Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 72 / 2 / i

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- Solution i

Solution

Follow successive photons from the same comoving source. Their comoving radial distance is fixed, so

χ = \int_{t_{e} (t_{0})}^{t_{0}} \frac{c d t}{a ( t )}, 0 = \frac{c}{a ( t _{0} )} - \frac{c}{a ( t _{e} )} \frac{d t _{e}}{d t _{0}} .

It follows that

d t_{e} / d t_{0} = a (t_{e}) / a (t_{0}) = 1/ (1 + z)

. Taking the logarithmic derivative of

1 + z = a (t_{0}) / a (t_{e})

, with both endpoints allowed to vary, gives

\frac{z ˙}{1 + z} = H (t_{0}) - H (t_{e}) \frac{d t _{e}}{d t _{0}} .

Therefore the observer-time redshift drift is

\overset{z}{˙} = H (t_{0}) (1 + z) - H (t_{e}) .

Holding the emission time fixed would omit the second term and would not compare successive observations of the same source.

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