# Gamma Factor Identities

In this note I will present three interesting (and useful) identities of the Lorentz factor (which most people call the gamma factor).

1. $$\frac{d\gamma}{dv} = \frac{{\gamma}^{3}v}{{c}^{2}}$$

2. ${\gamma}^{2} - 1 = {(\beta \gamma)}^{2}$

3. $\frac{d(\gamma v)}{dv} = {\gamma}^{3}$

Before we begin, I would like to make it clear that $\beta = \frac{v}{c}$ and $\gamma = \frac{1}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}$.

Identity 1

$\frac{d\gamma}{dv} = \frac{-1}{2}{\left(1-{\frac{{v}^{2}}{{c}^{2}}}\right)}^{-3/2}\frac{2v}{{c}^{2}}$

$\frac{d\gamma}{dv} = \frac{{\gamma}^{3}v}{{c}^{2}}$

Identity 2

${\gamma}^{2} - 1 = {\left(1-\frac{{v}^{2}}{{c}^{2}}\right)}^{-1} -1$

${\gamma}^{2} - 1 = {\left(1-\frac{{v}^{2}}{{c}^{2}}\right)}^{-1} -1$

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

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

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

${\gamma}^{2} - 1 = {(\beta \gamma)}^{2}$

Identity 3

$\frac{d(\gamma v)}{dv} = \frac{d\gamma}{dv}v+\gamma$

$\frac{d(\gamma v)}{dv} = \frac{{\gamma}^{3}{v}^{2}}{{c}^{2}} + \gamma$

$\frac{d(\gamma v)}{dv} = \gamma \left(\frac{{\gamma}^{2}{v}^{2}}{{c}^{2}} + 1\right)$

$\frac{d(\gamma v)}{dv} = \gamma \left(\frac{{v}^{2}}{{c}^{2} - {v}^{2}} + \frac{{c}^{2} -{v}^{2}}{{c}^{2} - {v}^{2}}\right)$

$\frac{d(\gamma v)}{dv} = \gamma \left(\frac{{c}^{2}}{{c}^{2} - {v}^{2}} \right)$

$\frac{d(\gamma v)}{dv} = \gamma ({\gamma}^{2})$

$\frac{d(\gamma v)}{dv} = {\gamma}^{3}$

Check out my other notes at Proof, Disproof, and Derivation Note by Steven Zheng
5 years, 11 months ago

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You need to fix up the LaTeX for the last line. Apart from that, awesome.

- 5 years, 11 months ago

What's wrong with it? I can't see it. By last line do you mean the very last line?

- 5 years, 11 months ago

Yeah, there was something wrong but it's fixed now.

- 5 years, 11 months ago

I was writing the note back then. Yeah, usually I write half the note, then I post it and continue.

- 5 years, 11 months ago

Ohh.

- 5 years, 11 months ago

Seems pretty straightforward, does these formulas have an importance in physics ? and please answer me in English LOL (I do not know much physics).

- 5 years, 11 months ago

Yes! If you read my derivation of E=mc^2 then identity 3 is used. They often occur in relativistic dynamics. Pretty much whenever calculus becomes important in relativity, these identities will be convenient.

- 5 years, 11 months ago

There's a negative sign missing in the first line of identity 1.

- 11 months, 2 weeks ago