special relativity

The equation explained

A'^2 = A^2 \frac{1 - \frac{v}{V}}{1 + \frac{v}{V}} .

In wordsFor an observer moving straight away from the source, the square of the amplitude in the moving system is the square of the amplitude in the stationary system times one minus the speed over the velocity of light, divided by one plus it.OverviewThe amplitude for an observer moving along the line from the source, the case

\varphi = 0

: its square falls by the same factor as the frequency, and rises without bound as an approaching observer's speed nears that of light.Full explanationWith

\varphi = 0

the factor

\frac{\left( 1 - \frac{v}{V} \right)^2}{1 - \left( \frac{v}{V} \right)^2}

simplifies to

\frac{1 - \frac{v}{V}}{1 + \frac{v}{V}}

, the square of the Doppler factor for this direction. At

v = 0.6 \, V

the squared amplitude is a quarter and the amplitude half of its value in

K

; approaching at the same speed (

v = -0.6 \, V

) the squared amplitude is four times as great. The paper draws the conclusion in the next sentence: for an observer approaching a light source with the velocity

V

, the source would have to appear infinitely intense.Every step

A'^2 = A^2 \frac{\left( 1 - \frac{v}{V} \cos \varphi \right)^2}{1 - \left( \frac{v}{V} \right)^2}

The squared amplitude for any angle, the previous display.

A'^2 = A^2 \frac{\left( 1 - \frac{v}{V} \right)^2}{\left( 1 - \frac{v}{V} \right) \left( 1 + \frac{v}{V} \right)}

Put

\varphi = 0

and factor the denominator.

A'^2 = A^2 \frac{1 - \frac{v}{V}}{1 + \frac{v}{V}}

Cancel one factor

1 - \frac{v}{V}

.AlsoOverviewFull explanationEvery step

Read it in the paper