# Precession of Mercury’s Orbit

by Bjarne
Tags: mercury’s, orbit, precession
P: 344
 Quote by DrGreg Image credit: ©Allen McCloud CC BY-SA 3.0 Actually Bill_K and Jonathan Scott are both right but they are talking about different things. The diagram above represents the geometry of a two-dimensional slice of space around the Sun, with the Sun at the centre. (See Flamm's paraboloid for the technical details.) The region between two concentric circles has approximately conical geometry. Imagine cutting such a region out the diagram with a pair of scissors. To make the region lie flat, you have to make another cut in the radial direction to open it out, and then you will have an annulus with a gap missing. If you now draw an ellipse on your almost-annulus, there will be a gap. This gap represents the precession.
Still I am not sure what is the cause of the precession.
1.) Is it because of higher speed by perihelion relative how strong the curvature of space is ?
2.) Is it because of inertia, - increasing resistance against motion the further a body approach c ? - I mean when accelerating 100 km/h + 100km/h is not 200 km/h but 199,999...etc..km/h,
3.) or how can the cause / result of it be described.
 Sci Advisor P: 5,142 The simplest model derives a general-relativistic 1/r³ correction to the usual 1/r potential which causes the angular deficit.
P: 3,176
 Quote by Bjarne Can someone explain the cause of the Precession of Mercury’s Orbit in a "language" so simple that my grandmother can understand it? I mean only the part (43 arc second) that is due to relativity.
Here is the simplest explanation of precession due to spatial curvature that I know of:

From: http://www.physics.ucla.edu/demoweb/...spacetime.html

However, I was told (and a quick estimation seemed to confirm it) that the purely spatial curvature causes just a small part of the 43 arc second relativistic contribution. While most of it is due to gravitational time dilation (time curvature) and effective potentials:

http://www.bun.kyoto-u.ac.jp/~suchii/eff.potent.html
PF Patron
P: 193
 Quote by Bjarne I read it here > http://milesmathis.com/merc.html Now, if we subtract Mercury's precession from the Earth's, we achieve the apparent precession of Mercury as seen from the Earth. This is the number we want. This gives us a difference of ΔP = .8 arcsec/yr Or 80 arc secs per century. Therefore we will see Mercury precess about 80 arc seconds per century, due to curvature of the field alone. This is almost double Einstein's 43, which is enough to disprove his math and postulates. It also means that we will have to re-figure the perturbation total. The number 528 from above cannot be correct, as I said, since that is a Newtonian number, not a Relativistic number.
That site seems to basically be all about saying everybody's "wrong" with science and math.

157. How to Build a Nucleus without the Strong Force. With simple logic and diagrams.

129. A Recalculation of the Roche Limit. Showing that current math is wrong, and how to find a different kind of limit.

175. The Extinction of π. Here I show that the true value of π, defined as the ratio of circumference to diameter, is 4.

So, I think that site is only good for humor.
P: 344
 So, I think that site is only good for humor.
I think that is correct, but I just wonder; is the 43 arc second deviation per century seen from a Mercury or Earth perspective?

I understand that space is curving (Fig.1+2+3+4+5+6+7).
But the confusion is with references to Fig 8 + 9.
Fig 8 - “Similar Mercury executes its Newtonian ellipse in local space”
Fig 9. ---but the global curvature causes the ellipse to precess.

I understand; that:
Fig 8.; “ Newtonian space is without any curvature” and --
Fig.9 is with curvature.

So what the illustration only say is; That the curvature causes the ellipse to precess, - not WHY.

 However, I was told (and a quick estimation seemed to confirm it) that the purely spatial curvature causes just a small part of the 43 arc second relativistic contribution. While most of it is due to gravitational time dilation (time curvature) and effective potentials:
http://www.bun.kyoto-u.ac.jp/~suchii/eff.potent.html

The essence from this link is so far I understand Fig 2 > “precession because of extra dwell time at inner part.

According to classic understanding I don’t think there should be any extra “dwell time” – WHY should that happen due to general relativity?

I still do not understand the cause of that anomaly, (and do not have a mathematical background to do so).
Is there someone what can explain the cause in simple words?

I simply cannot find any description in this thread, that explain the cause of that precision anomaly.

* * *

I have considered another option; - we know it requires more and more energy to get a diminishing increase in speed.

So each time when Mercury accelerate towards perihelion, the planet need more potential energy to be able to reach that speed it should according to classis orbit mechanics, to overcome the increasing inertia against acceleration (towards perihelion).

But where must that extra energy come from?

As I see it this must mean that Mercury will lose speed, and hence approach the Sun. (Circling inwards to the Sun, in the same way as Illustration 3 above; “Captured and then plunged”. -

But this does not fit with reality, - why not ? .
What prevent that from happen?
Maybe this question do not have anything with the precession anomaly to do, But I would anyway appreciate to get some suggestion why Mercury due to Inertia (by acceleration) not is losing speed / energy. - It must be a fact that it will not reach that speed it "should" according to classis orbit mechanics.
Mentor
P: 13,609
 Quote by Bjarne I think that is correct, but I just wonder; is the 43 arc second deviation per century seen from a Mercury or Earth perspective?
Neither. It is a calculated result.

From the perspective of the Earth, Mercury's observed apsidal precession is 5600 arcseconds/century if the observations are expressed in mean of date coordinates or 574 arcseconds/century if the observations are expressed in J2000 or ICRF coordinates. The difference between these two values results from the 5026 arcseconds/century general precession of the equinox. In either case, the observed precession is not 43 arcseconds/century. Let's call the observed precession 574 arcseconds/century.

Newtonian mechanics explains most but not all of this observed precession: The outer planets cause Mercury's orbit to precess by 531 arcseconds/century -- which is off by 43 arcseconds/century. That 531 arcseconds/century is a calculated result; all that can be observed is the 574 arcseconds/century value.
P: 191
 Quote by A.T. Here is the simplest explanation of precession due to spatial curvature that I know of: From: http://www.physics.ucla.edu/demoweb/...spacetime.html However, I was told (and a quick estimation seemed to confirm it) that the purely spatial curvature causes just a small part of the 43 arc second relativistic contribution. While most of it is due to gravitational time dilation (time curvature) and effective potentials: http://www.bun.kyoto-u.ac.jp/~suchii/eff.potent.html

I tried this. The ellipse on my piece of paper didn't precess. What's the trick?
PF Patron
P: 1,090
 Quote by MikeLizzi I tried this. The ellipse on my piece of paper didn't precess. What's the trick?
Draw the ellipse on the flat paper with an extra piece of paper in the gap to complete the shape. If you then close the gap, making the cone, and place the extra piece of paper adjacent to the join so that it still continues the ellipse on one side, it will not quite join up correctly on the other side. You could make it join to the next bit by keeping the extra piece of paper still and turning the cone under it to bring the join to the other side. This represents the precession from one orbit to the next (or at least the component of precession due to space curvature).
P: 191
 Quote by MikeLizzi I tried this. The ellipse on my piece of paper didn't precess. What's the trick?
I found the trick to the original exercise. Don’t draw the ellipse directly opposite the cutout. That’s the one place where any drawing will not change orientation when the disc is pulled into a cone.

I’m not stating this as an insult. I know no one in this forum invented that exercise. But it tells me nothing about relativity. I am a little better at Paper Mache though.
PF Patron
P: 1,090
 Quote by MikeLizzi I found the trick to the original exercise. Don’t draw the ellipse directly opposite the cutout. That’s the one place where any drawing will not change orientation when the disc is pulled into a cone. I’m not stating this as an insult. I know no one in this forum invented that exercise. But it tells me nothing about relativity. I am a little better at Paper Mache though.
Ah. If you do that, the drawing will still look similar to an ellipse, but it will have a slight point at the join.
P: 3,176
 Quote by MikeLizzi I found the trick to the original exercise. Don’t draw the ellipse directly opposite the cutout. That’s the one place where any drawing will not change orientation when the disc is pulled into a cone.
There is no trick. You can draw the ellipse directly opposite the cutout, just as shown in the picture. The lines will meet then, but not at zero angle. So the object will not continue on the old ellipse but a shifted one. That's precession.
P: 3,176
 Quote by Bjarne But the confusion is with references to Fig 8 + 9. Fig 8 - “Similar Mercury executes its Newtonian ellipse in local space” Fig 9. ---but the global curvature causes the ellipse to precess. I understand; that: Fig 8.; “ Newtonian space is without any curvature” and --
By "Newtonian ellipse" they mean the inverse square law, that accelerates the object towards the cone tip. In flat space (called Euclidean not Newtonian) that acceleration law creates a closed ellipse.

Note that in reality the trajectory caused by the time curvature is different but they want to show only the effect of purely spatial curvature on the precession.

 Quote by Bjarne Fig.9 is with curvature.
On the cone the curvature exists only if you include the tip. Locally, excluding the tip it is still Euclidean, so the shape of the ellipse doesn't change there. But because it goes around the tip, it doesn't close properly, because there is an angle defect.

Note that in reality the spatial geometry is not cone, and the curvature is everywhere, but the cone approximation is good to illustrate the principle.

 Quote by Bjarne So what the illustration only say is; That the curvature causes the ellipse to precess, - not WHY.
It does show WHY. The trajectory is not closed properly. When the object arrives at the starting point, it has a different direction, than last time it was there. So it will not take the same path again.
P: 191
 Quote by A.T. There is no trick. You can draw the ellipse directly opposite the cutout, just as shown in the picture. The lines will meet then, but not at zero angle. So the object will not continue on the old ellipse but a shifted one. That's precession.
I don't know what you are talking about.

If I draw a vertical ellipse on the upper part of a disc symetric to the y axis and make a cutout in the lower part of the disc symetric to the y axis and close the disc into a cone symetric to the y axis my ellipse is still vertical and symetric to the y axis. I don't get any precession.
PF Patron
P: 1,090
 Quote by MikeLizzi I don't know what you are talking about. If I draw a vertical ellipse on the upper part of a disc symetric to the y axis and make a cutout in the lower part of the disc symetric to the y axis and close the disc into a cone symetric to the y axis my ellipse is still vertical and symetric to the y axis. I don't get any precession.
It's not an ellipse then - it's got a slight point at the join (at least in theory). If you follow the direction of the line from one side of the join it will diverge slightly from the line on the other side. In practice, you'd have to have a highly eccentric ellipse and a large angle missing to make this visible with ordinary pencil and paper.
P: 3,176
 Quote by A.T. There is no trick. You can draw the ellipse directly opposite the cutout, just as shown in the picture. The lines will meet then, but not at zero angle. So the object will not continue on the old ellipse but a shifted one. That's precession.
 Quote by MikeLizzi If I draw a vertical ellipse on the upper part of a disc symetric to the y axis and make a cutout in the lower part of the disc symetric to the y axis and close the disc into a cone symetric to the y axis my ellipse is still vertical and symetric to the y axis.
Being symmetric doesn't make it an ellipse. An ellipse is a smooth shape, but this will have a sharp corner, where you stitch the cone together.

 Quote by MikeLizzi I don't get any precession.
You get it, if you continue the path smoothly over the seam (dashed line in the picture). Eventually you have to make the cut-out-angle bigger and layout the seam area flatten out to see how the path will continue.

But it's easy to show mathematically. Just consider the angles between your original ellipse and your cuts. On the tip side they total less than 180°, so when you weld the cut lines, you don't get a smooth path.
 PF Patron Sci Advisor P: 1,772 Here's a very exaggerated version (not to scale) with 120° precession to get the point across. Attached Thumbnails
P: 344
OK I begin to understand, we can say that the deformation of space happens along the orbit with different values, and the result is that the angle defects whereby the ellipse cannot close. Simple…. When first the confusion first is gone. Thank’s .

 Quote by A.T. However, I was told (and a quick estimation seemed to confirm it) that the purely spatial curvature causes just a small part of the 43 arc second relativistic contribution. While most of it is due to gravitational time dilation (time curvature) and effective potentials
Time and distance deformation should (as I understand it) not be able to cause the other part of the precession. Because time and distance dilation are (as I understand it) proportional to each other (?).

But what about the; - “relativistic resistances” ?.
It requires more and more energy to get a diminishing increase in speed.

So each time when Mercury accelerate towards perihelion, the planet need more potential energy to be able to reach that speed it should according to classis orbit mechanics, to overcome the increasing relativistic resistances (towards perihelion).
But Mercury do not get that extra speed from anywhere, which mean that each time heading perihelion, - speed must be a little too low, compared to what it must be according to classis orbit mechanics.
Does that too not also have an influence ?
P: 3,176
 Quote by Bjarne http://www.bun.kyoto-u.ac.jp/~suchii/eff.potent.html The essence from this link is so far I understand Fig 2 > “precession because of extra dwell time at inner part. According to classic understanding I don’t think there should be any extra “dwell time” – WHY should that happen due to general relativity? I still do not understand the cause of that anomaly, (and do not have a mathematical background to do so). Is there someone what can explain the cause in simple words?
I don't understand it well enough to explain it simply, yet still correctly. But here are more visualizations and explanations:
http://www.fourmilab.ch/gravitation/orbits/

 Related Discussions Special & General Relativity 2 Special & General Relativity 2 Special & General Relativity 38 General Astronomy 24 Special & General Relativity 1