Derivation of gamma by experiment

This animation shows how the factor gamma, γ, would be determined by experiment. From the point of view of the observer in system A, system B is moving along the x axis at speed v. From the point of view of the observer in system B, system A is moving along the x axis at speed -v. Therefore it is natural to write:

$\begin{array}{l} x_{A} = γ (x_{B} + v t_{B}) \\ x_{B} = γ (x_{A} - v t_{A}) \end{array}$

(1.1)

where t_B and t_A are times measured in the A and B moving systems and γ is a factor independent of the xs and ts that is to be determined. Equations 1 are the same as Galilean transformations except for the factor γ. And the fact that t_A and t_B may not be equal.
Thus x_A=0 when x_B=-vt_B and x_B=0 when x_A=vt_A. In order to derive γ from these equations the following experiment is performed. A light pulse is allowed to propagate distance x_A=c*t_A in system A and a similar light pulse is allowed to propagate a distance x_B=c*t_B in system B. Of course, the speed of propagation has to be c for both observer A and observer B whether the pulse observed is in the same system as the observer or not. Then the equations above become:

$\begin{array}{l} c t_{A} = γ t_{B} (c + v) \\ c t_{B} = γ t_{A} (c + v) \end{array}$

(1.2)

If we multiply the left sides and the right sides of these equations we obtain a single equation:

$c^{2} t_{A} t_{B} = γ^{2} t_{A} t_{B} (c^{2} - v^{2})$

(1.3)

Dividing each side of this equation by t_At_B we obtain:

$c^{2} = γ^{2} (c^{2} - v^{2})$

(1.4)

From which:

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

(1.5)

Most importantly, notice that, if v is non-zero, γ cannot be 1 and therefore the simple Galilean transformation which sets γ equal to 1 cannot be valid.