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. dγdv=γ3vc2\frac{d\gamma}{dv} = \frac{{\gamma}^{3}v}{{c}^{2}}

  2. γ21=(βγ)2{\gamma}^{2} - 1 = {(\beta \gamma)}^{2}

  3. d(γv)dv=γ3\frac{d(\gamma v)}{dv} = {\gamma}^{3}

Before we begin, I would like to make it clear that β=vc\beta = \frac{v}{c} and γ=11v2c2\gamma = \frac{1}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}.

Identity 1

dγdv=12(1v2c2)3/22vc2\frac{d\gamma}{dv} = \frac{-1}{2}{\left(1-{\frac{{v}^{2}}{{c}^{2}}}\right)}^{-3/2}\frac{2v}{{c}^{2}}

dγdv=γ3vc2\frac{d\gamma}{dv} = \frac{{\gamma}^{3}v}{{c}^{2}}

Identity 2

γ21=(1v2c2)11{\gamma}^{2} - 1 = {\left(1-\frac{{v}^{2}}{{c}^{2}}\right)}^{-1} -1

γ21=(1v2c2)11{\gamma}^{2} - 1 = {\left(1-\frac{{v}^{2}}{{c}^{2}}\right)}^{-1} -1

γ21=c2c2v21{\gamma}^{2} - 1 = \frac{{c}^{2}}{{c}^{2} -{v}^{2}} - 1

γ21=c2c2v2c2v2c2v2{\gamma}^{2} - 1 = \frac{{c}^{2}}{{c}^{2} -{v}^{2}} - \frac{{c}^{2} -{v}^{2}}{{c}^{2} -{v}^{2}}

γ21=v2c211v2c2{\gamma}^{2} - 1 = \frac{{v}^{2}}{{c}^{2}}\frac{1}{1 - \frac{{v}^{2}}{{c}^{2}}}

γ21=(βγ)2{\gamma}^{2} - 1 = {(\beta \gamma)}^{2}

Identity 3

d(γv)dv=dγdvv+γ\frac{d(\gamma v)}{dv} = \frac{d\gamma}{dv}v+\gamma

d(γv)dv=γ3v2c2+γ\frac{d(\gamma v)}{dv} = \frac{{\gamma}^{3}{v}^{2}}{{c}^{2}} + \gamma

d(γv)dv=γ(γ2v2c2+1)\frac{d(\gamma v)}{dv} = \gamma \left(\frac{{\gamma}^{2}{v}^{2}}{{c}^{2}} + 1\right)

d(γv)dv=γ(v2c2v2+c2v2c2v2)\frac{d(\gamma v)}{dv} = \gamma \left(\frac{{v}^{2}}{{c}^{2} - {v}^{2}} + \frac{{c}^{2} -{v}^{2}}{{c}^{2} - {v}^{2}}\right)

d(γv)dv=γ(c2c2v2)\frac{d(\gamma v)}{dv} = \gamma \left(\frac{{c}^{2}}{{c}^{2} - {v}^{2}} \right)

d(γv)dv=γ(γ2)\frac{d(\gamma v)}{dv} = \gamma ({\gamma}^{2})

d(γv)dv=γ3\frac{d(\gamma v)}{dv} = {\gamma}^{3}

Check out my other notes at Proof, Disproof, and Derivation

Note by Steven Zheng
5 years, 2 months ago

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

Sharky Kesa - 5 years, 2 months ago

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

Steven Zheng - 5 years, 2 months ago

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

Sharky Kesa - 5 years, 2 months ago

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

Steven Zheng - 5 years, 2 months ago

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@Steven Zheng Ohh.

Sharky Kesa - 5 years, 2 months ago

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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 - 5 years, 2 months ago

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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 - 5 years, 2 months ago

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

Devvrat Tiwari - 2 months, 2 weeks ago

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