Harrison B. Prosper Florida State University

YSP

  Island Universes  The Expanding Universe  The Universal Scale Factor  Models of the Universe  Summary

Topics

Henrietta Leavitt 1912

Luminosity-Period Relation of Cepheid Variables

Cepheid Variables

She deserved, but did not get, a Nobel Prize

Island Universes

1924 – Edwin Hubble Using the work of

Henrietta Leavitt, Hubble measured the distances to several galaxies and found that they are immense star systems very far from Earth

The Expanding Universe

The Expanding Universe

1 Mpc (Mega-parsec) = 3.26 x 106 light years (ly)

The Universal Scale Factor

a < 1

D(t1)

t1 = past

D(t0)

a = 1

t0 = today

a > 1 D(t2)

t2 = future

a(t) is the scale factor of the Universe

t is cosmic time

λe = a(t) λo

D(t) = a(t) D(t0)

z = (λo - λe) / λe a = 1

1+ z

The Hubble Parameter

a > 1 d(t2)

t2 = future

D(t) = a(t)D(t0 )D(t) = a(t)D(t0 )D(t)

D(t)=a(t)a(t)

≡ H (t)

H(t) is called the Hubble parameter The Hubble constant H0 is simply the Hubble parameter H(t0) at the present epoch

D(t) ≡ dD

dt= v(t)

Why Can We Assign a Cosmic Time?

We have learned that your now and my now do not coincide as we move relative to each other.

However, since our relative speeds are small relative to c, it is a very good approximation to take our nows to be the same.

The same is true for galaxies. Their motions relative to space are << c. Consequently, we can assign each galaxy approximately the same cosmic time.

9

Models of the Universe

Distribution of Galaxies

APM Galaxy Survey, Steve Maddox, Will Sutherland, George Efstathiou & Jon Loveday

Models of the Universe – II

ds2 = c2dt2 − a2 (t)dl2

D(t) = a(t)dl= a(t)D(t0 ) where t0 is the lifetime of

the Universe

How Far Is Far ?

D1, t1

D0, t0

L = c (t0 – t1)

Models of the Universe – III

E = 1

2 mv2 − GmMD

According to Newton’s laws, the total energy E of a galaxy of mass m at a

distance D is given by

v

D where M is the total mass enclosed within the sphere of radius D

M = ( 43πD3)ρ

D(t) = a(t)d0writing

aa

⎛⎝⎜

⎞⎠⎟

2

= 8π3

Gρ + 2Emd0

2

⎛

⎝⎜⎞

⎠⎟1a2gives

Models of the Universe – IV Alexander Friedmann 1888 - 1925

aa

⎛⎝⎜

⎞⎠⎟

2

= 8π3

Gρ − Kc2

a2

We can write this differential equation as

ρ0 ≡3H0

2

8πG, Ω0 ≡ ρ(1) / ρ0

Ω(a) ≡ [ρ(a) / ρ0 + (1− Ω0 ) / a2 ]

aa= H0 Ω(a)

where

Models of the Universe – IV

Models of the Universe – V

Since the Friedmann equation is a 1st order differential equation

we can re-write it as follows

where C is a constant determined by the initial conditions

aa= H0 Ω(a)

where

H0t =

daa Ω(a)∫ + C

Models of the Universe – VI Georges Lemâitre 1927

Assume Ω0 = 1 a(0) = 0 Ω(a) = Ω0 / a3

and remember that a(t0) = 1

H0t =

daa Ω(a)∫ + C

Summary

 Expansion of Space In 1929, Hubble discovered the expansion of the Universe.

 The Friedmann Equation This equation describes how the scale factor a(t) varies with cosmic time.

Different cosmological models give different predictions for a(t)