Experimental verification of GR: Light bending

[DrGreg]

I am not sure if you understand my question. I have no objection to treating r as the real distance but if we do that then how is space exactly curved in that situation?

The general explanation of the fact that the GR outcome is double the Newton outcome is that the extra half is due to curvature of space. Now if we assume r = rho then where is the curvature? Anyone care to explain my lack of understanding here?

I haven't been following this thread in detail, but I am assuming the result stated earlier, \theta=4GMc^2/r, is correct, where r is Schwarzschild r coordinate, and that result was calculated in curved spacetime, and the answer would have been different in flat spacetime. All that is being said that if you replace r by r+\epsilon, where \epsilon is very small compared with r, the answer is nearly the same. The fact that you've put the "wrong" radius into the final answer and got virtually the same answer does not imply that you could have assumed flat spacetime and got the same answer.

 

[Passionflower]

I haven't been following this thread in detail, but I am assuming the result stated earlier, \theta=4GMc^2/r, is correct, where r is Schwarzschild r coordinate, and that result was calculated in curved spacetime, and the answer would have been different in flat spacetime. All that is being said that if you replace r by r+\epsilon, where \epsilon is very small compared with r, the answer is nearly the same. The fact that you've put the "wrong" radius into the final answer and got virtually the same answer does not imply that you could have assumed flat spacetime and got the same answer.

If we take \rho = r (and thus g_{11} = 0) then how can space be curved since then C = 2 \pi r ?

 

[DrGreg]

If we take \rho = r (and thus g_{11} = 0) then how can space be curved since then C = 2 \pi r ?

But we didn't assume \rho = r during the derivation of the formula involving r, all we are doing at the very end of the calculation (performed in curved spacetime) is noticing that the correct formula with r turns out be approximately the same as the approximate formula with \rho.

There is a difference between saying two things are equal and two things are approximately equal. When they are approximately equal you have to look at the mathematical formula being used to see whether a small change in the input makes a small or large change to the output. In this case the change in output is small, but it might not be for some other formula.

 

[Passionflower]

But we didn't assume \rho = r during the derivation of the formula involving r, all we are doing at the very end of the calculation (performed in curved spacetime) is noticing that the correct formula with r turns out be approximately the same as the approximate formula with \rho.

There is a difference between saying two things are equal and two things are approximately equal. When they are approximately equal you have to look at the mathematical formula being used to see whether a small change in the input makes a small or large change to the output. In this case the change in output is small, but it might not be for some other formula.

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.

 

[DrGreg]

When two things are "approximately equal" then it depends on the context as whether the difference is small enough to be negligible or not. When you are deriving the formula \theta=4GMc^2/r, the curvature of space cannot be neglected in that context, you make use of it during the derivation. It is only when you have derived the final result that you can see that a small change in the value of r results in a small change in the value of \theta in this new context. This is nothing to do with the physics, it's a mathematical fact.

As an analogy, consider the function

f(x) = \frac{\epsilon}{x+\epsilon}​

where \epsilon is some small constant. Then, provided x is much, much larger than \epsilon, we can approximate the denominator by taking \epsilon = 0, but we certainly cannot approximate the numerator by taking \epsilon = 0.

So we can say

f(x) \approx \frac{\epsilon}{x}​

but we can't say

f(x) \approx 0​

 

[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[/color]

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.