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