Timeline for Numerically solving for cosmological scale factor in mathematica
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 3, 2019 at 16:24 | answer | added | Alfred | timeline score: 1 | |
| Dec 3, 2019 at 16:12 | comment | added | Alfred | @G. Smith Thanks ! | |
| Dec 3, 2019 at 15:36 | comment | added | G. Smith | @Alfred Please feel free to answer. | |
| Dec 3, 2019 at 6:15 | comment | added | Alfred | @G. Smith I put my answer as a comment so as not to "steal" if from you, but why don't you answer instead of commenting ? | |
| Dec 3, 2019 at 6:13 | comment | added | Alfred | Anyway using $GeV/\hbar$ as a unit for inverse time is a bit strange. But at least it is meaningful. | |
| Dec 3, 2019 at 6:09 | comment | added | Alfred | As for the units, $H_0$ has the dimension of the inverse of a time (velocity divided by distance). Therefore $H_0$ multiplied by the Planck constant would indeed be an energy. I did not check it, but a reasonable value for $H_0$ might be $1.44 10^{-42} Gev/\hbar$, but you need to divide by $\hbar$ to get a correct result. | |
| Dec 3, 2019 at 5:59 | comment | added | Alfred | G. Smith is right on several points. One is that when you multiplied by π(π‘)^2 you forgot the third term in the right-hand-side of your second equation. It should be $π(π‘)^2Ξ©_Ξ$. But the worst is the left-hand-side. It should be $(ππ/ππ‘)^2$, not $π^2π/ππ‘^2$. It is a first order equation (to the second degree), not a second order equation. So you need only one boundary condition, namely $a(t)=1$ now. | |
| Dec 3, 2019 at 4:59 | answer | added | seVenVo1d | timeline score: 1 | |
| Dec 2, 2019 at 16:58 | comment | added | G. Smith | shouldn't the boundary conditions be $-\infty$ and $0$? I donβt know what that means. Numbers are not boundary conditions. | |
| Dec 2, 2019 at 16:55 | comment | added | G. Smith | If all you did was multiply through by $a(t)^2$, why is $\Omega_\Lambda$ not multiplied by that? And how did the square of a first derivative turn into a second derivative? | |
| Dec 2, 2019 at 14:31 | comment | added | SAMCRO | All I have done in the second equation is multiply through by $a(t)^2$. Also I am following a particle physics paper which is why Hubble is in GeV. In the case where these are the units, shouldn't the boundary conditions be -$\infty$ and $0$? | |
| Dec 2, 2019 at 3:08 | comment | added | G. Smith | Your second equation looks wrong. | |
| Dec 2, 2019 at 3:06 | comment | added | G. Smith | Why do you want to express the Hubble constant in GeV? I suggest expressing it in inverse gigayears. | |
| Dec 2, 2019 at 0:42 | comment | added | G. Smith | Integrate forward and backward from now. | |
| Dec 2, 2019 at 0:04 | comment | added | G. Smith | $t=0$ is the Big Bang; $t=t_0=13.8$ Gy is now. | |
| Dec 1, 2019 at 23:47 | comment | added | G. Smith | $t_0$ doesnβt mean $t=0$. | |
| Dec 1, 2019 at 22:58 | history | asked | SAMCRO | CC BY-SA 4.0 |