How can we deduce the relation $m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$ between relativistic mass and rest mass in special relativity? I was watching a video on Youtube which deduce Einstein's relation $E=mc^2$ and the process of deduction used the relation between relativistic mass and rest mass, which is
$$m= \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}.$$ 
So I look for a nice deduction of this relation. 
 A: The question -- once you understand that $E=m$ (see [a] for a correct proof) -- is equivalent to asking "How does energy transform from the rest frame of the body to a moving frame?"
Consider the standard set-up in which $E=m$ is proven -- a body at rest gives out two flashes of light in opposite directions, then you analyse the frame from a relatively moving reference frame. In this frame, one of the beams' energy transforms as $\sqrt {\frac{{1 + v}}{{1 - v}}} \frac{E}{2}$ and the other transforms as $\sqrt {\frac{{1 - v}}{{1 + v}}} \frac{E}{2}$. By energy conservation, the energy lost -- and thus the mass reduction -- must equal the total of these quantities, which is $\gamma E$, as required.
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My original answer -- which you can see the comments are relevant to -- considered a bizarre approach where you write $m=m_0+\frac12m_0v^2$, then keep replacing the $m_0$ with the expansion for $m$ i.e. $m=m_0+\frac12mv^2$, which leads to $m=m_0/(1-v^2/2)$, which is wrong. But the approach is arbitrary, it is pretending that kinetic energy transforms as $\mathrm{KE}\to\gamma\,\mathrm{KE}$ (it doesn't, this is wrong), for no good reason.
A: $$
\int\vec{\rm F}\cdot{\rm d}\vec{r}
=
c^{2}\Delta m
$$
Just replace
$$
\vec{F}
=
{{\rm d} \over {\rm d}t}\left\lbrack{m\vec{v} \over \sqrt{1 -v^{2}/c^{2}}}\right\rbrack
\quad\mbox{and}\quad
{{\rm d}\vec{r} \over {\rm d}t} = \vec{v}
$$
A: The easiest way for me was to write down the relativistic momentum equation $p=mv\gamma(v)$. Then from this you can write down the relativistic second Newton's law: 
\begin{align}
\substack{\text{I start with a clasic}\\\text{second Newton's law}}\longrightarrow \boxed{Fdt = dp} \Longrightarrow  F &= \frac{dp}{dt} \longleftarrow\substack{\text{here we insert formula for}\\\text{a relativistic momentum $p=mv\gamma(v)$}}\\
F &= \frac{d [mv\gamma(v)]}{dt}\\
F &= \frac{d}{dt}[ m v \gamma(v)] \longleftarrow \substack{\text{here we name the product $m\gamma(v)$}\\\text{a "relativistic mass" $\tilde{m}$}}\\
F &= \frac{d}{dt}[ \tilde{m} v]
\end{align}
So we can clearly see the relativistic mass $\tilde{m}$ is the product of a rest mass $m$ and Lorentz factor (for speed) $\gamma(v)$: 
$$\tilde{m} = m\gamma(v) = \frac{m}{\sqrt{1 - v^2/c^2}}$$
The only time I used the relativistic mass was when dealing with a photon in a gravitational field. I don't know any other ways to deal with that topic. If anyone knows it, please do tell =)
