Find E' in FLP Volume 1 17.4: Why Does It Work?

[Special K]

In FLP Volume 1 section 17.4, Feynman derives the 4 momentum. He gives the expression for v'(velocity in the moving reference frame) then says to find E' we need to square v', subtract it from one, take the squad root, and take the reciprocal. He does this to get E' is simply mo times the above procedure performed to v'. Why is this procedure correct? I do not understand why those Operations on v' then multiplying by mass give E' necessarily? Unless it just worked out that way?

 

[BvU]

Hello K,

Do you agree with (17.6)$$ E = m = m_0/\sqrt{1-v^2} \quad ? $$ because that is what he does with $v'$ to get $E'$

 

[Special K]

Hello K,

Do you agree with (17.6)$$ E = m = m_0/\sqrt{1-v^2} \quad ? $$ because that is what he does with $v'$ to get $E'$

Oh I see, he was just taking v' and putting it in the same relation to E' as E is to v. That all makes sense, thank! I figured it had to be from some other formula, because the operation on its own didn't click