Question

For an extremely relativistic particle of rest energy e=mc² ,show that the momentum (P) is given by;
P=E[1-½(E/E)²] 

Answer 1

Relativistic momentum p is classical momentum multiplied by the relativistic factor $γ.$

$p=γmu,$

where m is the rest mass of the object, $u$ is its velocity relative to an observer, and the relativistic factor.

$γ= \large\frac{1}{\sqrt{1-\large\frac{u^2}{c^2}}}$ $p=\large\frac{mu}{\sqrt{1-\large\frac{u^2}{c^2}}}$ $E = mc^2 \to m = \large\frac{E}{c^2}$

is the three dimensional relativistic momentum of the object in the lab frame with magnitude |p| = p. The relativistic energy E and momentum p include the Lorentz factor defined by:

${\displaystyle \gamma _{(\mathbf {u} )}={\frac {1}{\sqrt {1-{\frac {\mathbf {u} \cdot \mathbf {u} }{c^{2}}}}}}={\frac {1}{\sqrt {1-\left({\frac {u}{c}}\right)^{2}}}}}$

Some authors use relativistic mass defined by:

${\displaystyle m=\gamma _{(\mathbf {u} )}m_{0}}$

although rest mass m₀ has a more fundamental significance and will be used primarily over relativistic mass m in this article.

Squaring the 3-momentum gives:

${\displaystyle p^{2}=\mathbf {p} \cdot \mathbf {p} ={\frac {m_{0}^{2}\mathbf {u} \cdot \mathbf {u} }{1-{\frac {\mathbf {u} \cdot \mathbf {u} }{c^{2}}}}}={\frac {m_{0}^{2}u^{2}}{1-\left({\frac {u}{c}}\right)^{2}}}}$

then solving for u² and substituting into the Lorentz factor obtains its alternative form in terms of 3-momentum and mass, rather than 3-velocity:

$\gamma = \sqrt{1 + \left(\frac{p}{m_0 c}\right)^2}$

Inserting this form of the Lorentz factor into the energy equation:

$E = m_0c^2\sqrt{1 + \left(\frac{p}{m_0 c}\right)^2}$ $\large(\frac{E}{E_o})^2$ $=1+\large\frac{p^2}{E_o^2} \to$ $p^2 = E_o^2 (E^2-E_o^2) \to p = E_o\sqrt{E^2-E_o^2}$