Usually, and that means in the past year, when you look at the false color MDI image from SOHO, you can look at the corresponding magnetogram and see some sort of disturbance going on, even it it is not visible as a sunspot, sunspeck, or plage area.
Not today.
Left: SOHO MDI “visible” image Right: SOHO Magnetogram
Click for larger image
Wherefore art though, cycle 24?
In contrast, September 28th, 2001
idlex wrote: I can’t make my satellite orbit a notional thing like the barycentre. Just like I can’t make my satellite orbit a notional thing like the Lagrangian points between the Sun and Jupiter.
Geoff Sharp replied: This might be a limitation with your simulator, perhaps it follows nature, but perhaps not?
No, it’s not a limitation of my simulation model. When it works out where the various planets in the solar system are going, it’s always working out the gravitational force exerted on them by all the other planets individually. That’s about 15 or 20 calculations for each time interval for each planet. That would be very tedious for me to work out, but it’s no trouble for my computer.
What I suspect you are saying is that, for convenience, we can lump together all these 15 masses and put them all at the solar system barycentre, and use this to perform our calculations for planetary motion.
Now, this is fine if all these bodies are a long way away. If we want to work out the gravitational effect of the Andromeda galaxy on the solar system, we don’t really have to calculate it for each and every star in that galaxy. We can instead lump all the stars in the galaxy together and put them at the centre of mass of the galaxy, and just calculate the effect of that single mass upon the solar system. The result will be a good approximation.
But this isn’t quite true once you’re in among some collection of stars or planets. And it becomes less and less true the nearer you get to the centre of mass. For example, for convenience the mass of the Earth is usually located at its centre. But if the Earth may be regarded as a spherical cloud of gas, a stone that was dropped into the Earth from its surface would have part of the Earth’s mass pulling it downwards into the Earth, and part of it pulling upwards. When the stone reached the centre of the Earth, the surrounding mass of the planet would pull it in all directions, and these forces would all cancel each other out. There would be no gravitational force due to the Earth’s mass acting on the stone once it got to the centre of the Earth. However, if we had continued to locate the entire mass of the Earth for convenience at its centre, the forces acting on the stone would be calculated to be infinite once it reached the centre of the Earth. And this would result in the wrong answers being found. What was a good approximation outside the planet becomes a worse and worse approximation once you’re inside it.
And the same goes for the solar system. Once you’re inside it, you really have to calculate the gravitational forces due to the separate individual bodies within it, and stop lumping them all together at the barycentre.
The solar system barycentre is a very useful thing, like a pair of boots, when being used outside the solar system. But, like a pair of wet and muddy boots, it should be left at the door once you’re back inside the house of the solar system.
Leif Svalgaard (08:04:37) :
http://www.bautforum.com/against-mainstream/72665-explaining-planetary-alignments-relationship-sunspot-cycle.html
I looked at it carefully and the whole discussion is just nonsense and has been dealt with adequately by the commenters on the thread.
Hi Leif,
I thought most of the critical commenters made fools of themselves by not understanding what he was showing them, but there you go. It took me a few readings to work out the relativistic argument, and I’d be interested to know what you think is nonsensical about his presentation of that part of the theory. He did link to another concurrent thread where the rate of movement of matter and energy from the core of the sun was under discussion. Obviously, the different figures coming out of that affected the outcome of his calculations. I’ll go and have a read of that thread and compare with the figures you have provided. Where are those from by the way?
One of the other things I found interesting about it was that the magnitude of the core displacement Ray calculated matches the size of the ‘corrugations’ which increase the solar diameter during the solar cycle you told us about a while ago.
tallbloke (23:07:55) :
“the sun has an internal oscillation period of around 10.5 years”
Here is how Ray gets the 10.5 years:
“Clearly this means that this proposal works only if the Sun has some natural resonance of 10.5 years and all of these forces are activating that resonance. Based on that further assumption, I could calculate sunspot numbers over several centuries with a correlation of r=0.66 from the planetary forces.”
Yet another astrologer that can calculate everything centuries in advance.
Geoff Sharp (06:16:50) :
jtom (04:55:13) :
“Can someone give me a simple answer to a simple question?”
It’s not a simple question right now, the answer has been buried for sometime, but we are digging.
The is a simple answer: ‘bogus’
E.M.Smith (05:22:22) :
That would serve as a simple proof that the solar orbital angular momentum was not available for fiddling with spin… Do we have numbers of good enough accuracy to make that measurement?
Let’s do it for the two major players: Sun [s] and Jupiter [j]. Then the angular momenta around the barycenter are respectively:
As = rs ms vs for the Sun and Aj = rj mj vj for Jupiter, where for simplicity we assume that the orbits are circular. rs is the Sun’s distance to the barycenter, ms is the Sun’s mass, and vs is the Sun’s speed in its orbit around the barycenter. similar for Jupiter but with ‘j’s: rj, mj, and vj. Now vs = 2pi rs / 12 years and vj = 2pi rj / 12 years. the two periods [12 years = T] are the same because s and j are always on opposite sides of the BC. By definition of the center of mass rs = a mj / (ms + mj) where a is the distance between s and j which we can take as constant because it does not vary much. we can simplify further: rs = a / (1 + ms/mj) ~ a mj /ms, because ms is 1000 times bigger than mj. rj is very close to ‘a’ [again because mj is 1000 times smaller than ms], so we have vs = 2pi/T a (mj/ms), rs = a (mj/ms), and vj = 2pi/T a. In words: the barycenter is 1000 times [=ms/mj] closer to the Sun than to Jupiter and therefore the Sun moves 1000 times slower than Jupiter, so the orbital angular momentum for the Sun would be 1000 times smaller than that of Jupiter. If this was the only thing going on, the angular momenta would not change as the bodies orbit. Therefore from this we would have no spin-orbit exchange. This is not a surprise, and we did this calculation only to get a feeling for the magnitudes involved.
But let us now introduce other bodies [e.g. Saturn or Uranus]. These would move the barycenter and rs and vs would change and hence As. Let us assume that rs doubles, then vs doubles too with good approximation because the other planets move much slower than Jupiter. Hence As approximately quadruples. We can write rj = (a – rs) and vs = 2pi (a-rs) / T, and thus Aj = (a-rs) mj (a-rs) 2pi/T = [a-rs]^2 mj 2pi/T or Aj = [1 – 2rs/a] a^2 mj 2pi/T because rs is much smaller than a. So, if rs increases, Aj will decrease [because of the -2rs]. In numbers, let rs1 be the old rs and rs2 = 2 rs1 be the new [after addition of Saturn etc]. now 2rs1/a = 2/1000 and thus 2rs2/a = 2 (2rs1)/a = 4/1000, so Aj will now be 2/1000 smaller than before or 2As [because As was a 1/1000 of Aj]. So when As increases by a factor of four, Aj decreases by half that. But as we have added another planet to move rs out by a factor of two, its angular momentum decreases too because Jupiter moving rs out first. So, what the Sun gains, the planets lose. All this can be made exact and horribly complicated, but the type of ‘back of the envelope’ calculation shown here [scientists are trained to do this] is enough to show the essential point, so everybody can see it without anybody having to rely on some complicated table downloaded from the Internet and calculated in ways that cannot easily be checked. A danger with this type of ‘quick’ calculation is that errors easily creep in, but they can be found if enough eyeballs are on the derivation. So there you have it.
tallbloke (10:10:40) :
I thought most of the critical commenters made fools of themselves by not understanding what he was showing them, but there you go.
Well, Ray was not the only fool 🙂
It took me a few readings to work out the relativistic argument, and I’d be interested to know what you think is nonsensical about his presentation of that part of the theory.
It is hard to know where to start. The standard way this goes is that I point out one flaw, then you respond, OK let that slide, but all the rest is good, and we slog our way through the thing, flaw by flaw, and in the end you say: ‘ok, but I still think there is something about it’. Let us start with the 5/3 number. As far as I can tell it comes from averaging the ‘result’ for 6 axes, 4 horizontal and 2 vertical. You don’t find the effective factor by averaging [over a nonsensical number of axes – should be three] but by integrating correctly the radiation over a sphere.
He says: “So taken over the whole sphere of directions for random photons we have 2x in 4 directions and 1x in two directions. That makes an average of 5/3x for random photon directions.”. This is just nonsense. Since you think you understood it, explain to me how the 5/3 comes about. If it is just by saying (2x * 4 + 1x *2) / (4+2) = 10x/6 = 5/3x then you have your nonsense there.
Paul Stanko (05:13:32) : Only if this minimum reaches an epic 3198 spotless days will it be at mean +1 sigma even if we include the Maunder minimum.
To what degree are we sure we are not measuring with a ‘rubber ruler’? Are the corrections and adjustments for differential observation methods certain to a degree sufficient to make that 1 sigma claim? (Or put differently, are we playing in the error bands of the data? Is it really, for example 3198 +/- 1000 ? so that sigma estimate is suspect? )
tallbloke (10:10:40) :
compare with the figures you have provided. Where are those from by the way?
Remind me what they were…
if the Sun has some natural resonance of 10.5 years and all of these forces are activating that resonance. Based on that further assumption, I could calculate sunspot numbers over several centuries with a correlation of r=0.66 from the planetary forces.”
Yet another astrologer that can calculate everything centuries in advance.
This sort of comment should be beneath you by now Leif, you’re a recidivist. 😉
Ray wasn’t trying to calculate sunspots centuries in advance and if you really had “looked at it carefully” . He was testing his theory by achieving a damn good hindcast with a r=0.66 correlation spanning several centuries.
Can your interpretation of the Leighton-Babcock solar dynamo theory do that?
tallbloke (11:42:38) :
from the planetary forces.
implies that it will work in the future, no?
“Yet another astrologer that can calculate everything centuries in advance.”
My definition of astrology may be different from yours. Mine covers the claim that the planets have an influence that does not follow known physical laws.
r=0.66. Can your interpretation of the Leighton-Babcock solar dynamo theory do that?
Dikpati and company claim a correlation coefficient of r=0.958 [figure 3 of http://www.leif.org/research/Dikpati-Prediction-2005GL025221.pdf ]. My grow-n-crash mock theory gets r=0.924 http://www.leif.org/research/Grow-N-Crash%20Prediction%20Model.pdf
beats the socks of 0.66.
@ur momisugly Leif Svalgaard (10:14:30) :
Thank you for your patience, and answers. (So far, kohai has missed hitting floor flat, while practicing unplanned shoulder rolls… 😉
“E.M.Smith (05:22:22) :
That would serve as a simple proof that the solar orbital angular momentum was not available for fiddling with spin… Do we have numbers of good enough accuracy to make that measurement?”
While I appreciate the “thought experiment” you posted in reply, what I was actually thinking about with this sentence was more of an existence proof (which is why I said “measurement”). I like existence proofs most since they are unarguable (and I’d really like a way to get this sticky trap to stop attracting me with slippery sweet arguments… one way or the other.)
So is the data on planetary position and velocity available such that one could actually do the calculation of actual orbital momenta at the two times (sun 2 radii out, vs sun 1/10 radius out). If this can be done with a few digits of precision I think it would be a clean, unarguable, existence proof that could be put up as a tombstone for all time on the spin-orbit couple argument. If the data are questionable, the math intractable, or there is some other complication that make this a dead end, fine. Just tell me to prune it and I’ll let go of the hope for an existence proof.
Back at the answer you posted. So we are doing an incremental calculation by adding each planet one by one, and ignoring the “center of mass” other than as a derived artifact. This argues to me that the basic flaw in the Solar Orbital angular momentum calculation via ‘center of mass of the solar system’ is in fact that it can not be (or ought not be) done that way. That the only correct way to calculate Orbital Angular Momentum for the largest object in a system of n-bodies is to do it as an n-body problem.
Is that a correct statement?
If this is true, and can be demonstrated (by, oh, calculating Solar Orbital Angular Momentum 2 ways, one using center of mass the other via n-body and showing they are not the same) this too could serve as a tombstone.
All my explorations of the n-body problem have led me to statements of the form “There are 6 equations for each body and when n>3 it can not be solved” but I think that an n-body of n=3 would likely be enough to be a proof. (Or, alternatively, if Solar Orbital angular momentum is found to be exactly the same using vectors rooted at the center of mass of a 3 body system vs vectors anchored between the 3 bodies, then the tombstone becomes “Angular Momentum is identical each way, and we can show conservation via 3 body vectors, so ‘center of mass’ does nothing to change momentum exchange between smaller bodies and largest body”
All this may be obvious to you, but it is not to the average non-physicist, and I think that is why the sticky trap hangs around. My interest here is to find a simple way to demonstrate it. (And if that is a sticky trap because there is no simple way to demonstrate an n-body angular momentum problem; well, that would be good to know too. I’d walk away frustrated, but I’d at least know that I would never be able to get a simple demonstration and ought to go do something more productive.)
Basically, I see this as being analogous to a pick-nick table with a big tub of cookies on it. You can put up a sign “Don’t Touch” and people still want to go grab the cookies… and you need to keep fending them off. But if you can put up a sign that says: “Poison. See the Green Ogre in the corner? He poisons cookies” folks will not be attracted to the cookies, no matter how sweet they look, if they can also see the Green Ogre.
I’m just trying to find what to write on the sign (or if the ogre does not really exist, either one.) Clearly saying “don’t touch” has not been enough…
Let us start with the 5/3 number. As far as I can tell it comes from averaging the ‘result’ for 6 axes, 4 horizontal and 2 vertical. You don’t find the effective factor by averaging [over a nonsensical number of axes – should be three] but by integrating correctly the radiation over a sphere.
He says: “So taken over the whole sphere of directions for random photons we have 2x in 4 directions and 1x in two directions. That makes an average of 5/3x for random photon directions.”. This is just nonsense.Since you think you understood it, explain to me how the 5/3 comes about. If it is just by saying (2x * 4 + 1x *2) / (4+2) = 10x/6 = 5/3x then you have your nonsense there.
First I will quote a couple of relevant parts of Ray’s thread, then comment.
“The important thing is that if you accept that there is a doubling (or 5/3 times) effect of gravity on radiation (which was proven in the Eclipse experiments) then the central part of the Sun experiences a different rate of acceleration from what the surface experiences. Because of rapid mixing of momentum between radiation and matter, this acceleration applies to the matter there.
Additionally, there is an argument that this increased acceleration also applies to the relativistic content of ordinary matter as well as to light. This was all explained by G D Birkhoff in 1927.”
“However in the vertical case there is no 2x factor, it is simply 1x the normal acceleration (I will show this soon with equations) which is why I say the effect over all random directions in space for photons in 5/3.”
“So let us start with a definition of a new variable which I will call “pull” and use the symbol “b” (being right next to “a” in the alphabet). Pull is defined as the time rate of change of momentum per unit mass.
b = 1/m.dp/dt
For the vertical photons, there is also a change in momentum predicted by Einstein. This is called “gravitational redshift” and Einstein’s formula may be manipulated by using E=mc^2 and E=hf to find that b = g in both the vertically up and down cases.
df/dt = fg/c (as given by Einstein)
so b = 1/m.dp/dt = 1/m/c.dE/dt = h/m/c.df/dt = h/m/c.fg/c = Eg/E = g
That means there is no 2x factor for vertical photons, just the normal Newtonian rate of change of momentum per unit mass.
So taken over the whole sphere of directions for random photons we have 2x in 4 directions and 1x in two directions. That makes an average of 5/3x for random photon directions.”
“I will add that GR experts do not all agree on the above equations, and this is the part that makes this ATM (against the mainstream) perhaps. Everything else is standard physics. However some GR experts claim that my 5/3 factor is really only 1 and at least one stated that it is actually 2. I cannot see how the bending of light by the sun during an eclipse is double the Newtonian value if the fact is 1. Anyone who disagrees with this factor being 5/3 or at least different from 1 will need to explain that.”
“In the case of vertically moving photons, there is no change in velocity, but a change in frequency. So it cannot be called acceleration. However that change in frequency does change the momentum of the photons and the net result on the whole system of matter and radiation is an acceleration.
So I am happy to accept that there might be a better way of expressing things, but I don’t think the actual concept needed does already exist. It is effectively acceleration but because of the radiation needs to be phrased as change in momentum per unit mass.”
tallbloke’s comment:
Tomes is a practical thinker who is prepared to work with imperfect and unresolved material in order to push ahead in the direction his ideas take him. He recognises and freely admits where there is still work to be done and even provides roadmaps for those who might wish to further investigate and refine the formulae.
It is always easy to rubbish a new theory by subjecting it to death by a thousand nitpicks, much harder to try to understand new frames of reference and grasp the novel ideas of others.
You say it is a nonsensical number of axes, but Tomes explains why he differentiates between X- X+ etc. It is because he treats the vertical case z+ z- separately because this is required in the output of his data. Nothing nonsensical about that.
When you say that “You don’t find the effective factor by averaging but by integrating correctly the radiation over a sphere.” , you are just parroting ‘Tusenfem’ and Tomes answers thusly:
” I agree. However from what Birkhoff wrote I believe that 5/3 is the correct result for the integration. However that remains to be proven.” and further that:
“As we agreed, it needs to be proven over the whole sphere, but the average is going to be somewhere between 1 and 3 and probably near 5/3.”
Now I’m sure a computer could fairly quickly integrate the random paths of the photons over a sphere, and I’m also sure the correct answer will turn out to be somewhere near Tomes finger in the air figure. So let’s talk about the beef, rather than arguing over the seasoning of the gravy.
My definition of astrology may be different from yours. Mine covers the claim that the planets have an influence that does not follow known physical laws.
Well there you have it. Ray Tomes is at pains to point out his theory is standard physics, and that the pull on the differential gradient of the relativistic matter in the sun’s core-surface makeup is at least approximately quantifiable, but you just write it off as astrology because it involves the gravitation of other chunks of matter spinning around the precious inviolable sun.
E.M.Smith (12:21:44) :
So is the data on planetary position and velocity available such that one could actually do the calculation of actual orbital momenta at the two times (sun 2 radii out, vs sun 1/10 radius out). If this can be done with a few digits of precision I think it would be a clean, unarguable, existence proof that could be put up as a tombstone
These things are known to incredible position, [both calculated from theory and measured from radar ranging and positional astronomy], see: http://ssd.jpl.nasa.gov/?orbits and links there.
We went over this last year, when the BC people claimed that the since both the Sun and the Earth orbit the BC, the distance between the Sun and the Earth should reflect that. On the JPL site they do this calculation for you with incredible precision and, of course, the distance came out with no regard for the BC. The BC crowd first tried to say that the calculations were faulty, based on ‘assumptions’, then in the end conceded that perhaps they were accurate, but never changed their tune in spite of that, just barreled on as usual, even going to such silly length as positing that only some of planets go around the BC, the others not. So much for ‘clean, unarguable, existence proof’ ! A simple fact of life is that no matter how clean, unarguable, etc, it is, if the ‘proof’ does not confirm people’s convictions, it is by definition no good.
This argues to me that the basic flaw in the Solar Orbital angular momentum calculation via ‘center of mass of the solar system’ is in fact that it can not be (or ought not be) done that way.
It can be done that way. It is the only way and the correct way, because the angular momentum of a collection of particles is simply the sum of the angular momentum of each particle. This is not an n-body problem at all.
All this may be obvious to you, but it is not to the average non-physicist, and I think that is why the sticky trap hangs around.
I don’t think that is the reason. I think the reason is that people of strong enough conviction are not interested in having that overturned, and are very vocally trying to foist their conviction onto others; to change the face of science, to teach NASA a lesson, to tell us that we should keep an open mind [yeah, but not so open that the brain falls out]… the list goes on.
And I realize [and anticipated] that my calculation would have no effect, but offer it anyway, so I can refer to it in the future should I find an occasion for doing so.
tallbloke (12:47:43) :
Well there you have it. Ray Tomes is at pains to point out his theory is standard physics, and that the pull on the differential gradient of the relativistic matter in the sun’s core-surface
There is no ‘relativistic matter’ in the sun’s core. His ideas are not ‘standard physics’ no matter how much pains he goes to claiming that.
The 5/3 debacle is just a sign that he does not have his stuff together. This is an elementary error, and he is not blazing new trails into unexplored territory. But, as I said: “It is hard to know where to start. The standard way this goes is that I point out one flaw, then you respond, OK let that slide, but all the rest is good, and we slog our way through the thing, flaw by flaw, and in the end you say: ‘ok, but I still think there is something about it’.”
There is something called the ‘smell test’. One can very quickly judge if something passes that, and Ray’s stuff stinks. Undoubtedly deeply felt and held by him and his adherents, but junk nevertheless. The simple referral to “pull on the differential gradient of the relativistic matter in the sun’s core-surface makeup” is enough to label the paper appropriately.
Leif Svalgaard (07:45:49) :
A dynamical system like that has only ONE barycenter defined by summing over all bodies
I remember a few months ago when I said the same thing, your reply was that it had infinitely many barycenters (centres of mass), because you introduced a pea in chosen positions millions of light years away.
But I agree with you now 🙂 The SSB is well defined.
There is something called the ’smell test’. One can very quickly judge if something passes that, and Ray’s stuff stinks.
How, er, quaint.
There is no ‘relativistic matter’ in the sun’s core.
And that would mean no energy either, according to Einstein and Birkhoff. It’s a wonder the sun continues to shine at all.
Thanks anyway Leif, your insights have been , invaluable.
anna v (08:28:41) :
Now lets take the sun and the center of mass of the whole solar system. Its own center of mass would be describing an ellipse with one of the focuses of the ellipse on the center of mass, and all the planets could be substituted by their center of mass position and total mass M ( center of mass of the planets) which planetary M would also be describing an ellipse . So the fact that the sun is revolving around the center of mass in an ellipse means that its velocity will be changing so as to keep the angular momentum constant.
Now there is the added complication that the center of mass M of the planets moves in time because the planets change position with respect to each other, and thus the overall center of mass moves. I do not think this invalidates the picture of two elliptic orbits each conserving angular momentum.
Thus the answers is: the sun describes an ellipse around the combined solar center of mass point.
No, this is clearly not the case. The orbit is considerably more complex, here is one example
http://arnholm.org/astro/sun/sc24/sim1/images/1430_1520_d.jpg
I have computed this using both simple centre of mass calculations given orbital element based planetary positions, as well as through gravity simulations. The result is the same. The solar orbit is very complex, that is a fact.
Whether it has any effect on solar activity in any way is another matter.
Carsten Arnholm, Norway (13:40:29) :
I remember a few months ago when I said the same thing, your reply was that it had infinitely many barycenters (centres of mass), because you introduced a pea in chosen positions millions of light years away.
But I agree with you now 🙂 The SSB is well defined.
I don’t think you do. The SSB is well defined once you have picked which bodies to include. If we pick different bodies then we’ll differ in the definition.
Geoff Sharp (20:10:25) :
“Leif Svalgaard (17:26:56) :
Consider a double star, each star with its complements of planets [make the distance between the stars large enough that the tidal effects are not too large]. The barycenter of that whole system lies halfway between the stars. What do the planets orbit?”
This is a minefield question with many variables, and maybe not even physically possible. It might be better if you get straight to your point.”
Maybe the orbit like fig 6 in:
http://www.scholarpedia.org/article/Three_body_problem
Transfer orbits can sure be exotic!
I think the “touchstone” for the barycenter / orbit business may be this:
The center of mass of the solar system is calculated as a linear average.
The center of gravitation experienced by any body is an inverse square function of it’s distance from all the individual masses of the solar system.
For most objects in the solar system, these two are dramatically different. Distance dominates.
For the sun, they are only a little bit different, which prompts the notion that the sun orbits the barycenter or that it’s orbital angular momentum can be calculated using a position vector rooted in the barycenter. This is at best a weak approximation and leads to errors (both numeric and errors of understanding as to what the barycenter is and does) and at worst is just flat out wrong as the sun does not orbit anything, it is perturbed a bit by everything and that is not an orbit.
In reality, the only way to calculate the solar orbital angular momentum (or perhaps changes in positional momentum if not really an orbit) is via solving an n-body problem and using the combined vectors from all the masses in the solar system sufficient to reach the desired level of precision.
Sensei, kohai asks: Floor, or no floor? (Nervously not looking at usual spot on the floor…) Does it all come down to just that: One is a linear average, the other a least squares function, never the twain ought to meet?
To clarify, I probably ought to have said:
“Distance dominates, inversely”
Leif Svalgaard (13:42:13) :
It is good form to stay on topic [if possible]. Try to google ‘dark matter’
I moved over here, as the conversation is already around this sort of thing, but does dark matter effect barycentre at all?
tallbloke (13:41:10) :
“There is no ‘relativistic matter’ in the sun’s core.”
And that would mean no energy either, according to Einstein and Birkhoff. It’s a wonder the sun continues to shine at all.
This has nothing to do with Einstein, but simply due to the Sun being hot inside. ‘Relativistic matter’ is matter moving at close to the speed of light and there is no such matter in the core of the Sun.
idlex (09:08:40) :
You have all the machinery to calculate the orbital angular momenta for all the planets and for the Sun. Compute for each time step the barycenter position. For each body, compute the angular momentum as the distance to the barycenter times the mass of the body times the speed around the barycenter. Plot the angular momentum on the same plot [or post a table] that show the AM for each as a function of time.
Leif Svalgaard (13:03:56) :
It can be done that way. It is the only way and the correct way, because the angular momentum of a collection of particles is simply the sum of the angular momentum of each particle. This is not an n-body problem at all.
Drat. Now I’m back where I started…
So I can use the ‘center of mass of the solar system’ as the root of my position vector and everything is fine. And the sun does orbit the ‘center of mass of the solar system’ and we do have a fleur-de-lies solar orbit.
But at least we have settled that to find where the angular momentum went we must look at the position vectors of the other planets? That is where it went, since they were the real gravitational force that changed the solar location relative to the ‘center of mass of the solar system’?
E.M.Smith (15:45:46) :
That is where it went, since they were the real gravitational force that changed the solar location relative to the ‘center of mass of the solar system’?
What the Sun gains in orbital momentum, the planets lose in orbital momentum, and vice versa. The sum stays constant, and the rotation or spin doesn’t enter at all..
Leif Svalgaard wrote:
idlex (09:08:40) : You have all the machinery to calculate the orbital angular momenta for all the planets and for the Sun. Compute for each time step the barycenter position. For each body, compute the angular momentum as the distance to the barycenter times the mass of the body times the speed around the barycenter. Plot the angular momentum on the same plot [or post a table] that show the AM for each as a function of time.
I will be delighted to do this.
[Now I’ll have to go back and read all the posts about angular momentum!]
I already routinely calculate the barycentre position, so that’s no problem. Nor is it a problem to find the distance of any body from the barycentre. And I know the masses of all the bodies. I also know the x,y,z velocities of all the bodies, where the x,y,z origin is the position of the Sun at the time my simulation starts. I don’t compute the speed of the barycentre, however. And I’m not exactly sure what you mean by “the speed [of the body] around the barycenter”. I suspect you simply mean how fast it’s going in whatever direction it happens to be going, relative to the barycentre.
However, since the barycentre is moving very slowly relative to any of the bodies in the solar system, I may be able to neglect the x,y,z velocity of the barycentre, and just use the x,y,z velocities relative to my origin, vx, vy, and vz, that I already have. In which case the speed you want may well be simply sqrt( vx^2 + vy^2 + vz^2).
If I’ve guessed correctly what is meant by “speed”, then it should be no problem to calculate the angular momentum of all the bodies in the solar system for each time interval. But at what intervals? And over what period of time?
Anyway, you’ll have to allow me a day or two to modify the program to calculate these values. It’s maybe best if I produce a table rather than a graph, so that people have raw numbers rather than wavy lines.
My simulation model is written in Java as a Java applet, and I’m planning to post it up on the Web at some point. It’s been – and continues to be – a great pleasure and a real eye-opener to write it.