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answer for problem 1.1.1. Consider a signal sent along an x, x' axis sent at t = 0, where the S' frames moves relative to the S frame in the + x direction with speed v. If the speed of the signal is the same speed c according to either frame, and assuming the coordinate transformation between the two takes the form

t= at' + bx'

x = ex' + ft'

y = y'

z = z'

derive the Lorentz coordinate transformation in the form of Eqn. 1.1.1 and the expression for g .

ANSWER

dt= adt' + bdx'

dx = edx' + fdt'

dx/dt = (edx'/dt' + f)/(a + bdx'/dt')

From the c invariance:

c = (ec + f)/(a + bc)

e = a + bc - f/c

t= at' + bx'

x = (a + bc - f/c)x' + ft'

dx/dt = [(a + bc - f/c)dx'/dt' + f]/(a + bdx'/dt')

Consider the origin of S' Þ f = bac

t= at' + bx'

x = (a + bc - ba)x' + bact'

dx/dt = [(a + bc - ba)dx'/dt' + bac]/(a + bdx'/dt')

Consider the origin of S Þ b = ab/c

ct= a(ct' + bx')

x = a(x' + bct')

Einstein's first postulate indicates that it does not matter if we say S is still and S' is moving in the + x direction or if we say S' is still and S is moving at the same speed in the other direction so the inverse transformations must from this symmetry be

ct'= a(ct - bx)

x' = a(x - bct)

Substitute into one above

ct= a[a(ct - bx) + ba(x - bct)]

ct= a²ct - b a²b ct

1 = a² - b²a²

a = (1 - b²) ^-1/2 = g

Finally:

ct'= g(ct - bx)

x' = g(x - bct)

ct= g(ct' + bx')

x = g(x' + bct')