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Something about calculating the Age of the Universe

  1. Apr 7, 2015 #1
    Good day all,


    First of all, I want to let you guys know that I'm quite new to the subject so please bear with me in case I'm asking a very basic question here. I have a struggle with the way the Age of the Universe is calculated by using Hubble's Law: V = H x D and I hope you guys could clear this up for me. Please note that my question is based on the following YouTube video where the Age of the Universe is being calculated at 5:18 : https://www.youtube.com/watch?v=pSqJD6KF0Rw

    Now, in the video he says that H has a constant of approximately 75 km/s/Mp (values may vary a bit depending on the source). I'm seeing this constant as an acceleration. An object that is within a Mp will move at 75km/s away. By the time that same object passes a distance of 1 Mp, it will move away at 150km/s. So for every Mp it reaches, it will go faster. This can be rewritten to calculate how much faster it goes for every meter which is 75000 m / 3.08567758 x 10^22 (1 Mp) = 2.43 x 10^-18 m/s/m. From this new information I can conclude that the object doesn't have a constant velocity of 75km/s over the whole Mp but that the velocity gradually increases over that Mp. By the time the object reaches the very last meter of the Mp, only then will it have reached a velocity of 75 km/s (in case that object started with velocity 0 that is)

    The guy in the mentioned video calculates the Age of the Universe by giving an example of an object being at distance D1 and has a velocity of V1. From that you can say that time T = D1 / V1. Since V1 could be rewritten as the Hubble's formula (H x D1), the formula is then T = D1 / (H x D1), the D1's then cancel each other which shows that T = 1 / H. From that he then calculates 1 / (2.43 x 10^-18 m/s/m) which then gives the Age of the Universe.

    Now, here's the thing I can't seem to wrap my head around. This Age calculation is based on a fact that the object had a constant velocity of V1. However, I just concluded that V1 differs for every meter an object goes, since it accelerates at 2.43 x 10^-18m/s for every meter it travels. When the guy rewrote V1 as H x D1, it will only give the velocity that the object had when it had that very distance of D1. But before it reached D1, it must have had different velocities since the velocity depends on the distance. Therefore, the time it needed for it to reach D1 would be longer (since it had smaller velocities before it reached D1). The guy in the video would therefore have underestimated the Age of the Universe.

    The thing is, other video explanations -even a lecture from Yale University on this- that I found basically show the same calculation. That’s when I thought I must be missing something here but I just can’t seem to notice what. Am I doing something wrong here? If so, what exactly?
     
    Last edited: Apr 7, 2015
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  3. Apr 7, 2015 #2

    marcus

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    Please give the link to the Yale lecture
    The one you linked to contains a bad error which I can't believe would also be in a Yale astrophysics lecture. I'd have to see the Yale lecture.

    the thing you linked to is a mixture of good and bad. He presents the "ant on the balloon" analogy, which is OK. The ant only understands 2d and his whole existence is concentrated on this 2d surface. It is analogous to our situation in 3d.

    That can be helpful. He also explains that distances between bound objects (held together by atomic, molecular, gravitational forces) do not expand. Only large-scale distances like between unrelated clusters of galaxies. That's good to point out.

    It's a mixture of good and bad, pedagogically.

    He assumes that H does not change. (wrong) and he confuses the age of the universe with a different quantity called the "Hubble time" which is 1/H (very wrong).

    Basically one has to only watch the first half. The last half where he does that phony "age of universe" thing one should never have watched and one should just try to forget everything he said.
     
    Last edited: Apr 7, 2015
  4. Apr 7, 2015 #3

    Bandersnatch

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    Hi JohnnyGui, welcome to PF!

    First of all, I feel compelled to add a caveat that I'm myself just a student of cosmology struggling with the subject, so I claim no expertise.


    As the author himself admits in the comments section, the calculations he makes are intended as a 'broad approximation' that is supposed to give order of magnitude results.

    However, this particular bit appears to be plain incorrect. He either uses it with faulty understanding, or makes use of the fact that the calculation gives an answer that is close to the actual answer - despite this being just a coincidence.

    ##1/H_0## that you see in the video is the Hubble radius [edit: sorry, that should be Hubble time - see my post below; treat every further instance of 1/H as if it were c/H] - it's the proper distance at a given time at which objects recede with the speed of light. It uses the Hubble constant, ##H_0##. The constant is 'constant' only in the sense of being the same everywhere (so, constant spatially) but it evolves with time.
    The current value of about 70km/s/Mpc nets the Hubble radius of about 14 Gly. If you were to ask how long it takes for light to travel that distance you'd get the 14 Gy that is about as much as the age of the universe.

    However, and as you've seen yourself, this cannot net you the age of the universe. This is easy to see if you consider the future value of the Hubble constant - which will asymptotically approach something like 63 km/s/Mpc (from memory, so it might be off a few km/s). A 100 billion years for now, the ##1/H_0## calculation would net you about 17 Gly, and the 'age of the universe' that you'd get from this result would be grossly incongruent with the actual age, and further increasing with age.
    A similar disparity emerges if you go back in time.
    That it currently nets about the right age is a coincidence related to how the rate of the expansion changed recently from decelerating to accelerating.

    To correctly calculate time, you need to use the Hubble parameter, which you can find defined e.g. here:
    http://cosmocalc.wikidot.com/advanced-user
    The calculation is nowhere near as simplistic as what you saw in that video, so I'm inclined to give the author the benefit of the doubt and suggest he sacrificed truthfulness for broad appeal. Although in this particular case I'm having a hard time justifying such 'dumbing down'.
     
    Last edited: Apr 7, 2015
  5. Apr 7, 2015 #4

    Bandersnatch

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    Hi marcus, I see you edited your post which made most of mine redundant.

    Here's what I've found after a quick search:
    http://teachers.yale.edu/curriculum/viewer/initiative_05.04.04_u
    While it's not a lecture per se, it is endorsed by Yale and it does have the same calculation as in the OP, which makes me worried.

    Just to make this clear, it should be 'Hubble time' as per marcus' post. Hubble radius is ##c/H_0##, but in a certain set of units where c=1 it reduces to ##1/H_0##, which is how it sneaked in there - sorry for the confusion.
     
  6. Apr 7, 2015 #5

    marcus

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    Sorry about that. I'm a slow typer and I didn't realize you had posted when I decided to add some more explanation to mine. If I had realized you were on to it, I would have just gone away and let you handle it.

    the problem here is A. there is this totally wrong way to infer age of U which by coincidence DOES give something that is roughly right, so it is very tempting to use the nonsense method. As you know, Hubble time is about 14.4 billion y and age is about 13.8 billion y.

    B. and people want to know the age of U, they are naturally curious. So if we debunk the bad estimate and take that tempting bit of hokum away then we should provide a replacement. It does not feel right to take something away and not give something else to take its place. But I don't know an easy derivation of U-age that does not involve some calculus. Do you? Is there a nice way to explain, and get 13.8?
     
  7. Apr 7, 2015 #6

    Bandersnatch

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    No I don't. I don't think there is one, but I'd be glad to be proven wrong.
     
  8. Apr 7, 2015 #7
    Thanks guys for your answers. Please bear in mind that I'm no cosmologist nor a cosmology student but just someone who's very interested in cosmology so there's a chance I might not understand everything you guys say ;P

    Here it is, you can see him calculating the age of U using the same calculation at 32:20 (he will start with an example first) : https://www.youtube.com/watch?v=E9bjdR_k5qg
    A side note: He'll make an error during the calculation saying that V = H / D but one of his students will correct this.

    So if I understand correctly, 1/H is the time it took at "distance over speed of light" time ago (D / c)? In other words, if an object is calculated to have a distance of 3 light years, then its calculated velocity and the Hubble constant are values that were 3 years ago?

    I didn't know that the Hubble constant indeed changes over time! Could you please explain to me how you calculated the age of the Universe by using this 70km/s/Mp and getting 14 Gly since I can't see the association there?

    The thing is, even if the Hubble's constant doesn't change over time, the calculation in the video would still be wrong. Here's an example I have made myself. Please bear in mind that my calculation could be totally wrong so correct me if it is. I'm really not sure if I'm calculating this the correct way:

    Say the Hubble's constant H is 0,5 m/s per meter.
    Say that we measure the distance of an object 4 seconds after its "Big Bang" starting with a velocity of 0.5m/s

    To calculate its distance one could do the following:

    - If 1 second passes (t=1) the object would have traveled 1 x 0.5 = 0.5 meters

    - If 2 seconds (t=2) pass the object would have traveled 2 x 0.5 = 1 meter
    From that distance and time point the object would now accelerate to a velocity of 1.0m/s since it has now reached a whole meter and it accelerates every meter with 0.5m/s extra.

    - If 3 seconds pass (t=3) the object would now have a distance of 1 meter (the already traveled distance) + 1 meter (which is traveld by the new velocity 1m/s) = 2 meters
    From that point and time the object would now accelerate to a velocity of 1.5m/s since it has yet again reached an extra meter (it accelerates every meter with 0.5m/s)

    - Here’s the tricky part (at least for me). How much will it travel from t=3 to t=4 with a velocity of 1.5m/s and an acceleration of 0.5m/s/m?:
    To travel from 2 meters (at which the object is now) to 3 meters at 1.5m/s, the object would need 2/3 of a second. So it would have traveled a total distance of 3 meters after 3.66667 seconds. After reaching this 3 meter it would accelerate to 1.5m/s + 0.5m/s = 2.0m/s since it has reached yet again another whole meter.

    - There is still a third of a second left to reach t=4 now (4 – 3.66667 seconds). The object with a velocity of 2.0m/s would travel in a third of a second a distance of 0.33333 x 2.0m/s = 0.666667 meters.

    So after 4 seconds, the object would have traveled a total of 3.66667 meters.

    If we now fill this in the Hubble formula, H being 0.5m/s/m and D being 3.66667m, it would give a total different time other than 4 seconds:
    D / (D x H) = (1 / H) = 1 / 0.5 = 2 seconds instead of 4. This is what I meant with underestimating the age of the object since it calculates the time using a velocity that was only that velocity when the object was in the very last moment at a distance of 3.66667 meters. I hope I'm making sense here (which I highly doubt :P)
     
    Last edited: Apr 7, 2015
  9. Apr 7, 2015 #8

    marcus

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    Well we can go thru the harder (some elementary calculus) treatment and maybe that will help us think up an easier way to say it.
    Everything I say right now is tentative, just don't want to have to keep repeating IMHO and "I think" and "Maybe" all the time.

    The easiest way to think of the Hubble growth rate is as a percent per billion years (how much the distance would grow in a billion years if the rate stayed the same all that time). You know the present rate is about 7% per billion years and it is declining towards a steady 6%.
    More precisely the numbers are 1/14.4 ≈ 0.07 and 1/17.3 ≈ 0.06.

    Astronomers have studied millions of data points and have learned somewhat about the HISTORY of how that growth rate has changed over time and we can PLOT the approximate curve showing what they have deduced about its change over time. It is basically the hyperbolic cotangent curve
    coth(1.5x) but to make small numbers I will measure time in billions of years and divide everything by 17.3. so x = time/17.3 billion years
    and I am plotting coth(1.5x)/17.3. In this picture the point x = 1 on the x axis should really be labeled "17.3 billion years".

    This curve shows how the rate tails out to a longterm level value of 0.06
    4aprilcoth.png
    So to take a bizarre or comic example, suppose that tomorrow we receive a radio message from some aliens and they tell us they measure time with a Cesium clock like we do and their year is the same number of seconds as ours and they say " we measured the large distance expansion rate and it is 40% per billion years but we don't know the age of the universe. " Then we can look at the curve and tell that they were living when the universe was 1.73 billion years old. That is the x=0.1 point. And that was the age of the universe back then, when they measured the expansion rate, and sent the message.

    And then the next day, which was Thursday, we got a message from some other aliens saying they have measured the expansion rate and it is 20% per billion years (by coincidence they use the same atomic clock second and length of year). Then we can look at the curve and deduce that they measured the expansion rate and sent the message at the x=0.2 point, when the universe was 3.46 billion years old.

    That is not a solution to the problem so much as it is CONTEXT for understanding. We know that the expansion rate, as a percent growth per billion years, changes over time. And we know that the LONGTERM rate tends toward something definite, namely about 0.06 per billion years. But I didn't say yet how we could deduce the age of U for us, at this particular point in time. I'm dodging the hard issues and just giving context.

    It's frustrating. It's about fitting a differential equation model to data. It's about fitting curves to data and checking that a bunch of different things are all consistent with the same picture. I'm basically stumped. What's a simple story of how we know the age of the U? George Jones sometimes answers questions like this.
     
    Last edited: Apr 7, 2015
  10. Apr 7, 2015 #9

    marcus

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    I know a way to quickly get the age, but I can't explain in a simple intuitive way how it works. I'm sure someone else can. So here is an UNINTUITIVE way to do it.
    We can measure the rate distances are now growing and we can estimate the longterm rate (in neither case do we need to first know the age of U, we can measure those rates independently of knowing the age).

    We find the rate now is 1/14.4 per billion y, and the longterm rate is 1/17.3 per billion years.
    What we want is the fraction which is the SUM OVER THE DIFFERENCE of these two rates.

    To prepare, let's first try for a bit of accuracy in the percentages. It's not 6% it is 1/17.3 = 0.05780... = 5.78%
    It's not 7%, it is 1/14.4 = 0.06944...= 6.944%

    Now let's take the sum over the difference of these growth rates.

    We make the fraction $$\frac{6.944+5.78}{6.944 - 5.78}$$
    and it comes out 10.931...
    and we take the CUBE ROOT of that and it comes out 2.2193... or about 2.22
    and we take the natural LOGARITHM of that and it comes out 0.7972... or about 0.8

    and that is the age on the time scale where you divide everything by 17.3 billion years, so to get the answer in familiar units you multiply that 0.797 or that 0.8 by 17.3 billion years, and you get the usual answer for the age: 13.8 billion years.

    If we were to do it with the two rough approximations for the growth rates, 7% and 6% the answer would be off but it would show the general idea:

    log of cube root of (7+6)/(7-6)

    A quicker way would be to take log of cube root of (173 + 144)/(173 - 144). Not bothering to calculate percentages just take the sum and difference of the numbers we started with.
     
    Last edited: Apr 7, 2015
  11. Apr 7, 2015 #10

    wabbit

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    But is the Hubble time such an awful first shot ? If we want a quick first estimate at an order of magnitude, a constant expansion at H0 might not be so bad, even tough it's really awful in detail, as for instance OP's objections show.

    What I'm wondering is, might there be an expression for the age, of which 1/H0 would be a kind of 0th order approximation ? Or is it really just sheer coincidence as others have said ? (I don t doubt that getting two significant digits is coincidence - but one ? )

    What's true in any case, is that the coincidence works against this idea - if we got say 8 bn yr this way, the message "actually closer to 14 but you need to work a bit more for that" would be fine. Here the correction is by chance so small it's misleading as to the status of the rough first estimate.
     
    Last edited: Apr 7, 2015
  12. Apr 7, 2015 #11
    I have one other question to make myself understand this. Is the whole approach of that guy calculating the Age of the Universie wrong because of the counter arguments I gave in my OP? Or is there something else which makes it wrong?
     
  13. Apr 7, 2015 #12

    marcus

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    Assuming there are still observant monkeys living in the solar system in year 17.3 billion, they will be AMAZED that the Hubble time 1/H at that era is almost EXACTLY equal to the age of the age of the universe!!!
    But then the universe will be getting older and older. and the Hubble time, by contrast, will stay essentially the same---an ever lasting 17.3 billion years.

    Because the rate of distance growth has leveled out at 1/17.3 = 0.0578 ≈ 0.06 = 6% per billion years
    (that is what the Hubble time is, by definition, the reciprocal of the growth rate)
    You can see it leveled out at that approximate 6% on the graph of the curve.

    Time keeps going up, the Hubble time 1/T levels off and stays ever the same

    So if those monkeys thought there was a hard and fast connection, because of the remarkable coincidence at around year 17.3 billion, they would eventually be surprised to discover there wasn't one.

    Wait till the age of the universe is 50 billion years and the Hubble time is still 17.3 billion, they'll be really puzzled and scratching their bewildered monkey heads. :oldbiggrin:

    If there is an approximation rule it would have to work in year 50 billion as well as now in year 13.8 billion.

    Mr. Johnny, you might enjoy getting acquainted with the Lightcone calculator that was developed by one of the PF members here and which makes histories of the universe (where you specify the limits) either in the form of tables or of graphs of curves.

    I have link to Lightcone in my signature. If you want I can describe how to use it, how to add more steps, how to narrow down the range of sizes, etc. Or you can just click on the link and discover how to use it
    It has expansion "speeds" labeled v, and real time labeled T, and Hubble radius R, which you can read as Hubble time in billions of years, and expansion factors from some point in the past, labeled S.
    It runs on the equations of the standard cosmic model that cosmologists use. Check it out :oldsmile:
     
    Last edited: Apr 7, 2015
  14. Apr 7, 2015 #13

    George Jones

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    It corresponds to a two-term Taylor expansion.

    Taylor expanding the scale factor gives

    $$\begin{align}
    a\left( t \right) &= a\left( t_0 \right) + \left( t - t_0 \right) \dot{a} \left( t_0 \right) + \ldots \\
    &= a\left( t_0 \right) + \left( t - t_0 \right) a \left( t_0 \right) H \left( t_0 \right) + \ldots . \\
    \end{align}$$

    Lopping off the ... (i.e., taking all the higher-order derivatives of the scale factor to be small), and taking ##t = 0## so that ##a\left( t \right) =0 ## gives

    $$ 0 = a\left( t_0 \right) - t_0 a \left( t_0 \right) H \left( t_0 \right) .$$

    Rearranging gives the result.
     
  15. Apr 7, 2015 #14

    marcus

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    Hi George, I was hoping you'd comment : ^)

    How would you explain to a beginner how to estimate the age? From things we can actually observe.
    Since the scale factor a(t) is accelerating, maybe it is not adequate to lop off the second derivative along with the higher order terms.
    Instead of a linear or first-order approximation, what would happen if you include second order?

    It seems to me that the method you just showed us only works if the truncated Taylor expansion is good all the way back to the start of expansion where we have t=0 and you use the fact that a(t=0) = 0
     
    Last edited: Apr 7, 2015
  16. Apr 7, 2015 #15

    wabbit

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    @marcus, did you just call me a monkey ?:biggrin:
     
  17. Apr 7, 2015 #16

    marcus

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    We are all, including Mozart, Gauss, and Riemann, evolved monkeys and we must try to be an honor to the universe
     
  18. Apr 7, 2015 #17

    wabbit

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    @George, thanks, that's interesting too because it starts to hint at further corrections, and evaluating the next term would I expect reveal the coincidence (2nd order might be much worse than first).
     
    Last edited: Apr 7, 2015
  19. Apr 7, 2015 #18

    marcus

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    Wabbit, any rule to calculate the age must not depend on knowing the age. That is, it must work at any era of universe history. If it does not work in year 50 billion but only works if applied around the present, then wouldn't you consider it spurious?
     
  20. Apr 7, 2015 #19

    wabbit

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    Yes and no... Sure a rule that would be universally valid would be better, but even one which is valid under some assumptions can be a start, especially if you then look at why it s wrong etc..

    In a de Sitter era this gives you a lower bound. But say we are pre-1998. We would use something still, not sure what was the estimate then - and that something wouldn't involve eventual exponential growth. Not good but still not worthless.

    There is one difficulty with exponential growth (the proper first approximation for our future monkeys), which is that is doesn't extrapolate back to zero, so the age you get is fully dependent on the scale you pick to start. Isn't then the result heavily dependents on how much (early) inflation you put in ? Maybe I m a little confused here...
     
    Last edited: Apr 7, 2015
  21. Apr 7, 2015 #20

    George Jones

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    Yes, could well be.

    I have some hard deadlines for work tomorrow, so I don't know if I can follow through, but here are a couple of suggestions (which might not lead anywhere).

    1) Adding another term leads to a quadratic for t_0.

    2) Use realistic numbers in the matter-##\Lambda## reasonable approximation for the universe (in another thread),

    $$a\left(t\right) = A \sinh^{\frac{2}{3}} \left( Bt \right) $$

    to look at the relative sizes of the terms.

    According to my notes, in SI units, ##B## is something like ##3 \times 10^{-18} \rm{s}^{-1}##, and each differentiation brings a factor of this small B outside.
     
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