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Relativistic Energy & Mass-Energy Equivalence

The most famous equation in physics, $E=mc^2$, reveals that mass and energy are equivalent. Even a particle at rest has energy $m c^2$. Nuclear reactions convert tiny amounts of mass into enormous energy, and pair production converts pure energy into matter.

Key Concepts

Rest Energy
E0=mc2E_0=mc^2. A particle at rest has energy mc2mc^2 purely from its mass. For a proton: E0=938.3E_0=938.3 MeV; for an electron: E0=0.511E_0=0.511 MeV.
Total Energy
E=γmc2E=\gamma mc^2. This includes both rest energy and kinetic energy: E=E0+K=(mc2)+(γ1)mc2E=E_0+K=(mc^2)+(\gamma-1)mc^2.
Mass-Energy Equivalence
ΔE=Δmc2\Delta E=\Delta mc^2. Mass can be converted to energy and vice versa. In nuclear fission, ~0.1% of mass is converted; in fusion, ~0.7%.
Pair Production & Annihilation
A photon with E2mec2=1.022E\ge2m_ec^2=1.022 MeV can produce an electron-positron pair (in the field of a nucleus). Conversely, an electron and positron annihilate to produce two photons of 0.511 MeV each.

Key Equations

Mass-energy equivalence
E0=mc2E_0 = mc^2
Rest energy; mass and energy are interconvertible at rate c².
Total relativistic energy
E=γmc2=(pc)2+(mc2)2E = \gamma mc^2 = \sqrt{(pc)^2 + (mc^2)^2}
Includes rest energy and kinetic energy.
Binding energy
B=[Zmp+NmnM(Z,N)]c2B = [Zm_p + Nm_n - M(Z,N)]c^2
Energy released when Z protons and N neutrons form a nucleus of mass M.
Worked Example

Energy Released in Nuclear Fission

Problem

When U-235 undergoes fission, approximately 0.090% of its mass is converted to energy. A nuclear plant uses 1.0 kg of U-235 per day. How much energy (in J) is released?

Solution

Mass converted: Δm=0.00090×1.0 kg=9.0×104\Delta m=0.00090\times1.0\text{ kg}=9.0\times10^{-4} kg.

E=Δmc2=9.0×104×(3×108)2E=\Delta m\cdot c^2 = 9.0\times10^{-4}\times(3\times10^8)^2
E=9.0×104×9×1016=8.1×1013 JE = 9.0\times10^{-4}\times9\times10^{16} = 8.1\times10^{13}\text{ J}

That is 81 terajoules — enough to power a city for several days.

Answer E=8.1×1013E=8.1\times10^{13} J.
Practice

Exercises

7 problems
1 of 7

Find the rest energy E0E_0 (in J) of a 1.0 kg object. (c=3×108c=3\times10^8 m/s)

J
2 of 7

An electron (mec2=0.511m_ec^2=0.511 MeV) and a positron annihilate. Each photon produced has energy EγE_{\gamma} (in MeV). (Both at rest.)

MeV
3 of 7

Pair production threshold: minimum photon energy to create an electron-positron pair (in MeV). mec2=0.511m_ec^2=0.511 MeV.

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4 of 7

A proton (mpc2=938.3m_pc^2=938.3 MeV) is accelerated to γ=1000\gamma=1000. Find its total energy EE (in GeV).

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5 of 7

The Sun converts Δm=4.0×109\Delta m=4.0\times10^9 kg to energy per second. Find the power (in W).

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6 of 7

Deuterium fusion: 2H+2H3He+n^2H+^2H\to^3He+n. Mass difference Δm=3.3×1030\Delta m=3.3\times10^{-30} kg. Find energy released (in MeV). (11 MeV =1.602×1013=1.602\times10^{-13} J.)

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7 of 7

A particle has kinetic energy K=3mc2K=3m c^2. Find its total energy EE in units of mc2mc^2.

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Key Takeaways

  • Rest energy E0=mc2E_0=mc^2: mass is a form of energy, interconvertible at rate c2c^2.
  • Total energy E=γmc2E=\gamma mc^2; kinetic energy K=(γ1)mc2K=(\gamma-1)mc^2 vanishes at rest.
  • Nuclear reactions (fission ~0.1%, fusion ~0.7%) convert mass to energy; this is the source of stars' luminosity.
  • Pair production (γe++e\gamma\to e^++e^-) and annihilation (e++e2γe^++e^-\to 2\gamma) demonstrate direct mass-energy conversion.