Separation of variables and integration

Question

I also have that $\rho(t_0)=\rho_0$ and $a(t_0) = 1$.

I'm trying to separate the variables here to reach the desired solution, with my working as follows:

\begin{align*} \frac{\dot{\rho}}{\rho} &= -3\frac{\dot{a}}{a}\left[ 1 + w(t) \right] \\ \int_{t_0}^a \frac{\dot{\rho}}{\rho} \mathrm{dt} &= -3\int_{t_0}^a\frac{\dot{a}'}{a'}\left[ 1 + w(t) \right]\mathrm{dt} \\ \ln{\rho(a)} - \ln{\rho(t_0)}&=-3\int_{1}^a\frac{\dot{a}'}{a'}\left[ 1 + w(t) \right]\mathrm{dt} \\ \rho(a)&=\rho_0 \exp \left(-3\int_{t_0}^a\frac{\dot{a}'}{a'}\left[ 1 + w(t) \right]\mathrm{dt}\right)\\ \rho(a)&=\rho_0 \exp \left(3\int_{a}^1\frac{\mathrm{da'}}{a}\left[ 1 + w(a') \right]\right) \end{align*}

I'm not convinced that my manipulation of limits is correct (changing $t_0$ to 1 when changing $\mathrm{dt}$ to $\mathrm{da'}$) and I also don't entirely understand how the dependencies have been written in the solution. Why is it valid to change $w(t)$ to $w(a')$? If the integral on the LHS is changing the dependency of $\rho$ from $t$ to $a$, mustn't the RHS be integrated before we can do the same to $w$? Is it even valid to have $a$ as a limit of integration if integrating w.r.t $t$?

Answer 1

Physically, it doesn't make sense to integrate with bounds of $t_0$ and $a$ because they don't have the same units ($t$ is time, $a$ is unitless). I assume they did something like this: $$\begin{aligned} \frac{\dot{\rho}}{\rho}&=-3\frac{\dot{a}}{a}[1+w(t)]\\ \int_{t_0}^t \frac{\dot{\rho}(t')}{\rho(t')}\,dt'&=-3\int_{t_0}^t \frac{\dot{a}(t')}{a(t')}[1+w(t')]\,dt'\\ \int_{\rho_0}^{\rho(t)} \frac{1}{\rho'}\,d\rho'&=-3\int_1^{a(t)}\frac{1}{a'}[1+w(a^{-1}(a'))]\,da'\\ \rho(t)&=\rho_0 \exp\left(3\int_{a(t)}^1 \frac{da'}{a'}[1+w(a^{-1}(a'))]\right). \end{aligned}$$ Since $\rho$ and $w$ are functions of $t$, and $a$ is a function of $t$, you can also express $\rho$ and $w$ as functions of $a$. You can then abuse notation a bit to write $$\rho(a)=\rho_0 \exp\left(3\int_a^1 \frac{da'}{a'}[1+w(a')]\right).$$ Here, $t'$, $\rho'$, and $a'$ are used only as "dummy variables" so that $a$, $\rho$, or $t$ do not appear in both the integrand and the limits.

These $\rho(a)$ and $w(a')$ are not the original functions $\rho(t)$ and $w(t)$ evaluated at $t=a$ and $t=a'$, respectively. They are functions of $a$ obtained by taking the original $\rho(t)$ and $w(t)$, substituting $t=f(a)$ (whatever $f$ may be), and then calling these new functions $\rho(a)$ and $w(a)$. Then $w(a')$ is $w(a)$ evaluated at $a=a'$.

It's probably easier to interpret $\rho(t)$, $\rho(a)$, $w(t)$, and $w(a)$ as the quantities $\rho$ and $w$ expressed as functions of $t$ or $a$, rather than the functions $\rho$ and $w$ evaluated at the values $t$ or $a$.

There is a bit of abuse of notation going on, so if you wanted to be clear, you could write the original functions as $\rho_1(t)$ and $w_1(t)$, and the final functions as $\rho_2(a)\equiv\rho_1(t(a))$ and $w_2(a)\equiv w_1(t(a))$.

Answer 2

Actually not a complete answer but way to long for a comment

Let us start on the LHS. As $\dot\rho$ commonly refers to $\frac{\mathrm d\rho(t)}{\mathrm dt}$ we may treat the occuring integral as $$\int_{t_0}^a\frac{\dot\rho}{\rho(t)}\mathrm dt=\int_{t_0}^a\frac{\frac{\mathrm d\rho(t)}{\mathrm dt}}{\rho(t)}\mathrm dt=\int_{t_0}^a\frac{\mathrm d\rho(t)}{\rho(t)}=[\ln\rho(t)]_{t_0}^a=\ln\left(\frac{\rho(a)}{\rho(t_0)}\right)$$ Which you did intuitively. Don't get tricked by the fact that $\rho$ was a function of $t$; after we integrated, and by assuming $a$ is a newly introduced variable, the dependency of $\rho$ changes since we intgrated w.r.t. $\rho(t)$.

Dealing with the RHS is a little bit more problematic. As far as I can tell there is a substitution applied on the RHS; but a rather weird one. I will try to rescue what is left. First of all, lets change the upper limit to $x$. Otherwise this is quite confusing to write. Now we may apply the subsitution $a(t)=a'$ to obtain

$$\int_{t_0}^x\frac{\dot a}{a(t)}[1+w(t)]\mathrm dt=\int_{t_0}^x\frac1{a(t)}[1+w(t)]\frac{\mathrm da(t)}{\mathrm dt}\mathrm dt\stackrel{a(t)=a'}=\int_{a(t_0)}^{a(x)}\frac1{a'}[1+w(a^{-1}(a'))]\mathrm da'$$

Clearly the lower limit becomes $1$ and the inner structure is right. However, the argument of the $w$ function and the upper limit do not match the solution and certainly won't with this approach. On the other hand it is the only one I can think of right now.

It is possible that there is somewhere an assumption regarding to some values of the occuring functions made, as physicists tend to bend mathematics into the shape they need...