# Ecliptic Longitude and Right Ascension of the Sun

1. Nov 9, 2008

### Helios

Here's a question. Where do the ecliptic longitude and the right ascension ( of the Sun )reach their maximum difference? What is the declination for these events? What calendar dates do you suppose this happens? I know the answer. I've never heard anyone make this observation except me. These events don't even have a name?! We can even name them here and now for the first time!

2. Nov 12, 2008

### giann_tee

What are you saying? Ecliptic longitude is measured on one great circle, and right ascension on another one. The sun travels the ecliptic in non-uniform way strictly speaking due to eccentricity of the orbit. Lets say its about a degree a day. Then, sun's coordinate point on the equator also travels like a shadowy projection. If obliquity was really big, the speed on the equator of that point (defined by intersection of great circle containing north celestial pole P and the sun) travels a little quicker starting from gamma point, then slower, then quicker and there I give up. What is the difference between the two?

3. Nov 12, 2008

### Helios

OK, here's a hint. The relationship between the two is

$$tan \alpha = tan \lambda cos \epsilon$$

Right Ascension ( alpha )
Ecliptic Longitude ( lambda )
Obilquity Angle ( epsilon ) = 23.439°

Find where $$\lambda - \alpha$$ reaches the maximum ( 4 times per year ).

4. Nov 13, 2008

### giann_tee

I wrote all equations for triangle declination-ascension-longitude but I don't know what to do next.
sin(alpha)=cos(e)sin(lambda)
Right? Lambda is changing at constant rate (pretend), alpha is whatever.

5. Nov 14, 2008

### Helios

That's not it.
write
$$f = \lambda - \alpha = \lambda - arctan( tan\lambda cos\epsilon)$$

then find $$df/d\lambda$$
set this to zero to find maximum
then values for $$\lambda$$
and $$\alpha$$ will follow.

6. Nov 14, 2008

### giann_tee

I haven't reverse engineered your equation, normally I'd trust it but it seems more complicated than necessary. I used program "Derive" to solve the first differential as an equation but all solutions seem to be complex! It appears absurdly complicated.

Clearly, equator and ecliptic are two great circles on same sphere at an angle. Great circle ecliptic_poles-sun is rotating and so is celestial_poles-sun circle which is tracing sun on equator and recording declination. All 3 coordinates are one-on-one and one on date of tropical year (short term). Hence lets see the tables with numbers...

7. Nov 14, 2008

### giann_tee

...and I get

for l=0 to 360
RA=acos[cos(l) / asin( sin(e)sin(l) )]

8. Nov 15, 2008

### giann_tee

May 31st 2009, angular difference worth almost 2.5 degrees. How about that?

9. Nov 15, 2008

### Helios

Well, the month is right, but you're still cold

10. Nov 16, 2008

### giann_tee

I protest! First its irrelevant thing to know. Second I didn't design these awful arc sine functions.

11. Nov 16, 2008

### Helios

Irrelevant? It's not irrelevant. Know that there's an "Equation of Time" that corrects sundial time to local mean time. The Sun runs fast or slow compared the mean time. Should we correct for obliquity, we must know
$$\lambda- \alpha$$
The fact that it has extrema should be interesting to any calculus student. Oh well.

So there's another feature of these same points!
Firstly, Ecliptic Slope (sigma) is related to declination by
$$cos\epsilon = cos d cos\sigma$$

Find where the declination equals the ecliptic slope or the negative of the ecliptic slope.
Then use
$$sin d = sin \lambda sin \epsilon$$

This approach is too easy, a dead give-away to the problem.