Mass-Energy Relation

The Mass-Energy Relation is one of the most profound and revolutionary discoveries in physics, proposed by Albert Einstein in his Special Theory of Relativity (1905). It fundamentally connects mass and energy, two quantities that were once considered distinct.

Einstein showed that mass is a form of energy, and they are interchangeable. This relationship is beautifully expressed by the famous equation:

$E=m{c}^2$

Here, E is the energy, m is the mass, and c is the speed of light in a vacuum ($3\times 10^{8}m/s$).

1.0The Formula: $E=m{c}^2$

The most famous equation in physics, $E=m{c}^2$, is the cornerstone of the mass-energy relation.

• E represents the total energy of the object.
• m represents the relativistic mass of the object, which increases with its velocity.
• c is the speed of light in a vacuum ($3\times 10^{8}m/s$).

A more general and complete formula from Special Relativity, which you'll need for JEE problems involving moving particles, is the relativistic energy-momentum relation:

$E^2=(pc{)}^2+(m_0{c}^2{)}^2$

• E is the total energy.
• p is the relativistic momentum.
• $m_0$ is the rest mass, the mass of the object when it's stationary.

From this, we can define a particle's total energy as the sum of its rest energy ($E_0=m_0{c}^2$) and its kinetic energy (KE).

$E=E_0+KE=m_0{c}^2+KE$

This leads to the expression for relativistic kinetic energy: 

$KE=(\gamma -1)m_0{c}^2\text{, where }\gamma \text{ is the Lorentz factor }(\gamma =\frac{1}{\sqrt{1-v^2/c^2}}).$

2.0Derivation of Mass-Energy Relation

A full, rigorous Mass-Energy Relation derivation is complex, but understanding the key steps is essential for a deeper grasp of the concept. The derivation connects the concepts of relativistic momentum and total energy.

Relativistic Momentum

In Special Relativity, the momentum of a particle is given by: 

$p=\gamma m_{0}v=\frac{m_{0}v}{\sqrt{1-\frac{v^2}{c^2}}}$

Relativistic Kinetic Energy

Kinetic energy is defined as the work done to accelerate a particle. Through a detailed integration of the force over distance, the relativistic kinetic energy is found to be: 

$KE=(\gamma -1)m_0{c}^2$

This expression is crucial. When a particle's velocity v is much smaller than c, the Lorentz factor γ can be approximated using a binomial expansion: 

$\gamma \approx 1+\frac{1}{2}\frac{v^2}{c^2}$

​Substituting this into the kinetic energy equation:

$K\approx \left(1+\frac{1}{2}\frac{v^2}{c^2}-1\right)m_0{c}^2=\frac{1}{2}{m}_0{v}^2$

This shows that the familiar classical kinetic energy formula is just a low-speed approximation of the relativistic formula.

Connecting Energy and Momentum

The total energy of a particle is the sum of its rest energy and kinetic energy.

$E=m_0{c}^2+(\gamma -1)m_0{c}^2=\gamma m_0{c}^2$

Deriving the Full Equation

Let's start with the expressions for total energy and momentum:

$E=\frac{m_0{c}^2}{\sqrt{1-v^2/c^2}}⟹E^2=\frac{m_0^2{c}^4}{1-v^2/c^2}$

$p=\frac{m_{0}v}{\sqrt{1-v^2/c^2}}⟹p^2=\frac{m_0^2{v}^2}{1-v^2/c^2}$

Now, let's manipulate the expression for $E^2$: 

$E^2(1-v^2/c^2)=m_0^2{c}^4$

$E^2-E^2\frac{v^2}{c^2}=m_0^2{c}^4$

$E^2-\frac{(\gamma m_0{c}^2{)}^2{v}^2}{c^2}=m_0^2{c}^4$

$E^2-(\gamma m_{0}v{)}^2{c}^2=m_0^2{c}^4$

$E^2-p^2{c}^2=m_0^2{c}^4$

Rearranging this gives us the final, powerful equation: 

$E^2=(pc{)}^2+(m_0{c}^2{)}^2$

This is the most complete and fundamental mass-energy relation formula.

3.0Applications of the Mass-Energy Relation

The mass-energy relation has countless applications — from understanding the universe to powering modern technology.

Nuclear Energy

In nuclear fission, such as in reactors, a small amount of mass is converted to energy:

$\Delta E=\Delta m{c}^2$

Example: In uranium-235 fission, about 0.1%0.1% of the total mass is transformed into energy.

Nuclear Fusion

Fusion of hydrogen into helium in the Sun releases massive amounts of energy due to the same principle.

Particle Accelerators

When particles are accelerated close to the speed of light, their relativistic mass increases, confirming $E=m{c}^2$.

Medical Imaging (PET Scans)

Positron Emission Tomography (PET) uses mass-to-energy conversion from matter-antimatter annihilation.

Cosmic Phenomena

High-energy astrophysical events, such as supernovae or black holes, involve conversion of enormous mass into radiant energy.

4.0Solved Problems on Mass-Energy Relation

1. In a nuclear fission reaction, the mass of the products is 0.1% less than the mass of the reactants. If the mass of the reactant is 1 kg, find the energy released in joules.

Solution: The mass defect (Δm) is the mass that gets converted into energy.

$\Delta m=0.1\%\text{ of }1\text{ kg}=\frac{0.1}{100}\times 1\text{ kg}=0.001\text{ kg}$

Using Einstein's mass-energy relation, $E=\Delta m{c}^2$:

This shows how a tiny amount of mass can release a huge amount of energy.

2. An electron is moving with a velocity such that its kinetic energy is equal to its rest mass energy. What is the speed of the electron?

Solution:

Rest mass energy of the electron, $E_0=m_0{c}^2$.

Given: Kinetic Energy,

$KE=E_0=m_0{c}^2$.

The formula for relativistic kinetic energy is $KE=(\gamma -1)m_0{c}^2$.

$m_0{c}^2=(\gamma -1)m_0{c}^2$

$1=\gamma -1⟹\gamma =2$

The Lorentz factor, $\gamma =\frac{1}{\sqrt{1-v^2/c^2}}$:

$$\begin{gathered} 2=\frac{1}{\sqrt{1-v^2/c^2}} \\ 4=\frac{1}{1-v^2/c^2} \\ 1-v^2/c^2=\frac{1}{4} \\ {v}^2/c^2=1-\frac{1}{4}=\frac{3}{4} \\ v=c\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}c \end{gathered}$$

The speed of the electron is​​ $\frac{\sqrt{3}}{2}$ times the speed of light, which is approximately 0.866c.

3. Calculate the energy released in MeV when a proton and an antiproton, both at rest, annihilate each other. 

(Given: rest mass of $proton{m}_p=1.672\times 10^{-27}kg)$

Solution: In annihilation, the total mass of both particles is converted into energy. Total mass converted,

$\Delta m=m_p+m_{\bar{p}}=2\times 1.672\times 10^{-27}\text{ kg}=3.344\times 10^{-27}\text{ kg}$

Energy Released, E = \Delta mc^2:

$$\begin{gathered} E=(3.344\times 10^{-27})\times (3\times 10^8{)}^2\text{ J} \\ E=3.0096\times 10^{-10}\text{ J} \end{gathered}$$

To convert to MeV, we use the conversion factor

: 

$$\begin{gathered} E=\frac{3.0096\times 10^{-10}}{1.6\times 10^{-19}}\text{ eV}=1.881\times 10^9\text{ eV}=1881\text{ MeV} \\ E=\frac{3.0096\times 10^{-10}}{1.6\times 10^{-19}}\text{ eV}=1.881\times 10^9\text{ eV}=1881\text{ MeV} \end{gathered}$$

4. A particle with rest mass m0​ has a total energy of $3m_0{c}^2$ Find the momentum of the particle.

Solution: We use the relativistic energy-momentum relation: 

$$\begin{gathered} E^2=(pc{)}^2+(m_0{c}^2{)}^2. \\ Given:E=3m_0{c}^2 \\ (3m_0{c}^2{)}^2=(pc{)}^2+(m_0{c}^2{)}^2 \\ 8(m_0{c}^2{)}^2=(pc{)}^2 \\ p=2\sqrt{2}{m}_{0}c. \end{gathered}$$

5. A particle's momentum is equal to its rest mass times the speed of light (p=m_0​c). Calculate the kinetic energy of the particle in terms of its rest mass energy.

Solution: Using the energy-momentum relation: 

$E^2=(pc{)}^2+(m_0{c}^2{)}^2$

$Substitutep=m_{0}c$

$E^2=(m_{0}c\cdot c{)}^2+(m_0{c}^2{)}^2=(m_0{c}^2{)}^2+(m_0{c}^2{)}^2$

$E^2=2(m_0{c}^2{)}^2$

$E=\sqrt{2}{m}_0{c}^2$

The total energy is

$E=KE+m_0{c}^2$

6. The Sun's power output is approximately $3.8\times 10^{26}W(J/s)$ How much mass does the Sun lose every year due to this energy emission? (Given: 1 year $\approx 3.15\times 10^{7}s$)

Solution: 

Energy emitted per second $(\frac{dE}{dt})$ is 3.8 \times $10^{26}$ J/s.

Total energy lost in one year ($\Delta E$) is:

$$\begin{gathered} \Delta E=(\text{Power})\times (\text{time})=(3.8\times 10^{26}\text{ J/s})\times (3.15\times 10^7\text{ s}) \\ \Delta E\approx 1.197\times 10^{34}\text{ J} \end{gathered}$$

Using the mass-energy relation, $\Delta m=\frac{\Delta E}{c^2}$

$$\begin{gathered} \Delta m=\frac{1.197\times 10^{34}\text{ J}}{(3\times 10^8\text{ m/s}{)}^2}=\frac{1.197\times 10^{34}}{9\times 10^{16}}\text{ kg} \\ \Delta m\approx 1.33\times 10^{17}\text{ kg} \end{gathered}$$

The Sun loses over 133 trillion kilograms of mass every year.