Mercury's precession

1. Apr 25, 2010

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 [Broken]

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

Last edited by a moderator: May 4, 2017
2. Apr 25, 2010

starthaus

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

Last edited by a moderator: May 4, 2017
3. Apr 25, 2010

zincshow

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...

4. Apr 25, 2010

starthaus

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.

Last edited: Apr 25, 2010
5. Apr 26, 2010

utesfan100

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

Last edited by a moderator: May 4, 2017
6. Apr 26, 2010

D H

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

7. Apr 26, 2010

D H

Staff Emeritus
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.

8. Apr 26, 2010

George Jones

Staff Emeritus
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 [Broken]

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.

Last edited by a moderator: May 4, 2017
9. Apr 26, 2010

starthaus

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.

Last edited: Apr 26, 2010
10. Apr 26, 2010

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.

Attached Files:

• Newton ring.GIF
File size:
2.6 KB
Views:
92
Last edited: Apr 26, 2010
11. Apr 26, 2010

D H

Staff Emeritus
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.

12. Apr 26, 2010

starthaus

Yes, it is :-)

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.

13. Apr 26, 2010

D H

Staff Emeritus
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.

14. Apr 26, 2010

starthaus

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".

15. Apr 26, 2010

D H

Staff Emeritus
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.

16. Apr 26, 2010

D H

Staff Emeritus
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.

17. Apr 26, 2010

starthaus

Of course I believe you.

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.

18. Apr 26, 2010

D H

Staff Emeritus
No, it does not. Show your work.

19. Apr 26, 2010

starthaus

I don't like your tone. Bye.

20. Apr 26, 2010

D H

Staff Emeritus
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

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

This is an elliptical integral.

21. Apr 27, 2010

starthaus

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 cancelled out the square roods (which give the integral its elliptic charracter).

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).

Last edited by a moderator: May 4, 2017
22. Apr 27, 2010

George Jones

Staff Emeritus
It seems to me that this not true.
Why isn't their integral an elliptic integral? Do you know the reason?
No, the authors have not "inadvertently cancelled out the square roots," so this isn't the reason why their integral is not an elliptic integral.

23. Apr 27, 2010

starthaus

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.

24. Apr 27, 2010

D H

Staff Emeritus
Do you know what perturbation techniques are?

25. Apr 28, 2010

starthaus

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

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook