Hubble's law and the expansion of the Universe.
In 1925 the
American astronomer Edwin Hubble (1889-1953) proposed a very simple relationship
between the distance of a galaxy and its velocity of recession or approach (v). The Hubble
formula provides a very powerful way of determining not only distances of remote galaxies
but also the age of the Universe itself.
He stated that they were related by the
formula:
Velocity of recession (v) = Hubble constant (H) x distance from Earth (r)
where H is Hubble's
constant and r is the distance of the galaxy from the Earth. Now clearly the value of Hubble's
constant is critical to the measurement of the distance of a given galaxy and therefore to the
measurement of the size of the Universe.
At present its
value if thought to be about 70 kms-1 Mpc-1.
This
means that:
The velocity of recession of a galaxy increases by 70 kms-1 for every 1 Mpc increase in distance.
The value of H
can be found by measuring the distance of another galaxy using the period-luminosity
relationship for a Cepheid variable star. The period of intensity variation is directly
proportional to the star's absolute brightness. This brightness variation is due to the balance
between gravitational attraction and the force due to radiation pressure.
It is
important to realise that the number quoted above as the value of H is the value at the
present time.
The value of H will have varied over the lifetime of the universe and will
continue to do so in the future.
The recession velocity of a given galaxy can be found
by measuring the Doppler shift of lines in its spectrum (see Figure 1). Once this is found the
distance of the galaxy can be calculated using the Hubble formula.
Example problem
For a galaxy in the Virgo cluster the Doppler shift at a wavelength of 500 nm is 2 nm.
If the velocity of light in free space is 3x108 ms-1 calculate the velocity of recession of the galaxy.
Δλ = λv/c and so v = cΔλ/λ = 3x108x2x10-9/500x10-9
v = 1.2x106 ms-1 = 1200 kms-1
Value of H in SI units
You can convert the value of H to SI units as follows.
Take the Hubble
constant H to be 70 kms
-1 Mpc
-1 and one light year to be 9.46x10
15
m
One Parsec = 3.26 light years = 3.0857x10
16 m therefore 1 Mpc =
3.0857x10
22 m
So 70 kms
-1Mpc
-1 = 70x10
3/
3.09x10
16x10
6 = 2.27 x 10
-18 ms
-1m
-
1Using the result worked out in the previous example we can now calculate the
distance of the galaxy in the Virgo cluster.
Example problems
Find the distance of the galaxy with a recession velocity of 1200 kms-1 if the Hubble constant is 70 kms-1 Mpc-1.
Using v = Hr
r = v/H = 1200/70 = 17.14 Mpc = 17.14x3.09x1022 = 5.29x1023 m = 5.6x107 light years.
The galaxy is therefore 56 million light years away.
The radius of the observable Universe
If we take the maximum velocity of recession (v
m) to be that of light (3x10
8
ms
-1) we can work out the maximum possible radius of the observable universe.
(R)
This can be found by using Hubble's formula:
v = HR
R =
3x10
8x3.09x10
22/70x10
3 = 1.33x10
26 m
This is equal to 1.4x10
10 light years or
fourteen thousand million light years.
The age of the universe
Assume that the radius of the universe (R) = velocity of recession of the most distant galaxies
(v) multiplied the age of the universe (to) and also that the galaxies have been moving
apart with constant velocity since the beginning of time. Written as a formula:
Age of the Universe (to) = Radius of the Universe (R)/Velocity of recession of the most distant galaxies (v)
Take R = 1.33x10
26 m and if the
maximum possible velocity of recession is that of light, the age of the universe (t
o) can be
found from the equation
t
o = R/v = 1/H = [1/70x10
3] x 3.09 x 10
22 = 4.43 x10
17 s =
1.40 x 10
10 years = 14 thousand million years
It seems that Hubble's constant is
probably 70 kms
-1 Mpc
-1 with an accuracy of +/- 10%.
Notice that since t
o =
1/H a large value of H implies a young universe.
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