Return to Chapter 1, Starting Special Relativity

 t= at' + bx'

x = ex' + ft'

Multiply the top through by c and use new constants

ct= gct' + gx'

x = ex' + ft'

Using these to describe the location of the S' origin x' = 0 is at x = vt at time t

ct= gct'

vt = ft'

Divide

c/v = gc/f

f = gv

Rewrite out the transformations

ct= gct' + gx'

x = ex' + gvt'

Useing these to describe the location of the S origin x = 0 is at x' = -vt' at time t'

0 = - evt' + gvt'

The second equation results in

e = g

Rewrite out the transformations

ct= gct' + gx'

x = g(x' + vt')

The S frame location of a pulse of light from the origin is at x = ct at time t and the S' frame location of it is at x' = ct' at time t' resulting in

ct= gct' + gct'

ct = gct' + gvt'

Eliminating ct results in

gct' + gct' = gct' + gvt'

simplified

g = gv/c

Call v/c b and rewrite out the transformations

ct= g(ct' + bx')

x = g(x' + bct')

All that is left is to solve for g in terms of v or b.

Doing some algebra

ct= gct' + gbx'

x = gx' + gbct'

gct' = ct - gbx'

gx' = x - gbct'

gct' = ct - b(x - gbct')

gx' = x - b( ct - gbx')

gct' = ct - bx + gb²ct'

gx' = x - bct + gb²x'

gct' - gb²ct' = ct - bx

gx' - gb²x' = x - bct

g(1 - b²)ct' = ct - bx

g(1 - b²)x' = x - bct

ct' = g ^-1 (1 - b²) ^-1 (ct - bx)

x' = g ^-1 (1 - b²) ^-1 (x - bct)

Notice that the inverse transformation here takes the same equation form at the last expression we had for the transformation

ct= g(ct' + bx')

x = g(x' + bct')

Now I invoke the first postulate of special relativity. Had we turned around our axis so that instead of looking at it as the S' frame moving in the +x direction it would have been seen as the S frame moving in what we chose to be the -x' direction the transformation must take the same equation form. So inverting the equations but be able to be accomplished by doing nothing more than switching which frame is primes and the sign on v or b. Therefor by the first postulate of special relativity

ct' = g_(b) ^-1 (1 - b²) ^-1 (ct - bx) = g_(-b)(ct - bx)

x' = g_(b) ^-1 (1 - b²) ^-1 (x - bct) = g_(-b)(x - bct)

Either results in

g_(b) ^-1 (1 - b²) ^-1 = g_(-b)

g_(-b)g_(b) = (1 - b²) ^-1

The right is an even function of b and so must the left be so g_(-b) = g_(b) resulting in

g ² = (1 - b²) ^-1

g = (1 - b²) ^-1/2

So now we have g as a function of b or v and the tranformations and their inverse equations result in

ct= g(ct' + bx')

x = g(x' + bct')

ct' = g(ct - bx)

x' = g(x - bct)

QED

Return to Chapter 1, Starting Special Relativity