Mecflu - a teoria especial da relatividade

Mecflu - a teoria especial da relatividade

The Special Theory of Relativity

14-1. Substitute Eq. (14.12) into Eqs. (14.9) and (14.10):

From (1)

From (2)

vx c γ


c γ γ or

14-2. We introduce cosh,sinhyyvcαα≅≅ and substitute these expressions into Eqs. (14.14); then cosh sinh

cosh sinh ; x ct x t a x x

Now, if we use cosh α = cos (iα) and i sinh α = sin (iα), we can rewrite (1) as ( ) ( )

cos sin sin cos x i ict i ict x i ict iαα α α

Comparing these equations with the relation between the rotated system and the original system in ordinary three-dimensional space,

cos sin sin cos x x x x x

We can see that (2) corresponds to a rotation of the 1xict− plane through the angle iα.

14-3. If the equation

xi ct xi ct is Lorentz invariant, then in the transformed system we must have

xi ct xi ct where

We can rewrite (2) as


Now, we first determine how the operator 2

We know the following relations:


2xx x xµν µλ µν µλ νλ ν λ xµ νλ ν


x x µν λ µ x∂ ∂µ νλ

Since µ and λ are dummy indices, we see that the operator 2x2µ∂∂∑ is invariant under a Lorentz transformation. So we have

This equation means that the function ψ taken at the transformed point (x′,ict′) satisfies the same equation as the original function ψ (x,ict) and therefore the equation is invariant. In a Galilean transformation, the coordinates become x v t y v t

z zv t

Using these relations, we have x y xt x x t x v t

y v t zy v t

Therefore, xy z

xy z x y z c t xyz c t v v v t vx t v y t v z t

This means that the function ψ (x′,ict′) does not satisfy the same form of equation as does (,)xictψ, and the equation is not invariant under a Galilean transformation.

14-4. In the K system the rod is at rest with its ends at and . The K′ system moves with a velocity v (along the x axis) relative to K. 1x 2x x x

If the observer measures the time for the ends of the rod to pass over a fixed point in the K′ system, we have v t

v t


14-5. The “apparent shape” of the cube is that shape which would be recorded at a certain instant by the eye or by a camera (with an infinitesimally short shutter speed!). That is, we must find the positions that the various points of the cube occupy such that light emitted from these points arrives simultaneously at the eye of the observer. Those parts of the cube that are farther from the observer must then emit light earlier than those parts that are closer to the observer. An observer, looking directly at a cube at rest, would see just the front face, i.e., a square.

When in motion, the edges of the cube are distorted, as indicated in the figures below, where the observer is assumed to be on the line passing through the center of the cube. We also note that the face of the cube in (a) is actually bowed toward the observer (i.e., the face appears convex), and conversely in (b).

(a)Cube moving toward the observer. (a)Cube moving away from the observer.


x x

We transform the time t at the points and in the K system into the K′ system. Then, 1x2x vx t

vx t

From these equations, we have

1xx t t v v x


K K′ v

Suppose the origin of the K′ system is at a distance x from the origin of the K system after a time t measured in the K system. When the observer sees the clock in the K′ system at that time, he actually sees the clock as it was located at an earlier time because it takes a certain time for a light signal to travel to 0. Suppose we see the clock when it is a distance A from the origin of the

K system and the time is t in K and 11t′ in K′. Then we have v t ct t

tv x

We eliminate A, t, and x from these equations and we find 1


This is the time the observer reads by means of a telescope.

14-8. The velocity of a point on the surface of the Earth at the equator is eRv

which gives

According to Eq. (14.20), the relationship between the polar and equatorial time intervals is

so that the accumulated time difference is

Supplying the values, we find


w dm′m + dm v + dv

The unsurprising part of the solution to the problem of the relativistic rocket requires that we apply conservation of momentum, as was done for the nonrelativistic case. The surprising, and key, part of the solution is that we not assume the mass of the ejected fuel is the same as the mass lost from the rocket. Hence

where –dm is the mass lost from the rocket, dm′ is the mass of the ejected fuel, ()()21wvVvVc≡−− is the velocity of the exhaust with respect to the inertial frame, and

2211wwcγ≡−. One can easily calculate 3ddγγββ=, ad after some algebra one obtains

where we of course keep infinitesimals only to first order. The additional unknown dm′ is unalarming because of another conservation law

and will finally come to its desired form after dividing by dt dv dm mV

The quantity dt can be measured in any inertial frame, but would presumably only make sense for the particular one in which we measure v. Interestingly, it is not important for the ejected fuel to have an especially large kinetic energy but rather that it be near light speed, a nontrivial distinction. For such a case, a rocket can reach 0.6c by ejecting half its mass.

Solving (1) for and substituting into (2) gives 1x

Solving (2) for t and substituting into (1) gives



θ x

From example 14.1 we know that, to an observer in motion relative to an object, the dimensions

while the perpendicular component will be unchanged: sinθA

So, to the observer in K′, the length and orientation of the stick are

sin tan cos 1 vc θθ θ

or tan tan θθ γ

14-12. The ground observer measures the speed to be

The length between the markers as measured by the racer is

The time measured in the racer’s frame is given by v t x

= The speed observed by the racer is


t γ µ

receiverv source

In K, the energy and momentum of each photon emitted are

0 and h Eh p c νν==

Using Eq. (14.92) to transform to K′:

; h Eh E vp p

v h


So 0

β βνν ββ which agrees with Eq. (14.31).

Since cλν=


S o the shift is 0.1 nm toward the red (longer wavelength).

θ θ′Earth star

Consider a photon sent from the star to the Earth. From Eq. (14.92)


E h p pcc

Substituting yields

1 22cos cos cos cos 1β θβ θ β θ θ β+− − = −′′ cos cos cos cosθ θβ θ θ β− −=′′ −

Solving for cos θ yields cos cos

where angle in earth’s frame angle in star’s frame vcβ θ θ

Since cνλ=,


θ observer source light

Proceeding as in example 14.1, we treat the light as a photon of energy hν.

c ν=


For the source receding from the observer (at a much later time) we have

c ν=− and


1 source approaching observer

1 source receding from observer

θ observer source v

Proceeding as in the previous problem, we have In

cos r rt

KE h h p

r rt rt r t h c c νβνν β βββ β βh or

For 0λλ>, we have

14-20. As measured by observers on Earth, the entire trip takes

4 lightyears80 2 y

The people on earth age 80

3 years. The astronaut’s clock is ticking slower by a factor of γ. Thus, the astronaut ages



Those on Earth age 26.7 years. The astronaut ages 25.4 years.

md Fm v v v v (1)

14-2. The total energy output of the sun is

The corresponding rate of mass decrease is dm dE

The mass of the sun is approximately 1., so this rate of mass decrease can continue for a time 309910 kg×

Actually, the lifetime of the sun is limited by other factors and the sun is expected to expire about years from now. 94.510× pc E E

p cT mc T

14-24. The minimum energy will occur when the four particles are all at rest in the center of the mass system after the collision.

Conservation of energy gives (in the CM system) or which implies γ = 2 or 32β=

To find the energy required in the lab system (one proton at rest initially), we transform back to the lab

The velocity of K′(CM) with respect to K(lab) is just the velocity of the proton in the K′ system. So u = v.


Substituting into (1)


2 2 The minimum proton energy in the lab system

7, of which 6 is kinetic energy.ppmcismc xyvv=+vij Then

0 xyyx q v v B ij k vB

dt dt γ=×==FvBpv gives

qBd v v dt m =−viγj

Define 0qBmωγ≡ Thus or


t t

So cos sin cos sinxy vA t B vC t D ω ω ω ω = +

Thus cossinvtvt=−vijωω

Then sin cosvv t=+ri jω ωωω

The path is a circle of radius v ω pvm v r qBmqBqBγγ===

From problem 14-2


c r

14-26. Suppose a photon traveling in the x-direction is converted into an e and as shown below − e+ before after

Cons. of energy gives

2pepcE= where momentum of the photon

energy of energy of pep

Cons. of gives xp

()2cosmomentum of ,peepppeθe+−== Dividing gives

THE SPECIAL THEORY OF RELATIVITY 479 cos p e p c E cppθ== or

An isolated photon cannot be converted to an electron-positron

This result can also be seen by transforming to a frame where 0xp= after the collision. But, before the collision, 0xpppc=≠ in any frame moving along the x-axis. So, without another object nearby, momentum cannot be conserved; thus, the process cannot take place.

14-27. The minimum energy required occurs when the p and p are at rest after the collision. By conservation of energy

Since E, 0.5 MeVe=


We desire rel classical


7 The classical kinetic energy will be within 1% of the correct for 03.510 m/sec, independent of mass.v×value≤≤

14-30. A neutron at rest has an energy of 939.6 MeV. Subtracting the rest energies of the proton (938.3 MeV) and the electron (0.5 MeV) leaves 0.8 MeV.

Ot her than rest energies 0.8 MeV is available.

0.98c θ

Conservation of energy gives 2pEEπ= where E energy of each photon (Cons. of p=yp implies that the photons have the same energy).


Th e energy of each photon is 339 MeV.

Conservation of gives xp mv2cos where momentum of each photonppppγθ==

2202ET T p c+= Using the quadratic formula (taking the + root since T ≥ 0) gives

electron proton

afterbefore p e n ν

Conservation of yp gives

Conservation of gives xp

eppppνp==≡ Conservation of energy gives



0.5 MeVeE= and solving for pc gives p = 0.554 MeV/c

Substituting into

210 MeV, or 200 eV


Using the Lorentz transformation this becomes ct x vt xv t x vtcsx vc v c

vx v xc t t c x ct x x x

14-35. Let the frame of Saturn be the unprimed frame, and let the frame of the first spacecraft be the primed frame. From Eq. (14.17a) (switch primed and unprimed variables and change the sign of v) uv u uv

Substituting v = 0.9 c

X x x x idd == µ µ ττFm ct we have dx d xd Fm m d d dxdx Fm F m

di ctdd Fm icm d d t τ τ

Thus () dx d Fm m x vt dx d t m v F i dx dx Fm m F F F

vxd Fi cm t dxdt icm i m

Fi F

Th us the required transformation equations are shown.

14-37. From the Lagrangian

we compute

L kx

L mc

Then, from (2) and (3), the Lagrange equation of motion is dm c kx

from which

Using the relation dv dv dx dv c

we can rewrite (4) as


This is easily integrated to give 2

where E is the constant of integration.

The value of E is evaluated for some particular point in phase space, the easiest being x = a; β = 0:

From (8) and (9), 2

Eliminating 2β from (10), we have mc mc k a x

k mc a x ka x k mc a x and, therefore, ka x mc k a xdx

The period will then be four times the integral of dt = dt(x) from x = 0 to x = a:

a k ax m mc dx

k k ax ax

Since x varies between 0 and a, the variable xa takes on values in the interval 0 to 1, and therefore, we can define

from which


Using (14) – (17), (13) transforms into

12 cos2 1c os a d

Since 221kamc for the weakly relativistic case, we can expand the integrand of (18) in a series of powers of κ :

21c os

3 1c os κφ (19)

Substitution of (19) into (18) yields

23 1c os a d

a c

Evaluating (20) and substituting the expression for κ from (17), we obtain



14-38. () ()(for constant) dp d Fm u dt dt

d mu m d u m u uc u u c

ducm dtuc


14-39. The kinetic energy is

For a momentum of 100 MeV/c,

In order to obtain γ and β, we use the relation

so that



This is a relativistic velocity.


This is a nonrelativistic velocity.

14-40. If we write the velocity components of the center-of-mass system as jv, the transformation of ,jpα into the center-of-mass system becomes

where 2 must be satisfied, we have


, j p cv

c E

Em cT

where T and E represent the kinetic and total energy in the laboratory system, respectively, the subscripts 0 and 1 indicate the initial and final states, and is the rest mass of the incident particle. 0m

The expression for in terms of 0E1γ is

1E can be related to (total energy of particle 1 in the center of momentum reference frame after the collision) through the Lorentz transformation [cf. Eq. (14.92)] (remembering that for the inverse transformation we switch the primed and unprimed variables and change the sign of v):


Then, from (1), (2), and (4), 2 2

For the case of collision between two particles of equal mass, we have, from Eq. (14.127),

and, consequently,

Thus, with the help of (6) and (7), (5) becomes

1c os 2 sγ γθ γ

We must now relate the scattering angle θ in the center of momentum system to the angle ψ in the lab system.

Squaring Eq. (14.128), which is valid only for 1mm2=, we obtain an equation quadratic in cos θ. Solving for cos θ in terms of tan2ψ, we obtain

2cos 1 1t an

One of the roots given in (9) corresponds to θ = π, i.e., the incident particle reverses its path and is projected back along the incident direction. Substitution of the other root into (8) gives


1 2cos 1 sin1t an 2

== (10)

An elementary manipulation with the denominator of (10), namely,

2 cos 1 sin 2 cos 1 cos sin

sin cos cos cos 1c os cos 1 cos γγ ψ ψ (1) provides us with the desired result:


T = 0.1 GeV

T = 1 GeV

T = 10 GeV ψ 14-42.

φ θ hν hν′ γm c y

From conservation of energy, we have

cos cosehh mv

Momentum conservation along the y axis gives sin sine h mv


In order to eliminate φ, we use (2) and (3) to obtain

1 cos cos

sin sin e h mv c c h mv

Then, h h h mv c c c c νν ν νco φ φθ γ


Since 2

h v

From (5) and (7), we can find the equation for ν′:

h h h h hm



eh mc or,

The kinetic energy of the electron is

1 cos e e c m c h h E Emc γν νTm θ

1c os

1 cose e

T Emc mc θ θ