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Recently, a friend gives me the Friedmann's equations under the following form:

\begin{gathered} \left(\frac{\dot{a}}{a}\right)^{2}=\frac{8 \pi G}{3 c^{4}} \rho+\frac{\Lambda}{3}-\frac{k}{3 a^{2}}\quad(1) \\ \frac{\ddot{a}}{a}=-\frac{4 \pi G}{c^{4}} P-\frac{1}{2}\left(\frac{\dot{a}}{a}\right)^{2}-\frac{k}{2 a^{2}}+\frac{\Lambda}{2}\quad(2) \end{gathered}

I have difficulties to convince myself that they are right since from side I know them under the form :

\begin{aligned} &\frac{R^{\prime \prime}}{R}=-\frac{4 \pi G}{3}\left(\rho+\frac{3 p}{c^{2}}\right)+\frac{\Lambda}{3}\quad(3) \\ &\left(\frac{R^{\prime}}{R}\right)^{2}=\frac{8 \pi G \rho}{3}+\frac{\Lambda}{3}-\frac{k}{R^{2}}\quad(4) \end{aligned}

In order to make the link, I think I have to take $a(t)=\dfrac{R(t)}{R_{0}}$ but even with this convention, I can't get to find the equations (1) and (2). There are factors and terms that differs between eq(2) and eq(3), like between eq(1) and eq(4).

EDIT : so finally, Could someone tell me if (1) and (2) are wrong and if (3) and (4) are right ?

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    $\begingroup$ Have you tried to take a look at the Wikipedia page about the Friedmann equations? $\endgroup$ Commented Mar 17, 2022 at 7:42
  • $\begingroup$ Substitute equation 1 into equation 2 to find a similar equation to that of equation 3. And they are both. If you use $a(t) = R(t)/R_0$, sometimes people let $R_0 = 1$ (like when setting $a_0 = 1$, and if you are still stuck, then Weinberg's textbook (the newer one) helps. $\textbf{Or}$, depending on your knowledge of tensor calculus, Ryden gives a great Newtonian derivation without a curvature term. If you want a thorough derivation, let me know. $\endgroup$
    – MathZilla
    Commented Mar 18, 2022 at 20:05
  • $\begingroup$ @JeanbaptisteRoux . So we can say that eq(1) and eq(2) are wrong, aren't they ? $\endgroup$
    – user87745
    Commented Mar 22, 2022 at 2:32

1 Answer 1

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(1) and (4) are almost the same, except for factors of $c^4$ and $3$ in the first and third terms on the right hand side.

If you add zero in the form $\frac12(8πGρ/3 + Λ/3 - k/R^2) - \frac12(R'/R)^2$ to the right hand side of (3), you get (2), except for a factor of $c^2$ difference in the first term.

So (1) and (2) are almost right, aside from a wrong factor of $3$ and some misplaced factors of $c$.

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  • $\begingroup$ Thanks for your quick answer. What do you mean by the term 'cs" in the sentence "and some misplaced cs" ? $\endgroup$
    – user87745
    Commented Mar 22, 2022 at 2:30
  • $\begingroup$ @youpilat13 Factors of $c$. I reworded. $\endgroup$
    – benrg
    Commented Mar 27, 2022 at 4:58

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