# Under which circumstances Relativistic Kinetic energy tends to Classical kinetic energy?

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?

• I don't think the second equation has a square root. – Andrei Geanta Jul 2 '18 at 17:08
• @AndreiGeanta Sorry and Thank you so much for pointing that out! – Man_57 Jul 2 '18 at 17:17
• @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 – Aaron Stevens Jul 2 '18 at 17:17

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:$$K.E=m_0c^2(\gamma -1) \approx m_0 c^2 (1/2) (v/c)^2 = (1/2)m_0v^2.$$

• 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". – Aaron Stevens Jul 2 '18 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 