Question

How to derive the Relativistic Energy-momentum relation.

Answer 1

Hint: We first define the Relativistic Energy-momentum relation in Physics, as the Relativistic-dispersion relation, which relates the total energy (also known as relativistic energy) to the rest mass of an object and its momentum. We will derive this equation with the help of some previously known equations, that are, total energy of a particle and the equation for relativistic momentum.

Complete step-by-step solution:
Let the total energy of the particle (not taking into consideration the potential energy of the particle) be E and the relativistic momentum of the particle be p. Then we can write:
$\Rightarrow E=K+m{{c}^{2}}$
Where,
$K$ is the kinetic energy of the particle. And,
$m{{c}^{2}}$ is the rest mass energy.
Also, this total energy E is written as:
$\Rightarrow E=\dfrac{m{{c}^{2}}}{\sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}}}$ [Let this expression be equation number (1)]
And, the relativistic momentum is given by:
$\Rightarrow p=\dfrac{mv}{\sqrt{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}}}$ [Let this expression be equation number (2)]
Now, we begin with squaring the energy equation, that is, number (1) as follows:
$\Rightarrow {{E}^{2}}=\dfrac{{{m}^{2}}{{c}^{4}}}{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}}$
Inserting, ${{v}^{2}}-{{v}^{2}}$ as shown and simplifying, we get:
$\begin{align}
  & \Rightarrow {{E}^{2}}=\dfrac{{{m}^{2}}{{c}^{2}}({{v}^{2}}-{{v}^{2}}+{{c}^{2}})}{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}} \\
 & \Rightarrow {{E}^{2}}=\dfrac{{{m}^{2}}{{c}^{2}}{{v}^{2}}}{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}}+\dfrac{{{m}^{2}}{{c}^{4}}-{{m}^{2}}{{c}^{2}}{{v}^{2}}}{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}} \\
\end{align}$
Using the relativistic momentum equation, that is equation number (2) in R.H.S., we get:
$\begin{align}
  & \Rightarrow {{E}^{2}}={{p}^{2}}{{c}^{2}}+{{m}^{2}}{{c}^{2}}\dfrac{\left( {{c}^{2}}-{{v}^{2}} \right)}{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}} \\
 & \Rightarrow {{E}^{2}}={{p}^{2}}{{c}^{2}}+{{m}^{2}}{{c}^{2}}\dfrac{\left( {{c}^{2}}-{{v}^{2}} \right)}{\dfrac{{{c}^{2}}-{{v}^{2}}}{{{c}^{2}}}} \\
 & \Rightarrow {{E}^{2}}={{p}^{2}}{{c}^{2}}+{{m}^{2}}{{c}^{4}} \\
\end{align}$
Thus, the final equation of the Relativistic Energy-Momentum relation has been derived. And it is given by the following equation:
$\Rightarrow {{E}^{2}}={{p}^{2}}{{c}^{2}}+{{m}^{2}}{{c}^{4}}$

Note: Here, the term ${{m}^{2}}{{c}^{4}}$ is a constant and it is equal to, ${{E}^{2}}-{{p}^{2}}{{c}^{2}}$. This states that the difference between the squares and relativistic momentum will remain constant in all reference frames. It should also be noted that relativistic momentum and relativistic energy changes if we change the inertial reference frame but the mass of the particle remains constant.