This lesson is beyond the Class 12 syllabus. Its one result that the syllabus does use is \(E = mc^2\), which the chapter on the nucleus needs for binding energy.
Momentum that stays conserved
The chapter on momentum showed that the total momentum of an isolated system is conserved. By the first postulate, this law must hold in every inertial frame. But clocks and rulers change from frame to frame, and so do measured velocities. Einstein found that \(m_0v\) is then conserved in one frame and not in another. The quantity that stays conserved in every frame is \(\gamma m_0 v\).
Here \(m_0\) is the rest mass, the mass measured by an observer at rest with the body. It is the same in every frame. Some books call \(\gamma m_0\) the relativistic mass and say the mass grows with speed. This book keeps one mass, \(m_0\), and lets \(\gamma\) sit in the formulas.
An electron of rest mass \(9.1 \times 10^{-31}\,\text{kg}\) moves at \(0.8c\). Find its relativistic momentum and compare it with \(m_0v\).
The speed is \(v = 0.8 \times 3 \times 10^8 = 2.4 \times 10^8\,\text{m/s}\), and \(\gamma = 5/3\).
Newton's value is \(m_0v = 9.1 \times 10^{-31} \times 2.4 \times 10^8 = 2.18 \times 10^{-22}\,\text{kg m/s}\).
The relativistic momentum is \(p = \gamma m_0v = \tfrac53 \times 2.18 \times 10^{-22} = 3.64 \times 10^{-22}\,\text{kg m/s}\), larger by the factor \(\gamma = 5/3\).
Values used: \(c = 3 \times 10^8\,\text{m/s}\), \(m_e = 9.1 \times 10^{-31}\,\text{kg}\), \(e = 1.6 \times 10^{-19}\,\text{C}\), \(1\,\text{u} = 931.5\,\text{MeV}/c^2\).
Energy at rest
Einstein found a matching change for energy. The total energy of a free body moving at speed \(v\) is \(E = \gamma m_0c^2\). At rest, \(\gamma = 1\), yet the energy is not zero. It is \(E_0 = m_0c^2\), the rest energy. Mass itself is a store of energy, and a large one, because \(c^2\) is huge. This is mass-energy equivalence.
The rest energy of an electron is \(m_0c^2 = 9.1 \times 10^{-31} \times (3 \times 10^8)^2 = 8.19 \times 10^{-14}\,\text{J}\). Divided by \(1.6 \times 10^{-19}\,\text{J}\) per electron volt, that is \(0.51\,\text{MeV}\). For nuclei the unit is the atomic mass unit, and \(1\,\text{u}\) of mass holds \(931.5\,\text{MeV}\). The chapter on the nucleus uses this to find how much energy holds a nucleus together.
One gram of matter is turned entirely into energy. How much energy is released, in units of \(10^7\) kilowatt hours? Take \(1\,\text{kWh} = 3.6 \times 10^6\,\text{J}\).
Kinetic energy, old and new
The kinetic energy is what the motion adds to the rest energy, \(K = (\gamma - 1)m_0c^2\). At low speed this must agree with Newton's \(\tfrac12 m_0v^2\), and it does. For small \(v/c\), \(\gamma \approx 1 + \tfrac12 v^2/c^2\), so \((\gamma - 1)m_0c^2 \approx \tfrac12 m_0v^2\). At high speed the two part company.
The two curves lie together up to about \(0.3c\). At \(0.6c\) Einstein's value is \(0.25\,m_0c^2\), while Newton's is \(0.18\,m_0c^2\). At \(0.8c\) they are \(0.67\) and \(0.32\). Near \(c\) the solid curve rises without limit, while the dashed one would reach only \(0.5\,m_0c^2\) at \(v = c\). Newton's formula undercounts the energy of a fast body.
Take the rest energy of an electron as \(0.5\,\text{MeV}\). The electron moves at \(0.6c\). Find its total energy and its kinetic energy, and compare the kinetic energy with Newton's value.
At \(0.6c\), \(\gamma = 1/\sqrt{1 - 0.36} = 1.25\).
\(E = \gamma m_0c^2\), \(K = (\gamma - 1)m_0c^2\), and Newton's \(\tfrac12 m_0v^2 = \tfrac12 (v/c)^2\,m_0c^2\).
\(E = 1.25 \times 0.5 = 0.625\,\text{MeV}\). \(K = 0.625 - 0.5 = 0.125\,\text{MeV}\). Newton: \(\tfrac12 \times 0.36 \times 0.5 = 0.09\,\text{MeV}\).
The electron has \(0.625\,\text{MeV}\) in all, of which \(0.125\,\text{MeV}\) is kinetic. Newton's formula gives only \(0.09\,\text{MeV}\), more than a quarter too little.
The rest mass of an electron is \(9.1 \times 10^{-31}\,\text{kg}\). Find its rest energy in MeV. Take \(1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}\).
One relation for energy and momentum
Momentum and energy both carry a \(\gamma\). Eliminating \(\gamma\) gives a relation between them that holds at every speed.
\[ E^2 - (pc)^2 = \gamma^2 m_0^2c^4 - \gamma^2 m_0^2 v^2 c^2 \]
Why. square \(E = \gamma m_0c^2\) and \(pc = \gamma m_0vc\), and subtract.
\[ E^2 - (pc)^2 = \gamma^2 m_0^2 c^4 \left(1 - v^2/c^2\right) \]
Why. take out the common factor \(\gamma^2 m_0^2c^4\).
\[ E^2 - (pc)^2 = m_0^2c^4 \]
Why. \(\gamma^2 = 1/(1 - v^2/c^2)\), so the bracket cancels \(\gamma^2\).
\[ E^2 = (pc)^2 + (m_0c^2)^2 \]
Read it. rearrange. The right side no longer contains the speed.
The energy-momentum relation is Pythagoras' theorem in disguise. Draw a right triangle with the rest energy along the base and \(pc\) up the side. The slanting side is the total energy \(E\). Fig 5.1 shows the triangle for an electron at three speeds, and for a photon.
The photon, a particle with no rest mass
A photon has zero rest mass. Then the triangle has no base, and \(E = pc\), so \(p = E/c\). This is the photon momentum of the chapter on photons, now derived. Dividing \(p = \gamma m_0v\) by \(E = \gamma m_0c^2\) gives \(v = pc^2/E\) for any particle. For a photon, \(E = pc\) makes \(v = c\). A particle with no rest mass always moves at \(c\), and a particle with rest mass never can.
A particle has rest energy \(m_0c^2 = 3\,\text{MeV}\) and momentum such that \(pc = 4\,\text{MeV}\). Find its total energy, its kinetic energy and its speed.
\(E^2 = 4^2 + 3^2 = 25\), so \(E = 5\,\text{MeV}\).
\(K = E - m_0c^2 = 5 - 3 = 2\,\text{MeV}\).
\(\gamma = E/m_0c^2 = 5/3\), and \(v/c = \sqrt{1 - 1/\gamma^2} = \sqrt{1 - 9/25} = 0.8\). Check: \(v/c = pc/E = 4/5\).
A particle has rest energy \(6\,\text{MeV}\) and momentum such that \(pc = 8\,\text{MeV}\). Find its kinetic energy, in MeV.
A photon has an energy of \(2\,\text{eV}\). Find its momentum, in units of \(10^{-27}\,\text{kg m/s}\). Take \(1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}\).
Why can no body with rest mass be accelerated to the speed of light?
The Sun shines with a power of about \(3.8 \times 10^{26}\,\text{W}\). By \(E = mc^2\), that energy leaves as mass: \(3.8 \times 10^{26} / 9 \times 10^{16}\), about four billion kilograms every second. The fusion of hydrogen into helium in the Sun's core turns this much mass into sunlight. The same equivalence gives the energy of the two reactors at Kudankulam, where the splitting of uranium nuclei turns a little mass into heat.
- Using \(\tfrac12 m_0v^2\) for a particle near the speed of light.Use \(K = (\gamma - 1)m_0c^2\); Newton's formula is good only while \(v \ll c\).
- Forgetting the rest energy when finding the kinetic energy from \(E\).\(K = E - m_0c^2\). The total energy includes the rest energy.
- Using \(E = pc\) for a particle with rest mass.\(E = pc\) holds only when \(m_0 = 0\). Otherwise use \(E^2 = (pc)^2 + (m_0c^2)^2\).