Experimental verification of GR: Light bending

[Passionflower]

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

 

[starthaus]

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

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%

 

[Passionflower]

one good source is this.

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?

 

[starthaus]

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?

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

 

[bcrowell]

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?

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 (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.

 

[Passionflower]

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

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?

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 (subsection 6.2.7).

Thanks for that reference.

 

[starthaus]

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.

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

 

[Passionflower]

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

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?

 

[starthaus]

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?

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

 

[Passionflower]

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

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?

 

[starthaus]

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?

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..."

 

[Passionflower]

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

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?

 

[starthaus]

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?

$$d\theta$$ 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 $$0<\rho<1$$

 

[Passionflower]

$$d\theta$$ 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 $$0<\rho<1$$

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?

 

[starthaus]

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?

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

 

[Passionflower]

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

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.

 

[starthaus]

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.

Why don't you look at the picture?

 

[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?

 

[starthaus]

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).

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

there is no additional curvature if one equates the Schwarzschild r with $rho$.

Or am I completely wrong?

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.

 

[Passionflower]

An excellent discussion related to this topic can also be found at: http://www.natscience.com/Uwe/Forum.aspx/physics-research/4056/Flatness-Newtonian-Gravity

 

[bcrowell]

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.

Starthaus was correct. In the case of the sun, the Schwarzschild coordinate r is basically the same as the metric distance from the center of the sun. The approximation would be poor in the case of light rays coming close to a black hole.

 

[Passionflower]

Starthaus was correct. In the case of the sun, the Schwarzschild coordinate r is basically the same as the metric distance from the center of the sun. The approximation would be poor in the case of light rays coming close to a black hole.

In that case you should perhaps read posting #18

 

[starthaus]

In that case you should perhaps read posting #18

You asked about the "Principe" (Eddington's experiment). The experiment deals with light bending by the Sun, right?

 

[bcrowell]

In that case you should perhaps read posting #18

I've read it, and I don't see your point, but maybe we're not understanding each other clearly.

In the limit of weak fields, the deflection of light is $\theta=4GMc^2/r$, where r is the distance of closest approach. What I'm claiming is that in the case of the sun, the result is approximately the same regardless of whether r is taken to be the Schwarzschild r coordinate or the distance from the center of the sun.

 

[Passionflower]

I've read it, and I don't see your point, but maybe we're not understanding each other clearly.

In the limit of weak fields, the deflection of light is $\theta=4GMc^2/r$, where r is the distance of closest approach. What I'm claiming is that in the case of the sun, the result is approximately the same regardless of whether r is taken to be the Schwarzschild r coordinate or the distance from the center of the sun.

Am I claiming otherwise? So we seem to agree.

In other words, the frequently heard statement that the double size, as compared to the Newtonian result, is "due to the curvature" is not true because space in such a modified Schwarzschild solution, where it is assumed that r = $rho$ (the physical distance), is obviously flat.

 

[bcrowell]

Am I claiming otherwise? So we seem to agree.

In other words, the frequently heard statement that the double size, as compared to the Newtonian result, is "due to the curvature" is not true because space in such a modified Schwarzschild solution, where it is assumed that r = $rho$ (the physical distance), is obviously flat.

I don't agree with this. I don't think you can make this type of far-reaching conclusion just because it's possible to make one approximation at one step.

 

[Passionflower]

I don't agree with this. I don't think you can make this type of far-reaching conclusion just because it's possible to make one approximation at one step.

That is fair, but could you indicate where the error is?

Would you agree that it is true that in the Schwarzschild solution for smaller values of r the space increasingly becomes curved but that for larger values of r, or in the extreme case, for an infinite value of r, the space becomes more and more flat?

If this is true, then does it not follow that for a given r, r = $rho$, the space, for which the shell r is its boundary, is flat?

 

[Passionflower]

That is fair, but could you indicate where the error is?

Would you agree that it is true that in the Schwarzschild solution for smaller values of r the space increasingly becomes curved but that for larger values of r, or in the extreme case, for an infinite value of r, the space becomes more and more flat?

If this is true, then does it not follow that for a given r, r = $rho$, the space, for which the shell r is its boundary, is flat?

Nobody?

 

[K^2]

Of course it's a small curvature approximation. The derivation assumes a hyperbolic trajectory in advance.

In general, light beam approaching a black hole can make some number of turns around it, and only then escape. No classical trajectory can do that.

General solution is incredibly complex and almost certainly non-algebraic. Best one could do is write down the differential equations for null geodesic in Schwarzschild Metric.

 

[Passionflower]

Of course it's a small curvature approximation. The derivation assumes a hyperbolic trajectory in advance.

In general, light beam approaching a black hole can make some number of turns around it, and only then escape. No classical trajectory can do that.

General solution is incredibly complex and almost certainly non-algebraic. Best one could do is write down the differential equations for null geodesic in Schwarzschild Metric.

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 misunderstanding here?

 

[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

$$f(x) \approx 0$$​