## Solution: The Doppler Factor

Exercise: From the figure derive the equation

λ(X2 − X1)  =  − ,

where λ is a scale factor and

D    ′ /  f

is the Doppler factor. Derive and discuss the result

$\Large{D\,\,\, = \,\,\,\frac{{f'}}{f}\,\,\, = \,\,\,\gamma \,\left( {1\,\, \pm \,\,v/c} \right)\,\,\, = \,\,\,\sqrt {\frac{{c \pm v}}{{c \mp v}}} \,\,\, = \,\,\,\frac{1}{{\gamma \,\left( {1\,\, \mp \,\,v/c} \right)}}}$

In the diagram above, the spacetime vector - cT' V has magnitude cT' since the source spacetime velocity V has an interval of 1. The dotted blue light directions and the blue spacetime vector X21 are lightlike. T' is the source period and T is the receiver period. The maps for both receding and approaching sources are drawn. Notice that for either case, we have a spacetime triangle to work with:

$\Large{\begin{array}{l}cT\,\, - \,\,cT'\,V\,\,\, = \,\,\,{X_{21}}\\\\\frac{1}{{cT'}}\,{X_{21}}\,\,\, \equiv \,\,\,\lambda \,{X_{21}}\,\,\, = \,\,\,\frac{T}{{T'}}\,\, - \,\,V\,\,\, = \,\,\,\frac{{f'}}{f}\,\, - \,\,V\,\,\, \equiv \,\,\,D\,\, - \,\,V\\\\0\,\,\, = \,\,\,{\lambda ^2}\,{X_{21}}\,\widetilde {{X_{21}}}\,\,\, = \,\,\,\left( {D\,\, - \,\,V} \right)\,\,\left( {D\,\, - \,\,\widetilde V} \right)\,\,\, = \,\,\,{D^2}\,\, - \,\,2\gamma \,D\,\, + \,\,1\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{since}}\,\,\,\,V\,\, + \,\,\widetilde V\,\,\, = \,\,\,\gamma \left( {1\,\, + \,\,{\bf{v}}/c} \right)\,\, + \,\,\gamma \left( {1\,\, - \,\,{\bf{v}}/c} \right)\,\,\, = \,\,\,2\gamma \\\\{\rm{Solving}}\,\,{\rm{the}}\,\,{\rm{quadratic}}:\,\,\,\,\,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,D\,\,\, = \,\,\,\frac{{f'}}{f}\,\,\, = \,\,\,\gamma \left( {1\,\, \pm \,\,v/c} \right)\,\,\, = \,\,\,\frac{{1\,\, \pm \,\,v/c}}{{\sqrt {1\,\, - \,\,{v^2}/{c^2}} }}\,\,\, = \,\,\,\sqrt {\frac{{c\,\, \pm \,\,v}}{{c\, \mp \,\,v}}} \\f\,({\rm{receiver}})\,\,\, = \,\,\,f'\,({\rm{source}})\,\sqrt {\frac{{c\,\, \mp \,\,v}}{{c\, \pm \,\,v}}} \,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{\rm{Top/bottom}}\,\,{\rm{signs:}}\,\,\,{\rm{receding/approaching}}\,\,{\rm{source.}}} \right\}\end{array}}$

Since the result came from solving a quadratic, how do we know which signs go with which case? Looking at the receding diagram on the left above,
cT ≥ c∆t = γcT', so the received period is larger than the transmitted period and the received frequency is lower (red shifted) than the source. For the approaching diagram on the right, γcT' = c∆t ≥ cT, so the received period is smaller and the received frequency is larger (blue shifted).

For further practice, we show alternative derivations using the above diagrams. For the diagram on the left (receding case), time dilation gives
c∆t = γcT'. The horizontal dotted line is part of an equilateral right triangle, with value = v∆t = c∆t v/c. Then,

$\Large{cT\,\,\, = \,\,\,c\Delta t\,\, + \,\,c\Delta t\,v/c\,\,\, = \,\,\,c\Delta t\,\left( {1\,\, + \,\,v/c} \right)\,\,\, = \,\,\,\gamma cT'\,\left( {1\,\, + \,\,v/c} \right)}$

Or, for the diagram on the right (approaching case),

$\Large{cT\,\,\, = \,\,\,c\Delta t\,\, - \,\,c\Delta t\,v/c\,\,\, = \,\,\,c\Delta t\,\left( {1\,\, - \,\,v/c} \right)\,\,\, = \,\,\,\gamma cT'\,\left( {1\,\, - \,\,v/c} \right)}$