Re: Galaxies without dark matter halos?

greywolf42 <[EMAIL PROTECTED]> wrote:
: Differentiating between observations and theoretical calculation is never
: "quibbling."  In this case, I was alluding to the fact that we do not get
: orbital speeds from astrometry (observation) until we include a theoretical
: calculation of distance (calculation).

Unfortunately, that has nothing to do with our discussion.  I never
used the word "speed" nor even "velocity," except in the phrase
"angular velocity" or in responding to your other irrelevancies.  We
do NOT need to know the distance to determine the angular velocity.

: Your statement, to which I was replying was: "What's observed
: astrometrically is the projected angular velocity."   We actually observe an
: angular velocity (not projected) on the celestial sphere.  WE then include
: the assumption or calculation of distance, to convert angular motions into
: projected angular velocity -- of the orbit.  I assumed the last three words,
: since we were discussing determinations of orbital parameters.

Wrong again.  The distance is irrelevant.  You can change the distance
all you like, but the apparent (or the true) angular velocity of one
star about the other remains the same.

: The key here is that we can't determine the true *orbital* angular velocity
: from observation (which I thought was your original claim).  Because we
: don't know the inclination of the orbit.

Wrong again.  We deduce the inclination from the astrometry, along
with the rest of the orbital parameters, as I've said over and over.

: You've provided no explanations, but only unsupported claims.  I belive you
: are incorrect.  However, I'll be happy to listen to an explanation.  How
: does the astrometry give you the orbital inclination?  Give equations,
: please.

Equations:

        M = nt
        M = E - e sinE

        x1 = a (cosE - e)
        x2 = a sqrt(1-e^2) sinE
        x3 = 0

where n=mean motion, t=time from periapse, M = mean anomaly,
e=eccentricity, E=eccentric anomaly, a=semimajor axis.

        Y = B X

where X is the vector (x1,x2,x3) and B is the rotation matrix composed
of the node, inclination, and periapse.  With astrometric data, you
have observations of y1 and y2 as a function of time.  Do a weighted
linearized least-squares fit to the data, and iterate to convergence.
>>From this solution, you get the six elliptic orbital elements.  If you
happen to know the distance to the binary, you can express "a" in
distance units; otherwise you have to use units of angle on the sky.
As I mentioned before, the system of equations does have an ambiguity
in the sign of the inclination, but the fit converges to whichever
solution is "closer" to the starting conditions.

: The same equation applies to astrometry (transverse velocity) as for radial
: velocity.

Wrong again.  The equation you copied is only for one-dimensional
data.  Astrometry gives two dimensions at the same time.  That makes a
crucial difference.  With radial velocity, the one-dimensional
observable scales equally (and, thus, degenerately) with "a" and "i".
With astrometry, one coordinate scales with "i", but the other (along
the line of nodes) is independent of "i".

-- 
John F. Chandler             [EMAIL PROTECTED]