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Description

Derpy’s take on relativistic kinetic energy.
 
Yup that’s actually how you get to E=mc^2

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Yet One More Idiot
Artist -

World's biggest idiot xD
@Cirrus Light
Also I don’t comprehend how (mc2)/((1-v2^/c2)(1/2)) - mc2^ = mc2(γ-1) because I’m sure that mc2((1/(γ))-1) would be the answer, otherwise I’m missing out something on the distributive law.
The reason for that is because γ = 1/sqrt(1-v2/c2). So then
 
(mc2)/((1-v2/c2)(1/2)) - mc2  
=(mc2)·(1/((1-v2/c2)(1/2)) - 1)  
=(mc2)·(γ-1)  
:)
SirLogarithm

@Cirrus Light
 
I’m baffled as to how most of this was derived, though some aspects I understand.
 
I don’t comprehend how you go from an equation to find work under conditions of varying force to the equation ʃ(madx)/(1-V2/c2)(3/2)) (x1 to x2).
 
Also I don’t comprehend how (mc
2)/((1-v2/c2)(1/2)) - mc2 = mc2(γ-1) because I’m sure that mc2((1/(γ))-1) would be the answer, otherwise I’m missing out something on the distributive law.
Cirrus Light
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Sciencepone of Science!
Ah, for a sec, I was going to correct how she seemed to have put the graph of the Lorentz Factor as crossing the origin; and I was gonna be like; “Deerpy, why’d you put that gamma equals zero when velocity is zero? Its minimum value is one!” but now I see it’s slightly above the x-axis.
 
:P
 
But how did we go from W = ʃFdx (x1 to x2) to = ʃ(madx)/(1-V2/c2)(3/2)) (x1 to x2)?
 
I mean, my question is why did we multiply by the square of the lorentz factor?
 
That, and how did we get to E
2 = (mc^2)^2 + (pc)2 from the previous step?
 
I missed class the day they did the e = mc
2 lecture. Knowing that makes me feel a little like crying sometimes :<
 
Also, you get a muffin.  
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