# Relativistic kinetic energy

The kinetic energy of an object is given by,

$$KE = m_0c^2\left(1/\sqrt{1-v^2/c^2} - 1\right)$$

Where $m_0$ is the rest mass.

If the total nuclear energy of the object ($E = m_0c^2$) is converted to kinetic energy of an identical object, then we have,

$$m_0c^2\left(1/\sqrt{1-v^2/c^2} - 1\right) = m_0c^2$$

Giving,

$$v/c = \sqrt3/2$$

Does this ratio (found by a simple calculation) relate to any theories in physics or is it of very little importance or relevance?

• To me, I don’t see any theoretical importance of this ratio. – Kevin Kwok May 5 '18 at 12:47
• It's not entirely clear what you mean by "total nuclear energy", but $m_0 c^2$ is not the amount of nuclear binding energy present in an object - it's the rest mass energy of that object - the energy due to the mass of the particles and their interactions. – Brionius May 5 '18 at 13:22
• And I'd say you've already stated its significance. If you give an object a kinetic energy equal to its rest mass energy, it will be moving with a speed of $\frac{\sqrt{3}}{2} c$ – Brionius May 5 '18 at 13:24
• The total nuclear energy would NOT be $E=m_{0}c^{2}$ - it would be $E = \gamma m_0c^2$. Hence, $E_{i}=\gamma m_0c^2$ and $E_{f} = (\gamma-1) m_0c^2+ m_{0}c^{2}$, hence $\Delta E=0$ – Cinaed Simson Jul 21 at 1:44

The number by itself is found as the cos(30 degrees) in a right triangle with hypotenuse 1 with legs 1/2 and sqrt(3)/2. Of course in special relativity, the relevant Minkowski-right triangle has hypotenuse 1 and the legs are 2 and sqrt(3). (This value for $\beta$ arises from the simpler equation that $\gamma=2$.)