In the relativistic mass equation $$m=\frac{m_0}{(1-v^2/c^2)^{1/2}},$$ if we put $v\ll c$ we get the rest mass. Likewise, in the equation $$v=\frac{{v_1}-{v_2}}{{{(1-\frac{v_1v_2}{c^2})}}},$$ if we put $v_1\ll c$, we get the classical limit. Same for the relativistic momentum $$p=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}},$$ we retain our classical limit.

But when we put $v\ll c$ in the equation $$KE=m_0c^2(\gamma-1)$$ with $\gamma$ is the Lorentz factor, $\gamma=1/(1-v^2/c^2)^{1/2}$, when we put $v<<c$ we get zero.

Furthermore, when I equate $$\frac{1}{2}mv^{2}=m_0c^2(\gamma-1)$$ to see what could be the velocity limit for which the relativistic kinetic energy approaches the classical limit, I get a biquadratic equation in $\beta$ (where $\beta=\frac{v}{c})$.
On solving it, I get no real solution.

Then what is the limit so that the relativistic kinetic energy can tend to the classical kinetic energy? Also, where did I go wrong in my calculations?

  • $\begingroup$ I don't think the second equation has a square root. $\endgroup$ Jul 2, 2018 at 17:08
  • $\begingroup$ @AndreiGeanta Sorry and Thank you so much for pointing that out! $\endgroup$
    – Man_57
    Jul 2, 2018 at 17:17
  • $\begingroup$ @AndersSandberg gives a great answer. The issue is that in your work you are (accidentally) taking the limit as $v$ goes to $0$ instead of looking at when $v<<c$. You need the series expansion like in the below answer already provided to look at when $v<<c$ correctly $\endgroup$ Jul 2, 2018 at 17:17

2 Answers 2


The trick is to use a series expansion. $$\gamma = 1 + (1/2)\beta^2 + (3/8)\beta^4 + (5/16)\beta^6 + \ldots$$ where $\beta=v/c$. So if we use the first two terms:$$K.E=m_0c^2(\gamma -1) \approx m_0 c^2 (1/2) (v/c)^2 = (1/2)m_0v^2.$$

  • 1
    $\begingroup$ This is a great, correct answer. To add to what the OP was confused about, looking at when $v<<c$ is not the same thing as taking the limit as $v$ goes to $0$. In some situations we get the same answer either way, but as we can see in this answer it is not always the same. Taking the limit as $v$ goes to $0$ always results in $\gamma$ approaching $1$. However saying $v<<c$ is much different, and depends on how "much less" we want $v$ to be compared to $c$. This is when you use the series expansion above to get $\gamma$ to whatever accuracy you desire in this "limit". $\endgroup$ Jul 2, 2018 at 17:15

In the first cases you took the correct limit $v \ll c$ on the whole expressions. In the last case you took the limit only in the $\gamma$ factor, when you would take the limit on the whole expression $m_0c^2 (\gamma -1)$, because the $c^2$ in the $m_0c^2$ cancels $c^2$ terms in $\gamma$.

Thus, using the series expansion given by Anders Sandberg we obtain

$$KE= \frac{1}{2} m_0 v^2 + \frac{3}{8} m_0 \frac{v^4}{c^2} + \frac{5}{16} m_0 \frac{v^6}{c^4} + \mathcal{O}({v^8}/{c^6})$$

and taking the limit on the whole expression for $KE$ gives the expected answer

$$ \lim_{v \ll c} KE = m_0 (1/2)v^2$$

Also the equation $({1}/{2})mv^{2}=m_0c^2(\gamma-1)$ has a real solution: $v=0$. This is correct because the relativistic and nonrelativistic energies coincide only for objects at rest

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