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Chapter 4: Mass-Energy Equivalence

E = mc². The most famous equation in physics states that mass and energy are equivalent and interconvertible. A small amount of mass contains an enormous amount of energy.

E = mc²

\( E = mc^2 \)

Rest energy of mass m

The Numbers

1 kg of mass = (1)(3×10⁸)² = 9×10¹⁶ J = 25 billion kWh!
This is the energy of about 20 megatons of TNT.

Nuclear Applications

Nuclear Fission

Heavy nuclei (uranium, plutonium) split into lighter fragments. The mass difference becomes kinetic energy. About 0.1% of mass is converted. Powers nuclear reactors and atomic bombs.

Nuclear Fusion

Light nuclei (hydrogen) fuse into heavier ones (helium). About 0.7% of mass is converted. Powers the Sun and stars. Potentially clean energy via fusion reactors.

Matter-Antimatter Annihilation

When matter meets antimatter, 100% of mass converts to energy (photons). 1 kg matter + 1 kg antimatter → 1.8×10¹⁷ J (43 megatons TNT equivalent).

Binding Energy and Mass Defect

A bound system has less mass than its components! The "missing" mass is the binding energy:

\( \Delta m = \frac{E_b}{c^2} \)

• Hydrogen atom: mass 13.6 eV/c² less than proton + electron
• Helium-4: mass 28.3 MeV/c² less than 2 protons + 2 neutrons
• This "mass defect" is measurable and confirms E = mc²

Interactive Simulations

E=mc^2 Energy Calculator: Mass-Energy Conversions

Python

script.py73 lines

import numpy as np

c = 299792458.0  # m/s
c2 = c**2

print("=" * 70)
print("E = mc^2: MASS-ENERGY EQUIVALENCE CALCULATOR")
print("=" * 70)

# Convert various masses to energy
masses = [
    ("Electron", 9.109e-31),
    ("Proton", 1.673e-27),
    ("1 gram", 1.0e-3),
    ("1 kg", 1.0),
    ("Human (70 kg)", 70.0),
    ("Earth (5.97e24 kg)", 5.97e24),
]

print(f"{'Object':>20} {'Mass (kg)':>14} {'Energy (J)':>14} {'Energy (equiv)':>20}")
print("-" * 70)

for name, mass in masses:
    E = mass * c2
    if E < 1e6:
        equiv = f"{E:.4e} J"
    elif E < 1e9:
        equiv = f"{E/1e6:.2f} MJ"
    elif E < 1e15:
        equiv = f"{E/1e9:.2f} GJ"
    elif E < 1e18:
        equiv = f"{E/4.184e15:.2f} Mt TNT"
    else:
        equiv = f"{E/4.184e15:.2e} Mt TNT"
    print(f"{name:>20} {mass:14.4e} {E:14.4e} {equiv:>20}")

print("\n" + "=" * 70)
print("NUCLEAR REACTIONS: Mass Defect")
print("=" * 70)

# Hydrogen fusion: 4H -> He + energy
m_H = 1.00794  # atomic mass units
m_He = 4.00260
amu_to_kg = 1.6605e-27
amu_to_MeV = 931.494

delta_m = 4 * m_H - m_He  # mass defect in amu
E_released = delta_m * amu_to_MeV  # MeV

print(f"\nHydrogen fusion: 4 x 1H -> 4He + energy")
print(f"  4 x m_H  = {4*m_H:.5f} u")
print(f"  m_He     = {m_He:.5f} u")
print(f"  Mass defect = {delta_m:.5f} u")
print(f"  Energy released = {E_released:.3f} MeV")
print(f"  Efficiency = {delta_m/(4*m_H)*100:.4f}% of rest mass")

# Uranium fission
m_U235 = 235.0439
m_fission_products = 234.8604  # approximate
delta_m_U = m_U235 - m_fission_products
E_fission = delta_m_U * amu_to_MeV

print(f"\nUranium-235 fission:")
print(f"  Mass defect = {delta_m_U:.4f} u")
print(f"  Energy per fission = {E_fission:.1f} MeV")
print(f"  Efficiency = {delta_m_U/m_U235*100:.4f}%")

# Matter-antimatter annihilation
print(f"\nMatter-antimatter (electron-positron):")
print(f"  Energy = 2 x {0.511:.3f} MeV = {2*0.511:.3f} MeV")
print(f"  Efficiency = 100% (total mass converted)")

Click Run to execute the Python code

Code will be executed with Python 3 on the server

Fortran: E=mc^2 Energy Equivalence for Various Objects

Fortran

program.f9072 lines

program mass_energy
  implicit none
  double precision :: c, c2, mass, energy
  double precision :: m_H, m_He, delta_m, E_MeV
  double precision :: amu_to_MeV

c = 299792458.0d0
  c2 = c * c
  amu_to_MeV = 931.494d0

print '(A)', repeat('=', 65)
  print '(A)', 'E = mc^2 CALCULATOR'
  print '(A)', repeat('=', 65)

! Various mass-energy conversions
  print '(A20, A14, A16)', 'Object', 'Mass (kg)', 'Energy (J)'
  print '(A)', repeat('-', 52)

mass = 9.109d-31
  print '(A20, ES14.4, ES16.4)', 'Electron', mass, mass*c2
  mass = 1.673d-27
  print '(A20, ES14.4, ES16.4)', 'Proton', mass, mass*c2
  mass = 1.0d-3
  print '(A20, ES14.4, ES16.4)', '1 gram', mass, mass*c2
  mass = 1.0d0
  print '(A20, ES14.4, ES16.4)', '1 kilogram', mass, mass*c2
  mass = 70.0d0
  print '(A20, ES14.4, ES16.4)', 'Human (70 kg)', mass, mass*c2

print '(/,A)', 'Note: 1 kg = 8.988 x 10^16 J = 21.5 megatons TNT'

! Nuclear binding energy
  print '(/,A)', repeat('=', 65)
  print '(A)', 'NUCLEAR MASS DEFECT AND BINDING ENERGY'
  print '(A)', repeat('=', 65)

! Deuterium formation: p + n -> D + gamma
  block
    double precision :: m_p, m_n, m_d, dm, BE
    m_p = 1.007276d0  ! proton mass (u)
    m_n = 1.008665d0  ! neutron mass (u)
    m_d = 2.014102d0  ! deuteron mass (u)

dm = m_p + m_n - m_d
    BE = dm * amu_to_MeV

print '(A)', 'Deuterium: p + n -> D'
    print '(A,F10.6,A)', '  m_p + m_n = ', m_p + m_n, ' u'
    print '(A,F10.6,A)', '  m_D       = ', m_d, ' u'
    print '(A,F10.6,A)', '  Delta m   = ', dm, ' u'
    print '(A,F10.4,A)', '  Binding E = ', BE, ' MeV'
  end block

! Helium-4 binding
  block
    double precision :: m_4p, m_4He, dm4, BE4
    m_4p = 2.0d0 * 1.007276d0 + 2.0d0 * 1.008665d0
    m_4He = 4.002602d0
    dm4 = m_4p - m_4He
    BE4 = dm4 * amu_to_MeV

print '(/,A)', 'Helium-4: 2p + 2n -> He-4'
    print '(A,F10.6,A)', '  Sum parts = ', m_4p, ' u'
    print '(A,F10.6,A)', '  m_He4     = ', m_4He, ' u'
    print '(A,F10.6,A)', '  Delta m   = ', dm4, ' u'
    print '(A,F10.4,A)', '  Binding E = ', BE4, ' MeV'
    print '(A,F10.4,A)', '  Per nucleon= ', BE4/4.0d0, ' MeV'
  end block

end program mass_energy

Click Run to execute the Fortran code

Code will be compiled with gfortran and executed on the server

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