Hubble's law- little bit of help

1. Oct 29, 2004

Cosmo16

Hey, I am really interested in cosmology especially the acceerating vs. decelerating universe debate and I was wondering if anyone could help me out with what the implications hubble's law acutually are.

I know that is states that somthing that is twice as far away then it is moving twice as fast. But, does it say why? Also, how does this help prove that the universe is accelerating?

2. Oct 29, 2004

Garth

Welcome to the forums Cosmo16!

The observations that go into Hubble's law are the red shift in the spectra of distant galaxies, and the distance to those galaxies.
As they are so very far away it is very difficult to measure these distances but it can be done by a variety of methods that combine in what is known as the distance ladder.

When Hubble first discovered his law he had little to go on to determine distance apart from the assumption that if spiral galaxies are roughly the same size and intrinsic brightness, or Absolute magnitude, (these are called standard rulers and standard candles respectively) then those that appeared smaller and fainter must be farther away. Using their angular diameter and distance modulus these distances can be calculated to about two significant places but the assumption that all galaxies are standard introduces a much greater systematic error.
Nevertheless, and using some pretty dodgy interpretation of the data Hubble concluded that z = Hd , where z is the red shift, d is the distance and H is a constant, now known as Hubble's constant. The data has got better in the past 75 years! The most recent determination of H gives it a value of around 70 km/sec/Megaparsec. It is now known to vary with the age of the universe so now it is correctly called the Hubble parameter.

The red shift of light coming from those galaxies can be explained by doppler shift, in which case these galaxies are rushing away from us, and the further they are the faster they go. So either we have intergalactic BO or there is some other explanation for their motion! However, about 10 years before Hubble made this discovery Einstein had realised that his theory of General Relativity predicted that the universe must be either expanding or contracting. He did not like this idea for philosophical reasons and introduced another term, the cosmological constant to hold the universe static. After Hubble's announcement it was realised that Einstein's theory had predicted Hubble's observation. GR explains these red shifts as the effect of the whole universe expanding, like dots on an inflating balloon all moving apart. Every galaxy therefore thinks all the others are rushing away from them. Einstein called his cosmological constant "his biggest blunder", but now we are not so sure, he might have been right in the first place!
So we have observation on the one hand and theory on the other, which interprets and explains the observation; that is the way good science progresses.

His theory in its bare form also predicts that this expansion must be slowing down, because the gravitational attraction of the galaxies upon each other slows down their mutual recession. So about 10 years ago it came as a complete surprise when distant Type Ia supernovae appeared to be fainter than they ought to! If these really are standard candles then the universe must be accelerating in its expansion. One theory to explain this is to invoke the cosmological constant again, but today there are several such competing explanations, for example perhaps these supernovae were simply fainter in the past.
I hope this helps - Garth

Last edited: Oct 29, 2004
3. Oct 29, 2004

Cosmo16

Thanks, also, does measurment of redshift take in to effect that we are not going the same direction as the Supernove we are measuring?

4. Oct 29, 2004

Garth

Each galaxy and therefore supernovae embedded in them have their own peculiar motion, however this is less important at cosmological distances than the Hubble flow, which is caused by the expansion of the space-time of the universe. According to GR it is not the galaxies that are moving in space-time, but it is space-time itself, which is expanding and carrying the galaxies with it. Again imagine dots on a balloon that is being blown up, they all move apart and yet they are not moving across the rubber sheet.
So there may be a component of a galaxy's or supernova's red-shift which is caused by their motion through space-time, but at distance the major part of this is caused by the cosmological expansion of space-time. On the other hand, for a nearby galaxy like Andromeda the peculiar motion overwhelms the cosmological and Andromeda even shows a blue shift, because it is heading towards us!
However for distant galxies/supernovae their peculiar motions are insignificant and are not taken into account.
Garth

Last edited: Oct 29, 2004
5. Oct 29, 2004

Cosmo16

Thanks, I've been wondering about that for awhile.

6. Oct 29, 2004

Chronos

The Hubble flow has enjoyed an interesting history. There has been a great deal of both theoretical and observational evidence suggesting it. Explaining what it is and why it is-what-it-is / was-what-it-was has proven much more difficult. The discovery that it appears to have abruptly changed billions of years ago, as Garth pointed out, does not simplify matters. It at least suggests the effect is complex, that it may not yield to a single, simple explanation.

7. Oct 29, 2004

hellfire

The Hubble law is a relation which gives the recession velocity (due to expansion of space) of an object:

$$v = H d$$

where v is usually in km/seg, d in Mpc (Megaparsec) and H is the Hubble parameter in Km/seg Mpc. H is defined as:

$$H = \frac {\dot{a}} {a}$$

where a is the scale factor (it tells you how much smaller the universe was in a past epoch).

The acceleration of the expansion of space is related to the deceleration parameter:

$$q = \frac {- a \ddot{a}} {\dot{a}^2}$$

It is equivalent to say that, whether the expansion accelerates or decelerates, is determined by the second time derivative of the scale factor, since if $\ddot{a} > 0$, then $q < 0$. The sign of the second time derivative of the scale factor tells you whether the expansion accelerates or decelerates.

The first time derivative of the Hubble parameter is:

$$\dot{H} = \frac {\ddot{a} a - \dot{a}^2} {a^2}$$

$$\dot{H} = - {H^2} (q + 1)$$

Thus $\dot{H} > 0$ only if $q < - 1$.

So, just to clarify the relation between acceleration and Hubble parameter: the sign of the first time derivative of the Hubble parameter does NOT tell you whether the expansion of space accelerates or decelerates. It tells you another thing: $q < - 1$ is a very low value of q (a very strong acceleration). As far as I know it can be proven that a positive time derivative of the Hubble parameter ($\dot{H}$) implies:

a. An equation of state with w < -1 is needed. The parameter w relates density to pressure -which appear in the Friedmann equations- of the content of the universe and w < -1 means the content behaves as a phantom energy, which leads to a big-rip.

b. Some of the energy conditions in general relativity are violated.

I dont think I can prove this. May be someone can (...would be nice).

Last edited: Oct 29, 2004
8. Nov 30, 2004

hellfire

I have found out how to do this. This is surprisingly simple and may be trivial for some of you, but I hope it might be interesting for others...

The time derivative of the Hubble parameter [1]:

$$\dot{H} = \frac {d}{dt}(\frac {\dot{a}} {a}) = \frac {\ddot{a} a - \dot{a}^2} {a^2} = \frac {\ddot{a}} {a} - (\frac {\dot{a}} {a})^2$$

The first Friedmann equation [2]:

$$(\frac {\dot{a}} {a})^2 = \frac{8 \pi G} {3 c^2} \rho - \frac{k c^2} {a^2} + \frac{\Lambda c^2} {3}$$

The second Friedmann equation [3]:

$$\frac {\ddot{a}} {a} = - \frac{4 \pi G} {3 c^2} (\rho + 3p) + \frac{\Lambda c^2} {3}$$

For a flat universe with k = 0, one obtains inserting [3] and [2] into [1] (lambda terms cancel out):

$$\dot{H} = - \frac{4 \pi G} {3 c^2} (\rho + 3p) - \frac{8 \pi G} {3 c^2} \rho$$

$$\dot{H} \frac{3 c^2} {4 \pi G} = - 3 \rho - 3p$$

Therefore (since all constants on the left are positive),

$$\dot{H} > 0$$

implies

$$- 3 \rho - 3p > 0$$

$$p < - \rho$$

or an equation of state:

$$p = w \rho$$

with w < -1.

Last edited: Nov 30, 2004