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View Full Version : Vectors of Canopus and Vega

Robert Tulip
2009-Jul-05, 01:53 AM
How do the direction and speed of proper motion of Canopus and Vega plot on to the following star maps?

Vega: http://commons.wikimedia.org/wiki/File:Precession_N.gif Vega is the large star at bottom and has proper motion (μ) RA: 201.03 mas/yr Dec.: 287.47 mas/yr

Canopus: http://commons.wikimedia.org/wiki/File:Precession_S.gif Canopus is the large star at upper left and has proper motion (μ) RA: 19.99 mas/yr Dec.: 23.67 mas/yr

Is this information sufficient to plot the vectors?

How do you calculate the direction and speed of right ascension and declination from these maps?

http://en.wikipedia.org/wiki/Vega#Kinematics states that the net proper motion of Vega is 327.78 mas/y, which results in angular movement of a degree every 11,000 years.

Do the numbers above indicate that the net proper motion of Canopus is about one tenth of that of Vega, resulting in angular movement for Canopus of a degree about every 125,000 years?

Just by Pythagorean formula, the figures for Vega give √(RAxRA + DecxDec) = 350 mas/y. Given the actual Vega figure of 327.78 this would result in a speed of 29 mas/y for Canopus, but I don’t know how to work it out properly or to plot the direction.

This is to assist in response to questions at ATM thread Canopus-Sol Relation.

Thanks.

tusenfem
2009-Jul-05, 09:53 AM
The proper motion of a star is determined through Doppler shift of lines in the spectrum of the star, and has to be determined for each star individually. You cannot just do some pythagorean math to get the proper motion of another star. The proper motion is the "true" motion of the star in space, where one takes out the effects of the orbit of the Earth and the rotation of the Earth etc etc.

Also, I haven't got a clue how you come up with that equation above, the square root of "Right Ascention" squared plus "Declination" squared? At least to have a result in arcseconds per year you should have a Δ or something in there.

Robert Tulip
2009-Jul-05, 10:46 AM
I haven't got a clue how you come up with that equation above, the square root of "Right Ascention" squared plus "Declination" squared? At least to have a result in arcseconds per year you should have a Δ or something in there.
Right ascension is movement across, and declination is movement up or down. If the heavens were a flat Euclidean sheet, the line of proper motion would be the hypotenuse of the triangle formed by the orthogonal lines of right ascension and declination, giving its length, the distance of proper motion, as the square root of the added squares of the other sides by geometry.

I assume it is the curvature of the celestial sphere that produces the 10% difference between the Pythagorean number and the actual result for Vega.

agingjb
2009-Jul-05, 11:01 AM
The celestial sphere is approximately flat at the scale of annual stellar proper motions.

But Right Ascension doesn't convert directly to degrees (it's like longitude and miles). Vega is about 40 N, so (I think) RA would need a factor of around 3/4.

(Not that I have a clue how the proper motions of Canopus and Vega relate to precession.)

tusenfem
2009-Jul-05, 04:03 PM
Okay, I think I made a mistake in writing about proper motion, it is the motion projected on the celestial sphere, with a nice descriptive image here (http://en.wikipedia.org/wiki/File:Proper_motion.JPG).

But you seem to mix up things. RA and Dec describe the location of a star on the celestial sphere with a descripitve image here (http://en.wikipedia.org/wiki/File:Equatorial_coordinates.png). So, to be clear, you have to at least talk about the RA-velocity or time-variation, in oder to get the proper motion of the star. Also, with RA and Dec being angles, you cannot just do a pythagorean addition. Naturally, when the proper motion is small, then sin(a) = a etc. and then you can come to some sort of equation like you wrote down.

But then, I don't understand how you want to translate that from Vega to Canopus.
But reading again your OP, it looks like you already have the proper motion of both stars, so on the celestial sphere you can easily plot those vectors. If you want then to compare the two, you just take a projection of the northern and southern polar region and plot the proper motion vector in RA-Dec space.

To use the images that you linked to, which are spherical coordinates in some way, you will have to figure out how the coordinate system of the figs compares to RA-Dec and then make the appropriate correction (which I cannot do at the moment here on the couch at home).