Newtonian light deflection

Question

It is often stated that the light deflection by a point mass $M$ of a photon passing within distance $b$ is according to Newton

$$ \theta = \frac{2GM}{bc^2}, $$

which is exactly half the value obtained by Einstein's relativity. To obtain this Newtonian value, one has to assume that the photon is gravitationally attracted and therefore, in Newtonian theory, has a mass. However, this is wrong, as we know. My question is therefore, how one can in Newtonian theory justify the above result.

Added in edit: A similar question has been asked before, but (1) it was not specifically about the Newtonian case and (2) John Rennie's answer specifically doesn't answer my question, since it obtains the Newtonian value from the acceleration $$ \vec{a} = - \frac{GM}{r^2}\frac{\vec{r}}{r} $$ for the deflected particle. However, this acceleration is obtained from the Newtonian $$ m\vec{a} = \vec{F} = - \frac{GMm}{r^2}\frac{\vec{r}}{r} $$ for a particle of mass $m$. But the photon has mass $m=0$, for which this relation doesn't provide anything useful.

In other words, to obtain the Newtonian acceleration for the photon, one has to take the limit $m\to0$, but it is well-known that taking this limit is not necessarily the same as setting $m=0$. So, the essence of my question is why/how taking this limit can be justified. Note that the recent paper (quoted in aforementioned answer) also assumes a massive light particle and considers the limit $m\to0$ w/o justification.

An alternative interpretation is that Newton's theory is incapable to make statements about massless particles, but that one can in anticipation of relativity consider every energy-carrying particle to have a mass, since it is necessarily affected by gravity. However, this is clearly beyond Newton's theory.

Added later This question arose in the context of Eddington's measurement of the deflection of light by the Sun, which Einstein predicted to be twice the above Newtonian value. So, I wonder what the physicists of that time thought about the alternative (Newtonian) value and how they justified its derivation.

Answer 1

An alternative interpretation is that Newton's theory is incapable to make statements about massless particles, but that one can in anticipation of relativity consider every energy-carrying particle to have a mass, since it is necessarily affected by gravity. However, this is clearly beyond Newton's theory.

Yes, Newtonian mechanics by itself does not make any predictions about light whatsoever.

The problem is, to say anything about light, you need a model of light to start with. Since it's not clear what model of light we should use in the context of Newtonian mechanics, it follows that it's not clear what prediction Newtonian mechanics should make about light.

The best classical theory for light as of this writing is Maxwell's electrodynamics. The problem is, this theory is inherently relativistic (it obeys special relativity not Galilean relativity), which means by its very nature it lies outside the domain of Newtonian mechanics. We can make low velocity approximations in some circumstances when we're looking at forces on massive particles (which we do all the time in introductory courses) so as to use bits and pieces of Newtonian mechanics, but there is no sensible low velocity approximation for the behavior of light itself.

So we have a couple of options for what we might want to do:

1. Light as an instantaneous ray. Because this model does not specify any structure about the ray itself and Newtonian mechanics doesn't say anything about instantaneous rays, we have to conclude that Newtonian mechanics can't say anything about this. Historically, Aristotle and Heron of Alexandria thought light traveled instantaneously, but it wasn't agreed upon and it was an open question.
2. Small mass corpuscle model of light. Perhaps light is a small classical particle with a very, very small mass. In that case, Newtonian mechanics works perfectly well, and we can apply John Rennie's answer with $m\approx 0$. The calculation of $$ \theta = \frac{2GM}{bc^2} $$ is then a robust prediction of Newtonian mechanics. Something that is really important to realize about this model is that if the corpuscles are to obey Newtonian mechanics in this model, then they will not have a constant speed (but the speed will be finite). This is a major conflict with our current understanding of light.
3. Massless corpuscle model of light. Perhaps light is a classical particle, but it has $m=0$. Then Newtonian mechanics does not have anything to say about this, because Newtonian mechanics is formulated specifically for massive bodies.
4. Wave in a medium model of light. The prediction is entirely dependent on the interaction between gravity and the aether, which would depend on the specific model of the aether. I'm not aware of anyone discussing this, but this could be due to my ignorance.
5. Electromagnetic wave model of light. As I said, Maxwell's theory of electromagnetism is inherently relativistic, but this wasn't realized until Einstein. In any case, there is no prediction for how gravity, as modeled by Newton, would influence the electromagnetic field.

The only sensible model to consider after (5) is

6. Electromagnetism generalized to curved spacetime as described by general relativity. In that model, light rays (which are approximations to light waves) travel along null geodesics of spacetime. When spacetime is curved due to gravity (or perhaps we might say the curvature is gravity rather than due to gravity but I don't have the insight to say which set of semantics is better), null geodesics are altered exactly to produce the deflection $$ \theta = \frac{4GM}{bc^2}. $$ This leads us right back to this page.

We see that models (1), (3), (4), and (5) don't say anything about the question of whether light is deflected or not by gravity per se. Only models (2) and (6) give robust predictions.

However, to justify John Rennie's post, maybe we can add an extra postulate to models (1) and (3): light takes a trajectory as if it were a massive particle with $m\rightarrow 0$. This is an additional postulate that can't be derived from (1) or (3) directly. Nonetheless, as a hypothesis I think it seems reasonable.

By Einstein's time (~1915), there were only three hypotheses known: $\theta = 0$, $\theta = \theta_{0}$, or $\theta = 2\theta_{0}$ where $$ \theta_{0} = \frac{2GM}{bc^2}. $$ No other possibility was proposed, really. It made sense to try and test which one of the three possibilities is true.

As an aside, it's interesting to track the history of this/these predictions. Amazingly, Isaac Newton himself was the first one to pose the question in The Queries, but he made no predictions. Afterwards, people such as Pierre Simon Laplace, John Michell, and Henry Cavendish have speculated on the possibility, with Henry Cavendish making the first unpublished calculations and Johann Georg von Soldner making the first published calculations using (one of the) the corpuscle model. See [1] and [2].

Between 1905 and 1915, Einstein used arguments in the context of special relativity to argue that gravity must have an interaction with the electromagnetic field, but they were outside the scope of what people would consider Newtonian mechanics. The arguments at that stage relied on things like the strong equivalence principle and mass-energy equivalence. He managed to obtain the prediction of $\theta = \theta_{0}$ by coincidence, and it took until 1915 to predict $\theta = 2\theta_{0}$ because he didn't take the curvature of space (of spacetime) into account. See [3].

Answer 2

1. By the way, it is perhaps helpful to mention that when something is massless $m= 0$ in the framework of Newtonian mechanics, it doesn't mean it travels with the speed of light; it usually refers to an approximation $m\to 0$.

2. When we speak of deflection of light with mass $m=0$ in Newtonian mechanics, we mean using Newton's 2nd law and Newton's gravitational law in a form that still makes sense, i.e. $$ \vec{a}~=~\vec{g}~\equiv~-GM\frac{\vec{r}}{r^3}. \tag{1}$$ This is equivalent to considering a massive corpuscle of light with $m>0$. The trajectory (which is a hyperbola) does not depend on $m$. [Here we are assuming that $m\ll M$.]

3. Furthermore, in the Newtonian theory, if the speed of the light particle far away from the mass $M$ is $v_0$, it would move faster than the speed of light at finite distances from $M$.

4. It is not hard to derive that the Newtonian angle of deflection is $$\theta~=~\frac{2GM}{bv_0^2} \tag{2}$$ for small $\theta\ll 1$, cf. e.g. footnote 1 in my Phys.SE answer here.

5. In general relativity it does in principle matter whether $m$ is massive or massless. It influences whether the geodesic is time-like [$\epsilon=1$] or null/light-like [$\epsilon=0$]. However, one may show that there is no discontinuity in the relativistic small angle result [which is twice the Newtonian formula (2)], i.e the discrete parameter $\epsilon\in\{0,1\}$ does not enter the result, cf. my Phys.SE answer here.

Answer 3

You’re right, setting $m=0$ is not always the same as taking a limit $m\to 0$, as we all know from elementary calculus, or merely the definition of a limit and continuity. Here, the $m=0$ case is clearly different from the limiting $m\to 0$ case; in the former, there is no dynamics to be predicted, while in the latter there is.

Now, just for the sake of raising more trouble, let’s suppose we decide to model this particle not merely by a $m\to 0$ limit, but by considering it as a limit of a massive charged particle, so we consider a $(m,q)\to (0,0)$ limit. The equations of motion are \begin{align} m\vec{a}&=\frac{-GMm}{r^2}\vec{r}+q(\vec{E}+\vec{v}\times\vec{B}), \end{align} so cancelling the $m$,

\begin{align} \vec{a}&=\frac{-GM}{r^2}\vec{r}+\frac{q}{m}(\vec{E}+\vec{v}\times\vec{B}). \end{align} Now, we’re going to have a doubly singular limit as $(q,m)\to (0,0)$; the result clearly depends on the specific manner in which we take the limit $(q,m)\to (0,0)$.

So, we see that even with such a slight change to our equations, the theory predicts such wildly different outcomes, i.e something is breaking down in our theory. Now, we have two ways of confronting this issue

• first, we can try to preserve our current theory as much as possible (this would be the approach when you have nothing better to try). So, you could try to cook up some specific combination of $(q,m)$ and take a limit along a very specific path and say with that you get some reasonable-looking answer (this is similar by analogy to, say, modelling a point particle as a limit of a ball as the radius goes to $0$, but the mass/charge stays finite.) which conforms to experiments as much as possible.
• we could say that we’re reaching the boundary of the domain of validity of our current theory, and that perhaps we need some new perspective to fully understand/describe the situation.

We now know with hindsight of course that the second option is much better. So, I would interpret the Newtonian $\theta$ result as motivation, rather than anything too concrete, to the idea that gravity affects things with energy-momentum. Also, the fact that the two $m$’s cancels out in the Newtonian law of gravitation already tells us there is something really different about the nature of gravity from electromagnetism. So, with all of this together, we should (atleast I would) interpret the $\theta$ from the Newtonian prediction as motivation to want to seek a deeper understanding, and it should come as no shock that the prediction doesn’t match with experiments. It may be disheartening (?) but not surprising; more surprising would be if we actually got the right answer, because then we’d wonder how did we arrive at a correct answer (i.e was it a fluke we got the right answer, or is there something deeper within Newton’s description which we have yet to understand). So, with all of this, the fact that null geodesics around Schwarzschild exhibit the behavior as in experiments then shows that yes, Einstein’s theory is a strict generalization of Newton’s, so that’s progress :)

Answer 4

(Note that I don't know the history of Newtonian light deflection befor GR, but that may be interesting -- we would see what was considered plausible before we knew what should come out.)

I would point out the following: In Newtonian mechanics and Newtonian gravity, we cannot really predict the value of light deflection, because massless particles don't really fit in, small-mass particles are different, the "speed of light" is not a constant or fundamental quatity etc. (This is also contained in the other answers.) Hence, one has to motivate the correct way to deal with light in a gravitational field, and inevitably, we are guided by the GR result.

However, it should not come as a surprise that -- with this guidance -- we get a quite similar formula, differing only by a constant factor: Plausibly, the angle $\theta$ will depend on $GM$, $b$, and $c$ (if we assume $c$ is special, or some other velocity $v_0$, as in Qmechanic's answer), and no other dimeniosnful parameters; also, the angle (in the small-$\theta$ approximation) should be linear in $GM$. Then, the expression is already fixed to be $$ \theta=\alpha \, \frac{GM}{bc^2} $$ with some dimensionless parameter $\alpha$. It remains to find this number; note that by pursuing this line of reasoning, it is already guaranteed that $\alpha \neq 0$.

On the other hand, one could have argued that no mass, no force, no deflection. Within classical mechanics, this ambiguity is just there.

Answer 5

To answer precisely to your question, the deflection of light $ \theta\simeq\frac{2GM}{c^2b} $ according to Newton gravitation can be calculated as follows:

• the eccentricity of the hyperbola of the photon is: $ e=\sqrt{1+(\frac{c^2b}{GM})^2}, $

• its axis is: $ \varphi_0=\arccos(-{1\over e}) $ which gives the deflection $ \theta=2\varphi_0-\pi. $

Because $ (\frac{c^2b}{GM})^2\gg 1 $, you have $ e\simeq\frac{c^2b}{GM} $ which leads to $ \varphi_0\simeq\arccos(-\frac{GM}{c^2b}). $

Since $ \frac{GM}{c^2b} $ is close to $ 0 $, you can write: $ \varphi_0=\arccos(-\frac{GM}{c^2b})\simeq\frac{\pi}{2}+\frac{GM}{c^2b} $ and then $ \theta=2\varphi_0-\pi\simeq 2\times(\frac{\pi}{2}+\frac{GM}{c^2b})-\pi $ which leads to:

$ \theta\simeq\frac{2GM}{c^2b}. $

Actually, the use of Newton gravitation for a photon is not relevant because with classical mechanics, you use the law of conservation of the angular momentum which includes the mass of the particle: since the photon has no mass, obviously you can not use this law.

Best regards.

Answer 6

We do not need to know the mass of an object moving at some speed, to know its trajectory through solar system, assuming that said mass is not so large that it affects the trajectories of planets, in the case when the mass is so large, we can calculate said mass from the magnitude of said effects.