\[z \simeq \frac{v}{c} \]
. This equation is easily derived from the exact relationship \[z= \sqrt{\frac{1+v/c}{1-v/c}} -1\]
.Use the binomial expansion and assume
\[v \ll c\]
so ignore powers of \[v/c\]
of 2 and higher.\[\begin{equation} \begin{aligned} z &= \sqrt{\frac{1+v/c}{1-v/c}} -1 \\ &= (1+v/c)^{1/2}(1-v/c)^{11/2}-1 \\ & \simeq (1+1/2v+...)(1+1/2v/c+...)-1 \\ &= 1+1/2v/c +1/2v/c +...-1 \\ &=v/c \end{aligned} \end{equation}\]
The exact expression is a form of the Doppler shift.
\[\frac{\Delta f}{f} =z = \sqrt{\frac{1+v/c}{1-v/c}} -1 \simeq v/c\]
(1) Using
\[c=f \lambda\]
, \[0= f \Delta | \lambda + (\Delta f) \lambda \rightarrow \| \frac{\Delta f}{f} \| = \| \frac{\Delta \lambda}{\lambda} \]
so we can also write (1) above as \[\frac{\Delta \lambda}{\lambda} =z \simeq v/c\]
.