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I have an example of an equation where I don't know how the right hand side is derived from left hand side. It looks very weird as the differential changes. Notice that $m$ and $c$ are constants.

$$\frac{m}{2} \int \frac{\textrm{d}v^2}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{-mc^2}{2} \int \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \mathrm d\left(1-\frac{v^2}{c^2}\right) $$

Which rules apply here? Could anyone explain this or provide me with links to any good websites? Than you.

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Since you say that the differential changes, there seems to be a typo -- the "differential" is actually missing on the left-hand side. –  joriki Nov 26 '12 at 17:29
    
It looks like the substitution $u=1-\frac{v^2}{c^2}$, but if so it is not done correctly, the $v^2$ on top has mysteriously disappeared. How does it finish, so one can check whether it is a typo or a mistake? –  André Nicolas Nov 26 '12 at 17:30
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Something looks wrong here: $$d\left(1-\frac{v^2}{c^2}\right)=-\frac{2v}{c^2}$$and the $\,c^2\,$ in the denominator and the minus sign are "balaanced" by the constant in the right side, yet the $\,2v\,$ doesn't appear anymore... –  DonAntonio Nov 26 '12 at 17:33
    
@joriki I did add the differential on the LHS. It was a typo yes. –  71GA Nov 26 '12 at 17:40
    
This is a part of a derivation i took from Wikipedia article "Relativistic kinetic energy" here –  71GA Nov 26 '12 at 17:46

3 Answers 3

up vote 2 down vote accepted

What you want is to get a diferential that contains whats inside the square root in the denominator, so you create a derivative that contains both $dv^2$ and $d\left(1-\frac{v^2}{c^2}\right)$. It is easy to do so, and we see that this one can be used: $$\frac{d\left(1-\frac{v^2}{c^2}\right)}{d(v^2)}=-\frac{1}{c^2}$$ Isolating $d(v^2)$ we get: $$d(v^2)=-c^2 d\left(1-\frac{v^2}{c^2}\right)$$

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In first equation, how did you know that LHS equals $-\frac{1}{c^2}$? –  71GA Nov 26 '12 at 20:10
    
Oh never mind i just nave to derive this as it was (please confirm) $$\frac{\textrm d \left(1-\frac{x}{c^2}\right)}{\textrm d x}$$ and this equals $$0 - \frac{1}{c^2} = - \frac{1}{c^2}$$ –  71GA Nov 26 '12 at 21:24
    
Yes, its just a simple derivative. –  Ivan Lerner Nov 27 '12 at 3:21

$$ \mathrm dv^2=c^2\mathrm d\frac{v^2}{c^2}=-c^2\mathrm d\left(-\frac{v^2}{c^2}\right)=-c^2\mathrm d\left(1-\frac{v^2}{c^2}\right)\;. $$

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Thank you. This is what i needed. And how i needed it. –  71GA Nov 26 '12 at 21:14
    
There is ony one disturbing thing to me... Why does it hold that $$\textrm d \left(-\frac{v^2}{c^2}\right) = \textrm{d} \left(1 - \frac{v^2}{c^2} \right )$$ –  71GA Nov 26 '12 at 21:19

Added: The answer belows tries to address the OP as it was, assuming it was $\,dv\,$ and not $\,dv^2\,$ as it happened to be later.

Imo, it must be

$$\frac{m}{2}\int\frac{v^2}{\sqrt{1-\frac{v^2}{c^2}}}dv=\frac{m}{2}\int\frac{v}{\sqrt{1-\frac{v}{c^2}}}\left(-\frac{2v}{c^2}\right)\left(-\frac{c^2}{2}\right)dv=$$

$$=-\frac{mc^2}{4}\int\frac{v}{\sqrt{1-\frac{v^2}{c^2}}}d\left(1-\frac{v^2}{c^2}\right)$$

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The differential on the LHS is $\mathrm d v^2$ and not $\mathrm d v$ you used. I am sorry for the confusion, but it was a typo i fixed now. –  71GA Nov 26 '12 at 17:42

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