Energy-Momentum Relation

Starting from the classical formula of the kinetic energy $$E = \frac{1}{2}mv^2$$ and inserting the formula for the momentum $$p = mv$$ leads to the following relation between the momentum and energy of an object $$\boxed{E = \frac{2^2}{2m}}$$ Switching now to special relativity, this formula cannot be applied anymore. Instead, we have to use the formula for the relativistic energy $$E = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}c^2$$ and the relativistic momentum $$p = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}v$$ where $m_0$ is the rest mass and $v$ is the speed of the object. In the first step, we can divide both equations solve the resulting equation for $v$: $$v = \frac{c^2p}{E}$$ Now we can multiply the energy formula with the square root in the denominator and get $$E\sqrt{1-\frac{v^2}{c^2}} = m_0c^2$$ Inserting now the derived formula for $v$ into the equation above results in $$E\sqrt{1-\frac{c^2p^2}{E^2}} = E_0$$ where $E_0$ was used to substitute the product of $m_0$ and $c^2$. In the last step, we only have to square this equation and receive with $$\boxed{E^2 = E_0^2 + p^2c^2}$$ a very important formula in the special theory of relativity. It shows how to calculate the sum up of the momentum and the rest energy of an object. Similar to classical mechanics, there is no simple linear dependency, but instead, it is required to calculate the squared sum of $E_0$ and $p$. We will see later that this formula gets very important when dealing with particle physics.