# Precession of Mercury

1. Dec 4, 2008

### Nickelodeon

Could someone explain in layman terms the current thinking on why the orbit of Mercury precesses? Presumably, it is not precessing in a gyroscope sense but the perihelion of the orbit just advances in the same plane.

2. Dec 4, 2008

### mathman

Precession of Mercury's orbit, as well as other planets, results from the gravitational effects of all planets on each other.

3. Dec 4, 2008

### Janus

Staff Emeritus
As mathman pointed out, the other planet's effect Mercury's orbit, as can the rotation of the Sun itself.

You can calculate these effect from equations such as found here:
http://scienceworld.wolfram.com/physics/OrbitalPrecession.html

Back in the 19th century such calculations were done, but it was found that Mercury's precession was greater than that which could be accounted for. This led to the idea that there was an undiscovered planet orbiting closer to the Sun than Mercury that accounted for the extra precession. They even went as far as giving this planet the name of "Vulcan". A search was made for this planet, and despite a few false alarms, no planet Vulcan was found.
Then Einstein, in his General Theory of Relativity, showed that the extra precession could be explained without an inner planet(in fact, the theory required the extra precession to exist), and the need for the planet Vulcan disappeared.(At least until the mid 1960's, where it provided a home for Mr. Spock)

4. Dec 5, 2008

### Nickelodeon

I'm having difficulty visualising how the planets could consistently nudge the Mercury's perihelion in the same direction. I would have thought that they could pull it one way then perhaps the other way depending on their relative positions.

The next thing I would like to know is regard to Einstein and how he explains the mechanics of what causes the small discrepancy. If the sun wasn't rotating would there be no discrepancy or is there something else going on?

5. Dec 5, 2008

### Jonathan Scott

The additional perihelion precession in Einstein's General Relativity is nothing to do with the rotation of the sun. It is to do with the way in which a massive body modifies the shape of space-time.

If you draw an ellipse on a flat sheet of paper, but then remove a segment of the paper (a narrow wedge extending out from the centre) and join it up again, making it slightly cone-shaped, the ellipse no longer meets up correctly, and if continued it does not repeat until slightly after going all the way round. The GR effect on Mercury's perihelion precession is very similar to this, although it involves curvature in time as well as space.

6. Dec 5, 2008

### Nickelodeon

This worries me because Janus says that the Sun's rotation does have an impact on Mercury's precession which introduces some doubt as to there being a definitive answer.

7. Dec 6, 2008

### Jonathan Scott

The Sun's rotation does indeed have a effect, but it only gives a very tiny correction to the original Newtonian plus Einstein calculation. It is only relatively recently that the Sun's rotation rate and the resulting oblateness were known well enough to make an accurate calculation, and up to that point there was still some speculation that Einstein might be wrong when the rotation was taken into account.

For some actual figures, see the Wikipedia entry on http://en.wikipedia.org/wiki/Tests_of_general_relativity" [Broken].

Last edited by a moderator: May 3, 2017
8. Dec 7, 2008

### Nickelodeon

Is it the oblateness of the sun or frame dragging due to the rotation or both which causes the discrepancy?

9. Dec 7, 2008

### D H

Staff Emeritus
If you followed the link in Janus' post you would have seen stuff involving J2, J4, etc. These are classical, non-spherical mass moments of the Sun about its center of mass. The Sun, because it rotates very slowly, is very close to spherical and thus has a very small J2 value: 2×10-7. There is no doubt that Mercury's relativistic recession results from relativity rather than the Sun's oblateness.

10. Dec 7, 2008

### Nickelodeon

My maths is appalling so I'm struggling with the formula in Janus's link. The only thing that the formula doesn't seem to take into account is the mass of the sun, big M ?, reducing each year. Do you think, over the century, this would have any effect on Mercuy's orbit?

*** Just looked up the sun's mass loss per year and it seems negligible so please disregard last question ***

Last edited: Dec 7, 2008
11. Dec 8, 2008

### Ian

Nickelodeon,
When calculating the precession of any planet make sure you calculate the length the body advances along the path of orbit - don't calculate the angular advance as Einstein did, or everyone else for that matter.
If the orbit of mercury was circular you would not observe an advance because Einstein's indicator (the perhelion) would not exist if the orbit was circular but the orbit would still advance.

12. Dec 8, 2008

### Nickelodeon

That sounds sense.

I'm still no nearer understanding the mechanical effect that causes the advance though. Einstein obviously had a feel for it and, presumably, devised his fomulae based on that insight. Does anyone know what that insight was?

13. Dec 8, 2008

### RandallB

Nothing wrong with converting to, or caculating for angular advance.
"Length" can be confusing as it requires defining where the distance was measured, periapsis or apoapsis, while both will convert to the same angular advance.

14. Dec 9, 2008

### Ian

RandallB,
If you look at Einstein's original equation when he calculated the advance of mercury's perhelion you will see that he first deduced for a circular orbit and then he added the 'correction' for an elliptical orbit.
The units of the advance for a circular orbit are 'metres', as shown below:

(24pi3r3)/(c2t2) Metres.

and after adding the r(1-e2) correction for elliptical motion the equation becomes:

If you calculate the advance for any planet assuming the orbit is circular then the advance must be expressed in metres along the length of orbit. This figure will always be the same (~27833 metres) as the sun's mass is the only contributing factor.
The correction factor and change of units have made you think that the advance is different for all planets.
Besides, Einstein didn't actually explain the advance, he only produced a calculation that agreed with observation.

15. Dec 9, 2008

### Jonathan Scott

Yes, the precession distance is the same for all circular orbits around the sun. I'm not quite clear on your units; I'd write it as 6 pi Gm/c^2 where m is the mass of the sun, and Google Calculator gives me 27.835km which agrees well enough with your numeric result.

A similar related feature applies to the curvature of space, in that in isotropic coordinates the "length deficit" in the circumference of any circle around the sun is one third of that precession distance.

That's absurd. He produced an equation which was extremely simple in concept but mathematically very tricky, saying that the curvature of space-time is determined by the distribution of matter, and the only free parameter (ignoring cosmological considerations) was effectively the strength of the gravitational potential, which is simply matched up against Newtonian theory in the limit. The perihelion precession of Mercury depends on a second-order term in the complicated maths from that theory, yet came out right first time. This may have been a fluke (and sometimes I subscribe to that point of view) but it certainly wasn't contrived to match the observations.

16. Dec 9, 2008

### Nickelodeon

The thinking man would look at the phenomenum (the precession discrepancy over and above planetary influencies) and decide on a probably cause, be it relative time, space curvature or relative mass variances then apply the maths accordingly. I'm sure Einstein did this but is there anyone out there who can shed some light on what those mechanical variance might be?

17. Dec 9, 2008

### Jonathan Scott

What do you mean "mechanical variance"?

Einstein's model is simply that free falling objects follow "geodesics" in space-time (the local equivalent of straight lines), but mass curves space-time with the result that near to a large mass those geodesics describe accelerated motion relative to a static coordinate system. Although this is completely different from Newton's description of gravity, it not only reproduces the same effects, but it also explains why Newton's theory doesn't quite get Mercury right.

The shape of the orbit can be calculated in Einstein's theory in a very similar way to Newton's, by describing the equation of motion relative to a coordinate system. Einstein's result is basically the same as Newton's except that relativistic corrections mean that any oscillation about the average orbit radius has a period which is a tiny fraction longer than the the time it takes to complete an orbit, so the shape of the orbit precesses forwards.

Since Einstein's theory can be approximated by Newton's, the conventional method of computing the total precession of Mercury's orbit is simply to add the Einstein correction to the classical Newtonian precession. However, it should be noted that Einstein's theory is effectively being used to calculate the whole orbit including the correction, not just the relativistic correction.

Last edited: Dec 9, 2008
18. Dec 9, 2008

### Nickelodeon

by the "mechanical variance(s)" I was suggesting that for a planet not to be where it should be, as calculated by euclidean type maths, there has to be some influence acting on it, some imbalance. Given that it is following its local geodesic path (all the way round) I can't see where this imbalance is coming from unless the intensity of the geodesics are slightly assymetric somewhere along the line.

19. Dec 9, 2008

### RandallB

I’ve never cared for the term “geodesic” often used as though it were a magic bullet that explains how GR works.
As you imply there is no reason not to call the Newton path a “geodesic” as well.

So in that context I think I understand what your looking for in what is the “variance” that accounts for GR giving the different result vs. Newton.
Whether you want to call it “mechanical” or not for GR you will have to look to the “Curve” or “Warp” in an additional dimension (unrealistic in Euclidean terms) of something loosely described as “Space-Time”. So to measure what ever is going on in this “unrealistic” dimension you are wanting understand it more or less in Euclidean terms.
What Einstein’s formulas say is that due the this “distortion of local space-time” no matter how unrealistic it might seem we can measure the inputs (mass of things) to this “unrealistic hidden from direct observation extra dimension or otherwise un-perceivable thing” to quantify in imaginary terms (literally imaginary numbers used in some versions of the math) the curve in that unknown whatever. Then with that imaginary curve defined GR can predict back into our Euclidean terms expectations outputs (how masses in orbits etc. behave) including “variances” to our normal Euclidean expectations.

In this case that variance would be a slightly stronger acceleration being applied to the orbiting bodies.
That stroger acceleration produces a few things at variance with our Euclidean expectations.
A slightly faster orbit than expected giving “precession”.
An added acceleration that continues to pull the orbit down into a small faster orbit.
As effect continues the orbit gets smaller and the precession becomes more noticeable.

You have to say the math of Einstein was built to match observations – The M&M experiments failing to find the ether and checked against the Mercury precision. But then what theory of physics does not do that? There cannot be a theory to explain something without observing something that needed explaining.

But you also have to accept that Einstein provided a context to define and explain how his theory worked physically by using the addition of a “space-time” demension.
However it is a reasonable complaint, to call that explanation “unrealistic”.
With our current understanding that GR is a “background independent” theory not even Einstein could deny it is unrealistic in Euclidean terms.
A little ironic since he was so adamant about QM not being complete due to not being realistic ( as he put it not Local & Realistic).

If we knew all the answers we would know which was "more correct" QM or GR; those two still don't get along.

20. Dec 9, 2008

### Nickelodeon

Thanks for that.

I think I've sort of figured it out now in my own mind, enough to sleep easier anyway. Space curvature is the key but I think the geodesic map is not as rigid as depicted in literature - I think it's moving. (not sure I'm allowed to say that)

cheers
Nick

21. Dec 10, 2008

### RandallB

I’ve only seen explanations that address different angles by planet, and did not know when measured as a distance in the circumference all would give the same value; very interesting.

I’ve not been able to find the original equation Einstein calculated the advance of mercury's perihelion with. I assume it also included a value that accounted for one full circumference or at some point a full circumference was removed from it to only show only the advancing distance portion.
Do either of you have a reference that shows the complete equation of calculating the Mercury orbit from the Einstein GR equations.

Also, do you have a reference that goes into some detail of what the isotropic circumference "length deficit" is or means? I don’t understand what that represents.

Last edited: Dec 10, 2008
22. Dec 10, 2008

### Jonathan Scott

This is a standard result; I can find it for example in Rindler "Essential Relativity" section 8.4 (as I don't feel strong enough to get MTW out at the moment). After reinserting factors of G and c to get back to conventional units, equation 8.62 says that the precession angle per orbit is as follows:

$$\frac{6 \pi \, G m}{c^2 \, a \, (1-e^2)}$$

where $m$ is the mass of the sun, $a$ is the semi-major axis and $e$ is the eccentricity. To convert this back to a distance for a circular orbit, multiply by $a$, giving $6 \pi \, G m/c^2$ as the distance.

Sorry, that's probably just what I call it. I mean the amount by which the proper circumference exceeds the coordinate circumference in the isotropic coordinate system. To first order, rulers are shrunk relative to the coordinate system by a fraction $Gm/rc^2$ so proper distances exceed coordinate distances by that fraction. For a circle of circumference $2 \pi r$, the excess length is $2 \pi \, Gm/c^2$ regardless of the value of $r$.

This obviously depends on the coordinate system, but there are similar effects in other coordinate systems. Basically, the potential (for weak fields) is like $-Gm/rc^2$, so when anything including that term is multiplied by $r$ the term becomes $Gm/c^2$ (half of the Schwarzschild radius).

23. Dec 10, 2008

### Jonathan Scott

P.S. what's probably more interesting (and probably the point I meant to remember in the first place) is that a "length deficit" occurs in proper space too, regardless of coordinate system, in that if you take the ratio of the increase in proper circumference to the corresponding increase in proper radius, you don't get the usual $2 \pi$ as for flat space but you get approximately $2 \pi \, (1 - Gm/rc^2)$. One way to model this is to assume that in proper space, there is a length $2 \pi \, Gm/c^2$ missing from the circumference at every radius, directly reflecting the curvature of space.

(I hope I got the signs right as it's quite easy to get confused).

This is however only a weak field result which relies on the following approximation:

$$(1 - Gm/rc^2) \approx \sqrt{1-2Gm/rc^2}$$

24. Dec 12, 2008

### RandallB

Thanks Jonathan seeing how GR treats the extra distance traveled really helps understand how the equations work.
I was concerned with the focus or barrycenter of the ellipse; but I see when looking at the distance added by GR to the orbit circumference you must think of it relative to the center of the ellipse not one of the focus points – hence the use of “a” Semi-Major-Axis.
Getting the visualization right helps.

Another piece to the puzzle:
The two body circular orbit is not in fact actually circular in GR, since velocity is slowly increasing making the orbit grow smaller. Is there a part of the equations that reduce to how the obit radius reduces with each orbit?
Might it also be changing by a fix distance per orbit or maybe per unit of time.
Whatever it is I’m sure it is directly related to how the velocity is increasing over time with each orbit.

Thanks RB

25. Dec 12, 2008

### Jonathan Scott

The only GR effect which would make the orbit grow smaller with time is gravitational radiation, where energy is lost in the form of gravitational waves. This is absolutely negligible on the scale of solar system effects, being orders of magnitude too small to detect, and much smaller than other tiny effects such as everyday collisions with dust, meteorites and so on.

If you want to try to calculate the effect for Mercury, see the Wikipedia article on http://en.wikipedia.org/wiki/Gravitational_wave" [Broken] which includes a calculation of the rate at which energy is lost from an orbiting body to gravitational radiation, using the Earth as a worked example.

Last edited by a moderator: May 3, 2017