Chapter 55Lorentz Contraction, Time Dilation, and the Transformation of Velocities

Example 55.1: The length of a structure passing by at 0.9 times the speed of light

A structure whose length at rest relative to the spaceship (proper length) is $100\ \mathrm{m}$ is placed along the direction of travel inside a spaceship flying at $v=0.9c$. To how many $\mathrm{m}$ does this structure appear to shrink, as seen from the ground (frame S)?

Solution. From $\beta=v/c=0.9$ we have $\beta^2=0.81$ and $1-\beta^2=0.19$. By Theorem 55.1 (Eq. \eqref{eq:55-thm-contraction}),

(With $\gamma=1/\sqrt{0.19}=2.294$, we also get $\ell/\gamma=100/2.294=43.6\ \mathrm{m}$, in agreement.) The length is about $44\ \mathrm{m}$; that is, the structure appears a little shorter than half its proper length. The directions perpendicular to the direction of motion (the width and height of the structure) are unchanged.

Example 55.2: Measuring a rod at rest in frame S′ from frame S (the contraction is “mutual”)

Let $\ell_0$ be the proper length of a rod at rest relative to frame S′. The rod lies along the $x'$ axis of S′, with one end A′ at $x'=0$ and the other end B′ at $x'=\ell_0$. Find the length $\ell$ of this rod in frame S, using the inverse Lorentz transformation $x=\gamma(x'+vt')$, $t=\gamma(t'+vx'/c^2)$ (Section 54.7 of the previous chapter).

Solution. Substituting the coordinate $x_{\mathrm{A}}'=0$ of A′ gives $x_{\mathrm{A}}=\gamma\,v\,t_{\mathrm{A}}'$ and $t_{\mathrm{A}}=\gamma\,t_{\mathrm{A}}'$. Substituting the coordinate $x_{\mathrm{B}}'=\ell_0$ of B′ gives

The length in frame S is the difference of the coordinates measured at the same time in S, $t_{\mathrm{A}}=t_{\mathrm{B}}$. From $\gamma t_{\mathrm{A}}'=\gamma(t_{\mathrm{B}}'+v\ell_0/c^2)$ we get $t_{\mathrm{A}}'-t_{\mathrm{B}}'=v\ell_0/c^2$. Hence

(in the last step we used $\gamma(1-\beta^2)=1/\gamma$). Therefore, even when viewed from frame S, the rod at rest in S′ appears shrunk by the same ratio, to $\sqrt{1-(v/c)^2}$ times its length. The result has exactly the same form as when a rod at rest in S is viewed from S′ (Eq. \eqref{eq:55-contraction}).

Example 55.3: How much is the length of a Shinkansen train contracted?

When a person on the ground observes a Shinkansen (bullet train) of total length $400\ \mathrm{m}$ (proper length) running at $300\ \mathrm{km}$ per hour, by how many $\mathrm{m}$ is it shortened by Lorentz contraction? Take $c=2.998\times10^8\ \mathrm{m/s}$, and compare the result with the size of an atom (about $10^{-10}\ \mathrm{m}$).

Solution. We have $v=300\ \mathrm{km/h}=300\times10^3/3600=83.3\ \mathrm{m/s}$, so $\beta=83.3/(2.998\times10^8)=2.78\times10^{-7}$. Since this value is much smaller than 1, we approximate $\gamma^{-1}=\sqrt{1-\beta^2}$ using $\sqrt{1-x}\approx1-x/2$ from the Taylor expansion (University Mathematics, Section 4.2). Since $\beta^2=7.72\times10^{-14}$,

The contraction is about $0.015\ \mathrm{nm}$ ($15\ \mathrm{pm}$), only about $1/7$ of the size of a single atom. We do not notice Lorentz contraction at everyday speeds because $\beta$ is extremely small and the contraction is proportional to $\beta^2$.

Example 55.4: One hour aboard a spaceship flying at 0.99c

While one hour ($3600\ \mathrm{s}$) elapses on the ground, how many minutes does the clock of a person aboard a spaceship flying at $v=0.99c$ advance?

Solution. One hour on the ground is a time interval $\Delta t=3600\ \mathrm{s}$ measured by clocks on the ground. The spaceship's clock measures events that occur at the same place inside the spaceship, so the time $\Delta\tau$ ticked off by the spaceship's clock is the proper time. By Theorem 55.2, $\Delta t=\gamma\Delta\tau$, that is, $\Delta\tau=\Delta t/\gamma=\Delta t\sqrt{1-\beta^2}$. For $\beta=0.99$ we have $\beta^2=0.9801$, $1-\beta^2=0.0199$, and $\sqrt{0.0199}=0.14107$ ($\gamma=7.0888$), so

Thus, during one hour on the ground, the clock on the spaceship advances only about $8.5$ minutes.

Example 55.5: The period of a light clock

A light clock has a mirror separation of $d=0.15\ \mathrm{m}$. (a) Find the period (the proper time) $\Delta\tau$ when the clock is at rest. (b) When this clock is seen moving at $v=0.6c$ in the direction parallel to the mirror surfaces, find the period $T$ and the distance $cT$ traveled by the light during that time. Take $c=3.00\times10^8\ \mathrm{m/s}$.

Solution. (a) $\Delta\tau=2d/c=2\times0.15\ \mathrm m/(3.00\times10^8\ \mathrm{m/s})=1.00\times10^{-9}\ \mathrm{s}=1.00\ \mathrm{ns}$. (b) For $\beta=0.6$ we have $\gamma=1/\sqrt{1-0.36}=1/0.8=1.25$, so by Eq. \eqref{eq:55-lc-result}, $T=\gamma\,\Delta\tau=1.25\ \mathrm{ns}$. The distance traveled by the light is $cT=3.00\times10^8\times1.25\times10^{-9}=0.375\ \mathrm m$, which is longer than the round-trip distance at rest, $2d=0.30\ \mathrm m$. As a check using Eq. \eqref{eq:55-lc-pyth}: the one-way distance is $cT/2=0.1875\ \mathrm m$, the displacement of the clock is $vT/2=0.6\times0.1875=0.1125\ \mathrm m$, and the perpendicular distance is $d=0.15\ \mathrm m$. Then $0.15^2+0.1125^2=0.0225+0.012656=0.035156$ and $0.1875^2=0.035156$, so the Pythagorean theorem is satisfied.

Example 55.6: Checking that length contraction and time dilation are compatible

A spaceship of proper length $\ell_0=100\ \mathrm m$ passes an observation point P (fixed on the ground) at speed $v=0.6c$. How much time, in $\mathrm{s}$, elapses between the moment the front of the spaceship passes P and the moment its rear passes P, as measured (a) by a clock on the ground and (b) by a clock on the spaceship? (c) Verify that (a) and (b) satisfy the time dilation formula (Theorem 55.2). Take $c=3.00\times10^8\ \mathrm{m/s}$.

Solution. We have $\gamma=1.25$. (a) Seen from the ground, the length of the spaceship is Lorentz-contracted to $\ell=\ell_0/\gamma=100/1.25=80\ \mathrm m$. Since the spaceship, with this length, passes the observation point P at speed $v=0.6c=1.80\times10^8\ \mathrm{m/s}$, the time required for the passage is

(b) Seen from the spaceship, the spaceship of length $\ell_0=100\ \mathrm m$ is at rest, and the observation point P passes by at speed $v$. The distance from the front to the rear is $100\ \mathrm m$, so

(c) The two events “the front passes P” and “the rear passes P” occur at the same place (P) on the ground, so $\Delta t_{\mathrm{G}}$, measured by the ground clock (a single clock placed at P), is the proper time. By Theorem 55.2, the time interval seen from the spaceship is $\Delta t_{\mathrm{S}}=\gamma\,\Delta t_{\mathrm{G}}=1.25\times4.44\times10^{-7}=5.56\times10^{-7}\ \mathrm s$, which agrees with the value in (b). Length contraction (the previous section) and time dilation are not separate phenomena: they are two manifestations of a single theory, both following from the same Lorentz transformation, and they do not contradict each other.

Example 55.7: A numerical check of “each sees the other's clock running slow”

Two clocks synchronized in frame S, C1 ($x=0$) and C2 ($x=L=180\ \mathrm m$), are placed in frame S. A clock K is at the origin of frame S′ and moves relative to S in the $+x$ direction at $v=0.6c$ ($\gamma=1.25$). At the instant K passes C1, K and C1 both read $0$. (a) From the standpoint of frame S, find the readings of C2 and K at the instant K comes alongside C2, and verify the statement “K is running slow.” (b) From the standpoint of frame S′, find the value shown by C2 at the instant K is alongside C1 ($t'=0$). (c) From the standpoint of frame S′, find by how much the reading of K and the reading of C2 each advance until K comes alongside C2, and verify the statement “the clocks of frame S are running slow.” Take $c=3.00\times10^{8}\ \mathrm{m/s}$.

Solution. We have $v=0.6c=1.80\times10^{8}\ \mathrm{m/s}$ and $L/c=180/(3.00\times10^{8})=0.600\ \mathrm{\mu s}$.

(a) In frame S, the time K takes to travel the distance $L$ from C1 to C2 is $t=L/v=180\ \mathrm m/(1.80\times10^{8}\ \mathrm{m/s})=1.00\ \mathrm{\mu s}$. Since C1 and C2 are synchronized in S, C2 reads $1.00\ \mathrm{\mu s}$ at the instant K comes alongside C2. On the other hand, K is a moving clock, so during an interval of $1.00\ \mathrm{\mu s}$ in frame S the proper time of K advances only $1.00/\gamma=1.00/1.25=0.800\ \mathrm{\mu s}$, and K reads $0.800\ \mathrm{\mu s}$. Since C2 reads $1.00\ \mathrm{\mu s}$ and K reads $0.800\ \mathrm{\mu s}$, K is running slow as seen from frame S.

(b) The event of K coming alongside C1 occurs at $t'=0$ in frame S′. The value $t_{\mathrm{C2}}$ shown by C2 at that same time $t'=0$ of frame S′ is the time $t$ of frame S for which $t'=\gamma\,(t-vL/c^2)=0$ at the position of C2, $x=L$. Hence

Seen from frame S′, at $t'=0$ C1 reads $0$ and C2 reads $0.360\ \mathrm{\mu s}$. Two clocks synchronized in S are not synchronized as seen from S′: C2, which is at the rear with respect to the direction in which the clocks of frame S move (the $-x'$ direction), is ahead by $vL/c^2$. (For this position $x'$, putting $t'=0$ and $x'=L/\gamma=144\ \mathrm m$ into the inverse Lorentz transformation $t=\gamma(t'+vx'/c^2)$ also gives $t=1.25\times0.6\times144/(3.00\times10^{8})=0.360\ \mathrm{\mu s}$, in agreement.)

(c) In frame S′, at $t'=0$ the clock C2 is at distance $L/\gamma=180/1.25=144\ \mathrm m$ from K (in S′ the distance between C1 and C2 is contracted), and it approaches K at speed $v$. It comes alongside K at $t'=144\ \mathrm m/(1.80\times10^{8}\ \mathrm{m/s})=0.800\ \mathrm{\mu s}$. Since K is at rest in S′, during this time the reading of K advances from $0$ to $0.800\ \mathrm{\mu s}$ (the same value as in (a)). Since C2 is moving in S′, during an interval of $0.800\ \mathrm{\mu s}$ in frame S′ it advances only $0.800/\gamma=0.640\ \mathrm{\mu s}$, and its reading goes from $0.360\ \mathrm{\mu s}$ to $0.360+0.640=1.000\ \mathrm{\mu s}$. This agrees with the reading of C2 at the instant it comes alongside K, $1.00\ \mathrm{\mu s}$, found in (a). Measured in the time of frame S′, while K advances $0.800\ \mathrm{\mu s}$, C2 advances only $0.640\ \mathrm{\mu s}$, so the clocks of frame S are running slow as seen from frame S′.

The conclusions of (a) and (c) are opposite, yet they do not contradict each other. In (a), the advance of a single clock, K, is measured by two synchronized clocks of frame S, C1 and C2. In (c), the advance of a single clock, C2, is measured by two synchronized clocks of frame S′ (the clock at $x'=144\ \mathrm m$ and K). The pairs of clocks being compared are different, and moreover “the two clocks of the other frame are not synchronized as seen from one's own frame” (the offset of $0.360\ \mathrm{\mu s}$ found in (b)); this is why each observer sees the other's clock running slow.

Illustration 55.1: Lifetime and flight distance when the muon's speed is $v=0.999c$

Taking the muon's speed to be $v=0.999\,c$, compute the muon's lifetime $T_\mu'$ as seen from the ground and the distance $L$ it travels during that time. (Take the mean lifetime of a muon at rest to be $T_\mu=2.197\times10^{-6}\ \mathrm s$.)

Setting up the coordinate systems. Let S be the coordinate system fixed to the ground (the stationary frame), and let S′ be the coordinate system that moves together with the muon in the $+x$ direction with speed $v$ (Figure 55.4). The muon is at rest at the origin of S′. In both frames, we take the event of the muon's birth to be at the origin and at time 0, and we take the event of its decay to be at the time $t'=T_\mu$ in S′. Since the muon does not move in S′, $x'=0$ at every time.

Solution. Since we want the coordinates as seen from S, we use the inverse Lorentz transformation from S′ to S (Section 54.7 of the previous chapter, with $v\to-v$):

First, substituting the event of birth ($x'=0$, $t'=0$), we obtain

so that the event is again at the origin and at time 0 in S (consistent with the condition we arranged at the start). Next, we substitute the event of decay ($x'=0$, $t'=T_\mu$). Since the muon does not move in S′, $x'(T_\mu)=0$:

Here $\sqrt{1-(0.999)^2}=\sqrt{1-0.998001}=\sqrt{0.001999}=0.044710$ ($\gamma=22.366$). Therefore the muon's lifetime as seen from the ground is the time $t(T_\mu)$ in S, and it works out to

(Rounding to $T_\mu=2.2\times10^{-6}\ \mathrm s$ gives $4.92\times10^{-5}\ \mathrm s$.) The distance $L$ traveled during that time is the decay position $x(T_\mu)$ in S, so that

This is about $22$ times the distance $cT_\mu=6.6\times10^{2}\ \mathrm m$ obtained by ignoring relativity. As a check, it is the same value as $L=v\,T_\mu'$ (seen from S, the muon travels straight ahead at speed $v$ for the time $T_\mu'$).

This average flight distance of $14.7\ \mathrm{km}$ is about the same size as the altitude of $15$–$20\ \mathrm{km}$ at which muons are born, so a considerable fraction of them reach the ground (if relativity were ignored, they could fly only $660\ \mathrm m$, and almost none would arrive). The fraction that actually arrives is calculated in Example 55.8.

Example 55.8: Fraction of muons reaching the ground (agreement of the two viewpoints)

A muon born at an altitude of $h=20\ \mathrm{km}$ descends straight toward the ground at a speed of (i) $v=0.999c$ or (ii) $v=0.9999c$. Find the fraction that reaches the ground without decaying. Also, for (i), verify that the calculation from the muon's point of view (contraction of the atmosphere) gives the same fraction. Take $T_\mu=2.197\ \mathrm{\mu s}$ and $c=2.998\times10^8\ \mathrm{m/s}$.

Solution. The fraction surviving until time $t$ is $\exp(-t/T_\mu)$ (the definition of the mean lifetime). Seen from the ground, the muon's clock runs slower by a factor of $\gamma$, so during the time $t$ of the ground frame, the proper time $t/\gamma$ elapses. Since the time the descent takes in the ground frame is $t=h/v$, the surviving fraction is

(i) With $\gamma=22.366$ and $\gamma vT_\mu=22.366\times0.999\times2.998\times10^{8}\times2.197\times10^{-6}=1.472\times10^{4}\ \mathrm m$, we get $P=\exp(-20000/14717)=\exp(-1.359)=0.257$, that is, about $26\%$ (the right end of the green solid line in Figure 55.5). (ii) With $\gamma=1/\sqrt{1-0.9999^2}=70.71$ and $\gamma vT_\mu=4.657\times10^{4}\ \mathrm m$, we get $P=\exp(-20000/46570)=\exp(-0.429)=0.651$, about $65\%$. Incidentally, if relativity is ignored, $P=\exp(-20000/658.7)=6.5\times10^{-14}$.

Next, we calculate (i) from the muon's point of view. The thickness of the atmosphere shrinks to $h/\gamma=894\ \mathrm m$, and the time for the descent (the proper time) is $\tau=894\ \mathrm m/v=2.985\times10^{-6}\ \mathrm s$, so $P=\exp(-\tau/T_\mu)=\exp(-2.985/2.197)=\exp(-1.359)=0.257$, which agrees with the result seen from the ground.

Example 55.9: The Lorentz factor needed for survival

A muon is born at an altitude of $h=15\ \mathrm{km}$ and descends straight down at nearly the speed of light, $v\approx c$. How large must $\gamma$ be, at least, for half ($50\%$) or more of such muons to reach the ground? Take $T_\mu=2.197\ \mathrm{\mu s}$ and $c=2.998\times10^8\ \mathrm{m/s}$.

Solution. Setting $v\approx c$ in the formula of Example 55.8, we have $P=\exp(-h/(\gamma cT_\mu))\ge\dfrac12$. Taking the natural logarithm of both sides gives $-h/(\gamma cT_\mu)\ge-\ln2$, that is,

(Here $cT_\mu=2.998\times10^{8}\times2.197\times10^{-6}=658.7\ \mathrm m$. Also, $\ln2=0.6931$ is the number that appears in the half-life $T_{1/2}=T_\mu\ln2$. One may read this as saying that half of the muons survive when the proper time of the descent equals the half-life.) We need $\gamma\gtrsim33$. In this case, the total energy of the muon (given by $E=\gamma mc^2$, which we study in Chapter 57) is $105.658\ \mathrm{MeV}\times33\approx3.5\ \mathrm{GeV}$ or more. Many actual cosmic-ray muons have energies of several $\mathrm{GeV}$, which is consistent with $\gamma$ having a value of about this size.

Illustration 55.2: When the light of rubidium atoms recedes

Rubidium atoms emit light of wavelength $\lambda_0=7.947\times10^{-7}\ \mathrm m$. When this light recedes from a stationary observer at a speed of $1/10$ of the speed of light, what wavelength $\lambda$ does the light have as seen by the observer?

Solution. The wavelength in the coordinate system of the source is $\lambda_0$, and the receding speed is $v=c/10$ ($\beta=0.1$). From the second equation of Eq. \eqref{eq:55-doppler}, the wavelength measured by the observer is

Therefore,

Light of $795\ \mathrm{nm}$ (near infrared) in the coordinate system of the source appears as light of $879\ \mathrm{nm}$, a wavelength about $10.6\%$ longer. The redshift parameter is $z=(\lambda-\lambda_0)/\lambda_0=0.1055$. In terms of frequency, $\nu_0=c/\lambda_0=3.77\times10^{14}\ \mathrm{Hz}$ drops to $\nu=\sqrt{0.9/1.1}\,\nu_0=0.9045\,\nu_0=3.41\times10^{14}\ \mathrm{Hz}$.

Example 55.10: Finding the receding speed from the redshift parameter

The wavelength of a spectral line of the light emitted by a certain galaxy is observed stretched to $1.10$ times the value measured in a laboratory on the ground ($z=0.10$). At what multiple of the speed of light is this galaxy receding from Earth? Compare with the value obtained from the classical Doppler formula $\lambda'=\lambda(1+\beta)$.

Solution. From Eq. \eqref{eq:55-doppler}, $\lambda'/\lambda=1+z=\sqrt{(1+\beta)/(1-\beta)}$. Squaring both sides gives $(1+z)^2=(1+\beta)/(1-\beta)$. Clearing the denominator and solving for $\beta$:

Substituting $z=0.10$, we get $\beta=(1.21-1)/(1.21+1)=0.21/2.21=0.0950$. That is, the galaxy is receding at about $9.5\%$ of the speed of light ($2.85\times10^{4}\ \mathrm{km/s}$). The classical formula $z=\beta$ gives $\beta=0.10$, so there is a difference of about $5\%$. Note that the redshift of very distant galaxies is due mainly not to the speed at which a galaxy moves through space, but to the stretching of the wavelength of light by the expansion of cosmic space itself (cosmological redshift), so for large $z$ this formula cannot be read directly as a “velocity.”

Example 55.11: Magnitude of the Doppler shift in Mössbauer spectroscopy

A $^{57}\mathrm{Fe}$ $\gamma$-ray source with $E_0=14.4\ \mathrm{keV}$ is moved toward an absorber at speed $v=1.0\ \mathrm{mm/s}$. By how much is the observed energy shifted? Also, is it necessary to use the relativistic formula (Eq. \eqref{eq:55-doppler-blue})? Take $c=2.998\times10^8\ \mathrm{m/s}$ and the width of the absorption line to be $5\ \mathrm{neV}$.

Solution. $\beta=v/c=1.0\times10^{-3}/(2.998\times10^{8})=3.34\times10^{-12}$, which is extremely small. Expanding Eq. \eqref{eq:55-doppler-blue} in $\beta$, we have $\sqrt{(1+\beta)/(1-\beta)}=1+\beta+\beta^2/2+\cdots$ (see University Mathematics, Section 4.2), so the term of order $\beta^2\sim10^{-23}$ can be neglected, and $\nu'\approx\nu(1+\beta)$. Since the energy is proportional to the frequency ($E=h\nu$),

This shift is about 10 times the width of the absorption line (about $5\ \mathrm{neV}$), so if the source is moved at a speed of the order of $1\ \mathrm{mm/s}$, one can scan across the whole absorption line. The relativistic correction (the $\beta^2/2$ term) is of the order of $10^{-23}$ and is entirely negligible; the first-order Doppler effect $\Delta E=E_0\,v/c$ alone is sufficient.

Illustration 55.3: Watching the clock of a rocket that departed from Earth through a telescope

At time $t=0$ (0:00), a rocket departed from Earth with velocity $v$. After some time had passed, we looked at the rocket through a telescope from Earth and saw that the rocket's clock read $t'=1$:00. What is the time $t$ on Earth at this moment? Also find the value for the case $v=0.6c$.

Solution. (1) Relation between Earth's position and time. Substitute Earth's position $x(t)=0$ into the Lorentz transformation, Eq. \eqref{eq:55-lorentz}. Since the origins were made to coincide at the moment of departure ($t=t'=0$), an event on Earth at time $t$ is, in frame S′,

These are the same as Eqs. \eqref{eq:55-dop-t} and \eqref{eq:55-dop-x}. That is, seen from frame S′ (the rocket), Earth is receding at speed $v$ in the $-x'$ direction.

(2) The event E at which the light departs. When the rocket's clock reads $t'_{\mathrm E}=3600\ \mathrm s$ (1 hour), the rocket is at the origin $x'=0$ of frame S′. At this moment (at the same time in frame S′), Earth is, by Eq. \eqref{eq:55-rocket-earth}, at the position $x'=-v\,t'_{\mathrm E}=-3600\,v$. At the time $t'_{\mathrm E}$ of frame S′, the rocket and Earth are separated by a distance $3600\,v\ [\mathrm m]$ (with $v$ measured in $\mathrm{m/s}$; this is the magnitude of the distance between Earth and the rocket in frame S′).

(3) The chase of the light. The light emitted at E travels in frame S′ in the $-x'$ direction at speed $c$, so the position of the light at time $t'$ is $x'_{\text{light}}(t')=-c\,(t'-t'_{\mathrm E})$. The position of Earth is $x'_{\text{Earth}}(t')=-v\,t'$. The light reaches Earth at the instant $t'=t'_{\mathrm R}$ when the two positions are equal:

Here $c-v\gt 0$ because $v\lt c$: since Earth flees more slowly than the light, it is eventually overtaken by the light.

(4) Converting to the reading of Earth's clock. Since R is an event that occurs on Earth ($x=0$), $t'=\gamma t$ of Eq. \eqref{eq:55-rocket-earth} can be used. If $t$ is the Earth time of R, then $t'_{\mathrm R}=\gamma t$, so

Here we used $c\sqrt{1-v^2/c^2}=\sqrt{c^2-v^2}=\sqrt{(c-v)(c+v)}$. Thus, the time on Earth at which we see through the telescope that “the rocket's clock reads $3600\ \mathrm s$” is

For $v=0.6c$ ($\beta=0.6$), $\sqrt{(1+0.6)/(1-0.6)}=\sqrt{1.6/0.4}=\sqrt4=2$, so $t=2\times3600\ \mathrm s=7200\ \mathrm s$, that is, 2 hours after departure (2:00).

Example 55.12: The clock we see and the actual clock when Earth's time is 2 hours

For a rocket receding from Earth at $v=0.8c$, at the instant when $2$ hours have elapsed on Earth's clock since departure, find (a) the value of the rocket's clock seen through a telescope (the value at the time the light was emitted), and (b) the value that the rocket's clock actually shows at that same instant (simultaneous in Earth's frame).

Solution. From $\beta=0.8$, $\gamma=1/\sqrt{1-0.64}=1/0.6=5/3$. (a) Using Eq. \eqref{eq:55-rocket-result} in reverse, the value we see is $\tau=t\sqrt{(1-\beta)/(1+\beta)}=2\ \mathrm h\times\sqrt{0.2/1.8}=2\ \mathrm h\times\dfrac13=40\ \mathrm{min}$. (b) At the same time $t=2\ \mathrm h$ in Earth's frame, the rocket's clock shows, by time dilation, $t/\gamma=2\ \mathrm h\times0.6=72\ \mathrm{min}$. Therefore, the telescope shows $40$ minutes, but at the same instant the rocket's clock actually reads $72$ minutes. The difference of $32$ minutes corresponds to the amount by which the rocket's clock advanced during the time the light took to travel from the rocket to Earth (the Earth time at which the light was emitted is $t_{\mathrm E}=t/(1+\beta)=2/1.8=1.11\ \mathrm h$, and the rocket was then at the position $x_{\mathrm E}=0.8c\times1.11\ \mathrm h=0.89$ light-hours).

Example 55.13: Composing 0.6c and 0.6c

A spaceship (frame S′) traveling at $v=0.6c$ as seen from the ground (frame S) fires a missile at speed $0.6c$, as seen from the spaceship, in the direction of travel. Find the speed of the missile as seen from the ground. Compare it with the value obtained by simple addition in high school physics.

Solution. In frame S′ the velocity is $u'_x=0.6c$ ($u'_y=u'_z=0$), and the velocity of S′ relative to S is $v=0.6c$, so the velocity in frame S is, by the first equation of the inverse transformation, Eq. \eqref{eq:55-vel-inverse},

The value from simple addition, $0.6c+0.6c=1.2c$, exceeds the speed of light, but the correct composition law gives $0.882c$, which does not exceed $c$. Incidentally, this agrees with the result obtained from the product of the matrices $\Lambda(0.6)$ in Exercise 54.8 of the previous chapter: the product $\Lambda(0.6)\Lambda(0.6)$ gives $\beta=15/17$.

Example 55.14: Transformation of a velocity perpendicular to the direction of motion

In frame S, a particle moves in the $y$ direction at speed $u_y=0.5c$ ($u_x=u_z=0$). Frame S′ moves in the $x$ direction at $v=0.6c$ relative to S. Find the velocity components $u'_x,u'_y$ and the speed $u'$ of the particle in frame S′. Also verify Eq. \eqref{eq:55-speed-limit} of Theorem 55.5.

Solution. From $\beta=0.6$ we get $\gamma=1.25$, and $1-vu_x/c^2=1-0=1$, so by Eq. \eqref{eq:55-vel-transform},

The speed is $u'=\sqrt{(0.6)^2+(0.4)^2}\,c=\sqrt{0.52}\,c=0.721\,c\lt c$. The speed in frame S was $u=0.5c$. ($u_y$ becomes smaller by the factor $1/\gamma=0.8$, and in its place a component $-v$ in the $x'$ direction is added.) Checking Theorem 55.5: the left-hand side is $c^2-u'^2=(1-0.52)c^2=0.48c^2$, and the right-hand side is $(1-0.36)(1-0.25)c^2/1^2=0.64\times0.75\,c^2=0.48\,c^2$, so the two agree.

Example 55.15: Light traveling through flowing water (Fizeau's experiment)

Water with refractive index $n=1.33$ flows at speed $v=5.0\ \mathrm{m/s}$ relative to the laboratory (frame S). Find the speed $u_x$, as seen from the laboratory, of light traveling in the same direction as the flow of the water. In the coordinate system S′ moving with the water, the speed of light is $u'_x=c/n$. Take $c=2.998\times10^8\ \mathrm{m/s}$.

Solution. The speed of the coordinate system S′ moving with the water is the flow speed of the water, $v=5.0\ \mathrm{m/s}$, and the speed of light in S′ is $u'_x=c/n=2.254\times10^{8}\ \mathrm{m/s}$. By the inverse transformation (Eq. \eqref{eq:55-vel-inverse}),

Since $v\ll c$, we can take $vu'_x/c^2\ll1$ and use $1/(1+x)\approx1-x$, which gives

(the $v^2$ term was neglected, and $u'_x=c/n$). The part dragged along by the water is $v(1-1/n^2)=5.0\times(1-1/1.33^2)=5.0\times0.4347=2.17\ \mathrm{m/s}$, which is only about $43\%$ of the Galilean addition ($v=5.0\ \mathrm{m/s}$). The quantity $1-1/n^2$ is called the “Fresnel drag coefficient”; Fizeau measured the speed of light in flowing water with an interferometer in 1851 and showed that the result agrees with this value. This is one of the experiments that showed, even before relativity was born, that the addition given by the Galilean transformation does not hold for light.

Example 55.16: Speed of a GPS satellite and the delay due to special relativity

For a GPS satellite in a circular orbit of radius $r=2.656\times10^{7}\ \mathrm m$, find (a) the speed $v$ and (b) the time by which the satellite's clock falls behind the clock on the ground in one day ($86\,400\ \mathrm s$), taking only the effect of special relativity into account. Take $GM=3.986\times10^{14}\ \mathrm{m^3/s^2}$ and $c=2.998\times10^{8}\ \mathrm{m/s}$.

Solution. (a) The centripetal force of the circular motion is universal gravitation, so $mv^2/r=GMm/r^2$ gives $v=\sqrt{GM/r}$. Substituting the numbers, $GM/r=3.986\times10^{14}/2.656\times10^{7}=1.5008\times10^{7}\ \mathrm{m^2/s^2}$ and $v=3.874\times10^{3}\ \mathrm{m/s}$ (about $3.87\ \mathrm{km/s}$). (b) By Eq. \eqref{eq:55-gps-sr}, the fractional delay is $v^2/(2c^2)$, and since $c^2=8.988\times10^{16}\ \mathrm{m^2/s^2}$,

The delay in one day is $8.35\times10^{-11}\times86\,400\ \mathrm s=7.2\times10^{-6}\ \mathrm s=7.2\ \mathrm{\mu s}$. (Even if $\gamma$ is computed exactly, for $\beta^2=1.67\times10^{-10}$ the terms of second and higher order in the Taylor expansion are of order $\beta^4\sim10^{-20}$, so the error of the approximation above is negligible.)

Example 55.17: Advance due to gravity, the total, and the error in position

For the satellite of Example 55.16, find (a) the advance per day due to general relativity, (b) the total together with the effect of special relativity, and (c) the error in distance, $c\,\Delta t$, after one day if no correction is made. Take the radius of the Earth to be $R=6.378\times10^{6}\ \mathrm m$.

Solution. (a) By Eq. \eqref{eq:55-gps-gr}, the fraction is the product of $GM/c^2=3.986\times10^{14}/8.988\times10^{16}=4.435\times10^{-3}\ \mathrm m$ and $1/R-1/r=1/(6.378\times10^{6})-1/(2.656\times10^{7})=1.568\times10^{-7}-3.765\times10^{-8}=1.191\times10^{-7}\ \mathrm{m^{-1}}$, namely $4.435\times10^{-3}\times1.191\times10^{-7}=5.28\times10^{-10}$. In one day the clock gains $5.28\times10^{-10}\times86\,400\ \mathrm s=4.56\times10^{-5}\ \mathrm s=45.6\ \mathrm{\mu s}$. (b) The total is $+45.6-7.2=+38.4\ \mathrm{\mu s}$ (gains time). (c) $c\,\Delta t=2.998\times10^{8}\ \mathrm{m/s}\times38.4\times10^{-6}\ \mathrm s=1.15\times10^{4}\ \mathrm m\approx11.5\ \mathrm{km}$. Without correction, an error of about $11.5\ \mathrm{km}$ arises in one day.

Exercise 55.1: Values of the Lorentz contraction

For the length ratio of a moving rod, $\ell'/\ell=\sqrt{1-(v/c)^2}$, find its value (a) for $v=0.5c$ and (b) for $v=0.9c$. (c) For what value of $v$ does the length shrink to exactly half? (d) For what value of $1-\beta$ (when $\beta$ is very close to 1) does a ruler of length $1\ \mathrm m$ appear to be $1\ \mathrm{mm}$ long?

Hint: For (c) and (d), solve $\sqrt{1-\beta^2}=\ell'/\ell$ for $\beta$. For (d), use $1-\beta^2=(1-\beta)(1+\beta)\approx2(1-\beta)$.

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Exercise 55.2: The youthfulness of a space traveler

A spaceship flying at the constant speed $0.8c$ as seen from Earth leaves Earth and arrives at a star $4$ light-years away (the distance light travels in 4 years). (a) How many years does the trip take by Earth's clocks? (b) How many years does it take by the spaceship's clock (use time dilation)? (c) From the spaceship's standpoint, calculate the distance between Earth and the star using Lorentz contraction, and verify that you get the same answer as in (b). (Ignore the periods of acceleration and deceleration.)

Hint: (b) $\Delta\tau=\Delta t/\gamma$. (c) As seen from the spaceship, the two bodies approach at speed $0.8c$.

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Exercise 55.3: The low-speed limit of the Doppler effect

(a) Show, by Taylor expansion (to first order in $\beta$), that the relativistic Doppler formula $\nu'=\sqrt{(1-\beta)/(1+\beta)}\,\nu$ becomes $\nu'\approx\nu\,(1-\beta)$ for $\beta\ll1$. (b) For $\beta=0.01$, find the difference between the exact value $\sqrt{(1-\beta)/(1+\beta)}$ and the approximate value $1-\beta$. (c) Expand up to the second-order terms, and verify that the difference in (b) is explained by approximately $\beta^2/2$.

Hint: $(1-\beta)^{1/2}\approx1-\beta/2$ and $(1+\beta)^{-1/2}\approx1-\beta/2$. To second order, $(1-\beta)^{1/2}(1+\beta)^{-1/2}=(1-\beta/2-\beta^2/8)(1-\beta/2+3\beta^2/8)$.

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Exercise 55.4: The rocket problem ($v=0.8c$)

A rocket leaves Earth at $v=0.8c$ (at the instant of departure, the rocket's clock and Earth's clock both read 0:00). (a) Find the reading of Earth's clock that is simultaneous, in Earth's frame, with the instant when the rocket's clock points to 1:00. (b) At what time by Earth's clock is the rocket's clock seen through a telescope to point to 1:00? (c) Verify that the difference between (a) and (b) (the time light takes to travel from the rocket to Earth) equals the distance between the rocket and Earth at that moment divided by $c$.

Hint: (a) $t=\gamma\tau$. (b) $t=\sqrt{(1+\beta)/(1-\beta)}\ \tau$. (c) The position of the rocket is $x=vt$.

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Exercise 55.5: Composing velocities in two steps

(a) An inertial frame S′ moves relative to S in the $+x$ direction with speed $v=\beta_1c$, and an object moves inside S′ in the $+x'$ direction with speed $u'_x=\beta_2c$. Express the object's speed $u_x=\beta c$ in frame S in terms of $\beta_1,\beta_2$. (b) Show that $1-\beta^2=\dfrac{(1-\beta_1^2)(1-\beta_2^2)}{(1+\beta_1\beta_2)^2}$, and explain why $\beta\lt 1$ whenever $0\le\beta_1,\beta_2\lt 1$. (c) Find $\beta$ for $\beta_1=\beta_2=0.9$.

Hint: (a) Use the first equation of the inverse transformation (Eq. \eqref{eq:55-vel-inverse}). (b) Factor $(1+\beta_1\beta_2)^2-(\beta_1+\beta_2)^2$.

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Exercise 55.6: A muon with speed $0.98c$

For a muon heading toward the ground with speed $v=0.98c$ (mean lifetime at rest $T_\mu=2.197\ \mathrm{\mu s}$), find (a) $\gamma$ and the mean lifetime as seen from the ground, (b) the mean flight distance $L$ as seen from the ground, and (c) the fraction of muons born at an altitude of $12\ \mathrm{km}$ that reach the ground without decaying. (d) From the muon's standpoint, calculate the thickness of the atmosphere and the descent time, and verify that the fraction is the same as in (c). Take $c=2.998\times10^{8}\ \mathrm{m/s}$.

Hint: See Example 55.8. The surviving fraction is $\exp(-\text{proper time of descent}/T_\mu)$.

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Exercise 55.7: The clock on the International Space Station

For a clock on the International Space Station, which orbits Earth in a circular orbit (orbital radius $r=6.778\times10^{6}\ \mathrm m$, altitude about $400\ \mathrm{km}$), find (a) the speed $v$, (b) the delay per day due to special relativity, (c) the gain per day due to general relativity (Eq. \eqref{eq:55-gps-gr}), and (d) the net result. Take the radius of Earth's surface to be $R=6.378\times10^{6}\ \mathrm m$, $GM=3.986\times10^{14}\ \mathrm{m^3/s^2}$, and $c=2.998\times10^{8}\ \mathrm{m/s}$. Compared with a GPS satellite, which of the two effects is the stronger for the space station?

Hint: Follow the same procedure as in Examples 55.16 and 55.17, changing only the numbers.

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Exercise 55.8: Apparent superluminal motion (the difference in light-travel times)

A blob of gas ejected from the center of a distant galaxy flies at the constant speed $\beta c$ in a direction tilted by an angle $\theta$ from the direction toward the observer (the line-of-sight direction); $\theta=0$ is the direction heading straight toward the observer. The blob emits light signals at times $t=0$ and $t=\Delta t$ (measured by the clock of the frame of the galaxy from which the blob departed). Assuming that the observer is far away (distance $D\gg\beta c\,\Delta t$), (a) express the difference $\Delta t_{\mathrm{obs}}$ between the times at which the two signals reach the observer in terms of $\Delta t,\beta,\theta$. (b) The observer sees the blob as having moved by $\beta c\,\Delta t\sin\theta$ in the direction perpendicular to the line of sight. Find its “apparent speed” $\beta_{\mathrm{app}}c$ (the displacement perpendicular to the line of sight divided by $\Delta t_{\mathrm{obs}}$). (c) Find $\beta_{\mathrm{app}}$ for $\beta=0.98$ and $\theta=10^\circ$. (d) Show that $\beta_{\mathrm{app}}$ is largest when $\cos\theta=\beta$ and that its maximum value is $\gamma\beta$. Also explain why $\beta_{\mathrm{app}}$ can exceed 1 if $\beta\gt1/\sqrt2$, and state that this does not contradict relativity.

Hint: Use the approach of Section 55.5 (take into account the time the light needs to arrive). (a) The position at which the second signal is emitted is nearer to the observer than the position of the first by $\beta c\,\Delta t\cos\theta$. (d) Differentiate $\beta_{\mathrm{app}}(\theta)$ with respect to $\theta$.

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Exercise 55.9: A fast-moving light clock

A light clock has a mirror separation of $d=0.30\ \mathrm m$. (a) Find its period (proper time) $\Delta\tau$ when it is at rest. (b) Find the period $T$ when this clock is seen moving at $v=0.8c$ in the direction parallel to the faces of the mirrors. (c) Find the distance $cT$ that the light travels during one round trip in this case, and verify that it satisfies the Pythagorean theorem $(cT/2)^2=d^2+(vT/2)^2$. (d) Suppose that Q, who is observing this clock, has a light clock of identical construction (at rest relative to Q). How long does the period of Q's clock appear to be as seen by P, the person who moved the clock? Take $c=3.00\times10^{8}\ \mathrm{m/s}$.

Hint: (b) $T=\gamma\,\Delta\tau$. (d) By the principle of relativity, P and Q are on an equal footing.

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Exercise 55.10: The color of an approaching light source

A light source emitting light of wavelength $500\ \mathrm{nm}$ approaches a stationary observer at $v=0.5c$. (a) Find the frequency ratio $\nu'/\nu$ measured by the observer and the wavelength $\lambda'$. (b) What wavelength results if we use the classical formula that does not include time dilation, $\nu'/\nu=1/(1-\beta)$? Find the difference from (a). (c) Find the wavelength when the source recedes at the same speed $0.5c$. Take $c=2.998\times10^{8}\ \mathrm{m/s}$.

Hint: (a) Use the formula for the approaching case, Eq. \eqref{eq:55-doppler-blue}. (b) $\lambda'=\lambda/(\nu'/\nu)$. (c) Use the formula for the receding case, Eq. \eqref{eq:55-doppler}.

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