# Question on advance of perihelion.

1. May 8, 2007

### Ian

I have never accepted Einstein's thought pattern on this and other matters. I have examined his equation to calculate the advamce of Mercury's perihelion and have also calculated the advance in the case of a circular orbit, using the same equation.
What I found is that if you remove the correction term for elliptical orbits ( r(1 - e2) from Einstein's equation, the result is not nil advance, but rather it remains the same but is expressed as a length along the circumference of orbit and not an angular measure.
What Einstein said was that the greater velocity during the closer approach at perhelion adds a little extra mass to the system which makes the orbit advance.
If the orbital velocity is constant as is for the circular case, then Einstein's equation ought to reflect his reasoning and show no advance at all.
So what makes the orbit advance if it is circular?
Is Einstein's reasoning right or wrong on this point?

2. May 8, 2007

### Chris Hillman

That is not what gtr says, I doubt you will be able to provide a citation supporting your contention that Einstein ever said otherwise.

Are you talking about his 1915 paper where he presented this computation, or what?

You've clearly misunderstood something, so I'd suggest you consult a modern textbook such as D'Inverno, which most students will probably find easier to follow than the original papers. See http://math.ucr.edu/home/baez/RelWWW/HTML/reading.html#gtrmoderntext [Broken] for a comparison of coverage of some standard gtr textbooks.

This doesn't even make sense as stated: a circular orbit has no pericenter. You no doubt meant to ask if the computation simplifies in the case when the orbit is almost circular. Indeed it does; try the cosmology textbook by Peacock, where the author uses the equation of simple harmonic motion to derive the pericenter advance for a nearly circular orbit. This will make more sense if you first study the notion of "effective potentials" in other standard textbooks (such as MTW or Schutz).

The result given by Einstein is correct. The reasoning he used in his 1915 paper can be criticized, which is one reason why modern textbooks use somewhat different approaches. (Briefly, the general idea is that modern textbook presentations use the exact Schwarzschild vacuum solution rather than the weak-field approximation used by Einstein, which turns out to be harder to analyze carefully. That's not surprising: one of the well-known benefits of exact solutions is that they avoid the delicate reasoning which is often required to obtain a correct approximation.) Years ago there was a thread which touched on Weinberg's remarsk on this issue in sci.physics.research, which you can look up after you've mastered the textbook computations.

Last edited by a moderator: May 2, 2017
3. May 8, 2007

### pervect

Staff Emeritus
There's a fairly simple calculation of the perihelion shift using approximate methods in Goldstein's "classical mechanics".

4. May 9, 2007

### Chris Hillman

It's been a while since I looked at that book--- is this the Newtonian shift (e.g. for Sun-Mercury-Jupiter system) or the gtr extra-Newtonian shift? (The former is harder to obtain carefully than the latter, at least to high accuracy.)

5. May 9, 2007

### pervect

Staff Emeritus
This analysis ignores the other planets, it's a purely GR analysis. Goldstein does not go into the GR aspects much, citing MTW and another older source for the equations of motion. The idea is to show of a practical example of pertubation theory.

6. May 9, 2007

### smallphi

A simple derivation of the effect (using Schwartzschild metric) is given in Exploring Black Holes by Taylor and Wheeler.

That book doesn't derive the metric but teaches you how to interpret it in terms of observers.

7. May 9, 2007

### Ian

Chris Hillman,
I don't speak too clearly sometimes so let me try again.
I can calculate and account for the 43 arc.sec advance observed every century in the case of mercury's orbit by using Einstein's equation - (24pi cubed, r squared, all over c squared, t squared by 1 minus e squared). The units are angular. Please pardon the layman's LaTex.
The above equation had to have been changed by Einstein initially to include the parameters for an eccentric orbit because pure math and theory dictate pure circumstances like circular orbits. So Einstein's original equation must have been the above minus 'r(1 - e2)' which is the correction term for an eccentric orbit.
Therefore Einstein's original equation certainly must have been (24pi cubed, r cubed, all over c squared, t squared). This equation is for a circular orbit that obviously has no pericenter, but the equation is just as valid, and very importantly the units are now of length.
In other words, if you calculate this there still remains a rotation of the orbit but it is now given as the length of an arc about the orbit. The length I calculate is ~ 27833 metres per revolution and is the same for all planets in the solar system.
Remember Chris, the advance of perihelion is only a flashing light that alerts us to the fact that the orbit is rotating. A circular orbit has no perihelion and thus no flashing light to alert us, but the math insists the orbit still rotates whether circular or eccentric.
I mean, for all the talk of "frame dragging" and the expensive attempts to observe it.......

P.S. the citation you asked for is in the january special edition of 'Cosmology'.

8. May 9, 2007

### pervect

Staff Emeritus
Aside from the references already mentioned (and probably almost any GR textbook), the differential equations that describe a test particle following a geodesic in the Scwarzschild space-time are online at

http://www.fourmilab.ch/gravitation/orbits/

It turns out that there are two conserved quantites for an orbit in the Scwarzschild geometry - "energy at infinity" and "angular momentum". The existence of these conserved quantites can be attributed to the existence of certain symmetries in the Schwarzschild space-time, called Killing vectors.

These are the quantities, normalized per unit mass of the test particle are $$\tilde{E}$$ and $$\tilde{L}$$

Specifying these two quantities and one point on the orbit is sufficient to specify the entire orbit, much as in the Newtonian case.

One can write the differential equation for r in geometric units as

$$\frac{dr}{d\tau}^2 + \left(1 - \frac{2M}{r}\right) \left(1+\frac{\tilde{L^2}}{r^2} \right) = \frac{dr}{d\tau}^2 + 1 - \frac{2M}{r} + \frac{\tilde{L}^2}{r^2} - \frac{2M\,\tilde{L}^2}{r^3} = \tilde{E}^2$$

The equation for $\phi$ is simple, and also serves as a defintion for $$\tilde{L}$$

$$\frac{d \phi}{d \tau} = \frac{\tilde{L}}{r^2}$$

These results are presented on the webpage and in standard textbooks - I'm not going to attempt to derive them in a post. I don't think the webpage would be sufficient for a person to derive these equations, but it should be sufficient for a person to see what the equations look like in geometric units. People interested in derivations should probably find a textbook and read it.

If one replaces proper time (in the GR equation above), the equations are formally similar to the equations for a Newtonian orbit of energy E and L in terms of Newtonian coordinate time, with the addition of a term of order 1/r^3.

Note that the similarity is only formal because proper time isn't a coordinate time, and the above equations are written in terms of proper time of the orbiting body.

As far as units go, in geometric units M and $$\tilde{L}$$, r, and $$\tau$$ all have units of cm, while $$\tilde{E}$$ and $$\frac{dr}{d\tau}$$ are dimensionless.

One can see that all the terms in the first equation for r above are dimensionless in geometric units. Units are also consistent in the second equation, with units of 1/cm. Geometric units are units where G and c both have values of unity. Conversion of geometric to standard units and vica-versa is described, for instance, in

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

assuming it hasn't been munged since the last time I looked at it (I took another glance just now, it seems OK, but I haven't checked some of the recent additions).

Last edited: May 9, 2007
9. May 9, 2007

### Chris Hillman

"Another one done hit!" (Robert Penn Warren, All the King's Men)

Ian,

I've been on the web from the beginning, and experience has taught me that opening gambits like "I don't agree with Einstein's way of thinking" rarely leads to good things on public physics discussion fora. After some good faith effort I managed to find an interpretation of your initial query which avoided the depressing conclusion that you plan to argue that all the textbooks are terribly wrong about the extra-Newtonian precession of pericenters predicted by Einstein. Unfortunately, I can't do the same for this post.

Actually, as I assume you know, it's a 43 second advance which remains after a Newtonian accounting for a much larger precession due to the perturbing effect of Jupiter (and to a lesser extent the other planets) on the orbit of Mercury. This extra-Newtonian precession was a mysterious anomaly in the second half of the nineteenth century; resolving it was gradually recognized as an important unsolved problem.

You can learn latex by left clicking with your mouse on any formatted formula in PF to see how the poster obtained it. It sounds like you might have been staring at an expression for the precession per orbit, such as (15.36) in D'Inverno's textbook
$$2 \pi \varepsilon = \frac{24 \pi^3 \, a^2}{c^2 \,T^2 \, (1-e^2)} + O(\epsilon^2)$$
in non-relativistic units where T is the period, e the eccentricity and a the semi-major axis of the unique Keplerian orbit which the test particle motion approximates. Other authors derive equivalent formulas (typically with T eliminated in favor of m, the mass of the central object, and often expressed in relativistic units).

You seem to be completely misunderstanding the significance of the limiting case e=0, so perhaps it's no surprise that I cannot make any sense of that.

By studying the textbooks I mentioned, you should be able to figure out where you went wrong. I'd advise that you begin by trying to follow a modern derivation and to stop worrying about what you (incorrectly) think Einstein himself said.

Sigh...so are you the "Ian Lungold" who posts about the Mayan calendar?

Last edited: May 9, 2007
10. May 9, 2007

### Chris Hillman

You should bookmark a permalink to a version you have studied as I did at http://en.wikipedia.org/wiki/User:Hillman/Archive
This particle article is in fact one of the physics-related articles which was quite often marred by cranky/fringe edits when I was active at WP. Indeed consulting the history shows someone very recently someone inserted some very weird fringe or even cranky stuff and was reverted.