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I Comoving mass density leads to linear cosmology?

  1. May 15, 2017 #1
    The Friedmann equation expressed in natural units (##\hbar=c=1##) is given by
    $$\left(\frac{\dot a}{a}\right)^2 = \frac{l_P^2}{3}\rho - \frac{k}{R^2}$$
    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 rename it 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 comoving mass density ##\rho_0##.

    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
    $$\left(\frac{a'}{a}\right)^2 = \frac{l_P^2}{3}\rho a^2 - \frac{k}{R_0^2}.$$
    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, in order to describe the mass density in terms of ##[\hbox{comoving length}]^{-4}## we make the substitution ##\rho \rightarrow \rho_0##.

    Thus the complete comoving Friedmann equation is given by
    $$\left(\frac{a'}{a}\right)^2 = \frac{l_P^2}{3}\rho_0 - \frac{k}{R_0^2}.$$

    Does this dimensional reasoning make sense?

    I can solve the comoving Friedmann equation for the 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.
     
    Last edited: May 15, 2017
  2. jcsd
  3. May 15, 2017 #2

    PeterDonis

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    2016 Award

    Staff: Mentor

    I don't see how this works. It looks to me like the dimensions don't match for all the terms.

    If ##a## is dimensionless, then ##\dot{a} / a## has dimensions of ##[\hbox{proper length}]^{-1}##, so its square has dimensions ##[\hbox{proper length}]^{-2}##, so that is the dimension of the LHS.

    If ##l_P## has dimensions ##[\hbox{proper length}]## and ##\rho## has dimensions ##[\hbox{proper length}]^{-2}## (length over length cubed), then the first term on the RHS is dimensionless. (Note, however, that I'm not sure ##l_P^2## should appear in this term in "natural" units for GR--in those units ##G = 1##. I think you are confusing "natural" units for GR with "natural" units for quantum field theory, in which ##G = 1## doesn't work; there ##G## has to have dimensions of length squared. But that's not applicable here. If you take out ##l_P^2## in this term, its dimensions match the others.)

    If ##k## is dimensionless and ##R## has dimension ##[\hbox{proper length}]##, then the second term on the RHS has dimension ##[\hbox{proper length}]^{-2}##.

    I think you need to recheck your derivation.
     
  4. May 15, 2017 #3
    I'm assuming natural units in terms of quantum field theory so that ##\hbar## and ##c## are set to dimensionless unity. I choose the remaining dimension to be length so that ##G## has dimension ##[\hbox{proper length}]^{2}##.

    Mass density ##\rho## has units of mass over length cubed. Mass has units of inverse length. Therefore the proper mass density has dimension ##[\hbox{proper length}]^{-1}/[\hbox{proper length}]^{3}=[\hbox{proper length}]^{-4}##.
     
  5. May 15, 2017 #4

    PeterDonis

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    2016 Award

    Staff: Mentor

    Ah, ok. You should be aware that that is a very uncommon choice in cosmology. One reason for that might be that it leads one to make mistakes that would not be made if the usual choice of cosmology units of ##c = G = 1## were made. To illustrate that, I'm going to run through a similar derivation to yours, but using the standard cosmology units.

    We start with what you call the "proper" Friedmann equation:

    $$
    \left( \frac{\dot{a}}{a} \right)^2 = \frac{8 \pi}{3} \rho - \frac{k}{R^2}
    $$

    We now make a coordinate transformation to conformal time, which works just as you show it. The result is (making no other substitutions):

    $$
    \left( \frac{a'}{a} \right)^2 = \frac{8 \pi}{3} \rho a^2 - \frac{k a^2}{R^2}
    $$

    Now: what are the units of this equation? The LHS is obviously ##[\hbox{comoving length}]^{-2}##. But the RHS looks like it's still ##[\hbox{proper length}]^{-2}##. Do we need to change the RHS to make the units match?

    The answer is: no, we don't. We don't need to do anything to the RHS because of units. The factor of ##a^2## (##a## is dimensionless so it doesn't change any units) is the only "correction" that needs to be made. Why? Because when you make the coordinate transformation from proper time to coordinate time, you don't change any of the space coordinates, so there is no change in the spatial "length units", only in the time "length units". So if only spatial quantities are involved, as they are on the RHS, then what you are calling "proper length" and "comoving length" are the same.

    Btw, you can also see the problem with your derivation by noting that it ends up with the same equation in "comoving" terms as you had in "proper" terms. That doesn't make sense since you only transformed one of the coordinates (time).
     
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