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I am currently trying to calculate apparent positions from raw JPL data.
I've got it pretty much figured out, but there is one thing that's bugging me: Has the light time correction of the moon to be done heliocentric or geocentric?

Light time correction is needed because the speed of light is finite and therefore the bodies of the solar system are not seen where they are, but where they have been before (See: Has everything we see happened in the past?)

A simple algorithm (discarding light retardation due to the sun) for the light time correction works like this:

  1. Calculate barycentric cartesian positions of earth and planet at time $T$
  2. Calculate distance $s$ and light time $\Delta T = \frac s c$
  3. Calculate position of planet at time $T' = T - \Delta T$
  4. Repeat steps 2 & 3 until $T'$ doesn't change significantly.

Note: The movement of the solar system is not taken into account because the calculation is done with the solar system as the inertial frame of reference.

If I calculate the same thing for earths moon, do I have to take the movement of the system earth-moon into account; Do I have to use a heliocentric frame of reference?
If not, the movement of the moon is a (well, actually not so) simple orbit around the earth that keeps the same radius - more or less.
But if so, the movement of the moon is a combination of moon orbit and earth orbit and calculating the position at time $T'$ therefore changes the distance of earth and moon.

The "Explanatory Supplement to the Astronomical Almanac" does not mention any special handling of the moon so I assume one has indeed to take the movement of the system earth-moon into account when calculating moons position at time $T'$ and do the calculation in a heliocentric frame of reference.

But why can't I do these calculations with "earth-moon" as the inertial frame of reference (please excuse if that's wrong nomenclature) instead of the solar system?

Edit
After Seans post I have noticed I was imprecise.
I know different frames of reference will give different results and as JPLs data is fitted to observations, so obviously one has to use exactly the calculations that JPL used to derive the planetary ephemerides.

To explain my question further, I will assume Moon is trying to send a light ray to Earth (center of Earth, to be precise):
If Earth is moving, Moon has to adjust the direction of the light ray so it will hit a moving target.
If Earth is not moving, Moon can send the light ray directly downwards from its orbit.

So, obviously, Earth is moving though the solar system and Moon has to hit a moving target.
But, the solar system itself is moving, too. Every light ray in the solar system has to hit a moving target.

This is not taken into account.

I think it all boils down to this:
Without gravitational influences, does every light ray move in a straight line with the speed of light $c$ in every valid frame of reference?
Wikipedia seems to say so.

If so, why is the moving solar system a valid frame of reference (so the movement of light can be calculated as written in the inital question) for earth and moon while earth-moon is not?

Edit 2

According to Sean, a frame of reference with the barycenter of earth+moon would be valid. But the results of both ways of calculation differ. This is because the position of earth at time $T'$ is never calculated, so in the solar system, the position of moon at time $T'$ relative to the position of earth at time $T$ changes with two components: Movement of the moon on its orbit around the earth + movement of the barycenter of earth and moon on its orbit around the sun. Calculated in the frame of reference of earth+moon however, the movement of moon only consists of its orbit around the earth.

Therefore, there is a difference in calculation.

This leads me to my confusion: Only one frame of reference can be valid (different results for the apparent position of moon) while it seems as if both frames of reference should be valid.

Edit 3
Why is it wrong to define the movement of the moon as an orbit around earth (and neglecting the orbit of earth+moon around the sun), when calculating apparent positions for an observer on earth?

Edit 4
According to Sean, it is not. Strange. I assume that the Astronomical Almanac is confirmed through observations and all other results therefore must be wrong.
The difference is 0" to 20".
Sadly I could not find any exact observational data of moon - its apparent diameter is ~30', so they would have to be exact to the level of 1% of its disc.
It's still a riddle to me.
If Sean is correct, the Explanatory Supplement is wrong, as are practically all other ephemerides. Which leads me to the suspicion there might be more to it than Sean (and I) think.

Community
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Imanuel
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  • Can you elaborate a little on which light time correction you're concerned with? "Systems that I look at" confuses me. What are you correcting? Finite speed of light or parallax, or bending...? – user6972 Aug 07 '13 at 05:45
  • I have updated the question and hopefully answered your questions :) – Imanuel Aug 07 '13 at 09:55
  • I would assume that the corrections for the moon are done geocentric and then transformed into other systems as the need arises, since that would by far be the easiest. May I also point out that the light-time delay for the moon would only be a bit over a second. So the Moon would only be about 1km from where we see it. 1km is almost small enough to be nullified by measurement error. So regardless, you should be able to use the data provided as properly corrected data – Jim Aug 07 '13 at 13:34
  • I would have assumed so, too - that's why I have asked. The correction is very small, yes, but I want to know which calculation is mathematically correct. – Imanuel Aug 07 '13 at 13:36
  • I think the light time correction from the JPL data could be done using the difference of sun-moon and sun-earth, but it would be quite small and probably not a big factor. – user6972 Aug 08 '13 at 03:53

1 Answers1

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I am not sure if i understood the question, but let us go like this. you are permitted to use a heliocentric refrence frame, and all motions relative to that frame are to be accounted for. So, please do it with the motion of earth and moon accounted for.

the reason thereof is, you can of course, calculate the barycenter of earth and moonin an earth moon reference system, and with time, this will trace a circle, ignoring the non-circularity of lunar orbit, traced on moon's orbital plane. (The moon is going around the earth's center, and the barycenter is not the earth's center, on the line joining the centers of mass of earth and moon. the only point in this line that does not move is the center of earth, in a geocentric coordinate system, because that is set to be the origin. had this been a selenocentric system, the non moving origin would be center of moon).

However, in a heliocentric system, the still origin is the center of sun, and in that system, the barycenter here traces a 3-D cycloid with winds on the orbital plane of moon, and reference circle on the orbital plane of earth. so you will have different results, if you reduce the problem to geocentric reference frame.

(What i understand from your question is the "why" behind forcing the calculation to a heliocentric system, and not on earth moon system. so i did not do the calculation itself here.

Sean
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  • Thank you for your answer. It's still not clear to me, though - I have edited the question. – Imanuel Aug 07 '13 at 13:34
  • when calculating ephimerides, generally, one does not think about the light beams, the Huygens principle says, that the light beam that has left its source, forgets it's source, so once the beam is sent, there is nothing more the source can do to influence it.

    yes, without gravity, the light rays would move in a straight force, as EM force or strong or weak interaction does not effect photons.

    as for validity, it depends on what we are looking at. a hundred years back, an earthbound astronomer would take the ephimerids calculated on earth as standard, as that was valid for his observations.

     – Sean Aug 07 '13 at 13:57
  • but, the JPL has a few other standards, and earthbased coordinates are also standard ones for J2000 system. The fact that often Heliocentric coordinates are choosen, is because the ephemerides are constantly updated from spacecraft borne ranging systems, that are on orbits which are defined heliocentric. As the device has to start at earth and go to another planet, it makes also sense to take heliocentric coordinates.

    To see the standards, see here: http://www.cv.nrao.edu/~rfisher/Ephemerides/ephem_descr.html

    (I think that your question asks the reasons behind choosing a particular system)

     – Sean Aug 07 '13 at 14:02
  • Whether coordinates are given in ICRF or geocentric is irrelevant (planetary formulas are easier in ICRF, sure). When calculating light rays, why is the barycenter of earth-moon moving while the barycenter of the solar system is not? In reality, both are moving, but the latter is neglected in the proposed algorithms. Why one and not the other? – Imanuel Aug 07 '13 at 14:13
  • once again, the light is not moving a hiting target, that is a continuous stream, in spherical astronomy. so the question is rather, where in the sky will i see object X if i am in this place, of course my place may also be moving. So i can choose my place to either be the sun, or the earth – Sean Aug 07 '13 at 14:20
  • I think you don't understand my question. If I assume the solar system to be moving, the distance the light has to travel from A to B increases (or decreases, depending on the positions of A and B). Why is this neglected? Because in an inertial frame of reference, light moves with $c$ in a straight line, so any movement of the frame in total is irrelevant, right? Why is the movement of Earths barycenter not neglected for the same reason? – Imanuel Aug 07 '13 at 14:28
  • because, the time light takes is not the time a single photon takes. while plotting the trajectory of a single photon, you will calculate all changes, otherwise, you will only consider the lengths of straight paths between objects, which gives the "stream length of a photon river", to put it poetically.

    now from superposition principle, you say :

    coordinate of A = planetary motion + motion within star system + motion OF star system. Had this been a quasar outside, you will consider the motion of the container, the solar system. since the light source, moon, is inside, this can be neglected

     – Sean Aug 07 '13 at 14:36
  • simply because, earth and moon are BOTH affect by the same way, by the total motion of solar system. so you dont need to add this effect on each on separately. – Sean Aug 07 '13 at 14:37
  • Thank you for this explanation. Earth and moon are affected by the movement of the earth-moon barycenter in the same way, right? Then why take is this movement relevant? – Imanuel Aug 07 '13 at 14:53
  • not exactly. the barycenter motion has two components. one component is due to the motion of earth and moon, in heliocentric frame. this affects them separately. the second component is due to the total motion of the solar system itself. this one causes same effect on both. therefore you ignore the motion of solar system on the barycenter. – Sean Aug 07 '13 at 14:56
  • The barycenter motion has only one component - for a given time, JPL gives the coordinates of the barycenter. With the geocentric coordinates of the moon (given by JPL), the barycentric coordinates of earth and moon can be calculated. So JPL gives geocentric positions of earth+moon and an offset of the frame of reference, if you like. Why do I need this offset if it affects both bodies in the same way? – Imanuel Aug 07 '13 at 14:59
  • the total motion is : motion of barycenter in geocentric coordinate (this is your exactly one component)

    the offset that you need to convert geocentric coordinates to heliocentric coordinates (this is due to the fact that for a spacecraft in solar system outside earth, this will generate a different effect in both objects, as the barycenter also orbits sun)

    the motion of solar system itself (this will make the motion of bodies to appear different to a quasar but for a ship inside the solar system, this will cause same effects on each,this can be ignored) + motion of universe + ...

     – Sean Aug 07 '13 at 15:25
  • out of them, if you are on earth or on the barycenter, you can ignore anything beyond it (same effect for everything). if you are in the solar system, and have the barycenter itself in heliocentric and earth moon in barycentric coordinates, then you change to the larger container, the heliocentric coordinates. Any larger moving container will have same effect on everything, in this case the moving solar system. hence the motion of the larger container is ignored. but now that you are in the solar system, the earth-moon system is a subcontainer, and it will make a difference. – Sean Aug 07 '13 at 15:30
  • Ok. Does that mean that if I would use a frame of reference with its origin in the barycenter of earth+moon to dump the first component ("geocentric") so the translation from inner to outer container is a simple offset, same for every body in the inner container, would that be valid? It would still give different results. Thanks for your patience. – Imanuel Aug 07 '13 at 15:46
  • Does that mean that if I would use a frame of reference with its origin in the barycenter of to dump the first component ("this will be barycentric" for earth and, like you get from JPL, "geocentric" for the barycenter ITSELF) so the translation from inner to outer container is a simple offset (correct), same for every body in the inner container (correct, for both earth and moon), would that be valid? . yes that is the plan, and that is why you dont have any prefect reference frame of perfect motion – Sean Aug 07 '13 at 16:00
  • Thank you. So it seems to me - both frames of reference seem valid (only I misnamed the second one as "geocentric"). But they produce different results - so only one can be valid. I have updated the answer. – Imanuel Aug 07 '13 at 16:25
  • your edit 2 is correct. now, the uniqueness of validity depends on, as you got it, context. who is watching it? a spacecraft? an earthbound amature? – Sean Aug 07 '13 at 16:37
  • Which frame of reference do I have to apply for an earthbound observer? Strictly like written in the Explanatory Supplement (solar system) or with the barycenter of earth+moon as origin? – Imanuel Aug 07 '13 at 16:46
  • for an earth bound observer, ignore corrections due to solar system's own motion, and take the origin in earth's center – Sean Aug 07 '13 at 17:21
  • Solar systems own movement is always ignored and after all corrections, the positions are always reduced to geocentric coordinates. The question is whether to ignore the orbit around the sun when calculating previous positions of the moon (since it's orbiting earth and the observer is on earth). Which would contradict the Explanatory Supplement and produce wrong results (compared to other ephemerides). My question is: Why is it wrong to define the movement of the moon as an orbit around earth, when calculating apparent positions for an observer on earth? – Imanuel Aug 07 '13 at 18:51
  • it is not. it is correct. but of course that would contradict ephemerides define otherwise. – Sean Aug 07 '13 at 19:44
  • Strange. This is exactly what I have thought in the first place, but since all other ephemerides do this differently, I assumed (and still assume) that this might be wrong. I have updated the question. – Imanuel Aug 08 '13 at 12:33
  • could you share a link to the ephemerides ? so that i can look up the contexts thereof and also the questions that you are calculating - i will check it in detail myself now. – Sean Aug 08 '13 at 17:36
  • Sure: I use IMCCE because it uses DE405 and can output ecliptical coordinates which I haven't found on Horizons - it has no nutation (for ecliptical coordinates), but enough to compare the results. (Settings: Moon, alone, DE405, geocenter, ecliptic, mean of date) – Imanuel Aug 08 '13 at 18:52
  • good, and from those output, what is the target you want to calculate? like the what of moon? – Sean Aug 08 '13 at 19:00
  • The apparent position of moon (i.e. its center) as visible from earth. Further corrections for topocentrical coordinates are planned but that is irrelevant for the question as the results already differ. The question is: Which algorithm is correct for the light time correction of moon; or: Why is it wrong to choose an earthbound frame of reference for this task? – Imanuel Aug 08 '13 at 19:09
  • okey, do you also have a set of answers against which i should check? – Sean Aug 08 '13 at 19:11
  • Excuse me? I don't understand your last comment. – Imanuel Aug 08 '13 at 20:52
  • Please continue the discussion in [chat]. – Manishearth Aug 09 '13 at 07:55
  • i dont know which chat room is the correct for this discussion. @pharaoh, after i do my calculation, i need to compare it with the values which you are trying to achieve. – Sean Aug 09 '13 at 16:19