**Further mathematical detail **

**Relativistic redshift formula**

If the velocity (v) is comparable with the speed of light (c), then the formula relating the velocity and redshift (z) is

Where θ is the angle the light source is moving with respect to the observer**. **θ = 0 degrees means the light source is moving directly away from the observer, θ = 180 degrees means the light source is moving directly towards the observer and θ = 90 degrees means the light source is moving at right angles to its direction from the observer.

**The deceleration parameter**

As discussed previously, the distance d(t) at a time t of an object moving away from us due to the expansion of the Universe, is given by:

d(t) = d_{o.}a(t)

where d_{o }is the distance of the object at the current age of the Universe (t_{o}) and a(t) is the cosmic scale factor.

The deceleration parameter is a dimensionless number and is defined as:

Where a’(t) is the first derivative of a(t), a’’(t) is the second derivative. The minus sign means that if the second derivative is ** negative**, which means the rate of increase of a(t) is slowing down, then q(t) will be

**.**

*positive*Over the last sixty years or so, most models of the Universe have had a deceleration parameter between -1 (which is a rapidly increasing exponential acceleration) and +3 which is a rapid deceleration. If the deceleration parameter is lower than -1 it cause some interesting challenges because it would mean that there is a faster than exponential expansion.

The Hubble parameter H(t) is equal to the recessional velocity divided by the distance. If we have an object a distance d(t) away then

d(t) = d_{o}a(t)

The recessional velocity v(t) at a distance d(t) is given simply by differentiating which gives

d’(t) = d_{o}a’(t)

Therefore, the Hubble parameter is given by.

If we differentiate the above expression, using the product rule, with u = a’(t) and v= 1 / a(t), then we get

Multiplying both sides of equation 3 by the following factor

gives

Using the definitions of q(t) and H(t) given in equations 1 and 2 gives

*A simple example*

I

*A simple example*

If we have a Universe, with an accelerating expansion in which the scale factor increases as the time squared:

a(t) = (t/t_{o})^{2 } *(In reality the scale factor will be a far more complex function of time than this).*

Then a’(t) = 2t/t_{o}^{2 } and a’’(t) = 2/t_{o}^{2 }

Then from equation 1 the deceleration parameter q(t) is:

Which simplifies to q(t) = – ½ . The value is negative indicating an accelerating expansion

The Hubble parameter H(t) from equation 2 is:

Which simplifies to H(t) = 2/t. As the value of the Hubble parameter is inversely dependent on t, its value decreases with time.