Relativistic Kinetic Energy


The work-energy theorem is a simple expression that relates the work done on a rigid body (W) to the change in kinetic energy of the body (\(\Delta K\)):

\[\begin{equation} \Delta K = W \end{equation}\]

Work is defined as the path integral of force. Writing out this path integral, we arrive at an expression that should be familiar to many students in physics.

\[\begin{equation} \Delta K = W = \int_{x_i}^{x_f} \vec{F} \cdot d\vec{x} \end{equation}\]

In the expression above, \(x_i\) and \(x_f\) are the initial and final positions of the rigid body. Although this expression is widely used in classical mechanics, we often forget that the work-energy theorem can be used in special relativity as well. With a bit of simple calculus, it is easy to solve for the kinetic energy of a relativistic particle using the formula above.

Relativistic Work Energy Theorem

Any work done on a point particle will change its kinetic energy - it does not matter if we are analyzing a relativistic or non-relativistic case. The only thing that changes in our analysis of kinetic energy in each case is the expression for force (\(\vec{F}\)). In the non-relativistic situation, we are familiar with Newton’s second law:

\[\begin{equation} \vec{F} = m\vec{a} \end{equation}\]

Recall that this is a simplification of the impulse momentum theorem when mass is constant:

\[\begin{equation} \vec{F} = \frac{d\vec{p}}{dt} = \frac{d}{dt} \left [ m\vec{v} \right ] = \vec{v} \frac{dm}{dt} + m\frac{d\vec{v}}{dt} = m\frac{d\vec{v}}{dt} = m\vec{a} \end{equation}\]

The impulse-momentum theorem can be used in the relativistic case as well. The only difference is that we will use the relativistic expression for momentum instead of the non-relativistic expression.

\[\begin{equation} \gamma = \frac{1}{\sqrt{1-(v/c)^2}} \end{equation}\] \[\begin{equation} \vec{p} = \gamma m \vec{v} \end{equation}\] \[\begin{equation} \vec{F} = \frac{d\vec{p}}{dt} = \frac{d}{dt} \left [ \gamma m \vec{v} \right ] \end{equation}\]

In the expressions above, \(\gamma\) is the Lorentz factor and \(m\) is the rest mass of the particle. The velocity \(\vec{v} = \beta c\) where \(c\) is the speed of light in a vacuum and \(\beta\) is the relativistic beta of the particle. We can solve for the kinetic energy of the particle in a number of ways. I am going to integrate with respect to velocity.

The Derivation

We will orient our axes so that the velocity and force are along the x-axis. Letting \(v_x \equiv v\) and \(F_x \equiv F\) yields:

\[\begin{equation} F = \frac{dp}{dt} = \frac{d}{dt} \left [ \gamma mv \right ] \end{equation}\]

Let’s plug this expression for force into the work-energy theorem and change our variable of integration to v. Since momentum and \(\gamma\) are both functions of speed, this substitution will simplify the analysis.

\[\begin{align} \Delta K &= \int_{x_i}^{x_f} \vec{F} \cdot d\vec{x}\\[0.1cm] &= \int_{x_i}^{x_f} F dx cos(0)\\[0.1cm] &= \int_{x_i}^{x_f} F dx\\[0.1cm] &= \int_{x_i}^{x_f} \frac{d}{dt} \left [ \gamma mv \right ] dx\\[0.1cm] &= \int_{t_i}^{t_f} \frac{d}{dt} \left [ \gamma mv \right ] v dt \end{align}\]

Now that everything is in terms of speed \(v\) and time \(t\) we can solve this integral by using integration by parts. Just to keep the notation clear, I will use a and b instead of u and v in my expressions. Let:

\[\begin{equation} a = v \Rightarrow da = \frac{dv}{dt} dt \end{equation}\] \[\begin{equation} db = \frac{d}{dt} \left [ \gamma m v \right ] dt \Rightarrow b = \gamma mv \end{equation}\]

We then get:

\[\begin{align} \Delta K &= \left [ ab \right ]_{t_i}^{t_f} - \int_{t_i}^{t_f} b da\\[0.1cm] &= \left [ \gamma m v^2 \right ]_{t_i}^{t_f} - \int_{t_i}^{t_f} \gamma mv \frac{dv}{dt} dt\\[0.1cm] &= \left [ \gamma m v^2 \right ]_{v_i}^{v_f} - \int_{v_i}^{v_f} (\gamma mv) dv\\[0.1cm] &= \left [ \frac{m v^2}{\sqrt{1-(v/c)^2}} \right ]_{v_i}^{v_f} - \int_{v_i}^{v_f} \frac{mv}{\sqrt{1-(v/c)^2}} dv \end{align}\]

In the last two lines above, we use the fact that the velocity of the system is equal to \(v_i\) at time \(t_i\) and \(v_f\) at time \(t_f\). We can solve the integral above using a simple u-substituion. Let \(u = 1 - (v/c)^2\). Then, \(du = -2v dv/c^2\).

\[\begin{align} \Delta K &= \left [ \frac{m v^2}{\sqrt{1-(v/c)^2}} \right ]_{v_i}^{v_f} - \int_{v_i}^{v_f} \frac{mv}{\sqrt{1-(v/c)^2}} dv\\[0.1cm] &= \left [ \frac{m v_f^2}{\sqrt{1-(v_f/c)^2}} - \frac{m v_i^2}{\sqrt{1-(v_i/c)^2}} \right ] + \frac{mc^2}{2} \int_{1 - (v_i/c)^2}^{1-(v_f/c)^2} \frac{du}{\sqrt{u}}\\[0.1cm] &= \left [ \frac{m v_f^2}{\sqrt{1-(v_f/c)^2}} - \frac{m v_i^2}{\sqrt{1-(v_i/c)^2}} \right ] + \frac{mc^2}{2} \left [ 2\sqrt{u} \right ]_{1 - (v_i/c)^2}^{1-(v_f/c)^2}\\[0.1cm] &= \left [ \frac{m v_f^2}{\sqrt{1-(v_f/c)^2}} - \frac{m v_i^2}{\sqrt{1-(v_i/c)^2}} \right ] + mc^2 \left [ \sqrt{1-(v_f/c)^2} - \sqrt{1-(v_i/c)^2} \right ]\\[0.1cm] &= \left [ \frac{m v_f^2}{\sqrt{1-(v_f/c)^2}} - \frac{m v_i^2}{\sqrt{1-(v_i/c)^2}} \right ] + mc^2 \left [ \frac{1-(v_f/c)^2}{\sqrt{1-(v_f/c)^2}} - \frac{1-(v_i/c)^2}{\sqrt{1-(v_i/c)^2}} \right ]\\[0.1cm] &= \left [ \frac{m v_f^2}{\sqrt{1-(v_f/c)^2}} - \frac{m v_i^2}{\sqrt{1-(v_i/c)^2}} \right ] + \left [ \frac{mc^2}{\sqrt{1-(v_f/c)^2}} - \frac{m v_f^2}{\sqrt{1-(v_f/c)^2}} - \frac{mc^2}{\sqrt{1-(v_i/c)^2}} + \frac{m v_i^2}{\sqrt{1-(v_i/c)^2}} \right ]\\[0.1cm] &= \frac{mc^2}{\sqrt{1-(v_f/c)^2}} - \frac{mc^2}{\sqrt{1-(v_i/c)^2}}\\[0.1cm] &= \gamma_f mc^2 - \gamma_i mc^2\\[0.1cm] &= \left (\gamma_f - \gamma_i \right )mc^2 \end{align}\]

Suppose that our particle starts from rest (or we want to use a speed of zero as a baseline). That means \(v_i = 0\) and \(\gamma_i = 1\). Thus, the final expression for the kinetic energy is:

\[\begin{equation} K_f = \left ( \gamma_f - 1 \right ) mc^2 \end{equation}\]

I plot the kinetic energy of a particle with mass m = 1 kg at different values of \(\beta\) in the graph below. Note that the kinetic energy of the moving particle is completely determined by its rest mass and the Lorentz factor (along with a couple of constants). As \(\beta \rightarrow 1\) (i.e. the speed of the particle approaches the speed of light) the energy tends to infinity.

Relativistic Kinetic Energy

The Low Speed Limit

How Does it Relate to Non-Relativistic Kinetic Energy?

In a previous blog post I showed how one can expand the Lorentz factor, drop higher order terms, and retrieve the low-speed limit of relativistic expressions. Why don’t we do that here? Recall that the expansion for \(\gamma\) is given by:

\[\begin{equation} \gamma = 1 + \frac{1}{2}\beta^2 + \frac{3}{8}\beta^4 + \frac{5}{16}\beta^6 + \frac{35}{128}\beta^8 + \dots \end{equation}\]

Substitute that into our expression for kinetic energy.

\[\begin{equation} K_f = \left ( \frac{1}{2}\beta_f^2 + \frac{3}{8}\beta_f^4 + \frac{5}{16}\beta_f^6 + \frac{35}{128}\beta_f^8 + \dots \right ) mc^2 \end{equation}\]

Drop all terms of \(\beta_f^3\) or higher.

\[\begin{equation} K_f \approx \frac{1}{2}\beta^2 mc^2 = \frac{1}{2}mv^2 \end{equation}\]

That looks familiar. We got the classical, non-relativistic kinetic energy when we took the low speed limit. Everything is self-consistent! In the figure below, I plot the non-relativistic and relativistic calculations for kinetic energy at different values of \(\beta\). Notice that at low speeds, they match up pretty well. But when \(\beta\) starts to increase, the non-relativistic kinetic energy is way off from the correct relativistic kinetic energy.

Relativistic vs. Non-Relativistic Calculations for Kinetic Energy


Whenever we break into a new area of physics, we need to remember that our previous knowledge is still valid. Here, we used a simple classical concept to derive the relativistic kinetic energy of a particle.

Until next week!