# Cosmo calculator-recession speed tutorial

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Cosmo calculator--recession speed tutorial

This thread could be a tutorial/discussion using Morgan's online calculator that calculates recession speeds, if you give it the redshift.

If you try out the calculator and have questions, or are surprised by anything, please ask. No telling what you get for answers maybe some that conflict :surprised: but by asking you at least start the process of getting clear about redshift and recession speed.

Lots of stuff is observed to have redshift 2 or more, some quasars we observe have redshift over 6, and the CMB has redshift 1100.

With Morgan's calculator you can enter a redshift, like 2 or 6, and find out the present distance to that thing, and how fast it is receding---how fast the distance to it is growing (assuming it is stationary in the space around it wrt CMB).

http://faculty.cns.uni.edu/~morgan/ajjar/Cosmology/cosmos.html [Broken]

You've heard that Dark Energy is 73% and the combined total for matter including ordinary and dark matter is 27% and you probably have seen the figure for the Hubble parameter that is commonly used, which is 71 km/second per Megaparsec. MORGAN BELIEVES YOU SHOULD WORK A LITTLE SO TO USE THE CALCULATOR YOU HAVE TO TYPE IN 0.27 and 0.73 and 71

Morgan makes you type in values of Omega matter density (.27) and Lambda dark energy density (.73) yourself. And the current best figure for the Hubble parameter 71. THEN YOU TYPE IN THE REDSHIFT, like say 6 for a distant quasar. So what you've typed in from top to bottom:
.27
.73
71
6

Then press "calculate" it will tell you the recession speeds. And how long ago it was that the quasar was radiating the light that we are now getting from it.

And how far away it used to be when the light left it on its way to us, and how far away the quasar is at the present time.

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Wallace
Nice tool that, thanks for the link.

Gold Member
Dearly Missed
Nice tool that, thanks for the link.

thought you'd like that if you hadn't seen it already Ned Wright has the same basic calculator at his website, but different options.
He actually has three slightly different cosmology calculators, but if I remember right none of them give recession speed.

they give other useful stuff but not that.

so Morgan's is noteworthy because of giving recession speeds.

If anyone has not checked out Wright's calculators here are links

standard model (gives angular distance as well as comoving and travel time)
http://www.astro.ucla.edu/~wright/CosmoCalc.html

advanced version CosmoCalc (lets you put in the dark energy equation of state and other parameters)
http://www.astro.ucla.edu/~wright/ACC.html

a BACKWARDS version of the CosmoCalc (you put in the travel time and it gives you back the redshift, instead of other way round)
http://www.astro.ucla.edu/~wright/DlttCalc.html

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Gold Member
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The URL for Morgan's CosmosCalculator changed. The one I gave in the first post no longer works. here is the new one:

http://www.uni.edu/morgans/ajjar/Cosmology/cosmos.html

If you use it be sure you type in 0.27, 0.73, and 71
in the first three boxes on the lefthand margin

then type in the redshift z that you want, and press 'calculate'

cristo
Staff Emeritus
Thanks for all the links, marcus-- they seem very useful!

George Jones
Staff Emeritus
Gold Member
The URL for Morgan's CosmosCalculator changed. The one I gave in the first post no longer works. here is the new one:

http://www.uni.edu/morgans/ajjar/Cosmology/cosmos.html

If you use it be sure you type in 0.27, 0.73, and 71
in the first three boxes on the lefthand margin

then type in the redshift z that you want, and press 'calculate'

Siobahn Morgan (University of Northern Iowa) has a page of astronomy, astrophysics, and cosmology applets, which I found from removing some stuff from the end of the above URL.

http://www.uni.edu/morgans/ajjar/

"This website is set up as a resource for instructors of astronomy at all levels. Items are presented in a variety of formats and levels of complexity, so that instructors at all levels from K-12 to college level could make use of the programs here."

stevebd1
Gold Member
I've recently been looking at how redshift is used to calculate distances to various objects in the cosmos and I now feel I have a fair undertsanding of the difference between relativistic redshift and cosmological redshift. Attached is a table I put together comparing the data from both types of redshift and my question is, would it be a reasonable assumption to say that the 'time/distance' from the relativistic redshift data is the actual time/distance the light itself has travelled? There are a few discrepencies when you add the 'age then' from the cosmo redshift data to the 'time/distance' from the relativistic redshift data which makes me suspect this might not be the case (unless it's just data from two different equations not entirely working together). If it isn't, what exactly is the relevance for the relativistic redshift data?

Steve

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Gold Member
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...would it be a reasonable assumption to say that the 'time/distance' from the relativistic redshift data is the actual time/distance the light itself has travelled? ...

I don't think so---or perhaps I don't understand your third column of numbers which you call
"time/distance" (units: Gy/Gly)

The special relativity doppler shift
could be calculated for a particle or other object which is at ZERO DISTANCE from us, at the instant it whizzes past us.
Therefore your fraction "time/distance" is indeterminant. It might very likely have zero denominator, and the time is not well defined. But maybe you can explain your third column so I can understand.

I understand your first column is just z values from 0.01 to 1089, which you are going to interpret as EITHER a dopplershift z, or as a cosmological redshift z.

YOUR SECOND COLUMN LOOKS GOOD to me. It looks like you just used the special rel doppler formula. I introduce a variable rho = 1+z for convenience in writing it:

$$\rho = 1+z = \sqrt{\frac{1+\beta}{1-\beta}}$$

$$\beta = \frac{\rho^2 - 1}{\rho^2 + 1}}$$

so in your second column you took z to be a doppler shift, and you just plugged into the second equation here and found beta, the speed in units of c.

Your third column, as I say, mystifies me. But the OTHER COLUMNS SEEM STRAIGHTFORWARD. You could have gotten them straight from Morgan's calculator.
In those columns you are interpreting z as a COSMOLOGICAL redshift, which is a totally different thing from a special rel Doppler shift. So there would probably be no relation between those other columns (four thru nine)

SR doppler formula is used when two objects are right close together in the same reference frame. In that case, expansion of largescale distances can normally be neglected.
Or it could be used for CORRECTION due to local motion in a receding frame. But that probably doesn't apply here.

The main thing you would need to explain, to make it clear to me, is what your time/distance column actually means.
When you observe a certain SR doppler shift (call it z if you want) then how do you decide on a distance and how do you decide on a time----so what is this ratio you call time/distance?

stevebd1
Gold Member
Thanks for the reply Marcus. In truth, I posted a question in another thread (https://www.physicsforums.com/showthread.php?t=206984) asking if the relativistic redshift equation incorporated the cosmological redshift (which I now understand is a seperate set of equations). I did some searching and found this thread and the other one called 'Hubble parameter in early universe' which went towards helping me understand the cosmological redshift. The figures you see in columns 4 to 9 are taken from the cosmo calculator, I was inquiring if there was some connection between them and the relativistic redshift figures (which it appears there is not).

The third colomn is based on figures produced in the following equations-

Doppler effect (which applies to < z = 0.1)

$$r (distance) = \frac{v}{H} = \frac{zc}{H}$$

if z = 0.03

$$r (distance) = \frac{v}{H} = \frac{zc}{H} = \frac{0.03c}{71} = \frac{8993.775}{71} = 126.673 Mpc = 407.25 Mly (or 0.407 Gly)$$

Relativistic Doppler effect (which applies to > z = 0.1) I've actually applied this to anything over z = 0.03

$$z = \sqrt{\frac{1 + v/c}{1 - v/c}} -1$$

which provides the following equation-

$$v/c = \frac{(z + 1)^{2} - 1}{(z + 1)^{2} + 1}$$

therefore if z = 0.1

$$v/c = \frac{(0.1 + 1)^{2} - 1}{(0.1 + 1)^{2} + 1} = 0.095c$$

$$r (distance) = \frac{v/c}{H} = \frac{0.095c}{71} = \frac{28480.2875}{71} = 401.13 Mpc = 1309 Mly (or 1.309 Gly)$$

with some help from the following calculator-

http://hyperphysics.phy-astr.gsu.edu/hbase/astro/hubble.html#c3

Hopefully the above sheds some light on column 3. It seemed odd that the relativistic redshift would produce a set of figures that appear to be describing time or distance (as implied at the hyperphysics website).

Steve

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Gold Member
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Steve, I see now how to read the third column. All the other columns made sense right away and I've commented already.

You've taken care to be clear and use LaTex which makes it very easy to read and understand your equations and calculations, which helps and I hope you are rewarded for this both now and in the hereafter.

You drew the right conclusion.
... I was inquiring if there was some connection between them and the relativistic redshift figures (which it appears there is not).

At very small redshift things moosh together and different things turn out to be approximately the same so there appears to be connection. One can used sloppy methods to calculate and it works approximately. You only get forced to be careful at larger redshifts.

What your table shows is that what you get by special rel doppler and a naive application of the Hubble equation does NOT correspond to the reality you get from a good redshift calculator. having made that point, the natural thing to ask is WHY.

WHY DO THE DISTANCES IN COLUMN 3 NOT MATCH UP TO ANYTHING?
Well for one thing the Hubble equation v = H0 d applies only to recession speed and distance at this instant right now in January 2008.

As long as z is small and v is small we can be sloppy and forget about differences between speed then, and speed now, and average speed during the time of flight---it is all about the same. But for larger z, watch out!

So when you take a COSMOLOGICAL REDSHIFT and apply the special rel inverse doppler formula to it, you do in fact get a speed. But is it the speed NOW?
Or is it the speed then, when the light set out on its journey to us? Or what is it supposed to mean? some average?

It isnt anything. It isnt well defined. Because special rel is not generally applicable. But whatever it might be, it surely is not the speed NOW. so you cannot apply the Hubble formula! The v = H d equation simply does not apply.

So when you apply it and get a distance that does not match anything, nobody should be surprised. That is my attempt to answer the question of why it doesn't match the cosmo calculator results.

BTW Ned Wright calculators seem to have more decimal place accuracy. I'm not an expert in computational stuff, at all! I just like Morgan a lot because it gives recession speeds, and Wright does not. But if you are doing some calculating where you need precision you might compare.

Another thing. At first when I read your column 3 i thought time/distance meant a RATIO of time divided by distance.
After your next post, I realized that you meant "time OR distance" and that the units were "Gy OR Gly". this is just the normal hazards of notation. that is probably the main reason I didn't see what you were doing immediately. I will be more alert to that possible reading another time.

stevebd1
Gold Member

Would it be a fair assumption to say -

The simple Doppler effect equation can be used when z < 0.01
I revised this from 0.03 which still had a degree of error when compared to the figure from the relativistic Doppler effect equation. z = 0.01 still covers a radius of 137.815 Mly which takes us well out of the Local 'Virgo' Supercluster.

The relativistic Doppler effect equation should be used when z < 0.1 to take in account special relativity.

The cosmological redshift equations are used when z > 0.1 to take into account cosmic expansion.

But as a whole, it's normally best to use the cosmological redshift equations in order to get the most accurate results (I'm aware there is a gravitational redshift but this applies locally so I've not looked too much into that).

I would be interested to see one of the equations that the Ned Wright & Morgan calculators use in order to work out the now and then figures. I noted there was an equation in 'Hubble parameter in early universe' thread-

$$dt = \frac{da}{H_0\sqrt{\Omega_m/a^3+\Omega_r/a^4 + \Omega_v}}$$

Would this be an example? If it is (though I could make a guess at what they were), I'd appreciate if some-one could confirm what the quantities are.

regards
Steve

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stevebd1
Gold Member
Thanks again for the posts, Marcus. One last question. Hyperphysics is a great website for basic astrophysics and I'm a little perturbed by the fact that they would provide a cosmic calculator based the SR doppler equation that would produce results for z all the way up to 100,00+ without any mention of the potential effect of the cosmological redshift. Is it possible the Hyperphysics web page is out of date or misinformative even.

Hyperphysics calculator-
http://hyperphysics.phy-astr.gsu.edu/hbase/astro/hubble.html#c3

regards
Steve

Jorrie
Gold Member
I noted there was an equation in 'Hubble parameter in early universe' thread-

$$dt = \frac{da}{H_0\sqrt{\Omega_m/a^3+\Omega_r/a^4 + \Omega_v}}$$

Would this be an example? If it is (though I could make a guess at what they were), I'd appreciate if some-one could confirm what the quantities are.
Sorry, it was a bit sloppy of me not to define all the parameters used there. Here they are: $a$ is the expansion parameter, $H_0$ Hubble's constant in appropriate units (inverse time), $\Omega_m$ is the matter density parameter, $\Omega_r$ the radiation energy density parameter and $\Omega_v$ the vacuum energy density parameter, all as fractions of the critical density.

Apparently, Morgan's calculator does not consider radiation energy density, which is only important in epochs z > 1000. I've used this equation to obtain information for z (or a) vs. t up to $z=10^6$, $t= t_0+10^{11}$ seconds, where it should still be accurate.

BTW, I think that column 3 of yours (post 7) describes look-back time or look-back distance in terms of light travel time, which are obviously equivalent.

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Gold Member
Dearly Missed
...
BTW, I think that column 3 of yours (post 7) describes look-back time or look-back distance in terms of light travel time, which are obviously equivalent.

it gets it about "Wright" as a rough approximation, doesn't it?
here are some sample lookback times using Ned Wright's calculator
http://www.astro.ucla.edu/~wright/CosmoCalc.html

Code:
z  Wright calc  special rel doppler approx (Steve's 3rd column)
1   7.731       8.268
2   10.324     11.025
3   11.476     12.160
4   12.094     12.721
5   12.469     13.036

I would not encourge anybody to think of what you get in Steve's 3rd column as the real lookback time (in a cosmology context)
because it could get them confused. Using SR doppler leads to misconceptions about the lookback time or anything else involving the cosmological redshift z, because the cosmological z is not a doppler. But if you want a rough approximation it is definitely the right order of magnitude. Maybe by adjusting parameters one could get it even closer. Thanks for pointing out the similarity, Jorrie!

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Jorrie
Gold Member
Lookback time

I would not encourge anybody to think of what you get in Steve's 3rd column as the real lookback time (in a cosmology context)
because it could get them confused. Using SR doppler leads to misconceptions about the lookback time or anything else involving the cosmological redshift z, because the cosmological z is not a doppler.

Yep, I agree. I shot from the hip and I've missed!

I have done a little work before on lookback times and found that it is more than just a little "model dependent". As a fair example, I've considered: (i) a flat universe with no cosmological constant against (ii) a flat universe with ~73% vacuum energy component, both for a Hubble constant of ~66 km/s/Mpc and a redshift z=1.

The resulting curves are shown in the attached graph, with widely different lookback times/distances, s1 and s2. Neither has anything to do with Doppler shift, because the redshifts are the same! The lookback distances depend on the shape of the expansion curve.

Lessons: 1. Don't shoot without aiming! 2. Don't confuse cosmological redshift with Doppler shift!

-J

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Gold Member
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Jorrie you impress me as having good knowledge and communication skill. I am not part of PF staff so I can't say anything besides personal opinion---don't speak for anybody but myself. But I would propose that you think about starting a thread which is a volunteer Jorrie basic cosmology tutorial thread.

See if we can't get half a dozen of the regular visitors who are interested in learning basic cosmology to participate in the thread. If you set it up, posted a short outline of what you planned to cover, if people showed interest, then I would certainly steer people to that thread.

You know how to start threads. You just start a thread with a name like
"Basic cosmology tutorial (if there's enough interest)"
and list a very brief "course outline"
and ask people to respond to your first post if they are interested, and if you think it's enough people then you proceed to the next topic of your outline.

I would say make each "lecture" quite brief, don't write a lot, and then if you get responses and questions it makes them work, instead of you working, and only continue following the outline as long as people respond.
I don't want to make it sound like it would be a lot of work for you.
(But it would still be too much work for me, otherwise I'd have done it already )

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stevebd1
Gold Member
While looking for more info regarding the cosmological redshift I found a cosmological calculator that also provided 'Omega Matter then', 'Omega Lambda then' and 'Omega radiation then' (which gets introduced at about z = 38). Unfortunately its inbedded in Yahoo geocities-

http://www.geocities.com/alschairn/cc_e.htm

Steve

hellfire
While looking for more info regarding the cosmological redshift I found a cosmological calculator that also provided 'Omega Matter then', 'Omega Lambda then' and 'Omega radiation then' (which gets introduced at about z = 38). Unfortunately its inbedded in Yahoo geocities-

http://www.geocities.com/alschairn/cc_e.htm
Yes, this is my humble calculator. Nice that you found it useful for you.

Jorrie
Gold Member
Yes, this is my humble calculator. Nice that you found it useful for you.

Nice calculator, Hellfire! Hope you don't mind a few questions in order to understand the interpretation of your results.

Using your default values, I requested redshift 1000 and got stumped by the last three values that your calculator gave, the "Omega-then"s. Since Omega is defined as a ratio of actual density to critical density and the flat universe must always stay at (or near) critical density, should the sum of those "Omega-then"s not have been ~1.0? The calculator yielded:

$$\Omega_m=0.67, \Omega_\Lambda=0, \Omega_r=0.14$$

$$\Omega_m=0.76, \Omega_\Lambda=0, \Omega_r=0.24$$

or am I misinterpreting the meanings?

I assume that "Radius of the observable universe then" does not refer to the radius of the present observable universe back then, but rather the radius of what was observable then?

Jorrie

hellfire
Using your default values, I requested redshift 1000 and got stumped by the last three values that your calculator gave, the "Omega-then"s. Since Omega is defined as a ratio of actual density to critical density and the flat universe must always stay at (or near) critical density, should the sum of those "Omega-then"s not have been ~1.0? The calculator yielded:

$$\Omega_m=0.67, \Omega_\Lambda=0, \Omega_r=0.14$$

$$\Omega_m=0.76, \Omega_\Lambda=0, \Omega_r=0.24$$

or am I misinterpreting the meanings?
Thanks for pointing this out.

Of course space must remain flat for such a model. I had an error in the calculation of $\Omega_r$. The correct formula for the calculation of $$\Omega[/itex] is: [tex]\Omega = \Omega_0 \frac{H^2_0}{H^2} \frac{\rho}{\rho_0}$$

and thus for $\Omega_r$:

$$\Omega_r = \Omega_{r,0} \frac{H^2_0}{H^2} \frac{1}{a^4}$$

I have corrected it now and I get:

$$\Omega_m=0.67, \Omega_\Lambda=0, \Omega_r=0.33$$

I assume that "Radius of the observable universe then" does not refer to the radius of the present observable universe back then, but rather the radius of what was observable then?
Correct.

Jorrie
Gold Member
...
and thus for $\Omega_r$:

$$\Omega_r = \Omega_{r,0} \frac{H^2_0}{H^2} \frac{1}{a^4}$$

I have corrected it now and I get:

$$\Omega_m=0.67, \Omega_\Lambda=0, \Omega_r=0.33$$

Ah, thanks! I thought I've missed something...

What value did you use for $\Omega_{r,0}$?

I've used $8.35 \times 10^{-5}$ and got the answers that I posted last time, quite different from yours, so I expect we've used a different $\Omega_{r,0}$. Maybe it's not a bad idea to put that into an input box as well?

hellfire
What value did you use for $\Omega_{r,0}$?

I've used $8.35 \times 10^{-5}$ and got the answers that I posted last time, quite different from yours, so I expect we've used a different $\Omega_{r,0}$. Maybe it's not a bad idea to put that into an input box as well?
I assumed $\Omega_{r, 0} = 0.0005 \Omega_{m, 0}$. An input field would be of course better. Also for the dark energy one could think of modelling the equation of state instead of assuming a cosmological constant...

stevebd1
Gold Member
$$dt = \frac{da}{H_0\sqrt{\Omega_m/a^3+\Omega_r/a^4 + \Omega_v}}$$
Sorry, it was a bit sloppy of me not to define all the parameters used there. Here they are: $a$ is the expansion parameter, $H_0$ Hubble's constant in appropriate units (inverse time), $\Omega_m$ is the matter density parameter, $\Omega_r$ the radiation energy density parameter and $\Omega_v$ the vacuum energy density parameter, all as fractions of the critical density.

Apparently, Morgan's calculator does not consider radiation energy density, which is only important in epochs z > 1000. I've used this equation to obtain information for z (or a) vs. t up to $z=10^6$, $t= t_0+10^{11}$ seconds, where it should still be accurate.

Jorrie, thanks for posting the info regarding the actual equations involved. Regarding the calculations, would it be acceptable to assume the following-

$a$ = z

$\Omega_m$ = 0.238x10^26 expressed as a fraction of the critical density (if $\Omega_m$ = 0.26 and critical density = 0.918x10^-26 kg/m^3). Same applies to $\Omega_r$ and $\Omega_v$

and what do the following equal-

$dt$ = ? (distance x time?)

$da$ = ? (distance x z?)

regards
Steve

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Jorrie
Gold Member
Regarding the calculations, would it be acceptable to assume the following-
$a$ = z
Actually a=1/(z+1) ~ 1/z, if z >> 1

$\Omega_m$ = 0.238x10^26 expressed as a fraction of the critical density (if $\Omega_m$ = 0.26 and critical density = 0.0.918x10^-26). Same applies to $\Omega_r$ and $\Omega_v$
Correct, but I guess you meant that $\rho_m \approx 0.238\times10^{-26}$ kg/m$^3$?

and what do the following equal-

$dt$ = ? (distance x time?) No, it is the time differential.

$da$ = ? (distance x z?) No, it is the expansion factor differential, so that da/dt is the instantaneous rate of change of the expansion factor.

The way I used it for the very early epochs is to take initial conditions for z and t at some point where it is reasonably well known, e.g., z=1080 and t ~ 380,000 years. Choose $\Delta z$ at some convenient value and step z upwards, meaning decreasing a ~ 1/z, while numerically integrating the equation "backwards" in time as far as you like.

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stevebd1
Gold Member
The way I used it for the very early epochs is to take initial conditions for z and t at some point where it is reasonably well known, e.g., z=1080 and t ~ 380,000 years. Choose $\Delta z$ at some convenient value and step z upwards, meaning decreasing a ~ 1/z, while numerically integrating the equation "backwards" in time as far as you like.

Everything else is pretty much clear but I can safely say you lost me there. Is there any chance you could cut and paste a working example (hopefully nothing that would be a hassle) or provide a link to a web page which goes into further detail.

regards
Steve

See if we can't get half a dozen of the regular visitors who are interested in learning basic cosmology to participate in the thread. If you set it up, posted a short outline of what you planned to cover, if people showed interest, then I would certainly steer people to that thread.

Jorrie,

I am interested in following along in a basic cosmology tutorial.

I'm trying to follow along in this one - very interesting.

I have always wanted to know how we measure / calculate the age of the universe, the recessional speed, how the Hubble constant is approximated, details about dark matter and energy and so on.

There is another that I've always wanted to investigate but I haven't found the time and that is using spherical harmonics to map / calculate the total density and the geometry (open closed, or flat) of the universe.

I'm sure I've butchered some of the terms.

-Sparky_

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Jorrie
Gold Member
Everything else is pretty much clear but I can safely say you lost me there. Is there any chance you could cut and paste a working example (hopefully nothing that would be a hassle) or provide a link to a web page which goes into further detail.

Hi Steve, it's a bit of a hassle to cut and paste from my "model", because it's a spreadsheet - and not designed for general use! I've developed it from what I learned from many textbooks and other sources, so even referencing is no easy task.

What I can do fairly easily is to provide you with a rough algorithm for calculating the basic elements of Friedman's first cosmological equation. If that does not help, I guess Marcus's idea of an introductory tutorial is an option. Let's take my cryptic paragraph in the last post and expand on it a little in 'algorithm style':
The way I used it for the very early epochs is to take initial conditions for z and t at some point where it is reasonably well known, e.g., z=1080 and t ~ 380,000 years. Choose $\Delta z$ at some convenient value and step z upwards, meaning decreasing a ~ 1/z, while numerically integrating the equation "backwards" in time as far as you like.

Start with z=1080, t=380,000 years (roughly the redshift and time of the release of the CMB)

Define the constants $$\Omega_m, \Omega_\Lambda, \Omega_r$$

Define an incremental z ($\Delta z$), e.g., 100

Repeat from here

$$a_1 = 1/(z+1)$$

$$z=z+\Delta z$$ (note: VB [Visual Basic] notation)

$$a = 1/(z+1)$$

$$\Delta a = a-a_1$$ (note: for positive $$\Delta z$$, this will be negative)

$$\Delta t = \frac{\Delta a}{aH_0\sqrt{\Omega_m/a^3+\Omega_r/a^4 + \Omega_\Lambda}}$$ (note: assuming a flat cosmos  note the aH_0 correction)

$$t=t+\Delta t$$ (note: numerical integration of time against expansion factor a)

Repeat until $z > 1,0000$ (or whatever value you want)

This will give you a curve of $a$ against $t$ for the redshift range that you specified. Once this is working, the rest of the cosmological parameters are easily obtained from the algorithm.

Remember that $H_0$ must be in inverse time here, e.g., $H_0=0.0724$/Gy, which is the same as 71 km/s/Mpc. Also watch out for the units of time! I use Gy, but you can convert everything to My or years, whatever you want.

There may be better algorithms, but this one is particularly easy to understand and use.

Regards, Jorrie

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Jorrie
Gold Member
I am interested in following along in a basic cosmology tutorial.

Thinking about it.... stevebd1
Gold Member
I've applied the equation but I'm almost certain I've misinterpreted something as the answer I'm getting is about a third of what it should be. I'd appreciate it if someone could take a look and possibly point out at what point the equation goes astray.

Working with z = 1080, t = 383,000, increment of change ($$\Delta z$$)= 100 (a redshift of 1180 provides a time of 330,000 yrs so we're looking for an answer in the region of $$\Delta t = -53,000\,\text{yrs}$$)-

$$for\,a_1$$

$$a_1 = 1/(z+1)$$

$$a_1 = 1/(1080+1) = 0.000925$$

$$for\,a$$

$$z = z+\Delta z$$

$$z = 1080+100$$

$$a = 1/(1180+1) = 0.000847$$

$$\Delta a = a-a_1$$

$$\Delta a = 0.000847 - 0.000925 = -7.8\text x10^{-5}$$

The parameters for $$\Omega_m, \Omega_\Lambda, \Omega_r$$ I'm assuming are the quantities relavent to that time so-

$$\Omega_m = 0.65\rho_{c} = 0.65\,\text x\,0.918\text x10^{-26}\,\text{kg/m}^{3} = 0.5967\text x10^{-26}\,\text{kg/m}^{3}$$

$$\Omega_r = 0.35\rho_{c}= 0.35\,\text x\,0.918\text x10^{-26}\,\text{kg/m}^{3} = 0.3213\text x10^{-26}\,\text{kg/m}^{3}$$

$$\Omega_\Lambda = 0$$

$$H_0$$= inverse Hubble time $$1/13.78\,\text{Gyr} = 0.07256\,\text{Gyr}$$

The above parameters are put into the equation-

$$\Delta t = \frac{\Delta a}{H_0\sqrt{\Omega_m/a^3+\Omega_r/a^4 + \Omega_\Lambda}}$$

$$\Delta t = \frac{-7.8\text x10^{-5}}{0.07256\sqrt{\frac{0.5967\text x10^{-26}}{0.000847^3}+\frac{0.3213\text x10^{-26}}{0.000847^4} + 0}}$$

$$\Delta t = \frac{-7.8\text x10^{-5}}{0.07256\sqrt{\frac{0.5967\text x10^{-26}}{6.0765\text x10^{-10}}+\frac{0.3213\text x10^{-26}}{5.1468\text x10^{-13}} + 0}}$$

$$\Delta t = \frac{-7.8\text x10^{-5}}{0.07256\sqrt{0.0982\text x10^{-16} + 0.0624\text x10^{-13} + 0}}$$

$$\Delta t = \frac{-7.8\text x10^{-5}}{0.07256\sqrt{6.2498\text x10^{-15}}}$$

$$\Delta t = \frac{-7.8\text x10^{-5}}{0.07256\,\text x\,7.9056\text x10^{-8}}$$

$$\Delta t = \frac{-7.8\text x10^{-5}}{5.74\text x10^{-9}}$$

$$\Delta t = -1.359\text x10^{4} = -13,590$$ (answer in years?)

The answer should be approx. -53,000 years. I tried the same equation with the current values for $$\Omega_m, \Omega_\Lambda, \Omega_r$$ (0.27, 0.73 and 0 respectively) but the answer came out incorrect also.

regards
Steve

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Jorrie
Gold Member
The parameters for $$\Omega_m, \Omega_\Lambda, \Omega_r$$ I'm assuming are the quantities relavent to that time so-

$$\Omega_m = 0.65\rho_{c} = 0.65\,\text x\,0.918\text x10^{-26}\,\text{kg/m}^{3} = 0.5967\text x10^{-26}\,\text{kg/m}^{3}$$ Wrong; must be today's value, 0.27
$$\Omega_r = 0.35\rho_{c}= 0.35\,\text x\,0.918\text x10^{-26}\,\text{kg/m}^{3} = 0.3213\text x10^{-26}\,\text{kg/m}^{3}$$ Wrong; must be today's value, ~8.35E-05

$$\Omega_\Lambda = 0$$ Wrong; must be today's value, 0.73

You must put them just like that into the equation; no multiplication by $\rho_{c}$ , because you need a dimensionless value. Hellfire used a different value for $\Omega_r$; check above if you want to correlate with his values.

The answer should be approx. -53,000 years. I tried the same equation with the current values for $$\Omega_m, \Omega_\Lambda, \Omega_r$$ (0.27, 0.73 and 0 respectively) but the answer came out incorrect also.

The -53,000 years is in the right ballpark! Just use the right values and you should get it. For an actual program, I've used
$\Delta z = 10$, to be more accurate.

Keep trying! Jorrie

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