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The hypersine cosmic model

  1. Jun 20, 2015 #1

    marcus

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    This function $$\frac{e^x - e^{-x}}{2}$$ is called the hyperbolic sine. I'll refer to it as "hypersine" for short. You could say it "splits the difference" between the rising exponential function e^x and the exponential function run backwards, e^-x which slopes downwards---you take the difference between upwards and downwards sloping exponentials and divide by two.

    It's a nice function to get to know, if you aren't familiar with it already. It turns out that in our universe distances, areas, and volumes expand over time according to powers of hypersine.

    Distances grow according to the 2/3 power ##(\frac{e^x - e^{-x}}{2})^{2/3}##
    Areas grow according to the 4/3 power ##(\frac{e^x - e^{-x}}{2})^{4/3}##
    Volumes grow according to the square of the hypersine ##(\frac{e^x - e^{-x}}{2})^2##

    The hypersine has a nice symmetry which the ordinary exponential function ex does not have. If you flip it right to left, over the y-axis, and then flip it top to bottom over the x-axis, you wind up with the original function. It is the blue curve in this picture.
    sinh.png
     
    Last edited: Aug 11, 2015
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  3. Jun 20, 2015 #2

    marcus

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    One thing to get used to, with this model, is the natural time scale on which the hypersine tracks universe expansion.
    Instead of plugging the age t, measured on this scale, directly into the function you first multiply it by 3/2.
    So what goes in for x, in the above formulas, is 3/2 t. This may seem an unnecessary complication but right now I don't see any good way to avoid it. In our universe, cosmological distances grow according to
    $$\Big (\frac{e^{\frac{3}{2}t} - e^{-\frac{3}{2}t}}{2}\Big )^{2/3}$$
    On the natural time scale the present age is 0.8.
    And the changeover from a slight deceleration to a slight acceleration happened right around age 0.44
    Here is the raw distance growth function, you can see its present-day value is 1.3 (look up from .8 on the time axis)
    sinh^(2:3).png

    For many purposes it is convenient to DIVIDE BY 1.3 so that the distance growth function will be normalized to equal 1 at the present day. The normalized function is called the "scale factor" and denoted a(t). It tells us how big a distance is at some given time, compared with its present size. Here's the normalized version a(t). You can see that a(now) = a(.8) = 1.
    a(x)27Apr.png
    After a moment's inspection you can probably see the place around time 0.44 in our universe's history when distance growth stopped decelerating and gradually began to accelerate. The size of the normalized scale factor a(t) there is right about 0.6. When acceleration began, distances were about 60% of their present size.
    Please don't completely forget about the UNnormalized size function, before we divided it by its current value of 1.3 to force it to equal 1 at present. There is something appealing about letting the expansion history draw its own curve, so to speak. And it will turn up later. This happens in a formula where we actually have to multiply the normalized a(t) by 1.3 to undo the damage of divided by 1.3. When you see "1.3a" in a formula that is the raw unnormalized scale factor that was plotted first.
     
    Last edited: Jun 21, 2015
  4. Jun 20, 2015 #3

    marcus

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    Side comment: you can already calculate some things using just what we've talked about so far. How big were distances when the universe was a quarter of its present age?
    Well the present age is 0.8 so that just requires finding a(.2). The google window has a calculator function in addition to search, you just type stuff, in format suitable for the calculator, into the window and press "enter".
    There is a handy ABBREVIATION "sinh" for the hyperbolic sine, which makes this a snap. Remember to multiply the age .2 by 3/2. that gives .3, and then put this into google:
    sinh(.3)^(2/3)/1.3
    That will give the size of distances then, compared with their size now.
    The raw scale factor is sinh(.3)^(2/3) but we have to divide by 1.3 to normalize it and make the present value equal 1.
    When I put sinh(.3)^(2/3)/1.3 into the google window and press enter I get 0.35.
    Distances back then were 35% of their present size.

    Incidentally that means they have expanded by a factor of 1/.35 ≈ 2.9 since that time, and that expansion affected wavelengths of light too. So light emitted by a galaxy back then would, by the time it reaches us, have its lightwaves stretched out by the same factor 1/.35. the wavelengths would be almost 3 times longer, enlarged by a factor of 2.9. The convention in astronomy is to call that "redshift 1.9". Astronomers subtract 1 from the actual enlargement factor and call what they get "z." So the actual expansion factor (which works for both distances and wavelengths) is z+1 and that is also the reciprocal 1/a of the scale factor.
     
    Last edited: Jun 21, 2015
  5. Jun 20, 2015 #4

    marcus

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    I'm trying to think through a really basic concrete treatment of cosmology that goes through this "hypersine" picture of the actual expansion history.
    Using that "sinh" abbreviation the history is
    $$a(t) = \frac{\sinh^{2/3}(\frac{3}{2}t)}{1.3} $$
    What I'd like to do is introduce the derivative of a(t). The time rate of increase "da/dt" ...
    basically a number per unit time. We can visualize it as the slope of the a(t) curve.
    A common notation for the derivative is simply to put a prime or apostrophe on the function: a'(t).

    And the next step is the fractional change in a(t) per unit time. a'(t)/a(t), the change (per unit time) as a fraction of the whole--in other words the gain (per unit time) as a percentage of the present size.

    That instantaneous fraction rate of increase of a(t) is really important. That is actually what the Hubble rate is, H(t) which we see all the time is actually defined as a'(t)/a(t), the fractional rate of increase of the scale factor a(t) at a given moment in time.

    And since H(t) is a number per unit time (the ratio of the infinitesimal fractional increase per infinitesimal unit of time) its reciprocal is a time TH(t) = 1/H(t) called the Hubble time.

    The reciprocal TH is a convenient handle on H itself. Because in cosmology the fractional rate of increase of distance is so slow, so small---like the present H(now) is only 1/144 of a percent per million years---the reciprocal is large by human standards: TH(now) = 14.4 billion years.

    Maybe a picture would help. this time from Jorrie's Lightcone calculator. The blue curve is the scale factor a(t) which we have already seen. The present is 0.8, a(now) = a(.8) = 1.
    H(t), the gold curve, is very big at first, because the slope of a(t) is steep, it rises sharply at first.
    The red curve shows how the reciprocal of H(t) behaves.
    zeit20Jun.png
    To illustrate how H is the fractional growth rate of a(t), take for instance time t=1.1. I judge the slope of a(t) at that point to be 3/2. Between 1.0 and 1.2 it goes up 1 and 1/2 squares. So the slope a'(t) at time 1.1 is 1.5 and the height a(t) at that time is 1.4. The ratio a'/a is 1.5/1.4 ≈ 1.07 which agrees with H(1.1) as shown by the gold curve.
    You can also see how at time 0.6 the gold curve H(.6) is about 1.4 and the reciprocal of 1.4 is about 0.7, which is where the red curve is. On the other hand the current value of H is H(.8) = 1.2, and the reciprocal of that is about 0.8333..., shown by the red curve. The relations among three evolving cosmological quantities is shown visually.
    H and its reciprocal converge to 1 in the far future. This corresponds to our having taken H and unit growth rate and its reciprocal 1/H = 17.3 billion years as the unit of time. That is why the present age is given as 0.8.
    (This may be out of order but recall that H = a'/a so in the longterm when H≈ 1 we have a'(t) ≈ a(t). This is characteristic of exponential growth of the form a(t) = et.)
     
    Last edited: Jun 21, 2015
  6. Jun 21, 2015 #5

    marcus

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    I want now to say how the growth rate H arises from the UNnormalized scale factor 1.3a which we saw as a function of time a couple of posts ago.
    It's pretty amazing how simple the relation is. You recall from the graph that H is steadily decreasing and levels off at 1. While a increases with time. So H is going to have to depend on the RECIPROCAL ##\frac{1}{1.3a}## which decreases as a gets large. In fact it depends very strongly--it depends on the CUBE of that reciprocal. $$H = \sqrt{(\frac{1}{1.3a})^3 + 1}$$ Let's check that at time 0.8.

    It's the present, so there the normalized scale factor is a = 1 and the raw one is 1.3. When we cube 1/1.3 we get around 0.44. Then add one and get 1.44. Taking the square root we get 1.2.
    1.2 is right! You can see that from the gold curve in the graph in the previous post. The gold curve goes right through 1.2.

    Let's check it also for time 0.6. In that case a is about 0.8, so 1.3a ≈ 1.04 and the cube of 1/1.04 ≈ .9.
    The square root of 1.9 is roughly 1.4. And that's right! The gold curve of H goes right through 1.4.

    As a side remark this means that from the colors in a galaxy you can tell what the Hubble expansion rate was back when the light was emitted.
    If the wavelengths in the light are stretched out to twice their original length that means distances were HALF their present size then, and a=1/2. So 1.3a= 0.65 and you can take it from there.
    We can tell from the incoming light how much the wavelengths have been stretched because hot hydrogen has a distinctive red wavelength, hot sodium has a distinctive yellow, and so on. The spectral lines form recognizable patterns.

    I don't expect we'll need such exactitude in this discussion, but if more precision were required that number 1.3 we're using all the time could be improved to 1.3115. But the values of H we are getting just using 1.3 are fairly close, and what I want to do now is say how if you know the expansion rate H at the time some light was emitted you can tell what the age of the universe was at that time.
     
    Last edited: Jun 21, 2015
  7. Jun 21, 2015 #6

    marcus

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    We are coming full circle with our information about the universe expansion history. First we depicted the expansion as a function of time. I like the unnormalized scale factor 1.3a(t) so I'll write it this way $$1.3a(t) = \Big (\frac{e^{\frac{3}{2}t} - e^{-\frac{3}{2}t}}{2}\Big )^{2/3}$$ So if you know the time t (the expansion age of our universe) you can get to a(t) the scale factor, the size of a generic distance then compared with what it is at present.
    t ⇒ a
    But sometimes what you can OBSERVE is the scale factor! E.g. studying a galaxy's light, if the waves are triple the length they were when emitted then the light started towards us when a = 1/3 (when distances were 1/3 the size they are now). And we just saw a way to calculate H from that. $$H = \sqrt{(\frac{1}{1.3a})^3 + 1}$$ So if you know what the scale factor was when something occurred you can figure out what the expansion rate was then.
    a ⇒ H
    Now we want to come full circle. If we know what the expansion rate H was at some instant in the past, what time was it? what was the expansion age then? It turns out there is a simple formula that closes the circle and allows the time to be calculated: $$t = \frac{1}{3} \ln\frac{H+1}{H-1}$$ H ⇒ t
    Let's check that, referring to our blue-gold-red graph a couple of posts back. Suppose H = 1.4.
    then ##\frac{H+1}{H-1} = 2.4/0.4 = 6##, so put (ln 6)/3 into google. It comes out 0.6 which is right!
    You can see from the graph that when the expansion age of our universe was 0.6 the Hubble rate was, in fact, 1.4.
    Let's check again, suppose H = 1.2.
    2.2/0.2=11
    Put (ln 11)/3 into google.
    It comes out 0.8 which is right! (actually it comes out 0.799 but I'm rounding what google calculator says.)
    In fact when the age is t = 0.8 the expansion rate is H = 1.4 and vice versa.
    t ⇒ a ⇒ H ⇒ t
     
    Last edited: Jun 22, 2015
  8. Jun 22, 2015 #7

    Drakkith

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    Oh Marcus, I know of no one else who makes talking several hundred stories above my head look so easy. :wink:
     
  9. Jun 22, 2015 #8

    marcus

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    If it looks easy try "coming up". It may turn out it is no climb at all. My struggle is with organization. Exploring to find the right order and choose the right definitions among possible ones. Jorrie and Wabbit help enormously.

    There is a problem I love that I want to put in, but I haven't quite got to the point where it fits. A guy wakes up and is surprised to find the CMB is cooler than it was when he went to sleep. What time is it?

    It's a chance to use the two formulas introduced in the preceding post.
    $$H = \sqrt{(\frac{1}{1.3a})^3 + 1}$$
    $$t = \frac{1}{3} \ln\frac{H+1}{H-1}$$
    Maybe the CMB is only 1/10 the present temperature.
    So the guy says "I'm sometime in the future when distances are 10 times what they were.
    In other words, a(tunknown) = 10
    so 1.3a = 13, ahah!"
    $$H = \sqrt{(\frac{1}{13})^3 + 1} = 1.00023$$
    $$t_{unknown} = \frac{1}{3} \ln\frac{2.00023}{.00023} = 3.02$$
    We are measuring time in units of 17.3 billion years, so if the guy wants to convert the 3.02 it translates into around 51.9 billion years.
    But he might not want to bother with the Earth-bound unit and he might just say
    "Hmmm, I went to sleep at 0.8 and awoke in 3.02!"
     
    Last edited: Jun 22, 2015
  10. Jun 22, 2015 #9

    marcus

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    It's nice that the CMB temperature is such a handy expansion gauge. If you can measure the CMB temperature, that tells you the scale factor a.
    The temperature goes as 1/a. An expansion of distance by 2 cuts the temperature in half.
    And knowing the scale factor a lets you calculate the age of the universe, i.e. what time it is.

    Suppose you could hop back into the past and you landed at a time when the CMB temperature was TWICE today's. What would the time be?
     
  11. Jun 22, 2015 #10

    Drakkith

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    I got 4.23 billion years. Do you set a = 0.4?
     
  12. Jun 22, 2015 #11

    marcus

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    Yes! I did not work it out yet, but in answer to your question yes, to make the temp twice as big you set distances half the size, so a=0.5.
    Oh wait! You set a=0.4.
    I was busy just now and didn't have time to respond. Let me work it out with a=0.5 and see what we get.

    First I have to multiply a by 1.3
    1.3 x 0.5 = 0.65
    Then I have to take the square root of .65-3 + 1

    google says (.65^(-3) + 1)^(1/2) = 2.15. So the expansion rate was nearly twice today's rate of 1.2

    How far in the past was that? What was the age then?

    google says ln(3.15/1.15)/3 = 0.336

    So that gives me a rough idea, a third of a time unit, somewhere between year 5 and 6 billion. If I want to know more exactly I can multiply 0.336 by 17.3.
     
    Last edited: Jun 22, 2015
  13. Jun 22, 2015 #12

    marcus

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    The reciprocal 1/a of the normalized scale factor is just too useful not to have a symbol tag of its own. In Jorrie's calculator it is denoted S for "stretch factor" S = z+1 the factor by which wavelengths are enlarged while they are on their way to us. and by which distances are enlarged.
    It may seem like an unnecessary complication to have s = 1/a when we already have a. I may have to retract this. Anybody reading please let me know how it seems to you. Is this needless duplication? an encumbrance? Or will it prove convenient?

    So now there's an alternative way to write that formula for H.
    $$H = \sqrt{(\frac{1}{1.3a})^3 + 1}$$
    and also
    $$H = \sqrt{(\frac{s}{1.3})^3 + 1}$$

    I'll use that form in showing how to calculate distances.
     
  14. Jun 22, 2015 #13

    Drakkith

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    Didn't you set a equal to 0.8 for present time? Why wouldn't it be 0.4 for half that?
     
  15. Jun 22, 2015 #14

    marcus

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    You are helping me see that my language must not have been clear enough. t is the time, a(t) is a measure of size at that time.
    The present time is 0.8 (in a rather natural time unit related to the longterm growth rate, namely a unit that is 17.3 billion years in size)

    a (size, scale...) is normalized to equal one at the present time. So a(.8) = 1

    so when distances are half their present size, a = 0.5, but we don't right off know what time that was.

    To answer the question "Didn't you set a equal to 0.8 for present time?", no I just did the customary thing in cosmology which is to set a equal to 1 at the present time. The scale factor is almost always normalized to equal one at present, so that it is related to redshift z by z+1 = 1/a
    That way, at present, since a=1 we have 1/a = 1 and z, the redshift must equal zero. That makes sense, light emitted today and received today would not have time to be redshifted : ^)
     
  16. Jun 22, 2015 #15

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    On the contrary, I think overall your language is fine. It's just that it's very difficult to dive into a subject you have very little knowledge about and try to comprehend the math and all. At least for me. I have a hard time remembering and connecting all the different concepts until I've gone through all the steps myself, preferably on paper (which I haven't done yet).
     
  17. Jun 22, 2015 #16

    marcus

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    Maybe it's a good time to summarize. I have only one more formula to add, a formula for distance (the present distance of the source of some light that comes in with a given wave stretch s, how much distance that light has covered IOW) but before that we could review what we have.
    $$1.3a(t) = \Big (\frac{e^{\frac{3}{2}t} - e^{-\frac{3}{2}t}}{2}\Big )^{2/3}$$
    $$H = \sqrt{(\frac{1}{1.3a})^3 + 1}$$
    $$t = \frac{1}{3} \ln\frac{H+1}{H-1}$$

    The formulas are comparatively simple and they work because time is being measured in a convenient rather natural unit instead of in billions of years. It's determined by the longterm distance growth rate that the current Hubble rate seems to be converging towards. The reciprocal of a growth rate is a time, and we use that time as our unit--in Earth terms it happens to be 17.3 billion years. that makes the present age of the universe, 0.8, equal to about 13.8 billion years.

    So there are three quantities in our model: time t, scale factor a(t) normalized to equal one at present, and H(t) the instantaneous fractional growth rate at time t.

    I have to go out, but I think this summary could be fleshed out some, and might be useful at this point.
    I'll see if I can do that when I get back. In the meanwhile if anybody reading has suggestions of what brief definitions explanations overview might be useful, suggestions are welcome!

    Then I want to go on and give the distance formula for the light's distance from its source. that is a little tricky because expansion helps to put distance between the light and its source, so it is not simply the elapsed time multipled by c.
     
    Last edited: Jun 23, 2015
  18. Jun 22, 2015 #17

    Drakkith

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    By the way, it may help to label your graph axes. It took me a bit to realize that the X-axis was the time axis. Although, now that I think about it, when you're graphing a function with respect to a variable, the variable is always the X-axis...
    Still, sometimes its the little things. :wink:
     
  19. Jun 23, 2015 #18

    marcus

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    You are right that it would help to have the axes labeled in the two graphs in post #2. In fact I made them in late April while I was learning how to use https://www.desmos.com/calculator and had not figured out how to label axes yet. there is a little icon at the upper right corner that looks like a monkey wrench. If you click on it it gives you the opportunity to label the axes. Maybe I'll get around to re-drawing those graphs with the free online utility https://www.desmos.com/calculator and label them this time. It's a nice utility. Anybody can go there and have functions you type in plotted and then take a screen shot of whatever section of the graph you like. the screen shot shows up on your desktop and you can upload it to PF posts to illustrate what you're saying.

    I like online math utilities that are free and open for everyone to use. Besides Desmos.com there are several that you get if you google "definite integral calculator". The ones I've tried are surprisingly easy to use.

    Imagine you are studying a galaxy whose light is stretched by a factor of 3--the wavelengths are 3 times the length they were when the light was emitted. How far is that galaxy right now?
    You go to one of the online "definite integral calculators" and paste this into the box:
    ((s/1.3)^3+1)^(-1/2)
    then you put "s" in for the variable and 1 and 3 in for the limits and press calculate.

    Or if the stretch factor is 4, you make the limits 1 and 4.
     
    Last edited: Jun 23, 2015
  20. Jun 23, 2015 #19

    marcus

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    Speaking of "definite integral" integrating is a system of adding up little bits (which it's often easy to get an online calculator to do) and I want to try to explain something about distance.
    Suppose at time 0.7 you have a little bit of distance, say the amount some light traveled in a little bit of time. How big will that be NOW, at time 0.8?
    Think about what you want to call this bit of distance, how you want to denote it.
    I want to call it "cdt" a little bit of time multiplied by the speed of light. Whatever you call it, its size NOW will be that divided by a(.7).

    remember that 1/a(.7) is the factor by which distances and wavelengths get stretched between time .7 and the present .8.

    so if you ask me how big that little bit of distance cdt is now, I would say cdt/a(.7)

    How far does light travel between time 0.6 and the present 0.8?

    You have to add up all the little steps the light made and remember to put a(t) in the denominator to enlarge them according to the time t they were made. The integral sign evolved from an antique letter S for "sum". It means you add up all the little steps. $$D(\text{time .6 up to now}) = \int_.6^.8{\frac{cdt}{a(t)}}$$ We can actually get online integrators to do the adding up for us.
    You do not have to have taken a college calculus course to use an online integrator. You just need to be able to type the function to be integrated into the box, and specify the variable (like "t" that is being advanced in little steps) and specify the limits it goes between. (like .6 and .8).
     
    Last edited: Jun 23, 2015
  21. Jun 23, 2015 #20

    marcus

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    There is another distance integral which takes more thought to understand why it works. At the risk of putting some potential readers off, I want to show it. Imagine you are an astronomer and some light from an interesting galaxy come in at your telescope and says "I've been stretched by a factor of 3."
    And you say "You've had a long journey! I know how far you are now from your source."

    How did you figure out, just from the number 3, how far away the light's source galaxy is now (including the effect of expansion)?

    Remember that stretch s=1 signifies the present (a wavelength that gets multiplied by 1 is not enlarged at all.) The stretch factor reaches back in time, from 1 to larger and larger amounts of stretch. We could take it in little steps and run the integral that way.

    $$D(\text{from stretch 1 back to S}) = \int_1^S{\frac{ds}{H(s)}} = \int_1^S{\frac{ds}{\sqrt{(s/1.3)^3+1}}}$$

    For reference, the stretch quantity was introduced back in post #12. Since it is just the reciprocal of the scale factor s = 1/a it seems at first unreasonable to have a separate notation for it. It's handy to have something that increases going back in time (both t and a(t) increase going forward) and I mentioned it would come up when we consider distances.
    How to find the distance from home, of some light that comes in stretched s=3
    ==quote post#18==
    You go to one of the online "definite integral calculators" and paste this into the box:
    ((s/1.3)^3+1)^(-1/2)
    then you put "s" in for the variable and 1 and 3 in for the limits and press calculate.
    ==endquote==
    I like the "definite integral calculator" at the "Number Empire" site. There are several others but here's link to that one.
    http://www.numberempire.com/definiteintegralcalculator.php
    I pasted that above thing into the box and changed the variable from "x" to "s" to match the variable in the formula, and set the limits, and pressed calculate. And the answer came out 0.99 almost one Zeit!
    Actually since the answer is a distance in this case, it is 0.99 Light Zeit.
    If you like the answer in billions of lightyears, then that is basically one percent less than 17.3 billion light years.

    The light wouldn't have been able to travel that far on its own, in the allotted time, but expansion helped it.
     
    Last edited: Jun 23, 2015
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