$m=m_0+(1/2)m_0v^2/c^2$ vs $m=m_0/\sqrt{1-v^2/c^2}$ Confused what equation I should use,
$$m=m_0+\frac12\frac{m_0v^2}{c^2}\tag{1}$$
or
$$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\tag{2}$$
when solving for relativistic mass. When I plugged in $0.833c$ for velocity and $5\ \mathrm{kg}$ for rest mass, the equations gave two different answers about $3\ \mathrm{kg}$ apart. I can find $(2)$ all over the internet, but finding $(1)$ is harder. Is $(1)$ a true relationship?
 A: The correct expression for kinetic energy is not the Newtonian $\frac{1}{2}mv^2$, but the relativistic
$$\left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1\right)m_0c^2.$$
These two expressions give nearly identical answers for speeds much less than the speed of light. So, the total energy of an object is the sum of its rest energy ($m_0c^2$) and kinetic energy, yielding
$$E = m_0c^2 + \left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1\right)m_0c^2 = \frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}.$$
At this point, if you want to identify the relativistic mass as $m_r = E/c^2$, then you get
$$m_r = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}.$$
However, most physicists today regard relativistic mass as a quantity that is not useful to calculate since it's identical to total energy.
A: Your second equation is true, (in so far as relativistic mass is a useful concept...) and your first equation follows from this in the low speed limit. As
$$
m = \frac{m_0}{\sqrt{1 - v^2 / c^2}} = m_0(1 - v^2/c^2)^{-1/2}
$$
which we can write as a binomial series
$$
m = m_0(1 + \frac{v^2}{2c^2} + \frac{3v^4}{8c^4} + \cdots)
$$
As long as our velocities are small ($v \ll c$) we can drop the higher order terms to get
$$
m \approx m_0 + \frac{m_0v^2}{2c^2}
$$
as you have.
