special relativity

The equation explained

\varphi(v) \cdot \varphi(-v) = 1 .

In wordsThe unknown factor for the speed v, times the same factor for the speed −v, equals one.OverviewGoing from

K

k

and back must return the original forces. The way there multiplies by the factor for

v

, the way back by the factor for

-v

, so their product is 1.Full explanationThe paper introduced the factor as

\psi(v)

on p. 908 and writes it

\varphi(v)

from here on (p. 909); it is the same factor, and the site keeps both letters as printed. The argument inverts the relations of the previous display in two ways. First, solve them for

K

's forces in terms of

k

's. Second, apply the same relations to the transformation from

k

back to

K

, which is the transformation with velocity

-v

. The two answers must be identical, and comparing them gives the condition. The first component shows it most simply; the others give the same result.Every step

X' = \varphi(v) X

The first relation of the previous display, with the factor written as it is printed from p. 909 on.

X = \frac{X'}{\varphi(v)}

The first way of inverting: solve for

X

X = \varphi(-v) X'

The second way: going from

k

back to

K

is the same transformation with velocity

-v

, so the same relation holds with the two systems exchanged and

v

replaced by

-v

\varphi(v) \cdot \varphi(-v) = 1

Both give

X

, so

1 / \varphi(v) = \varphi(-v)

, which is the display. Solving the relations for

Y

and

N

gives the same condition, because

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

and

\beta

is the same for

v

and

-v

.AlsoOverviewFull explanationEvery step

Read it in the paper