Mass-Energy Equivalence Derivation

Introduction

The derivation of Einstein’s most famous mass-energy equivalence from scratch is not trivial. However, this mass-energy equivalence equation is just a derivation based on special relativity and my previous blog posts on the topic of special relativity have discussed all the prerequisites to derive this equation.

In this blog post, I would like to derive Einstein’s mass-energy equivalence equation in detail.

Mass-Energy Equivalence

Newton’s second law states that the rate of change of momentum of a body is directly proportional to the force applied, and this change in momentum takes place in the direction of the applied force.

$$
\begin{align}
F = \frac{dp}{dt} = \frac{d(mv)}{dt}
\end{align}
$$

where $p$ is the momentum, $m$ is the relativistic mass, $v$ is the velocity, and $t$ is time.

The kinetic energy is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. An infinitesimal work, as well as infinitesimal kinetic energy $d E_k$, could be expressed as follows

$$
\begin{align}
d E_k = F ds = F v dt = \frac{d(mv)}{dt} v dt = v d(mv)
\end{align}
$$

The total kinetic energy of the body which starts from velocity zero to $v’$ is

$$
\begin{align}
E_k &= \int d E_k = \int_{0}^{v’} v d (mv) = \int_{0}^{v’} v d \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}} \\
&= m_0 \int_{0}^{v’} v d \frac{v}{\sqrt{1 - \frac{v^2}{c^2}}}
\end{align}
$$

where $m_0$ is the rest mass.

The total kinetic energy could be further solved using integral by parts easily.

$$
\begin{align}
E_k &= m_0 \Bigg( v \frac{v}{\sqrt{1 - \frac{v^2}{c^2}}} \Bigg\rvert_{0}^{v’} - \int_{0}^{v’} \frac{v}{\sqrt{1 - \frac{v^2}{c^2}}} dv \Bigg) \\
&= \frac{m_0 v’^2}{\sqrt{1 - \frac{v’^2}{c^2}}} - m_0 \Bigg( -c^2 \sqrt{ 1 - \frac{v^2}{c^2}} \Bigg\rvert_{0}^{v’} \Bigg) \\
&= \frac{m_0 v’^2}{\sqrt{1 - \frac{v’^2}{c^2}}} + m_0 \Bigg( c^2 \sqrt{ 1 - \frac{v’^2}{c^2}} - c^2 \Bigg) \\
&= \Bigg( \frac{m_0 v’^2}{\sqrt{1 - \frac{v’^2}{c^2}}} + m_0 c^2 \sqrt{ 1 - \frac{v’^2}{c^2}} \Bigg) - m_0 c^2 \\
&= \frac{m_0 c^2}{\sqrt{1 - \frac{v’^2}{c^2}}} - m_0 c^2 \\
&= m’ c^2 - m_0 c^2 \\
&= \gamma m_0 c^2 - m_0 c^2 \\
\end{align}
$$

where $m’$ is the relativistic mass at the velocity of $v’$, and $\gamma$ is the Lorentz factor.

We define the total energy $E$ as the sum of $E_k$ and $m_0 c^2$. We have the mass-energy equivalence.

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

Although it is not recommended, people sometimes use relativistic mass in the above equivalence, and we would get the familiar mass-energy equivalence expression.

$$
E = m c^2
$$

where $m$ represents the relativistic mass in this particular place.

Significance

If something could spontaneously lose mass via some kind of form, it would emit energy. Because $c^2$ is a very large value, the energy it will emit might be extremely significant. For example, radioactive substances which lose mass by ejecting $\alpha$ and $\beta$ particles. We already know $\alpha$ and $\beta$ particles have high energy. Actually, long time before this famous mass-energy equivalence equation was born, we already knew that there are things that could emit energy spontaneously. Our sun is one of the best examples which could see daily.

References

• Relativistic Momentum and Relativistic Mass