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# ChR ISM

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```The relativistic kinetic energy of
the particle is given by: ( ) \u239f
\u239f
\u23a0
\u239e
\u239c\u239c\u239d
\u239b
\u2212
\u2212
= 1
1
1
2
2
icrelativist
cv
mcK

A spreadsheet program to graph Krelativistic and Kclassical is shown below. The
formulas used to calculate the quantities in the columns are as follows:

Cell Formula/Content Algebraic Form
A8 A7+0.05 v/c + 0.05
B7 0.5*\$B\$3*A7^2 2
2
1 mv
C7 \$B\$3*(1/((1\u2212A7^2)^0.5)\u22121)
( ) \u239f
\u239f
\u23a0
\u239e
\u239c\u239c\u239d
\u239b
\u2212
\u2212
1
1
1
2
2
cv
mc

A B C
1
2
3 mc2= 100 MeV
4
5
v/c
2
2
1 mv ( )
2
21
1 mc
cv \u239f
\u239f
\u23a0
\u239e
\u239c\u239c\u239d
\u239b
\u2212
6 Kclassical Krelativistic
7 0.00 0.00 0.00
8 0.05 0.13 0.13
9 0.10 0.50 0.50
10 0.15 1.13 1.14

23 0.80 32.00 66.67
24 0.85 36.13 89.83
25 0.90 40.50 129.42
26 0.95 45.13 220.26

The solid curve is the graph of the relativistic kinetic energy. The relativistic
formula, represented by the continuous curve, begins to deviate from the classical
curve around v/c \u2248 0.4.

Special Relativity

1075
0
50
100
150
200
250
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
v /c
E
(M
eV
)
Kclassical
Krelativistic

45 \u2022\u2022 [SSM] (a) Show that the speed v of a particle of mass m and total
energy E is given by ( ) 212
22
1
\u23a5\u23a5\u23a6
\u23a4
\u23a2\u23a2\u23a3
\u23a1 \u2212=
E
mc
c
v and that when E is much greater than
mc2, this can be approximated by ( )2 2221 Emccv \u2212\u2248 . Find the speed of an electron
with kinetic energy of (b) 0.510 MeV and (c) 10.0 MeV.

Picture the Problem We can solve the equation for the relativistic energy of a
particle to obtain the first result and then use the binomial expansion subject to
E >> mc2 to obtain the second result. In parts (b) and (c) we can use the first
expression obtained in (a), with E = E0 + K, to find the speeds of electrons with
the given kinetic energies. See Table 39-1 for the rest energy of an electron.

(a) The relativistic energy of a
particle is given by Equation R-15:
2
2
2
1
c
v
mcE
\u2212
=

Solving for v/c yields:

( ) 21
2
22
1
\u23a5\u23a5\u23a6
\u23a4
\u23a2\u23a2\u23a3
\u23a1 \u2212=
E
mc
c
v (1)

Expand the radical expression binomially to obtain:

( ) ( ) sorder term-higher
2
111 2
22
2
22
+\u2212=\u2212=
E
mc
E
mc
c
v

Chapter R

1076
Because the higher-order terms are
much smaller than the 2nd-degree
term when E >> mc2:

( )
2
22
2
1
E
mc
c
v \u2212\u2248

(b) Solve equation (1) for v: ( )
2
22
1
E
mccv \u2212=

Because E = E0 + K:
( ) 2
0
2
0
2
0
1
111
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b +
\u2212=+\u2212=
E
K
c
KE
Ecv

For an electron whose kinetic
energy is 0.510 MeV:

( )
c
cv
866.0
MeV0.511
MeV510.01
11MeV510.0 2
=
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b +
\u2212=

(c) For an electron whose kinetic
energy is 10.0 MeV:

( )
c
cv
999.0
MeV0.511
MeV0.101
11MeV0.10 2
=
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b +
\u2212=

46 \u2022\u2022 Use the binomial expansion and Equation R-17 to show that when
pc << mc2, the total energy is given approximately by E \u2248 mc2 + p2/(2m).

Picture the Problem We can solve Equation R-17 for E and factor ( )22mc from
under the resulting radical to obtain an expression to which we can apply the
binomial expansion to write the radical expression as a power series. Finally, we
can invoke the condition that pc << mc2 to complete the argument that the total
energy is given approximately by ( )mpmcE 222 +\u2248 .

From Equation R-17 we have: ( )2222 mccpE +=

Factor ( )22mc under the radical and simplify to obtain:

( ) ( ) ( ) 2
2
2
22
22
2
22
22
22 111
mc
pmc
mc
cpmc
mc
cpmcE +=+=\u23a5\u23a5\u23a6
\u23a4
\u23a2\u23a2\u23a3
\u23a1 +=
Special Relativity

1077
Expand the radical factor binomially to obtain:

sorder term-higher
2
111 22
221
22
2
++=\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b +
cm
p
cm
p

For pc << mc2:
22
221
22
2
2
111
cm
p
cm
p +\u2248\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b +

Substituting for
21
22
2
1 \u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b +
cm
p and
simplifying yields:
m
pmc
cm
pmcE
22
11
2
2
22
2
2 +=\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b +=

47 \u2022\u2022 Derive the equation E2 = p2c2 + m2c4 (Equation R-17) by eliminating v
from Equations R-10 and R-16.

Picture the Problem We can solve Equation R-16 for v and substitute in
Equation R-10 to eliminate v. Simplification of the resulting expression leads
to ( ) .22222 mccpE +=

Express the relativistic momentum of
a particle using Equation R-10: 221 cv
mvp \u2212=

From Equation R-16 we have:
E
pc
c
v = \u21d2
E
pcv
2
=

Substitute for v and simplify to
obtain:
( )
2
22
2
1
c
Epc
E
pcm
p
\u2212
=
or
2
22
2
1
1
E
cpE
mc
\u2212
=

Multiply both sides of the
equation by 2
22
1
E
cpE \u2212 :

2
2
22
1 mc
E
cpE =\u2212
Chapter R

1078
Square both sides of the equation: ( )222 222 1 mcEcpE =\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b \u2212

Solve for E2 to obtain Equation R-17: ( )22222 mccpE +=

48 \u2022\u2022 The rest energy of a proton is about 938 MeV. If its kinetic energy is
also 938 MeV, find (a) its momentum and (b) its speed.

Picture the Problem (a) We can solve the relation for total energy, momentum,
and rest energy of the proton for its momentum and evaluate this expression for
K = E0. In Part (b) we can start with the expression for the relativistic energy of
the proton, solve for u, and evaluate the resulting expression for K = E0.

(a) Relate the total energy,
momentum, and rest energy of a
proton:

( )22222 mccpE +=
Solving for p yields: ( )
c
KEK
c
EKE
c
EE
p
0
2
2
0
2
0
2
0
2
2+=
\u2212+=\u2212=

If K = E0:
c
E
c
EE
p EK 0
2
0
2
0 32
0
=+==

Substitute numerical values and
evaluate p:
( )
c
c
p EK
GeV/ 62.1
MeV 9383
0
=
==

(b) The relativistic energy of a
particle is given by Equation R-15:

( )2
2
1 cv
mcE
\u2212
= \u21d2 2
2
01
E
Ecv \u2212=

Because E = E0 + K:
( )20
2
01
KE
Ecv +\u2212=

Special Relativity

1079
For K = E0:
( )
cc
c
EE
Ecv EK
866.0750.0
4
111 2
00
2
0
0
==
\u2212=+\u2212==

49 \u2022\u2022 What percentage error is made in using 221 mv for the kinetic energy of
a particle if its speed is (a) 0.10c and (b) 0.90c?

Picture the Problem We can use the expressions for the classical and relativistic
kinetic energies of a particle to obtain a general expression for the fractional error
made in using 221 mv for the kinetic energy of a particle as a function of its speed.

The percentage error made in using
2
2
1 mv for the kinetic energy of a
particle is given by:

rel
classical
rel
classicalrel
rel
1
K
K
K
KK
K
K \u2212=\u2212=\u394

The relativistic kinetic energy of the
particle is given by Equation R-14:

( )
( )
2
2
2
2
2
rel
1
1
1
1
mc
cv
mc
cv
mcK
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b
\u2212
\u2212
=
\u2212
\u2212
=

Substitute for Krel in the expression for
relK
K\u394 and simplify to obtain:

( ) ( )
2
2
2
2
2
2
2
1
rel
1
1
12
1
1
1
1
1
c
cv
v
mc
cv
mv
K
K
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b
\u2212
\u2212
\u2212=
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b
\u2212
\u2212
\u2212=\u394

(a) Evaluate f for v = 0.10c:
( ) ( )
( )
%75.0
1
10.01
12
10.0110.0
2
2
2
=
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b
\u2212
\u2212
\u2212=
c
ccf

Chapter R

1080
(b) Evaluate f for v = 0.90c:

( ) ( )
( )
%69.0
1
90.01
12
90.0190.0
2
2
2
=
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b
\u2212
\u2212
\u2212=
c
ccf

General Problems

50 \u2022 A spaceship departs from Earth for the star Alpha Centauri, which is
4.0 c\u22c5y away in the reference frame of Earth. The spaceship travels at 0.75c. How
long does it take to get there (a) as measured on Earth and (b) as measured by a
passenger on the spaceship?

Picture the Problem We can find the duration of the trip, as measured on earth,
using the definition of average speed; that is, \u394t = L/v. The elapsed time measured
by the passenger is the proper time and is related to \u394t through the time dilation
equation (Equation R-3).

(a) Relate the time \u394t for the trip as
measured on earth to its length L and
the speed u of the spaceship:

v
Lt =\u394
Substitute numerical values and
evaluate \u394t: y3.5y33.575.0
y0.4
\u394 ==\u22c5=
c
ct

(b) Use Equation R-3, the time
dilation equation, to express the
proper time measured by a
passenger:

2
0 1\u394\u394 \u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212=
c
vtt

Substitute numerical values and
evaluate \u394t0: ( ) ( ) y5.375.01y33.5\u394 2
2
p =\u2212= c
ct

51 \u2022 The total energy of a particle is three```