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Hubble's law and the age of the universe

  1. Sep 4, 2012 #1
    Hubble's law states that the speed of a distant object is proportional to it's distance (this implies an expanding universe). By finding the distance and speed (via redshift) of various distant galaxies you can produce a graph displaying the linear relationship. The gradient of the line is Hubble's constant.
    In school I've been told that Hubble's constant is the inverse of the age of the universe, this doesn't make sense to me, because in order to arrive at that solution you must first assume that the velocity of a moving body is constant, and that defies the entire point of Hubble's law. What sounds better to me is something along these lines:
    If Velocity = Hubble's constant × Displacement, and a galaxy 470 Mpc (1 Mpc = 3 × 1019 km) away is travelling at 30000 km/s, what's the age of the universe.
    Now this doesn't really seem that complicated at first glance, just find the time it takes for a body who's velocity is proportional to its displacement to reach its current position from originally being at the same spot as us (for that means that everything was at the same spot, ie. The Big Bang)
    So, using calculus which is at the greatest extent of my yr 12 knowledge I came up with this little differential equation:
    [tex]V = Hs \\ \frac{ds}{dt} = Hs \\ \int \frac{1}{s} \ ds = \int H \ dt \\ \ln|s| + c = Ht \\ t = \frac{\ln|s| + c}{H}[/tex]
    The problem is: 1. this doesn't actually answer the question, 2. I have no idea what I've actually done; integrating has something to do with area under the curve, but which curve? I started with a straight line graph and ended up with a differential equation. By substituting in H and s you cannot get a value because of that constant sitting in there. So I was wondering if anyone here knows how to get an answer to this problem. I dunno, can you integrate definitely? Can you use common sense or context to eliminate the constant? Is there a better way of going about it? Am I completely and utterly wrong?
     
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  3. Sep 4, 2012 #2

    marcus

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    Good you are starting to work with differential equations! Keep this up! They are the backbone of understanding all processes in nature.

    But it may be premature for you to be applying them to Hubble law expansion of distances.

    Much of the confusion is caused by misleading language---terminology that got established early in the history of the field.

    Galaxies do not "travel" except at relatively slow speeds which in most cases can be neglected. Traveling means going somewhere. They mostly sit still and the distances between them increase. There is no speed limit on that kind of distance increase.

    You can have a pair of galaxies which are both at rest with respect to the Background of ancient light, and the expansion process itself, and the distance between them can be increasing at 2c or 3c. this is fairly normal. Most of the galaxies which we can observe with telescope are, in fact, receding faster than c in this sense. They are not moving but the distances to them are increasing.

    Hubble constant is not in fact constant but has changed enormously over the course of expansion. So you see how the (historically rooted) terminology can be misleading. It is better to tell yourself "Hubble growth rate, Hubble distance growth rate, Hubble distance growth rate..." over and over until you no longer think "constant" when you hear the word "Hubble".

    Hubble law is about a type of distance called "proper" distance which is a very intuitive natural idea of distance. It is what you would measure by any direct normal means like radar (or string or yardstick) if you could stop the expansion process at a particular moment in universe time and measure directly. "proper" is the distance at a particular moment.

    Hubble law is about the fractional (or percentage) rate that proper distances are growing at a particular moment.

    Currently they are growing at about 1/139 of one percent per million years.

    A more convenient handle on the rate of expansion is the RECIPROCAL of H(t) called the *Hubble time*

    It is currently 13.9 billion years. One percent of that (namely 139 million years) is the time it would take a typical distance to increase by one percent.

    Try typing this into google. Or simply paste it in:
    1/(70.4 km/s per Mpc)

    Sometimes then you then need to type a space or an equal sign to get it to calculate. Or it will give the answer as a suggestion. That is the reciprocal of the standard estimate of H(now). You should get 13.9 billion years. If it gives it to you in seconds then insist on years by typing or pasting in:

    1/(70.4 km/s per Mpc) in years

    GR is our theory of geometry (as well as gravity). It explains why under the present circumstances geometry is nearly Euclidean and why under other circumstances it is not Euclidean. Distances can grow, or shrink, triangles can add to more than 180 degrees, or less than 180 degrees,depending on the density of matter. And so on. Geometry is a part of nature, and interacts with the rest of nature. And geometry even has something which is a little bit analogous to "momentum". Once it has started changing in a certain way (e.g.expanding) it tends to persist doing that, although other things can gradually slow the change down.

    In recent years some proposals for a quantum version GR, and in one or more of the quantum GR theories, geometry can even "bounce". Geometry can contract down until density reaches a critical level and gravity becomes repellent and then it can rebound. So geometry is not a static stage, it is one of the actors in the play of nature. It interacts with matter. This was the basic message from GR, in 1915. Almost a hundred years now!
     
    Last edited: Sep 4, 2012
  4. Sep 4, 2012 #3

    Simon Bridge

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    Naively interpreting Hubbles law ... and keeping H a constant...

    IF speed of recession is proportional to the distance to the object.

    v=Hd

    Thus - [H] = (L/T)/L = 1/T
    ... i.e. Hubbles constant has dimensions of inverse time ... what time is this?
    The assertion is that it is the age of the Universe - so take the estimated age of the Universe (by some other method) and compare.


    As for the calculus...

    from v=Hd

    dv/dt=Hv

    solve the differential equation for v(t)

    v(t)=Ce^Ht

    Then [itex]d=\int_0^T v(t)dt[/itex] and solve for T ... this will give the time that two objects ended up a distance d apart by expansion alone - assuming they started out in the same place.

    The time they were in the same place is the start of the Universe.
    So time T is the age of the Universe...

    It is, however, more complicated than that.
     
  5. Sep 6, 2012 #4
    Thanks for the replies. So Hubble's constant is far from constant? That makes sense, cool. I was thinking a bit after making that little equation and thinking that it didn't seem to work because other objects further away would indicate on older universe than closer ones and very low displacements give a negative time value so that the universe would seem to have... done.... something. But yeah, thanks.
     
  6. Sep 6, 2012 #5
    If you choose two points in intergalactic space, you can measure the distance in between them and call it ##d_0##. Then, later at some time t, you can take a measurement of the distance in between the two points and call it ##d(t)##. Then $$d(t) = a(t)d_0$$ where ##a(t)## is the 'scale factor', which determines how large a region of the universe is today compared with an earlier time.

    So, you can see that an expanding universe has a growing scale factor, since distances are getting larger. In other words, the time derivative of the scale factor, ## \dot a(t)##, is positive.

    Hubble's constant is then defined as $$ \dot a(t) \over a(t)$$ So, as the universe gets larger, the scale factor increases. However, ##\dot a(t)## increases by only a very small amount due to the acceleration of the universe. So, Hubble's constant is decreasing with time.
     
  7. Sep 10, 2012 #6

    Simon Bridge

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    The weird results from the little equation you made are not due to the Hubble constant not being constant in time though.

    Remember, the "speed of separation" depends on the separation. Close objects got there in the same time as distant objects even in a time-constant model.
     
  8. Sep 10, 2012 #7

    marcus

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    Hi Orcinus, if you are still around. You really should go here
    http://www.einsteins-theory-of-relativity-4engineers.com/CosmoLean_A20.html
    and check this out.

    It shows you the history of the changing Hubble's constant.

    It shows the Hubble constant by way of the HUBBLE TIME which is its reciprocal 1/H.

    You know that google doubles as a calculator so if you pick some estimate of the current H, like 70.4 km/s per Mpc then to ask google for 1/H you simply type or paste this into the window:

    1/(70.4 km/s per Mpc)
    and it will tell you a time, either in years or seconds. I think it gives the answer in years.

    And you can take the current figure for the Hubble time of 13.9 billion years and get its reciprocal, just paste in:

    1/(13.9 billion years) in km/s per Mpc

    When you say IN to google calculator it tells the calculator that you want the answer to be IN terms of whatever unts you specify (like km/s per Mpc) and the calculator will in many cases be able to comply with that request. It does in this case.

    So Jorrie's A20 calculator that I gave you the link to is in effect telling you the history of the evolution of the Hubble growth rate. It certainly has not been a constant, as you can see from the A20 table.
     
    Last edited: Sep 10, 2012
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