Newton as a First Order Taylor Approximation of Einstein

One of my favourite proofs from undergraduate physics is the derivation of the classical kinetic energy formula from Einstein's energy-momentum relation. Specifically, we'll see how the classical expression for kinetic energy emerges as a first-degree Taylor approximation of the relativistic energy equation when velocities are much smaller than the speed of light ($c$).

According to Einstein's theory of relativity, the total energy $E$ of a particle with rest mass $m_0$ is given by:

$$
E = \gamma m_0 c^2,
$$

where $\gamma$, the Lorentz factor, is defined as:

$$
\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}
$$

where $v$ is the velocity of the particle and $c$ is the speed of light in a vacuum. The relationship simplifies significantly when velocities are much smaller than $c$.

The kinetic energy $K$ of a particle is defined as the difference between its total energy and its rest energy:

$$
K = E - E_0,
$$

where $E_0 = m_0 c^2$ is the rest energy of the particle. Plugging in the expression for $E$:

$$
K = \gamma m_0 c^2 - m_0 c^2 = (\gamma - 1) m_0 c^2.
$$

Now we need to simplify $\gamma$ for small $v$.

To obtain a first-order approximation, we can expand $\gamma$ using a Taylor series around $v = 0$. The Taylor expansion of $\gamma$ at $v = 0$ is given by:

$$
\gamma(v) = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \approx 1 + \frac{1}{2} \frac{v^2}{c^2} + \text{higher order terms}.
$$

For our purposes, since we are interested in the first-order approximation, we can ignore the $\frac{1}{2} \frac{v^2}{c^2}$ term and retain just the constant term:

$$
\gamma(v) \approx 1.
$$

Thus, the kinetic energy expression simplifies to:

$$
K \approx \left(1 - 1\right) m_0 c^2 = 0 + \frac{1}{2} m_0 \frac{v^2}{c^2} = \frac{1}{2} m_0 v^2.
$$

We can now express the kinetic energy $K$ for small velocities as:

$$
K \approx \frac{1}{2} m_0 v^2
$$

which is precisely the classical formula for the kinetic energy of an object. Good fun!