special relativity

The equation explained

\xi = a \frac{V^2}{V^2 - v^2} x' .

In wordsThe distance along the motion in the moving system is the factor still to be found times the square of the velocity of light over that square less the square of the speed of the moving system, times the distance along the motion.OverviewThe moving system's distance along the motion, in terms of a stationary distance alone: the time has gone. Apart from the factor still to be found, this is the stretching of the first coordinate that the whole section has been working towards.Full explanationThe stationary time is taken out of the expression for

\xi

by putting in what the previous display found it to be for this ray. What is left depends on

x'

alone, with a coefficient built from the two speeds. The arithmetic is worth doing slowly, because the

V^2 - v^2

in the denominator of the time expression and the

V - v

from the ray's own progress combine into a single factor rather than cancelling. The paper prints the result and no working; the steps here supply it. When

a

is settled at the end of the section as

\frac{1}{\beta}

times a function that turns out to be one, this line becomes

\xi = \beta (x - v t)

, the form the section closes with.Every step

\xi = a V \left( t - \frac{v}{V^2 - v^2} x' \right)

The expression for the moving distance that the section has reached.

\xi = a V \left( \frac{x'}{V - v} - \frac{v}{V^2 - v^2} x' \right)

Put in the stationary time this ray has taken, from the previous display.

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

Put both terms over the one denominator

V^2 - v^2

\frac{V + v}{V^2 - v^2} - \frac{v}{V^2 - v^2} = \frac{V}{V^2 - v^2}

Subtract: the two terms in

v

cancel and

V

is left.

\xi = a V \cdot \frac{V}{V^2 - v^2} x' = a \frac{V^2}{V^2 - v^2} x'

The

V

outside and the

V

from the bracket make

V^2

, which is the line as printed.AlsoOverviewFull explanationEvery step

Read it in the paper