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

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

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

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

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

## Comments

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TopNewestSeems pretty straightforward, does these formulas have an importance in physics ? and please answer me in English LOL (I do not know much physics). – Haroun Meghaichi · 2 years, 10 months ago

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– Steven Zheng · 2 years, 10 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.Log in to reply

You need to fix up the LaTeX for the last line. Apart from that, awesome. – Sharky Kesa · 2 years, 10 months ago

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– Steven Zheng · 2 years, 10 months ago

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

– Sharky Kesa · 2 years, 10 months ago

Yeah, there was something wrong but it's fixed now.Log in to reply

– Steven Zheng · 2 years, 10 months ago

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

– Sharky Kesa · 2 years, 10 months ago

Ohh.Log in to reply