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Hubble's constant and decelaration parameter

  1. Jan 13, 2010 #1
    1. The problem statement, all variables and given/known data
    I've been measuring hubble's constant by calcultaing the redshift on a star and it's distance from earth using a simple method of assuming galaxies have a standard size, then measuring their apparent size to gauge their distance away.

    To refine the model i have been asked to include a deceleration parameter q0. I have found an equation relating q0 to Hubble's constant but i'm confused by one of the terms, the 'angular diameter distance'. What is it and what is it actually measuring?


    2. Relevant equations
    [tex]d_a= \frac{c}{q_0h_0} \frac{zq_0+(q_0-1)(\sqrt{(2zq_0+1)}-1)}{(1+z)^2}[/tex]

    3. The attempt at a solution
    it tells me to use q0=0.1 or 0.5 or 0.8 and plot da against z and check which value of H0 fits the data. q0 seems to just be an arbitrary number? is it just an expansion coefficient for the universe expansion?
    Now i can see that as i increase q i increase the "bend" of the line produced, but as i change H0 i just change the gradient, which i'm not sure really helps me at all?

    questions in summary:
    What is "angular diameter distance"?
    surely there is only one value for H0 so how do i "see which fits the data"?
    what is q0? it seems to just be arbitrary?

    Sorry for a lot of confusion, but i've not taken any astronomy before but have changed courses and have been set this as "catch-up" over, i also have no astronomy textbooks since i hadn't been doing it until recently.
     
  2. jcsd
  3. Jan 13, 2010 #2
    The angular diameter distance is a unit of distance used often in astronomy. Wikipedia has a good article explaining it further, but it basically is that

    [tex]
    d_A=\frac{x}{\theta}
    [/tex]

    where [itex]d_A[/itex] is the angular diameter distance, [itex]x[/itex] is the actual size of the of the object in question and [itex]\theta[/itex] is the angular size of the object.
     
  4. Jan 14, 2010 #3
    Aaaah i get it now

    Since i have already worked out x from measurement i am in fact using that as the relation to z.

    I think my problem previously wad that i was using the value i had calculated for Ho to work out dA rather than the other way about. You're explanation was very simple but it made it click , thanks :D
     
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