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Inclination of Equatorial North Pole relative to Galactic Equator

by fizixfan
Tags: coordinates, equatorial, galactic
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fizixfan
#1
Jun16-14, 08:06 PM
P: 27
If the axis of the Earth is tilted at 23.4 degrees away from the North Ecliptic Pole, and the Ecliptic Plane is tilted at 60.2 degrees away from the North Galactic Pole, then why is the Earth's rotational axis inclined at 27.13 degrees relative to the Galactic Equator instead of (90-(60.2+23.4)) = 6.4 degrees? Is there a formula that will allow me to calculate this, and if so, can anyone show me how to do it? I know it involves spherical trigonometry.

I'm somewhat familiar with transformations of galactic, ecliptic and equatorial coordinate systems and spherical trigonometry. But I can't find the formula that will allow me to transform Equatorial coordinates to Galactic coordinates.

Here is a diagram which may or may not help. It at least illustrates my dilemma.

Thanks!
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Equatorial, Ecliptic and Galactic Planes and Poles_highlight.jpg  
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Matterwave
#2
Jun16-14, 11:49 PM
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Can you double check your statement "The Ecliptic plane is tilted at 60.2 degrees away from the North Galactic Pole"? In your picture the Ecliptic plane is tilted 60.2 degrees away from the Galactic plane, not the North Galactic pole.

It seems though that the Wikipedia page on the Milky Way suggests that what your drawing, and not your statement is correct about that point though. So it might be that you just misstated that fact.

The only other thing I can think of right now is that the Earth's 23 degree tilt might be in the other direction relative to the galactic plane (i.e. it might be 23 degrees counter-clockwise from the perpendicular line to the ecliptic, rather than 23 degrees clockwise from this line), but applied to your drawing this would get a declination of the galactic north pole of 53 degrees rather than 27 degrees...
Nugatory
#3
Jun17-14, 12:09 AM
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Quote Quote by fizixfan View Post
If the axis of the Earth is tilted at 23.4 degrees away from the North Ecliptic Pole, and the Ecliptic Plane is tilted at 60.2 degrees away from the North Galactic Pole, then why is the Earth's rotational axis inclined at 27.13 degrees relative to the Galactic Equator instead of (90-(60.2+23.4)) = 6.4 degrees? Is there a formula that will allow me to calculate this, and if so, can anyone show me how to do it? I know it involves spherical trigonometry.
The angle ABD is not uniquely determined from the angles ABC and CBD so the problem is underspecified. Even if all four points lie in the same plane, ABD is not uniquely determined; it can be either ABC+CBD or ABC-CBD. Allow D to lie in a different plane than A, B, and C and ABD can take on any value between those two extremes.

Matterwave
#4
Jun17-14, 12:12 AM
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Inclination of Equatorial North Pole relative to Galactic Equator

Yes, Nurgatory is right! I was looking at this in a 2-D point of view as well. The tilt might not only be above and below the line you drew, but might be into the page or out of the page as well.

So in this case, the relative angle between the Galactic north pole and the celestial north pole might be anywhere between 6 degrees and 53 degrees. It happens to be 27 degrees.
D H
#5
Jun17-14, 08:03 AM
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P: 15,066
Quote Quote by fizixfan View Post
If the axis of the Earth is tilted at 23.4 degrees away from the North Ecliptic Pole, and the Ecliptic Plane is tilted at 60.2 degrees away from the North Galactic Pole, then why is the Earth's rotational axis inclined at 27.13 degrees relative to the Galactic Equator instead of (90-(60.2+23.4)) = 6.4 degrees?
You're not working two dimensional space. You can't simply add or subtract angles anymore.


Is there a formula that will allow me to calculate this, and if so, can anyone show me how to do it? I know it involves spherical trigonometry.
One approach is to use cartesian coordinates as an intermediary.

On January 1, 2000, the galactic north pole had a right ascension ##\alpha## of 12h 51m 26.00s and a declination ##\delta## of 27 7' 42.0". Note that right ascension and declination are expressed in celestial (earth equatorial) coordinates. The unit vector along the north galactic pole has celestial coordinates of ##\hat z_g = \bigl[ \cos\delta\cos\alpha, \cos\delta \sin\alpha, \sin\delta\bigr]##.

The celestial coordinates of the celestial north pole is just ##\hat z_c = \bigl[0, 0, 1\bigr]##. The inner product between ##\hat z_g## and ##\hat z_c## is ##\sin\delta##. Since these are both unit vectors, this is the cosine of the angle between these two vectors. Since ##\arccos(\sin x) = 90^{\circ}-x##, the angle between the north celestial pole and north galactic pole is 90 - (27 7' 42.0") = 62 52' 18" (or 62.8717 degrees).

The ecliptic plane is formed by rotating the celestial plane about the x-axis by the obliquity of the ecliptic ##\epsilon## of 23 26' 21.406", making the celestial coordinates of the ecliptic north pole ##\hat z _e = [0, -\sin\epsilon, \cos\epsilon]##. The inner product between ##\hat z_e## and ##\hat z_c## is ##\cos\epsilon##. Thus the obliquity of the ecliptic is the angle between the celestial and ecliptic north poles.

What about the angle between the galactic and ecliptic north poles? Now we want to find the inner product between ##\hat z_g## and ##\hat z_e## and take the inverse cosine of that inner product. The inner product is ##\hat z_g \cdot \hat z_e = \sin\delta \sin\epsilon - \sin\alpha \cos\delta \cos\epsilon##. Letting WolframAlpha do the math for me (http://www.wolframalpha.com/input/?i...6%2F60%29%2F60), the angle is 60.1889 degrees.
fizixfan
#6
Jun17-14, 10:03 AM
P: 27
Quote Quote by Matterwave View Post
Can you double check your statement "The Ecliptic plane is tilted at 60.2 degrees away from the North Galactic Pole"? In your picture the Ecliptic plane is tilted 60.2 degrees away from the Galactic plane, not the North Galactic pole.

It seems though that the Wikipedia page on the Milky Way suggests that what your drawing, and not your statement is correct about that point though. So it might be that you just misstated that fact.

The only other thing I can think of right now is that the Earth's 23 degree tilt might be in the other direction relative to the galactic plane (i.e. it might be 23 degrees counter-clockwise from the perpendicular line to the ecliptic, rather than 23 degrees clockwise from this line), but applied to your drawing this would get a declination of the galactic north pole of 53 degrees rather than 27 degrees...
You're right Matterwave, it should be the Ecliptic Plane is tilted at 60.2 degrees away from the Galactic Equator. My drawing is also in 2D, but to properly represent the relative orientation of the three planes (Equatorial, Ecliptic and Galactic), it would have to be in 3D. I wasn't visualizing this correctly.
fizixfan
#7
Jun17-14, 12:57 PM
P: 27
Quote Quote by D H View Post
You're not working two dimensional space. You can't simply add or subtract angles anymore.

One approach is to use cartesian coordinates as an intermediary.

On January 1, 2000, the galactic north pole had a right ascension ##\alpha## of 12h 51m 26.00s and a declination ##\delta## of 27 7' 42.0". Note that right ascension and declination are expressed in celestial (earth equatorial) coordinates. The unit vector along the north galactic pole has celestial coordinates of ##\hat z_g = \bigl[ \cos\delta\cos\alpha, \cos\delta \sin\alpha, \sin\delta\bigr]##.

The celestial coordinates of the celestial north pole is just ##\hat z_c = \bigl[0, 0, 1\bigr]##. The inner product between ##\hat z_g## and ##\hat z_c## is ##\sin\delta##. Since these are both unit vectors, this is the cosine of the angle between these two vectors. Since ##\arccos(\sin x) = 90^{\circ}-x##, the angle between the north celestial pole and north galactic pole is 90 - (27 7' 42.0") = 62 52' 18" (or 62.8717 degrees).

The ecliptic plane is formed by rotating the celestial plane about the x-axis by the obliquity of the ecliptic ##\epsilon## of 23 26' 21.406", making the celestial coordinates of the ecliptic north pole ##\hat z _e = [0, -\sin\epsilon, \cos\epsilon]##. The inner product between ##\hat z_e## and ##\hat z_c## is ##\cos\epsilon##. Thus the obliquity of the ecliptic is the angle between the celestial and ecliptic north poles.

What about the angle between the galactic and ecliptic north poles? Now we want to find the inner product between ##\hat z_g## and ##\hat z_e## and take the inverse cosine of that inner product. The inner product is ##\hat z_g \cdot \hat z_e = \sin\delta \sin\epsilon - \sin\alpha \cos\delta \cos\epsilon##. Letting WolframAlpha do the math for me (http://www.wolframalpha.com/input/?i...6%2F60%29%2F60), the angle is 60.1889 degrees.
Thanks D H, this is one of the most complete answers I've received. I now realize I was looking at these three coordinates in 2D rather than 3D, and it's not a simple matter of adding or subtracting the angles.

I plugged your formula into an Excel spreadsheet (which uses radians for angles, which then have to be converted degrees), and this is what I got:

acosd(sind(d)*cosd(eps)-sind(a)*cosd(d)*sind(eps))

where
a = 192.858 degrees (3.366007089 radians)
d = 27.1283 degrees (0.473478155 radians)
eps = 23.4393 degrees (0.409092959 radians )

= 1.085552547 radians = 62.1976 degrees

instead of 60.1889 degrees, which is odd. Maybe I got something wrong, or maybe an error crept in through all the conversions.

Another calculation I've seen for the inclination between the ecliptic and the galactic plane is this:

= arccos (sin (delta1) * sin (delta2) + cos (delta1) * cos (delta2) * cos (alpha1-alpha2))

where
alpha1 = 270 degrees
delta1 = 66.5607 degrees

alpha2 = 192.8 degrees
delta2 = 27.1 degrees

Plug these values into the equation and you get 60.1891 degrees (very close to your estimate).

I found this calculation at http://answers.yahoo.com/question/in...3110530AAMkEYV

This whole quest began when I first wondered how we on Earth are oriented relative to the Galactic Equator. A simple estimate is the declination of Polaris above the GE (27.4 degrees), but living in the city, the Milky Way is not visible, so it's not possible to just look up and see it. Instead, one has to resort to visualizations and calculations. I've been chewing on this question for a long time, and I'm just beginning to wrap my thoughts around it.
D H
#8
Jun17-14, 01:30 PM
Mentor
P: 15,066
Quote Quote by fizixfan View Post
I plugged your formula into an Excel spreadsheet (which uses radians for angles, which then have to be converted degrees), and this is what I got:

acosd(sind(d)*cosd(eps)-sind(a)*cosd(d)*sind(eps))

where
a = 192.858 degrees (3.366007089 radians)
d = 27.1283 degrees (0.473478155 radians)
eps = 23.4393 degrees (0.409092959 radians )

= 1.085552547 radians = 62.1976 degrees

instead of 60.1889 degrees, which is odd. Maybe I got something wrong, or maybe an error crept in through all the conversions.
I can't recreate your error, even if I use your numbers:
http://www.wolframalpha.com/input/?i...09092959%29%29

Note that I did not input 192.858 degrees and such into wolfram alpha. My input was
acosd(sind(d)*cosd(eps)-sind(a)*cosd(d)*sind(eps))
where a=(12+(51+26.00/60)/60)*15, d=27+(7+42.00/60)/60, eps=23+(26+21.406/60)/60
In other words, I used a=12h 51m 26.00s, d=27 7' 42.00", and eps=23 26' 21.406"


Another calculation I've seen for the inclination between the ecliptic and the galactic plane is this:

= arccos (sin (delta1) * sin (delta2) + cos (delta1) * cos (delta2) * cos (alpha1-alpha2))

where
alpha1 = 270 degrees
delta1 = 66.5607 degrees

alpha2 = 192.8 degrees
delta2 = 27.1 degrees

Plug these values into the equation and you get 60.1891 degrees (very close to your estimate).
That's the same formula I used, just with slightly different numerical values and the declination of the north ecliptic pole instead of the obliquity of the ecliptic. The declination of the north ecliptic pole and the obliquity of the ecliptic add to 90 degrees, by definition. That means your sin(delta1) and my cos(eps) are one and the same, and similarly for your cos(delta1) and my sin(eps). Finally, cos(270-alpha2) reduces to -sin(alpha2). That formula is essentially the same as mine.
fizixfan
#9
Jun17-14, 07:06 PM
P: 27
Quote Quote by D H View Post
I can't recreate your error, even if I use your numbers:
http://www.wolframalpha.com/input/?i...09092959%29%29

Note that I did not input 192.858 degrees and such into wolfram alpha. My input was
acosd(sind(d)*cosd(eps)-sind(a)*cosd(d)*sind(eps))
where a=(12+(51+26.00/60)/60)*15, d=27+(7+42.00/60)/60, eps=23+(26+21.406/60)/60
In other words, I used a=12h 51m 26.00s, d=27 7' 42.00", and eps=23 26' 21.406"



That's the same formula I used, just with slightly different numerical values and the declination of the north ecliptic pole instead of the obliquity of the ecliptic. The declination of the north ecliptic pole and the obliquity of the ecliptic add to 90 degrees, by definition. That means your sin(delta1) and my cos(eps) are one and the same, and similarly for your cos(delta1) and my sin(eps). Finally, cos(270-alpha2) reduces to -sin(alpha2). That formula is essentially the same as mine.
Got it! I used the good old Excel spreadsheet, and found out how to convert radians to degrees.

This is the formula I used :

=DEGREES(ACOS(SIN(d)*COS(eps)-SIN(a)*COS(d)*SIN(eps)))
[noting that Excel will only let you use radians], and where

a = 192.8583333 degrees = 3.3660129065754 radians
d = 27.12833333 degrees = 0.473478737245195 radians
eps = 23.43927944 degrees = 0.409092600600583 radians

The result is 60.18894305 degrees (1.05049523 radians). Thanks for your help, this is great. I'm finally getting somewhere after all these years.

Now if I could only figure out how to back-calculate these numbers so the result would be "d"... is it possible to do this?


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