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Passionflower
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Can anyone point me to a good source using the Schwarzschild metric that gives Einsteins predicted result of the measurement in Principe?
Passionflower said:Can anyone point me to a good source using the Schwarzschild metric that gives Einsteins predicted result of the measurement in Principe?
I am confused, the writer first talks about a FLRW metric for the "lens" and then around this lens is a Schwarzschild metric. OK.starthaus said:one good source is this.
Passionflower said: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?
Passionflower said: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?
Thanks for that reference.starthaus said: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.bcrowell said: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).
Passionflower said:For the Schwarzschild case at one point he defines:
[tex]
\rho = r_0/r
[/tex]
Why?
Does he imply that [itex] \rho[/itex] is some kind of physical distance?
Thanks for that reference.
Ok, assuming that is correct, then when does he do [itex]\rho=\int (1-2m/r)^{-1/2}dr[/itex] to relate the physical distance to the Schwarzschild r?starthaus said:No, it is simply a change of variable allowing him to evaluate the integral in an easier way.
Passionflower said:Ok, assuming that is correct, then when does he do [itex]\rho=\int (1-2m/r)^{-1/2}dr[/itex] to relate the physical distance to the Schwarzschild r?
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 said:He doesn't. His [tex]\rho[/tex] has nothing to do with [tex]\sqrt{1-2m/r}[/tex] . It is a somewhat unfortunate naming convention, this is all.
Passionflower said: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?
I suppose I am not very successful in explaining to you my question.starthaus said:He doesn't need to, he simply calculates [tex]\theta[/tex]:
Passionflower said:I suppose I am not very successful in explaining to you my question.
Would you agree with me that [itex]\theta[/itex] depends on r? If so, how do we get r?
I follow that, but don't we need to know the value of [itex]r_0[/itex] to get a physical result? If so, how do we get [itex]r_0[/itex], or even more to the point what was the value for [itex]r_0[/itex] in the Principe experiment?starthaus said:[tex]d\theta[/tex] depends on r .
[tex]\theta[/tex] does not depend on r, since r is the integration variable. You can only say that [tex]\theta[/tex] depends on [tex]r_0[/tex] but even this dependency is taken away by the change of variable [tex]\rho=r_0/r[/tex] that makes [tex]0<\rho<1[/tex]
Passionflower said:I follow that, but don't we need to know the value of [itex]r_0[/itex] to get a physical result? If so, how do we get [itex]r_0[/itex], or even more to the point what was the value for [itex]r_0[/itex] in the Principe experiment?
No, that is completely incorrect.starthaus said:The experiment uses rays of light "grazing" the Sun. Therefore, a very good approximation for [tex]r_0[/tex] is the radius of the Sun.
Passionflower said:No, that is completely incorrect.
The formula speaks about [itex]r[/itex] and [itex]r_0[/itex] the both refer to r values in the Schwarzschild solution.
Note that those r values are not physical radii.
Passionflower said:It seems the truth of the matter is that r (as defined in the Schwarzschild solution) is simply assumed to be the same as [itex]rho[/itex] (e.g the physical distance).
there is no additional curvature if one equates the Schwarzschild r with [itex]rho[/itex].
Or am I completely wrong?
Passionflower said:No, that is completely incorrect.
The formula speaks about [itex]r[/itex] and [itex]r_0[/itex] the both refer to r values in the Schwarzschild solution.
Note that those r values are not physical radii.
In that case you should perhaps read posting #18bcrowell said: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 said:In that case you should perhaps read posting #18
Passionflower said:In that case you should perhaps read posting #18
Am I claiming otherwise? So we seem to agree.bcrowell said: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 [itex]\theta=4GMc^2/r[/itex], 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 said: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 = [itex]rho[/itex] (the physical distance), is obviously flat.
That is fair, but could you indicate where the error is?bcrowell said: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.
Nobody?Passionflower said: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 = [itex]rho[/itex], the space, for which the shell r is its boundary, is flat?
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?K^2 said: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 haven't been following this thread in detail, but I am assuming the result stated earlier, [itex]\theta=4GMc^2/r[/itex], 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 [itex]r[/itex] by [itex]r+\epsilon[/itex], where [itex]\epsilon[/itex] is very small compared with [itex]r[/itex], 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 said: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?
If we take [itex]\rho = r[/itex] (and thus [itex]g_{11} = 0[/itex]) then how can space be curved since then [itex]C = 2 \pi r[/itex] ?DrGreg said:I haven't been following this thread in detail, but I am assuming the result stated earlier, [itex]\theta=4GMc^2/r[/itex], 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 [itex]r[/itex] by [itex]r+\epsilon[/itex], where [itex]\epsilon[/itex] is very small compared with [itex]r[/itex], 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.
But we didn't assume [itex]\rho = r[/itex] 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 [itex]\rho[/itex].Passionflower said:If we take [itex]\rho = r[/itex] (and thus [itex]g_{11} = 0[/itex]) then how can space be curved since then [itex]C = 2 \pi r[/itex] ?
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?DrGreg said:But we didn't assume [itex]\rho = r[/itex] 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 [itex]\rho[/itex].
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.