Experimental verification of GR: Light bending

[sf222]

Are you saying they are approximately equal but the difference is still an explanation for the fact the GR shows a double result compare to the Newtonian explanation? And the reason is because of the spatial curvature?

If that is so then I am sorry I am not getting it, if we say $\rho$ is approximately $r$ then space is obviously approximately flat.

I am sure I miss something but I fail to see what.

The effect of spatial curvature on the rate of deflection of the path of light is not proportional to the slight difference between r and rho at any given point, it is proportional to how that difference changes along the path. Notice that, at the point of tangency, that difference isn't changing along the path, and the rate of deflection at that point is the same as the Newtonian rate. The doubling of the total deflection comes about because of how the rate differs at other points along the path. The best explanation of all this is here:

http://www.mathpages.com/home/kmath115/kmath115.htm

 

[Passionflower]

The effect of spatial curvature on the rate of deflection of the path of light is not proportional to the slight difference between r and rho at any given point, it is proportional to how that difference changes along the path. Notice that, at the point of tangency, that difference isn't changing along the path, and the rate of deflection at that point is the same as the Newtonian rate. The doubling of the total deflection comes about because of how the rate differs at other points along the path. The best explanation of all this is here:

http://www.mathpages.com/home/kmath115/kmath115.htm

I see, now that proposition seems to make a lot of sense.

Thank you so much for that information. I will study the link you provided. Are there any references to this in the literature?

However I remain to have some questions. It is claimed that the Newtonian result will give a factor 1/2 or the GR result. However take a look at this paper:
http://arxiv.org/pdf/gr-qc/0309072v3

It appears that if we apply Newtonian Gravity and Galilean Relativity we get the Schwarzschild metric with a difference that r represents a physical radius for which $r = c / 2 \pri$.

So, how can Newton's theory give a discrepancy of a factor 2?

By the way, I know someone who claims that the integral Einstein used to calculate the deflection does not give the proper results when integrated numerically, however this integral is different from the one referenced in http://www.mathpages.com/home/kmath115/kmath115.htm

I like to see which integral Einstein actually used to calculate his prediction. Which is the paper I have to look at?Edited to add

Ok, I looked at the mathpages website and now it get clearer.

So considering the Schwarzschild metric in this form:

$$\color{blue}(d\tau)^2 = \left(1 - \frac{2m}{r}\right)(dt)^2 - (dx)^2 - (dy)^2 - (dz)^2 \color{red} - \frac{1}{r^2}\left(\frac{2m}{r-2m}\right)(xdx + ydy + zdz)^2$$

It is clear there are two components:

• The blue part reflects the g_tt part (e.g. the Newtonian first half)
• The red part reflects "the space differentials" (the other half)

Very nice.

Again sf222, thanks for pointing me in the right direction.