3
$\begingroup$

Can we derive the analytic Friedmann Equations without general relativity, starting from completely classical/nonrelativistic arguments? (If we consider sufficiently small volumes.)

$\endgroup$
6
  • $\begingroup$ GR is considered the apogee of classical thinking in the sense that it doesn't use quantum concepts. $\endgroup$ Commented Nov 8, 2018 at 21:32
  • 1
    $\begingroup$ I think the word you want is "nonrelativistic," not "classical." Yes, there are nonrelativistic pseudoderivations of this sort of thing, but they're not rigorous. Why would anyone downvote this? This is a perfectly reasonable question. $\endgroup$
    – user4552
    Commented Nov 8, 2018 at 21:37
  • $\begingroup$ @BenCrowell Yes, that makes more sense. I edited the question. Any idea about where I could see the rigorous pseudo derivations? $\endgroup$ Commented Nov 8, 2018 at 21:46
  • $\begingroup$ You can find it in Coles and Lucchin textbook called cosmology, Weinberg's cosmology has a derivation in section 1.5 but it wasn't as easy to follow as Coles Lucchin. $\endgroup$ Commented Nov 8, 2018 at 22:12
  • 1
    $\begingroup$ As a side note, I think you should approve the answers given to some of your questions, like the one about the velocity of stars, it may not seem like but it require time and focus to craft a decent answer. $\endgroup$ Commented Nov 8, 2018 at 23:12

1 Answer 1

4
$\begingroup$

A Newtonian approach is possible but of of course it is not rigorous.

We're in a newtonian approximation now, and we want to describe the motion of a unit mass at a point P on the surface of a sphere. So let's take this spherical distribution of matter, take a point at a distance $l$ from the origin. The equation of motion of that point is

$$ \frac{d^2l}{dt^2} = - \frac{GM}{l^2} \tag{1} $$

$$ \Rightarrow \, \, \, \ddot{l}\, \dot{l} = \frac{d}{dt} \frac{\dot{l}^2}{2}= - \frac{GM}{l^2} \dot{l} = \frac{d}{dt} \frac{GM}{l} \tag{2} $$ from which you can find

$$ \frac{d}{dt} \Big(\frac{\dot{l}^2}{2} - \frac{GM}{l} \Big) =0 \tag {3} $$

$$ \Rightarrow \, \, \dot{l}^2 = \frac{2GM}{l} + constant \tag{4} $$

This is the equation of conservation of energy per unit mass.

Now in cosmology you have $$l= d_c \frac{a}{a_0} = \tilde{D} a \tag {5}$$

where $d_c$ is the comoving distance and $a$ is the usual scale factor. We put this in $(4)$ and remembering that $\tilde{D}$ is not affected by the time derivative and $M = \frac{4\pi}{3} \rho (\tilde{D} a \, ) ^3$, we get

$$ ( \,\tilde{D} \dot{a} \, )^2 = \frac{2G \frac {4\pi}{3} \rho (\tilde{D} a \, ) ^3}{\tilde{D}a} \tag{6} $$

From which we have

$$ \dot{a} = \frac{8\pi G}{3} \rho a^2 + constant \tag{F2} $$

that is the second Friedmann equation when $constant = -kc^2$.

Now we derive the first Friedmann equation simply replacing $\rho$ with $\rho_{eff} = \rho + \frac{3p}{c^2}$ that is we put a relativistic term in the density by hand. It goes like this, starting from $(1)$

$$ \frac{d^2l}{dt^2} = - \frac{GM}{l^2} = -\frac{G}{l^2} \frac{4\pi}{3} \rho_{eff} l^3 = - \frac{4 \pi G}{3} l \big( \, \rho + \frac{3p}{c^2} \, \big) $$

Using again $(5)$ one obtains the first Friedmann equation

$$ \ddot{a} = - \frac{4 \pi}{3} G \big( \, \rho + \frac{3p}{c^2} \, \big) a \tag{F1} $$

You can find this kind of treatment in Coles-Luccin: Cosmology-The origin and evolution of cosmic structure. I don't know in which chapter cause I don't have the book and I used my notes for this answer, but I'm pretty sure my notes were taken from that book.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.