Waste less time on Facebook — follow Brilliant.
×

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
2 years, 7 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

Seems 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, 7 months ago

Log in to reply

@Haroun Meghaichi 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. Steven Zheng · 2 years, 7 months ago

Log in to reply

You need to fix up the LaTeX for the last line. Apart from that, awesome. Sharky Kesa · 2 years, 7 months ago

Log in to reply

@Sharky Kesa What's wrong with it? I can't see it. By last line do you mean the very last line? Steven Zheng · 2 years, 7 months ago

Log in to reply

@Steven Zheng Yeah, there was something wrong but it's fixed now. Sharky Kesa · 2 years, 7 months ago

Log in to reply

@Sharky Kesa I was writing the note back then. Yeah, usually I write half the note, then I post it and continue. Steven Zheng · 2 years, 7 months ago

Log in to reply

@Steven Zheng Ohh. Sharky Kesa · 2 years, 7 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...