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Critical density of the universe

According to my revision guide we can calculate the critical density of the universe using the equation ρ0=3H028πG \rho_0 = \frac {3H_0^2}{ 8\pi G} but surely as this uses Hubble's constant (and the fact that it's ρ0 \rho_0 not ρc \rho_c ) then this is finding the density of our universe, not the critical density? Or am I getting it all mixed up? Either way ρc=ρ0 \rho_c = \rho_0 to give us a flat universe supposedly, but I don't understand which one I'm finding! :redface:

Edit: Just checked and in the textbook it's down as ρc \rho_c in the equation and in the formula booklet it's ρ0 \rho_0 ... Now I'm even more confused :eek:

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(edited 9 years ago)
Original post by furryface12
According to my revision guide we can calculate the critical density of the universe using the equation ρ0=3H028πG \rho_0 = \frac {3H_0^2}{ 8\pi G} but surely as this uses Hubble's constant (and the fact that it's ρ0 \rho_0 not ρc \rho_c ) then this is finding the density of our universe, not the critical density? Or am I getting it all mixed up? Either way ρc=ρ0 \rho_c = \rho_0 to give us a flat universe supposedly, but I don't understand which one I'm finding! :redface:

Edit: Just checked and in the textbook it's down as ρc \rho_c in the equation and in the formula booklet it's ρ0 \rho_0 ... Now I'm even more confused :eek:

Posted from TSR Mobile


I think probably you're being taught/asked to assume that the universe is critical and therefore that ρc=ρ0

obviously the value you get out depends on the value of H0 you're using and of course IRL there's still rather a lot of uncertainty about what it's value is.
Original post by Joinedup
I think probably you're being taught/asked to assume that the universe is critical and therefore that ρc=ρ0

obviously the value you get out depends on the value of H0 you're using and of course IRL there's still rather a lot of uncertainty about what it's value is.


Yeah that might be it- it does say that further down the page but is fairly unhelpful about it if I'm honest :tongue: True, in the exam questions there are several different values for H0 given so that's the point they're trying to make I think :smile:


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