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Yet One More Idiot ![Artist - Artist -](https://derpicdn.net/badges/2016/8/23/540676fb2fd6546ee45a1c1.svg)
World's biggest idiot xD
"@SirLogarithm":/653650#comment_2687403
[bq="SirLogarithm"]"@Cirrus Light":#comment_539e0ba663687240cdd60200
Also I don't comprehend how (mc^2)/((1-v^[==2==]^/c^[==2==]^)^(1/2)) - mc^[==2==]^ = mc^[==2==]^(γ-1) because I'm sure that mc^[==2==]^((1/(γ))-1) would be the answer, otherwise I'm missing out something on the distributive law.[/bq]The reason for that is because γ = 1/sqrt(1-v^[==2==]^/c^[==2==]^). So then
(mc^[==2==]^)/((1-v^[==2==]^/c^[==2==]^)^[==(1/2)==]^) - mc^[==2==]^
=(mc^[==2==]^)·(1/((1-v^[==2==]^/c^[==2==]^)^(1/2)) - 1)
=(mc^[==2==]^)·(γ-1)
:)
[bq="SirLogarithm"]"@Cirrus Light":#comment_539e0ba663687240cdd60200
Also I don't comprehend how (mc^2)/((1-v^[==2==]^/c^[==2==]^)^(1/2)) - mc^[==2==]^ = mc^[==2==]^(γ-1) because I'm sure that mc^[==2==]^((1/(γ))-1) would be the answer, otherwise I'm missing out something on the distributive law.[/bq]The reason for that is because γ = 1/sqrt(1-v^[==2==]^/c^[==2==]^). So then
(mc^[==2==]^)/((1-v^[==2==]^/c^[==2==]^)^[==(1/2)==]^) - mc^[==2==]^
=(mc^[==2==]^)·(1/((1-v^[==2==]^/c^[==2==]^)^(1/2)) - 1)
=(mc^[==2==]^)·(γ-1)
:)