameter. As $m \to 0$, the trajectory of the object converges, so we can take that as a solution for the path of light. (The trajectory it converges to is that of a "test particle" that has no gravitational influence of its own.) I don't know of any significant applications of such a theory.

Today we know that gravity is manifested as the curvature of spacetime, and so affects light, which travels along null geodesics in spacetime. The effects of the gravity of entire galaxies on light traveling through them can be very dramatic. This is what people study in the field of gravitational lensing.

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Would it work to say that you want to talk about Newtonian mechanics and the quantum theory of light? –  yakiv Jun 9 '12 at 22:16
@yakiv: It has exactly the same problem of incompatible symmetries. The quantum theory of light retains the Lorentz invariance of the classical theory. –  dmckee Jun 9 '12 at 23:48
It seems that the paper has now moved here. –  Kitchi Nov 23 '13 at 17:44

The answer depends on whether light is a particle or a wave. If you imagine light is a particle of some mass travelling at speed c, then you get a Newtonian deflection of light, half of Einstein's value. This is discussed extensively in many places.

But if you think light is a wave, a wave doesn't fall, it only refracts. In order to get a wave of light to bend, you need the frequency of the light to change in different places. In Newtonian gravity, there is no change in the frequency of light waves due to their motion, because there is no coupling of gravity to electromagnetism, and there is no time-dilation which changes the frequency of waves.

So in the wave theory, it was expected that light would be unaffected by gravity. The two theories are reconciled in modern physics, since the time-dilation of gravity is the cause of both the deflection of light (by changing the frequency of light waves) and of the deflection of matter (by changing the frequency of matter waves).

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If we interpret "Newtonian" very literally, you might say that light is made up of particles. –  yakiv Jul 13 '12 at 1:20
You might also be able to say (at least if gravitational mass and inertial mass are assumed to be the same thing) that the Newtonian equation modeling gravity predicts bending of space . . . –  yakiv Jul 13 '12 at 1:22
@yakiv: Newton doesn't predicts bending of space. It isn't bending of space, in the sense of a space-space metric tensor, that is important in GR at low velocities, only the time dilation effect. In Newtonian gravity viewed as a theory of local Newtonian rest frames linked by a connection (this is the Newtonian equivalence principle), there is no time dilation, it's not geometrical in this way, there is a curvature in the local free-falling frame, but a wave will not deviate like a particle, since time doesn't dilate. At the beginning of the answer, I did intepret Newtonian literally. –  Ron Maimon Jul 13 '12 at 6:47
Hmmm . . . it seems you did interpret it literally. My bad. –  yakiv Jul 13 '12 at 22:28

Yes, Newtonian physics does predict the bending of light by gravity. In fact, using Newtonian theory only, a geologist in 1783 noted that if the sun were 500 times larger in diameter but with the same density, light would not be able to escape it. That is, it would be a black hole.

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You've got to be careful here to distinguish between Newtonian gravitational mass (gravitational "charge") and Newtonian inertial mass (measure of inertia).

Could there be, in Newtonian mechanics, a particle with non-zero inertial mass but no gravitational mass? I think the answer is, in principle, yes. It would simply be a particle that does not gravitate.

Could there be, in Newtonian mechanics, a particle with no inertial mass? Such a particle would have zero momentum always (unless you allow for actually infinite speeds...). Since, the particle's momentum is constant, by $\vec F = \frac{d\vec p}{dt}$, there is no, and can be no, force acting on the particle; the particle is a "ghost" cipher*.

So if, by photon, you mean a gravitationally massless particle then a Newtonian photon doesn't gravitate.

If, by photon, you mean an inertially massless particle, then a Newtonian photon is a cipher*.

*nonentity

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So does it mean its impossible to have "zero inertial mass" particle in Newtonian mechanics? That means by newtonian mechanics, photon has to have somekind of mass which can be measured by inertial methods. Another thing is, as far as i know, the equivalence of gravitational and inertial mass in newtonian mechanics is not something demanded by the theory. that means the equivalence is merely accidental. in GR its different. –  Aftnix Jul 11 '12 at 18:54
It's not impossible as in logically impossible, i.e., its existence implies a contradiction. Rather, it's that an inertially massless particles can't be observed, can't be detected in principle under Newtonian mechanics. Think of Newton's laws of motion; zero inertial mass implies zero momentum, zero force, zero action and reaction. It doesn't interact at all. It's a dynamical "ghost". Massless particles exist in the relativistic context because there is an invariant speed. In Newtonian mechanics, there is no invariant speed or, sloppily, the invariant speed is infinity. –  Alfred Centauri Jul 11 '12 at 19:17
@AlfredCentauri: The word "ghost" has a specialized technical meaning, the photon is never a ghost, a ghost is a particle with negative sign norm states, and it doesn't make sense outside of quantum mechanics. If you want to say a particle that passes unimpeded through everything, say this, but it isn't true that the zero mass limit is the vanishing force limit, rather it's the infinite acceleration limit. In the zero mass limit, the acceleration due to gravity is unchanged, although the speed goes to infinity so the deflection from a point source vanishes. –  Ron Maimon Jul 13 '12 at 6:46
@RonMaimon, (1) this is not a QFT context so "ghost" is fine and makes sense here. (2) This is not a quantum context so "photon" is ambiguous. (3) The OP asked questions about massless particles in general (see final two questions). In Newtonian mechanics, regardless of whether you choose to call it a "photon" or not, if you consider a zero inertial mass particle, either you accept there are actual infinities, or you accept that the particle has zero momentum always, zero KE always. Sure gravity may accelerate it but what is it? –  Alfred Centauri Jul 13 '12 at 11:16
@RonMaimon, as a follow up, would you prefer the word "cipher" rather than "ghost"? From Griffith's "Introduction to Elementary Particles": "In classical mechanics, there's no such thing as a massless particle; its momentum ($mv$) and its KE ($\frac{mv^2}{2}$) would be zero, it could sustain no force, since $F=ma$ - it would be a dynamical cipher –  Alfred Centauri Jul 13 '12 at 11:51
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