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Hubble Velocity/Acceleration?

  1. May 12, 2014 #1
    Hubble’s law tells us that the velocity of distant objects increases with distance as v = Hr. Does this mean that the farther a distant object gets the more rapidly that object will recede away from us? That is, are these distant objects accelerating away from us? Can we say
    v = Hr, and so Acceleration = dv/dt =H dr/dt + r dH/dt = Hv = (H^2)r for constant H? Has such an acceleration been observed/measured?

  2. jcsd
  3. May 12, 2014 #2


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    H(t) has been declining since the first few seconds of expansion (as far as we know) and according to best fit model matching observations (LambdaCDM, the standard one people use) has been declining for past billions of years and WILL CONTINUE to decline *but at a slowing rate*. This slowing of the rate that H(t) is declining corresponds to what they call "acceleration" or "dark energy effect"or "cosmological constant effect".

    It is not expected to make H(t) ever increase, only slow decline and gradually makeH(t) essentially level outy to be constant. Then your equation is right! Constant H(t) would lead to exponential expansion. i.e. fixed percentage growth

    The projected percentage growth rate of large-scale distances (not expected to affect galaxy scale distances) according to LCDM model is 1/173 of one percent per million years.

    You might take a look at the tabular calculator that one of PF members made that shows past and future history. It will show the change o over time of 1/H or c/H (which grows as H decreases)

    It is "Lightcone" calculator link in my sig. click on it, you can adjust the parameters at will.

    You can get it to show the acceleration in distance growth that started around year 7 billion. but that takes a little work, so ask questions if you want on how to use it.
  4. May 14, 2014 #3
    Thanks for the info. It will take a while to fully digest...but the fact the distant objects are moving from us at v=Hr and ACCELERATING away from us as/if H(t) is constant is very interesting.
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