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I am trying to deduce the traveled distance of, e.g. a galaxy, as a function over time $D(t)$ in scenarios in which the Hubble value stays fixed or increases proportionally with time.

For the fixed Hubble scenario, I did this by deducing the time needed to travel an infinitesimally small distance $dD$ in which the speed doesn't increase (caused by of the fixed Hubble value). Starting from an initial distance $D_0$ and adding all these time intervals until a certain distance $D$ has been reached gives: $$t(D) = \int_{D_0}^{D} \frac{1}{H\cdot D}\cdot dD = \ln\bigg(\frac{D}{D_0}\bigg)\cdot \frac{1}{H}$$ $$D(t) = e^{Ht}\cdot D_0$$ I am struggling with the scenario in which the Hubble value increases proportionally with time according to $H(t)=H_0+C\cdot t$, where $H_0$ is the value at $t=0$. I know that in this case $$t(D) = \int_{D_0}^{D} \frac{1}{(H_0+Ct)\cdot D}\cdot dD$$ The $t$ within the integrand must be expressed in terms of $D$ to do the integration but then I'd have to know the relationship between $D$ and $t$ which I am trying to deduce in the first place. Rewriting the integration over $dD$, $dt$ or $dH$ always leaves me with an unknown function within the integrand. For example, knowing that $dt=\frac{dH}{C}$ and wanting to integrate over $dH$ until $H_t$ has been reached: $$D(H)=D_0+\int_0^{\frac{H_t-H_0}{C}}H\cdot D\cdot \frac{dH}{C}$$ Here, $D$ should increase with $H$ during the integration but merely knowing that $D(H)=\frac{HD}{H}$ doesn't help me.

Perhaps I'm missing something obvious but is there a way to deduce a definite formula for $D(t)$ when $H$ increases proportionally with time?

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I suggest that you differentiate both sides of your 2nd integral equation, and reposition some terms to get: $$(H_0 + Ct) dt = \frac{dD}{D}.$$ If you then integrate both sides, you get the result for $D(t)$.

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  • $\begingroup$ @AFG Thank you for the edit. $\endgroup$
    – Buzz
    Mar 14 at 0:31
  • $\begingroup$ This slipped me! Thanks a lot, I was able to deduce that $$D(t)=D_0\cdot e^{\frac{1}{2}t(2H_0+Ct)}$$ $\endgroup$
    – Phy
    Mar 14 at 22:01

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