Deriving EDE Equation: Need Opinions

[jordy1113]

Summary:: Need opinions on how to go about deriving EDE equation

Hello, I am an undergrad starting cosmo research under one of my professors and he assigned me to derive eqn 6 below. My plan now is to use eqn 4 to find the right side of eqn 5 and solve for omega d. I haven't tried this yet but was wondering if this is the right way to go or if there another way anyone sees, thank you : )

 

[LastScattered1090]

Hi @jordy1113 : this is an interesting question! I think it might be useful to step back for a minute and see where all this is coming from, rather than just blindly substituting one equation into another. For any constituent of the "perfect" (largely isotropic & homogenous) "fluid" that we consider to be permeating the Universe, we can write its equation of state as
$$P = w \rho $$
Where $P$ is the pressure of this fluid constituent, and $\rho$ is its density (if we take $c = 1$, then $\rho$ is the energy density). Again, these quantities could describe any individual constituent, whether it be, matter, radiation, or dark energy. Now consider at time $t$, (when the scale factor was $a(t)$), any finite volume $V$ in the Universe. It has internal energy (due to this constituent) of $U = \rho V$. We can consider it to be expanding adiabatically*, so that
$$dU = -P dV $$
Using just the above two equations, you can derive the following first-order differential equation for $\rho(a)$, which is the variation of our constituent's energy density with scale factor.
$$\frac{d\rho}{da} + \frac{3}{a}\rho(1+w) = 0 $$
I strongly encourage you to derive this ODE for yourself. Hint: from the definition of the scale factor, the volume is just given by: $V = V_0a^3$, where $V_0$ is the volume of our region now (at the present day).

*Note that the "adiabatic expansion" argument is a bit hokey, but it does give the right answer. A more rigorous GR-like way to derive this ODE is to consider that the "covariant derivative" of the stress-energy tensor must equal 0, for a perfect fluid. See e.g. equation 2.55 of Modern Cosmology by Scott Dodelson.

You can solve this ODE by method of integrating factors (or whatever) to find that the variation of density with scale factor for any constituent is given by:
$$\boxed{\rho(a)=\rho_0a^{-3(1+w)}}$$
where $\rho_0 = \rho(a=1)$ is the density at the present day.

For matter (baryonic or dark), $w = 0$, giving us a $\rho_m(a) = \rho_m^0 a^{-3}$ dependence as expected, since the number density of matter particles in any given volume just dilutes with the expansion. Here $\rho^0_m$ is just the present-day matter density, and I've just shifted the "naught" from subscript to superscript (like in your notes above) to make room for a subscript telling you what constituent we're dealing with.

For radiation (specifically for an isotropic blackbody radiation field), it can be shown that the relation between radiation pressure and energy density is given by $w = 1/3$, leading to a $\rho_r = \rho_r^0a^{-4}$ dependence: again this as expected, because number density of photons in a given volume dilutes as $a^{-3}$, but then energy per photon decreases by a further factor $a^{-1}$ due to increase of photon wavelength linearly with $a$ (cosmological redshift).

For the simplest form of dark energy, equivalent to a cosmological constant, we take $w = -1$, which from the boxed equation above, clearly leads to $\rho_{d}(a) = \rho^0_d = \mathrm{const}$. However, the whole point of your exercise is to allow for the possibility of more complicated time-dependence of the dark energy density, since the exact time dependence is unknown right now (we need more observational data to constrain it). We introduce this time-dependence as a scale-factor dependence of the equation-of-state parameter for dark energy: $w = w(a) = w_0 +$ (some function of $a$). To be honest, I would assume that $w_0 = -1$ here, since the whole point of this parameterization of $w$ is to separate out its time-dependent and time-independent components. Anyway, for now, let's just keep $w$ (for dark energy) fully general and see how far we can get without specifying a functional form for it. Again from the boxed equation:
$$\rho_d(a)=\rho_d^0a^{-3(1+w)}$$
By definition, the density parameter $\Omega_X$ for constituent $X$ is just the ratio of this constituent's energy density to the critical density. Therefore:
$$\Omega_d(a) \equiv \frac{\rho_d(a)}{\rho_\mathrm{crit}(a)} = \frac{\rho_d^0a^{-3(1+w)}}{\rho_\mathrm{crit}(a)}$$
But what is $\rho_\mathrm{crit}$, and how does it vary with $a$? Well, take a look at the First Friedmann Equation. We know that if the total density $\rho_\mathrm{tot}$ of all constituents is equal to the critical density, then the Universe will be geometrically flat (Euclidean). We can therefore set the curvature term in the First Friedman Equation to zero, just giving us:
$$H^2 = \frac{8\pi G}{3}\rho_\mathrm{tot} = \frac{8\pi G}{3}\rho_\mathrm{crit}$$
From this it follows that at any time in the Universe's history when the scale factor was $a$:
$$\rho_\mathrm{crit} = \frac{3H^2}{8\pi G}$$
And of course, evaluating the Friedmann equation at the present day:
$$\rho^0_\mathrm{crit} = \frac{3H_0^2}{8\pi G}$$
Therefore:
$$\frac{\rho_\mathrm{crit}}{\rho^0_\mathrm{crit}} = \frac{H^2}{H_0^2}$$
As a trick, I figured we could substitute this into the $\Omega_d$ equation above:
$$\Omega_d(a) = \frac{\rho_d^0a^{-3(1+w)}}{\rho^0_\mathrm{crit}\frac{\rho_\mathrm{crit}}{\rho^0_\mathrm{crit}}} = \frac{\Omega_d^0a^{-3(1+w)}}{\frac{\rho_\mathrm{crit}}{\rho^0_\mathrm{crit}}} = \frac{\Omega_d^0a^{-3(1+w)}}{\left(\frac{H}{H_0}\right)^2}$$
This last equality is useful, because the denominator is just the First Friedmann Equation (lefthand side) again, albeit written in a slightly different form:
$$\left(\frac{H}{H_0}\right)^2 = \Omega_r^0a^{-4} + \Omega_m^0a^{-3} + \Omega_k^0a^{-2} + \Omega_d^0a^{-3(1+w)}$$
Again, I strongly encourage you to derive this version of the First Friedmann Equation for yourself. You have all the info you need to above. Hint: bear in mind that $\rho_\mathrm{tot} = \rho_r + \rho_m + \rho_d$. And don't forget the curvature term and its $a$ dependence!
Anyway, we can substitute this righthand side into the denominator of our Omega_d equation to produce:
$$\boxed{\Omega_d(a) = \frac{\Omega_d^0a^{-3(1+w)}}{\Omega_r^0a^{-4} + \Omega_m^0a^{-3} + \Omega_k^0a^{-2} + \Omega_d^0a^{-3(1+w)}}}$$
Now we're getting somewhere! What I mean by that is that I seem to be edging closer to something that looks like your equations 4 and 6, but in a way that is fully general, derived from first principles, without any gobbledygook.

A Few Questions

1. Can you consider the Universe to be flat for the purposes of this exercise? I.e. can you set the curvature density parameter $\Omega_k^0 = 0$?
2. What form exactly is your prof using to parameterize $w(a)$? Is it something like $w = w_0 + \alpha/a$, where alpha is an ad hoc parameter to be fitted to the data? Whatever it is, it would need to be substituted in everywhere you see $w$ above.
3. Where the heck does the Epoch of Matter-Radiation Equality come in? I think I can see a possible benefit to using it as a reference epoch where the Friedmann equations take a simpler form. After all, presumably the whole point of considering the time of equality would be that $\Omega_r(a_\mathrm{eq}) = \Omega_m(a_\mathrm{eq})$, potentially allowing for some simplifications in the last boxed equation above.

If I had more time to play around with this, then what I said in point 3 would be my next direction to take. I.e. I might play around with ratios like $\Omega_d(a) / \Omega_d(a_\mathrm{eq})$

But unfortunately I don't have any more time. Can you take it further from here?

 

[jordy1113]

Hi @jordy1113 : this is an interesting question! I think it might be useful to step back for a minute and see where all this is coming from, rather than just blindly substituting one equation into another. For any constituent of the "perfect" (largely isotropic & homogenous) "fluid" that we consider to be permeating the Universe, we can write its equation of state as
$$P = w \rho $$
Where $P$ is the pressure of this fluid constituent, and $\rho$ is its density (if we take $c = 1$, then $\rho$ is the energy density). Again, these quantities could describe any individual constituent, whether it be, matter, radiation, or dark energy. Now consider at time $t$, (when the scale factor was $a(t)$), any finite volume $V$ in the Universe. It has internal energy (due to this constituent) of $U = \rho V$. We can consider it to be expanding adiabatically*, so that
$$dU = -P dV $$
Using just the above two equations, you can derive the following first-order differential equation for $\rho(a)$, which is the variation of our constituent's energy density with scale factor. I strongly encourage you to derive this ODE for yourself. Hint: from the definition of the scale factor, the volume is just given by: $V = V_0a^3$, where $V_0$ is the volume of our region now (at the present day).

*Note that the "adiabatic expansion" argument is a bit hokey, but it does give the right answer. A more rigorous GR-like way to derive this ODE is to consider that the "covariant derivative" of the stress-energy tensor must equal 0, for a perfect fluid. See e.g. equation 2.55 of Modern Cosmology by Scott Dodelson. Anyway, the ODE IS
$$\frac{d\rho}{da} + \frac{3}{a}\rho(1+w) = 0 $$
You can solve this ODE by method of integrating factors (or whatever) to find that the variation of density with scale factor for any constituent is given by:
$$\boxed{\rho(a)=\rho_0a^{-3(1+w)}}$$
where $\rho_0 = \rho(a=1)$ is the density at the present day.

For matter (baryonic or dark), $w = 0$, giving us a $\rho_m(a) = \rho_m^0 a^{-3}$ dependence as expected, since the number density of matter particles in any given volume just dilutes with the expansion. Here $\rho^0_m$ is just the present-day matter density, and I've just shifted the "naught" from subscript to superscript (like in your notes above) to make room for a subscript telling you what constituent we're dealing with.

For radiation (specifically for an isotropic blackbody radiation field), it can be shown that the relation between radiation pressure and energy density is given by $w = 1/3$, leading to a $\rho_r = \rho_r^0a^{-4}$ dependence: again this as expected, because number density of photons in a given volume dilutes as $a^{-3}$, but then energy per photon decreases by a further factor $a^{-1}$ due to increase of photon wavelength linearly with $a$ (cosmological redshift).

For the simplest form of dark energy, equivalent to a cosmological constant, we take $w = -1$, which from the boxed equation above, clearly leads to $\rho_{d}(a) = \rho^0_d = \mathrm{const}$. However, the whole point of your exercise is to allow for the possibility of more complicated time-dependence of the dark energy density, since the exact time dependence is unknown right now (we need more observational data to constrain it). We introduce this time-dependence as a scale-factor dependence of the equation-of-state parameter for dark energy: $w = w(a) = w_0 +$ (some function of $a$). To be honest, I would assume that $w_0 = -1$ here, since the whole point of this parameterization of $w$ is to separate out its time-dependent and time-independent components. Anyway, for now, let's just keep $w$ (for dark energy) fully general and see how far we can get without specifying a functional form for it. Again from the boxed equation:
$$\rho_d(a)=\rho_d^0a^{-3(1+w)}$$
By definition, the density parameter $\Omega_X$ for constituent $X$ is just the ratio of this constituent's energy density to the critical density. Therefore:
$$\Omega_d(a) \equiv \frac{\rho_d(a)}{\rho_\mathrm{crit}(a)} = \frac{\rho_d^0a^{-3(1+w)}}{\rho_\mathrm{crit}(a)}$$
But what is $\rho_\mathrm{crit}$, and how does it vary with $a$? Well, take a look at the First Friedmann Equation. We know that if the total density $\rho_\mathrm{tot}$ of all constituents is equal to the critical density, then the Universe will be geometrically flat (Euclidean). We can therefore set the curvature term in the First Friedman Equation to zero, just giving us:
$$H^2 = \frac{8\pi G}{3}\rho_\mathrm{tot} = \frac{8\pi G}{3}\rho_\mathrm{crit}$$
From this it follows that at any time in the Universe's history when the scale factor was $a$:
$$\rho_\mathrm{crit} = \frac{3H^2}{8\pi G}$$
And of course, evaluating the Friedmann equation at the present day:
$$\rho^0_\mathrm{crit} = \frac{3H_0^2}{8\pi G}$$
Therefore:
$$\frac{\rho_\mathrm{crit}}{\rho^0_\mathrm{crit}} = \frac{H^2}{H_0^2}$$
As a trick, I figured we could substitute this into the $\Omega_d$ equation above:
$$\Omega_d(a) = \frac{\rho_d^0a^{-3(1+w)}}{\rho^0_\mathrm{crit}\frac{\rho_\mathrm{crit}}{\rho^0_\mathrm{crit}}} = \frac{\Omega_d^0a^{-3(1+w)}}{\frac{\rho_\mathrm{crit}}{\rho^0_\mathrm{crit}}} = \frac{\Omega_d^0a^{-3(1+w)}}{\left(\frac{H}{H_0}\right)^2}$$
This last equality is useful, because the denominator is just the First Friedmann Equation (lefthand side) again, albeit written in a slightly different form:
$$\left(\frac{H}{H_0}\right)^2 = \Omega_r^0a^{-4} + \Omega_m^0a^{-3} + \Omega_k^0a^{-2} + \Omega_da^{-3(1+w)}$$
Again, I strongly encourage you to derive this version of the First Friedmann Equation for yourself. You have all the info you need to above. Hint: bear in mind that $\rho_\mathrm{tot} = \rho_r + \rho_m + \rho_d$
Anyway, we can substitute this righthand side into the denominator of our Omega_d equation to produce:
$$\boxed{\Omega_d(a) = \frac{\Omega_d^0a^{-3(1+w)}}{\Omega_r^0a^{-4} + \Omega_m^0a^{-3} + \Omega_k^0a^{-2} + \Omega_da^{-3(1+w)}}}$$
Now we're getting somewhere! What I mean by that is that I seem to be edging closer to something that looks like your equations 4 and 6, but in a way that is fully general, derived from first principles, without any gobbledygook.

A Few Questions

1. Can you consider the Universe to be flat for the purposes of this exercise? I.e. can you set the curvature density parameter $\Omega_k^0 = 0$?
2. What form exactly is your prof using to parameterize $w(a)$? Is it something like $w = w_0 + \alpha/a$, where alpha is an ad hoc parameter to be fitted to the data? Whatever it is, it would need to be substituted in everywhere you see $w$ above.
3. Where the heck does the Epoch of Matter-Radiation Equality come in? I think I can see a possible benefit to using it as a reference epoch where the Friedmann equations take a similar form. After all, presumably the whole point of considering the time of equality would be that $\Omega_r(a_\mathrm{eq}) = \Omega_m(a_\mathrm{eq})$, potentially allowing for some simplifications in the last boxed equation above.

If I had more time to play around with this, then what I said in point 3 would be my next direction to take. I.e. I might play around with ratios like $\Omega_d(a) / \Omega_d(a_\mathrm{eq})$

But unfortunately I don't have any more time. Can you take it further from here?

Wow thank you so much, I plan to take a look at this later this evening, and if I have any more questions I'll come ask. Thank you so much for taking the time to help me

 

[jordy1113]

Hi @jordy1113 : this is an interesting question! I think it might be useful to step back for a minute and see where all this is coming from, rather than just blindly substituting one equation into another. For any constituent of the "perfect" (largely isotropic & homogenous) "fluid" that we consider to be permeating the Universe, we can write its equation of state as
$$P = w \rho $$
Where $P$ is the pressure of this fluid constituent, and $\rho$ is its density (if we take $c = 1$, then $\rho$ is the energy density). Again, these quantities could describe any individual constituent, whether it be, matter, radiation, or dark energy. Now consider at time $t$, (when the scale factor was $a(t)$), any finite volume $V$ in the Universe. It has internal energy (due to this constituent) of $U = \rho V$. We can consider it to be expanding adiabatically*, so that
$$dU = -P dV $$
Using just the above two equations, you can derive the following first-order differential equation for $\rho(a)$, which is the variation of our constituent's energy density with scale factor.
$$\frac{d\rho}{da} + \frac{3}{a}\rho(1+w) = 0 $$
I strongly encourage you to derive this ODE for yourself. Hint: from the definition of the scale factor, the volume is just given by: $V = V_0a^3$, where $V_0$ is the volume of our region now (at the present day).

*Note that the "adiabatic expansion" argument is a bit hokey, but it does give the right answer. A more rigorous GR-like way to derive this ODE is to consider that the "covariant derivative" of the stress-energy tensor must equal 0, for a perfect fluid. See e.g. equation 2.55 of Modern Cosmology by Scott Dodelson.

You can solve this ODE by method of integrating factors (or whatever) to find that the variation of density with scale factor for any constituent is given by:
$$\boxed{\rho(a)=\rho_0a^{-3(1+w)}}$$
where $\rho_0 = \rho(a=1)$ is the density at the present day.

For matter (baryonic or dark), $w = 0$, giving us a $\rho_m(a) = \rho_m^0 a^{-3}$ dependence as expected, since the number density of matter particles in any given volume just dilutes with the expansion. Here $\rho^0_m$ is just the present-day matter density, and I've just shifted the "naught" from subscript to superscript (like in your notes above) to make room for a subscript telling you what constituent we're dealing with.

For radiation (specifically for an isotropic blackbody radiation field), it can be shown that the relation between radiation pressure and energy density is given by $w = 1/3$, leading to a $\rho_r = \rho_r^0a^{-4}$ dependence: again this as expected, because number density of photons in a given volume dilutes as $a^{-3}$, but then energy per photon decreases by a further factor $a^{-1}$ due to increase of photon wavelength linearly with $a$ (cosmological redshift).

For the simplest form of dark energy, equivalent to a cosmological constant, we take $w = -1$, which from the boxed equation above, clearly leads to $\rho_{d}(a) = \rho^0_d = \mathrm{const}$. However, the whole point of your exercise is to allow for the possibility of more complicated time-dependence of the dark energy density, since the exact time dependence is unknown right now (we need more observational data to constrain it). We introduce this time-dependence as a scale-factor dependence of the equation-of-state parameter for dark energy: $w = w(a) = w_0 +$ (some function of $a$). To be honest, I would assume that $w_0 = -1$ here, since the whole point of this parameterization of $w$ is to separate out its time-dependent and time-independent components. Anyway, for now, let's just keep $w$ (for dark energy) fully general and see how far we can get without specifying a functional form for it. Again from the boxed equation:
$$\rho_d(a)=\rho_d^0a^{-3(1+w)}$$
By definition, the density parameter $\Omega_X$ for constituent $X$ is just the ratio of this constituent's energy density to the critical density. Therefore:
$$\Omega_d(a) \equiv \frac{\rho_d(a)}{\rho_\mathrm{crit}(a)} = \frac{\rho_d^0a^{-3(1+w)}}{\rho_\mathrm{crit}(a)}$$
But what is $\rho_\mathrm{crit}$, and how does it vary with $a$? Well, take a look at the First Friedmann Equation. We know that if the total density $\rho_\mathrm{tot}$ of all constituents is equal to the critical density, then the Universe will be geometrically flat (Euclidean). We can therefore set the curvature term in the First Friedman Equation to zero, just giving us:
$$H^2 = \frac{8\pi G}{3}\rho_\mathrm{tot} = \frac{8\pi G}{3}\rho_\mathrm{crit}$$
From this it follows that at any time in the Universe's history when the scale factor was $a$:
$$\rho_\mathrm{crit} = \frac{3H^2}{8\pi G}$$
And of course, evaluating the Friedmann equation at the present day:
$$\rho^0_\mathrm{crit} = \frac{3H_0^2}{8\pi G}$$
Therefore:
$$\frac{\rho_\mathrm{crit}}{\rho^0_\mathrm{crit}} = \frac{H^2}{H_0^2}$$
As a trick, I figured we could substitute this into the $\Omega_d$ equation above:
$$\Omega_d(a) = \frac{\rho_d^0a^{-3(1+w)}}{\rho^0_\mathrm{crit}\frac{\rho_\mathrm{crit}}{\rho^0_\mathrm{crit}}} = \frac{\Omega_d^0a^{-3(1+w)}}{\frac{\rho_\mathrm{crit}}{\rho^0_\mathrm{crit}}} = \frac{\Omega_d^0a^{-3(1+w)}}{\left(\frac{H}{H_0}\right)^2}$$
This last equality is useful, because the denominator is just the First Friedmann Equation (lefthand side) again, albeit written in a slightly different form:
$$\left(\frac{H}{H_0}\right)^2 = \Omega_r^0a^{-4} + \Omega_m^0a^{-3} + \Omega_k^0a^{-2} + \Omega_d^0a^{-3(1+w)}$$
Again, I strongly encourage you to derive this version of the First Friedmann Equation for yourself. You have all the info you need to above. Hint: bear in mind that $\rho_\mathrm{tot} = \rho_r + \rho_m + \rho_d$. And don't forget the curvature term and its $a$ dependence!
Anyway, we can substitute this righthand side into the denominator of our Omega_d equation to produce:
$$\boxed{\Omega_d(a) = \frac{\Omega_d^0a^{-3(1+w)}}{\Omega_r^0a^{-4} + \Omega_m^0a^{-3} + \Omega_k^0a^{-2} + \Omega_d^0a^{-3(1+w)}}}$$
Now we're getting somewhere! What I mean by that is that I seem to be edging closer to something that looks like your equations 4 and 6, but in a way that is fully general, derived from first principles, without any gobbledygook.

A Few Questions

1. Can you consider the Universe to be flat for the purposes of this exercise? I.e. can you set the curvature density parameter $\Omega_k^0 = 0$?
2. What form exactly is your prof using to parameterize $w(a)$? Is it something like $w = w_0 + \alpha/a$, where alpha is an ad hoc parameter to be fitted to the data? Whatever it is, it would need to be substituted in everywhere you see $w$ above.
3. Where the heck does the Epoch of Matter-Radiation Equality come in? I think I can see a possible benefit to using it as a reference epoch where the Friedmann equations take a simpler form. After all, presumably the whole point of considering the time of equality would be that $\Omega_r(a_\mathrm{eq}) = \Omega_m(a_\mathrm{eq})$, potentially allowing for some simplifications in the last boxed equation above.

If I had more time to play around with this, then what I said in point 3 would be my next direction to take. I.e. I might play around with ratios like $\Omega_d(a) / \Omega_d(a_\mathrm{eq})$

But unfortunately I don't have any more time. Can you take it further from here?

Also if you are interested, here is the paper my question is coming from https://arxiv.org/pdf/astro-ph/0601544.pdf

 

[PeterDonis]

the "covariant derivative" of the stress-energy tensor must equal 0, for a perfect fluid

The "for a perfect fluid" qualifier is not necessary. The covariant divergence of the stress-energy tensor must vanish for any stress-energy tensor, because of the Einstein Field Equation.

 

[PeterDonis]

@jordy1113 please do not quote someone else's entire post in your responses. Just quote the particular parts you are responding to. If you're not responding to any particular part, just reference the post by number instead of quoting all of it.

 

[yucheng]

@jordy1113

Hi!

Just curious, what text/source did your professor assign you for this. Is it a textbook, some journal paper or just his personal notes?

I am asking because I am thinking of doing some "research" myself, maybe by searching up some interesting articles and deriving every equation therein, or to explore some advanced applications found in more specialized textbooks and to work through everything; this is just as an exercise to improve my problem solving $: )$

Thanks!

 

[yucheng]

Hi!
Just curious, what text/source did your professor assign you for this. Is it a textbook, some journal paper or just his personal notes?

I am asking because I am thinking of doing some "research" myself, maybe by searching up some interesting articles and deriving every equation therein, or to explore some advanced applications found in more specialized textbooks and to work through everything; this is just as an exercise to improve my skills $: )$

Thanks!