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Eccentricity in binaries - simulation 2 (Read 11539 times)
Tony
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Re: Eccentricity in binaries - simulation 2
Reply #15 - 10/30/06 at 23:44:41
 
There's nothing in the code that tells it to do inclination cycles.  The code only tells the objects to follow Newton's laws.  Inclination cycles, just like ecc cycles should be a natural consequence of obeying Newton's laws.
 
Gravity Simulator is giving you all your orbital elements based on the ecliptic, or more specifically, the x-y plane.
 
The inclination you're reading for the planet is from the ecliptic, not from the plane of the secondary's orbit.  And the cycle you're expecting to see is probably based on the plane of the secondary.  But you can't just subtract one from the other.  Converting coordinates from one plane to the other requires some tricky geometry.  (It's probably not that tricky, but off the top of my head I don't know how to do it).
 
What if you tried putting the planet in a 75 degree inclination orbit and the secondary in a 0 degree orbit.  Then you wouldn't have to do any conversions since the secondary would be in the ecliptic, and the inclination numbers you plot would now be the planet's plane with respect to the secondary's plane.  I'd bet you'd get the cycles you were looking for.
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Re: Eccentricity in binaries - simulation 2
Reply #16 - 10/30/06 at 23:52:37
 
Quote from Tony   on 10/30/06 at 23:44:41:
There's nothing in the code that tells it to do inclination cycles. The code only tells the objects to follow Newton's laws. Inclination cycles, just like ecc cycles should be a natural consequence of obeying Newton's laws.

 
Ok then.  
 
Quote:
Gravity Simulator is giving you all your orbital elements based on the ecliptic, or more specifically, the x-y plane.

The inclination you're reading for the planet is from the ecliptic, not from the plane of the secondary's orbit. And the cycle you're expecting to see is probably based on the plane of the secondary. But you can't just subtract one from the other. Converting coordinates from one plane to the other requires some tricky geometry. (It's probably not that tricky, but off the top of my head I don't know how to do it).

 
I didn't get this last time you mentioned it, but now you say that it's relative to the x-y plane it makes more sense. But still... in this sim the planet starts in the x-y plane and the companion is inclined at 75° to it. As the sim progresses the planet inclination starts increasing, but that's still measured from the x-y plane isn't it? So if at one point the planet inclination is 35° and the companion inclination is 75° doesn't that mean the relative inclination is 40°, since they're both measured relative to the same x-y plane?
 
 
Quote:
What if you tried putting the planet in a 75 degree inclination orbit and the secondary in a 0 degree orbit. Then you wouldn't have to do any conversions since the secondary would be in the ecliptic, and the inclination numbers you plot would now be the planet's plane with respect to the secondary's plane. I'd bet you'd get the cycles you were looking for.

 
Well I can try that next, I'll leave the current sim running for tonight to see where it goes. It's going to stop recording at 500,000 years anyway.
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Re: Eccentricity in binaries - simulation 2
Reply #17 - 10/31/06 at 00:06:39
 
Only if your longitude of the ascending nodes are equal in both orbits can you simply add or subtract inclinations and get a useful number.  Consider the following:

 
All three orbits have inclinations of 30 degrees.  The purple and green orbits have the same longitude of ascending node.  Therefore you can do 30-30=0 and conclude that the purple orbit is inclined 0 degrees to the green orbit.  
 
But the 30-30 trick doesn't work for the orange orbit.   It is clearly 60 degrees inclined from the green and purple orbits. Instead, you need 30+30 = 60.
 
The green and purple orbits have a longitude of ascending node of 90 degrees and the orange orbit has a longitude of 270, 180 degrees away from the purple and green.
 
You may be able to express the inclinations relative to each other in excel with this formula:
 
i1+cos(deltaLAN)*i2
 
Excel likes radians instead of degrees, so...
=E2+cos((G2-O2)/180 * 3.14159)*M2
 
Assuming my conversion logic is correct.
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« Last Edit: 10/31/06 at 01:11:57 by Tony »  
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Re: Eccentricity in binaries - simulation 2
Reply #18 - 10/31/06 at 07:30:01
 
420,000 years (planet):
 

 
So I guess Arg of Peri, LAN, and Inclination are doing weird looking things because they're not in the same plane as the companion? Maybe it'd look more sensible if they were calculated relative to the companion instead?
 
 
Here's what the companion's been doing:

 
LAN and Arg of Peri aren't shown, because they're oscilating slightly around 0 degrees - so they're going from 1 to 359 degrees and messing up the graph.
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Re: Eccentricity in binaries - simulation 2
Reply #19 - 10/31/06 at 15:33:09
 
Quote from Tony   on 10/31/06 at 00:06:39:
You may be able to express the inclinations relative to each other in excel with this formula:

i1+cos(deltaLAN)*i2

Excel likes radians instead of degrees, so...
=E2+cos((G2-O2)/180 * 3.14159)*M2

This is wrong.  I think the correct way is this:
 
i3 = acos(sin(i1)sin(i2)cos(deltaLAN)+cos(i1)cos(i2))
 
and for Excel:
 
=ACOS(SIN(E2/180*PI())*SIN(M2/180*PI())*COS((G2-O2)/180*PI())+COS(E2/180*PI())*C
OS(M2/180*PI()))/PI()*180
 
 
Assuming:
i1 is in the E column
i2 is in the M column
LAN1 is in the G column
LAN2 is in the O column.
 
I'll run a sim to verify.
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« Last Edit: 11/01/06 at 16:34:28 by Tony »  
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Re: Eccentricity in binaries - simulation 2
Reply #20 - 10/31/06 at 18:03:48
 
536,700 years:
 

 
Two peaks now, but the odd thing is that they're about 218,000 years apart - I thought the calculation said that the cycle should be 335,795 years?!
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Re: Eccentricity in binaries - simulation 2
Reply #21 - 10/31/06 at 18:22:13
 
If instead of measuring peak-to-peak, your measure start to first peak and multiply by 2, the accuracy improves a little.  I think in ideal conditions (ie, much slower timestep), the eccentricity should have gone all the way back down to 0.
 
Also keep in mind that the formulas in the paper don't use the equal sign.  They use the =~ sign.
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Re: Eccentricity in binaries - simulation 2
Reply #22 - 11/01/06 at 16:40:59
 
I verified the formula I gave above by running 2 simulations.:
 
1:  planet 2.5 AU, 1 Mj
     Secondary 200 AU 0.9 Ms, 75 degrees
 
2:  planet 2.5 AU, 1 Mj, 75 degrees
     secondary 200 AU 0.9 Ms
 
Time step 32K for 2 million years.
 
With a small modification the formula works flawlessly.  The inclination of the planes of the planet and the secondary are identical when compared to each other instead of against the ecliptic by using the Excel Formula.
 
I edited the formula in the above post by placing a few missing multiplication symbols (*) and converting the answer to degrees.  Again, the Excel formula is:
 
=ACOS(SIN(E2/180*PI())*SIN(M2/180*PI())*COS((G2-O2)/180*PI())+COS(E2/180*PI())*C
OS(M2/180*PI()))/PI()*180
 
Assuming:  
i1 is in the E column  
i2 is in the M column  
LAN1 is in the G column  
LAN2 is in the O column.  
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Re: Eccentricity in binaries - simulation 2
Reply #23 - 11/01/06 at 17:19:58
 
789,800 years
 
I'm calling it quits here. Top graph is the planet, bottom is the companion. I got no idea what all the properties of the planets' orbit that aren't eccentricity are doing...  
 

 

 
I'll try those formulas when I get a chance...
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Re: Eccentricity in binaries - simulation 2
Reply #24 - 11/01/06 at 20:07:43
 
Quote:
I got no idea what all the properties of the planets' orbit that aren't eccentricity are doing...

Your inclination is rising and falling with a period that is twice the period of the eccentricity's.
 
Your longitude of ascending node is changing with a period approximately equal to the period of your eccentricity.
You might want to create an additional column in your spreadsheet and add 180 to your LAN to center it on the graph so you don't get those vertical lines.  There's nothing special about 0 degrees or 90 degrees, etc.  They're simply degrees from the Vernal Equinox, which is a manmade reference point based on Earth's orbit.  Adding 180 degrees would give you the angle from the Spring Equinox, while still giving you the same trend, but unbroken.
 
It's a cool looking graph nonetheless.
 
Your graph for the secondary is very tight.  Ecc only ranges from .79 to .81.  This probably isn't too different than if you ran a simulation with only the star and the secondary, eliminating the planet.  Theoretically, in such a 2-body configuration these values should remain perfectly constant.  But noise generated by the numerical method will give you some fluxuation.  I suspect you're zoomed in on the noise.
 
Another orbital element you may be interested in is Longitude of Periapsis.  This differs from Argument of Periapsis in that "Argument" means degrees from the ascending node, and "Longitude" means degrees from the Vernal Equinox which is a fixed point.  Gravity Simulator doesn't output it, but its easy to compute.  It's Longitude of Ascending Node + Argument of Periapsis.  So you could make another column in Excel for Longitude of Periapsis.
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Re: Eccentricity in binaries - simulation 2
Reply #25 - 11/02/06 at 07:33:00
 
Quote from Tony   on 11/01/06 at 20:07:43:

Your inclination is rising and falling with a period that is twice the period of the eccentricity's.

 
I don't think it is. It looks like the inclination's period is the same as the eccentricity's, each time there's an ecc peak the inclination starts going in the other direction on the graph. You can see a cycle more clearly on the companion graph (where the bigger spikes are).  
 
 
Quote:
Your longitude of ascending node is changing with a period approximately equal to the period of your eccentricity.

 
Don't think so (see below)
 
 
Quote:
Your graph for the secondary is very tight. Ecc only ranges from .79 to .81. This probably isn't too different than if you ran a simulation with only the star and the secondary, eliminating the planet. Theoretically, in such a 2-body configuration these values should remain perfectly constant. But noise generated by the numerical method will give you some fluxuation. I suspect you're zoomed in on the noise.

 
There's definitely something corresponding to the eccentricity peaks in the companion inc and ecc graphs though (where the biggest spikes are).  
 
 
Quote:
Another orbital element you may be interested in is Longitude of Periapsis. This differs from Argument of Periapsis in that "Argument" means degrees from the ascending node, and "Longitude" means degrees from the Vernal Equinox which is a fixed point. Gravity Simulator doesn't output it, but its easy to compute. It's Longitude of Ascending Node + Argument of Periapsis. So you could make another column in Excel for Longitude of Periapsis.

 
Is there any reason why GravSim couldn't output this? It's a fairly useful orbital parameter after all.  
 
Here's a graph with the LAN+180° and the Long of Peri too.  
 

 
 
And the period of the eccentricity is the peak-to-peak or trough-to-trough distance, which looks like it's about 200,000 years. It looks to me on this graph that the period of oscillation for the ecc and inc are about 200,000 years, and the period of oscillation of the LAN and LoP is double that (about 400,000 years). I can't tell much from the Arg of Peri because it's just going all over the place.
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« Last Edit: 11/02/06 at 18:57:07 by EDG »  

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