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,

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


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

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

  • $\begingroup$ To me, I don’t see any theoretical importance of this ratio. $\endgroup$ – Kevin Kwok May 5 '18 at 12:47
  • $\begingroup$ 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. $\endgroup$ – Brionius May 5 '18 at 13:22
  • 2
    $\begingroup$ 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 $\frac{\sqrt{3}}{2} c$ $\endgroup$ – Brionius May 5 '18 at 13:24
  • $\begingroup$ 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 $\Delta E=0$ $\endgroup$ – 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 $\gamma=2$.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.