0
$\begingroup$

The Friedmann equation expressed in natural units ($\hbar=c=1$) is given by $$\left(\frac{\dot a}{a}\right)^2 = \frac{l_P^2}{3}\rho(t) - \frac{k}{R^2}$$ where $t$ is the proper time measured by a comoving observer, $a(t)$ is the dimensionless scale factor, $l_P=\sqrt{8\pi G\hbar/c^3}$ is the reduced proper Planck length, $\rho(t)$ is the proper mass density, the curvature parameter $k=\{-1,0,1\}$ and $R=R_0a$ is the proper spatial radius of curvature.

Now each quantity in this equation has dimensions of powers of $[\hbox{proper length}]$ therefore it seems reasonable to refer to it as the proper Friedmann equation.

I wish to find the corresponding comoving Friedmann equation that is defined solely in terms of quantities with dimensions of powers of scale-free $[\hbox{comoving length}]$. I want to then solve this equation to find the constant comoving mass density $\rho_0$ from first principles.

In order to achieve this goal I define conformal or scale-free time $\eta$ using $$dt=a\ d\eta$$ so that \begin{eqnarray*} \frac{da}{dt}&=&\frac{da}{d\eta}\frac{d\eta}{dt},\\ \dot{a} &=& \frac{a'}{a}. \end{eqnarray*} The Friedmann equation then becomes $$\left(\frac{a'}{a}\right)^2 = \frac{l_P^2}{3}\rho(\eta) a^2 - \frac{k}{R_0^2}.$$ Now the LHS of the equation and the second term on the RHS have dimensions of $[\hbox{comoving length}]^{-2}$ as required. But the remaining term involving $l_P^2\rho$ still has dimensions of $[\hbox{proper length}]^{-2}$. In order that it has dimensions of $[\hbox{comoving length}]^{-2}$ we need to express the Planck length squared, $l_P^2$, in terms of $[\hbox{comoving length}]^2$ and the mass density in terms of $[\hbox{comoving length}]^{-4}$. In order to express the proper Planck length in terms of $[\hbox{comoving length}]$ we need to divide by the scale factor so that $l_P \rightarrow l_P/a$. Finally, we replace the proper mass density $\rho(\eta)$ with the comoving mass density $\rho_0(\eta)$ which has dimensions $[\hbox{comoving length}]^{-4}$.

Thus the complete comoving Friedmann equation is given by $$\left(\frac{a'}{a}\right)^2 = \frac{l_P^2}{3}\rho_0(\eta) - \frac{k}{R_0^2}.$$ Now each quantity in this equation has dimensions of powers of $[\hbox{comoving length}]$.

Does this dimensional reasoning make sense?

Now let us find a cosmological solution for a Universe with a constant comoving mass density $\rho_0(\eta)=\rho_0$. This Universe would obey the so-called perfect cosmological principle in that it is homogeneous and isotropic in both time and space provided one uses spacetime coordinates that are themselves independent of scale.

I can solve the comoving Friedmann equation for the constant comoving mass density $\rho_0$ by defining a constant time $t_0$ such that $$\left(\frac{a'}{a}\right)^2=\frac{1}{t_0^2}$$ which has the solution $$a(\eta)=e^{\eta/t_0}.$$ By substituting $a(\eta)$ back into the comoving Friedmann equation one finds that the comoving mass density $\rho_0$ is given by $$\rho_0=\frac{3}{l_P^2t_0^2}\left(1+\frac{k t_0^2}{R_0^2}\right).$$ By substituting the $a(\eta)$ expression into $dt=a\ d\eta$ and integrating I find that the scale factor as a function of proper time $t$ takes the simple linear form $$a(t)=\frac{t}{t_0}.$$ By substituting $a(t)$ back into the proper Friedmann equation one finds that the proper mass density $\rho$ is given by $$\rho=\frac{\rho_0}{a^2}.$$ Therefore the comoving Friedmann equation implies a unique functional form for the proper mass density that does not depend on assumptions about the constituents of the Universe.

$\endgroup$
1
$\begingroup$

No, you're not right.

  1. Your first Friedman equation with $\dot a$ is right. If you assume it includes the cosmological constant term. So is your second Friedman equation, with a'.

  2. But then you made a substitution of $\rho a^2$ with $\rho_0$. Unfortunately that is not a constant. You just assume it is. $\rho$ is a function of t, and it is not necessarily proportional to 1/$a^2$. It depends on whether the mass energy is dust, radiation or any other equation of state, or constant for the cosmological constant term, or the combination thereof. See the Friedmann equations at https://en.m.wikipedia.org/wiki/Friedmann_equations

  3. Conformal length and proper length have the same units. They are not different dimensions. Comoving simply factors out the scale factor, which has no dimension. Possibly this led you to the confusion.

$\endgroup$
  • $\begingroup$ Ok - I do make the assumption that the comoving mass density $\rho_0$ is a constant. That seems like a reasonable assumption to me. However my derivation of the comoving Friedmann equation itself does not depend on that assumption but rather on the general dimensional argument that I have given. In answer to your 3rd point I accept that conformal length and proper length have the same units but I still maintain that they can be distinguished as conformal length is scale-free whereas proper length includes a scale factor. $\endgroup$ – John Eastmond May 16 '17 at 9:35
  • $\begingroup$ @JohnEastmond Thanks the fix. Note, comments about your edit should go into the comment of your edit, and not into the post. $\endgroup$ – user259412 May 16 '17 at 10:02
  • $\begingroup$ @John Eastmond. The comoving density as you call it is not constant. Just go by the equations. The density does not scale as 1/$a^2$ as I said in my post. The proof is you get an a(t) that is not right for either dust, radiation or the cosmological constant term. Check some references. Friedman equations are two, and depend on equation of state. Your assumption is not right. $\endgroup$ – Bob Bee May 16 '17 at 17:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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