fisica

1. Chapter 39
Relativity
Conceptual Problems
1 • The approximate total energy of a particle of mass m moving at speed
u << c is (a)mc2
+ 1
2
mu2
, (b) 1
2
mu2
, (c) cmu, (d) mc2
, (e) 1
2
cmu .
Determine the Concept The total relativistic energy E of a particle is defined to
be the sum of its kinetic and rest energies.
The sum of the kinetic and rest
energies of a particle is given by:
22
2
12
mcmumcKE +=+=
and )(a is correct.
Estimation and Approximation
7 •• The most distant galaxies that can be seen by the Hubble telescope are
moving away from us and have a redshift parameter of about z = 5. (The redshift
parameter z is defined as (f – f’)/f’, where f is the frequency measured in the rest
frame of the emitter, and f’ is the frequency measured in the rest frame of the
receiver.) (a) What is the speed of these galaxies relative to us (expressed as a
fraction of the speed of light)? (b) Hubble’s law states that the recession speed is
given by the expression v = Hx, where v is the speed of recession, x is the
distance, and H, the Hubble constant, is equal to 75 km/s/Mpc , where
1 pc = 3.26 c⋅y. (The abbreviation for parsec is pc.) Estimate the distance of such
a galaxy from us using the information given.
Picture the Problem (a) We can use the definition of the redshift parameter and
the relativistic Doppler shift equation to show that, for light that is Doppler-
shifted with respect to an observer, ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
−
=
1
1
2
2
u
u
cv , where u = z + 1, and to find the
ratio of v to c. In Part (b) we can solve Hubble’s law for x and substitute our result
from Part (a) to estimate the distance to the galaxy.
(a) The red-shift parameter is defined
to be: f'
'ff
z
−
= 0
The relativistic Doppler shift for
recession is given by:
cv
cv
f'f
+
−
=
1
1
0
37

2. Chapter 3938
Substitute for f ′ and simplify to
obtain:
1
1
1
1
1
1
11
0
00
−
−
+
=
+
−
+
−
=
cv
cv
cv
cv
f
cv
cv
ff
z
Letting u = z + 1 and simplifying
yields:
cv
cv
zu
−
+
=+=
1
1
1 ⇒
1
1
2
2
+
−
=
u
u
c
v
Substitute for u to express v/c as a
function of z:
( )
( ) 11
11
2
2
++
−+
=
z
z
c
v
Substituting the numerical value of z
and evaluating v/c gives:
( )
( )
946.0
115
115
2
2
=
++
−+
=
c
v
(b) Solving Hubble’s law for x
yields: H
v
x =
Substitute numerical values and
evaluate x:
yG3.12
Mpc
y103.26
Mpc
km/s
75
946.0946.0 6
⋅=
⋅×
×==
c
cc
H
c
x
Time Dilation and Length Contraction
13 •• Unobtainium (Un) is an unstable particle that decays into normalium
(Nr) and standardium (St) particles. (a) An accelerator produces a beam of Un
that travels to a detector located 100 m away from the accelerator. The particles
travel with a velocity of v = 0.866c. How long do the particles take (in the
laboratory frame) to get to the detector? (b) By the time the particles get to the
detector, half of the particles have decayed. What is the half-life of Un? (Note:
half-life as it would be measured in a frame moving with the particles) (c) A new
detector is going to be used, which is located 1000 m away from the accelerator.
How fast should the particles be moving if half of the particles are to make it to
the new detector?

3. Relativity 39
Picture the Problem The time required for the particles to reach the detector, as
measured in the laboratory frame of reference is the ratio of the distance they
must travel to their speed. The half life of the particles is the trip time as measured
in a frame traveling with the particles. We can find the speed at which the
particles must move if they are to reach the more distant detector by equating their
half life to the ratio of the distance to the detector in the particle’s frame of
reference to their speed.
(a) The time required to reach the
detector is the ratio of the distance to
the detector and the speed with
which the particles are traveling:
c
x
v
x
t
866.0
Δ
=
Δ
=Δ
Substitute numerical values and
evaluate Δt: ( )
s385.0
m/s10998.2866.0
m100
Δ 8
μ=
×
=t
(b) The half life is the trip time as
measured in a frame traveling with
the particles:
2
1 ⎟
⎠
⎞
⎜
⎝
⎛
−Δ=
Δ
=Δ
c
v
t
t
t'
γ
Substitute numerical values and
evaluate Δt′: ( )
s193.0
866.0
1s385.0
2
μ
μ
=
⎟
⎠
⎞
⎜
⎝
⎛
−=Δ
c
c
t'
(c) In order for half the particles to
reach the detector:
v
c
v
x'
v
x'
t'
2
1 ⎟
⎠
⎞
⎜
⎝
⎛
−Δ
=
Δ
=Δ
γ
where Δx′ is the distance to the new
detector.
Rewrite this expression to obtain:
t'
x'
c
v
v
Δ
Δ
=
⎟
⎠
⎞
⎜
⎝
⎛
−
2
1
Squaring both sides of the equation
yields:
2
2
2
1
⎟
⎠
⎞
⎜
⎝
⎛
Δ
Δ
=
⎟
⎠
⎞
⎜
⎝
⎛
−
t'
x'
c
v
v

4. Chapter 3940
Substitute numerical values for Δx′
and Δt′ and simplify to obtain: ( )2
2
2
2
3.17
s193.0
m1000
1
c
c
v
v
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
⎟
⎠
⎞
⎜
⎝
⎛
−
μ
Divide both sides of the equation by
c2
to obtain: ( )2
2
2
2
3.17
1
=
⎟
⎠
⎞
⎜
⎝
⎛
−
c
v
c
v
Solving this equation for v2
/c2
gives: ( )
( )
9967.0
3.171
3.17
2
2
2
2
=
+
=
c
v
Finally, solving for v yields: cv 998.0=
The Lorentz Transformation, Clock Synchronization, and
Simultaneity
17 •• A spaceship of proper length Lp = 400 m moves past a transmitting
station at a speed of 0.760c. (The transmitting station broadcasts signals that
travel at the speed of light.) A clock is attached to the nose of the spaceship and a
second clock is attached to the transmitting station. The instant that the nose of
the spaceship passes the transmitter, the clock attached to the transmitter and the
clock attached to the nose of the spaceship are set equal to zero. The instant that
the tail of the spaceship passes the transmitter a signal is sent by the transmitter
that is subsequently detected by a receiver in the nose of the spaceship. (a) When,
according to the clock attached to the nose of spaceship, is the signal sent?
(b) When, according to the clocks attached to the nose of spaceship, is the signal
received? (c) When, according to the clock attached to the transmitter, is the
signal received by the spaceship? (d) According to an observer that works at the
transmitting station, how far from the transmitter is the nose of the spaceship
when the signal is received?
Picture the Problem Let S be the reference frame of the spaceship and S′ be that
of Earth (transmitter station). Let event A be the emission of the light pulse and
event B the reception of the light pulse at the nose of the spaceship. In (a) and (c)
we can use the classical distance, rate, and time relationship and in (b) and (d) we
can apply the inverse Lorentz transformations.
(a) In both S and S′ the pulse travels
at the speed c. Thus:
s76.1
0.760c
m400p
A μ===
v
L
t

5. Relativity 41
(b) The inverse time transformation
is:
⎟
⎠
⎞
⎜
⎝
⎛
−= 2B
c
vx
t't γ
where
( )
54.1
760.0
1
1
1
1
2
2
2
2
=
−
=
−
=
c
c
c
v
γ
Substitute numerical values and
evaluate :'tB
( ) ( )( )
( ) ( )(
( )
)
s76.4
m/s10998.2
m400760.0
s09.354.1
m400760.0
s09.354.1
28
2B
μ
μ
μ
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×
−
−=
⎟
⎠
⎞
⎜
⎝
⎛ −
−=
c
c
't
(c) The elapsed time, according to
the clock on the ship is:
A
shipoflength
traveltopulseB ttt +=
Find the time of travel of the pulse to
the nose of the ship:
s33.1
m/s102.998
m400
8
shipoflength
traveltopulse
μ=
×
=t
Substitute numerical values and
evaluate tB:
s09.3s76.1s33.1B μμμ =+=t
(d) The inverse transformation for x
is:
( )vtxx' −= γ
Substitute numerical values and evaluate x′:
( ) ( )( )( )[ ] km70.1s1009.3m/s10998.2760.0m40054.1 68
=××−−= −
x'
The Relativistic Doppler Shift
27 •• A clock is placed in a satellite that orbits Earth with an orbital period
of 90 min. By what time interval will this clock differ from an identical clock on
Earth after 1.0 y? (Assume that special relativity applies and neglect general
relativity.)
Picture the Problem Due to its motion, the orbiting clock will run more slowly
than the Earth-bound clock. We can use Kepler’s third law to find the radius of

6. Chapter 3942
the satellite’s orbit in terms of its period, the definition of speed to find the orbital
speed of the satellite from the radius of its orbit, and the time dilation equation to
find the difference δ in the readings of the two clocks.
Express the time δ lost by the clock:
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−Δ=
Δ
−Δ=Δ−Δ=
γγ
δ
1
1p t
t
ttt
Because v << c, we can use Part (b)
of Problem 10: 2
2
2
1
1
1
c
v
−≈
γ
Substitute to obtain:
t
c
v
c
v
t Δ=⎥
⎦
⎤
⎢
⎣
⎡
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−−Δ= 2
2
2
2
2
1
2
1
11δ (1)
Express the square of the speed of
the satellite in its orbit: 2
222
2 42
T
r
T
r
v
ππ
=⎟
⎠
⎞
⎜
⎝
⎛
= (2)
where T is its period and r is the radius
of its (assumed) circular orbit.
Use Kepler’s third law to relate the
period of the satellite to the radius of
its orbit about Earth:
3
2
E
2
3
E
2
2 44
r
gR
r
GM
T
ππ
== ⇒ 3
2
22
E
4π
TgR
r =
Substitute numerical values and evaluate r:
( )( ) ( ) m1065.6
4
s/min60min90km6370m/s81.9 63
2
222
×=
×
=
π
r
Substitute numerical values in
equation (2) and evaluate v2
:
( )
( )
227
2
262
2
s/m1099.5
s/min60min90
m1065.64
×=
×
×
=
π
v
Finally, substitute for v2
in equation (1) and evaluate δ:
( )( )
( )
ms11
m/s10998.2
Ms/y31.56y0.1s/m1099.5
2
1
28
227
=
×
××
=δ
31 •• A particle moves with speed 0.800c in the + ′′x direction along the ′′x
axis of frame , which moves with the same speed and in the same direction
along the x′ axis relative to frame S′. Frame S′ moves with the same speed and in
′′S

7. Relativity 43
the same direction along the x axis relative to frame S. (a) Find the speed of the
particle relative to frame S′. (b) Find the speed of the particle relative to frame S.
Picture the Problem We can apply the inverse velocity transformation equation
to express the speed of the particle relative to both frames of reference.
(a) Express in terms of'ux :''ux
2
1
c
''vu
v''u
'u
x
x
x
+
+
=
where v of S ′, relative to S″, is 0.800c.
Substitute numerical values and
evaluate :'ux ( )
c
c
c
c
cc
'ux
976.0
64.1
60.1
800.0
1
800.0800.0
2
2
=
=
+
+
=
(b) Express ux in terms of :'ux
2
1
c
'vu
v'u
u
x
x
x
+
+
= where v, the speed of S,
relative to S ′, is 0.800c.
( )( )
c
c
c
cc
cc
ux
997.0
781.1
776.1
976.0800.0
1
800.0976.0
2
=
=
+
+
=Substitute numerical values and
evaluate ux:
Relativistic Momentum and Relativistic Energy
37 •• In reference frame S’, two protons, each moving at 0.500c, approach
each other head-on. (a) Calculate the total kinetic energy of the two protons in
frame S′. (b) Calculate the total kinetic energy of the protons as seen in reference
frame S, which is moving with speed 0.500c relative to S′ so that one of the
protons is at rest.
Picture the Problem The total kinetic energy of the two protons in Part (a) is the
sum of their kinetic energies and is given by ( ) 012 EK −= γ . Part (b) differs from
Part (a) in that we need to find the speed of the moving proton relative to frame S.

8. Chapter 3944
(a) The total kinetic energy of the
protons in frame S′ is given by:
( ) 012 EK −= γ
Substitute for γ and E0 and evaluate
K:
( )
( )
MeV290
MeV28.9381
500.0
1
1
2
2
2
=
⎟
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎜
⎝
⎛
−
−
=
c
c
K
(b) The kinetic energy of the moving
proton in frame S is given by:
( ) 01 EK −= γ (1)
where
2
1
1
c
uv
−
=γ
Express the speed u of the proton
in frame S:
2
1
c
'vu
v'u
u
x
x
+
+
=
( )( ) c
c
cc
cc
u 800.0
500.0500.0
1
500.0500.0
2
=
+
+
=
Substitute numerical values and
evaluate u:
Evaluate γ:
( )( )
67.1
800.0800.0
1
1
2
=
−
=
c
cc
γ
( )( )
MeV629
MeV28.938167.1
=
−=KSubstitute numerical values in
equation (1) and evaluate K:
General Problems
45 •• Frames S and S′ are moving relative to each other along the x and x′
axes (which superpose). Observers at rest in the two frames set their clocks to
t = 0 when the two origins coincide. In frame S, event 1 occurs at x1 = 1.0 c⋅y and
t1 = 1.00 y and event 2 occurs at x2 = 2.0 c⋅y and t2 = 0.50 y. These events occur
simultaneously in frame S′. (a) Find the magnitude and direction of the velocity of
S′ relative to S. (b) At what time do both these events occur as measured in S′?

9. Relativity 45
Picture the Problem We can use Equation 39-12, the inverse time transformation
equation, to relate the elapsed times and separations of the events in the two
systems to the velocity of S′ relative to S. We can use this same relationship in
Part (b) to find the time at which these events occur as measured in S′.
(a) Use Equation 39-12 to obtain:
( ) ( )
⎥⎦
⎤
⎢⎣
⎡
Δ−Δ=
⎥⎦
⎤
⎢⎣
⎡
−−−=−=Δ
x
c
v
t
xx
c
v
tt't'tt'
2
1221212
γ
γ
Because the events occur
simultaneously in frame S′,
Δt′ = 0 and:
x
c
v
t Δ−Δ= 2
0 ⇒
x
tc
v
Δ
Δ
=
2
Substitute for Δt and Δx and evaluate
v:
( ) c
cc
c
v 50.0
y1.00y0.2
y1.00y50.02
−=
⋅−⋅
−
=
Because :y50.012 −=−= tttΔ direction.in themoves' xS −
(b) Use the inverse time
transformation to obtain:
2
2
2
2
2
2
2
22
1
c
v
c
vx
t
c
vx
t't
−
−
=⎟
⎠
⎞
⎜
⎝
⎛
−= γ
( )( )
( )
y7.1
50.0
1
y0.250.0
y50.0
2
2
2
12
=
−
−
⋅−
−
==
c
c
c
cc
't't
Substitute numerical values and
evaluate t2′ and t1′:
49 •• Using a simple thought experiment, Einstein showed that there is mass
associated with electromagnetic radiation. Consider a box of length L and mass M
resting on a frictionless surface. Attached to the left wall of the box is a light
source that emits a directed pulse of radiation of energy E, which is completely
absorbed at the right wall of the box. According to classical electromagnetic
theory, this radiation carries momentum of magnitude p = E/c (Equation 32-13).
The box recoils when the pulse is emitted by the light source. (a) Find the recoil
velocity of the box so that momentum is conserved when the light is emitted.
(Because p is small and M is large, you may use classical mechanics.) (b) When
the light is absorbed at the right wall of the box the box stops, so the total

10. Chapter 3946
momentum of the system remains zero. If we neglect the very small velocity of
the box, the time it takes for the radiation to travel across the box is Δt = L/c. Find
the distance moved by the box in this time. (c) Show that if the center of mass of
the system is to remain at the same place, the radiation must carry mass m = E/c2
.
Picture the Problem We can use conservation of energy to express the recoil
velocity of the box and the relationship between distance, speed, and time to find
the distance traveled by the box in time Δt = L/c. Equating the initial and final
locations of the center of mass will allow us to show that the radiation must carry
mass m = E/c2
.
(a) Apply conservation of momentum
to obtain:
0i ==+ pMv
c
E
⇒
Mc
E
v −=
c
vL
tvd =Δ=
(b) The distance traveled by the
box in time Δt = L/c is:
Substitute for v from (a) to obtain:
2
Mc
LE
Mc
E
c
L
d −=⎟
⎠
⎞
⎜
⎝
⎛
−=
mM
mL
x
+
−
= 2
1
CM
(c) Let x = 0 be at the center of the
box and let the mass of the photon be
m. Then initially the center of mass
is at:
When the photon is absorbed at the
other end of the box, the center of
mass is at: mM
Mc
EL
Lm
Mc
MEL
x
+
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
−+
−
=
22
1
2
CM
Because no external forces act on the
system, these expressions for xCM
must be equal: mM
Mc
EL
Lm
Mc
MEL
mM
mL
+
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
−+
−
=
+
− 22
1
2
2
1
Solving for m yields:
⎟
⎠
⎞
⎜
⎝
⎛
−
=
2
2
1
Mc
E
c
E
m
Because Mc2
is of the order of 1016
J
and E = hf is of the order of 1 J for
reasonable values of f, E/Mc2
<< 1
and:
2
c
E
m =

11. Relativity 47
51 ••• When a moving particle that has a kinetic energy greater than the
threshold kinetic energy Kth strikes a stationary target particle, one or more
particles may be created in the inelastic collision. Show that the threshold kinetic
energy of the moving particle is given by
Kth =
Σmin + Σmfin( ) Σmfin + Σmin( )c2
2mtarget
Here Σmin is the sum of the masses of the particles prior to the collision, Σmfin is
the sum of the masses of the particles following the collision, and mtarget is the
mass of the target particle. Use this expression to determine the threshold kinetic
energy of protons incident on a stationary proton target for the production of a
proton–antiproton pair; compare your result with the result of Problem 38.
Picture the Problem Let mi denote the mass of the incident (projectile) particle.
Then ∑min = mi + mtarget and we can use this expression to determine the threshold
kinetic energy of protons incident on a stationary proton target for the production
of a proton–antiproton pair.
Consider the situation in the center
of mass reference frame. At
threshold we have:
∑=− 2
fin
222
cmcpE
Note that this is a relativistically
invariant expression.
t,0ttarget EEE ==In the laboratory frame, the target is
at rest so:
We can, therefore, write: ( ) ( )22
fin
22
i
2
t,0i ∑=−+ cmcpEE
For the incident particle: 2
i,0
22
i
2
i EcpE =−
and
thi,0i KEE +=
where is the threshold kinetic
energy of the incident particle in the
laboratory frame.
thK
Express in terms of the rest
energies:
thK ( ) ( )22
fint,0th
2
i.0t,0 2 ∑=++ cmEKEE
where
∑=+ 2
fini.0t,0 cmEE
and
2
targett,0 cmE =

12. Chapter 3948
( ) ( 22
fin
2
targetth
22
fin 2 ∑∑ =+ cmcmKcm )Substitute to obtain:
Solving for gives:thK
( )( )
target
2
infinfinin
th
2m
cmmmm
K
∑ ∑∑ ∑ −+
=
For the creation of a proton -
antiproton pair in a proton - proton
collision:
∑ = pin 2mm , ∑ = pfin 4mm and
ptarget mm =
Substituting for the sums and
simplifying yields:
( )( )
( )( ) 2
p
p
2
pp
p
2
pppp
th
6
2
26
2
2442
cm
m
cmm
m
cmmmm
K
==
−+
=
in agreement with Problem 38.
55 ••• For the special case of a particle moving with speed u along the y axis
in frame S, show that its momentum and energy in frame S′ are related to its
momentum and energy in S by the transformation equations
⎟
⎠
⎞
⎜
⎝
⎛
−= 2
'
c
vE
pp xx γ , , , andyy pp ='
zz pp ='
⎟
⎠
⎞
⎜
⎝
⎛
−=
c
vp
c
E
c
E x
γ
'
.
Compare these equations with the Lorentz transformation equations for x′, y′, z′,
and t′. Notice that that the quantities px, py, pz, and E/c transform in the same way
as do x, y, z, and ct.
Picture the Problem We can use the expressions for p
r
and E in S together with
the relations we wish to verify and the inverse velocity transformation equations
to establish the condition ( ) ( ) ( ) 2
2
22222
γ
u
v'u'u'uu' zyx +=++= and then use this
result to verify the given expressions for px′, py′, pz′ and E′/c.
In any inertial frame the momentum
and energy are given by:
2
2
1
c
u
m
−
=
u
p
r
r
and
2
2
2
1
c
u
mc
E
−
=
where u
r
is the velocity of the particle
and u is its speed.

13. Relativity 49
The components of p
r
in S are:
2
2
1
c
u
mu
p x
x
−
= ,
2
2
1
c
u
mu
p
y
y
−
= , and
2
2
1
c
u
mu
p z
z
−
=
Because ux = uz = 0 and uy = u:
0== zx pp and
2
2
1
c
u
mu
py
−
=
Substituting zeros for px and pz in
the relations we are trying to
show yields:
22
0
c
vE
c
vE
'px γγ −=⎟
⎠
⎞
⎜
⎝
⎛
−= , ,yy p'p =
0='pz , and
c
E
c
E
c
E'
γγ =⎟
⎠
⎞
⎜
⎝
⎛
−= 0
In S′ the momentum components
are:
2
2
1
c
u'
'mu
'p x
x
−
= ,
2
2
1
c
u'
'mu
'p
y
y
−
= , and
2
2
1
c
u'
'mu
'p z
z
−
=
The inverse velocity transformations
are:
2
1
c
vu
vu
'u
x
x
x
−
−
= ,
2
1
c
vu
u
'u
y
y
y
−
= , and
2
1
c
vu
u
'u
z
z
z
−
=
Substitute ux = uz = 0 and uy = u to
obtain:
v'ux −= , , andu'uy γ= 0='uz
Thus: ( ) ( ) ( )
2
2
2
2222
γ
u
v
'u'u'uu' zyx
+=
++=

14. Chapter 3950
First we verify that pz′ = pz = 0: ( ) 0
1
0
2
2
==
−
= zz p
c
u'
m
'p
Next we verify that py′ = py:
y
y
y
y
p
c
v
c
u
c
v
c
v
c
u
c
v
p
c
v
c
u
c
v
c
v
c
u
c
u
mu
c
u
c
v
c
u
c
u
mu
c
u
c
v
mu
c
u'
'mu
'p
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−−−
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−−−
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−−−
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
−
=
−−
−
−
=
−−
=
−
=
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
22
2
2
2
2
2
2
2
22
2
2
2
2
2
11
11
11
11
1
1
1
111
γ
γ
γ
γ
Next, we verify that ⎟
⎠
⎞
⎜
⎝
⎛
−= 2
c
vE
p'p xx γ :
E
c
v
c
v
c
u
c
v
c
v
c
u
c
v
E
c
v
c
v
c
u
c
v
c
v
c
u
E
c
v
c
u
c
v
c
u
c
u
mc
c
v
c
u
c
v
mv
c
u'
'mu
'p x
x
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
22
2
2
2
2
2
1
2
2
2
2
22
2
2
2
2
2
11
11
11
11
1
1
111
γ
γγ
γ
γ
γ
γ
γ
−=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−−−
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−−−
−=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−−−
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
−=
−−
−
−
−=
−−
−
=
−
=
−

15. Relativity 51
Finally, we verify that :or, EE'
c
E
c
vp
c
E
c
E' x
γγγ ==⎟
⎠
⎞
⎜
⎝
⎛
−=
E
c
v
c
u
c
v
c
v
c
u
c
v
E
c
v
c
u
c
v
c
v
c
u
E
c
u
c
v
c
u
E
c
u'
c
u
c
u
mc
c
u'
mc
E'
γ
γγ
γ
γ
γ
γ
γ
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−−−
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−−−
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−−−
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
=
−−
−
=
−
−
−
=
−
=
−−
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
22
2
2
2
2
2
1
2
2
2
2
1
2
2
2
2
2
2
11
11
11
11
1
1
1
1
11
The x, y, z, and t transformation
equations are:
( )vtxx' −= γ , yy' = , zz' =
and
⎟
⎠
⎞
⎜
⎝
⎛
−= 2
c
vx
tt' γ
The x, y, z, and ct transformation
equations are: ⎟
⎠
⎞
⎜
⎝
⎛
−= ct
c
v
xx' γ , yy' = , zz' =
and
⎟
⎠
⎞
⎜
⎝
⎛
−= x
c
v
ctct' γ
The px, py, pz, and E/c transformation
equations are: ⎟
⎠
⎞
⎜
⎝
⎛
−=
c
E
c
v
p'p xx γ , ,yy p'p = zz p'p =
and
⎟
⎠
⎞
⎜
⎝
⎛
−= xp
c
v
c
E
c
E'
γ
Note that the transformation equations for x, y, z, and ct and the transformation
equations for px, py, pz, and E/c are identical.