# $m = m_0 + (½) m_0v^2/c^2$ vs $m=m_0/\sqrt{1-v^2/c^2}$

Confused what equation I should use,
$\displaystyle{m = m_o + \frac{1}{2} \frac{m_ov^2}{c^2}}$ or $\displaystyle{m=m_o\sqrt{1-\frac{v^2}{c^2}}}$,
when solving for relativistic mass. When I plugged in $0.833c$ for velocity and $5kg$ for rest mass, the equations gave two different answers about $3 kgs$ apart. I can find $\displaystyle{m=m_o\sqrt{1-\frac{v^2}{c^2}}}$ all over the internet, but finding $\displaystyle{m = m_o + \frac{1}{2} \frac{m_ov^2}{c^2}}$ is harder. Is $\displaystyle{m = m_o + \frac{1}{2} \frac{m_ov^2}{c^2}}$ a true relationship?

• One of these is a low (compared to $c$) velocity approximation to the other. Obviously you should use the exact expression when the velocity is not low (compared to $c$). – dmckee --- ex-moderator kitten Sep 25 '16 at 22:10
• And while I suppose this is in response to an assignment meaning that you have no choice, you might look around the site for one of the many place where people explain that "relativistic mass" is a concept rarely used by working scientists and active discouraged by many. – dmckee --- ex-moderator kitten Sep 25 '16 at 22:11
• – user108787 Sep 26 '16 at 0:00

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.