Talk:Friedmann equations

Context

The author of this article says that the Friedmann equations relate certain cosmological parameters in the context of general relavity. I would like to modify this statement to say that these equations define certain cosmological models in general relativity, usually called the Friedmann dusts (or matter dominated Friedmann models) and Friedmann radiation fluids (or radiation dominated Friedmann models). The equations themselves arise in the course of deriving these models in a comoving coordinate chart.

Omega

I just hunted for 20 minutes and didn't find an article specifically on Omega, the density parameter. I put in a blurb about it on the Omega page, and created two new redirect articles, density of the universe and density parameter. I feel Omega deserves its own page, but I can't do it. --zandperl 13:42, 21 March 2006 (UTC)

Or it could go under either Omega (cosmology) or Omega (astronomy). --zandperl 13:47, 21 March 2006 (UTC)

I agree that we need an Omega (cosmology) article, and density parameter should redirect to it instead of this page. I also can't do it now, but maybe at the end of this semester. --Keflavich 01:45, 16 April 2006 (UTC)

Shouldn't "Density of the Universe" have its own page with discussion of:

-Variation in density.

-large scale variation and structure ~ link to page.

-Talk about steady state universe's that call for creation of matter to maintain constant density of universe.

-Why we think we know what the density is and how we found it. -Why we could be wrong.

Alejandr013 21:05, 4 August 2006 (UTC)

Development

There should be some more development of where the equations come from. I may add some straightforward derivation. Alejandr013 21:02, 4 August 2006 (UTC)

It would be nice to include the actual value of the cosmic density of matter here. I've been searching the web for days trying to find a value, it seems to be one of the values that everybody else knows but never bothers to write down. AJH —Preceding unsigned comment added by 210.98.27.91 (talk) 10:21, 11 April 2008 (UTC)

speed of light

I may be braindead, but I can't for the life of me see how ρ and $p/c^{2}$ have the same units. –Joke 02:27, 14 August 2006 (UTC)

ρ is density, Kilograms over Meters cubed ρ=:${\frac {Kg}{m^{3}}}$

$p/c^{2}$ is Pressure over velocity squared, or Force over area over velocity squared so

${\frac {p}{c^{2}}}={\frac {F}{Ac^{2}}}={\frac {F/A}{(m/s)^{2}}}={\frac {\frac {Kg*(m/s^{2})}{m^{2}}}{(m/s)^{2}}}={\frac {kg}{m^{3}}}$

Differentiating ρ, P, and p in physics can be difficult (did you catch the pun?). I thought it was standard to put pressure P capitalized and p little as momentum, and rho as density. What are the conventions on Wikipedia? Alejandr013 20:49, 14 August 2006 (UTC)

I think you left out a $c^{2}$ factor at the beginning -- I've taken the liberty of correcting it. Yes, it all comes down to the naming conventions for GR. Is $\rho$ the mass density or the energy density? In non-GR physics $\rho$ is normally mass density, but since c=1 is normally set by Geometrized unit system it doesn't matter most of the time. Of course a general audience will not necessarily know of these unit conventions -- and according to WP policy articles should written with the general public in mind as well as the experts.

There is an unoffical set of naming conventions in GR: User:Hillman/WikiProject GTR/Policies. --Michael C. Price ^talk 01:25, 15 August 2006 (UTC)

The $\rho$ which appears in the equations is mass-density, as noted above, but later in the article there is talk of energy density and vacuum energy. This is confusing, at least it was to me; the first time I read it I assumed $\rho$ was indeed energy density, but then noticed the dimensions of the equation wouldn't balance. I think it should be clarified by specifically saying 'mass density' when ρ is introduced. Or, perhaps by changing to $\rho$ being energy density which seems to be used on other pages (e.g. cosmological constant). Then the term becomes

${\frac {4\pi G}{c^{2}}}\left(\rho _{e}+3p\right)$

The subscript e for energy density would be nice, but I guess this is no place to introduce new notation? E4mmacro 04:17, 9 April 2007 (UTC)

mistake

I think, there is a mistake in the equations from the beginning. Instead of a² in the last term on the r.h.s. there should be R², where R = R(t) = a(t).R₀. The dimension of the first equation is then s^-2- same for all terms on the left and right hand side of the equation...

Whoplaysdice 09:56, 29 May 2007 (UTC)

If you are speaking of this equation:

$H^{2}\equiv \left({\frac {\dot {a}}{a}}\right)^{2}={\frac {8\pi G\rho +\Lambda }{3}}-K{\frac {c^{2}}{a^{2}}}$,

then the right-most term already has units of reciprocal seconds squared. The Gaussian curvature when a=1, K has units of reciprocal meters squared. c has units of meters per second. a has units of one (i.e. arbitrary units or unit-less). So it is: m^-2·m²s^-21 = s^-2. Is that OK? JRSpriggs 10:26, 29 May 2007 (UTC)

Thanks for the explanation very much. I haven't noticed that K has units of reciprocal meters squared... I'm sorry. Whoplaysdice 10:32, 29 May 2007 (UTC)

As a matter of fact, if the curvature is constant within a polygon, the area of the polygon times that curvature is just the "spherical" excess (in radians) of the sum of the interior angles of the polygon relative to the value in Euclidean space. JRSpriggs 04:38, 30 May 2007 (UTC)

Expansion lowers temperature

E.pajer (talk · contribs) added a new section Friedmann equations#Useful solutions. The equation of state for the perfect fluid appeared to be the ideal gas law. But then I realized that the "constant" w is proportional to the absolute temperature. However, the temperature of a gas decreases when it expands, see Adiabatic cooling. So the equation should probably be corrected to account for the decrease in temperature. JRSpriggs (talk) 14:10, 23 April 2008 (UTC)

Uh, no, w is not proportional to T. For simple models ("dust", "radiation", etc) w is constant and luckily these models are quite good approximations to the real universe for quite long periods of its history. Only when w is independent of T (or equivalently of a) do we get nice closed-form solutions. Of course the real universe is a mixture of dust, radiation, dark energy etc but usually one dominates. Only when there is a cross-over between cases does the effective value of w change significantly, which happens relatively quickly (especially on a log-log plot!). PaddyLeahy (talk) 12:12, 25 April 2008 (UTC)

To Paddy: Thanks for clarifying that. JRSpriggs (talk) 13:14, 25 April 2008 (UTC)

Simpler 2nd equation

The 2nd equation:

${\dot {H}}+H^{2}\equiv {\frac {\ddot {a}}{a}}=-{\frac {4\pi G}{3}}\left(\rho +{\frac {3p}{c^{2}}}\right)+{\frac {\Lambda c^{2}}{3}},$

could be replaced by:

${\dot {\rho }}=-3H(\rho +{\frac {p}{c^{2}}}),$

which eliminates $\Lambda ,$

The original 2nd equation can be recovered, when combined with the 1st equation. Any objections to replacing the 2nd equation with this? --Michael C. Price ^talk 10:13, 26 April 2008 (UTC)

That is equivalent, so we might want to mention it in the article. However, the original second equation has historical significance; and it is named the "Friedman's acceleration equation". So I think we must retain it. Remember that the equations given emerge more or less directly from Einstein field equations. JRSpriggs (talk) 11:46, 26 April 2008 (UTC)

Mistake

There appears to be a minor error near the beginning of the article where it states that k is the normalized curvature parameter at a=1. It is not defined at a=1 at all; k is the ratio of the modulus squared of a, divided by a squared. That is how it appears in my lecture notes from Imperial College. If my notes are wrong then someone should correct me before next Wednesday when I have a 'Particle Cosmology' exam there! In the meantime, I'll simply remove the "at a=1" bit of the sentence. Dazza79 (talk) 21:06, 23 May 2008 (UTC)

I reverted your edit because the text was correct as it stood. Obviously, I cannot read your notes, so I cannot comment on them. But you should bear in mind that different authors use different notations, different definitions for a, k, etc.. It is imperative that the notation in the article be consistent with itself, not that it be consistent with your notes. JRSpriggs (talk) 00:04, 24 May 2008 (UTC)

I agree with Dazza79; k = +/-1 or 0 is how k is normalised by rescaling a, according to whether the universe is closed, open or flat. --Michael C. Price ^talk 05:45, 24 May 2008 (UTC)

To MichaelCPrice: One can only choose k to be one of those three specific values by sacrificing the connection of a to the present which also means that R₀ would not be the present day radius of curvature. This is not helpful, in my opinion. Adjusting the variables so that the present has a special status is appropriate because, for us who live in the present, it is special. JRSpriggs (talk) 21:30, 24 May 2008 (UTC)

I don't see how renormalising 'k' removes the connection of 'a' to the present? What exactly do you mean by this phrase? The FRW metric for example contains both a 'k' and an 'a(t)'. So surely I am free to choose any value of a, at any time t, including the present, irrespective of my choice of +/-1 or 0 for k. If I choose a=1 to be the present, and k happens to be -1 say, then at a=2 k will still be -1. So how does my renormalised choice of -1 for the value of k affect the value of a? It should be totally independent. Or put another way, If I live in an open universe today I will still be living in an open universe tomorrow. 'a' can be as small or large as you like and the rescaled value of k remains the same. Yet the way I'm reading the article I get the impression that, with the wording as it stands, 'k' has the value it has now only at the present time. So if it is -1 today, it might be 0 at some point in the future? Dazza79 (talk) 22:54, 25 May 2008 (UTC)

To Dazza79: Let the cosmological constant and the equation of state for the contents of the universe both be fixed; and assume that the universe is homogeneous and isotropic, i.e. it has a Friedmann-Lemaître-Robertson-Walker metric. Then the possible instantaneous states (time slices) of the universe form a two-parameter class. The first parameter amounts to choosing a time for the time-slice. You may use any of $t,a,H\,$ (among others) for it. The second parameter describes the spacetime continuum as a whole. When the spatial curvature is positive, the second parameter amounts to choosing the minimal spatial curvature, i.e. the spatial curvature when $H=0\,$. Since the second parameter varies continuously, it cannot be represented correctly by k if k is restricted to have values in the discrete set {-1, 0, +1}.

You asked what can stop you from choosing a and k to be anything you want? Well, what if I want k to be a real number other than -1, 0, or +1; perhaps +4.37? JRSpriggs (talk) 05:14, 27 May 2008 (UTC)

OK. Then I think the word 'normalised' in the article is out of place. If 'k' can vary continuously, then it is not the normalised spatial curvature is it? Seeing the word normalised made me assume that we were taking the often used convention of putting k = +/-1 or 0. Let's simply call it the spatial curvature. I will edit the article accordingly. If you disagree, then my next question is how would you distinguish between the normalised spatial curvature as described above, and the k=+/-1, 0 spatial curvature, which I would certainly describe as normalised? Dazza79 (talk) 07:03, 27 May 2008 (UTC)

Looks good to me: now reads "k is the spatial curvature when a = 1 (i.e. today)". As currently written, k is clearly a dimensional quantity (inverse length squared), so it is only "normalised" in the sense that it applies now, which is already stated. Therefore 'normalised' was at least redundant if not wrong, and certainly misleading. PaddyLeahy (talk) 17:28, 27 May 2008 (UTC)

OK. I agree with dropping "normalised". JRSpriggs (talk) 05:38, 28 May 2008 (UTC)

Hawking & Ellis' Large Scale Structure of Space-Time, have k = +1, 0 or -1 and called this value "normalized".--Michael C. Price ^talk 07:16, 29 May 2008 (UTC)

Surely we can remove all references to R and R₀ by simply stating that a is the hyper-radius of the model universe? To explain the origin of the k term in the equation we need simply note that for a static hypersphere the Riemann curvature term is $6k{\frac {c^{2}}{a^{2}}}$, where k = +/-1 or 0 etc. This makes it very easy to see why k = 0 in the flat example, since a -> infinity when flat, and k = -1 is simply a -> ia for the hyperbolic case. --Michael C. Price ^talk 07:49, 28 May 2008 (UTC)

Why would we want to remove such references? IMHO the article would be incomplete without making clear what these terms mean, since very frequently (as in Dazza79's lecture notes) the Friedman equation is written in terms of R rather than a. PaddyLeahy (talk) 14:17, 28 May 2008 (UTC)

I personally find it clearer to just talk in terms of a and not R, but each according to their taste. There's no reason why we can't have both explanations. BTW I think the current explanation of k is totally confusing (and non-standard -- see note below about MWT and HE). The current explanation states that k is the spatial curvature today, leaving unanswered what it was yesterday and will be tomorrow. (You may think that the next two sentences clears everything up, but I disagree.) Anyway, I leave that clarification to others.

Any objections to a statement that a can also be interpreted as the hyper-radius, in which case k is restricted to +/-1 or 0? --Michael C. Price ^talk 17:35, 28 May 2008 (UTC)

The standard references, Misner, Wheeler & Thorne's Gravitation and Hawking & Ellis' Large Scale Structure of Space-Time, have k = +1, 0 or -1; HE even called its value "normalized".--Michael C. Price ^talk 07:14, 29 May 2008 (UTC)

How nice of JRSpriggs to revert my changes, without explanation here. Charming as ever. I trust an explanation of quite how my changes were "mistaken" will be appearing at some point.--Michael C. Price ^talk 19:16, 31 May 2008 (UTC)

My GR notes by Ray d'Inverno include the comment in the Robertson-Walker metric section 6, subsection 2, "k=+1 ..... The space is closed .... This is why in this case R(t) is often referred to as the 'radius of the universe'." before going on to give the Friedman equations in terms of same k & R. --Michael C. Price ^talk 07:59, 1 June 2008 (UTC)

MichaelCPrice's mistakes

• "$k \over a^{2}$ is one-sixth of the spatial component of the scalar curvature"
  

The scalar curvature, R, is a scalar and as such it has only one component which cannot reasonably be called "spatial".

• "$a$ and $k$ have two possible equivalent definitions"
  

Even if the definitions which follow this are inter-convertible, they are not equivalent because they yield different values for k and a.

• It is not reasonable to call a the radius of curvature of the universe when k=0 because in that case the radius of curvature is either infinite or, more correctly, undefined.

JRSpriggs (talk) 13:38, 1 June 2008 (UTC)

• Well of course they yield different values for k and a, as explained in the text, but $k \over a^{2}$ is invariant and is what appears in the equation. Your point is what precisely?
• For an isotropic homogeneous universe, modeled as a 3-sphere with radius a, we have :$R={\frac {6}{a^{2}}}({\ddot {a}}a+{\dot {a}}^{2}+kc^{2})$. k = 1 for sphere, -1 for a hyperboloid. The spatial component refers to the last term. Perhaps you can suggest a better description?
• It is well understood, even by beginners, that Euclidean geometry corresponds to a surface of a sphere of infinite radius.

--Michael C. Price ^talk 15:59, 1 June 2008 (UTC)

Ambiguous Spelling: Friedmann or Friedman

Throughout the article, his name is spelled two different ways. I think it would be better if we settled on a single spelling; ideally, the same way he did. Apparently, he spelled it with a single 'n', "Friedman". I've hunted around on the web, a bit, and found what I think could be a good reference. It appears to be an image of the title page of one of his papers: http://www.springerlink.com/content/l23864w241673530/fulltext.pdf?page=1 Michael McGinnis (talk) 19:25, 15 March 2009 (UTC)

Of his two famous papers, the earlier one (which you linked) spells it Friedman, and the later one ([1]) spells it Friedmann. The "Friedmann" spelling is overwhelmingly more common in modern physics papers, and it's also the spelling used in every online encyclopedia I checked except Wikipedia, so I think we should stick with "Friedmann", and I think the article on the man himself should be moved as well. -- BenRG (talk) 13:13, 16 March 2009 (UTC)

Actually, his name is "Алекса́ндр Алекса́ндрович Фри́дман" in Russian. The final three characters "ман" transliterate as "man". However, we are supposed to use the name by which most English speaking readers would know him, according to Wikipedia:Naming conventions#Use the most easily recognized name. JRSpriggs (talk) 01:39, 17 March 2009 (UTC)

Thank you, BenRG and JRSpriggs, for clearing this up for me. I do feel that a consistent spelling is needed, if only to reduce the possibility of confusion for people who might think that "Friedman" and "Friedmann" are two different people. I'm going to change all references to "Friedmann".Michael McGinnis (talk) 19:58, 19 March 2009 (UTC)

Mistake on Copernican Principle

Hi all!

I suspect the statement "There are serious consequences if homogeneity and isotropy (the Copernican Principle) are not quite true, [...]" is incorrect. The Copernican Principle states that the Earth does not occupy any particular position in the Universe. What the author refers to is the Cosmological Principle, which states that the Universe is isotropic about all locations and homogeneous. For a reference, please see Peacock's "Cosmological Physics" on page 66:

"[...] most scientists believe that it is not reasonable to adopt a cosmological model in which the universe is simply a joke played for the benefit of mankind. This attitude is encapsulated in the Copernican principle, which states that humans are not privileged observers."

This definition is much more compatible with what Copernicus did actually discover, namely that the Sun and not the Earth was at the center of the Solar System.

There is a certain confusion about this topic in the scientific community. Me too I was confused about the various definitions, until I read pages 65-66 of Peacock's book. They are an eye-opener.

Cheers,

Guido