Example, please (Scale Factor)

[revprez]

Looking at the Friedman solution for a fluid with a given pressure and energy density, we have:

 

[math]H^2 = \left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3}\rho + \frac{\Lambda}{3} - \frac{k}{a^2}[/math]

 

Would someone mind walking me through how we arrive at a?

 

Rev Prez

[Martin]

 

Would someone mind walking me through how we arrive at a?

 

Rev Prez

 

Rev, k is a constant that selects out 3 separate cases, positive curvature, negative curvature, flat

 

observations suggest that the universe is exactly or nearly flat

at a large scale, that is the angles of a triangle add up to 180 degrees (as long as you are not coming too close to some very massive body like a star)

 

so when a student first encounters the Friedmann eqn they should probably set k = 0, which selects the flat case

makes life easier.

 

the next thing they often do is to set the present value of the scale factor equal to one.

 

when you solve any differential equation, whether it is by machine cranking it out tinystep by tinystep, or whether it is by clever calculus methods, you have to start with some initial conditions and so you can start with

a(present) = 1

 

and the present value of the Hubble parameter is a measured thing

H(present) = 71 km/s per megaparsec (in the hideous conventional units they like to use)

 

another equivalent form is

H(present) = 1/(13.8 billion years)

because if you work out the 71 km/second dealie it comes to the reciprocal of a time---it just works out to the reciprocal of 13.8 billion years.

 

now you can imagine doing a machine solution of the thing where you start with a(present) = 1

and you work back in time little by little

because the size of H is telling you HOW FAST a CHANGES!

 

You go back a million years, and use the equation and H to recalculate what a(present - million years) would have been, and then you recalculate what H (present - million years) would have been, again using the equation.

and then you go back another million years, the same way.

 

gradually a curve takes shape, which is the scalefactor curve running thru 1 at the present time.

they call it either a(t) or R(t)

and there is this picture of what it looks like when you solve the Friedman eqn:

 

http://nedwww.ipac.caltech.edu/level5/March03/Lineweaver/Figures/figure14.jpg

 

here the curve is labeled R(t) instead of a(t) but it is the same thing

 

here is the context of that figure in Lineweaver's "Inflation and the CMB"

http://nedwww.ipac.caltech.edu/level5/March03/Lineweaver/Lineweaver7_7.html

 

here is some more stuff about it in the astronomy sticky thread

http://www.scienceforums.net/forums/showthread.php?p=140696#post140696

[Martin]

in the figure, the solid line corresponds to the case where the cosmological constant or dark energy is 0.73 of critical at present.

which is what they think it is

and the density of matter including dark or unseen matter is 0.27

and those add up to 1, which is the flat case (total density of energy is equal to critical value)

 

so in the figure the best line to follow is the dark solid like that is labeled

0.73 and 0.27

 

the other lines are other cases like no dark energy etc etc.

 

these things have to do with the LAMBDA that you plug into the Friedman equation, and with the fact that we are focusing on the flat case where the two parameters have to add up to one (the way 0.73 and 0.27)

 

so that figure gives a little taste of how varying the parameters in the equation, the Lambda and the k, are going to influence the outcome when you do the solution.

 

Personally I do not know any way to tackle this equation except by a machine solution. In effect, write a fortran or BASIC program or whatever.

I do not know any explicit solution that gives a formula for it in any realistic case like 0.73 and 0.27. maybe you can figure one. or maybe you can solve differential equations by a numbercrunch program.

[Rev Prez]

Don't I need an initial condition for [math]\dot{a}[/math] as well?

 

Rev Prez

[Rev Prez]

One other question, all of a sudden I can't post new threads or edit my posts. What's up?

[Martin]

Don't I need an initial condition for [math]\dot{a}[/math] as well?

 

Rev Prez

 

the answer IMO is you dont need it because you already have it' date=' but your question makes me think you know a bit about doing numerical solutions of differential equations (or have some sense for it) and that makes me happy

 

the reason you already have it is that at any given moment t the hubble parameter is BY DEFINITION equal to

a'(t)/a(t)

 

I have written a' instead of a-dot because its a bother, but i mean the timederivative a-dot

 

and when you start, you have set a(t[sub']0[/sub]) = 1

 

and you have an observed measured value for H(t₀)

so that very thing is the same as a'(t₀)/a(t₀)

=a'(t₀)/1 = a'(t₀)

 

===========

 

you cant edit and post?

probably just a meaningless temporary glitch

but I am not even remotely a member of the staff of this establishment

and basically know zero, so cannot say anything definite

[revprez]

the answer IMO is you dont need it because you already have it, but your question makes me think you know a bit about doing numerical solutions of differential equations (or have some sense for it) and that makes me happy.

 

Just some background. I've got multivariate calc and diff eq under my belt, along with linear algebra, abstract algebra and real analysis. I'm trying to pick up differential geometry on my own, and while I can now laboriously understand the algebra, I'm still very iffy with the actual derivations and calculations. Also, I'm not anywhere near the point where I can take an interesting problem like this one and simply write down and follow through mathematically. I'll probably have some more questions after I digest this, but thanks for the help.

 

you cant edit and post?

 

My bad. I logged in as "Rev Prez" instead of "revprez." "Rev Prez" was never validated (the activation email never got to me).

 

Rev Prez

[Martin]

Just some background. I've got multivariate calc and diff eq under my belt, along with linear algebra, abstract algebra and real analysis. I'm trying to pick up differential geometry on my own,...

 

I'm very glad you showed up here. it may turn out we are interested in some of the same things. don't count on me to answer a lot of questions but there are several highly competent people here who do regularly reply

on technical points. they include Severian, Swansont, Tom Mattson.

[Johnny5]

I'm trying to pick up differential geometry on my own

 

I want to learn differential geometry as well, because someone mentioned that it would help me do something which I am trying to do. What is it?

 

Regards

[revprez]

I want to learn differential geometry as well' date=' because someone mentioned that it would help me do something which I am trying to do. What is it?

 

Regards[/quote']

 

Differential geometry is the study of differential manifolds, that is smooth, continuous surfaces with interesting curvature.

 

Rev Prez