Homog. Lorentz Transformations

The Lorentz transformations relate space-time co-ordinates of an event as observed from two inertial frames S and S’ moving relative to each other. We consider S as a stationary frame and S’ moving relative to S with velocity v, along the X-axis as seen from S. We can choose the X-axis along the direction of motion so that X’-axis and X-axis always coincide. Hence Y’-Z’ plane moves parallel to Y-axis, so that Z’-axis remains parallel to Z-axis.

Let the event occur at space point (x, y, z) at time t as seen from S. Let (x’, y’, z’) and t’ be the space-time co-ordinates for the same event as seen from S’. Since space-time is uniform, the relation between space-time co-ordinates in S and S’ must be linear. That is, we expect that for a given (x, y, z, t) there exists only one set of values of (x’, y’, z’, t’). Writing the most general linear transformations, we get

x’ = a_xx x + a_xy y + a_xz z + a_xt t + b₁

y’ = a_yx x + a_yy y + a_yz z + a_yt t + b₂

z’ = a_zx x + a_zy y + a_zz z + a_zt t + b₃

t’ = a_tx x + a_ty y + a_tz z + a_tt t + b₄

where, a_xx, a_xy, ….., b₄ are constants, independent of x, y, z and t.

If we assume the initial condition that the origins of two co-ordinate systems fixed with S and S’ coincide at t = t’ = 0, i.e. (x = 0, y = 0, z = 0) implies (x’ = 0, y’ = 0, z’ = 0), then we have

b₁ = b₂ = b₃ = b₄ = 0.

If we take b’s to be zeros, the corresponding transformation equations are called homogeneous transformations, otherwise inhomogeneous. We shall be dealing with homogeneous transformations only, without loss of any physical consequence.

Further, since x’ – z’ plane has been chosen such that it always coincides with x-z plane, y = 0 implies y’ = 0. Hence, putting y = y’ = 0, we get

a_yx x + a_yz z + a_yt t = 0

for any values of x, z, t. Since x, z and t are linearly independent, the above relation implies,

a_yx + a_yz = a_yt = 0

Hence, we find,

y’ = a_yy y

The constant ayy may depend upon relative velocity v between S and S’, the only parameter which characterizes the two frames. However, Einstein’s first postulate asserts that all inertial frames are equivalent. Thus, we could as well assume S’ to be stationary and S moving in opposite direction relative S’. Since it is relative velocity that matters, we must now get,

y = a_yy y’

We find that

a²_yy = 1,   a_yy = ± 1

Proper solution is a_yy = ± 1. Because we expect that by continuously decreasing v to v = 0, we should get y’ = y (i.e. identity transformation). The value a_yy = – 1 implies y’ =  – y. Hence by taking v to zero, we cannot get y’ = y. (In fact y’ = – y implies space inversion and leads to improper transformation equations.) We shall only consider proper transformations, i.e. we have

y’ = y

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