Introduction to Special Relativity

Abstract

The relativity principle, first formulated by Galileo, states that the laws of Nature are the same in all inertial frames.

Problems

1.1

A spaceship of length $L_0 = 150\,\mathrm{m}$ is moving with respect to a space station with a speed $v = 2\,{\times }\,10^8\,\mathrm{m/s}$. What is the length L of the spaceship as measured by the space station?

Answer: $\quad L=L_0\sqrt{1-{v^2/c^2}}\simeq 112\ \mathrm{m}\, .$

1.2

How many years does it take for an atomic clock (with a precision of one part over $10^{15}$), which is placed at rest on Earth, to lose one second with respect to an identical clock placed on the Sun?

Answer: We can take the Sun as an inertial frame, in which Earth moves on an approximately circular orbit at constant speed $v \simeq 3{\times }10^4\ \mathrm{m/s} \simeq 10^{-4}\ c$. We must find a proper time interval on Earth, $\varDelta \tau$, such that it is one second less with respect to Sun time. Adopting Eq. (1.36) with v constant and setting $\delta t = 1\, \mathrm{s}$, we have $(\gamma (v) - 1) \varDelta \tau = \delta t$, hence

$$\varDelta \tau = \frac{\delta t}{1/\sqrt{1-{v^2/ c^2}} - 1 } \simeq 2\, \delta t\, {c^2 \over v^2}\, \simeq 6.34\,\mathrm{years}.$$

Our computation is actually incomplete, since it neglects gravitational field effects: their analysis would require a treatment in the framework of general relativity.

1.3

Two spaceships, moving along the same course with the same velocity $v = 0.98\ c$, pass space station Alpha, which is placed on their course, at the same hour of two successive days. On each of the two spaceships a radar permits to know the distance from the other spaceship: what value does it measure?

Answer: In the reference frame of space station Alpha, the two spaceships stay at the two ends of a segment of length $L = v T$, where $T = 1\ \mathrm{day}$. That distance is reduced by a factor $\sqrt{1 - v^2/c^2}$ with respect to the distance $L_0$ among the spaceships as measured in their rest frame. We have therefore

$$L_0 = \frac{1}{\sqrt{1 - v^2/c^2}} v T \simeq 1.28{\times }10^{14}\,\mathrm{m}.$$

1.4

Spaceship A is moving with respect to space station S with a velocity $2.7{\times }10^8 \,\mathrm{m/s}$. Both A and S are placed in the origin of their respective reference frames, which are oriented so that the relative velocity of A is directed along the positive direction of both x axes; A and S meet at time $t_A = t_S = 0$. Space station S detects an event, corresponding to the emission of luminous pulse, in $x_S = 3{\times }10^{13}\,\mathrm{m}$ at time $t_S = 0$. An analogous but distinct event is detected by spaceship A, with coordinates $x_A=1.3{\times }10^{14}\,\mathrm{m}\ ,\ t_A=2.3{\times }10^3\,\mathrm{s}$. Is it possible that the two events have been produced by the same moving body?

Answer: The two events may have been produced by the same moving body only if they have a time-like distance, otherwise the unknown body would go faster than light. After obtaining the coordinates of the two events in the same reference frame, one obtains, e.g. in the spaceship frame, $\varDelta x \simeq 6.09{\times }10^{13}\ \mathrm{m}$ and $c \varDelta t \simeq 6.27{\times }10^{13}\ \mathrm{m} > \varDelta x$: the two events may indeed have been produced by the same body, moving at an average speed $\varDelta x / \varDelta t \simeq 0.97\ c$.

1.5

Fizeau’s Experiment

In the experiment described in the figure, a light beam of frequency $\nu = 10^{15}\,\mathrm{Hz}$, produced by the source S, is split into two distinct beams which go along two different paths belonging to a rectangle of sides $L_1 = 10\,\mathrm{m}$ and $L_2 = 5\,\mathrm{m}$. They recombine, producing interference in the observation point O, as illustrated in the figure. The rectangular path is contained in a tube T filled with a liquid having refraction index $n = 2$, so that the speed of light in that liquid is $v_c \simeq 1.5{\times }10^{8}\,\mathrm{m/s}$ . If the liquid is moving counter-clockwise around the tube with a velocity $0.3\,\mathrm{m/s}$, the speed of the light beams along the two different paths changes, together with their wavelength, which is constrained by the equation $v_c= \lambda \nu$ (the frequency $\nu$ instead does not change and is equal to that of the original beam). For that reason the two beams recombine in O with a phase difference $\varDelta \phi$, which is different from zero (the phase accumulated by each beam is given by $2 \pi$ times the number of wavelengths contained in the total path). What is the value of $\varDelta \phi$? Compare the result with what would have been obtained using Galilean transformation laws .

Answer: Calling $L= L_1 + L_2 = 15\ \mathrm{m}$ the total path length inside the tube for each beam, and using Einstein laws for adding velocities, one finds

$$\quad \varDelta \phi =4\pi \nu L v \frac{n^2-1}{c^2 - n^2v^2}\simeq 4\pi \nu L v(n^2-1)/c^2 \simeq 1.89\ \mathrm{rad}\, .$$

Instead, Galilean laws would lead to

$$\quad \varDelta \phi =4\pi \nu L v \frac{n^2}{c^2 - n^2v^2}\, ,$$

a result which does not make sense, since it is different from zero also when the tube is empty: in that case, indeed, Galilean laws would imply that the tube is still filled with a “rotating ether”.

The experiment described above is very similar to the one performed by Fizeau. The only significant difference is that Fizeau made the two light beams going exactly along the same path, even if in two opposite directions, in order to eliminate any possible systematic effect related to different conditions (e.g., in temperature) in the various pieces of the tube. Fizeau observed a non-zero effect for water, and a null effect for air, for which $n \simeq 1$: this is in agreement with the relativistic prediction, but was misinterpreted at the time of Fizeau and associated with the partial drag hypothesis by Fresnel.

1.6

Relativistic Aberration of Light

In the reference frame of the Sun, Earth moves with a velocity of modulus $v = 10^{-4}\ c$ which forms, at a given time, an angle $\theta = 60^\circ$ with respect to the position of a given star. Compute the variation of this angle when it is measured by a telescope placed on Earth.

Answer: By applying the relativistic transformation laws for velocities to the photons coming from the star, one easily obtains that

$$ \tan \theta ' = \sqrt{1 - \beta ^2} \frac{\sin \theta }{\cos \theta + \beta } \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \mathrm{or\,\, equivalently} \,\,\,\,\,\,\,\,\,\,\,\, \cos \theta ' = \frac{\cos \theta + \beta }{1 + \beta \cos \theta }\, $$

to be compared with the classical expression $\tan \theta ' = \sin \theta /(\cos \theta + \beta )$. In the relativistic case all angles, apart from $\theta = \pi$, get transformed into $\theta ' = 0$ as $\beta \rightarrow 1$.

In the present case $\beta = v/c = 10^{-4}$ and it is sensible to expand in Taylor series obtaining, up to second order:

$$\theta ' = \theta - \beta \sin \theta \ + \beta ^2 \sin \theta \cos \theta + O(\beta ^3) $$

to be compared with result obtained by Galilean transformation laws:

$$\theta ' = \theta - \beta \sin \theta \ + 2 \beta ^2 \sin \theta \cos \theta + O(\beta ^3) \, .$$

It is interesting to notice that relativistic effects show up only at the second order in $\beta$. In the given case $\delta \theta = - 17.86216''$ to be compared with $\delta \theta =-17.86127''$ for the classical computation. Relativistic effects are tiny in this case and not appreciable by a usual optical telescope; an astronomical interferometer, which is able to reach resolutions of the order of few micro-arcseconds at radio wavelengths, should be used instead.

1.7

A particle is moving with a speed of modulus v and components $(v_x,v_y,v_z)$. What is the modulus $v'$ of the velocity for an observer moving at a speed w along the x axis? Comment the result as v and/or w approach c.

Answer: Applying the relativistic laws for the addition of velocities we find

$$v'^2 = v'^2_x + v'^2_y + v'^2_z = \frac{(v_x - w)^2 + (1 - w^2/c^2) (v_z^2 + v_y^2)}{(1 - v_x w /c^2)^2}$$

and after some simple algebra

$$ v'^2 = c^2 \left( 1 - \frac{(1 - v^2/c^2) (1 - w^2/c^2)}{(1 - v_x w /c^2)^2} \right) \, . $$

It is interesting to notice that as v and/or w approach c, also $v'$ approaches c (from below): that verifies that a body moving with $v = c$ moves with the same velocity in every reference frame (invariance of the speed of light).

1.8

During a Star Wars episode, space station Alpha detects an enemy spaceship approaching it from a distance $d = 10^8\ \mathrm{km}$ at a speed $v = 0.9\ c$; at the same time the station launches a missile of speed $v' = 0.95\ c$ to destroy it. As soon as the enemy spaceship detects the electromagnetic pulses emitted by the missile, it launches against space station Alpha the same kind of missile, therefore moving at a speed $0.95\ c$ in the rest frame of the spaceship. How much time do the inhabitants of space station Alpha have, after having launched their missile, to leave the station before it is destroyed by the second missile?

Answer: Let us make computations in the reference frame of the space station. Setting to zero the launching time of the first missile, the enemy spaceship detects it and launches the second missile at time $t_1 = d/(v + c)$ and when it is at a distance $x_1 = c d/(v + c)$ from the space station. The second missile approaches Alpha with a velocity $V=(v+v')/(1+vv'/c^2)\simeq 0.9973\ c$, therefore it hits the space station at time $t_2 = t_1 + x_1/V \simeq 352\ \mathrm{s}.$

1.9

Two particles move at the same speed v along two orthogonal axes of the laboratory, x and y. Assuming that the rest frames of the two particles are connected to the laboratory frame by a pure boost, i.e. that their axes are parallel to those of the laboratory, compute the components and the modulus of the relative velocities of the two particles, i.e. the velocity of each particle in the rest frame of the other. Are the two relative velocities opposite to each other? Are their moduli equal? Do you have any explanation?

Answer: In the laboratory we have $\varvec{{v}}_1 = (v,0)$ and $\varvec{{v}}_2 = (0,v)$. Let us call $O'$ and $O''$ the rest frames of particles 1 and 2, respectively, and $\varvec{{v}}_2'$ ($\varvec{{v}}_1''$) the speed of particle 2 (1) in the rest frame of particle 1 (2). Then, setting as usual $\beta = v/c$ and $\gamma = 1/\sqrt{1 - v^2/c^2}$, a simple application of the transformation law for velocities shows that $\varvec{{v}}_2' = (-v , v/\gamma )$ and $\varvec{{v}}_1'' = (v\gamma , -v)$. While the modulus is the same, $|\varvec{{v}}_2'| = |\varvec{{v}}_1''| = c \sqrt{\beta ^2 (2 - \beta ^2)}$, one has that, contrary to naïve expectations, $\varvec{{v}}_2' \ne - \varvec{{v}}_1''$. The reason is easy to understand: while the axes of the two particle frames are both parallel to the laboratory frame, they are not parallel to each other: in order to go from reference frame $O'$ to reference frame $O''$ we need the combination of two boosts along two different axes, x and y. As the reader can verify by composing the corresponding matrices, that is not a pure boost, but involves a rotation around the z axis, therefore the discrepancy is only due to a mutual rotation of the frame axes, i.e. the actual velocities are in fact opposite to each other. Such a rotation is a purely relativistic effect which leads to new phenomena, like Thomas precession.

1.10

Two particles proceed with four-velocities $u^\mu$ and $w^\mu$ in the laboratory frame. What is the relative speed $v_r$ of the two particles?

Answer: The relative speed is defined as the velocity of one particle measured in the rest frame of the other, hence it is clear that it must be a Lorentz invariant quantity (it is, in some sense, a private fact of the two particles). Therefore we must look for an invariant quantity to be built in terms of $u^\mu$ and $w^\mu$: the only non-trivial invariant combination of these two quantities is $u^\mu w_\mu$, since we have $u^\mu {u}_\mu = w^\mu {w}_\mu = c^2$. Since we are looking for an invariant quantity, we can work in any frame we prefer, for instance in the reference frame of particle 1, where, setting $\beta _r = v_r/c$ and $\gamma _r = 1/\sqrt{1 - \beta _r^2}$, we have

$$u = (c,0); \ \ \ \ \ w = (\gamma _r c, \gamma _r \varvec{{v}}_r); \ \ \ \ \ u^\mu {w}_\mu = c^2 \gamma _r.$$

We deduce

$$v_r = c \beta _r = c \sqrt{1 - {1 \over \gamma _r^2}} = c \sqrt{1 - {c^2 \over u^\mu {w}_\mu }}.$$

Last expression is explicitly Lorentz invariant and can be used to compute the relative velocity in any frame.

1.11

We are on the course of a spaceship moving with a constant speed v while emitting electromagnetic pulses which, in the rest frame of the spaceship, are equally spaced in time. We receive a pulse every second while the spaceship is approaching us, and a pulse every two seconds while the spaceship is leaving us. What is the speed of the spaceship?

Answer: Let us call $\varDelta \tau$ the time spacing of pulse emissions in the rest frame of the spaceship. By time dilation, such spacing becomes $\varDelta t = \gamma \varDelta \tau$ in our frame. While the spacehip is approaching us, pulses travelling towards us are at a distance of $\varDelta t (c - v)$ from each other: indeed, each pulse and the spaceship travel respectively $c \varDelta t$ and $v \varDelta t$, in the same direction, before the next pulse is emitted; on the contrary, when the spaceship is leaving, pulses towards us travel in the direction opposite to the spaceship and their mutual distance is $\varDelta t (c + v)$. Therefore the arrival time intervals are respectively

$$\varDelta t (c \mp v)/c = \gamma (1 \mp \beta ) \varDelta \tau = \sqrt{1 \mp \beta \over 1 \pm \beta } \varDelta \tau , $$

in agreement with the longitudinal Doppler effect. Combining the two effects, we deduce that

$$\frac{1 + v/c}{1 - v/c} = 2 \,$$

hence $v = 1/3\ c$.

1.12

A spacecraft is moving towards a plane mirror at rest with a constant velocity $v=5\,{\times }\,10^{7}\,\mathrm{m/s}$, which is orthogonal to the mirror. The spacecraft casts a laser pulse of frequency $\nu =10^{15}\,\mathrm{Hz}$, which is reflected by the mirror and comes back with a new frequency $\nu '$. What is the value of $\nu '$?

Answer: The pulse proceeds parallel to the spacecraft velocity both before and after reflection. Since reflection leaves the frequency unchanged only in the rest frame of the mirror, two different longitudinal Doppler effects have to be taken into account, that of the original pulse with respect to the mirror and that the reflected pulse with respect to the spacecraft, therefore $\nu '=\nu \ (1+{v/c})/(1-{v/c}) = 1.4\,{\times }\,10^{15}\,\mathrm{Hz}\, .$

1.13

Consider a problem similar to the one above, with the difference that now the spacecraft is moving parallel to the mirror surface. What is the value of $\nu '$ in this case?

Answer: In this case, if we require that the pulse hits the mirror surface and catches the spacecraft on its way back, we obtain, assuming standard reflection laws, that the pulse makes an angle with respect to the spacecraft velocity, which in the mirror frame is $\pm \theta$ respectively before and after reflection, with $\cos \theta = v/c$ (the value of $\theta$ is actually irrelevant in the following).

Let us put $\omega = 2 \pi \, \nu$, $\omega ' = 2 \pi \, \nu '$, and let us call $\tilde{\omega }$ the angular frequency of the pulse as observed in the mirror frame. We can apply equation (1.107) twice to transform frequencies, however, since the angle $\theta$ is already known in the mirror frame, the easiest way is to apply the transformation starting in both cases from the mirror frame. We then obtain:

$$ \omega = \gamma (v) \tilde{\omega }( 1 - \beta \cos \theta ) $$

$$ \omega ' = \gamma (v) \tilde{\omega }( 1 - \beta \cos \theta ) $$

which immediately gives $\omega ' = \omega$, i.e. the frequency the spacecraft gets back is unmodified in this case.

1.14

Spacecraft A passes very close to an astrophysical object S, which emits electromagnetic radiation of frequency $\nu = 10^{14}$ Hz in all directions, $\nu$ being measured in S rest frame. In our inertial rest frame, both A and S move at constant velocity v along two orthogonal axes in the xy plane, their world lines being respectively $\varvec{{x}}_A (t) = (vt,0)$ and $\varvec{{x}}_S (t) = (0,vt + d)$, with $v = c/5$ and $d = 10^3$ km.

What is, according to spacecraft A, the minimum distance reached from object S? What is the frequency spacecraft A receives when it sees S at such closest distance?

Answer: Let us take S world line, expressed in our reference frame, $\varvec{{x}}_S (t) = (0,vt + d)$, and transform it to reference frame A by a standard Lorentz transformation along the x axis

$$ t' = \gamma t; \ \ \ x_S' = -\gamma v t; \ \ \ y_S' = vt + d $$

where $\gamma = 1/\sqrt{1 - v^2/c^2}$. We could re-express everything in terms of $t'$, to find the world line of S in A, however this is not strictly necessary, since we just need to compute the point of the world line for which

$$ x_S'^2 + y_S'^2 = \gamma ^2 v^2 t^2 + (vt + d)^2 = v^2 t^2 (1 + \gamma ^2) + 2 v d t + d^2\, $$

is minimum. To that aim it is completely irrelevant whether the world line is parametrized in terms of t or $t'$. The condition for the minimum gives:

$$2 t v^2 (1 + \gamma ^2) + 2vd = 0 \ \ \rightarrow \ \ \ t_{min} = - \frac{d}{v (1 + \gamma ^2)}$$

from which we find

$$d_{min}^2 = \frac{d^2}{1 + \gamma ^2} - 2 \frac{d^2}{1 + \gamma ^2} + d^2 = \frac{d^2}{2 - v^2/c^2} = \frac{25 d^2}{49}$$

hence $d_{min} = 5 d /7 \simeq 714$ km. It is interesting to compare with the minimum distance as observed from our reference frame, which is $d/\sqrt{2} \simeq 707$ km.

In A frame, the photon leaving S when it is at minimum distance and reaching A travels orthogonal to the direction of S, hence we have to apply the trasverse Doppler effect law. However we have to be careful about the exact transformation we need to do: the photon is transverse in frame A, hence we have to transform from frame A to frame S, i.e. we can write

$$ \nu = \gamma ' \nu ' $$

where $\gamma ' \equiv 1/\sqrt{1 - v_S'^2}$ and $v_S'$ is S velocity as seen by A. Applying the relativistic law for the transformation of velocities we obtain $\varvec{{v}}_S' = (-v, v/\gamma )$ hence

$$ v_S' = \sqrt{v^2 + (1 - v^2/c^2) v^2} = \sqrt{\frac{1}{25} \left( 2 - \frac{1}{25} \right) }\, c = \frac{7}{25}\, c ; \ \ \ \gamma ' = 25/24$$

and

$$ \nu ' = \nu /\gamma ' = \frac{24 \nu }{25} \simeq 0.96 \times 10^{14}\ \mathrm{Hz} \, .$$

1.15

A particle moves in one dimension with an acceleration which is constant and equal to a in the reference frame instantaneously at rest with it. Determine its world line x(t) in the reference frame where it is placed at rest in $x = 0$ at time $t = 0$.

Answer: One can make use either of the inertial frame time t or of the particle proper time $\tau$ to parametrize the world line, which are related by $d \tau = \sqrt{1 - v^2(t)/c^2}\ dt$ and will be assumed to be synchronized so that $t = 0$ when $\tau = 0$. There are various way to solve this problem, let us discuss two of them. First, to derive the equation of motion, let us notice that, in the frame instantaneously at rest with the particle, the velocity goes from 0 to $a d \tau$ in the interval $d \tau$, so that, by relativistic addition of velocities, v changes into $(v + a d \tau )/(1 + v a d \tau /c^2)$ in the same interval, meaning that ${dv / d \tau } = a ( 1 - {v^2 / c^2} )$. We can now make use of the spatial component of the four-velocity, $u = dx/d\tau$, which is related to v by the following relations:

$$u = \frac{v}{\sqrt{1 - v^2/c^2}} \, , \,\,\,\,\,\,\,\,\,\,\,\, v = {u \over \sqrt{1+{u^2/c^2}} }\, , \,\,\,\,\,\,\,\,\,\,\,\, \sqrt{1+{u^2\over c^2}} \sqrt{1-{v^2\over c^2}} = 1\, . $$

Using previous equations, it is easy to derive

$$ {d u\over d t} = {d u \over d v}\ {d v \over d \tau }\ {d \tau \over d t} = a $$

which can immediately be integrated, setting $u(0) = 0$, as $u(t) = a t$. Using the relation between u and v we have

$$v(t) = {a t \over \sqrt{1 + {a^2 t^2 / c^2}}}. $$

That gives the velocity, as observed in the laboratory, for a uniformly accelerated motion: for $t \ll c/a$ one recovers the non-relativistic result, while for $t \gg c/a$ the velocity reaches asymptotically that of light. The dependence of v on t can be finally integrated, using the initial condition $x(0) = 0$, giving

$$x(t) = {c^2 \over a} \left( \sqrt{1 + {a^2 t^2 \over c^2}} - 1 \right) . $$

This is usually known as hyperbolic motion, since the world line defines an hyperbola in the xt plane:

$$\left( \frac{ax}{c^2} + 1\right) ^2 - \frac{a^2t^2}{c^2} = 1 \, .$$

A simpler way to solve the same problem makes use of the rapidity $\chi$, which is defined by $\beta = \tanh \chi$. We have seen that, contrary to velocity, rapidity is an additive quantity for collinear boosts. By virtue of that, the infinitesimal change of rapidity in the rest frame, $d \chi = a d \tau /c$, goes unchanged to the laboratory frame, i.e. we can write, in the laboratory:

$$ {d \chi \over d \tau } = a/c \ \ \ \rightarrow \ \ \ \chi (\tau ) = a \tau /c$$

leading immediately to $\beta (\tau ) = \tanh (a \tau /c)$ and $\gamma (\tau ) = \cosh (a \tau /c)$. Then $dt = \gamma d \tau$ can be immediately integrated to obtain $t = (c/a) \sinh (a \tau /c)$ and finally

$$\beta = \tanh (a \tau /c) = \frac{\sinh (a \tau /c)}{\cosh (a \tau /c)} = {a t/c \over \sqrt{1 + {a^2 t^2 / c^2}}}. $$

1.16

Two identical spacecrafts of mass M and initially at rest at a relative distance L, start moving at the same time $t = 0$ under the influence of a constant force F acting along their relative distance, i.e. such that $dp/dt = F$ where p is the relativistic spatial momentum. The two spacecrafts are initially connected by a thin and inextensible cable. Determine the world lines of both spacecrafts and say whether the cable breaks or not.

Answer: Regarding the motion of each spacecraft, this problem is identical to the previous one, even if this is not obvious at a first sight. The equation $dp/dt = F$ can be immediately integrated yielding $p = F t$ ($p = 0$ at $t = 0$), hence $E = \sqrt{F^2 c^2 t^2 + M^2 c^4}$, from which we obtain

$$ v(t) = \frac{p c^2}{E} = \frac{F c^2 t}{\sqrt{F^2 c^2 t^2 + M^2 c^4}} = \frac{a t}{\sqrt{1 + (a/c)^2 t^2}} $$

where $a \equiv F/M$, i.e. exactly the same solution of Problem 1.15. Another way to look at the correspondence between the two problems is to consider how the quantity dp/dt transforms, in the case of one-dimensional motions, for longitudinal boosts. Going to a new inertial frame $O'$, moving with velocity V with respect to the original one and setting $\tilde{\gamma }= 1/\sqrt{1 - V^2/c^2}$, we have

$$dt' = \tilde{\gamma }\left( dt - {V \over c^2} dx \right) = \tilde{\gamma }dt \left( 1 - {V v \over c^2} \right) $$

$$dp' = \tilde{\gamma }\left( dp - {V \over c^2} dE \right) = \tilde{\gamma }dp \left( 1 - {V v \over c^2} \right) $$

where in the last equation we have used the relation

$$dE = d(\sqrt{p^2 c^2 + M^2 c^4}) = {p c^2 \over E} dp = v dp.$$

It is now evident that $dp'/dt' = dp/dt$, i.e. such a quantity is invariant under any longitudinal boost: that includes the boost bringing to the frame istantaneously at rest with the spacecraft, hence a constant dp / dt implies a constant acceleration in the aircraft rest frame, which is exactly Problem 1.15.

Let us now address the cable issue. From Problem 1.15 we obtain the world lines of the two spacecrafts

$$ x_1(t)= {c^2 \over a} \left( \sqrt{1 + {a^2 t^2 \over c^2}} - 1 \right) ; \ \ \ \ \ \ \ \ x_2(t)= L + {c^2 \over a} \left( \sqrt{1 + {a^2 t^2 \over c^2}} - 1 \right) \, $$

where integration constants have been set such that $x_1 (0) = 0$ and $x_1 (0) = L$. In the original frame, the two spacecrafts keep at distance L at every time: one would deduce that the cable does not break, however, since the cable is moving as well, we must compute its length in the aircraft rest frame. Let us consider, in particular, an inertial frame $O'$ which is istantaneously at rest with aircraft 1, and let us fix its origin so that $x_1' = t_1' = 0$ corresponds to the event in which aircraft 1 reaches speed v in the original frame O; for convenience, from now on we translate the origin of O so that also in this frame $x_1 = t_1 = 0$, while $x_2 = L$ and $t_2 = 0$ corresponds to the event for which also spacecraft 2 reaches the same speed v. After a Lorentz transformation, we obtain that, according to $O'$, spacecraft 2 is at rest in $x_2' = \gamma L$ at $t_2' = - \beta \gamma L$. In order to understand where spacecraft 2 is at time $t' = 0$, thus obtaining the relative distance as seen from aircraft 1, we need to integrate again the equation of the hyperbolic motion, which tells us that, in a time interval $\beta \gamma L$ and starting from rest, aircraft 2 will travel a distance

$$\frac{c}{a}\sqrt{c^{2}+a^{2}(\gamma \beta L)^{2}} - \frac{c^2}{a}$$

so that the total distance will be

$$\gamma L + \frac{c}{a} \left( \sqrt{c^2 + (a \gamma \beta L)^2} - 1 \right) $$

which is surely greater than L, hence the cable will break.

1.17

A muon, which is a particle (usually indicated by the Greek letter $\mu$) of mass $m=1.89\,{\times }\,10^{-28}\ \mathrm{kg}$ and carrying the same electric charge as the electron, has a mean life time $\tau = 2.2\,{\times }\,10^{-6}\ \mathrm{s}$ when it is at rest. The particle is accelerated instantaneously through a potential gap $\varDelta V = 10^{8}\ \mathrm{V}$. What is the expected life time t of the particle in the laboratory after the acceleration? What is the expected distance D traveled by the particle before decaying?

Answer:

$$\quad t=\tau \ {(mc^2+e \varDelta V) \over mc^2}\simeq 4.28\,{\times }\,10^{-6}\ \mathrm{s}\, $$

$$D = {\sqrt{(e \varDelta V)^2 + 2 e m c^2 \varDelta V} \over mc^2} c \tau = 1.1\,{\times }\,10^3\ \mathrm{m}.$$

Notice that the average traveled distance D is larger than what expected in absence of time dilation, which is limited to $c \tau$ due to the finiteness of the speed of light. The increase in the traveled distance for relativistic unstable particles is one of the best experimental proofs of time dilation, think e.g. of the muons created when cosmic rays collide with the upper regions of the atmosphere: a large fraction of them reaches Earth’s surface and that is possible only since their life times appear dilated in Earth’s frame.

1.18

The energy of a particle is equal to $2.5\,{\times }\,10^{-12}\ \mathrm{J}$, its momentum is $7.9\,{\times }\,10^{-21}\ \mathrm{N\ s}$. What are its mass m and velocity v?

Answer: $m={\sqrt{E^2-c^2p^2} / c^2}\simeq 8.9\,{\times }\,10^{-30}\ \mathrm{kg}, v = {pc^2}/{E} \simeq 2.84\,{\times }\,10^8 \,\mathrm{m/s}.$

1.19

A particle of mass $M= 10^{-27}\ \mathrm{kg}$ decays, while at rest, into a particle of mass $m= 4\,{\times }\,10^{-28}\ \mathrm{kg}$ plus a photon. What is the energy E of the photon?

Answer: The two outgoing particles must have opposite momenta with an equal modulus p to conserve total momentum. Energy conservation is then written as $Mc^2 = \sqrt{m^2 c^4 + p^2 c^2} + pc$, so that

$$E = pc = \frac{M^2 - m^2}{2M} c^2 = 0.42\ Mc^2 \simeq 3.78\,{\times }\,10^{-11}\ \mathrm{J} \simeq 2.36\,{\times }\,10^8\ \mathrm{eV},$$

while the energy of the massive particle is $c^2 (M^2 + m^2) / (2M)$.

1.20

Consider a system composed of two photons of momenta $\varvec{{k}}_1$ and $\varvec{{k}}_2$. Determine its invariant mass $M_{inv}$ and the velocity $\varvec{{v}}_{cm}$ of its center of mass frame.

Answer: Since $E_1 = k_1\, c$ ed $E_2 = k_2\, c$, where $k_1 = |\varvec{{k}}_1|$ and $k_2 = |\varvec{{k}}_2|$, we have

$$ M_{inv}^2 c^2= (k_1 + k_2)^2 - |\varvec{{k}}_1 + \varvec{{k}}_2|^2 = 2 k_1 k_2 (1 - \cos \theta ) $$

where $\theta$ is the angle formed by the directions of the two photons. Moreover

$$ \varvec{{v}}_{CM} = \frac{\varvec{{k}}_1 + \varvec{{k}}_2}{k_1 + k_2}\, c; \ \ \ v_{CM} = c\, \sqrt{1 - (1 - \cos \theta ) \frac{ 2 k_1 k_2}{(k_1 + k_2)^2} }. $$

Therefore only in the case of collinear photons ($\theta = 0$) we have a vanishing invariant mass and $v_{cm} = c$.

1.21

A spaceship of initial mass $M = 10^{4}\,\mathrm{kg}$ is boosted by a photonic engine: a light beam is emitted opposite to the direction of motion, with a power $W=10^{13}\,\mathrm{W}$, as measured in the spaceship rest frame. What is the derivative of the spaceship rest mass with respect to its proper time? What is the spaceship acceleration a in the frame instantaneously at rest with it?

Answer: The engine power must be subtracted from the spaceship energy, which is $M c^2$ in its rest frame, hence ${d M / dt}= - W/c^2\simeq 1.1\,{\times }\,10^{-4}\ \mathrm{kg/s}$. Since the particles emitted by the engine are photons, they carry a momentum equal to 1 / c times their energy, hence

$$ a (\tau ) = \frac{W}{c (M - W \tau /c^2)}.$$

1.22

Consider again Problem 1.21. If the spaceship moves along the positive x direction and leaves the space station at $\tau = 0$, compute its velocity with respect to the station reference frame (which is assumed to be inertial) as a function of the spaceship proper time.

Answer: According to the solution of previous Problem 1.21, the spaceship acceleration a in the frame instantaneously at rest with it is

$$ a(\tau )= {W \over c (M - \tau W /c^2)} = {\alpha c \over 1 - \alpha \tau }; \ \ \ \ \ \ \alpha = \frac{W}{Mc^2} \simeq 1.1\,{\times }\,10^{-8}\ \mathrm{s}^{-1}. $$

Recalling from the solution of Problem 1.15 that

$$ v={u\over \sqrt{1+{u^2\over c^2}} }, \,\,\,\,\,\,\,\,\,\,\, {d u\over d\tau }=a(\tau ) \sqrt{1+{u^2\over c^2}}, $$

where u is the x-component of the four-velocity we can easily integrate last equation

$$ {d u \over \sqrt{1+{u^2\over c^2}}} = {\alpha c\, d \tau \over 1 - \alpha \tau } $$

obtaining

$$ {u \over c} = \sinh \left( - \ln (1 - \alpha \tau ) \right) . $$

Expressing v/c as a function of u/c we finally get

$$ {v \over c} = \tanh \left( - \ln (1 - \alpha \tau ) \right) = {1 - (1 - \alpha \tau )^{2} \over 1 + (1 - \alpha \tau )^{2}}. $$

A much simpler way to obtain the same result is the following. Since all photons are emitted by the engine in the same direction (they are collinear), from Problem 1.20 we know that, independently of their energies, they can be considered as a single photon, i.e. a system of zero invariant mass. Therefore, from the point of view of the spaceship, what happens after a proper time $\tau$ can be described by the result of Problem 1.19: a spaceship of initial mass M and which is initially at rest decays into a photon plus a residual spaceship of residual mass $M(1 - \alpha \tau )$; the fact that photons are emitted continuously over a finite time interval is irrelevant, since, from the kinematic point of view, we are only interested in the initial and final states. Then, from the solution to Problem 1.19 and setting $m = M(1 - \alpha \tau )$, we easily obtain:

$$ {v(\tau ) \over c} = {p(\tau ) c \over E(\tau )} = \frac{M^2 - M^2(1 - \alpha \tau )^2}{M^2 + M^2(1 - \alpha \tau )^2} = {1 - (1 - \alpha \tau )^{2} \over 1 + (1 - \alpha \tau )^{2}} \, $$

which coincides with previous result. The solution is defined, of course, only for $\alpha \tau < 1$ since after that the spaceship mass vanishes. Notice that, for $\alpha \tau = 1$, the spaceship does not disappear, since that would not be consistent with kinematic constraints (the total invariant mass must stay M forever): it becomes a photon of momentum Mc/2. However, from $\tau = 1/\alpha$ on, the engine must stop, since a photon cannot emit another photon in the opposite direction (that would violate four-momentum conservation, leading in particular to a different invariant mass).

One can also obtain time t in the station frame as a function of proper time $\tau$:

$$ t = \int _0^\tau d \tau ' \gamma (\tau ') = \int _0^\tau d \tau ' \frac{1 + (1 - \alpha \tau )^2}{2 (1 - \alpha \tau )} = \frac{1}{2 \alpha } \left[ \frac{1 - (1 - \alpha \tau )^2}{2} - \ln (1 - \alpha \tau ) \right] $$

which diverges as $\alpha \tau \rightarrow 1$.

1.23

A spaceprobe of mass $M = 10\ \mathrm{kg}$ is boosted by a laser beam of frequency $\nu = 10^{15}\ \mathrm{Hz}$ and power $W = 10^{12}$ W, which is directed from Earth against an ideal reflecting mirror (i.e. reflecting all incoming photons) placed in the back of the probe. Assuming that the probe is initially at rest with respect to Earth, and that the laser beam is always parallel to the spaceprobe velocity and orthogonal to the mirror, determine the evolution of the spaceprobe position in the Earth frame and compute the total time $\varDelta t$ for which the laser must be kept switched on, in order for the spaceprobe to reach a velocity $v = 0.5\, c$.

Answer: In the reference frame of the spaceship, every photon gets reflected from the mirror with a negligible change in frequency ($\varDelta \lambda '/\lambda ' \sim h \nu ' / (M c^2) < 10^{-37}$, see Problem 1.33), hence it transfers a momentum $2 h \nu '/c$ to the spaceprobe, where $\nu '$ is the photon frequency in the spaceprobe frame. The acceleration of the probe in its proper frame is therefore

$$ a' = \frac{2 h \nu '}{M c} \frac{d N_\gamma '}{d t'} = \frac{2 W'}{Mc} $$

where $W'$ is the power of the beam measured in the proper frame, which is given by the photon energy, $h \nu '$, times the rate at which photons arrive, i.e. the number of photons hitting the mirror per unit time, ${d N_\gamma '}/{d t'}$. The above result is twice as large as that obtained if the light beam is emitted directly by the spaceprobe, as in Problem 1.21, since in this case each reflected photon transfers twice its momentum.

Both $\nu '$ and the rate of arriving photons are frequencies, hence they get transformed by Doppler effect and we can write

$$W' = h \nu ' \frac{d N_\gamma '}{d t'} = \sqrt{\frac{1-\beta }{1+\beta }} h \nu \sqrt{\frac{1-\beta }{1+\beta }} \frac{d N_\gamma }{d t} = \frac{1-\beta }{1+\beta }\ W$$

which gives us the transformation law for the beam power. From the acceleration in the proper frame, $a' = 2 W (1-\beta )/[(1+\beta ) Mc]$, one obtains the derivative of $\beta$ with respect to the proper time $\tau$ (see Problem 1.15)

$$\frac{d \beta }{d \tau } = \frac{d v}{c d \tau } = \frac{a'}{c} \left( 1 - \frac{v^2}{c^2} \right) = \alpha (1 -\beta )^2$$

where we have set $\alpha = 2 W /(M c^2)$. Last equation, after integration with the initial condition $\beta (0) = 0$, leads to $\alpha \tau = \beta /(1 - \beta )$, i.e.

$$ \beta (\tau ) = \frac{\alpha \tau }{1 + \alpha \tau } \, . $$

Regarding the position of the probe, we have

$$ d x = c \beta dt = c \beta \frac{d \tau }{\sqrt{1 - \beta ^2}} = \frac{\alpha \tau }{\sqrt{1 + 2 \alpha \tau }}\ d \tau $$

which gives, after integration and using $x(0) = 0$

$$ x = \frac{c}{3 \alpha } \left( (\alpha \tau - 1 ) \sqrt{2 \alpha \tau + 1} + 1 \right) . $$

Setting $\beta = 0.5$ we obtain $\tau = \alpha ^{-1} = M c^2/(2 W) = 9\,{\times }\,10^5\ \mathrm{s}$. The corresponding time in the Earth frame can be obtained by integrating the relation

$$ dt = \frac{d \tau }{\sqrt{1 - \beta ^2}} = \frac{1 + \alpha \tau }{\sqrt{1 + 2 \alpha \tau }}\ d \tau $$

yielding

$$ t = \frac{1}{3 \alpha } \left( (\alpha \tau + 2) \sqrt{2 \alpha \tau + 1} - 2 \right) . $$

In order to compute the total time $\varDelta t$ that the laser must be kept switched on, we have to consider that a photon reaching the spaceprobe at time t has left Earth at a time $t - x(t)/c$, where x is the probe position, hence the emission time is

$$ t_{em} = t - x/c = \frac{1}{\alpha } \left( \sqrt{ 2 \alpha \tau + 1} - 1 \right) .$$

Setting $\alpha \tau = 1$, we obtain $\varDelta t = (\sqrt{3} - 1)/\alpha \simeq 6.59\,{\times }\,10^5\ \mathrm{s}.$

It is interesting to notice that, analogously to Problem 1.22, also in this case the solution can be obtained in a much simpler way by mapping the problem onto a photon-particle diffusion one. Indeed, all the photons emitted by the laser are collinear to each other and can be considered as one single incident photon; the same applies to reflected photons, which can be treated as one single reflected photon. Then the problem reduces to a one-dimensional problem in which a photon of momentum $k = W t_{em}/c$ hits a particle of mass M which is initially at rest: we have to tune k so that after the collision the particle has a velocity $v = c/2$, i.e. a momentum $p = M v / \sqrt{1 - v^2/c^2} = M c / \sqrt{3}$. If we call $-k'$ the momentum of the reflected photon, conservation of four-momentum implies

$$ k = p - k'; \ \ \ \ \ k c + M c^2 = k'c + \sqrt{p^2c^2 + M^2 c^4} $$

which combined together lead, after some algebra, to

$$k = \frac{M c}{2} \left( {p \over Mc} - 1 + \sqrt{1 + \left( p \over Mc\right) ^2} \right) = \frac{M c}{2} \left( \sqrt{3} - 1 \right) $$

hence to $t_{em} = M c^2 (\sqrt{3} - 1)/(2 W)$ which coincides with the result found previously.

1.24

An electron–proton collision can give rise to a fusion process in which all available energy is transferred to a neutron . As a matter of fact, there is a neutrino emitted whose energy and momentum in the present situation can be neglected. The proton rest energy is $0.938\,{\times }\,10^9\,\mathrm{eV}$, while those of the neutron and of the electron are respectively $0.940\,{\times }\,10^9\,\mathrm{eV}$ and $5\,{\times }\,10^5\,\mathrm{eV}$. What is the velocity of the electron which, knocking into a proton at rest, may give rise to the process described above?

Answer: Notice that we are not looking for a minimum electron energy: since, neglecting the final neutrino, the final state is a single particle state, its invariant mass is fixed and equal to the neutron mass. That must be equal to the invariant mass of the initial system of two particles, leaving no degrees of freedom on the possible values of the electron energy: only for one particular value $v_e$ of the electron velocity the reaction can take place.

A rough estimate of $v_e$ can be obtained by considering that the electron energy must be equal to the rest energy difference $(m_n - m_p)\ c^2 = (0.940-0.938)\,{\times }\,10^9\,\mathrm{eV}$ plus the kinetic energy of the final neutron. Therefore the electron is surely relativistic and $(m_n - m_p) c$ is a reasonable estimate of its momentum: it coincides with the neutron momentum which is instead non-relativistic ($(m_n - m_p) c \ll m_n c$). The kinetic energy of the neutron is thus roughly ${(m_n-m_p)^2 c^2 / (2 m_n)}$, hence negligible with respect to $(0.940-0.938)\,{\times }\,10^9\,\mathrm{eV}$. The total electron energy is therefore, within a good approximation, $E_e\simeq 2\,{\times }\,10^6\ \mathrm{eV}$, and its velocity is $v_e= c \sqrt{1-{m_e^2/ E_e^2}}\simeq 2.9\,{\times }\,10^8 \,\mathrm{m/s}$. The exact result is obtained by writing $E_e=(m_n^2-m_p^2-m_e^2) c^2 / (2 m_p)$, which differs by less than 0.1 % from the approximate result.

1.25

A system made up of an electron and a positron, which is an exact copy of the electron but with opposite charge (i.e. its antiparticle), annihilates, while both particles are at rest, into two photons. The mass of the electron is $m_e \simeq 0.9\,{\times }\,10^{-30}\ \mathrm{kg}$: what is the wavelength of each outgoing photon? Explain why the same system cannot decay into a single photon.

Answer: The two photons carry momenta of modulus m c which are opposite to each other in order to conserve total momentum: their common wavelength is therefore, as we shall see in next Chapter, $\lambda =h/(m_e c)\simeq 2.4\,{\times }\,10^{-12}\ \mathrm{m}$. A single photon should carry zero momentum since the initial system is at rest, but then energy could not be conserved; more in general the initial invariant mass of the system, which is $2 m_e$, cannot fit the invariant mass of a single photon, which is always zero.

1.26

A piece of copper of mass $M = 1\ \mathrm{g}$, is heated from $0\,^{\circ }\mathrm{C}$ up to $100\,^{\circ }\mathrm{C}$. What is the mass variation $\varDelta M$ if the copper specific heat is ${C_\mathrm{Cu} = 0.4\ \mathrm{J/g ^{\circ }C}}$?

Answer: The piece of copper is actually a system of interacting particles whose mass is defined as the invariant mass of the system. That is proportional to the total energy if the system is globally at rest, see Eq. (1.75). Therefore $\varDelta M= C_\mathrm{Cu}\ \varDelta T/c^2\simeq 4.45\,{\times }\,10^{-16}\ \mathrm{kg}\ .$

1.27

Consider a system made up of two point-like particles of equal mass $m = 10^{-20}\,\mathrm{kg}$, bound together by a rigid massless rod of length $L = 2\,{\times }\,10^{-4}\,\mathrm{m}$. The center of mass of the system lies in the origin of the inertial frame O, in the same frame the rod rotates in the x–y plane with an angular velocity $\omega = 3\,{\times }\,10^{10}\,\mathrm{s^{-1}}$. A second inertial reference frame $O'$ moves with respect to O with velocity $v = 4c/5$ parallel to the x axis. Compute the sum of the kinetic energies of the particles at the same time in the frame $O'$, disregarding corrections of order $\omega ^3$.

Answer: There are two independent ways of computing the sum of the kinetic energies of the particles. The first way, which is the simplest one, is based on the assumption that the kinetic energy of the system coincides with the sum of the kinetic energies of the particles, hence one can compute the total energy of the system in the reference O, which, in the chosen approximation, is: $E_t=m\, (2c^2+\omega ^2L^2/4)$. In $O'$ the total energy is computed by a Lorentz transformation $E'_t=E_t/\sqrt{1-v^2/c^2} = (5/3) m\, (2c^2+\omega ^2L^2/4)$, hence the sum of the kinetic energies is $E'_t-2mc^2= 4/3 mc^2+5m \omega ^2L^2/12$. Numerically one has $12\,{\times }\,10^{-4}(1+1.25\,{\times }\,10^{-4})\ \mathrm{J}$. Alternatively, we can compute the velocities of both particles in a situation corresponding to equal time in $O'$. The simplest such situation is when the rod is parallel to the y axis and the particles move with velocity $v_\pm =\pm \omega L/2$ parallel to the x axis. Using Einstein formula one finds in $O'$ the velocities $v'_\pm =c(4/5\pm \omega L/(2c))/(1\pm 2\omega L/(5c))$ and hence the kinetic energies $E'_\pm =mc^2(1/\sqrt{1-(v'_\pm /c)^2}-1)= mc^2\left( (5\pm 2\omega L/c)/(3\sqrt{1-(\omega L/(2c))^2})-1\right)$. It is apparent that the sum $E'_++E'_-$ gives the known result up to corrections of order $(\omega L/c)^4$.

1.28

A spinning top, which can be described as a rigid disk of mass $M= 10^{{-1}}\,\mathrm{kg}$, radius $R= 5\,{\times }\,10^{{-2}} \ \mathrm{m}$ and uniform density, starts rotating with angular velocity $\varOmega = 10^{{3}}\,\mathrm{rad/s}$. What is the energy variation of the spinning top due to rotation, as seen from a reference frame moving with a relative speed $v= 0.9\ c$ with respect to it?

Answer: The speed of the particles composing the spinning top is surely non-relativistic in their frame, since it is limited by $\varOmega R = 50 \,\mathrm{m/s}\ \simeq 1.67\,{\times }\,10^{-7}\ c$. In that frame the total energy is therefore, apart from corrections of order $(\varOmega R/c)^2$, $E_{tot} = M c^2 + I \varOmega ^2/2$, where I is the moment of inertia, $I = M R^2/2$. Lorentz transformations yield in the moving frame

$$E'_{tot} = \frac{1}{\sqrt{1 - v^2/c^2}} E_{tot} = \frac{1}{\sqrt{1 - v^2/c^2}} \left( M c^2 + \frac{1}{2} I \varOmega ^2 \right) , $$

to be compared with the energy observed in absence of rotation, ${M c^2}/{\sqrt{1 - v^2/c^2}}$. Therefore, the energy variation due to rotation in the moving frame is $\varDelta E' = I \varOmega ^2/ (2 \sqrt{1 - v^2/c^2}) \simeq 143\,\mathrm{J}.$

1.29

A photon of energy E knocks into an electron at rest producing a final state composed of an electron–positron pair plus the initial electron: all three final particles have the same momentum. What is the value of E and the common momentum p of the final particles?

Answer: $\quad E=4\, m c^2\simeq 3.3\,{\times }\,10^{-13}\ \mathrm{J}, \,\,\,\,\, p = {E}/{3c} = {4}/{3}\ mc \simeq 3.6\,{\times }\,10^{-22}\ \mathrm{N/m}.$

1.30

A particle of mass $M = 1\,\mathrm{GeV}/c^2$ and energy $E = 10\,\mathrm{GeV}$ decays into two particles of equal mass $m = 490\,\mathrm{MeV}$. What is the maximum angle that each of the two outgoing particles may form, in the laboratory, with the trajectory of the initial particle?

Answer: Let $\hat{x}$ be the direction of motion of the initial particle, and x–y the decay plane: this is defined as the plane containing both the initial particle momentum and the two final momenta, which are indeed constrained to lie in the same plane by total momentum conservation. Let us consider one of the two outgoing particles: in the center of mass frame it has energy $\epsilon = M c^2/2 = 0.5\,\mathrm{GeV}$ and a momentum $p_x = p \cos \theta$, $p_y = p \sin \theta$, with $\theta$ being the decay angle in the center of mass frame and

$$ p = c \sqrt{{M^2 \over 4} - m^2} \simeq {0.1\ \mathrm{GeV} \over c}.$$

The momentum components in the laboratory are obtained by Lorentz transformations with parameters $\gamma = (\sqrt{1 - v^2/c^2})^{-1} = E / (M c^2) = 10$ and $\beta = v/c = \sqrt{1 - 1/\gamma ^2} \simeq 0.995$,

$$p_y' = p_y = p \sin \theta $$

$$p_x' = \gamma (p \cos \theta + \beta \epsilon ). $$

It can be easily verified that, while in the center of mass frame the possible momentum components lie on a circle of radius p centered in the origin, in the laboratory $p_x'$ and $p_y'$ lie on an ellipse of axes $\gamma p$ and p, centered in $(\gamma \beta \epsilon , 0)$. If $\theta '$ is the angle formed in the laboratory with respect to the initial particle trajectory and defining $\alpha \equiv \beta \epsilon / p \simeq 5$, we can write

$$\tan \theta ' = {p_y ' \over p_x '} = {1 \over \gamma } {\sin \theta \over \cos \theta + \alpha }. $$

If $\alpha > 1$ the denominator is always positive, $\tan \theta '$ is limited and $|\theta '| < \pi /2$, i.e. the particle is always forward emitted, in the laboratory, with a maximum possible angle which can be computed by solving $d \tan \theta ' / d \theta = 0$; that can also be appreciated pictorially by noticing that, if $\alpha > 1$, the ellipse containing the possible momentum components does not contain the origin. Finally one finds

$$ \theta _{max} ' = \tan ^{-1} \left( {1 \over \gamma } {1 \over \sqrt{\alpha ^2 - 1}} \right) \simeq 0.02\ \mathrm{rad}. $$

1.31

A particle of mass $M= 10^{-27}\,\mathrm{kg}$, which is moving in the laboratory with a speed $v=0.99\, c$, decays into two particles of equal mass $m=3\,{\times }\,10^{-28}\,\mathrm{kg}$. What is the possible range of energies (in $\mathrm{GeV}$) which can be detected in the laboratory for each of the outgoing particles? Supposing that in the center of mass frame the final particles are emitted isotropically, i.e. with equal probability in all directions, what is the probability, in the laboratory, of detecting a particle of energy in the range $[E,E + dE]$?

Answer: In the CM frame the outgoing particles have equal energy and modulus of the momentum completely fixed by the kinematic constraints: $\epsilon = M c^2 /2$ and $p = c\ \sqrt{M^2/4 - m^2}$. The only free variable is the decaying angle $\theta$, measured with respect to the initial particle trajectory, which however results in a variable energy E in the laboratory frame. Indeed by Lorentz transformations $E = \gamma (\epsilon + {v} p \cos {\theta } )$, with $\gamma = (1 - {v^2/c^2})^{-1/2}$, hence ignorance about $\theta$ in the CM frame results in ignorance about E in the laboratory. The maximum/mininum values of E are obtained as $\cos \theta$ becomes maximum/minimum, hence

$$E_{max/min} = \gamma \left( {M c^2 \over 2} \pm v p \right) \,\,\, \Rightarrow \,\,\, E_{max} \simeq 3.56\ \mathrm{GeV} \, , \, E_{min} \simeq 0.41\ \mathrm{GeV}.$$

Since the infinitesimal solid angle in spherical coordinates is $\sin \theta \, d \theta \, d \phi$, where $\phi$ is the azimuthal angle, an isotropic distribution in the CM frame means that the probability for one of the two final particles to be emitted with an angle in the range $[\theta ,\theta + d \theta ]$ is $P_\theta (\theta ) d \theta = \sin \theta \, d \theta /2$. The energy in the laboratory is a function of $\theta$, hence, calling $P_E dE$ the probability for the energy to be in the range $[E,E + dE]$, we have $P_E dE = P_\theta (\theta ) d \theta = \sin \theta \, d \theta /2$. Since, by differentiating $E(\theta )$, we get $d E = \gamma v p \sin \theta \, d \theta$, we finally obtain

$$ P_E dE = {dE \over 2 \gamma v p} $$

i.e. a flat distribution between the minimum and maximum possible values.

1.32

A particle of rest energy $M c^2 = 10^9\,\mathrm{eV}$, which is moving in the laboratory with momentum $p = 5\,{\times }\,10^{-18}\,\mathrm{N\ s}$, decays into two particles of equal mass $m=2\,{\times }\,10^{-28}\,\mathrm{kg}$. In the center of mass frame the decay direction is orthogonal to the trajectory of the initial particle. What is the angle between the trajectories of the outgoing particles in the laboratory?

Answer: Let $\hat{x}$ be the direction of the initial particle and $\hat{y}$ the decay direction in the center of mass frame, $\hat{y} \perp \hat{x}$. For the process described in the text, $\hat{x}$ is a symmetry axis, hence the outgoing particles will form the same angle $\theta$ also in the laboratory. For one of the two particles we can write $p_x = p/2$ by momentum conservation in the laboratory, and $p_y = c \sqrt{M^2/4 - m^2}$ by energy conservation. Finally, the angle between the two particles is $2 \theta = 2 \; \mathrm{atan} (p_y/p_x) \simeq 0.207 \; \mathrm{rad}.$

1.33

Compton Effect

A photon of wavelength $\lambda$ knocks into an electron at rest. After the elastic collision, the photon moves in a direction forming an angle $\theta$ with respect to its original trajectory. What is the change $\varDelta \lambda \equiv \lambda ' - \lambda$ of its wavelength as a function of $\theta$?

Answer: Let $\varvec{{q}}$ and $\varvec{{q}} '$ be respectively the initial and final momentum of the photon, and $\varvec{{p}}$ the final momentum of the electron. As we shall discuss in next chapter, the photon momentum is related to its wavelength by the relation $q \equiv |\varvec{{q}}| = h/\lambda$, where h is Planck’s constant. Total momentum conservation implies that $\varvec{{q}}$, $\varvec{{q}} '$ and $\varvec{{p}}$ must lie in the same plane, which we choose to be the x–y plane, with the x-axis parallel to the photon initial trajectory. Momentum and energy conservation lead to:

$$ p_x = q - q' \cos \theta , $$

$$ p_y = q' \sin \theta , $$

$$ q c + m_e c^2 - q'c = \sqrt{m_e c^4 + p_x^2 c^2 + p_y^2 c^2}. $$

Substituting the first two equations into the third and squaring both sides of the last we easily arrive, after some trivial simplifications, to $m_e (q - q') = q q' (1 - \cos \theta )$, which can be given in terms of wavelengths as follows

$$ \varDelta \lambda \equiv \lambda ' - \lambda = \frac{h}{m_e c} (1 - \cos \theta ); \,\,\,\,\,\,\,\,\,\, \frac{\varDelta \lambda }{\lambda } = \frac{h \nu }{m_e c^2} (1 - \cos \theta ). $$

The difference is always positive, since part of the photon energy, depending on the diffusion angle $\theta$, is always transferred to the electron. This phenomenon, known as Compton effect, is not predicted by the classical theory of electromagnetic waves and is an experimental proof of the corpuscular nature of radiation. Notice that, while the angular distribution of outgoing photons can only be predicted on the basis of the quantum relativistic theory, i.e. Quantum Electrodynamics, the dependence of $\varDelta \lambda$ on $\theta$ that we have found is only based on relativistic kinematics and can be used to get an experimental determination of h. The coefficient $h / (m_e c)$ is known as Compton wavelength, which for the electron is of the order of $10^{-12}\ \mathrm{m}$, so that the effect is not detectable ($\varDelta \lambda /\lambda \simeq 0$) for visible light.

1.34

A particle of mass M decays, while at rest, into three particles of equal mass m. What is the maximum and minimum possible energy for each of the outgoing particles?

Answer: Let $E_1,\ E_2,\ E_3$ and $\varvec{{p}}_1,\ \varvec{{p}}_2,\ \varvec{{p}}_3$ be respectively the energies and the momenta of the three outgoing particles. We have to find, for instance, the maximum and minimum value of $E_1$ which are compatible with the constraints $\varvec{{p}}_1 + \varvec{{p}}_2 + \varvec{{p}}_3 = 0$ and $E_1 + E_2 + E_3 = M\, c^2$. The minimum value is realized when the particle is produced at rest, $E_{1min} = m\, c^2$, implying that the other two particles move with equal and opposite momenta. Finding the maximum value requires some more algebra.

From $E_1^2 = m^2\, c^4 + p_1^2\, c^2$ and momentum conservation we obtain

$$E_1^2 = m^2\, c^4 + |\varvec{{p}}_2 + \varvec{{p}}_3|^2\, c^2 = m^2\ c^4 + (E_2 + E_3)^2 - \mu ^2\ c^4$$

where $\mu$ is the invariant mass of particles 2 and 3, $\mu ^2\, c^4 = (E_2 + E_3)^2 - |\varvec{{p}}_2 + \varvec{{p}}_3|^2\, c^2$. Applying energy conservation, $E_2 + E_3 = (M\, c^2 - E_1)$, last equation leads to

$$E_1 = \frac{1}{2 M\, c^2} \left( m^2\, c^4 + M^2\, c^4 - \mu ^2\, c^4 \right) .$$

We have written $E_1$ as a function of $\mu ^2$: $E_{1max}$ corresponds to the minimum possible value for the invariant mass of the two remaining particles. On the other hand it can be easily checked (see Eq. (1.79)) that, for a system made up of two or more massive particles, the minimum possible value of the invariant mass is equal to the sum of the masses and is attained when the particles are at rest in their center of mass frame, meaning that the particles move with equal velocities in any other reference frame. Therefore $\mu _{min} = 2 m$ and $E_{1max} = (M^2\ - 3 m^2)\ c^2 / (2 M )$: this value is obtained in particular for $\varvec{{p}}_2 = \varvec{{p}}_3$.

1.35

A particle of mass M decays, while at rest, into N particles of masses $m_i$, $i = 1,N$ ($N \ge 3$). What is the maximum and minimum possible energy for each of the outgoing particles?

Answer: The problem is very similar to the previous one, we work it out for particle 1 but the solution can be trivially generalized. The minimum energy is always $E_{1min} = m_1\, c^2$, since the particle can be produced at rest with the remaining particles taking care of momentum conservation. This is true even if $m_1 = 0$, like for a photon: in this case the minimum energy is just a lower bound, since a photon with exactly zero energy means no photon at all.

As for the maximum energy, let us call $\varvec{{p}}_i$ the momenta of the various emitted particles, $\varvec{{q}}$ the total momentum of all particles except particle 1 and $\epsilon$ their energy ($\varvec{{q}} \equiv \sum _{i > 1} \varvec{{p}}_i$, $\epsilon = \sum _{i > 1} E_i$). Then $\varvec{{p}}_1 = - \varvec{{q}}$ and we can write

$$E_1^2 = m_1^2\, c^4 + |\varvec{{q}}|^2\, c^2 = m_1^2\ c^4 + \epsilon ^2 - \mu ^2\ c^4$$

where $\mu$ is the invariant mass of particles $2,3,\dots N$. Applying energy conservation, $\epsilon = (M\, c^2 - E_1)$, we have again

$$E_1 = \frac{1}{2 M\, c^2} \left( m_1^2\, c^4 + M^2\, c^4 - \mu ^2\, c^4 \right) $$

which is maximum when $\mu$ attains its minimum, which by Eq. (1.79) is $\mu _{min} = \sum _{i > 1} m_i$. Finally we can write

$$ E_{1max} = \frac{M^2 + m_1^2 - \left( \sum _{i = 2}^N m_i\right) ^2}{2 M} c^2 \, . $$

1.36

A particle at rest, whose mass is $M=750\ \mathrm{MeV}/c^2$, decays into a photon and a second, lighter, particle of mass $m=135\ \mathrm{MeV}/c^2$. Subsequently the lighter particle decays into two further photons. Considering all the possible decay angles of the second particle, compute the maximum and the minimum values of the possible energies of the three final photons.

One can ask the same question in the case of a direct decay of the first particle into three photons whose energies are constrained only by energy-momentum conservation. What are the maximum and minimum energies of the final photons in the direct decay?

Answer: This problem is analogous to Problem 1.34. We first consider the direct decay. If $E_i$ with $i=1,2$ are the energies of two, arbitrarily chosen, final photons, $p_i=E_i/c$ are their momenta. If $\theta$ is the angle between these momenta, on account of momentum conservation, we can compute the momentum of the third photon as the length of the third side of a triangle whose other sides have lengths $p_1$ and $p_2$ and form an angle $\pi - \theta$. Therefore energy conservation gives: $\sqrt{E_1^2+E_2^2+2E_1E_2\cos \theta }+E_1+E_2=Mc^2$ from which we get: $M^2c^4-2Mc^2(E_1+E_2)=-2E_1E_2(1-\cos \theta )$. Since $0\le 1-\cos \theta \le 2$ one has the inequalities:

$$Mc^2/2\le (E_1+E_2) \,\,\,\,\,\,\,\,\,\,\, \mathrm{and} \,\,\,\,\,\,\,\,\,\,\,\, (Mc^2-2E_1)(Mc^2-2E_2)\ge 0 \, . $$

If we interpret $E_1$ and $E_2$ as cartesian coordinates of a point in a plane, we find that the point must lie in a triangle with vertices in the points of coordinates $(0,Mc^2/2)$, $(Mc^2/2,0)$ and $(Mc^2/2,Mc^2/2)$. This shows that each of the three photon energies can range between 0 and $Mc^2/2$. In Particle Physics the distribution of points associated with a sample of decay events (in this case the distribution of points inside the triangle described above) is called Dalitz plot.

Now consider the indirect decay and assume that photons 1 and 2 are the decay products of the second particle with mass m. In this case there is a constraint for the third photon with energy $E_3=Mc^2-E_1-E_2$. Indeed its momentum $p_3=E_3/c$ must be equal, and opposite, to that of the light particle whose energy is $E_1+E_2$. Thus we have $(E_1+E_2)^2-(Mc^2-E_1-E_2)^2=m^2c^4$ which implies $E_1+E_2=(M^2+m^2)c^2/(2M)$, which can be read as the equation of a line intersecting the above mentioned triangle. It is apparent that the boundaries of the intersection segment give the maximum and minimum possible values for $E_1$ and $E_2$, which are respectively $m^2c^2/(2M) = 12.15\ \mathrm{MeV}$ and $Mc^2/2 = 375\ \mathrm{MeV}$. The energy of the third photon is instead fixed and equal to $E_3=(M^2-m^2)c^2/(2M)$: we have $E_3 < Mc^2/2$ and, for the given values of M and m, also $E_3 > m^2c^2/(2M)$.

Another significant difference for the indirect decay is that two of the emitted photons will always form a system with an invariant mass equal to m. Therefore, if one observes several examples of such decays and measures in each case all kinematic parameters of the emitted photons, thus reconstructing the invariant mass for each couple of photons and making an histogram for its probability distribution, one would obtain a smooth distribution for the direct decay, and instead a smooth distribution plus a sharp peak located at m for the indirect decay, thus “discovering” the presence of the intermediate particle m, even if it is not directly revealed by the detector. This is a common strategy for the discovery of new particles, the last renowned example being the Higgs boson.

1.37

A proton beam is directed against a laser beam coming from the opposite direction and having wavelength $0.5\,{\times }\,10^{-6}\ \mathrm{m}$. Determine what is the minimum value needed for the kinetic energy of the protons in order to produce the reaction (proton + photon $\rightarrow$ proton + $\pi$), where the $\pi$ particle has mass $m \simeq 0.14\ M$, the proton mass being $M \simeq 0.938\ \mathrm{GeV} / c^2$.

Answer: Let p and k be the momenta of the proton and of the photon respectively, $k = h / \lambda \simeq 2.48\,\mathrm{eV}/c$. The reaction can take place only if the invariant mass of the initial system is larger or equal to $(M + m)$: that is most easily seen in the center of mass frame, where the minimal energy condition corresponds to the two final particles being at rest. In particular, if E is the energy of the proton, we can write

$$ (E + kc)^2 - (p - k)^2 c^2 \ge (M + m)^2 c^4, \,\,\,\,\,\,\, \mathrm{hence} \,\,\,\,\,\,\,\, E + pc \ge \frac{m c^2}{kc} (M + m/2) c^2.$$

Taking into account that $m c^2 \simeq 0.13\,\mathrm{GeV}$ and $k c \simeq 2.48\,\mathrm{eV}$ we deduce that $E + pc \sim 5\,{\times }\,10^7\ Mc^2$, so that the proton is ultra-relativistic and $E \simeq pc$. The minimal kinetic energy of the proton is therefore $p_{min} c \simeq (m c / k) (M + m/2) c^2/2 \simeq 2.6\,{\times }\,10^7\ \mathrm{GeV}$.

1.38

A particle of mass M decays into two particles of masses $m_1$ and $m_2$. A detector reveals the energies and momenta of the outgoing particles to be $E_1 = 2.5\,\mathrm{GeV}$, $E_2 = 8\,\mathrm{GeV}$, $p_{1x} = 1\,\mathrm{GeV}/c$, $p_{1y} = 2.25\,\mathrm{GeV}/c$, $p_{2x} = 7.42\,\mathrm{GeV}/c$ and $p_{2y} = 2.82\,\mathrm{GeV}/c$. Determine the masses of all involved particles, as well as the velocity $\varvec{{v}}$ of the initial particle.

Answer: $M \simeq 3.69\,\mathrm{GeV}/c^2$, $m_1 \simeq 0.43\,\mathrm{GeV}/c^2$, $m_2 \simeq 1\,\mathrm{GeV}/c^2$, $v_x = 0.802\ c$, $v_y = 0.483\ c$.

1.39

A particle of mass $\mu =0.14\,\mathrm{GeV}/c^2$ and momentum directed along the positive z axis, knocks into a particle at rest of mass M. The final state after the collision is made up of two particles of mass $m_1=0.5\,\mathrm{GeV}/c^2$ and $m_2=1.1\,\mathrm{GeV}/c^2$ respectively. The momenta of the two outgoing particles form an equal angle $\theta =0.01\,\mathrm{rad}$ with the z axis and have equal magnitude $p=10^4\,\mathrm{GeV}/c$. What is the value of M?

Answer: Momentum conservation gives the momentum of the initial particle, $k = 2 p \cos \theta$. The initial energy is therefore $E_{in} = \sqrt{\mu ^2 c^4 + k^2 c^2} + M\ c^2$ and must be equal to the final energy $E_{fin} = \sqrt{m_1^2 c^4 + p^2 c^2} + \sqrt{m_2^2 c^4 + p^2 c^2}$, hence

$$ M c^2 = \sqrt{m_1^2 c^4 + p^2 c^2} + \sqrt{m_2^2 c^4 + p^2 c^2} - \sqrt{\mu ^2 c^4 + 4 p^2 \cos \theta ^2 c^2}. $$

The very high value of p makes it sensible to apply the ultra-relativistic approximation,

$$\begin{aligned} M c^2\simeq & {} p c \left( 1 + { m_1^2 c^2 \over 2 p^2} \right) + p c \left( 1 + {m_2^2 c^2 \over 2 p^2} \right) - 2 p c \cos \theta \left( 1 + {\mu ^2 c^2 \over 8 p^2 \cos \theta ^2} \right) \nonumber \\&\nonumber \\\simeq & {} pc \left( \theta ^2 + (2 m_1^2 + 2 m_2^2 - \mu ^2) c^2 / (4 p^2) \right) \simeq pc\ \theta ^2 = 1\ \mathrm{GeV}. \nonumber \end{aligned}$$

1.40

A flux of particles, each carrying an electric charge $q = 1.6\,{\times }\,10^{-19}\ \mathrm{C}$, is moving along the x axis with a constant velocity $v = 0.9\ c$. If the total carried current is $I = 10^{-9}\ \mathrm{A}$, what is the linear density of particles, as measured in the reference frame at rest with them?

Answer: If $d_0$ is the distance among particles in their rest frame, the distance measured in the laboratory appears contracted and equal to $d = \sqrt{1 - v^2/c^2}\ d_0$. The electric current is given by $I = d^{-1} v q$, hence the particle density in the rest frame is

$$d_0^{-1} = \sqrt{1 - v^2/c^2} \frac{I}{vq} \simeq 10.1 \; \mathrm{particles/m}\, .$$

An alternative approach is to apply the transformation properties of the four-current, whose components are $(\rho c, I)$ in the laboratory and $(\rho _0 c, 0)$ in the rest frame of the particles, then from Eq. (1.92):

$$ \rho _0 c = \gamma \left( \rho c - {v \over c} I \right) $$

which coincides with the result above considering that $I = \rho v$ and $\rho _0 = q/d_0$.

1.41

Transformation Laws for Electromagnetic Fields

Our inertial reference frame moves with respect to a conducting rectilinear wire with velocity $v=0.9\ c$ parallel to the wire. In its reference frame the wire appears neutral and one has an electric current $I=1\ \mathrm{A}$ through the wire in the direction of our velocity. We adopt a simplified scheme in which the current is carried by electrons with a linear density $\rho _{wire}=6\,{\times }\,10^{16}\ \mathrm{m^{-1}}$, moving with an average uniform velocity $V=10^2\ \mathrm{m/s}$ in the opposite direction with respect to our velocity. The wire is made neutral by protons at rest, having the same linear density as the electrons. Coming back to our reference frame, do we detect any electric field? If the answer to our question is positive, what is the absolute value of the electric field at a distance $r=1\ \mathrm{cm}$ from the wire?

Answer: The answer could be obtained straightforwardly by applying the transformation laws of electromagnetic fields that we have derived (see Eq. (1.97)). However, let us proceed in a different way, starting from the sources observed in the different inertial frames: the reader is invited to verify that results coincide with those obtainable from Eq. (1.97).

If the electrons in the wire are uniformly distributed the distance between two neighboring electrons is $d=\rho _{wire}^{-1}=1.66\,{\times }\,10^{-17}\ \mathrm{m}$ in the wire frame. Due to the length contraction the same distance is $d/\sqrt{1-(V/c)^2}$ in the electron frame, that is, in a frame moving with respect to the wire with the average electron velocity. Using Einstein formula we compute our velocity with respect to the electron frame $v'=(v+V)/(1+vV/c^2)$ and the electron density in our frame: $\rho _{moving,e}= \sqrt{1-(V/c)^2}/(d\sqrt{1-(v'/c)^2})= (1+vV/c^2)/(d\sqrt{1-(v/c)^2})$ while the proton density is $\rho _{moving,p}= 1/(d\sqrt{1-(v/c)^2})\equiv \gamma /d$. Thus the resulting charge density is $\rho _{moving,tot}=-evV /(c^2d\sqrt{1-(v/c)^2})=-I\beta \gamma /c$ where we have set $\beta =v/c$. Since Maxwell equations are the same in every inertial frame, we conclude that in our frame we have an electric field with absolute value $E_{moving}=\beta \gamma I/(2\pi \epsilon _0 rc)=\beta \gamma c B_{wire}=1.23\,{\times }\,10^4\ \mathrm{V/m}$ and directed towards the wire. $B_{wire}$ is the absolute value of the magnetic induction at the same distance from the wire in the wire frame; the electric field measured in our frame is orthogonal with respect to the original magnetic field (in particular it is directed like $\varvec{{v}} \wedge \varvec{{B}}_{wire}$). It is also easy to verify that the electric current in our frame is $\gamma I$, so that we measure a magnetic field of absolute value $B_{moving} = \gamma B_{wire}$ and parallel to the original magnetic field.

The complete set of field transformation rules, Eq. (1.97), could be obtained through the analysis of similar gedanken experiments. To that purpose, the reader is invited to compute the electric and magnetic fields felt by an observer moving: (a) parallel to a wire having a uniform charge density and zero electric current; (b) parallel to an infinite plane carrying zero charge density and a uniform current density orthogonal to the observer velocity; (c) orthogonal to an infinite plane carrying uniform charge density and zero electric current.

1.42

A relativistic particle with mass m and charge q moves in a time independent magnetic field B. In cylindrical coordinates $z , \ \rho , \ \phi$ the magnetic field components are $B_\phi =B_\rho =0$ and $B_z=f(\rho )$, those of the vector potential are $A_z=A_\rho =0$ and $A_\phi =\rho a(\rho )$ with $a(\rho )$ a positive increasing function of $\rho$. The Lagrangian equations of the particle (see (1.104)) are easily reduced to three prime integrals. Compute them.

Answer: The Lagrangian is $\mathcal{L}=-mc^2\sqrt{1-v^2/c^2} +q\rho ^2 a \dot{\phi }$, we denote the time derivative by a dot and $v^2= (\dot{z})^2 +(\dot{\rho })^2+\rho ^2 (\dot{\phi })^2$. We have three prime integrals because the Lagrangian is independent of $z, \ \phi$ and t. The corresponding prime integrals are: the energy $mc^2/\sqrt{1-v^2/c^2}\equiv E$, the z-momentum component $P_z=m\dot{z}/\sqrt{1-v^2/c^2}$ and the z-angular momentum component $L_z=m\rho ^2\dot{\phi }/\sqrt{1-v^2/c^2}+q\rho ^2 a(\rho )$. Therefore we have

$$\dot{z}= P_z c^2/E\ \ ,\ \dot{\phi }=(L_z/\rho ^2 -qa(\rho ))c^2/E$$

$$ \dot{\rho }=\pm c\sqrt{1-c^2(m^2c^2+P_z^2+(L_z/\rho ^2 -qa(\rho ))^2)/E^2}.$$

These equations must be integrated starting from suitable initial coordinates. A typical ultra-relativistic case corresponds to $(P_zc /E)^2=1-2\epsilon ^2$ and $m^2c^4/E^2=\epsilon ^2$, this implies $\epsilon ^2\ge c^2(L_z/\rho ^2 -qa(\rho ))^2)/E^2$. This sets bounds on the range of $\rho$ which cannot vanish if $L_z$ does not, while is bound by the increasing behavior of $a(\rho )$. The values of $\rho$ for which one has equality correspond to spiral orbits of the particle.