<p>Yes, because the Milky Way galaxy and the Earth's equator are in different planes, currently about 63 degrees apart. To avoid confusion, rename the galactic Cartesian coordinates to (u,v,w). The transformation from (u,v,w) to (x,y,z) can be expressed as multiplication by a rotation matrix.</p>

CC BY-SA 3.0
2016-04-08T21:50:30.527Z
Okay yes labelling them both as x,y, and z is why I think I was so confused! Thanks a lot! Just another question, since I am working with galaxies in the Local Group with these proper motions, where in galactic coordinates the x direction had to be shifted by 8.5 kph to move to the Milky Way centre, how will the equatorial coordinates have to be altered?
R Thompson
CC BY-SA 3.0
2016-04-08T22:40:06.873Z
I guess you'd use the same rotation matrix to transform the galactic translation vector (8.5, 0, 0) to an equatorial translation vector. Good luck!
Mike G
CC BY-SA 3.0
2016-04-08T21:39:53.373Z
2016-04-08T21:39:53.373Z
Mike G
<p>That is expected as the x,y,z axes of the equatorial coordinate system don't point in the same direction as those of the galactic coordinate system. To compare you would have to change basis from equitorial to galactic  coordinates, which is effectively what NED does.</p>

CC BY-SA 3.0
2016-04-08T21:16:59.243Z
Thanks, so the conversion that NED is providing me with is correct?
R Thompson
CC BY-SA 3.0
2016-04-08T21:11:43.510Z
2016-04-08T21:11:43.510Z
James K
<p>I'm currently working with proper motions and have needed to convert my galactic longitude and latitude into right ascension and declination (ie from galactic to equatorial coordinates). This is fine since NED have a tool to do this, however once I convert into Cartesian coordinates I do not get the same results as I had for the galactic-cartesian conversion.</p>

<p>I understand that the conversion from galactic to cartesian is</p>

<p>x = Dcos(l)cos(b)</p>

<p>y = Dcos(b)sin(l)</p>

<p>z = Dsin(b)</p>

<p>where D is the distance to the galaxy in question in my case.</p>

<p>Also for the equatorial to cartesian transformation</p>

<p>x = Dcos(RA)cos(DEC)</p>

<p>y = Dcos(DEC)sin(RA)</p>

<p>z = Dsin(DEC)</p>

<p>I have done this for both my galactic coordinates and converted equatorial coordinates, but I do not get the same x,y and z. Should this be happening?</p>

<coordinate><positional-astronomy>
Question on Equatorial and Galactic coordinate systems
R Thompson
# Question on Equatorial and Galactic coordinate systems

(Asked by: username_0 on 2016-04-08)

I'm currently working with proper motions and have needed to convert my galactic longitude and latitude into right ascension and declination (ie from galactic to equatorial coordinates). This is fine since NED have a tool to do this, however once I convert into Cartesian coordinates I do not get the same results as I had for the galactic-cartesian conversion.

I understand that the conversion from galactic to cartesian is

x = Dcos(l)cos(b)

y = Dcos(b)sin(l)

z = Dsin(b)

where D is the distance to the galaxy in question in my case.

Also for the equatorial to cartesian transformation

x = Dcos(RA)cos(DEC)

y = Dcos(DEC)sin(RA)

z = Dsin(DEC)

I have done this for both my galactic coordinates and converted equatorial coordinates, but I do not get the same x,y and z. Should this be happening?

**Tags:** coordinate, positional-astronomy


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## 2 answers

### Answer by username_1 on 2016-04-08. Score: 1 | Accepted answer

Yes, because the Milky Way galaxy and the Earth's equator are in different planes, currently about 63 degrees apart. To avoid confusion, rename the galactic Cartesian coordinates to (u,v,w). The transformation from (u,v,w) to (x,y,z) can be expressed as multiplication by a rotation matrix.

> **Comments:**
> **username_0**: Okay yes labelling them both as x,y, and z is why I think I was so confused! Thanks a lot! Just another question, since I am working with galaxies in the Local Group with these proper motions, where in galactic coordinates the x direction had to be shifted by 8.5 kph to move to the Milky Way centre, how will the equatorial coordinates have to be altered?  
> **username_1**: I guess you'd use the same rotation matrix to transform the galactic translation vector (8.5, 0, 0) to an equatorial translation vector. Good luck!  

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### Answer by username_2 on 2016-04-08. Score: 0

That is expected as the x,y,z axes of the equatorial coordinate system don't point in the same direction as those of the galactic coordinate system. To compare you would have to change basis from equitorial to galactic coordinates, which is effectively what NED does.

> **Comments:**
> **username_0**: Thanks, so the conversion that NED is providing me with is correct?  

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