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?
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4$\begingroup$ 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 $c$). $\endgroup$– dmckee --- ex-moderator kittenCommented Sep 25, 2016 at 22:10
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2$\begingroup$ 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. $\endgroup$– dmckee --- ex-moderator kittenCommented Sep 25, 2016 at 22:11
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2$\begingroup$ Essential: profmattstrassler.com/articles-and-posts/… $\endgroup$– user108787Commented Sep 26, 2016 at 0:00
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1$\begingroup$ (1) is the binomial expansion of (2) without higher order terms $\endgroup$– bemjanimCommented Mar 14, 2020 at 20:45
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2 Answers
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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.
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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.
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1$\begingroup$ I take the third term (or second, if zero-indexing) should be $\frac {3v^4}{8c^4}$? $\endgroup$ Commented Mar 14, 2020 at 22:36
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