# Talk:Friedmann equations

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## 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.

Isn't "relating practical real parameters within a theoretical context" the same as a "model"? All anyone needs to do is be careful when referring to the "FLRW metric", as derived from GR assuming the Cosmological Principle, or when referring to the "FLRW model", when certain additional assumptions are made in order to arrive at practical solutions within the metric.

Kentgen1 (talk) 13:10, 3 November 2010 (UTC)

## 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.

## 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 ${\displaystyle p/c^{2}}$ have the same units. –Joke 02:27, 14 August 2006 (UTC)

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

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

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

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 ${\displaystyle c^{2}}$ factor at the beginning -- I've taken the liberty of correcting it. Yes, it all comes down to the naming conventions for GR. Is ${\displaystyle \rho }$ the mass density or the energy density? In non-GR physics ${\displaystyle \rho }$ 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 ${\displaystyle \rho }$ 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 ${\displaystyle \rho }$ 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 ${\displaystyle \rho }$ being energy density which seems to be used on other pages (e.g. cosmological constant). Then the term becomes

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

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

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

could be replaced by:

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

which eliminates ${\displaystyle \Lambda ,}$

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 ${\displaystyle t,a,H\,}$ (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 ${\displaystyle H=0\,}$. 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 ${\displaystyle 6k{\frac {c^{2}}{a^{2}}}}$, 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

• "${\displaystyle k \over a^{2}}$ 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".

• "${\displaystyle a}$ and ${\displaystyle k}$ 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 ${\displaystyle k \over a^{2}}$ 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 :${\displaystyle R={\frac {6}{a^{2}}}({\ddot {a}}a+{\dot {a}}^{2}+kc^{2})}$. 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 —Preceding unsigned comment added by Coccoinomane (talkcontribs) 08:26, 24 May 2009 (UTC)

I implemented the above proposed change. I also added the following clarifying line:
"Notice, however, that the mainstream of Cosmological research endorses a homogeneous and isotropic Universe on scales larger than ~100 Mpc." I think it is correct to inform the reader of this.
Cheers,
Guido
Coccoinomane (talk) 08:13, 3 June 2009 (UTC)

To Kentgen1 (talk · contribs): See Help:Section. Heading for sections, subsections, subsubsections, etc. should be created using equals signs on either side of the title. For example, "==Your title here==" generates a section title "Your title here" when left-justified on a line. Thus your use of bold-face text to create a title (e.g. "Caveats to the Use of the Friedmann Equations to Model the Universe") is inappropriate. Also, it is preferred to use tripled single quotation marks to delimit bold-face rather than <b> and </b>.

We do not assume that the matter in the universe is an ideal gas. I do not know how you reached the conclusion that we so assume.

If the Earth were at the center of a large spherical void within a denser surrounding, that would not cause the matter near the edge of the void to accelerate away from us. Thus this is not an alternative explanation for the acceleration due to the cosmological constant.

The deviation of gravitational potential field from its background value is proportional to 1/r. The resulting gravitational force field (gradient of the potential) is proportional to 1/r2. This holds for ordinary celestial bodies as well as super-massive black-holes (outside the near field effect); there is no distinction between types of bodies in this respect. I do not see how you get that this would cause a violation of the cosmological principle which has to do with average behavior. That is, the homogeneity and isotropy are understood to be approximations which hold best at large scales.

Yes, the universe is adiabatic because there is no external system to exchange heat with or do work upon. JRSpriggs (talk) 07:16, 21 June 2010 (UTC)

Third opinion: Edits like this are unacceptable because of their disruptive nature. There is simply no reason to put Wikipedia links inline like that. But even beyond that, there's a lack of reliable sourcing overall. Consider the text:
"The equations also rely on the simplification that the substance of the universe may be treated like an ideal fluid or gas. (http://en.wikipedia.org/wiki/FLRW_metric , http://en.wikipedia.org/wiki/Adiabatic_process , http://en.wikipedia.org/wiki/Ideal_gas_law )"
Wiki articles are not considered reliable, so how is the conclusion that the equations rely on that "simplification" drawn? If we could find some actual sources to back that up then at least the issue could be discussed, but until then it seems to be Kentgen's original research on the topic. — HelloAnnyong (say whaaat?!) 19:14, 1 July 2010 (UTC)
Fourth opinion : the ideal gas law is not assumed. If it was then the pressure would not appear as a free variable to be solved for. Kentgen's original research is incorrect. --Michael C. Price talk 22:57, 24 July 2010 (UTC)

## Derivation of simplified Friedmann equations

Since issues have repeatedly arisen concerning the origin of the equations and the correct values of the constants in them, I think it would be helpful to derive them here. For simplicity, I will focus on the special case when k2≈0 and Λ=0. Any components of tensors or pseudo-tensors omitted below are zero. Assume that

${\displaystyle ds^{2}=-c^{2}dt^{2}+{a(t)}^{2}(dx^{2}+dy^{2}+dz^{2}+dw^{2})\,}$

with

${\displaystyle w={\sqrt {k^{-1}-(x^{2}+y^{2}+z^{2})}}\,.}$

Then

${\displaystyle dw={\frac {-2xdx-2ydy-2zdz}{2w}}\,.}$
${\displaystyle dw^{2}={\frac {(xdx+ydy+zdz)^{2}}{w^{2}}}={\frac {k(xdx+ydy+zdz)^{2}}{1-k(x^{2}+y^{2}+z^{2})}}\approx k(xdx+ydy+zdz)^{2}\,.}$

So the metric tensor (general relativity) is given by

${\displaystyle g_{00}=-c^{2}\,;\quad g_{ij}\approx a^{2}(\delta _{ij}+kx^{i}x^{j})\,}$

where it is understood that a is a function of time only. Thus

${\displaystyle g^{00}=-c^{-2}\,;\quad g^{ij}\approx a^{-2}(\delta _{ij}-kx^{i}x^{j})\,.}$

The partial derivatives of the metric are

${\displaystyle g_{ij,0}\approx 2a{\dot {a}}(\delta _{ij}+kx^{i}x^{j})\,}$
${\displaystyle g_{ij,m}\approx ka^{2}(\delta _{im}x^{j}+x^{i}\delta _{jm})\,.}$

The Christoffel symbol is defined by

${\displaystyle \Gamma _{\beta \gamma }^{\alpha }={1 \over 2}g^{\alpha \sigma }(g_{\sigma \beta ,\gamma }+g_{\gamma \sigma ,\beta }-g_{\beta \gamma ,\sigma })\,.}$

So

${\displaystyle \Gamma _{ij}^{0}\approx c^{-2}a{\dot {a}}(\delta _{ij}+kx^{i}x^{j})\,.}$
${\displaystyle \Gamma _{0j}^{i}=\Gamma _{j0}^{i}\approx a^{-1}{\dot {a}}\delta _{ij}\,.}$
${\displaystyle \Gamma _{jm}^{i}\approx kx^{i}\delta _{jm}\,.}$
${\displaystyle \Gamma _{\rho 0}^{\rho }\approx 3a^{-1}{\dot {a}}\,.}$
${\displaystyle \Gamma _{\rho i}^{\rho }\approx kx^{i}\,.}$

The Ricci curvature tensor is defined by

${\displaystyle R_{\alpha \beta }=\partial _{\rho }{\Gamma _{\beta \alpha }^{\rho }}-\partial _{\beta }\Gamma _{\rho \alpha }^{\rho }+\Gamma _{\rho \lambda }^{\rho }\Gamma _{\beta \alpha }^{\lambda }-\Gamma _{\beta \lambda }^{\rho }\Gamma _{\rho \alpha }^{\lambda }\,.}$

So

${\displaystyle R_{00}\approx 0-3(a^{-1}{\ddot {a}}-a^{-2}{\dot {a}}^{2})+0-3a^{-2}{\dot {a}}^{2}=-3a^{-1}{\ddot {a}}\,.}$
${\displaystyle R_{0i}=R_{i0}\approx 0-0+ka^{-1}{\dot {a}}x^{i}-kx^{i}a^{-1}{\dot {a}}=0\,.}$
${\displaystyle R_{ij}\approx [c^{-2}({\dot {a}}^{2}+a{\ddot {a}})(\delta _{ij}+kx^{i}x^{j})+3k\delta _{ij}]-k\delta _{ij}+3c^{-2}{\dot {a}}^{2}(\delta _{ij}+kx^{i}x^{j})-2c^{-2}{\dot {a}}^{2}(\delta _{ij}+kx^{i}x^{j})=c^{-2}(2{\dot {a}}^{2}+a{\ddot {a}})(\delta _{ij}+kx^{i}x^{j})+2k\delta _{ij}\,.}$
${\displaystyle R=g^{\alpha \beta }R_{\alpha \beta }\approx 3c^{-2}a^{-1}{\ddot {a}}+3c^{-2}(2a^{-2}{\dot {a}}^{2}+a^{-1}{\ddot {a}})+6a^{-2}k=6c^{-2}(a^{-2}{\dot {a}}^{2}+a^{-1}{\ddot {a}})+6a^{-2}k\,.}$

The Einstein field equations (without Λ) are

${\displaystyle R_{\mu \nu }-{1 \over 2}g_{\mu \nu }\,R={8\pi G \over c^{4}}T_{\mu \nu }\,}$

which give

0 0: ${\displaystyle -3a^{-1}{\ddot {a}}-{1 \over 2}(-c^{2})[6c^{-2}(a^{-2}{\dot {a}}^{2}+a^{-1}{\ddot {a}})+6a^{-2}k]=3a^{-2}{\dot {a}}^{2}+3c^{2}a^{-2}k\approx {8\pi G \over c^{4}}\rho c^{4}\,.}$
0 i or i 0: 0 ≈ 0.
i j: ${\displaystyle c^{-2}(2{\dot {a}}^{2}+a{\ddot {a}})(\delta _{ij}+kx^{i}x^{j})+2k\delta _{ij}-{1 \over 2}(a^{2}(\delta _{ij}+kx^{i}x^{j}))[6c^{-2}(a^{-2}{\dot {a}}^{2}+a^{-1}{\ddot {a}})+6a^{-2}k]=c^{-2}(-{\dot {a}}^{2}-2a{\ddot {a}})(\delta _{ij}+kx^{i}x^{j})-k\delta _{ij}\approx {8\pi G \over c^{4}}pa^{2}(\delta _{ij}+kx^{i}x^{j})\,.}$

These simplify to

${\displaystyle a^{-2}{\dot {a}}^{2}+c^{2}a^{-2}k\approx {8\pi G \over 3}\rho \,}$ ( which is the first Friedmann equation )
${\displaystyle (-a^{-2}{\dot {a}}^{2}-2a^{-1}{\ddot {a}})-c^{2}a^{-2}k\approx {8\pi G \over c^{2}}p\,.}$

If we add these together, we get

${\displaystyle -2{\frac {\ddot {a}}{a}}\approx {8\pi G \over 3}\left(\rho +{\frac {3p}{c^{2}}}\right)\,}$

that is

${\displaystyle {\frac {\ddot {a}}{a}}\approx {-4\pi G \over 3}\left(\rho +{\frac {3p}{c^{2}}}\right)\,}$

which is the second Friedmann equation. JRSpriggs (talk) 08:42, 22 March 2013 (UTC)

thank for writing this new section. but how does dw arise in metric? spherical coordinates transfom to Cartesian coordinates?Earthandmoon (talk) 14:16, 21 March 2013 (UTC)
The assumption of Friedmann–Lemaître–Robertson–Walker metric is that our three dimensional space is a hyper-sphere in a four dimensional space (with time added later). w is the extra dimension in that four dimensional space and it is characterized by ${\displaystyle k^{-1}=({\text{radius of curvature of spatial slice when a=1}})^{2}=x^{2}+y^{2}+z^{2}+w^{2}\,.}$ Which coordinate system we use for the three-space does not affect the result, so I just chose what I thought would be the most convenient one. The coordinates I chose are the Cartesian coordinates t,x,y,z associated with the reduced-circumference polar coordinates. JRSpriggs (talk) 14:31, 22 March 2013 (UTC)
Greek indices vary over 0,1,2,3. Roman indices vary over 1,2,3. x1,x2,x3 refer to x,y,z respectively.
If you doubt any step or do not understand it, please indicate the first such questionable step and I will attempt to explain it or fix it. JRSpriggs (talk) 08:04, 26 March 2013 (UTC)
thank you, i got it. Your derivation is nice. :)Earthandmoon (talk) 10:57, 26 March 2013 (UTC)

The trace of the Einstein field equations (which is independent of the choice of the spatial coordinates) is

${\displaystyle -R={8\pi G \over c^{4}}T\,}$
${\displaystyle -[6c^{-2}(a^{-2}{\dot {a}}^{2}+a^{-1}{\ddot {a}})+6a^{-2}k]={8\pi G \over c^{4}}(-c^{2}\rho +3p)\,}$
${\displaystyle {\frac {{\dot {a}}^{2}}{a^{2}}}+{\frac {\ddot {a}}{a}}+{\frac {kc^{2}}{a^{2}}}={-4\pi G \over 3}\left(-\rho +{\frac {3p}{c^{2}}}\right)\,}$

which can be combined with the first Friedmann equation to get the second. JRSpriggs (talk) 08:06, 28 March 2013 (UTC)

### Deriving the continuity equation (conservation of energy)

The time derivative of (00 component)

${\displaystyle {\frac {{\dot {a}}^{2}}{a^{2}}}+{\frac {kc^{2}}{a^{2}}}={\frac {8\pi G}{3}}\rho \,}$

is

${\displaystyle {\frac {2{\dot {a}}{\ddot {a}}}{a^{2}}}-{\frac {2{\dot {a}}^{3}}{a^{3}}}-{\frac {2kc^{2}{\dot {a}}}{a^{3}}}={\frac {8\pi G}{3}}{\dot {\rho }}\,.}$

Substituting in

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

gives

${\displaystyle -H{\frac {8\pi G}{3}}\left(\rho +{\frac {3p}{c^{2}}}\right)-2H{\frac {8\pi G}{3}}\rho ={\frac {8\pi G}{3}}{\dot {\rho }}\,.}$

Throwing out the common factor gives

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

that is,

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

which is the continuity equation. This says that the reduction in the mass in a volume is equal to the amount that moves out its sides due to the expansion of the universe plus the mass equivalent of the energy expended by pressure pushing that mass out. JRSpriggs (talk) 08:44, 28 March 2013 (UTC)

As a check, I will derive the continuity equation by another method. In general, it takes the form

${\displaystyle 0=T^{\mu \nu }{}_{;\nu }=T^{\mu \nu }{}_{,\nu }+T^{\sigma \nu }\Gamma ^{\mu }{}_{\sigma \nu }+T^{\mu \sigma }\Gamma ^{\nu }{}_{\sigma \nu }\,}$

as indicated at stress–energy tensor. In our case, the stress-energy of a fluid at rest is

${\displaystyle T^{00}=\rho \,;\quad T^{ij}=p\,g^{ij}\approx p\,a^{-2}(\delta _{ij}-kx^{i}x^{j})\,.}$

Thus the time component is

${\displaystyle 0=T^{0\nu }{}_{;\nu }=T^{0\nu }{}_{,\nu }+T^{\sigma \nu }\Gamma ^{0}{}_{\sigma \nu }+T^{0\sigma }\Gamma ^{\nu }{}_{\sigma \nu }\,}$
${\displaystyle 0\approx {\dot {\rho }}+p\,a^{-2}(\delta _{ij}-kx^{i}x^{j})\,c^{-2}a{\dot {a}}(\delta _{ij}+kx^{i}x^{j})+\rho \,3a^{-1}{\dot {a}}\,}$
${\displaystyle 0\approx {\dot {\rho }}+3pc^{-2}a^{-1}{\dot {a}}+3\rho a^{-1}{\dot {a}}\,}$
${\displaystyle {\dot {\rho }}\approx -3H(pc^{-2}+\rho )\,}$

which agrees with the continuity equation found above. Let us also examine the i component which is

${\displaystyle 0=T^{i\nu }{}_{;\nu }=T^{i\nu }{}_{,\nu }+T^{\sigma \nu }\Gamma ^{i}{}_{\sigma \nu }+T^{i\sigma }\Gamma ^{\nu }{}_{\sigma \nu }\,}$
${\displaystyle 0\approx -4kpa^{-2}x^{i}+3kpa^{-2}x^{i}+kpa^{-2}x^{i}\,}$
${\displaystyle 0\approx 0\,.}$

This identity means that conservation of linear momentum is automatic in FLRW. JRSpriggs (talk) 06:32, 2 April 2013 (UTC)

## Can the Friedmann equations be solved?

The Friedmann equations and the conserved energy equation are differential equations, can we solve them for a(t) function? How does the Friedmann equations describe the evolution of Universe in time? It describes the expansion of universe, right? but How about initial conditions (t=0)? Could you tell more about Cosmological principle, and why from it we can derive FLRW metric? Thank you for your attention. Earthandmoon (talk) 12:47, 28 March 2013 (UTC)

i read the article and found some useful information about a(t) in case k=0. how does a(t) look with k=1 or -1, and Λ Earthandmoon (talk) 12:52, 28 March 2013 (UTC)
Given an additional formula for the pressure, a numerical solution of the Friedmann equations is possible. You may specify freely k and the values of a and ρ at an initial time. Then a can be updated with the first Friedmann equation; and ρ with the continuity equation. Pressure p must be provided by the added equation.
Exact solutions are possible in some special cases (see Friedmann equations#Useful solutions). If we assume k=0 and
${\displaystyle p=w\rho c^{2}\,}$
for some constant w ≠ −1, then
${\displaystyle {\dot {\rho }}=-3H\left(\rho +{\frac {w\rho c^{2}}{c^{2}}}\right)=-3H(1+w)\rho \,}$
${\displaystyle {\frac {d\ln \rho }{dt}}=-3(1+w){\frac {d\ln a}{dt}}\,}$
${\displaystyle \rho =\rho _{0}\left({\frac {a}{a_{0}}}\right)^{-3(1+w)}\,.}$
So the first Friedmann equation becomes
${\displaystyle {\frac {{\dot {a}}^{2}}{a^{2}}}={\frac {8\pi G}{3}}\rho _{0}\left({\frac {a}{a_{0}}}\right)^{-3(1+w)}\,}$
${\displaystyle \left(a^{{\tfrac {3}{2}}(1+w)-1}{\dot {a}}\right)^{2}={\frac {8\pi G\rho _{0}}{3}}\,a_{0}^{3(1+w)}\,}$
${\displaystyle a^{{\tfrac {3}{2}}(1+w)-1}{\dot {a}}={\sqrt {\frac {8\pi G\rho _{0}}{3}}}\,\,a_{0}^{{\tfrac {3}{2}}(1+w)}.}$
Integrating with respect to t gives
${\displaystyle a^{{\tfrac {3}{2}}(1+w)}=(t-t_{0})\,{\tfrac {3}{2}}(1+w){\sqrt {\frac {8\pi G\rho _{0}}{3}}}\,\,a_{0}^{{\tfrac {3}{2}}(1+w)}+a_{0}^{{\tfrac {3}{2}}(1+w)}\,.}$
${\displaystyle a^{{\tfrac {3}{2}}(1+w)}=\left[(t-t_{0})\,(1+w){\sqrt {6\pi G\rho _{0}}}+1\right]a_{0}^{{\tfrac {3}{2}}(1+w)}\,.}$
${\displaystyle a=a_{0}\left[1+(t-t_{0})\,(1+w){\sqrt {6\pi G\rho _{0}}}\,\right]^{\frac {2}{3(1+w)}}.}$
${\displaystyle \rho =\rho _{0}\left[1+(t-t_{0})\,(1+w){\sqrt {6\pi G\rho _{0}}}\,\right]^{-2}.}$
On the other hand, if w = −1 (i.e. p = −ρc2), then
${\displaystyle \rho =\rho _{0}\,.}$
${\displaystyle a=a_{0}\exp {\left[(t-t_{0})\,{\sqrt {\frac {8\pi G\rho _{0}}{3}}}\,\right]}.}$
JRSpriggs (talk) 10:15, 1 April 2013 (UTC)
The exact solutions above assume that the universe is expanding. So that one takes the positive square-root. If the universe were contracting, then just change the sign of the square-root.
For an expanding universe, w > −1 implies that the universe begins with a big bang at
${\displaystyle t_{\text{Big Bang}}=t_{0}-\left[(1+w){\sqrt {6\pi G\rho _{0}}}\,\right]^{-1}\,.}$
If w < −1, then the universe ends with a big rip at
${\displaystyle t_{\text{Big Rip}}=t_{0}+\left[(-1-w){\sqrt {6\pi G\rho _{0}}}\,\right]^{-1}\,.}$
If w = −1, then neither occurs. JRSpriggs (talk) 10:56, 1 April 2013 (UTC)
What does t mean here? It is our local time, or of an infinite observer, or of the entire universe? Earthandmoon (talk) 12:50, 1 April 2013 (UTC)
Because in both FLRW metric and this article I have not found the definition of t, and ${\displaystyle \tau }$. Earthandmoon (talk) 12:59, 1 April 2013 (UTC)
The physical meaning of the coordinates is implied by the metric equation. In some articles, such as Schwarzschild metric, explanations are provided, but they are actually redundant because they follow from the metric equation. The proper time, τ, is the time measured by a clock moving along the same world line with the test particle. In FLRW, t is time measured by clocks which have been free-falling since the Big Bang and thus are at rest relative to the average motion of local matter (using the assumption of homogeneity and isotropy). This assumes they were synchronized at the Big Bang and thus remain synchronized locally. JRSpriggs (talk) 05:16, 2 April 2013 (UTC)
Thank you for all!Earthandmoon (talk) 15:00, 2 April 2013 (UTC)

The expansion of the universe will turn into a contraction if and when

${\displaystyle {\dot {a}}=0\,;\quad {\ddot {a}}<0,}$

that is, if and when

${\displaystyle \rho \,a^{2}={\frac {3kc^{2}}{8\pi G}}\,}$

(which requires positive curvature of space) and

${\displaystyle p>-{\tfrac {1}{3}}\rho c^{2}\,.}$

JRSpriggs (talk) 05:16, 2 April 2013 (UTC)

## Free fall in FLRW

Particles which have a proper motion in an expanding universe should decelerate relative to their immediate surroundings. To examine this effect, I use the formula for acceleration at User:JRSpriggs/Force in general relativity#Acceleration applied to a free-falling particle

${\displaystyle {\ddot {x}}^{\beta }=-\Gamma _{\alpha \mu }^{\beta }{\dot {x}}^{\alpha }{\dot {x}}^{\mu }+\Gamma _{\alpha \mu }^{0}{\dot {x}}^{\alpha }{\dot {x}}^{\mu }{\dot {x}}^{\beta }\,.}$
${\displaystyle {\ddot {x}}^{i}=-2\Gamma _{0j}^{i}{\dot {x}}^{j}-\Gamma _{jm}^{i}{\dot {x}}^{j}{\dot {x}}^{m}+\Gamma _{jm}^{0}{\dot {x}}^{j}{\dot {x}}^{m}{\dot {x}}^{i}\,.}$
${\displaystyle {\ddot {x}}^{i}\approx -2a^{-1}{\dot {a}}\delta _{ij}{\dot {x}}^{j}-kx^{i}\delta _{jm}{\dot {x}}^{j}{\dot {x}}^{m}+c^{-2}a{\dot {a}}(\delta _{jm}+kx^{j}x^{m}){\dot {x}}^{j}{\dot {x}}^{m}{\dot {x}}^{i}\,.}$
${\displaystyle {\ddot {x}}^{i}\approx -2H{\dot {x}}^{i}-kx^{i}{\dot {x}}^{j}{\dot {x}}^{j}+c^{-2}a^{2}H{\dot {x}}^{j}{\dot {x}}^{j}{\dot {x}}^{i}+c^{-2}a^{2}Hkx^{j}{\dot {x}}^{j}x^{m}{\dot {x}}^{m}{\dot {x}}^{i}\,.}$

JRSpriggs (talk) 11:44, 7 April 2013 (UTC)

## Friedmann qquations from R

We can use ${\displaystyle R=6/a^{2}({\ddot {a}}a+{\dot {a}}^{2}+k)}$ to make Lagrange function ${\displaystyle L={\sqrt {-g}}(R-2\Lambda )}$ to find Friedmann equations using variational principle. But we find only one equation, not both.95.55.17.46 (talk) 14:45, 30 January 2014 (UTC)

See Einstein–Hilbert action for the derivation of the Einstein field equations. Your argument assumes that variations in the metric δgμν which violate the symmetry of the Friedmann–Lemaître–Robertson–Walker metric can be safely ignored. That is false. JRSpriggs (talk) 10:38, 31 January 2014 (UTC)

## Is Omega-k correctly defined?

I can't find any other source for Omega-k "spatial curvature density". It seems intuitive that the definition should be Omega-k = Omega - 1. Which is correct? Further, how is Omega-k related to k? Asgrrr (talk) 12:27, 5 June 2016 (UTC)

## Critical value range and it's Big Bang overrun

The Universe permanently maintains an allowed critical value[2] range of energy with five constituents, a. expansion, b. mean time flow, c.dark matter, d. dark energy, e. leptohadronic matter. All these constituents change proportionally but are extant inside an allowed critical value range. Big Bang launches when there isn't any other method to maintain that equilibrium. We know that the Big Bang event occurred in the past, at approximately 13.8 billion years ago, because it deleted most data of our predecessor universe (all data except very few under the noise level of the low energy range that cause some low energy anisotropy, but under the noise level, although even under the noise level officially aren't detectable we can still see some non random neither homogenous patterns at that range). The maximum anisotropy of the full spectrum occurred before the Big Bang, but due to inflation shifted all paleosympantic (concerning our predecessor universe) energy spectrum to the subnoise (under the threshold level of noise) of the ultra low energy part of our spectrum. A simplistic popular belief is that each universe exists among two critical value enents, the paleosympantic (concerning our predecessor universe) Big Bang that produced our universe, and the universal death meta-Big Bang that will generate our offspring universe. Others focus only on the critical value part of the overall range of possible analogies among the universal constituents. Ω isn't functional at any level. The critical density occurs from a critical value allowed range. Thus the Ω parameter is permanently close to 1, but due to quantum noise never one. At extreme analogies among the energetic contributing universal constituents Ω cannot be close to 1 except if the universe explodes; and exactly that is what it does in order to maintain the Ω critical density inside the allowed critical value. The homogeneity of the thorough universal energy range isn't something other than the Big Bang goal. The Ω critical value range of the critical density itself forces homogeneity by expanding more or less universal regions during the Big Bang to achieve it, but the homogenous full spectrum of energy result, buries the anisotropic noise of the previous universe underneath the ultra low energy part of the spectrum at subthreshold levels, officially non detectable - but due to quantum mechanical probabilistic diffusion of information, will be discovered in the future (with some uncertainty because it is a subthreshold semi-anisotropy, it isn't a complete quantum decohesion or a pure anisotropy, because the Big Bang prevented that from happening, and that makes our job more difficult, but not impossible).

I understand what you mean but your English isn't correct. If the absolute value of the fullrange spectrum of energy is lower than the critical value, the universe is no longer possible to maintain Ω and and thus Big Bang occurs. -- 2.84.220.243  05:59, 26 July 2016‎ (UTC)

## Imaginary radius of curvature

I am not sure what the meaning of the following sentence is:

If k = −1, then (loosely speaking) one can say that i· a is the radius of curvature of the universe.

I would suggest to remove the sentence if there is no easy way to supplement it with a simple clarification. --Falktan (talk) 10:34, 25 September 2016 (UTC)

See Gaussian curvature. If the two principal curvatures have the same sign so that the Gaussian curvature is positive, then the radius of curvature, a, is the reciprocal of their geometric mean. What is meant here is that if the two principal curvatures have opposite signs (e.g. the Gaussian curvature is negative), then the reciprocal of the geometric mean of their absolute values is a. JRSpriggs (talk) 14:06, 25 September 2016 (UTC)