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I read in reliable sites that GR and classical physics calculate the angle of deflection in the same manner. The formula is almost identical: $$\theta = \frac{4GM}{c^2*r} \rightarrow \frac{4GM}{c*r} = v *\frac{1}{c}$$ GR modifies the formula adding a factor $ \theta = \frac{4GM}{c^2r}\times \frac{1+\gamma}{2}$ where gamma has value 1 for energy and 0 for mass, so that the classical laws apply to orbits $ \theta = \frac{2GM}{c^2r}. $ Now if I did not misinterpret the procedure, they find the g-pull at point of impact R $g = \frac{GM}{R^2} $ and then integrating over time they get the normal velocity the ray acquires from gravity $v = \frac{[2*]2GM}{R\times c}$ In the case of the Sun this value is [2*]127400 cm/s. Then, considering v and c the legs of a right triangle, they consider v/c (254800/3*10^10) the tangent of the angle theta (the angle of deflection in radians) and since theta is a small angle, they identify the value of the tangent with the angle itself. Also John Rennie did this here:(Deflection of light by the Sun)

the equation I gave is an approximate equation that works when θ is small. At these small angles the difference between θ and tanθ is negligible.

If my report is correct (please forgive any blunders), I have a couple of questions:

  1. If you consider the result of the integral (254800 cm/s) a vector of velocity v, why don't you follow the regular addition of vectors?

  2. Why should the angle between two vectors be intended in radians and not in degrees. What is the rationale?

  3. If you think that $v/c$ is really the tangent of the angle of deflection, why use the subterfuge of approximation and not find the exact angle using the appropriate function of $\mathrm{arctan}$, $\mathrm{tan}^{-1}(v/c)$?

@NaturalPhilosopy, that "doesn't make sense" I wrote in the question. Of course you are only reporting the literature, but I hope you, or someone here, realize that that is not rational, scientific. Why sould you assume anything? why should you deviate from the standard optimal math procedure? what is the logic, the rationale? you know the tangent, want to find the angle : arctan. And what if the angle is not so small? suppose v = c, then tan theta = 1, what do you do? Lastly, if you calculate the effective angle theta with tan= 254800/3*10^10, you discover that the difference is not that negligible.

@Jerry Schirner, sure, radians has a meaning but so do many other things in math. The fact that it is meaningful doesn't mean that it is relevant, appropriate here. That is not enough to justify, logically and physically that the ratio between two normal vectors corresponds to as many radians. Has anybody ever proved that? Can you do that?

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  • $\begingroup$ but I don't understand what you mean by classical physics giving a similar result. How so? Is that when we take the momentum of light to give it mass? $\endgroup$ – lucky-guess Jul 20 '17 at 7:44
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The small angle approximation is very reasonable here. The angle of deflection predicted by the approximation is ${4GM \over rc^2}=8.49\times10^{-6}$ radians. Writing $\tan{\theta}$ instead of $\theta$ is just clutter. The difference between the two is $2.039\times10^{-16}$. In order to resolve this difference in angle in visible light, you would need a telescope about $10^{9}~\text{m}$ wide.

Now you may object to making any approximation at all on principle, but using $\tan(\theta)$ instead of $\theta$ won't get you very far. This expression, $\tan(\theta)={GM \over rc^2}$, is just the first term in a Taylor expansion of a more complicated formula, which you could read about here.

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    $\begingroup$ " you may object to making any approximation at all on principle.." it is not a matter of approximation, it is a matter of being scientific or arbitrary: if we know that the value of the angle is tan^-1, that is the value. Period. Any other criterion is an arbitrary manipulation. It is not a matter of going far. $\endgroup$ – bobie Oct 1 '14 at 6:33
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    $\begingroup$ My point is that neglecting the other terms in the Taylor series will introduce a larger error than the small angle approximation. Taking the arctangent of this expression does not make it exact. Any result in physics is always an approximation. If you want to get an exact result, you might as well just point to the Einstein field equations and say 'the answer is in there, somewhere.' That will still not be exact because it fails to describe quantum gravity. Somewhere, you will have to draw a line to make an approximation. $\endgroup$ – George G Oct 1 '14 at 16:54
  • $\begingroup$ As Einstein (maybe) said, "as far as the laws of mathematics refer to reality, they are not certain, and as far as they are certain, they do not refer to reality" $\endgroup$ – lucky-guess Jul 20 '17 at 7:40
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  1. I am not sure what you mean by this, there are many ways to calculate vectors, some of which are quite complicated (especially in General Relativity when you are dealing with non-Euclidian geometry). If you rephrase it I may be able to answer.
  2. As Schirmer states, "Degrees do not. No one doing any serious geometry would work in degrees." This is due to the impracticality. For example, in calculus (assuming you may have had some very basic calculus experience) when computing derivatives of trigonometric functions, you won't have to deal with factors of pi over 180. This notion is also applicable in a physics concept known as "rotational motion". If you used degrees in this instance, the units for the rotational velocity would be in meters*degrees/seconds, instead of meters/second (which is clearly much more practical).
  3. Again, as Schirmer states, we assume that theta is so small that it actually ≈ tan of itself (this is the same for sin (see below for small angle approximation of cos). If you want a more basic application of this, do some research on the derivation of the "thin lens equation". The geometry here will let you have a stronger grasp of this approximation(which is used all the time in physics).
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    $\begingroup$ The small angle approximation for cosine is $1 - \frac{1}{2}\theta^{2}$ $\endgroup$ – Jerry Schirmer Aug 5 '14 at 19:39
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    $\begingroup$ 1)- there are many ways to calculate vectors :if you are considering the 2 normal vectors, there is only one way to add them. One vector is C and the other is the result of the integral of the acceleration 254800cm/s. $\endgroup$ – bobie Aug 12 '14 at 11:10
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    $\begingroup$ No, when one considers a curved metric, the classical vector space axioms fall apart. There is no natural notion concerning how to "add" two points on a sphere,and end up with a third point. Thus, we must consider the limit of infinitesimal displacements about a point. This is the basis for the calculus of manifolds. $\endgroup$ – Gödel Aug 12 '14 at 19:58

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