# 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.

Lei Mao

03-28-2020

03-28-2020