Mercury Precession & GR Calculation Impact

[zincshow]

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

How does this square with the idea of the shell theorem that says you do not feel a gravitational force from the inside of a hollow sphere of mass :

http://en.wikipedia.org/wiki/Shell_theorem

Thanks

 

[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

How does this square with the idea of the shell theorem that says you do not feel a gravitational force from the inside of a hollow sphere of mass :

http://en.wikipedia.org/wiki/Shell_theorem

Thanks

The authors of the paper abstract the other planets as circular rings , not as spherical shells

 

[zincshow]

The authors of the paper abstract the other planets as circular rings , not as spherical shells

Thanks. For some reason it seemed to me that if there is no force inside a 3d sphere of mass that there would be no force inside a 2d ring of mass. I will have to think about that for a while...

 

[starthaus]

Thanks. For some reason it seemed to me that if there is no force inside a 3d sphere of mass that there would be no force inside a 2d ring of mass. I will have to think about that for a while...

You are absolutely right, look at the calculations in paragraph II, in the opinion of the authors, the ring does not behave like a shell, the resultant force is non-null. I think that the paper is wrong, and I think I know where the authors introduced an error. AmJPhys is not a model of correctness, quite the opposite.

 

[utesfan100]

Thanks. For some reason it seemed to me that if there is no force inside a 3d sphere of mass that there would be no force inside a 2d ring of mass. I will have to think about that for a while...

Here is a link to how a ring's gravitational field can be derived.
"[URL
[/URL]
Here is an application to the planets.
http://farside.ph.utexas.edu/teaching/336k/Newton/node128.html"

 

[D H]

look at the calculations in paragraph II, in the opinion of the authors, the ring does not behave like a shell, the resultant force is non-null. I think that the paper is wrong, and I think I know where the authors introduced an error.

No error here. That a ring does not act like a shell is correct.

 

[D H]

The force is not null except at the center of the ring. Take the gradient of the potential and you will see that this is the case.

Rings are used to prove the shell theorem, but the rings are carefully constructed so that the point in question lies on the axis of each ring.

 

[George Jones]

Yes, this appears to be the correct solution to the problem, by using the non-null potential rather than the force. The paper is still wrong, the link you cite is correct.

I have just run through the (analytical) calculations. To the level of approximation in

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

the results in this reference and in

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

are the same.

Again, you have jumped too quickly to a conclusion.

 

[starthaus]

The force is not null except at the center of the ring. Take the gradient of the potential and you will see that this is the case.

Yes, the analytic formula of the infinitesimal mass, "dm" used in calculating the resultant force for a spherical shell is different from the formula for the "dm" in a ring. This is why rings behave differently from shells. This is why a ring does not exhibit zero force, though it has the same radial symmetry as a shell.

 

[yuiop]

Here is a simple visual demonstration that the gravity inside a ring is not null as inside a hollow sphere (in Newtonian terms):

In the above image the particle at point P inside the hollow sphere shell is at different distances from the shell but the shell mass contained in the circular intersection of the cone on the left is greater than the shell mass intersected on the right and this difference in intersected shell mass exactly compensates for the differences in distance. The regions outside the intersecting cones are identical and trivially cancel each other out, no matter what arbitary angle is chosen for the intersecting cones. See http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/GravField.htm for the source of the diagram.


If a section of the shell above and below point P (the green sections in the attached diagram) are removed, leaving an equatorial ring, then some of the compensating shell mass is removed (the light blue regions in the attached diagram) and the forces no longer cancel out. Anything in the plane of the ring, but not exactly at the centre, moves outwards as if there is a repulsive force at the centre of the ring.

 

[D H]

In short, to re-echo the claims of MIT students at the 1971 World Science Fiction Convention, "The Ringworld is unstable!"

It is also important to remember that Mercury's perihelion precession was known to be anomalously large well before Einstein. (The anomaly was first reported in 1859.) Those physicists were quite adept with calculation; they had to be because they did not have those electronic marvels that we rely on (possibly overly so) these days.

 

[starthaus]

In short, to re-echo the claims of MIT students at the 1971 World Science Fiction Convention, "The Ringworld is unstable!"

Yes, it is :-)

It is also important to remember that Mercury's perihelion precession was known to be anomalously large well before Einstein. (The anomaly was first reported in 1859.) Those physicists were quite adept with calculation; they had to be because they did not have those electronic marvels that we rely on (possibly overly so) these days.

Sure, the advancement of the perihelion was never in debate, just the approach used by the paper cited in the OP for calculating the force inside the ring. I find their approach inferior to the approach used for the shell calculation. The wiki calculation for the shell is quite elegant, makes a much better use of radial symmetry.

 

[D H]

The proof of the shell theorem such as that in wikipedia constructs the rings so that the point in question is along each of the rings axes. That construction ensures radial symmetry.

There is no radial symmetry in this problem to take advantage of. Mercury is not located along the axis of the outer planet's orbits. It is more or less located on the orbital planes of those orbits. The projection of Mercury's location onto each of those orbital planes is not at the center of those orbits.

 

[starthaus]

The projection of Mercury's location onto each of those orbital planes is not at the center of those orbits.

This is quite obvious. Nevertheless, it is easy to adapt the wiki proof for the shell to the ring. The only change required is a recalculation of the infinitesimal mass element "dm".

 

[D H]

Try it. You're going to get something rather ugly. In fact, the integral that results does not have a simple solution. You will get what are called elliptical integrals.

 

[D H]

This isn't homework, so if you want I can give you a link to a solution. (You probably wouldn't believe me if I the words came from my fingers. Fortunately, the solution is out there on the 'net. Besides, I don't particularly want to typeset the LaTeX with my fingers when the solution is out there on the 'net.)

Or you can try it yourself. When/if you get stuck I'll be glad to help.

 

[starthaus]

This isn't homework, so if you want I can give you a link to a solution. (You probably wouldn't believe me if I the words came from my fingers.

Of course I believe you.

Fortunately, the solution is out there on the 'net. Besides, I don't particularly want to typeset the LaTeX with my fingers when the solution is out there on the 'net.)

Or you can try it yourself. When/if you get stuck I'll be glad to help.

Thank you, I appreciate the offer, what I was saying is that I have already worked it out. In the wiki solution you need to replace the spherical mass element with the ring one. Everything else in the solution stays same.

 

[D H]

No, it does not. Show your work.

 

[starthaus]

No, it does not. Show your work.

I don't like your tone. Bye.

 

[D H]

Here's the calculation for the potential.

First some definitions. We want to calculate the gravitational potential due to a uniform ring of mass at some point. Define z as the distance between the point and the ring plane and x as the distance between the point and the ring axis. Without loss of generality, define

• \hat x is the direction from the center of the ring to the projection of the point in question onto the ring plane,
• \hat z is parallel to the ring axis, with positive z in the direction of the point in question, and
• \hat y completing a right-hand system (\hat y \equiv \hat z \times \hat x).

We also need some definitions regarding the ring itself.

• a is the radius of the ring,
• m is the mass of the ring,
• \rho is the linear density of the ring (\rho \equiv m/(2\pi a)).

Potential obeys the superposition principle. Thus

\phi(\mathbf x) = \int \frac{G\,dm}{r} =<br /> \int_0^{2\pi} \frac {G \rho a}{\sqrt{z^2+x^2+a^2-2ax\cos\theta}} d\theta

This is an elliptical integral.

 

[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 above paper is definitely wrong. The resultant integral in paragraph II needs to be an elliptic integral , therefore it will not have a closed expression. The authors have inadvertently canceled out the square roods (which give the integral its elliptic charracter).

How does this square with the idea of the shell theorem that says you do not feel a gravitational force from the inside of a hollow sphere of mass :

http://en.wikipedia.org/wiki/Shell_theorem

Thanks

You will get the desired elliptic integral that gives the correct answer by simply noticing that, for a ring:

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

Once you insert the modified expression for dM, you will get the elliptic integral that represents the resultant force along the direction that connects the center of the ring with the probe mass (in our case, the planet Mercury).

 

[George Jones]

The above paper is definitely wrong.

It seems to me that this not true.

The resultant integral in paragraph II needs to be an elliptic integral , therefore it will not have a closed expression.

Why isn't their integral an elliptic integral? Do you know the reason?

The authors have inadvertently canceled out the square roods (which give the integral its elliptic charracter).

No, the authors have not "inadvertently canceled out the square roots," so this isn't the reason why their integral is not an elliptic integral.

 

[starthaus]

It seems to me that this not true.Why isn't their integral an elliptic integral? Do you know the reason?

I explained why. I also explained how you can modify the solution on the wiki page in order to get the correct solution for the force exerted by a ring. The solution is an elliptic integral (with no closed form). The paper authors obtained a totally different integral.

 

[D H]

Do you know what perturbation techniques are?

 

[starthaus]

Do you know what perturbation techniques are?

Yes.
What does your question have to do with figuring out the expression of the integral that defines the resultant force?

 

[D H]

What does your question have to do with figuring out the expression of the integral that defines the resultant force?

The planetary effects on Mercury's orbit are perturbative. The exact integral isn't needed. All that are needed to estimate these perturbative effects are first-order terms. That is what the author of the paper cited in the original post developed in equations (2) to (5) of that paper.

 

[D H]

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."

I don't know what Dicke did in Gravitation and Relativity, but I suspect he used Lagrange's Planetary Equations or some similar formulation (e.g., Dalaunay's or Hill's Planetary Equations). Computing the Poisson brackets of orbital elements is not a lower level undergraduate treatment.

 

[starthaus]

The planetary effects on Mercury's orbit are perturbative. The exact integral isn't needed. All that are needed to estimate these perturbative effects are first-order terms. That is what the author of the paper cited in the original post developed in equations (2) to (5) of that paper.

This is not what we were discussing, I just showed that the correct answer is obtained by using dM=\frac{d \theta}{\pi}M in the approach used on the wiki page. Do you have an argument with that?

 

[D H]

This is not what we were discussing

That is [exactly[/i] what we are discussing. That paper is the central topic of this thread.

I just showed that the correct answer is obtained by using dM=\frac{d \theta}{\pi}M in the approach used on the wiki page. Do you have an argument with that?

Yes. First, a minor correction, that should be dM=M/(2\pi)\,d\theta.

More importantly, your result is only correct when angle is measured from the center of the ring. It is not correct when the angle is measured with respect to some non-central point such as the location of Mercury.

 

[starthaus]

That is [exactly[/i] what we are discussing. That paper is the central topic of this thread.

The paper produced the incorrect formula for the force.

Yes. First, a minor correction, that should be dM=M/(2\pi)\,d\theta.

Incorrect: dM=M/(\pi)\,d\theta

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

More importantly, your result is only correct when angle is measured from the center of the ring. It is not correct when the angle is measured with respect to some non-central point such as the location of Mercury.

It is precisely the case described in the paper and in the wiki page. The wiki description of the solution is very good, I suggest that you read it.

 

[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 &amp; (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.