Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I ran across THIS Newtonian explanation for the bending of light. Does it have any merit?

share|improve this question
1  
From a quick read-through, my impression is no, it does not have any merit. But I would have to spend some more time going through the details to come up with a proper answer. If I get a chance to do that, I'll come back and post an answer here. –  David Z Aug 11 '11 at 2:50
1  
Why don't the general science journal guys add LaTeX tutorial to their site? –  Pratik Deoghare Aug 11 '11 at 12:49
    
One is indeed intrigued to say "look at which program they used to create this document, and you know what to think of it". Unfortunately, this argumentation is not quite valid anymore since quite a lot of otherwise trustworthy scientists actually do use Microsoft Word nowadays (urgh), but these will at least know how to utilize a formula editor to get somewhat readable results. –  leftaroundabout Aug 11 '11 at 15:09
    
The wonderful thing about the Internet is that it empowers you to research the qualifications of the author of papers of this kind. I find no evidence that he has any, nor evidence of any association with any reputable academic institution. I do find evidence of how coherently he can express himself though(network54.com/Forum/304711/thread/1232960084/1233255575/…). Make up your own mind. –  David Wallace May 26 '12 at 9:47
add comment

3 Answers

up vote 6 down vote accepted

As Ben says, it's nonsense. The most nonsensical claim is this: "both Special relativity and Galilean relativity work." Though what the author has in mind is more specific than arbitrary light orbits, and he is not completely wrong in seeing that the Newtonian calculation is hand-waving, but does not address the actual problems and certainly does not fix them. Usually, what is done in the Newtonian "toy model" exercise is to treat light as a (massive) corpuscle with velocity far from the Sun being $c$. Of course, that means that somewhere along the orbit, the light will have velocity higher than $c$, but there's no mechanism in Newtonian gravitation have a constant-speed unbound orbit, so that's a bit of a necessary evil. That's what is usually taken to be the Newtonian prediction for the deflection of light.

(Note: the author explicitly says that by "Newtonian theory", he means "Newtonian theory obeying Galilean relativity". That's exactly what standard Newtonian mechanics is, so I feel confident if there's any misunderstanding, it is due to the author.)

One of the author's mistakes is that he considers the path of the light ray trajectory to be parabolic just as it should be in Newtonian theory under a constant gravitational field. But for a parabolic orbit around a point, the speed far away from the gravitating body would be zero, which obviously cannot be the case for light. On the other hand, the orbit far from the gravitating body should be straight, so in the far limit are lines. Thus, the orbit must be hyperbolic.

At first glance, one might think: aha, what if one takes the velocity somewhere else along the orbit to be $c$, say at Earth or closest distance to the Sun, or whatever? But that can't affect the situation too much, because if the corpuscle has velocity $v$~$c$, then its kinetic energy is so utterly dwarfs the gravitational potential energy due to the Sun, that it will simply be irrelevant just where on the orbit the velocity is exactly $c$, because it will be extremely close to it everywhere on the orbit.

We can check this more explicitly. In the Newtonian assumption, the light corpuscle will have a hyperbolic orbit, and we can talk about the the angle $\theta$ between the asymptotes of that hyperbola. If the corpuscle is unaffected by gravity, then this angle be $\pi$ radians, so a sensible measure of deflection is its deviation from $\pi$; for convenience let's call this $2\psi$. The relevant relationship is: $$e = 1+\frac{v_\infty^2R_p}{GM} = \frac{1}{\cos\frac{\theta}{2}} = \frac{1}{\sin\psi} \approx \frac{1}{\psi} + \frac{\psi}{6} + \ldots,$$ where $e$ is orbital eccentricity, $v_\infty$ is velocity at infinity, $R_p$ is the distance to gravitating body (Sun) at periapsis. The last bit is an approximation for small $\psi$. (Note also that a parabola would have eccentricity $e = 1$.)

So we have $v_\infty = c$ giving us some eccentricity and some $\psi$. Let's try velocity at periapsis $v_p = c$ instead, and compare the result. We know from conservation of mechanical energy that the velocity at $r = R_p$ and velocity at infinity are related by $\frac{1}{2}v_p^2 - \frac{GM}{R_p} = \frac{1}{2}v_\infty^2$, so plugging in $v_p=c$ gives: $$e' = 1+\left(c^2 - 2\frac{GM}{R_p}\right)\frac{R_p}{GM} = e - 2.$$ Conclusion: taking the velocity at closest approach to the Sun to be $c$ rather than velocity at infinity hardly makes a difference at all, since in that regime $e$ is very large and $\psi$ is very small ($R_p$ ~ solar radius).

With $e$ so large (~$10^5$ or higher), the addition of $1$ is similarly unimportant, thus we can also approximate the total deflection to be $$2\psi \approx e^{-1} \approx \frac{2GM}{c^2R_p},$$ which is exactly what the author says is wrong. Thus, his claim that Newtonian gravity predicts $4GM/{c^2R_p}$ is simply not correct, because no matter how much one finagles with making the speed $c$ anywhere on the orbit, a factor of $2$ is much too large to be compensated for under the constraints of the Sun's mass and $R_p\geq\text{solar radius}$.


Even without the Newtonian toy models, there's a very straightforward and intuitive way that GTR's effect on light is twice that of Newton, and it can be seen clearly in the linearized approximation. Like for Newton, GTR's gravity couples to energy, which would be the relativistic mass (though that term is depreciated). But unlike it, GTR's gravity also couples to pressure, and for a light beam the pressure in the direction of motion is exactly equal to its energy, in units of $c = 1$.


Addendum: A brief explanation of questions from the comments. In linearized gravity, one begins with the assumption that the metric is a perturbation of Minkowski spacetime, imposes a gauge condition, and proceeds in analogy to the retarded potential of electromagnetism. Using approximations appropriate to a nearly-Newtonian regime (energy dominates stress, negligible retardation), the result applicable here is: $$ds^2 = -\left(1+2\frac{\Phi}{c^2}\right)c^2dt^2 + \left(1-2\frac{\Phi}{c^2}\right)\underbrace{(dx^2+dy^2+dz^2)}_{dS^2},$$ where $\Phi$ is the Newtonian potential due to the gravitating source, and I'm calling the Euclidean distance element $dS^2$ for convenience. This is what some authors call the static, weak-field metric, and the derivation can be found in most GTR textboooks. Light travels along null geodesics, so $ds = 0$, giving $$\frac{1}{c}\frac{dS}{dt} = \sqrt{\frac{1+2\Phi/c^2}{1-2\Phi/c^2}} \approx 1 + 2\frac{\Phi}{c^2} + \mathcal{O}\left(\frac{\Phi^2}{c^4}\right).$$ Therefore, the effect of the gravitational field on light can be interpreted as follows: light travels "as if" it was in a medium with refractive index $n=c/v = 1-2\Phi/c^2$.

How it is associated with the "spatial curvature" can also been derived from the above. What we want is the Newtonian $d^2S/dt^2 = -\nabla\Phi$, and to do that, we can explicitly extremize the proper time of an orbiting particle, and again make use of Taylor series approximations, keeping terms only to $\mathcal{O}((v/c)^2,\Phi/c^2)$: $$\tau = \int dt\sqrt{\left(1+2\frac{\Phi}{c^2}\right) - \frac{1}{c^2}\left(1-2\frac{\Phi}{c^2}\right)\frac{dS^2}{dt^2}} \approx \int dt\left(1-\frac{1}{c^2}{\left(\frac{1}{2}\frac{dS^2}{dt^2} - \Phi\right)}\right),$$ which is exactly the action integral for Newtonian gravitation with Newtonian Lagrangian $\frac{1}{2}v^2 - \Phi$, except for the irrelevant addend of $1$ and rescaling by $c^2$ both of which drop out of the Euler-Lagrange equations to give the usual $d^2S/dt^2 = -\nabla\Phi$.

The interesting part in the above is this: since we're dropping the higher-order terms, the deviation from Euclidean $dS^2$ becomes completely irrelevant--it only contributes terms that are dropped this Newtonian limit. Newtonian behavior can just as well be described by the much simpler metric $$ds^2 = -\left(1+2\frac{\Phi}{c^2}\right)c^2dt^2 + \underbrace{(dx^2 + dy^2 + dz^2)}_{dS^2},$$ which produces the right Newtonian action integral in a more straightforward way. But of course light orbits in this metric follow $$\frac{1}{c}\frac{dS}{dt} = \sqrt{1+2\frac{\Phi}{c^2}} \approx 1 + \frac{\Phi}{c^2} + \mathcal{O}\left(\Phi^2/c^4\right),$$ in other words, light travels "as if" it was in a medium with refractive index $n = c/v = 1-\Phi/c^2$.

Without even calculating the exact amount of angular deviation, the effective indices of refraction make the following pretty clear: the effect of the spatial curvature is to double the deviation of light it would be without it, and since Newtonian gravitation can be reproduced entirely by the purely temporal component of the metric, this is can be reasonably be interpreted as why GTR's effect is twice that of Newtonian gravity.

share|improve this answer
    
Nice answer! But I have doubts about the final paragraph's interpretation of the factor of 2, or rather, I have doubts about such interpretations in general. I have also seen it interpreted as an effect in which there are equal contributions from the equivalence principle and from spatial curvature: einstein-online.info/spotlights/equivalence_deflection The fact that one can propose very different interpretations of the factor of 2, both seemingly equally valid, suggests to me that there is no unambiguous, simple explanation. –  Ben Crowell Aug 11 '11 at 15:38
    
You loose me so many times. Why can't we just assume some flat space, classical gravity, and apply the special relativity of light? You keep writing that the photon has to be treated as having $v \neq c$, but there is no justification for that. A photon has a momentum vector $\vec{p}$ that implies a velocity vector that has a direction and magnitude $c$, and $\vec{F}=dp/dt$. An orbit follows directly from that. I'm considering making a new question about this since no one seems to answer and they keep waving off my questions. –  AlanSE Aug 11 '11 at 16:53
1  
@Zassounotsukushi: "Why can't we just assume some flat space, classical gravity, and apply the special relativity of light?" Then you'd have a relativistic theory of gravity without having to use GR. That's impossible "You keep writing that the photon has to be treated as having v≠c, but there is no justification for that." In the Newtonian picture, a force acts on the photon and has a component parallel to the photon's direction of motion, so the photon's speed is not constant. –  Ben Crowell Aug 11 '11 at 20:54
    
@Ben: Those answers are consistent. The only difference is interpretative rather than mathematical or predictive: it's possible to interpret linearized gravity as a symmetric rank-2 field in flat spacetime rather than perturbation of flat spacetime, and in this picture the Sun's field couples to $p^\mu p^\nu$ of the light corpuscle and picks out the diagonal terms--in other words, energy and pressure of the light corpuscle's stress-energy tensor. This interpretation goes against the equivalence principle, but it is most appropriate if we're comparing to Newtonian gravitation. –  Stan Liou Aug 11 '11 at 22:31
1  
@AlanSE: I should also say that you can't assume Galilean relativity AND ordinary E&M. They are famously inconsistent. –  Jerry Schirmer May 25 '12 at 21:21
show 4 more comments

It's total nonsense. They claim this: "Contrary to what many people believe, Newtonian theory gives the same value for light bending as General relativity." As a counterexample, GR allows exotic knife-edge trajectories for light moving in a Schwarzschild metric, where the light can come in, orbit n times, where n is any integer, and then fly off. In Newtonian gravity, this is impossible, because the only possible orbits are conic sections

share|improve this answer
    
Is that due to a different rate of deflection or due to the curvature of space? If we limit the discussion to Minkowski space does your answer change? –  AlanSE Aug 11 '11 at 13:18
2  
In Minkowski space there's no gravity, so the question doesn't arise. –  Ben Crowell Aug 11 '11 at 15:28
    
lol, I wondered if I was getting that wrong. What about in the linearized gravity approximation. That is what I should have said. –  AlanSE Aug 11 '11 at 15:38
    
Linearized gravity is typically used to describe weak gravitational waves. IIRC it doesn't have bound systems, so it's probably not suitable for describing this. If you're asking about the weak-field limit, then Stan's answer covers that. –  Ben Crowell Aug 11 '11 at 16:18
add comment

It's a brilliant paper; where these replies go wrong is that they assume light ray trajectories to be parabolic, and the paper points out that this is not the case.

If dealing with a uniform gravitational field then it is possible to treat the trajectory as parabolic. But it's not a uniform field, it is a field directed at a point-source for the gravitational force, hence doubling of the deflection.

In general relativity this distortion of the gravitational field is treated as spacetime curvature. But in Newtonian physics it's treated as the gravity field distorted in Euclidean space.

So replies here suffer from people who don't bother to read the paper properly and make up nonsense.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.