Viewing last 25 versions of comment by Yet One More Idiot on image #653650

"[@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)
 
:)

"@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)
:)

"@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)
:)

"@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)
:)

"@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)
:)