Something about calculating the Age of the Universe

[marcus]

Exponential at constant percent growth rate.

an expression like "70 km/s per Mpc" is just an unfortunate, inept, and confusing way of expressing a percentage growth rate. It has done a lot of damage to people's minds.
Makes them think of ordinary speed of ordinary motion and picture stuff moving thru space, instead of the uniform pattern of distance expansion at a uniform percent "interest rate" that is constantly being gradually reduced by the bank but is now beginning to level off.

 

[wabbit]

I agree with marcus. Such a percentage should always be expressed as nautical knots per furlong.

Edit - which needless to say is H0=0.9 femtoknot per furlong.

 

[marcus]

...
Edit - which needless to say is H0=0.9 femtoknot per furlong.

Heh heh, excellent plan. Astronomers who use awkward units that confuse the rest of us should be required to use even worse units, so they will understand what it's like for us! And they should have to wear hair shirts too, with collars and neckties. : ^)

You are right about the 0.9! On the principle that satirical arithmetic should be checked for correctness, I typed this into google:
"0.07 per billion years in knots per furlong"
And google came back with:
"0.07 per (billion years) = 8.67408684 × 10^-16 knots per furlong"

 

[wabbit]

Well to be fair I got that 0.9 from google too - I know I said I'm a pen and paper guys, but sometimes one can't resist the appeal of technology : )

I asked "70 km/s per Mpc in knots per furlong" and got "8.87090465 × 10-16"

And I don't even know how long a furlong is...

 

[marcus]

It's rather fur

 

[JohnnyGui]

Exponential at constant percent growth rate.

an expression like "70 km/s per Mpc" is just an unfortunate, inept, and confusing way of expressing a percentage growth rate. It has done a lot of damage to people's minds.
Makes them think of ordinary speed of ordinary motion and picture stuff moving thru space, instead of the uniform pattern of distance expansion at a uniform percent "interest rate" that is constantly being gradually reduced by the bank but is now beginning to level off.

I'm sorry for being so stubborn but I'd REALLY like to understand how a Hubble constant like 70km/s per Mpc would translate into a certain percentage growth rate. If it doesn't translate to that at all, then how is did people discover that the universe is expanding at a certain percentage growth rate (with the growth rate eventually getting influenced)? I understand that it's no ordinary motion but it's at least something that creates more space in our dimension between galaxies which can be measured, so I think it can be said that the expansion is creating a certain distance of space in a certain amount of time.

 

[marcus]

I'm sorry for being so stubborn but I'd REALLY like to understand how a Hubble constant like 70km/s per Mpc would translate into a certain percentage growth rate.

Google calculator converts units. It can tell you how many kilometers a Megaparsec is. Just type in:
"Mpc in km"
It will say "1 Mpc =3.08567758 × 10¹⁹ kilometers"

70 km/s per Mpc is nothing else than a fractional growth rate because the two distances cancel, leaving only time (seconds) in the denominator.

$$\frac{70 km}{s \times Mpc} = \frac{70 km}{second \times 3.085 \times 10^{19} km} = \frac{70}{second \times 3.085 \times 10^{19}}$$

the kilometers in numerator and denominator cancel and you just have some numbers and "per second", that is seconds in the denominator.

It boils down to a very small number per second. A number per unit time is either a frequency or a growth rate, or something similar, some kind of count or fraction per unit time.

If you want to see what google calculator makes of it, you can type "70 km/s per Mpc" into google. You will probably get a frequency, a number per second.

Or try this, type in "70 km/s per Mpc in year^-1"

that is like asking it to give you the answer in terms of a number "per year" rather than a number per second. Here is what it told me when I typed that in:

70 ((km / s) per Mpc) = 7.15883225 × 10^-11 year^(-1)

That is numerically the same as saying 0.07158... per billion years.

 

[JohnnyGui]

- EDIT: Marcus! So sorry I didn't notice your reply before posting all this. I have actually done this way of rewriting the formula myself but doesn't this still yield a growth rate in the form of a constant velocity? Or is the velocity increasing by this very small growth rate that comes up? Btw, could you please check if my conclusions at the end of this post are correct or not?

Ok guys, believe it or not but out of desperation to try out the Hubble constant I have actually grabbed a rubber band to simulate the stretching scenario of the universe, put some points with pen on fixed distances measured from one side from the band, gave the stretching a certain velocity and measured the distance after each stretch. Here are the results (and God do I feel stupid for having to understand it this way):

In the following link you see on the left the Observer who measures the distances of the objects (planets, galaxies, etc.) and on the right the Edge of the Universe (probably by light?) which stretches the line. Hubble constants are on the right between the time frames.
Let's say that at Time (T) = 0 the Observer measures the distances of the objects and then waits for 1 second to pass to measure their velocity.

https://www.dropbox.com/s/0450u9w7k9vqiqk/Hubble Constant.jpg?dl=0

From this I can draw the following conclusions and please correct me of one (or more) of them are wrong:

1. The Hubble constant doesn't give any acceleration at all. If Object A (which goes at 1cm/s) passes a 2 cm distance, it still travels further at 1 cm/s/cm. Same for Object B and C, they're all having a constant speed no matter how far they have traveled. This means that the Edge of the Universe is also going at the same velocity. The Hubble constant doesn't translate into an acceleration.

2. The Hubble constant actually changes depending on the time when you measure it.

3. Over time, the Hubble constant gradually decreases, in a decreasing rate. This means that BEFORE T= 0, H was greater than 1cm/s/cm. So what was H at the time of the Big Bang?

4. When you look at a FIXED distance, over time, every object that passes that fixed distance will have a lesser velocity than the one who passed it before it.

 

[marcus]

You said you really wanted to understand something:

I'm sorry for being so stubborn but I'd REALLY like to understand how a Hubble constant like 70km/s per Mpc would translate into a certain percentage growth rate.

Do you understand it now.
The Hubble "constant" is not about motion, or speed of objects moving thru space. It is a badly expressed rate of distance growth. by rate I mean a fraction or number per unit time.

I've explained why there is no motion in the ordinary sense, in Hubble law expansion. Nobody gets anywhere by it, everybody just becomes farther apart. It would;t make sense to consider it as motion thru space because the distances to most galaxies we see are growing faster than c. that would be illegal if it were due to motion.

What I want to know now, Johnny, is do you understand this. Has my explanation gotten you anywhere? Or has it failed?
Are you now able to stop talking about Hubble law expansion as if it involved motion of objects thru space, and can you stop thinking of it that way?
Can you picture geometric expansion and percentage rate growth of distances between stationary (unmoving) objects.
that is what spacetime curvature (GR) is about, and cosmology is based on GR. I want to know whether I have completely failed.

 

[JohnnyGui]

You said you really wanted to understand something:

Do you understand it now.
The Hubble "constant" is not about motion, or speed of objects moving thru space. It is a badly expressed rate of distance growth. by rate I mean a fraction or number per unit time.

I've explained why there is no motion in the ordinary sense, in Hubble law expansion. Nobody gets anywhere by it, everybody just becomes farther apart. It would;t make sense to consider it as motion thru space because the distances to most galaxies we see are growing faster than c. that would be illegal if it were due to motion.

What I want to know now, Johnny, is do you understand this. Has my explanation gotten you anywhere? Or has it failed?
Are you now able to stop talking about Hubble law expansion as if it involved motion of objects thru space, and can you stop thinking of it that way?
Can you picture geometric expansion and percentage rate growth of distances between stationary (unmoving) objects.
that is what spacetime curvature (GR) is about, and cosmology is based on GR. I want to know whether I have completely failed.

I think I've got it! The key word for me understanding this is the Hubble constant being a fraction or growth rate per unit time that adds up to the distance and not the velocity itself of any object. I thank you so much for your effort for helping me understand this.

 

[Jorrie]

- or rather it gives the equivalent form
$$T_0=\frac{2}{3}\frac{1}{H_\infty}\tanh^{-1}( \frac{H_\infty}{H_0}) $$
This is nice.

Yup, this is the more standard form. I suppose if we want to use only 'standard' terminology/simbology, we could also write:

$$T_0\approx\frac{2}{3}\frac{1}{\sqrt{\Omega_\Lambda} H_0} \tanh^{-1}(\sqrt{\Omega_\Lambda}) $$

Although not as 'nice looking', it may be slightly easier to guide a newcomer into coping with this expression.

 

[Chronos]

Isn't a furlong the distance a cat skin can be stretched by the full effort of a man using a spud wrench?

 

[JohnnyGui]

I have one other question after looking at the image I made here: https://www.dropbox.com/s/0450u9w7k9vqiqk/Hubble Constant.jpg?dl=0
I was able to conclude out of this image that H = (ΔD / Δt) / D_t where D_t is the original distance you're calculating from.
I then looked at how H changes with time and how it relates to the shape of the time-distance graph:

Suppose the time-distance graph is showing a linear relation. In other words, the Universe (or object for that matter) is expanding at a constant velocity. This would be the scenario of the image in the link above. The corresponding Hubble parameter would then show an asymptote with time: https://www.dropbox.com/s/tl0z3jeaw1zudw9/Linear.jpg?dl=0

However, if the time-distance graph is showing an exponential relation. In other words, the Universe (or object for that matter) is expanding at an increasing velocity, then the Hubble parameter would remain constant over time: https://www.dropbox.com/s/n7mhj1l1qnksli5/Exponential.jpg?dl=0

If these graphs are correct and one would "calculate" the age of the Universe with 1/H₀, that would mean that he's considering H to be constant over time and therefore he's saying that the Universie was expanding at an exponential rate over time (since a constant H parameter corresponds to an exponential growth of the Universe). He would therefore calculate a maximum Age of the Universe since an exponential graph takes a lot of time to extrapolate back to a distance of 0.1 (0 isn't possible in an exponential case obviously).

Now, the fact that 1/H₀ coincides closely with the real age of the Universe today means that the expansion of the Universe had a somewhat exponential time-distance relation back then. In other words, it was accelerating and it therefore had a somewhat constant H parameter over a period in time back then.
Since you guys said that the time-distance graph now is showing a more linear relation, that means the Hubble parameter-over-time relation is transforming from a straight line into a negative exponential relation. Something like this: https://www.dropbox.com/s/2z0hc5akwsg9gql/Transition.jpg?dl=0

Is this conclusion correct?

 

[wabbit]

Not quite. If you assume exponential growth the age you get is infinite (you can truncate at some scale but then the result depends on the scale and I think it is best to keep this out of the picture in this discussion).
Also the current expansion is not linear, it is somewhat close to exponential. What looks broadly linear is the whole curve, because it curves down first, then up.

 

[JohnnyGui]

Not quite. If you assume exponential growth the age you get is infinite (you can truncate at some scale but then the result depends on the scale and I think it is best to keep this out of the picture in this discussion).
Also the current expansion is not linear, it is somewhat close to exponential. What looks broadly linear is the whole curve, because it curves down first, then up.

Oh right, totally forgot about the infinity problem in an exponential growth. How about my conclusion about the change of the Hubble parameter over time in relation with the distance-time graphs? Is that correct?

 

[wabbit]

OK looking at these graphs now. I see you are using comoving coordinates, seems like a good idea here.
The first one is exponential expansion, OK so far.
Second and third one, forget what I just said, I need to check that.

 

[JohnnyGui]

OK looking at these graphs now. I see you are using comoving coordinates, seems like a good idea here.
The first one is exponential expansion, OK so far.
Second and third one, forget what I just said, I need to check that.

What do you mean by "first one"? By first graph I meant the one with the linear relation of time-distance.

Also, what I meant by negative exponential relation was an asymptote. English is not my native language, sorry for the confusion.

 

[wabbit]

The difficulty I'm having is that I'm not quite sure exactly which distance (and time) you are referring to. Assuming distance is our current distance to a given galaxy, as in the labels in what I called your first chart, plotted over cosmological time, then yes, qualitately both your charts are correct.
For https://www.dropbox.com/s/tl0z3jeaw1zudw9/Linear.jpg?dl=0 , the relation is $d\propto t, H\propto 1/t$ and in the next one (exponential) H is indeed flat.

 

[JohnnyGui]

The difficulty I'm having is that I'm not quite sure exactly which distance (and time) you are referring to. Assuming distance is our current distance to a given galaxy, as in the labels in what I called your first chart, plotted over cosmological time, then yes, qualitately both your charts are correct.
For https://www.dropbox.com/s/tl0z3jeaw1zudw9/Linear.jpg?dl=0 , the relation is $d\propto t, H\propto 1/t$ and in the next one (exponential) H is indeed flat.

Thanks! The distance I meant was indeed the distance to any galaxy/object/planet that's moving away over time because of the expansion. However, couldn't this distance be extrapolated to the distance to the "edge" of the universe since a galaxy moving away from us is a reflection of the expansion of the "edge"?

Another question, is the acceleration of the expansion a constant acceleration over time?

 

[wabbit]

Not sure what you mean by edge. The model used has no edge, it is infinite flat space (or a sphere or or hyperbolic space, with very small curvature), so presumably you are referring to a horizon, such as the Hubble radius (a galaxy outside that radius is receding from us at more that light speed and will never be seen by us) or other. But this is not comoving : the Hubble radius increases more slowly than distances to galaxies do - i.e. galaxies are leaving that Hubble sphere gradually. Eventually, a very long time from now, it will approach a fixed radius but expansion will continue.

 

[JohnnyGui]

Not sure what you mean by edge. The model used has no edge, it is infinite flat space (or a sphere or or hyperbolic space, with very small curvature), so presumably you are referring to a horizon, such as the Hubble radius (a galaxy outside that radius is receding from us at more that light speed and will never be seen by us) or other. But this is not comoving : the Hubble radius increases more slowly than distances to galaxies do - i.e. galaxies are leaving that Hubble sphere gradually. Eventually, a very long time from now, it will approach a fixed radius but expansion will continue.

I am indeed biased by the thought of the universe having an edge. I should read more about the Hubble radius.

I'm sorry for bumping this old thread but there's something I have concluded (yet again) for which I need verification.

Previously I was able to conclude the relationship of the Hubble constant, at a specific time, with the distance of an object/galaxy if the velocity of that object was constant.
The formula is the following: H = (ΔD / Δt) / Dt where Dt is the original distance of an object you're calculating from.

I took it a step further and tried to conclude the relation of the H parameter if the distance of an object increases exponentially over time. In this case, I considered H to be constant over time since the following graph shows that an exponential increase in distance of an object/galaxy translates to a constant H over time: https://www.dropbox.com/s/n7mhj1l1qnksli5/Exponential.jpg?dl=0

Now, I know that the universe hasn't been expanding exponentially ever since it began but I thought I might try this out. After thinking this through I concluded that the formula for the relationship between a constant H and time (in the scenario of an exponential distance increase of a galaxy) is the following:

D_Δt+t = (H+1)^Δt • Dt

D_Δt+t is the new distance that the object will have after a particular time Δt
Δt is the time between the original distance Dt and the new distance D_Δt+t the object/galaxy will have.

I have tried this formula with several examples and it gave me correct answers. Rewriting the formula to get H would give:

H = ((D_Δt+t / Dt)^(1/Δt)) - 1

Again, this formula would give an estimated age if the universe has been expanding exponentially by a constant H ever since it began. I know this is wrong since this would give an infinite age of the universe. You can't give Dt a value of 0 and divide by that but if one would give Dt a very small number, perhaps an estimated age would come out of Δt from the point the universe has been expanding exponentially.

My questions is, are these formulas correct if one would mathematically describe the relationship of a constant H over time with an exponential expansion of the universe?

 

[wabbit]

Not sure about the exact formulas you write (they seem OK for small $\Delta t$ but in that case they can also be simplified), but yes, exponential expansion corresponds to constant $H$. Strictly speaking, this describles a vacuum with a cosmological constant and no matter or radiation - but it is also a close approximation to how the universe expands at very late times in the standard model of cosmology, and you could also say that our universe results from adding some matter and radiation to such a solution. Early on, the matter and radiation are concentrated and they have a lot of influence, inducing extra curvature - but after they thin out enough due to expansion, (almost) all that remain is the intrinsic curvature from the cosmological constant, and the result is (approximate) exponential expansion.

 

[JohnnyGui]

Not sure about the exact formulas you write (they seem OK for small $\Delta t$ but in that case they can also be simplified), but yes, exponential expansion corresponds to constant $H$. Strictly speaking, this describles a vacuum with a cosmological constant and no matter or radiation - but it is also a close approximation to how the universe expands at very late times in the standard model of cosmology, and you could also say that our universe results from adding some matter and radiation to such a solution. Early on, the matter and radiation are concentrated and they have a lot of influence, inducing extra curvature - but after they thin out enough due to expansion, (almost) all that remain is the intrinsic curvature from the cosmological constant, and the result is (approximate) exponential expansion.

Thanks for the explanation. Reading your post, can I say that my mentioned formulas roughly describe a vacuum cosmological constant that expands exponentially?

 

[wabbit]

I think so, though I find it clearer written as $D(t+\Delta t)=D(t) e^{H\Delta t}\simeq D(t)(1+H\Delta t)$

 

[JohnnyGui]

I think so, though I find it clearer written as $D(t+\Delta t)=D(t) e^{H\Delta t}\simeq D(t)(1+H\Delta t)$

I have never seen that "e" before. When I use your formula in my examples, I calculated that it has a constant value of around 1,732. Is this correct?

Your other formula D(t)(1 + HΔt) however doesn't give any correct answers in my examples, neither are they even near my values. Am I doing something wrong?:

Suppose there is a constant H of 2m/s/m over time.
An object that starts at 3m distance, and thus a velocity of 6m/s (H x D), will have the following distances in the following 3 seconds:
Start distance is 3m at t=0, at which it will have a velocity of 6 m/s
Distance is 9m at t=1, at which it will then have a velocity of 18m/s (H x D)
Distance is 27m at t=2, at which it will then have a velocity of 54 m/s (H x D)
Distance is 81m at t=3

If I fill in the formula you gave, for example D(t) being 3m and I want to calculate the new distance D(t + Δt) after 2 seconds (thus at t=2), it will give me 3(1 + 2 x 2) = 15m instead of 27m. Using your other formula which contains that "e" symbol however does give me the correct values according to the example.

 

[Xyooj]

calculating the age of the universe is like pinpointing the position of an atom, by the time you thought you knew it's position it already changed?

perhaps it's not the universe that's expanding, but our own imagination that it is expanding so we develop the tools to design what we want to see?

 

[wabbit]

You are right, your formula $D_{Δt+t}= (H+1)^{Δt} • Dt$ is not correct, I looked at it too quickly.

$e\simeq 2.71828$ is http://en.m.wikipedia.org/wiki/E_(mathematical_constant)

It is a convenient basis for expressing exponential growth, precisely because it it the only number $x$ such that $x^{t+\delta }\simeq x^t (1+\delta)$ for small values of $\delta$

 

[wabbit]

calculating the age of the universe is like pinpointing the position of an atom, by the time you thought you knew it's position it already changed?

No, that age changes at the rate of one second per second, and this is not going to materially alter the result, which is currently estimated to be around 14 billion years.
It is not a figment of our imagination either, but a result of precise models of how things move. Of course these models can be superseded and the answer may change, as our understanding improves, but calling that an effect of imagination is a stretch.

 

[JohnnyGui]

You are right, your formula $D_{Δt+t}= (H+1)^{Δt} • Dt$ is not correct, I looked at it too quickly.

$e\simeq 2.71828$ is http://en.m.wikipedia.org/wiki/E_(mathematical_constant)

It is a convenient basis for expressing exponential growth, precisely because it it the only number $x$ such that $x^{t+\delta }\simeq x^t (1+\delta)$ for small values of $\delta$

I'm baffled... How come my formula isn't correct while it gives me correct answers in my examples? Is my given example wrongly executed?

 

[wabbit]

I think the reason is that H is (currently) a very small number, and $(1+H)^{\Delta t}\simeq e^{H\Delta t}\simeq 1+H\Delta t$ in this case.

So actually I was wrong again, your formula is OK (but only when $H$ is small, it doesn't work at all for $H=1$ for instance)... Sorry, should think more before writing:)

 

[JohnnyGui]

I think the reason is that H is (currently) a very small number, and $(1+H)^{\Delta t}\simeq e^{H\Delta t}\simeq 1+H\Delta t$ in this case.

So actually I was wrong again, your formula is OK (but only when $H$ is small, it doesn't work at all for $H=1$ for instance)... Sorry, should think more before writing:)

The thing is, even when I give H a small number such as 2 and Δt for example 3, (1 + H)^Δt would give me 27 while e^HΔt would give me a value of ≈403

 

[wabbit]

Right, and this is the case where your formula breaks down. "Small value of $H$" here means much smaller than one (I was assuming you used $H$ is SI units, in which case the current value is very small)

Note that your formula has a unit problem, is you change the time unit, the formula changes since 1 is dimensionless but H has units of inverse time.

 

[JohnnyGui]

Right, and this is the case where your formula breaks down. "Small value of $H$" here means much smaller than one.

Note that your formula has a unit problem, is you change the time unit, the formula changes.

Sorry for being so stubborn but wouldn't a value of H=2 in my formula in that case then give me a wrong answer than shown in my example? GIving H a value of 1 also give me correct answers according to my example.

 

[wabbit]

Yes it should. I need to look at your example again, this cannot work for H of order one or larger.

Edit:: actually I do not see a numerical examle in your post, so I don't know what is telling you that your formula works.

 

[JohnnyGui]

Yes it should. I need to look at your example again, this cannot work for H of order one or larger.

Edit:: actually I do not see a numerical examle in your post, so I don't know what is telling you that your formula works.

I was referring to my post #60:

Suppose there is a constant H of 2m/s/m over time.
An object that starts at 3m distance, and thus a velocity of 6m/s (H x D), will have the following distances in the following 3 seconds:
Start distance is 3m at t=0, at which it will have a velocity of 6 m/s
Distance is 9m at t=1, at which it will then have a velocity of 18m/s (H x D)
Distance is 27m at t=2, at which it will then have a velocity of 54 m/s (H x D)
Distance is 81m at t=3
Regarding units, I don't think (and correct me if I'm wrong on this) this is a unit problem since only very large values give me wrong answers. I randomly gave H a value of 1500, made an example of that, and my formula indeed gave me wrong answers. When giving H a value of 50 and making an example out of that, the anwers my formula gives are still correct.

I think one should plot my formula against your mentioned formula and see at which value of H they start to deviate from each other..