Why is the relativistic Lorentz factor defined this way?

Asked Feb 28 '15 at 16:35

Active Feb 28 '15 at 17:25

Viewed 315 times

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?

edited Feb 28 '15 at 17:25

Qmechanic

asked Feb 28 '15 at 16:35

Henry

1

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) – ACuriousMind Feb 28 '15 at 16:37
Ok but Where does it comes from? Someone woke up and decided that Gamma is defined like that? – Henry Feb 28 '15 at 16:41
3

possible duplicate of How do I derive the Lorentz contraction from the invariant interval? – Kyle Kanos Feb 28 '15 at 16:41
3

http://physics.stackexchange.com/q/114372/ – Feb 28 '15 at 16:43
2

You can derive it from the assumption that the speed of light is the same for all observers; wikipedia has an accessible derivation. – jabirali Feb 28 '15 at 16:50

0 Answers0