Correct Answer - Option 1 : E/c
Relation between Relativistic E and \(\vec{P}\)
1) E = K + moc2
2) \(E = \dfrac{m_oc^2}{\sqrt{1-\dfrac{r^2}{c^2}}}\)
3) \(P = \dfrac{m_o}{\sqrt{1-\dfrac{r^2}{c^2}}}\)
Derivation:
We begin by squaring equation (2) on both sides, i.e.
\(E^2 = \dfrac{m_o^2c^4}{1-r^2/c^2}\)
Next, we insert the term (V2 - V2) as shown:
\(E^2 = \dfrac{m_o^2 c^2 (V^2 - V^2 +c^2)}{1-r^2/c^2} \)
\(=\dfrac{m_o^2c^2V^2}{1-r^2/c^2} - \dfrac{m_o^2 c^2 V^2}{1-r^2 /c^2} + \dfrac{m_o^2 c^4}{1-r^2/c^2}\)
\(E^2=\underset{= \ p}{\underbrace{\left(\dfrac{m_oV}{\sqrt{1-r^2/c^2}}\right)^2}}c^2 + \dfrac{m_oc^2 c^4 - m^2 c^2 r^2}{1-r^2/c^2}\)
\(E^2 =p^2 c^2 + m_o^2 c^2 \left(\dfrac{c^2 - r^2}{1-r^2/c^2}\right)\)
\(E^2 = p^2 c^2 + m_o^2 c^2 \left(\dfrac{c^2 - V^2}{\dfrac{c^2 - V^2}{C^2}}\right)\)
\(E^2 = p^2 c^2 + m_o^2 c^4\) where E is the total energy of the particle
\(m_o^2 c^4 = E^2 - p^2 c^2\) ← will be identical in all inertial reference frames.
