Why is velocity equal to $c^2 \vec{p} \over E$ in special relativity? I read that velocity for a particle with four vector ($\vec{p}$, E) is $c^2 \vec{p} \over E$. Why is this so?
 A: The energy-momentum four-vector is $\left(\vec{p},\frac{E}{c}\right)$.
Relativistic momentum $\vec{p}=\gamma m\vec{v}$, while relativistic energy $E=\gamma m c^2 $, with $\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$, 
then $\frac{\vec{p}}{E}=\frac{\vec{v}}{c^2}$ 
Reference:
D. Griffiths, Introduction to Elementary Particles, John Wiley & Sons, 1987, p. 89.
A: The energy-momentum four-vector has to be parallel to the velocity four-vector. (If it weren't, then we could transform into the frame where the particle is at rest, and it would still have some momentum, which would violate rotational invariance.)
Velocity is by definition the ratio of the spacelike part of the velocity four-vector to its timelike part. Since the two vectors are parallel, the same holds for the energy-momentum four-vector.
A: 4-velocity defined as, 
$$U = (c\gamma, v^x\gamma, v^y\gamma, v^x\gamma)$$
Where $\vec{v} = (v^x, v^y, v^z)$ and $\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$ 
Thus the 4- momentum becomes, 
$$\vec{P} = m\vec{U}$$
Where $$\vec{P} = (mc\gamma, mv^x\gamma, mv^y\gamma, mv^x\gamma)$$
By setting $E=mc^2\gamma$ we have
$$\vec{P} = (\frac{E}{c}, \vec{p})$$
where $\vec{p} = m\vec{v}\gamma$
Here one can see that, 
$$\frac{\vec{p}}{E} = \frac{m\vec{v}\gamma}{mc^2\gamma}$$
Thus $$\vec{v} = \frac{\vec{p}c^2}{E}$$ 
A: The total relativistic energy in Special Relativity could be written as $E=m c^2$ with $m$ the relativistic mass. As for $\vec{p}= m \vec{v}$ using again the relativistic mass, not the at rest mass. You can see that $\vec{v}$ is left as the result of your expression.
