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Well, as I have read, a massive body can cause light to bend around itself due to its gravitational attraction. What I don't understand is how, as the Newtonian formula for the force of gravitational attraction is $$F = \frac{Gm_1m_2}{r^2}.$$ As photons do not have (rest) mass, shouldn't there be no attraction between light and the celestial bodies, and therefore no bending? Surely I am missing a key point. Please point out my logical flaw?

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marked as duplicate by John Rennie, Brandon Enright, DavePhD, Kyle Oman, Kyle Kanos Jun 27 '14 at 15:11

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    $\begingroup$ As far as we know, the "right" theory of gravitation is General Relativity, not Newtonian gravity, and General Relativity predicts every body/particle bearing energy experiences gravitational effects, not just massive bodies, as it was in Newtonian gravity. $\endgroup$ – giordano Jun 27 '14 at 9:00
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    $\begingroup$ possible duplicate of Why can lights(photons) bends in a curve through space without mass? $\endgroup$ – giordano Jun 27 '14 at 9:11
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The flaw is that you are trying to mix classical with relativistic concepts.

Gravitational lensing (this is the phenomenon you are referring to) is best described in terms of general relativity. Massive bodies bend spacetime, inducing a curvature, which is described by Einstein's equations:

$$G_{\mu\nu}=8\pi T_{\mu\nu},$$

where on the left hand side is the Einstein tensor which contains information about curvature and on the right hand side there is the energy-momentum tensor, containing information about energy and matter. From this formalism, it is possible to derive so-called geodesics, which are the paths objects will take through curved spacetime.

Photons feel this curvature and have to move according to it, resulting in the phenomenon we see as "bending". Below, you can find a visualization of the effect:

enter image description here

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  • $\begingroup$ "Massive bodies bend spacetime, inducing a curvature, which is described by Einstein's equations: $G_{\mu \nu} = 8\pi T_{\mu \nu}$, where on the left hand side is the Einstein tensor which contains information about curvature". This sentence is in direct contradiction with this answer here: "So Einstein's equations in vacuum mean exactly that: $G_{\mu \nu} = 8\pi T_{\mu \nu} = 0$ in a region without mass-energy." Obviously, if $G_{\mu \nu}=0$ everywhere in vacuum (also around massive bodies), then it tells nothing about curvature. $\endgroup$ – bright magus Jun 28 '14 at 8:55
  • $\begingroup$ @brightmagus You should read the next sentence in that answer as well. You have obviously misunderstood something. $\endgroup$ – Frederic Brünner Jul 3 '14 at 21:26
  • $\begingroup$ Are you saying that zero 1 thousand kilometers from the surface of the Earth is different than the zero 1 thousand kilometers from the surface of the Sun? $\endgroup$ – bright magus Jul 4 '14 at 7:12
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As others have said, general relativity is the right theory of gravity (as far as we know). But even in Newtonian gravity, we predict light bending. The thing is, for a photon of mass $m$ moving near the sun of mass $M$, it's true that $F=GMm/r^2$; but also for the photon $F=ma$. And what we care about is the acceleration of the photon, which is zero force divided by zero mass. So it's better to combine the two equations and get that for any object, its acceleration in the Sun's gravity is given by $a=GM/r^2$. This predicts light bending by the Sun.

Unfortunately, the amount of light bending it predicts turns out not to be the amount of light bending we observe (by looking at star positions during eclipses, and using radio interferometry). In fact, if you work it through with Einstein's theory of gravity, you find that light bends by exactly twice as much as Newton predicted - and this agrees with experiment. The extra bending comesabout because space is bent (distorting the geometry around the Sun), in addition to the bending of time into space that makes things (including light) fall.

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    $\begingroup$ Is that mathematically sound, though? You're cancelling a factor of zero from each side. $\endgroup$ – JohnnyMo1 Jun 27 '14 at 13:53
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    $\begingroup$ if you really want to be precise, take the limit as $m$ goes to zero, then it works. $\endgroup$ – Snowbody Jun 27 '14 at 14:23
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The situation is a little bit more complex. In cases of velocities approaching light speed, Newtonian mechanics doesn't really work. Such situations require using the equations of special relativity. But, even considering special relativity effects is not sufficient to solve problems where gravity is also acting. Special relativity conditions in the presence of gravity can only be correctly handled by general relativity.

General relativity doesn't work by producing forces on point-like masses to generate acceleration. Rather, general relativity works on masses moving linearly in a curved spacetime. The linear movements on a curved spacetime are named "geodesics". Practically, it requires using very complex differential equations to compute the trajectory. As non-physicists we can imagine this as such:

enter image description here

Of course, you can calculate the bending of light as if it were mass points moving with light speed in a non-relativistic universe. But the calculation will not correspond with the experimental results.

Practically, the first experimental evidence confirming general relativity was that the bending of starlight by the sun's gravity was not as we calculate it by Newtonian Mechanics. The Einsteinian results differ around 50% from the Newtonian results, and with this the experiments are consistent.

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Your logic is not flawed, you're simply using the wrong formulas. According to Newton's law of gravity

$$F = G \frac{m_1 m_2}{r^2}$$

only massive bodies feel the gravitational force and since the particles of light - photons - do not have any mass, they do not feel the gravitational force.

Einstein generalized the concept of gravity in a geometrical way, called general relativity. In general relativity, you cannot distinguish if you are in an accelerated reference frame or if you are under the influence of gravity. Basically, you cannot say, if you are in a resting elevator on earth or if you are in an accelerating elevator far away from any gravitationally interacting object. Since light would bend in an accelerating reference frame, it also has to bend due to gravity. See e.g. the picture here:

enter image description here

This is a famous consequence of general relativity. Shortly after the theory, the effect of bending photons was measured during a solar eclipse and this aspect of general relativity was confirmed. We should not forget, though, that Newton's law of gravity is still a very good approximation, but it cannot explain the bending of light particles, as you have correctly pointed out.

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  • $\begingroup$ So if we can accelerate a rocket/elevator fast enough, we would see light bending downwards? A laser beam would bend downwards in an arc form? And one more question, if we take Einstein's general relativity, does Newton's 3rd law of motion still hold up? As the acceleration of the elevator is causing light to bend downwards, thus exerting a force on the photons, will the photons also exert an equal and opposite force on the elevator's acceleration? Will they cause the acceleration to reduce? $\endgroup$ – Gummy bears Jun 27 '14 at 9:01
  • $\begingroup$ There is no force acting on the photons. They would travel in a straight way, but since we are in an accelerated reference frame, we would see the light bent (as in the right picture above). $\endgroup$ – pfnuesel Jun 27 '14 at 9:07
  • $\begingroup$ So its a pseudo force? It doesn't really exist but we just see it? Strange though... Something that came to my mind: Shouldn't there be a certain acceleration/gravitational pull, that will cause light to start an orbital around itself? $\endgroup$ – Gummy bears Jun 27 '14 at 9:13
  • $\begingroup$ Imagine yourself being in an elevator with a window. Outside the window, a friend of yours is standing. Now you are accelerated upwards. From your perception, in your reference frame, it looks like your friend is accelerated downwards, although there is no force acting on him. There is nothing strange here: It's just what you observe when you are in an accelerated reference frame. $\endgroup$ – pfnuesel Jun 27 '14 at 9:17

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