# Experimental verification of GR: Light bending

1. Jun 6, 2010

### Passionflower

Can anyone point me to a good source using the Schwarzschild metric that gives Einsteins predicted result of the measurement in Principe?

2. Jun 6, 2010

### starthaus

If you want the theoretical derivation, one good source is this.
If you want experimental verification, then VLBI based experiments have reached an accuracy of 0.04%

3. Jun 7, 2010

### Passionflower

I am confused, the writer first talks about a FLRW metric for the "lens" and then around this lens is a Schwarzschild metric. OK.

But at any rate how do we get to the formula listed under item (5)? Do I perhaps miss a step?

4. Jun 7, 2010

### starthaus

OK, now that I understand what you are after, see the detailed derivation here. Start from the middle, right after the Schwarzschild metric definition.

Last edited: Jun 7, 2010
5. Jun 7, 2010

### bcrowell

Staff Emeritus
I don't think it's an actual derivation. They're just reiterating a result that was derived somewhere else.

For a numerical treatment (valid for angles that aren't necessarily small), see http://www.lightandmatter.com/html_books/genrel/ch06/ch06.html#Section6.2 [Broken] (subsection 6.2.7).

In the limit of small angles, it's easy to constrain the result strongly by dimensional analysis. The only unitless parameter you can construct here is $Gm/c^2 r$, where r is the distance of closest approach (which is the same as the impact parameter in this limit). That means that in this limit, we have to have $\theta=A Gm/c^2 r$, where A is a unitless constant. The numerical treatment shows that A=4.0.

For a proof that A is exactly equal to 4, see Rindler, Essential Relativity, 2nd ed, p. 146.

Last edited by a moderator: May 4, 2017
6. Jun 7, 2010

### Passionflower

Thanks for that reference.

In the Newtonian case he considers two cases:
• The speed of light is c at the perigee, and the speed of light is assumed less at infinity.
• The speed of light is c at infinite and it is assumed that it is less at the perigee.
Unfortunately he does not calculate the result if we assume c slows down closer to a gravitational field.

For the Schwarzschild case at one point he defines:

$$\rho = r_0/r$$

Why?
Does he imply that $\rho$ is some kind of physical distance?

Thanks for that reference.

Last edited by a moderator: May 4, 2017
7. Jun 7, 2010

### starthaus

No, it is simply a change of variable allowing him to evaluate the integral in an easier way.

8. Jun 7, 2010

### Passionflower

Ok, assuming that is correct, then when does he do $\rho=\int (1-2m/r)^{-1/2}dr$ to relate the physical distance to the Schwarzschild r?

9. Jun 7, 2010

### starthaus

He doesn't. His $$\rho$$ has nothing to do with $$\sqrt{1-2m/r}$$ . It is a somewhat unfortunate naming convention, this is all.

10. Jun 7, 2010

### Passionflower

I understand that that is what you say, and I assume that for the moment, but the question remains: where does he go from the physical distance to the Schwarzschild r coordinate?

11. Jun 7, 2010

### starthaus

He doesn't need to, he simply calculates $$\theta$$:

"Integrating this from $$r = r_0$$ to oo gives the mass-centered angle swept out by a photon as it moves from the perihelion out to an infinite distance. If we define $$\rho= r_0/r$$ the above equation can be written in the form...."

12. Jun 7, 2010

### Passionflower

I suppose I am not very successful in explaining to you my question.

Would you agree with me that $\theta$ depends on r? If so, how do we get r?

Last edited: Jun 7, 2010
13. Jun 7, 2010

### starthaus

$$d\theta$$ depends on r .
$$\theta$$ does not depend on r, since r is the integration variable. You can only say that $$\theta$$ depends on $$r_0$$ but even this dependency is taken away by the change of variable $$\rho=r_0/r$$ that makes $$0<\rho<1$$

14. Jun 7, 2010

### Passionflower

I follow that, but don't we need to know the value of $r_0$ to get a physical result? If so, how do we get $r_0$, or even more to the point what was the value for $r_0$ in the Principe experiment?

Last edited: Jun 7, 2010
15. Jun 8, 2010

### starthaus

The experiment uses rays of light "grazing" the Sun. Therefore, a very good approximation for $$r_0$$ is the radius of the Sun.

16. Jun 8, 2010

### Passionflower

No, that is completely incorrect.

The formula speaks about $r$ and $r_0$ the both refer to r values in the Schwarzschild solution.
Note that those r values are not physical radii.

17. Jun 8, 2010

### starthaus

Why don't you look at the picture?

18. Jun 8, 2010

### Passionflower

It seems the truth of the matter is that r (as defined in the Schwarzschild solution) is simply assumed to be the same as $rho$ (e.g the physical distance).

This is not a big deal as the curvature of the Sun is not that high. But, and this I think is important, this brings into question the frequently heard statement that the bending of light is proof of GR because of the curvature. Beyond the curvature found in the Newton-Cartan theory, which is found when a geometric solution to Newton's equations of a point mass is used, there is no additional curvature if one equates the Schwarzschild r with $rho$.

Or am I completely wrong?

Last edited: Jun 8, 2010
19. Jun 9, 2010

### starthaus

The above is true only if $$r_0=1$$ since $$\rho=r/r_0$$.

There is no compelling reason to have $$\rho=r$$.

$$r_0$$ is equal to the perigee distance and, for rays of light grazing the Sun, is equal to the Sun radius.

20. Jun 9, 2010