<p>Newtonian treatments of the bending of light go back to Laplace who, in 1798, wrote about light escaping from massive bodies, ie: black holes!  See Appendix A of Hawking and Ellis "Large Scale Structure of Space-Time"  where there is a nice translation of Laplace's paper.</p>

<p>Newtonian treatments cannot properly deal with all aspects of light bending.  Notably, the important difference between the 'Luminosity Distance' and the 'Angular Diameter Distance' of a cosmological object is only a feature of Einstein-style theories of gravitation.  It is that difference which, for example, enables us to test different cosmological models (as in the Alcock-Paczynski test).  See for example the paper of Anderson etal. in arXiv:1303.4666 for technical details.</p>

<p>(This should probably have been a comment - but I don't have the rep for that).</p>

CC BY-SA 3.0
2015-08-26T09:39:52.877Z
2015-08-26T09:39:52.877Z
JonesTheAstronomer
<p>You can say that if a object has kinetic energy in the vicinity of a massive object it has potential energy.</p>
<p>And if the angle bitwext the radial vector and velocity vector
Is not 0 or 180 secs</p>
<p>The object has angular momentum</p>
<p>And a path with angular momentum
Is curved.</p>
<p>Light has kinetic energy so it can be bent by Newtonian gravity.</p>
<p>In fact I can’t swing by a massive object is not be deflected by gravity</p>

CC BY-SA 4.0
2021-09-22T05:27:21.767Z
*"a path with angular momentum Is curved"* Not necessarily. Even a body moving uniformly in a straight line has angular momentum. Please see https://physics.stackexchange.com/q/272688/123208
PM 2Ring
CC BY-SA 4.0
2021-09-21T17:42:47.740Z
2021-09-21T17:42:47.740Z
Physics Human
<p>A photon is an entity defined in the context of a relativistic field theory, and so it doesn't really make sense to talk about the Newtonian bending of a photon. Necessarily, we need to substitute an analogous question that's sensible in the Newtonian framework. To do so, we can imagine a classical <em>corpuscle</em> of light--appropriately enough, a theory of light advanced by Newton himself.</p>

<p>There are many problems with the Newtonian conception of light, but that's an issue for electromagnetism rather than gravity.</p>

<p>Critically, for a test particle, the trajectory depends on the initial velocity only, and not the mass. So we don't even have to address the mass of the light corpuscle at all to talk about its trajectory, as the acceleration is the gradient of the gravitational potential and if considering force (or gravitational potential <em>energy</em>), the mass cancels out anyway. If you wish to bring in the mass explicitly, we can still think of the trajectory of a massless particle as a limit of trajectories of particles with masses tending to zero but having the same velocity &mdash; a trivial limit because they're all the same trajectory in the Newtonian theory. On the other hand, if we recognize that light carries momentum, a Newtonian light corpuscle <em>shouldn't</em> have zero mass, so the question of what to do with a genuinely massless particle evaporates.</p>

<p>When people talk about the Newtonian deflection of light, they are typically considering a hyperbolic trajectory of test particle at the speed of light under Newtonian gravity. If the angle between the asymptotes is $\theta$, then $\theta = \pi$ represents a completely straight trajectory unaffected by gravity, and the eccentricity is
$$e = 1 + \frac{v_\infty^2R_p}{GM} = \frac{1}{\cos\left(\frac{\theta}{2}\right)} = \frac{1}{\sin\psi}\text{,}$$
where $R_p$ is the periapsis distance and $2\psi = \pi-\theta$ is the measure of deflection. On the scale of the solar system, it doesn't matter whether we set $v_\infty = c$ or anywhere else along the trajectory. For example, if the velocity at peripasis is $c$ instead, then $e\mapsto e-2$, and so is negligible for light deflection due to the Sun, $e &gt; 10^{5}$. The total deflection is approximately
$$2\psi \approx e^{-1} \approx \frac{2GM}{c^2R_p}\text{,}$$
which is half the correct general-relativistic prediction.</p>

<p>Note that here we did <em>not</em> assume that the speed is constant along the hyperbolic orbit. That would not be consistent with Newtonian gravity. Rather, what we have is a situation where if $v = c$ <em>anywhere</em> along the trajectory, then the speed along any other point is so close to $c$ that it doesn't practically matter for considering light deflection.</p>

CC BY-SA 3.0
2015-08-19T17:32:39.783Z
2015-08-27T22:21:14.817Z
2015-08-27T22:21:14.817Z
Stan Liou
<p>Famously, when Eddington attempted to measure gravitational lensing during the Eclipse, it was the measured <em>magnitude</em> of the lensing that gave gravity [pun, obviously, intended] to General Relativity - not the measurement of lensing itself. That is, Newtonian physics also predicted a lensing deflection, but only half of the deflection predicted by GR. </p>

<p>Question is: Why? I've read a lot about this, and I can see how when one integrates the Newtonian acceleration along the photon path with constant |v|=c this gives the 'correct' Newtonian value - but intuitively, I can't wrap my head around it. Why should a massless photon be affected by gravity at all in Newtonian gravity?</p>

<general-relativity><gravitational-lensing><newtonian-gravity>
Gravitational lensing in Newtonian physics
user1991
# Gravitational lensing in Newtonian physics

(Asked by: username_0 on 2015-08-19)

Famously, when Eddington attempted to measure gravitational lensing during the Eclipse, it was the measured *magnitude* of the lensing that gave gravity [pun, obviously, intended] to General Relativity - not the measurement of lensing itself. That is, Newtonian physics also predicted a lensing deflection, but only half of the deflection predicted by GR.

Question is: Why? I've read a lot about this, and I can see how when one integrates the Newtonian acceleration along the photon path with constant |v|=c this gives the 'correct' Newtonian value - but intuitively, I can't wrap my head around it. Why should a massless photon be affected by gravity at all in Newtonian gravity?

**Tags:** general-relativity, gravitational-lensing, newtonian-gravity


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## 3 answers

### Answer by username_4 on 2015-08-19. Score: 11 | Accepted answer

A photon is an entity defined in the context of a relativistic field theory, and so it doesn't really make sense to talk about the Newtonian bending of a photon. Necessarily, we need to substitute an analogous question that's sensible in the Newtonian framework. To do so, we can imagine a classical *corpuscle* of light--appropriately enough, a theory of light advanced by Newton himself.

There are many problems with the Newtonian conception of light, but that's an issue for electromagnetism rather than gravity.

Critically, for a test particle, the trajectory depends on the initial velocity only, and not the mass. So we don't even have to address the mass of the light corpuscle at all to talk about its trajectory, as the acceleration is the gradient of the gravitational potential and if considering force (or gravitational potential *energy*), the mass cancels out anyway. If you wish to bring in the mass explicitly, we can still think of the trajectory of a massless particle as a limit of trajectories of particles with masses tending to zero but having the same velocity — a trivial limit because they're all the same trajectory in the Newtonian theory. On the other hand, if we recognize that light carries momentum, a Newtonian light corpuscle *shouldn't* have zero mass, so the question of what to do with a genuinely massless particle evaporates.

When people talk about the Newtonian deflection of light, they are typically considering a hyperbolic trajectory of test particle at the speed of light under Newtonian gravity. If the angle between the asymptotes is $\theta$, then $\theta = \pi$ represents a completely straight trajectory unaffected by gravity, and the eccentricity is
$$e = 1 + \frac{v\_\infty^2R\_p}{GM} = \frac{1}{\cos\left(\frac{\theta}{2}\right)} = \frac{1}{\sin\psi}\text{,}$$
where $R\_p$ is the periapsis distance and $2\psi = \pi-\theta$ is the measure of deflection. On the scale of the solar system, it doesn't matter whether we set $v\_\infty = c$ or anywhere else along the trajectory. For example, if the velocity at peripasis is $c$ instead, then $e\mapsto e-2$, and so is negligible for light deflection due to the Sun, $e > 10^{5}$. The total deflection is approximately
$$2\psi \approx e^{-1} \approx \frac{2GM}{c^2R\_p}\text{,}$$
which is half the correct general-relativistic prediction.

Note that here we did *not* assume that the speed is constant along the hyperbolic orbit. That would not be consistent with Newtonian gravity. Rather, what we have is a situation where if $v = c$ *anywhere* along the trajectory, then the speed along any other point is so close to $c$ that it doesn't practically matter for considering light deflection.


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### Answer by username_1 on 2015-08-26. Score: 5

Newtonian treatments of the bending of light go back to Laplace who, in 1798, wrote about light escaping from massive bodies, ie: black holes! See Appendix A of Hawking and Ellis "Large Scale Structure of Space-Time" where there is a nice translation of Laplace's paper.

Newtonian treatments cannot properly deal with all aspects of light bending. Notably, the important difference between the 'Luminosity Distance' and the 'Angular Diameter Distance' of a cosmological object is only a feature of Einstein-style theories of gravitation. It is that difference which, for example, enables us to test different cosmological models (as in the Alcock-Paczynski test). See for example the paper of Anderson etal. in arXiv:1303.4666 for technical details.

(This should probably have been a comment - but I don't have the rep for that).


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### Answer by username_2 on 2021-09-21. Score: -3

You can say that if a object has kinetic energy in the vicinity of a massive object it has potential energy.

And if the angle bitwext the radial vector and velocity vector
Is not 0 or 180 secs

The object has angular momentum

And a path with angular momentum
Is curved.

Light has kinetic energy so it can be bent by Newtonian gravity.

In fact I can’t swing by a massive object is not be deflected by gravity

> **Comments:**
> **username_3**: *"a path with angular momentum Is curved"* Not necessarily. Even a body moving uniformly in a straight line has angular momentum. Please see https://physics.stackexchange.com/q/272688/123208  

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