Relativistic Kinetic Energy

In classical mechanics, the kinetic energy of an object depends on the mass of a body as well as its speed. The kinetic energy is equal to the mass multiplied by the square of the speed, multiplied by the constant 1/2. The equation is given as:

$E_{k} = \frac{1}{2}mv^{2}$,

where $m$ is the mass and $v$ is the speed (or the velocity) of the body.

The classical kinetic energy of an object is related to its momentum by the equation:

$E_{k} = \frac{p^{2}}{2m}$,

where $p$ is momentum.

If the speed of a body is a significant fraction of of the speed of light, it is necessary to employ special relativity to calculate its kinetic energy. It is important to know how to apply special relativity to problems with high speed particles. In special relativity, we must change the expression for linear momentum. Using $m$ for rest mass, $v$ and $\nu$ for the object's velocity and speed respectively, and $c$ for the speed of light in vacuum, the relativistic expression for linear momentum is:

$p = m\gamma v$, where $\gamma$ is the Lorentz factor:

$\gamma = \frac{1}{\sqrt{1 - (v/c)^{2}}}$.

Since the kinetic energy of an object is related to its momentum, we intuitively know that the relativistic expression for kinetic energy will also be different from its classical counterpart. Indeed, the relativistic expression for kinetic energy is:

$E_{k} = \frac{mc^{2}}{\sqrt{1 - (v/c)^{2})}} - mc^{2}$.

The equation shows that the energy of an object approaches infinity as the velocity $v$ approaches the speed of light $c$. Thus it is impossible to accelerate an object across this boundary.

The mathematical by-product of this calculation is the mass-energy equivalence formula (referred to in ). The body at rest must have energy content equal to:

Time Magazine - July 1, 1946

The popular connection between Einstein, E = mc2, and the atomic bomb was prominently indicated on the cover of Time magazine (July 1946) by the writing of the equation on the mushroom cloud itself.

$E_{rest} = E_{0} = mc^{2}$.

The general expression for the kinetic energy of an object that is not at rest is:

$KE = mc^2-m_0c^2$, where m is the relativistic mass of the object and m₀ is the rest mass of the object.

At a low speed ($v << c$), the relativistic kinetic energy may be approximated well by the classical kinetic energy. We can show this to be true by using a Taylor expansion for the reciprocal square root and keeping first two terms of the relativistic kinetic energy equation. When we do this we get:

$E_k \approx m c^2 \left(1 + \frac{1}{2} v^2/c^2\right) - m c^2 = \frac{1}{2} m v^2$.

Thus, the total energy can be partitioned into the energy of the rest mass plus the traditional classical kinetic energy at low speeds.