# Sun coordinates

1. Dec 14, 2009

### zero.kalvin

I am sorry if I had to post this in another section. But I am writing a program, and it's kind of essential for me to know what are the sun coordinates in the sky. From what I understood, is that the sun have fixed position in the galactic coordinates system, is this correct?
if so, what are these coordinates(numerical values for example) ?
I know I will have to do other coordinates changes after that, but at least i must have something.
(in few words, I have to correlate some events coming from the sky with the sun position, and determine if they came from the sun or not, so i need to calculate the angle distance between them).

Any help is appreciated.

2. Dec 14, 2009

### blkqi

In the galactic coordinate system we take Sun to be the center. So, yes.

You can calculate the angular separation between events in the sky (relative to Earth) if you put them on the equatorial coordinate system.

I'm not sure exactly what you're trying to do.

Last edited: Dec 14, 2009
3. Dec 14, 2009

### zero.kalvin

Sorry, I need to be very sure of what you are saying.
Are you telling me that the sun coords are (0,0) ?? because from what I understood from reading on this, is that the galactic center is on (0,0), but the map is just centered on the sun.
yes, I will change my coordinates in the end. what I have is events generate by a Monte Carlo generator(neutrinos), and study my background noise in the direction of the sun. So i need to know where is the sun in the sky.

4. Dec 14, 2009

### blkqi

you are right, galactic center is (0,0), intersected by the "prime meridian". The coordinate system is a sphere centered ON Sun. You don't use galactic coordinates to locate objects in our solar system.

5. Dec 14, 2009

### D H

Staff Emeritus
You need to model the orientation, rotation, and orbit of the Earth. A very simple model:
• Pretend the Earth moves in a circular orbit.
• Pretend a year is 365.2425 days long (i.e., the calendar with leap years).
• Thus the Earth's moves by 360/365.2425 degrees per day with this simple model.
• Pick as an epoch time noon on some vernal equinox.
• At this epoch time, orient the Earth such that
• The angle between the rotation axis and the radial vector from the sun is 90 degrees.
• The angle between the rotation axis and the velocity vector is 90-23.44 degrees.
• The Greenwich meridian is oriented sunward.
• The Earth rotates at one revolution per 1 day less 1/365.2425 days in this simple model.

You won't get an analemma with this simple model, and you most certainly will not get high precision. Make the orbit elliptical and you will get the analemma and the equation of time. Since perihelion is on January 3 this coming year, you can, without too much damage, pretend that perihelion occurs at winter solstice. Make the winter solstice rather than the vernal equinox your epoch point, and adjust the initial orientation appropriately.

Things are going to get *very* complex very fast if you need more accuracy than these simplistic models provide.

6. Dec 14, 2009

### D H

Staff Emeritus
You don't want galactic coordinates here. They are irrelevant to this problem.

7. Dec 14, 2009

### zero.kalvin

I already thought of you said, but the problem is that I need to study the galactic center as well, along with other galactic objects, so the galactic coordinates are important for me.
I already have a code that can change from the Galactic Coordinates to Azimuth and Elevation(which i need at the end to calculate how much events i have in my cone of sight).
But the problem is that, from what I've seen from searching around, is that I couldn't find these numerical values(the sun coords).

8. Dec 14, 2009

### ideasrule

As DH said, the galactic coordinate system is centered on the Sun, so you can't get coordinates for the Sun itself. That would be like standing on the north pole and asking which direction the north pole is in.

That said, do you just want to calculate the azimuth and elevation of the Sun? Use any star chart program. YourSky (http://www.fourmilab.ch/yoursky/) would do just fine.

9. Dec 14, 2009

### D H

Staff Emeritus
That doesn't matter. He can get the coordinates for the Earth, and from this determine the position of the Sun. Simply describe the Earth's orbit in galactic coordinates. Alternatively, (and this is what I suggest doing) describe the Earth's orbit (and orientation) in some other frame and convert to galactic coordinates.

The reason for doing things this way is that one can build in complexity if needed. Aiming an amateur telescope at some object doesn't require an inordinate amount of precision. Aiming a radio telescope array at some object does. For example, the uneven rotation of the Earth and the undulations of the Earth due to Earth tides affect the aiming of a radio telescope array.

10. Dec 14, 2009

### ideasrule

The OP doesn't understand why, if he can get the altitude & azimuth of other stars based on their galactic coordinates, he can't do likewise for the Sun. I don't think he was asking whether it's possible to express Earth's apparent position in galactic coordinates.

11. Dec 15, 2009

### zero.kalvin

No I am not searching for the apparent earth position, I am searching for the sun coordinates in the galactic coordinates.
Now correct me if I am wrong here:
I know how to calculate the declination of the sun as a function of which day of the year it is. I know the longitude of the sun in the galactic coordinates(it should be 0), and the latitude is a function of the longitude and the declination only(using the equatorial to galactic transformation).
However at the end, after my calculation, the sun apparently has no defined latitude during the period of the last 64 and the first 45 days of the year.

But what I don't understand, like the previous poster said, "Why am I not allowed to talk about the sun in this coordinates?" are you referring to something like the when the zenith is 0 or 180 degrees, the azimuth is not defined ?
if so, what are these formulas that show me why it's like this?

12. Dec 15, 2009

### zero.kalvin

Sorry I made a mistake(code error), it's defined all year long(latitude), but it's not constant.
it moves in the beginning of the year from -97 to -170 and then again to -97 at the end.

13. Dec 15, 2009

### D H

Staff Emeritus
Simple: Identically zero.

What exactly are you trying to do, zero.kalvin? It would help to know what the target application is, and how accurate you need to be.

14. Dec 15, 2009

### zero.kalvin

Ok, what I am trying to do is the following:
I have some MC neutrinos events filling my sky, each with it's own Azimuth and and Zenith. My angular resolution is around 2 degrees. And I am not looking up, I am looking down, through the earth. I need to look with a precision of an hour(time) the position of the sun in my sky, and see how much neutrinos I have in my cone of sight over a whole year as a function of how wide i open my cone of sight(centered around the sun).