My question is: why has the Lorentz factor the form

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

and not, for example one of those:

$$\gamma = \dfrac{1}{\sqrt{1 - \dfrac{v}{c}}}$$


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


$$\gamma = \dfrac{1}{\sqrt[3]{1 - \dfrac{v^2}{c^2}}}$$

Or one of the other many possibilities?

I mean: nobody ever told me where did the definition of $\gamma$ came from. Can you explain it to me?

  • 1
    $\begingroup$ Well, it appears in so many equations in special relativity that it appears convenient to define a shorthand for it, no? (In contrast to your other possibilities, which never appear) $\endgroup$
    – ACuriousMind
    Feb 28 '15 at 16:37
  • $\begingroup$ Ok but Where does it comes from? Someone woke up and decided that Gamma is defined like that? $\endgroup$
    – Henry
    Feb 28 '15 at 16:41
  • 3
    $\begingroup$ possible duplicate of How do I derive the Lorentz contraction from the invariant interval? $\endgroup$
    – Kyle Kanos
    Feb 28 '15 at 16:41
  • 3
    $\begingroup$ physics.stackexchange.com/q/114372 $\endgroup$
    – user12029
    Feb 28 '15 at 16:43
  • 2
    $\begingroup$ You can derive it from the assumption that the speed of light is the same for all observers; wikipedia has an accessible derivation. $\endgroup$
    – jabirali
    Feb 28 '15 at 16:50