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For flat FLRW cosmology, we can write down two Friedman equations and one matter equation:

(1) $H^2=\frac{8 \pi G}{3} \rho$

(2) $\frac{\ddot{a}}{a} = -\frac{4 \pi G}{3} (\rho +3p)$

(3) $\dot{\rho} = -3 H (\rho +p)$

It is well known that, given an equation of state, (1)+(3) implies (2) [also (1)+(2) implies (3)]. My question is: Why are (2) and (3) (plus equation of state) not a complete set of equations?

I'm sure it's probably completely obvious to some of you, but not to me.

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1 Answer 1

up vote 4 down vote accepted

I suppose a lot of what constitutes the "Friedmann Equation(s)" is just up to definitions. However, with the 3 equations you listed, there will be redundancy.

Here is the normal derivation I usually see (feel free to skip the first two paragraphs if you aren't familiar with tensors/GR):

Given einstien's equation in the form $R_{\mu\nu}=-8 \pi G S_{\mu \nu}$, (where $S_{\mu \nu}$ is related to the stress energy tensor by $S_{\mu \nu}=T_{\mu \nu} - \frac{1}{2}g_{\mu \nu}T^\lambda_\lambda$) and with some playing around with the spatial portion of the Robertson Walker Metric (see Weinburg Cosmology), we can get a Riemann Tensor $R_{ij}=\tilde{R}_{ij}-2\dot{a}^2 \tilde{g}_{ij}-a\ddot{a}\tilde{g}_{ij}$ (tilde means the spacial metric and it's curvature tensor) to $R_{ij}=-[2K+2\dot{a}^2+a\ddot{a}]\tilde{g}_{ij}$ (where $K$ is the curvature constant (-1,0,+1)).

We then decide on a stress energy tensor, using the principles of homogeneity and isotropy (we don't want there to be some strange asymmetry), to get one of the form $T_{00}=\rho$, $T_{i0}=0$, and $T_{ij}=a^2p\tilde{g}_{ij}$. Which gives us $S_{ij}=\frac{1}{2}(\rho-p)a^2\tilde{g}_{ij}$ and $S_{00}=\frac{1}{2}(\rho+3p)$.

Using all this, plugging back into the EFEs, we get two equations (one for i=j=0, and one for the rest):

(1) $-\frac{2K}{a^2}-2\frac{2\dot{a}^2}{a^2}-\frac{\ddot{a}}{a}=-4\pi G (\rho - p)$

(2) $\frac{3\ddot{a}}{a}=-4\pi G(3p+\rho)$

Then, cause the first equation is sorta unwieldy, we can add three times the first equation to the second to get the nicer (and more familiar):

(3) $\dot{a}^2+K=\frac{8}{3}\pi G a^2$

We can also get the following equation from (1) and (2):

(4) $\dot{\rho}=-\frac{3\dot{a}}{a}(\rho + p)$

Which is really no surprise, as this conservation law is found in any solution to the EFEs.

So really what equations you decide to call the "Friedman Equations" is up to whatever your doing. (1) and (2) are the direct consequences of derivation, which you can then use to derive (3) and (4). It just turns out for most cosmology calculations, we don't want to use (1), and the last 3 ((2)(3)(4)) are more useful. However there will be redundancy within these 3 equations (after all, they came from just two equations)!


EDIT: Just to adress the OPs question:

Lets take (2) and (4):

(2) $\frac{3\ddot{a}}{a}=-4\pi G(3p+\rho)$

(4) $\dot{\rho}=-\frac{3\dot{a}}{a}(\rho + p)$

Lets multiply (2) by $\frac{2a\dot{a}}{3}$.

$\frac{2a\dot{a}}{3}[\frac{3\ddot{a}}{a}]=\frac{2a\dot{a}}{3}[-4\pi G(3p+\rho)] => 2\dot{a}\ddot{a}=-\frac{8}{3}\pi G(3p a\dot{a} + \rho a\dot{a})$

Some tricky algebra here:

$=>2\dot{a}\ddot{a} = \frac{8}{3}\pi G(-3 a\dot{a}(\rho+p) + 2\rho a\dot{a})$

We now substitute in the conservation of energy equation.

(5) $2\dot{a}\ddot{a} = -\frac{8}{3}\pi G(\dot{\rho}a^2+2\rho a \dot{a})$

But hey! This looks sorta like what you would get if you differentiated the Friedmann Equation (3), note that $K$ is not time dependent (if it were, that would be crazy!). Since your integrating instead of differentiating you will have constants of integration, but you can just tie that into the $K$ term (which is a sorta interesting way to view what $K$ means).

EDIT: Doing a similar process will show you a way to derive (4) to begin with.

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I guess what I really want to know is: since you can get (2) from (3)&(4) [i.e. you can get the ii Einstein equation from the 00 Einstein equation and conservation of energy] why can't you get (3) from (2)&(4) [i.e. why can't you get the 00 Einstein equation from the ii Einstein equation plus conservation of energy]. –  Joss L Aug 4 '11 at 19:42
    
I added in an explanation of how to do what you ask! :) –  Benjamin Horowitz Aug 5 '11 at 17:12
    
Thanks! If I understand correctly, for the case of scalar field matter (for example) eqns (2) and (4) are equivalent to Hamilton's eqns of motion for (a,phi). You have shown that they imply that the Hamiltonian is conserved (eqn (3) is Hamiltonian=0), but it is not possible to use them to derive the Hamiltonian constraint (Hamiltonian=0), as that is an extra ingredient. I.e. the "constant of integration" you talk about (you can't fold it into K since that is 0,+1,-1 only) is set by the Hamiltonian constraint. –  Joss L Aug 6 '11 at 0:21
    
The (0,+1,-1) is essentially a convention that is made possible by the fact that you can choose $r$ (in the Robertson-Walker Metric) to be unit less, and scale appropriately. If you choose $r$ to have units of length, $k$ represents the Gaussian curvature of the space at present day ($a(t)=1$) in which case it can take on any value. So I would say that using this method (the one in my answer) take the second definition and define the constant of integration as the Gaussian curvature of the space. Someone who is more familiar with classical DG could probably think of some clever reason for this –  Benjamin Horowitz Aug 6 '11 at 0:38
    
I have a feeling this won't work, since you should not be able to derive the Hamiltonian constraint (the 00 Einstein eqn) from Hamilton's equations of motion (the ii Einstein eqn and the matter equation), since the latter hold everywhere in phase space while the former holds only on a lower dimensional constraint surface. –  Joss L Aug 6 '11 at 0:51

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