Mercury Precession & GR Calculation Impact

[D H]

The paper produced the incorrect formula for the force.

No, it did not. It used a first-order approximation, which is a perfectly valid thing to do in perturbation theory. You are making a strong claim here. You need to show that this claim is true.

There are two symmetrical elements participating in the force calculation n(see the drawing on the wiki page), so :

dM=\frac{2 R d \theta}{2 \pi R}M

I hope you are not talking about this *lousy* diagram:

The thing that looks like a ring in that diagram is not a ring. It is an orthogonal projection of the spherical shell. The bluish band represents a ring. The angle theta in that diagram is a polar angle. The azimuthal angle is not shown; the wiki article hand-waves over the azimuthal integration. The rings in that wiki article are constructed so that the point in question always lies on the ring axes. Mercury does not lie along the axis of any the planetary rings in the article cited in the original post. The construction in the wiki article does not apply to the topic of thread.

The wiki description of the solution is very good, I suggest that you read it.

I disagree. The wiki description of the shell theorem is very bad. It has long stood as one of my prototypical examples of bad wikipedia articles. No references, lousy diagrams, incomplete math, and far too much unjustified/behind-the-scenes hand-waving.

 

[starthaus]

No, it did not. It used a first-order approximation, which is a perfectly valid thing to do in perturbation theory. You are making a strong claim here. You need to show that this claim is true.

It is very simple, the integral that describes the force should be an elliptical integral, with no closed form. The authors inadvertently canceled out the square root part of the expression, thus obtaining a closed form which is wrong. I think I have told you this about three times already.

I hope you are not talking about this *lousy* diagram:

The thing that looks like a ring in that diagram is not a ring.

It is a transverse section through a sphere. I see that you figured that out. I am also glad you figured out your basic mistake in calculating dM

It is an orthogonal projection of the spherical shell. The bluish band represents a ring. The angle theta in that diagram is a polar angle. The azimuthal angle is not shown;

It is easy to figure out that the azimuthal angle is zero since the test probe is placed in the equatorial plane of the sphere.

the wiki article hand-waves over the azimuthal integration.

It doesn't. It is plain to see that the azimuthal angle is zero.

The rings in that wiki article are constructed so that the point in question always lies on the ring axes.

Right.

Mercury does not lie along the axis of any the planetary rings in the article cited in the original post. The construction in the wiki article does not apply to the topic of thread.

May I suggest that you read the paper again? Pay attention to paragraph II. Mercury is embedded in the plane of the ring, the azimuthal angle is ZERO.

 

[D H]

It is very simple, the integral that describes the force should be an elliptical integral, with no closed form.

Only if you want an exact answer. An exact answer is not needed here. For one thing, the planet are not rings of material. Requiring an exact answer when the whole setup is a simplifying approximation is, to be blunt, silly. For another, the authors of that paper are using perturbation theory. Do you know what that is?

The authors inadvertently canceled out the square root part of the expression, thus obtaining a closed form which is wrong. I think I have told you this about three times already.

The authors used an approximation and clearly indicated so in equation (2) and in the narrative describing that equation.

It is a transverse section through a sphere. I see that you figured that out. I am also glad you figured out your basic mistake in calculating dM

The only error here is yours. That transverse section / orthogonal projection is not the ring. The bluish rectangle represents the ring.

[QUOTE}It is easy to figure out that the azimuthal angle is zero since the test probe is placed in the equatorial plane of the sphere.[/QUOTE]
Describing a point on surface of a sphere requires two angles. Certainly you are familiar with spherical coordinates. The angle shown in that diagram is a polar angle (better stated, elevation angle). The azimuthal angle is not shown.

 

[starthaus]

Only if you want an exact answer.

You are changing the argument. All I was telling you is that the integral in the paper is the wrong one.
You never aknowledged the fact that you were also wrong about Mercury being offset from the plane of the ring. In the paper, Mercury IS placed in the plane of the ring.

For another, the authors of that paper are using perturbation theory. Do you know what that is?

I already told you that I know what "perturbation theory" is. If you want this subjected treated correctly, you can read here

The authors used an approximation and clearly indicated so in equation (2) and in the narrative describing that equation.

Yes, and the approximation integral is evaluated incorrectly. A correct evaluation would be an elliptic integral.

The only error here is yours.

You mean that you made a basic error and you got dM to be half of its correct expression? It is very simple, really, in the case of a ring the problem reduces to only two dimensions. Instead of the spherical mass element calculated on the wiki page , you need to calculate the ring mass element

\frac{2Rd\theta}{2\pi R}M=\frac{d\theta}{\pi}M

Plug the mass element in the integral and you get your correct result.

That transverse section / orthogonal projection is not the ring.

I did not tell you that.I told you precisely: "It is a transverse section through a sphere". Please try reading what I told you.

The bluish rectangle represents the ring.

The bluish rectangle is a trapezoid really. And it represents the section through the sphere surface perpendicular to the direction connecting the test probe position with the center of the sphere. By using this clever trick, the person that did the derivation for the wiki page gets the resultant of the forces exerted by the sphere on the test probe to line up with the direction sphere center -test probe. A very clever trick.
You do not even understand the wiki page, yet you are quick to criticize it .

 

[starthaus]

When calculation GR, the component due to other planets is removed. In that calculation outer planets are assumed to be rings of mass and the effect of that mass is seen to have an effect on mercury precession :

http://www.physics.princeton.edu/~mcdonald/examples/mechanics/price_ajp_47_531_79.pdf

The error starts right from equation (2) where the authors decide that:

ds_i=l_i*d\alpha

This is obviously not true. It would have been true for the trivial case when the Mercury was located in the center of the circle but this is obviously not the case. So, their "proof" is wrong from the beginning. The authors wave their hands by appending a note that aknowledges that their hack is invalid for a~R. In reality, it is never valid because it defies elementary geometry.
If one wants to read a decent treatment, here is a very good one.

 

[D H]

That is a strawman, starthaus. The authors said

If ds_1 and ds_2 are the arc lengths subtended by differential angular element d\alpha and if a is small compared to R¹¹

dm_i = \lambda ds_i \simeq l_i \lambda d\alpha \qquad\qquad(2)

The astute reader will note that Eq. (2) will not be valid for values of a\sim R, since the angle d\alpha would subtend a mass larger than ds_i by a factor of (\cos \alpha)^{-1}. Since the orbits of Venus and Mercury differ by approximately a factor of 2, we repeated the calculation including the (\cos \alpha)^{-1} term. An exact solution could not be found, but a series expansion to terms of order (a/R) showed that the errors introduced by ignoring the (\cos \alpha)^{-1} term were of order of 2.3%.​

In other words, the authors explicitly acknowledged that this is an approximation and they checked the validity of the approximation.

 

[starthaus]

That is a strawman, starthaus. The authors said

If ds_1 and ds_2 are the arc lengths subtended by differential angular element d\alpha and if a is small compared to R¹¹

dm_i = \lambda ds_i \simeq l_i \lambda d\alpha & (2)

The astute reader will note that Eq. (2) will not be valid for values of a\sim R, since the angle d\alpha would subtend a mass larger than ds_i by a factor of (\cos \alpha)^{-1}. Since the orbits of Venus and Mercury differ by approximately a factor of 2, we repeated the calculation including the (\cos \alpha)^{-1} term. An exact solution could not be found, but a series expansion to terms of order (a/R)[/itex[ showed that the errors introduced by ignoring the (\cos \alpha)^{-1} term were of order of 2.3%.​

<br /> In other words, the authors explicitly acknowledged that this is an approximation and they checked the validity of the approximation.

<br /> <br /> Yes, I pointed that out. They acknowledged that it is a hack that isn&#039;t valid for a~R. It is a hack that it is never valid since it defies elementary geometry (the kind they teach in 9-th grade). You know where you can find a decent treatment.

 

[George Jones]

The author's intent in this paper is "to show that the component of the precession of Mercury's perihelion due to the outer planets can be calculated easily at the undergraduate level." He did so quite admirably IMO. The mathematics is quite simple, and yet comes remarkably close to those obtained by "more advanced treatments."

Yes, I agree.

The authors have inadvertently canceled out the square roods

You have not acknowledged that you made a mistake here.

Yes, I pointed that out. They acknowledged that it is a hack that isn't valid for a~R. It is a hack that it is never valid since it defies elementary geometry (the kind they teach in 9-th grade). You know where you can find a decent treatment.

I don't like your tone. The authors have not made a mistake, they have made a useful approximation. This is the kind of skill to which students should be exposed, because this type of thing is ubiquitous in physics. If an approximation were equal to the unapproximated expression, it wouldn't be an approximation.

It is a useful exercise to express the difference between

http://www.physics.princeton.edu/~mcdonald/examples/mechanics/price_ajp_47_531_79.pdf

and

http://farside.ph.utexas.edu/teaching/336k/Newton/node127.html

in terms of powers of r/a.

 

[starthaus]

It is a useful exercise to express the difference between

http://www.physics.princeton.edu/~mcdonald/examples/mechanics/price_ajp_47_531_79.pdf

and

http://farside.ph.utexas.edu/teaching/336k/Newton/node127.html

in terms of powers of r/a.

Sure, it is. Or you can use the algorithm that I outlined in calculating the exact integral describing the resultant force exerted by a ring. Your choice.