but, hubble was thinking linearly. no cosmologists think that way anymore.
No one would still use it as evidence should it have been refuted
It's still right.
another thing is that hubble even reflected on his own work with skepticism. he said that the only general criterion of great distances is the apparent faintness which REQUIRES one determine a nebulae to be stationary or receeding. how do you determine that? that determines the answer! and, if it is mostly by visual aid that we determine this information - of space between nebulae as 'measured' - does that in any way determine a specific size of the universe?
We now have redshift evidence supporting his claim.
where does he get this measurement from ?
You really want to know? Well here you go:
'Derivation of the Hubble parameter
Start with the Friedman equation:
H^2 \equiv \left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G \rho}{3} - \frac{kc^2}{a^2}+ \Lambda
where H is the Hubble parameter, a is the scale factor, G is the gravitational constant, k is the geometry of the universe and equal to -1,0 or 1, and Λ is the cosmological constant.
[edit] Matter dominated universe (with a cosmological constant)
If the universe is matter dominated then the energy density of the universe ρ can just be taken to include matter so:
\rho = \rho_m (a)= \rho_{m0}a^{-3}\frac{1}{1}
where ρm is the density of matter today. We know for non-relativistic particles their energy density decreases proportional to the volume of the universe so the equation above must be true. We can also define (see Friedmann equations page for an explanation of the density parameter Ωm):
\rho_c = \frac{3 H^2}{8 \pi G}
\Omega_m \equiv \frac{\rho_{m0}}{\rho_c} = \frac{8 \pi G}{3 H_0^2}\rho_{m0}
so ρ = ρcΩma − 3. Also by definition
\Omega_k \equiv \frac{kc^2}{(a_0H_0)^2}
and
\Omega_{\Lambda} \equiv \frac{\Lambda H_0^{-2}}{3}
where the subscript zero refers to the values today, and a0 = 1. Substituting all this in into the Friedman equation at the start of this section and replacing a with a = 1 / (1 + z) gives:
H^2(z)= H_0^2 \left( \Omega_M (1+z)^{3} + \Omega_k (1+z)^{2} + \Omega_{\Lambda} \right)
Matter and dark energy dominated universe
If the universe is both matter dominated and dark energy dominated then the above equation for the Hubble parameter will also be a function of the equation of state of dark energy. So now:
\rho = \rho_m (a)+\rho_{de}(a)\frac{1}{1}
where ρde is the energy density of the dark energy. By definition an equation of state in cosmology is P = wρ, and if we substitute this into the fluid equation, which describes how the density of the universe evolves with time:
\dot{\rho}+3\frac{\dot{a}}{a}\left(\rho+P\right)=0
\frac{d\rho}{dt}=-3\frac{da}{dt}\frac{1}{a}\left(\rho+w\rho\right)
\frac{d\rho}{\rho}=-3\frac{da}{a}\left(1+w\right)
If w is constant
\ln{\rho}=-3\left(1+w\right)\ln{a}
\rho=a^{-3\left(1+w\right)}e^{constant}
Therefore for dark energy with a constant equation of state w, \rho_{de}(a)= \rho_{de0}a^{-3\left(1+w\right)}. If we substitute this into the Friedman equation in a similar way as before, but this time set k = 0 which is assuming we live in a flat universe,
H^2(z)= H_0^2 \left( \Omega_M (1+z)^{3} + \Omega_{de}(1+z)^{-3\left(1+w \right)} \right)
If dark energy does not have a constant equation of state w then:
\rho_{de}(a)= \rho_{de0}e^{-3\int\frac{da}{a}\left(1+w(a)\right)}.
and to solve this we must parameterise w(a), for example if w(a) = w0 + wa(1 − a), giving:
H^2(z)= H_0^2 \left( \Omega_M a^{-3} + \Omega_{de}a^{-3\left(1+w_0 +w_a \right)}e^{-3w_a(1-a)} \right)
Proof for it.'
There you go. Good enough?
NB: Not my proof!!
Topic: Religion