Special Relativity- Calculating Momentum from Relativistic Kinetic Energy

Special Relativity is one of the most important inventions in Physics involving principles of Mass-Energy Equivalence, Relativistic Kinetic Energy, Momentum conservation under Special theory of Relativity mentioned in Einstein Relativity.

Here we describe above concepts and derive a formula for calculating the momentum from relativistic kinetic energy using the principles of special relativity. We also solve one practice problem for understanding the practical applicability of this Special relativity in physics.

Relativity and Special Relativity – Problem -1

Derive the formula for Momentum using Relativistic Kinetic Energy principle and Calculate the momentum when the ratio of the particle’s Relativistic Kinetic Energy to its Rest Mass Energy is 5.

Solution:

Let the Rest Mass Energy = $K_0$

Let Relativistic Kinetic Energy = K

Let Ratio of The particle’s Relativistic Kinetic Energy to its Rest Mass Energy = $\frac{K}{K_0}=n$

Relative Kinetic Energy = $K={\mathrm{nK}}_0$

Let c= speed of light

$m_0=\mathrm{rest}\mathrm{mass}$

From above and the definition of Rest mass energy,

$\mathrm{Rest}\mathrm{mass}\mathrm{energy}=K_0=m_0{c}^2$ , hence,

Relative Kinetic Energy = $K={\mathrm{nK}}_0={\mathrm{nm}}_0{c}^2$

Total Energy = TE= Relativistic Kinetic Energy + Rest Mass Energy = ${\mathrm{nm}}_0{c}^2+m_0{c}^2=(n+1)m_0{c}^2$

But, from the Einstein Theory of relativity,

$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$

From the familiar equation of Energy – Mass relationship,

$E={\mathrm{mc}}^2=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}{c}^2$ , comparing this with

E=$(n+1)m_0{c}^2$

Mass-Energy Equivalence Application

From above two equations,

$(n+1)=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ , hence,

$(n+1{)}^2=\frac{1}{1-\frac{v^2}{c^2}}\Rightarrow \frac{v^2}{c^2}=1-\frac{1}{(n+1{)}^2}$, hence,

$\frac{v}{c}=\sqrt{1-\frac{1}{(n+1{)}^2}}$ ,

Hence,

Momentum Calculations

$\mathrm{Momentum}=p=\mathrm{mv}=(n+1)m_{0}v$

By substituting the value of v in the above equation , we get,

$p=m_{0}c(n+1)\sqrt{1-\frac{1}{(n+1{)}^2}}=\frac{m_{0}c(n+1)}{(n+1)}\sqrt{(n+1{)}^2-1}$

From above we get the formula for the Momentum as,

Numerical problem solution on Special Relativity

Here it is given that n=5.

By substituting the value of n as 5, in above equation of momentum we get,

$p=m_{0}c\sqrt{n(n+2)}=m_{0}c\sqrt{5(5+2)}=\sqrt{35}{m}_{0}c$

Conclusion

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