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H tells us the slope of the R-t curve, therefore A less steep curve means a lower expansion rate. The Hubble constant (H) will change with time and with 1+z. Mini quiz. De-acceleration. - 1. k = ?. 0. = ?. W. R. 1. k = ?. >1. = ?. W. t. W = 0 => k = - 1 and r = 0;=>

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H tells us the slope of the R-t curve, therefore

A less steep curve means a lower expansion rate

The Hubble constant (H) will change with time and with 1+z.


Mini quiz



k = ?


= ?




k = ?


= ?




W = 0 => k = -1 and r = 0;=>

(R/R)2R2 = -kc2 or R = constant = slope of R vs t, R = H0t (as in y = a+ bx with a = 0, x= R b = H0




Einstein-de Sitter is the name for a

k = 0 universe, Wt = 1 universe

See table 11.1, page 305, but we don’t know what q0 is yet!

Physical concepts needed to derive q0 and what it mean to follow

adiabatic expansion dq 0 needed to calculate acceleration change of expansion h with time
Adiabatic Expansion(dQ = 0)needed to calculate acceleration (change of expansion [H] with time)

Also need/use : the first law of thermodynamics is dU = dQ-pdV,

U = total “internal energy”

Q = “heat”

p = pressure, V = volume.

Energy = mass x c2 plus internal kinetic energy (negligible in our case) of the “gas” of the Universe


U is “internal” => don’t include total [bulk] KE (expansion of Universe) or PE (gravity term) of Universe in U.


Adiabatic expansion (dQ = 0) + first law of thermodynamics + equation for pressure (p) =>

derive the relationship between R and t:

Just before brick wall and smaller 1+z (matter dominated era), p = 0 !

Can’t assume p = 0 when in radiation dominated era or if have a cosmological constant

Won’t derive the equations for these case.

adiabatic exp dq 0



q0 = -R0R0/R02

Adiabatic Exp. (dQ = 0)

=> dU = -pdV, needed to derive how the Hubble “constant” changes with time.

U = Mc2. Remember M = r4pR3 /3

combine dU equation with escape equation to derive how H changes with time.

dU = c2d(r4pR3)/3, both r and R depend on t, dU = 0 => can derive De-acceleration parameter (q0; today) = W0/2

Where R with 2 dots over it

= acceleration


The sign of q tells us if universe de-accelerating or accelerating.

Since for here 2q0 = W0, we know the answer (see next slide) :





q0 = -R0R0/R02



From R0 > 0 , R02 > 0 and 2q0 = W0 > 0 =>

R < 0

Matter dominated, L = 0 universe, we’re de-accelerating


going back in time
Going back in time:

Why don’t the curves in the book look “funny” at low t where t is low enough (100,000 yrs after the BB) for radiation to be important?

Because this is at at time when the universe was about 100,000 years old R(t) is too small to be seen on the scale of the plots

one final detail
One Final detail:

Today and back as far as we can see (brick wall):

The “matter dominated era,” p = 0, and W0 = 2q0.

Old books use W0 and 2q0 interchangeably, but these are physically different things:

W0 = density ratio

q0 = de-acceleration parameter


GR/surface geometry

  • one universe
  • self gravitating
  • totally isolated universe
  • matter dominated
  • measure “parameters” of our model: curvature (k) critical density rc, W0, H0 q0.
  • => predict the fate of the universe: expand forever etc. see table 11.1
  • Start from geometry and demand that light travel a geodesic on the surface

Use a metric with curvature (k)

Require that gravity (and GR) apply

relate the concept of an initial “Bang” to future; assume matter dominated (M = (4/3)pR3 x r )

=> rc, H0 , W0 ,and k


pressure = 0 (part of matter dominated model)

Adiabactic expansion

first law of thermodynamics

derive de-acceleration parameter (q) relationship to W


H0 = the Hubble constant = about 50 km/(sec- Mpc)

Mpc = 3 x 1024 cm = 3 x 1019 km

Now to “back fill ” with FAQ:

If the universe started from a point:

Why don’t we see a direction for the explosion?

Why didn’t all the matter end up in one or many black holes?


1. Because everywhere in the universe had its own big bang. There was no preferred spot!

2. (a) Each place in the universe did start from the DECAY of black hole

(b) Matter didn’t come into existence until the density was less than needed to make a black hole

R = 2GM/c2 ; radius of black hole


H02 + kc2/R20 = G8pr0/3

Looking at the escape equation we see we’d like to measure density and the Hubble constant.


Looking at the Robertson-Walker metric we can see how to relate the Hubble constant to the expansion rate. This is a little “fudge” but gives the concept.


Change c to v, dt to t and set k = 0 because we’re so close, and space isn’t curved, set rR = D, divide by t and set R/t to R and then multiply top and bottom by R =>

RrR/R = v or DH = v !



entropy heat death and the like
Entropy, Heat Death, and the like

Book says universe is cooling off and we’re losing energy…

No, because the volume is expanding as well, the Universe is not losing energy.

Rather the energy/unit density is what is decreasing.

entropy heat death and the like1
Entropy, Heat Death, and the like

The book also says we’re headed for more and more disorder (or increased entropy).

On a local scale this is correct.

There is clearly and arrow to time. (None of us is getting any younger)

Absurd to think if the universe started to contract ( k = +1 case, L=0), we’d get younger and eventually turn into babies.

entropy heat death and the like2
Entropy, Heat Death, and the like

The but on the scale of the Universe, we’ve assumed for our calculations there is no change in (total) entropy; the “cognesetti” would recognize this as

dS = dQ /T where S = the entropy and T = the temperature. => Unless T = 0, dQ = 0 (what we assumed in deriving 2q0 = W0 implies dS = 0! Or, entropy remains constant! So we’ve “fudged,” but it works!

entropy heat death and the like3
Entropy, Heat Death, and the like

The answer is we don’t know. The book makes a good point that we probably won’t “bounce” because we will have gained entropy (disorder) therefore be so “disordered”, the universe won’t start “afresh.” But maybe at high temperatures and density, our “normal” definition of entropy fails! So really, we don’t know!