The Special Theory of Relativity

Concept Of Physics

H C Verma

1   The guru of a yogi lives in a Himalyan cave, $1000\ km$ away from the house of the yogi. The yogi claims that whenever he thinks about his guru, the guru immediately knows about it. Calculate the minimum possible time interval between the yogi thinking about the guru and the guru knowing about it.

Solution :

$\\$S = 1000 km = $10^6$ m $\\$ The process requires minimum possible time if the velocity is maximum. $\\$ We know that maximum velocity can be that of light i.e. = $3 \times 10^8 m/s$. $\\$ So, time = $\frac{Distance}{Speed} = \frac{10^6}{3 \times 10^8} = \frac{1}{300}s.$

2   A suitcase kept on a shop's rack is measured $50\ cm \times 25\ cm \times 10\ cm$ by the shop's owner. A traveller takes this suitcase in a train moving with velocity $0\ 6c$. If the suitcase is placed with its length along the train's velocity, find the dimensions measured by (a) the traveller and (b) a ground observer.

Solution :

$\\$ $l = 50\ cm,\ b = 25\ cm,\ h = 10\ cm,\ v = 0.6\ C$ $\\$ a) The observer in the train notices the same value of $l,\ b,\ h$ because relativity are in due to difference in frames. $\\$ b) In $2$ different frames, the component of length parallel to the velocity undergoes contraction but the perpendicular components remain the same. So length which is parallel to the $x$-axis changes and breadth and height remain the same. $\\$ $e' = e\sqrt{1-\frac{V^2}{C^2}} = 50\sqrt{1-\frac{(0.6)^2 C^2}{C^2}}$ $\\$ = $50\sqrt{1-0.36} = 50 \times 0.8 = 40 cm.$ $\\$ The lengths observed are $40 cm \times 25 cm \times 10 cm.$

3   The length of a rod is exactly $1\ m$ when measured at rest. What will be its length when it moves at a speed of (a) $3 \times 10 ^5\ m/s$, (b) $3 \times 10^6\ n/s$ and (c) $3 x 10^7\ m/s\ ?$

Solution :

$\\$L = 1 m $\\$a) v $3 \times 10^5 \frac{m}{s}$ $\\$ L' = $1\sqrt{1-\frac{9 \times 10^{10}}{9\times 10^{10}}} = \sqrt{1-10^{-6}} = 0.9999995 m.$ $\\$v = $3 \times 10^6 \frac{m}{s}$ $\\$ L' = $1\sqrt{1-\frac{9 \times 10^{12}}{9\times 10^{10}}} = \sqrt{1-10^{-4}} = 0.99995 m.$ $\\$v = $3 \times 10^7 \frac{m}{s}$ $\\$ L' = $1\sqrt{1-\frac{9 \times 10^{14}}{9\times 10^{10}}} = \sqrt{1-10^{-2}} = 0.9949 m = 0.995 m.$

4   A person standing on a platform finds that a train moving with velocity 0'6c takes one second to pass by him. Find (a) the length of the train as seen by the person and (b) the rest length of the train.

Solution :

$\\$ $v = 0.6\ cm/sec ;\ t = 1\ sec$ $\\$ a) length observed by the observer = vt $\Rightarrow 0.6 \times 3 \times 10^6 \Rightarrow 1.8 \times 10^8 m.$ $\\$ b) l = $l_0 \sqrt{1-\frac{V^2}{C^2}} \Rightarrow 1.8 \times 10^8 = l_0\sqrt{1-\frac{(0.6)^2 C^2}{C^2}}$ $\\$ $l_0 = \frac{1.8 \times 10^8}{0.8} = 2.25 \times 10^8 \frac{m}{s}$.

5   An aeroplane travels over a rectangular field $100\ m \times 50\ m$, parallel to its length. What should be the speed of the plane so that the field becomes square in the plane frame ?

Solution :

$\\$The rectangular field appears to be a square when the length becomes equal to the breadth i.e. 50 m. i.e. $L' = 50 ;\ L = 100 ;\ v = ?$ $\\$ $C = 3 \times 10^8 m/s$ $\\$ We know, L' = $L\sqrt{1-\frac{V^2}{C^2}}$ $\\$ $\Rightarrow 50 = 100 \sqrt{1-\frac{V^2}{C^2}} \Rightarrow = \sqrt{\frac{3}{2C}} = 0.866 C.$

6   The rest distance between Patna and Delhi is $1000\ km$. A nonstop train travels at $360\ km/h$. (a) What is the distance between Patna and Delhi in the train frame ? (b) How much time elapses in the train frame between Patna and Delhi ?

Solution :

7   A person travels by a car at a speed of $180\ km/h$. It takes exactly $10\ hours$ by his wristwatch to go from the station $A$ to the station $B$. (a) What is the rest distance between the two stations ? (b) How much time is taken in the road frame by the car to go from the station $A$ to the station $B$ ?

Solution :

$L_0 = 1000\ km = 106\ m$ $\\$ $v = 360\ km/h = \dfrac{(360 \times 5)}{18} = 100\ m/sec.$ $\\$ a) $h' = h_0 \sqrt{1 - \dfrac{v^2}{c^2}} = 10^6 \sqrt{1 - \Big(\dfrac{100}{3 \times 10^8})^2} = 10^6 = \sqrt{1 - \dfrac{10^4}{9 \times 10^6}} = 10^9$ $\\$ Solving change in length $= 56\ nm.$ $\\$ b) $\Delta{t} = \dfrac{\Delta{L}}{v} = \dfrac{56\ nm}{100\ m} = 0.56\ ns.$

8   A person travels on a spaceship moving at a speed of $5c/13$. (a) Find the time interval calculated by him between the consecutive birthday celebrations of his friend on the earth, (b) Find the time interval calculated by the friend on the earth between the consecutive birthday celebrations of the traveller.

Solution :

$v = 180\ km/hr = 50\ m/s$ $\\$ $t = 10\ hours$ $\\$ let the rest dist. be $L.$ $\\$ $L' = L \sqrt{1 - \dfrac{v^2}{c^2}} \Rightarrow L' = 10 \times 180 = 1800\ k.m.$ $\\$ or, $1800 \times 1800 = L(1 – 36 \times 10^{–14})$ $\\$ or, $L = \dfrac{3.24 \times 10^6}{1 - 36 \times 10^{-14}} = 1800 + 25 \times 10^{–12}$ $\\$ or $25\ nm$ more than $1800\ km$. $\\$ b) Time taken in road frame by Car to cover the dist $= \dfrac{1.8 \times 10^6 + 25 \times 10^{-9}}{50}$ $\\$ $= 0.36 \times 10^5 + 5 \times 10^{–8} = 10\ hours + 0.5\ ns.$

9   According to the station clocks, two babies are born at the same instant, one in Howrah and other in Delhi, (a) Who is elder in the frame of $2301$ $Up$ Rajdhani Express going from Howrah to Delhi ? (b) Who is elder in the frame of $2302$ $Dn$ Rajdhani Express going from Delhi to Howrah.

Solution :

$u = \dfrac{5c}{13}$ $\\$ $\Delta{t} = \dfrac{t}{\sqrt{1 - \frac{v^2}{c^2}}} = \dfrac{1y}{\sqrt{1 - \dfrac{25c^2}{169c^2}}} = \dfrac{y \times 13}{12} = \dfrac{13}{12}y.$ $\\$ The time interval between the consecutive birthday celebration is $\dfrac{13}{12} y.$ $\\$ b) The fried on the earth also calculates the same speed.

10   Two babies are born in a moving train, one in the compartment adjacent to the engine and other in the compartment adjacent to the guard. According to the train frame, the babies are born at the same instant of time. Who is elder according to the ground frame ?

Solution :

The birth timings recorded by the station clocks is proper time interval because it is the ground frame. That of the train is improper as it records the time at two different places. The proper time interval $\Delta{T}$ is less than improper. i.e. $\Delta{T'} = v \Delta{T}$ $\\$ Hence for – (a) up train $\rightarrow$ Delhi baby is elder $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ (b) down train $\rightarrow$ Howrah baby is elder.

11   Suppose Swarglok (heaven) is in constant motion at a speed of $0\cdot9999c$ with respect to the earth. According to the earth's frame, how much time passes on the earth before one day passes on Swarglok ?

Solution :

The clocks of a moving frame are out of synchronization. The clock at the rear end leads the one at from by $L_0\ V/C^2$ where $L_0$ is the rest separation between the clocks, and $v$ is speed of the moving frame. Thus, the baby adjacent to the guard cell is elder.

12   If a person lives on the average $100$ years in his rest frame, how long does he live in the earth frame if he spends all his life on a spaceship going at $60$% of the speed of light.

Solution :

$\Delta{t'} = v \Delta{t}$ ; $\\$ Hence, $\Delta{t'} = 70.7$ days in heaven.

$v = 0.9999\ C ;\ \Delta{t} =$ One day in earth ; $\Delta{t}' =$ One day in heaven $\\$ $v = \dfrac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}} = \dfrac{1}{\sqrt{1 - \dfrac{(0.9999)^2C^2}{c^2}}} = \dfrac{1}{0.014141782} = 70.712$

13   An electric bulb, connected to a make and break power supply, switches off and on every second in its rest frame. What is the frequency of its switching off and on as seen from a spaceship travelling at a speed $0\cdot8c\ ?$

Solution :

$t = 100\ years ;\ V = \dfrac{60}{100}\ K ;\ C = 0.6\ C.$ $\\$ $\Delta{t} = \dfrac{t}{\sqrt{1 -\dfrac{v^2}{c^2}}} = \dfrac{100}{\sqrt{1 - \dfrac{(0.6)^2C^2}{C^2}}} = \dfrac{100y}{0.8} = 125\ y$

14   A person travelling by a car moving at $100\ km/h$ finds that his wristwatch agrees with the clock on a tower A. By what amount will his wristwatch lag or lead the clock on another tower $B,\ 1000\ km$ (in the earth's frame) from the tower A when the car reaches there ?

Solution :

We know $\\$ $f' = f \sqrt{1 - \dfrac{V2}{C2}}$ $\\$ $f' =$ apparent frequency ;$\\$ $f =$ frequency in rest frame $\\$ $v = 0.8\ C$ $\\$ $f = \sqrt{1 - \dfrac{0.64C^2}{C^2}} = \sqrt{0.36} = 0.6\ s^{-1}$

15   At what speed the volume of an object shrinks to half its rest value ?

Solution :

$\Delta{t'} – \Delta{t} = 10 \times 3600\ \Big[\dfrac{1}{1 - (\frac{1000}{36 \times 3 \times 10^8})}\Big]$ $\\$ By solving we get, $\Delta{t'} – \Delta{t} = 0.154\ ns$. $\\$ $\therefore$ Time will lag by $0.154\ ns.$

$V = 100\ km/h,\ \Delta{t} =$ Proper time interval $= 10\ hours$ $\\$ $\Delta{t'} = \dfrac{\Delta{t}}{\sqrt{1 - \dfrac{V^2}{C^2}}} = \dfrac{10 \times 3600}{\sqrt{1 - \Big(\dfrac{1000}{36 \times 3 \times 10^8}\Big)}}$

16   A particular particle created in a nuclear reactor leaves a $1\ cm$ track before decaying. Assuming that the particle moved at $0\cdot995c$, calculate the life of the particle (a) in the lab frame and (b) in the frame of the particle.

Solution :

Let the volume (initial) be $V.$ $\\$ $V' = \dfrac{V}{2}$ $\\$ So, $\dfrac{V}{2} = v \sqrt{1 - \dfrac{V^2}{C^2}}$ $\\$ $\Rightarrow \dfrac{V}{2} = \sqrt{C^2 - V^2} \Rightarrow \dfrac{C^2}{4} = C^2 – V^2$ $\\$ $\Rightarrow V^2 = C^2 - \dfrac{C^2}{4} = \dfrac{3}{4}C^2 \Rightarrow V = \dfrac{\sqrt{3}}{2}C$

17   By what fraction does the mass of a spring change when it is compressed by $1\ cm ?$ The mass of the spring is $200\ g$ at its natural length and the spring constant is $500\ N/m$

Solution :

$d = 1\ cm,\ v = 0.995\ C$ $\\$ a) time in Laboratory frame $= \dfrac{d}{v} = \dfrac{1 \times 10^{-2}}{0.995C}$ $\\$ $\dfrac{1 \times 10^{-2}}{0.995 \times 3 \times 10^8} = 33.5 \times 10^{–12} = 33.5\ PS$ $\\$ b) In the frame of the particle $\\$ $t' = \dfrac{t}{\sqrt{1 - \dfrac{V^2}{c^2}}} = \dfrac{33.5 \times 10^{-12}}{\sqrt{1 (0.995)^2}} = 335.41\ PS.$

18   Find the increase in mass when $1\ kg$ of water is heated from $0°C$ to $4100°C$. Specific heat capacity of water $= 4200\ JAg-K.$

Solution :

$x = 1\ cm = 1 \times 10^{–2} m ;\ K = 500\ N/m,\ m = 200\ g$ $\\$ Energy stored $= \dfrac{1}{2} Kx^2 = \dfrac{1}{2} \times 500 \times 10^{–4} = 0.025 J$ $\\$ Increase in mass $= \dfrac{0.025}{C^2} = \dfrac{0.025}{9 \times 10^{16}}$ $\\$ Fractional Change of max $= {0.025}{9 \times 10^{16}} \times \dfrac{1}{2 \times 10^{-1}} = 0.01388 \times 10^{–16} = 1.4 \times 10{–8}.$

19   Find the loss in the mass of $1$ mole of an ideal monatomic gas kept in a rigid container as it cools down by $10°C.$ The gas constant $R - 8\cdot3\ J/mol-K.$

Solution :

20   By what fraction does the mass of a boy increase when he starts running at a speed of $12\ km/h ?$

Solution :

Energy possessed by a monoatomic gas $= \dfrac{3}{2} nRdt.$ $\\$ Now $dT = 10,\ n = 1\ mole,\ R = 8.3\ J/mol-K$. $\\$ $E = \dfrac{3}{2} \times t \times 8.3 \times 10$ $\\$ Loss in mass $= \dfrac{1.5 \times 8.3 \times 10}{C^2} = \dfrac{124.5}{9 \times 10^{15}}$ $\\$ $= 1383 \times 10^{–16} = 1.38 \times 10^{–15}\ Kg.$

21   A $100\ W$ bulb together with its power supply is suspended from a sensitive balance. Find the change in the mass recorded after the bulb remains on for $1$ year.

Solution :

Let initial mass be $m$ $\\$ $\dfrac{1}{2} mv^2 = E$ $\\$ $\Rightarrow E = \dfrac{1}{2}m \Big(\dfrac{12 \times 5}{18}\Big)^2 = \dfrac{m \times 50}{9}$ $\\$ $\Delta{m} = {E}{C^2}$ $\\$ $\Rightarrow \Delta{m} = \dfrac{m \times 50}{9 \times 9 \times 10^{16}} \Rightarrow \dfrac{\Delta{m}}{m} = \dfrac{50}{81 \times 10^{16}}$ $\\$ $\Rightarrow 0.617 \times 10^{–16} = 6.17 \times 10^{–17}.$

22   The energy from the sun reaches just outside the earth's atmosphere at a rate of $1400\ W/m^2$. The distance between the sun and the earth is $1\cdot5 \times 10^{11}\ m$. (a) Calculate the rate at which the sun is losing its mass. (b) How long will the sun last assuming a constant decay at this rate ? The present mass of the sun is $2 \times 10^{30} kg.$

Solution :

Given : Bulb is $100\ Watt = 100\ J/s$ $\\$ So, $100\ J$ in expended per $1\ sec.$ $\\$ Hence total energy expended in $1\ year = 100 \times 3600 \times 24 \times 365 = 3153600000\ J$ $\\$ Change in mass recorded $= \dfrac{Total\ energy}{C^2} = \dfrac{315360000}{9 \times 10^{10}}$ $\\$ $= 3.504 \times 10^8 \times 10^{–16}\ kg = 3.5 \times 10^{–8} Kg.$

23   An electron and a positron moving at small speeds collide and annihilate each other. Find the energy of the resulting gamma photon.

Solution :

a) $\dfrac{E}{t} = \dfrac{\Delta{mC^2}}{t} = \dfrac{\Delta{m}}{t} = \dfrac{E/ t}{C^2}$ $\\$ $C^2 = \dfrac{1400 \times 4\pi \times 2.25 \times 10^{22}}{9 \times 10^{16}} = 1696 \times 10^{66} = 4.396 \times 10^9 = 4.4 \times 10^9$ $\\$ b) $4.4 \times 10^9\ Kg$ disintegrates in $1\ sec$. $\\$ $2 \times 10^{30} Kg$ disintegrates in $\dfrac{2 \times 10^{30}}{4.4 \times 10^9}\ sec$ $\\$ $= \Big(\dfrac{1 \times 10^{21}}{2.2 \times 365 \times 24 \times 3600}\Big) = 1.44 \times 10^{–8} \times 10^{21} y = 1.44 \times 10^{13}\ y.$

$I = 1400\ w/m^2$ $\\$ Power $= 1400\ w/m^2 \times A$ $\\$ $= (1400 \times 4\pi R^2)w = 1400 \times 4\pi \times (1.5 \times 10^{11})^2$ $\\$ $= 1400 \times 4\pi \times (1/5)^2 \times 10^{22}$ $\\$

24   Find the mass, the kinetic energy and the momentum of an electron moving at $0\cdot8e.$

Solution :

Mass of Electron = Mass of positron $= 9.1 \times 10^{–31}\ Kg$ $\\$ Both are oppositely charged and they annihilate each other. $\\$ Hence, $\Delta{m} = m + m = 2 \times 9.1 \times 10{–31}\ Kg$ $\\$ Energy of the resulting $\gamma$ particle $= \Delta{m}\ C^2$ $\\$ $= 2 \times 9.1 \times 10^{–31} \times 9 \times 10^{16}\ J = \dfrac{2 \times 9.1 \times 9 \times 10^{-15}}{1.6 \times 10^{19}}\ ev$ $\\$ $= 102.37 \times 10^4\ ev = 1.02 \times 10^6\ ev = 1.02\ Mev.$

25   Through what potential difference should an electron be accelerated to give it a speed of (a) $0\cdot6c$, (b) $0\cdot9c$ and (c) $0\cdot99c\ ?$

Solution :

$m_e = 9.1 \times 10^{–31},\ v_0 = 0.8\ C$ $\\$ a) $m' = \dfrac{M_e}{\sqrt{1 - V^2 /C^2}} = \dfrac{9.1 \times 10^{-31}}{\sqrt{1 - 0.64C^2 /C^2}} = \dfrac{9.1 \times 10^{-31}}{0.6}$ $\\$ $= 15.16 \times 10^{–31}\ Kg = 15.2 \times 10^{–31} Kg.$ $\\$ b) $K.E.$ of the electron $: m'C^2 – m_eC^2 = (m' – m_e) C^2$ $\\$ $= (15.2 \times 10^{–31} – 9.1 \times 10^{–31})(3 \times 10^8)^2 = (15.2 \times 9.1) \times 9 \times 10^{–31} \times 10^{18}\ J$ $\\$ $= 54.6 \times 10^{–15}\ J = 5.46 \times 10^{–14}\ J = 5.5 \times 10^{–14}\ J$. $\\$ c) Momentum of the given electron = Apparent mass $\times$ given velocity $\\$ $= 15.2 \times 10^{–31} – 0.8 \times 3 \times 10^{8}\ m/s = 36.48 \times 10{–23}\ kg\ m/s$ $\\$ $= 3.65 \times 10^{–22}\ kg\ m/s$

26   Find the speed of an electron with kinetic energy (a) $1\ eV,$ (b) $10\ keV$ and (c) $10\ MeV.$

Solution :

$= \dfrac{9.1 \times 9 \times 10^{-15}}{2 0.8} \Rightarrow eV\ –\ 9.1 \times 9 \times 10^{–15} = \dfrac{9.1 \times 9 \times 10^{-15}}{1.6}$ $\\$ $\Rightarrow eV = \Big(\dfrac{9.1 \times 9}{1.6} + 9.1 \times 9 \Big) \times 10^{-15} = eV \Big(\dfrac{81.9}{1.6} + 81.9\Big) \times 10^{-15}$ $\\$ $eV = 133.0875 \times 10^{–15} \times V = 83.179 \times 10^{4} = 831\ KV.$

$\Rightarrow eV = 372.18 \times 10^{–15} \Rightarrow V = \dfrac{372.18 \times 20^{-15}}{1.6 \times 10^{-19}} = 2.76 \times 10^4$ $\\$ $\Rightarrow V = 2.726 \times 10^6 = 2.7\ MeV.$ $\\$

a) $ev\ –\ m_0C^2 = \dfrac{m_0C^2}{2\sqrt{1 - \dfrac{V^2}{C^2}}} \Rightarrow ev – 9.1 \times 10^{–31} \times 9 \times 10^{16}$ $\\$ $= \dfrac{9.1 \times 9 \times 10^{-31} \times 10^{16}}{2\sqrt{1 - \dfrac{0.36C^2}{C^2}}} \Rightarrow eV – 9.1 \times 9 \times 10^{–15}$

$ev\ –\ m_0C^2 = \dfrac{m_0C^2}{2\sqrt{1 - \dfrac{V^2}{C^2}}} \Rightarrow eV – 9.1 \times 9 \times 10^{–19} times 9 \times 10^{16} = \dfrac{9.1 \times 9 \times 10^{-15}}{2\sqrt{1 - \dfrac{0.36C^2}{C^2}}}$ $\\$ $\Rightarrow eV – 81.9 \times 10^{–15} = \dfrac{9.1 \times 9 \times 10^{15}}{2 \times 0.435}$ $\\$ $\Rightarrow eV = 12.237 \times 10^{–15}$ $\\$ $\Rightarrow V = \dfrac{12.237 \times 10^{-15}}{1.6 \times 10^{-19}} = 76.48\ kV$ $\\$ $V = 0.99\ C = ev\ –\ m_0C^2 = \dfrac{m_0C^2}{2\sqrt{1 - \dfrac{V^2}{C^2}}}$ $\\$ $\Rightarrow eV = \dfrac{m_0C^2}{2\sqrt{1 - \dfrac{V^2}{C^2}}} + m_0C^2 = \dfrac{9.1 \times 10^{-31} \times 9 \times 10^{16}}{2\sqrt{1 -(0.99)^2}} + 9.1 \times 10^{-31} \times 9 \times 10^{16}$ $\\$

27   What is the kinetic energy of an electron in electronvolts with mass equal to double its rest mass ?

Solution :

$\Delta{m} = m – m_0 = 2m_0 – m_0 = m_0$ $\\$ Energy $E = m_0c^2 = 9.1 \times 10^{–31} \times 9 \times 10^{16}\ J$ $\\$ E in e.v. = $\dfrac{9.1 \times 9 \times 10^{-15}}{1.6 \times 10^{-19}} = 51.18 \times 10^4 ev\ = 511\ Kev.$

28   Find the speed at which the kinetic energy of a particle will differ by $1$% from its nonrelativistic value $\dfrac{1}{2}m_0\ v^2$

Solution :

Neglecting the $v^4$ term as it is very small $\\$ $\Rightarrow \dfrac{3}{4} \dfrac{V^2}{C^2} = 0.01 \Rightarrow \dfrac{V^2}{C^2} = \dfrac{0.04}{3}$ $\\$ $\Rightarrow \dfrac{V}{C} = \dfrac{0.2}{\sqrt{3}} = V = \dfrac{0.2}{\sqrt{3}}C = \dfrac{0.2}{1.732} \times 3 \times 10^8$ $\\$ $= 0.346 \times 10^8\ m/s = 3.46 \times 10^7\ m/s.$