The Workbook

Light of energy \(E\) carries momentum \(E/c\). A beam of power \(P\) delivers momentum \(P/c\) each second.

1. Absorbed: \(F = \dfrac{P}{c} = \dfrac{6}{3 \times 10^{8}} = 2 \times 10^{-8}\) N.

2. Reflected: the momentum reverses, so the change is \(2P/c\) per second. \(F = 4 \times 10^{-8}\) N.

1. \(p = \dfrac{h}{\lambda} = \dfrac{6.6 \times 10^{-34}}{660 \times 10^{-9}} = 10^{-27}\) kg m/s.

2. The photon goes back with the same momentum. Change \(= p - (-p) = 2p = 2 \times 10^{-27}\) kg m/s.

3. Force \(=\) momentum change per second \(= 10^{20} \times 2 \times 10^{-27} = 2 \times 10^{-7}\) N.

1. Momentum per second \(= \dfrac{P}{c} = \dfrac{3}{3 \times 10^{8}} = 10^{-8}\) N. The beam is fully absorbed, so the plate receives all this momentum. The force acts along the beam.

2. Normal component: \(F\cos 60^\circ = 10^{-8} \times \dfrac{1}{2} = 5 \times 10^{-9}\) N.

   Along the plate: \(F\sin 60^\circ = 10^{-8} \times \dfrac{\sqrt{3}}{2} = 5\sqrt{3} \times 10^{-9} \approx 8.7 \times 10^{-9}\) N.

On reflection the normal component of the light's momentum reverses. The component along the mirror does not change.

Force \(= 2 \times \dfrac{P}{c}\cos 60^\circ = 2 \times 10^{-8} \times \dfrac{1}{2} = 10^{-8}\) N, along the normal, into the mirror.

The tangential momentum of the light is the same before and after, so no tangential momentum is given to the mirror.

Absorbed part (\(40\%\)): force \(= 0.4\dfrac{P}{c}\). Reflected part (\(60\%\)): force \(= 2 \times 0.6\dfrac{P}{c} = 1.2\dfrac{P}{c}\).

Total \(F = 1.6\dfrac{P}{c} = 1.6 \times 10^{-8}\) N. In general \(F = (1 + r)\dfrac{P}{c}\) with \(r\) the fraction reflected.

1. A beam of cross-section \(A_0\) that meets the surface at angle \(\theta\) covers an area \(A = A_0/\cos\theta\). So a surface of area \(A\) intercepts power \(IA\cos\theta\). This is the first \(\cos\theta\).

   The momentum per second is \(\dfrac{IA\cos\theta}{c}\) along the beam. Its normal component has another \(\cos\theta\): \(F_n = \dfrac{IA\cos^2\theta}{c}\).

   Pressure \(= F_n/A = \dfrac{I}{c}\cos^2\theta\).

2. \(\dfrac{300}{3 \times 10^{8}} \times \dfrac{1}{4} = 2.5 \times 10^{-7}\) Pa.

\(\dfrac{P}{c} = \dfrac{6}{3 \times 10^{8}} = 2 \times 10^{-8}\) N. Each half of the beam carries \(10^{-8}\) N of momentum per second.

1. Absorbed half: normal component \(10^{-8}\cos 60^\circ = 5 \times 10^{-9}\) N.

   Reflected half: normal momentum reverses, so \(2 \times 10^{-8}\cos 60^\circ = 10^{-8}\) N.

2. Total normal force \(= 5 \times 10^{-9} + 10^{-8} = 1.5 \times 10^{-8}\) N.

   Only the absorbed half gives a tangential force: \(10^{-8}\sin 60^\circ = 5\sqrt{3} \times 10^{-9} \approx 8.7 \times 10^{-9}\) N.

1. \(\lambda = \dfrac{h}{mv} = \dfrac{6.6 \times 10^{-34}}{0.2 \times 33} = \dfrac{6.6 \times 10^{-34}}{6.6} = 10^{-34}\) m.

2. Wave effects such as diffraction need an opening of about the size of \(\lambda\). Nothing is \(10^{-34}\) m across; the wavelength is far smaller than any object or gap. So the ball behaves as a particle.

1. \(\lambda = 12.27/\sqrt{100} = 1.227 \approx 1.23\) Å.

2. \(\lambda = 12.27/\sqrt{10^4} = 12.27/100 = 0.123\) Å.

3. \(\lambda \propto 1/\sqrt{V}\). To halve \(\lambda\), \(\sqrt{V}\) must double, so \(V\) must become four times.

1. \(K = \dfrac{p^2}{2m}\), so \(p = \sqrt{2mK}\). Then \(\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2mK}}\).

2. The work done by the field is \(eV\), so \(K = eV\). Put this in (i): \(\lambda = \dfrac{h}{\sqrt{2meV}}\).

3. \(2me = 2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} = 2.91 \times 10^{-49}\). \(\sqrt{2me} = 5.4 \times 10^{-25}\).

   \(\dfrac{h}{\sqrt{2me}} = \dfrac{6.63 \times 10^{-34}}{5.4 \times 10^{-25}} = 1.23 \times 10^{-9}\) m V\(^{1/2}\) \(= 12.3\) Å V\(^{1/2}\). With more exact constants this is \(12.27\).

For an electron \(K = eV\) means \(V\) in volt equals \(K\) in eV. So \(\lambda = \dfrac{12.27}{\sqrt{K}}\) Å with \(K\) in eV.

1. \(\lambda = 12.27/\sqrt{150} = 12.27/12.25 \approx 1.0\) Å.

2. \(\lambda = 12.27/\sqrt{600} = 12.27/24.5 \approx 0.5\) Å. Four times the energy gives half the wavelength.

1. Same \(v\): \(\lambda = \dfrac{h}{mv} \propto \dfrac{1}{m}\). \(\lambda_e : \lambda_p = m_p : m_e = 1836 : 1\).

2. Same \(K\): \(\lambda = \dfrac{h}{\sqrt{2mK}} \propto \dfrac{1}{\sqrt{m}}\). \(\lambda_e : \lambda_p = \sqrt{1836} : 1 \approx 43 : 1\).

3. Same \(p\): \(\lambda = h/p\) is the same for both. \(1 : 1\).

1. \(E = \dfrac{hc}{\lambda} = \dfrac{12400}{1} = 12400\) eV \(= 12.4\) keV.

2. \(\lambda = \dfrac{12.27}{\sqrt{K}}\) Å, so \(K = \left(\dfrac{12.27}{1}\right)^2 \approx 150\) eV.

3. The photon carries about \(12400/150 \approx 80\) times more energy. For the same wavelength the momenta are equal, but the photon's energy is \(pc\) while the electron's is only \(p^2/2m\).

1. \(K = eV = 400\) eV \(= 400 \times 1.6 \times 10^{-19} = 6.4 \times 10^{-17}\) J.

2. \(v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2 \times 6.4 \times 10^{-17}}{9.1 \times 10^{-31}}} = \sqrt{1.41 \times 10^{14}} \approx 1.2 \times 10^{7}\) m/s.

3. \(\lambda = \dfrac{12.27}{\sqrt{400}} = \dfrac{12.27}{20} \approx 0.61\) Å.

1. The number of electrons scattered per second into the detector is measured. It is plotted against the scattering angle, for several accelerating voltages.

2. At \(54\) V the count showed a strong peak at one angle, \(50^\circ\). Particles bouncing off a surface would not pick out one angle. Waves diffracted by the regular rows of atoms do. So electrons behave as waves, with the wavelength de Broglie predicted.

1. \(\lambda = \dfrac{12.27}{\sqrt{54}} = \dfrac{12.27}{7.35} \approx 1.67\) Å.

2. \(\lambda = 2d\sin\theta = 2 \times 0.91 \times 0.906 \approx 1.65\) Å.

3. The wavelength found from the diffraction pattern matches the wavelength predicted by \(\lambda = h/p\). This confirms de Broglie's formula for electrons.

1. The peak is at \(\phi = 50^\circ\).

2. \(\theta = 90^\circ - 25^\circ = 65^\circ\).

3. A higher voltage gives a higher momentum and a smaller \(\lambda\). From \(2d\sin\theta = \lambda\), a smaller \(\lambda\) needs a smaller glancing angle \(\theta\). Since \(\phi = 180^\circ - 2\theta\), the peak moves to a larger \(\phi\).

1. \(\lambda = 12.27/\sqrt{150} \approx 1.0\) Å. \(\sin\theta = \dfrac{\lambda}{2d} = \dfrac{1.0}{2.0} = 0.5\), so \(\theta = 30^\circ\).

2. \(\lambda = 12.27/\sqrt{600} \approx 0.5\) Å. \(\sin\theta = \dfrac{0.5}{2.0} = 0.25\), so \(\theta \approx 14.5^\circ\).

1. \(V\) fixes the kinetic energy \(eV\) of the electrons, and so their momentum and their de Broglie wavelength \(\lambda = 12.27/\sqrt{V}\) Å.

2. The detector counts the electrons scattered per second in its direction. It moves along a circular arc around the crystal, so the angle \(\phi\) can be varied.

3. Air molecules would scatter and absorb the slow electrons before they reached the crystal or the detector.

4. Waves scattered from regular rows of atoms add up only in directions where the path difference is a whole number of wavelengths. Only a wave gives such a sharp maximum at one angle.

1. Power on the sail \(P = IA = 1500 \times 100 = 1.5 \times 10^{5}\) W. Perfect reflection gives \(F = \dfrac{2P}{c} = \dfrac{3 \times 10^{5}}{3 \times 10^{8}} = 10^{-3}\) N.

2. \(a = \dfrac{F}{m} = \dfrac{10^{-3}}{100} = 10^{-5}\) m/s\(^2\).

3. \(v = at = 10^{-5} \times 86400 = 0.864\) m/s. Small, but it acts day after day without any fuel.

1. \(p = \dfrac{h}{\lambda}\) for both: \(p = \dfrac{6.6 \times 10^{-34}}{10^{-9}} = 6.6 \times 10^{-25}\) kg m/s.

2. Photon: \(E = \dfrac{hc}{\lambda} = \dfrac{1240}{1} = 1240\) eV.

3. Electron: \(1\) nm \(= 10\) Å, so \(\sqrt{K} = \dfrac{12.27}{10}\) and \(K = 1.227^2 \approx 1.5\) eV. The same momentum, but the electron carries far less energy.

An inertial frame is one in which Newton's first law holds: a body with no force on it keeps a constant velocity. Any frame moving at constant velocity relative to an inertial frame is also inertial.

1. Steady speed on a straight track: no acceleration. Inertial.

2. Braking means acceleration. Not inertial. A loose object slides forward with no force on it.

3. Rotation means acceleration towards the centre. Not inertial.

First postulate (principle of relativity): the laws of physics are the same in every inertial frame. No experiment inside a closed box can tell whether the box is at rest or moving uniformly.

Second postulate (constancy of \(c\)): the speed of light in vacuum is the same, \(c\), for every observer in every inertial frame. It does not depend on the motion of the source or of the observer.

1. A measures \(c\). In A's frame the lamp is at rest and light always moves at \(c\).

2. B also measures \(c\), not \(c + v\). This is the second postulate: the speed of light does not depend on the motion of the source.

3. For slow objects the ordinary rule works: \(20 + 10 = 30\) m/s. (The exact relativistic answer differs from \(30\) m/s by less than \(10^{-13}\) m/s.)

The relativistic effects depend on \(v^2/c^2\). For everyday speeds, even \(300\) m/s, this ratio is about \(10^{-12}\). The correction to "speeds add" is far too small to measure with ordinary instruments. Only for speeds near \(c\), or for light itself, does the difference show.

1. \(\gamma = \dfrac{1}{\sqrt{1 - 0.36}} = \dfrac{1}{\sqrt{0.64}} = \dfrac{1}{0.8} = 1.25 = \dfrac{5}{4}\).

2. \(\gamma = \dfrac{1}{\sqrt{1 - 0.64}} = \dfrac{1}{0.6} = \dfrac{5}{3} \approx 1.67\).

3. \(\gamma = \dfrac{1}{\sqrt{1 - 0.9216}} = \dfrac{1}{\sqrt{0.0784}} = \dfrac{1}{0.28} = \dfrac{25}{7} \approx 3.6\).

From \(\gamma = \dfrac{1}{\sqrt{1 - v^2/c^2}}\): \(1 - \dfrac{v^2}{c^2} = \dfrac{1}{\gamma^2}\), so \(\dfrac{v}{c} = \sqrt{1 - \dfrac{1}{\gamma^2}}\).

1. \(\gamma = 1.25\): \(\dfrac{v}{c} = \sqrt{1 - 0.64} = 0.6\).

2. \(\gamma = 2\): \(\dfrac{v}{c} = \sqrt{1 - \dfrac{1}{4}} = \dfrac{\sqrt{3}}{2} \approx 0.87\).

1. The dotted lines give \(\gamma = 1.25\) at \(0.6c\) and \(\gamma \approx 1.67\) at \(0.8c\).

2. The curve reaches \(\gamma = 2\) at about \(v/c = 0.87\).

3. As \(v \to c\), \(1 - v^2/c^2 \to 0\), so \(\gamma = 1/\sqrt{1 - v^2/c^2} \to \infty\). The dashed line at \(v = c\) is never reached.

4. For \(v < 0.2c\), \(\gamma < 1.021\). Time dilation and length contraction are then under \(2\%\). For cars, planes and rockets (\(v/c < 10^{-4}\)) they are completely negligible, and Newton's mechanics is exact enough.

1. \(\gamma = 25/7\) at \(0.96c\). \(\Delta t = \gamma\,\Delta t_0 = \dfrac{25}{7} \times 2.2 \approx 7.9\ \mu\)s.

2. Distance \(= v\,\Delta t = 0.96 \times 3 \times 10^{8} \times 7.86 \times 10^{-6} \approx 2263\) m \(\approx 2.26\) km.

3. Without dilation: \(0.96 \times 3 \times 10^{8} \times 2.2 \times 10^{-6} \approx 634\) m. Most muons would decay before reaching the ground. In fact many do reach it, which confirms time dilation.

The ship's clock measures the proper time \(\Delta t_0 = 1\) h between two ticks at the same place on the ship.

Earth measures \(\Delta t = \gamma\,\Delta t_0 = 1.25 \times 1\) h \(= 1.25\) h \(= 75\) min. Moving clocks run slow.

The astronaut measures the proper time between beats: \(\Delta t_0 = 1\) s. Earth measures \(\Delta t = \gamma\,\Delta t_0 = \dfrac{5}{3}\) s per beat.

Beats per minute \(= \dfrac{60}{5/3} = 36\).

1. Longer. \(\Delta t = \gamma\,\Delta t_0\) and \(\gamma \ge 1\). The proper time is the shortest time any observer can measure between the two events.

2. In the muon's own rest frame the creation and the decay happen at the same place (where the muon is). So the muon's frame measures the proper time, \(2.2\ \mu\)s.

1. The flash covers the distance \(d\) at speed \(c\): \(\Delta t_0 = \dfrac{d}{c}\).

2. In the moving frame the clock shifts by \(v\,\Delta t\) during one tick. The flash travels the slant distance \(c\,\Delta t\) (second postulate: still speed \(c\)).

   Pythagoras: \((c\,\Delta t)^2 = d^2 + (v\,\Delta t)^2\), so \(\Delta t^2 (c^2 - v^2) = d^2\).

   \(\Delta t = \dfrac{d}{\sqrt{c^2 - v^2}} = \dfrac{d/c}{\sqrt{1 - v^2/c^2}} = \gamma\,\Delta t_0\).

\(L = \dfrac{L_0}{\gamma}\). At \(0.8c\), \(\gamma = \dfrac{5}{3}\), so \(L = 5 \times \dfrac{3}{5} = 3\) m.

1. \(\gamma = 1.25\) at \(0.6c\). Length \(= \dfrac{100}{1.25} = 80\) m. Width \(= 20\) m, unchanged.

2. Only lengths along the direction of motion contract. The width is perpendicular to the motion, so both frames agree on it.

1. A. The proper length is measured in the frame in which the object is at rest. The car is at rest for A.

2. \(L = \dfrac{L_0}{\gamma} = \dfrac{60}{1.25} = 48\) m.

3. For A, the ground and B's stick move at \(0.6c\). A measures the stick as \(\dfrac{1}{1.25} = 0.8\) m. Each observer sees the other's lengths contracted.

1. In the muon's frame the atmosphere rushes past at \(0.96c\), so the \(2263\) m is contracted: \(\dfrac{2263}{25/7} = 2263 \times \dfrac{7}{25} \approx 634\) m.

2. Time to cover \(634\) m at \(0.96c\): \(\dfrac{634}{0.96 \times 3 \times 10^{8}} \approx 2.2\ \mu\)s, exactly the muon's lifetime. Both frames agree that the muon reaches the ground; one calls it time dilation, the other length contraction.

\(m_0c^2 = 9.1 \times 10^{-31} \times (3 \times 10^{8})^2 = 9.1 \times 10^{-31} \times 9 \times 10^{16} = 8.19 \times 10^{-14}\) J.

In eV: \(\dfrac{8.19 \times 10^{-14}}{1.6 \times 10^{-19}} = 5.1 \times 10^{5}\) eV \(= 0.51\) MeV.

\(E = mc^2 = 10^{-3} \times 9 \times 10^{16} = 9 \times 10^{13}\) J.

In kWh: \(\dfrac{9 \times 10^{13}}{3.6 \times 10^{6}} = 2.5 \times 10^{7}\) kWh, the yearly electricity of thousands of homes.

1. \(\gamma = \dfrac{5}{3}\) at \(0.8c\). \(p = \gamma m_0 v = \dfrac{5}{3} \times 9.1 \times 10^{-31} \times 2.4 \times 10^{8} = 3.64 \times 10^{-22}\) kg m/s.

2. \(m_0 v = 9.1 \times 10^{-31} \times 2.4 \times 10^{8} = 2.18 \times 10^{-22}\) kg m/s. The ratio is \(\gamma = \dfrac{5}{3}\).

1. \(\gamma = 1.25\). \(E = 1.25 \times 0.5 = 0.625\) MeV.

2. \(K = E - m_0c^2 = 0.625 - 0.5 = 0.125\) MeV.

3. Classical: \(\dfrac{1}{2}m_0 v^2 = \dfrac{1}{2}(0.6)^2 m_0c^2 = 0.18\,m_0c^2 = 0.09\) MeV. The true kinetic energy, \(0.25\,m_0c^2\), is larger. The classical formula undercounts at high speed.

1. \(E^2 = 4^2 + 3^2 = 25\), so \(E = 5\) MeV.

2. \(K = E - m_0c^2 = 5 - 3 = 2\) MeV.

3. \(E = \gamma m_0c^2\), so \(\gamma = \dfrac{5}{3}\). From Question 5, \(\gamma = 5/3\) means \(v = 0.8c\). Check: \(pc = \gamma m_0 v c = \dfrac{5}{3} \times 3 \times 0.8 = 4\) MeV.

1. P. The relativistic kinetic energy grows faster than \(v^2\) and becomes infinite at \(v = c\). The classical curve is a simple parabola that stays finite.

2. At \(0.8c\): P gives \((\gamma - 1)mc^2 = \left(\dfrac{5}{3} - 1\right)mc^2 = \dfrac{2}{3}mc^2 \approx 0.67\,mc^2\). Q gives \(\dfrac{1}{2}(0.8)^2 mc^2 = 0.32\,mc^2\).

3. \(K = (\gamma - 1)mc^2\) and \(\gamma \to \infty\) as \(v \to c\). It would take infinite energy to reach \(c\).

Put \(m_0 = 0\): \(E^2 = (pc)^2\), so \(E = pc\) and \(p = \dfrac{E}{c}\).

\(E = 2\) eV \(= 3.2 \times 10^{-19}\) J. \(p = \dfrac{3.2 \times 10^{-19}}{3 \times 10^{8}} \approx 1.07 \times 10^{-27}\) kg m/s. This agrees with \(p = h/\lambda\) for a \(620\) nm photon.

As \(v \to c\), \(\gamma = 1/\sqrt{1 - v^2/c^2} \to \infty\). The momentum \(p = \gamma m_0 v\) and the energy \(E = \gamma m_0 c^2\) both grow without limit.

To reach \(v = c\) a body with \(m_0 \ne 0\) would need infinite energy, which no source can supply. Each extra push adds less speed and more energy. Only massless particles, such as photons, move at \(c\).