User:Ddcampayo/Lorentz

From physical principles

The problem is usually restricted to two dimensions by using a velocity along the x axis such that the y and z coordinates do not intervene. It is similar to that of Einstein. As in the Galilean transformation, the Lorentz transformation is linear since the relative velocity of the reference frames is constant as a vector; otherwise, inertial forces would appear. They are called inertial or Galilean reference frames. According to relativity no Galilean reference frame is privileged. Another condition is that the speed of light must be independent of the reference frame, in practice of the velocity of the light source.

Galilean reference frames

In classical kinematics, the total displacement x in the R frame is the sum of the relative displacement x′ in frame R' and of the distance between the two origins x-x'. If v is the relative velocity of R' relative to R, the transformation is: x = x′ + vt, or x′ = x − vt. This relationship is linear for a constant v, that is when R and R' are Galilean frames of reference.

In Einstein's relativity, the main difference with Galilean relativity is that space is a function of time and vice-versa: t ≠ t′. The most general linear relationship is obtained with four constant coefficients, A, B, γ, and b:

$x'=\gamma x+bt\,$

$t'=Ax+Bt.\,$

The Lorentz transformation becomes the Galilean transformation when γ = B = 1 , b = -v.

An object at rest in the R frame at position x′=0, will be seen as moving with constant velocity v. Hence the transformation must satisfy x′=0 if x=vt. Therefore, b=-γ v and it may written as:

$x'=\gamma (x-vt)\,$

Principle of relativity

According to the principle of relativity, there is no privileged Galilean frame of reference. Therefore, the inverse transformation for the position from frame R′ to frame R must be

$x=\gamma \left(x'+vt'\right),$

with the same value of γ (which must therefore be an even function of v).

Speed of light independent of the velocity of the source

If the speed of light must be independent of the reference frame, the transformation must ensure that x = ct if x′ = ct′. In other words, the light emitted at t=t′=0 moves at velocity c in both frames. Replacing x and x′ in the preceding equations, one has:

$ct'=\gamma \left(c-v\right)t,$

$ct=\gamma \left(c+v\right)t'.$

Multiplying these two, one finds

$c^{2}tt'=\gamma ^{2}\left(c^{2}-v^{2}\right)tt'.$

From which

$\gamma ={\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}},$

called the "Lorentz factor".

Transformation of time

The factors A and B in the transformation for time can now be obtained. Substituting the derived expression for x′

$x=\gamma (x-vt)\,$

in the inverse transformation equation

$x=\gamma \left(x'+vt'\right),$

gives

$x=\gamma \left(\gamma (x-vt)+vt'\right).$

Solving for t′, this results in

$t'={\frac {1-\gamma ^{2}}{\gamma v}}x+\gamma t$

and identification with the general transformation

$t'=Ax+Bt\,$

results in

$A=-{\frac {\gamma v}{c^{2}}}\,$

$B=\gamma ,\,$

and thus finally in

$t'=\gamma \left(t-{\frac {v}{c^{2}}}x\right).$