Relativistic Energy

• The relativistic total energy of an object is given as

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

• The energy of an object as rest is therefore described by

$$E_0 = m_0c^2$$

• As a result, the kinetic energy can be defined as

$$E_\mathrm{kin} = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}c^2 - m_0c^2 = \left(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} - m_0\right)c^2$$ For large speeds, the previously derived formula $$E_\mathrm{kin} = \frac{1}{2}m_0v^2$$ for the kinetic energy of an object cannot be applied anymore, because for solving the integral it had been assumed that the mass $m_0$ is constant throughout the full acceleration process. Before we start the complete derivation, we want to take a look at a heuristic approach to get an idea of how to calculate the energy of an object relativistically. We have already seen that the dynamic mass of an object which moves with the speed $v$ is given as $$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$ If we multiply $m$ with $c^2$, we receive a term with the energy Joule and assume that this would be our formula for the relativistic energy: $$E = mc^2 = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$ We can now simplify the square root using the following mathematical approximation: $$\frac{1}{\sqrt{1-x^2}} \approx 1+ \frac{1}{2}x + \dots$$ where $x$ replaces the fraction $v^2/c^2$. For this derivation, we want to ignore higher-order terms behind the second plus sign which is true for $x \ll 1$. Applying this formula to the equation $mc^2$ leads to $$E \approx m_0c^2 + \frac{1}{2}m_0v^2c^2 + \dots$$ The resulting formula says that the total energy of an object (if the speed of the object is not too large), can be calculated as the sum of the rest mass multiplied by the squared speed of light plus a term that describes the classical kinetic energy. In general, we can now assume the following formula for the total energy $E$: $$\boxed{E = mc^2 = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}c^2}$$ which is a very famous equation found by Albert Einstein. We start again from the formula $$E_\mathrm{kin} = \int F'\,\mathrm{d}s$$ and replace $F'$ with $\mathrm{d}p/\mathrm{d}t$: $$E_\mathrm{kin} = v\,\mathrm{d}p'$$ Now, we can replace $p'$ with $\gamma m v$ as shown before: $$E_\mathrm{kin} = \int v\,\mathrm{d}(\gamma m_0 v)$$ In order to solve this integral, we first have to calculate the derivation of $\gamma m_0 v$ with respect to $v$: $$\mathrm{d}(\gamma m_0 v) = \frac{m_0}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} + \frac{m_0 v^{2}}{c^{2} \left(1 - \frac{v^{2}}{c^{2}}\right)^{\frac{3}{2}}}\,\mathrm{d}v$$ This makes it possible to solve the integral with the help of partial integration. After some algebraic rearrangements, this results in $$E_\mathrm{kin} = \frac{m_0c^2}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} + C$$ where $C$ is simply the integration constant from the indefinite integral. This constant can be determined when setting $E=0$ for $v=0$ which leads to $$C = -m_0c^2$$ This constant can therefore be interpreted as the rest energy $$\boxed{E_0 = m_0c^2}$$ of an object if it does not move relative to an observer. The total energy of an object is then given by rearranging the derived equation as $$\boxed{E' = E + E_\mathrm{kin}}$$ This equation says that the total energy $E'$ of an object equals the sum of the rest energy $E_0 = m_0c^2$ and its kinetic energy $E_\mathrm{kin}$. Additionally, the total energy $E'$ can be written as $$\boxed{E' = \gamma m_0 c^2}$$ The correlation between the rest energy and rest mass of a body makes it also possible to note down its mass in units of energy per squared speed of light by rearranging the formula according to $$m = \frac{E}{c^2}$$ For very small and light particles such as elementary particles, it is very common to use the unit Electronvolt instead of Joule. The conversion factor between these both units is simply the elementary charge $e = 1.602\cdot 10^{-19}\,\mathrm{C}$: $$\boxed{E[\mathrm{J}] = e E[\mathrm{eV}]}$$ Keeping this in mind, the rest mass of an electron with $m=9,109\cdot 10^{-31}\,\mathrm{kg}$ can then be written as $511\,\mathrm{keV}/c^2$ which is an important number to memorize. Sometimes the mistake is made to insert the relativistic mass into the classical formula for the kinetic energy. This, however, does not lead to the correct result, because it misses the integration as explained above.