Download PHYS 110B - HW #8

1
PHYS 110B - HW #8
due to gravity.
Fall 2005, Solutions by David Pace
Equations referenced as ”Eq. #” are from Griffiths
Problem statements are paraphrased
1
x − x0 = vo t + at2
|{z} 2
(2)
0
1
−h = − |g|t2
2
s
2h
t =
|g|
[1.] Problem 11.10 from Griffiths
An electron is released from rest and allowed to fall under the influence of gravity. During the first centimeter
of its free fall what fraction of the lost potential energy
is dissipated through radiation?
(4)
The radiated power is given by,
Solution
Recall the kinematic equations in one dimension with
constant acceleration.
v − vo = at
(3)
1
x − xo = vo t + at2
2
P =
µo q 2 a2
6πc
Eq. 11.61
(5)
(1)
These equations underlie a physical discrepancy with
this problem. The kinematic equations above rely on an
ideal system in which all of the potential energy of the
falling object is converted into kinematic energy. There
are corrections to these equations in the case where radiation is significant compared to the kinetic energy. It
turns out that the energy lost due to radiation in this
situation pales in comparison to the kinetic term and
therefore these equations are valid as a very good approximation. If you considered this initially, then it is
still best to assume the radiation term will be very small
and use the equations above. If the radiation term turns
out to be very large, then you must start over and think
about the new physical issues.
Let the distance fallen by the electron be equal to h. The
actual value of this parameter is given as 1 cm, but it
is easier to use the variable throughout the algebra and
then plug in the numerical value for the final calculation. The goal of this problem is to determine the fraction Urad /∆Upot where Urad is energy radiated during
the 1 cm fall, and ∆Upot is the total amount of gravitational potential energy that is lost during this fall. Treat
∆Upot as a positive value because we are admitting that
the energy is decreasing and we only wish to know how
this absolute amount relates to the radiated amount.
Power is a measure of energy per unit time. The total
energy radiated by this falling electron is the power radiated multiplied by the time it takes for it to fall the
specified distance. Solving for this time requires the use
of (1). Since the electron is dropped from rest we have
vo = 0. Setting the ground as my reference requires that
the acceleration be a = −g, where g is the acceleration
where this is known as the Larmor formula. Griffiths
explains in section 11.2.1 that this expression is valid for
v c. This condition is met here because the final velocity of any object falling (from rest) through a distance
of 1 cm in the Earth’s gravitational field is not going to
be appreciable compared to the speed of light.
Multiplying (4) and (5) gives the total energy radiated
by the electron in this measured fall.
Urad
(6)
=
Pt
=
µo q 2 a2
·
6πc
=
µo q 2 g 2
6πc
s
s
2h
|g|
2h
|g|
(7)
(8)
The total change in gravitational potential energy is
(leaving this value positive for the reason given previously),
∆Upot
=
mgxf − mgxo
(9)
=
mg(xf − xo )
(10)
=
mgh
(11)
2
so the time is now written in terms of parameters given
in the problem statement.
The desired ratio is,
2 2
Urad
∆Upot
=
=
=
µo q g
6πc
q
2h
|g|
(12)
mgh
µo q 2
6πcm
r
The total radiated power, Urad , is given from (5) multiplied by this time.
2|g|
h
(13)
Urad
(4π × 10−7 )(1.6 × 10−19 )2
6π(3.0 × 108 )(9.11 × 10−31 )
≈ 2.8 × 10−22
r
2(9.8)
0.01
(14)
(15)
The energy lost due to radiation is insignificant compared to the general kinetic loss.
=
µo q 2 a2 vo
·
6πc
|a|
(21)
=
µo q 2 vo |a|
6πc
(22)
Now to solve for the desired fraction,
Urad
K.E.i
=
µo q 2 vo |a|
6πc
1
2
2 mvo
(23)
=
2
µo q 2 vo |a|
·
6πc
mvo2
(24)
=
µo q 2 |a|
3πcmvo
(25)
[2.] Problem 11.13 from Griffiths
(a) Consider an electron decelerating (a = constant)
from initial velocity vo to final velocity v = 0. Assuming vo c so that you may apply the Larmor formula,
what fraction of the initial kinetic energy of this electron
is radiated away?
(b) Apply the following values to the problem in part
(a).
vo ≈ 105 m/s
d = 30 Å
(16)
where d is the total distance traveled. What does this
result tell us about radiation losses for electrons in ordinary conductors?
(b) In this part we want to apply given values to (25).
This requires solving for the absolute value of the acceleration because it is not given.
Using (1), (16), and (20),
xf − xi
=
Solution
1
vo t + at2
2
The initial kinetic energy of the electron is,
vo
|a|
a
−
2
(26)
vo2
a2
(27)
d
=
vo
where m is the electron mass.
d
=
vo2
1 vo2
−
|a| 2 |a|
(28)
The total power radiated away is found using (6). The
time it takes for the electron to come to rest is found using (1). The a term is written negative because this is a
deceleration.
d
=
1 vo2
2 |a|
(29)
|a|
=
vo2
2d
(30)
K.E.i
=
1
mv 2
2 o
(17)
v − vo
=
−at
(18)
−vo
−a
=
t
(19)
t
=
vo
|a|
(20)
where this derivation also assumes the initial location is
the origin and the velocity is v ≥ 0.
Solving for the numerical fraction of kinetic energy lost
3
to radiation,
2
vo2
Urad
µo q
=
·
K.E.i
3πcmvo 2d
(31)
=
µo q 2 vo
6πcmd
=
(4π × 10−7 )(1.6 × 10−19 )2 (105 )
(33)
6π(3.0 × 108 )(9.11 × 10−31 )(30 × 10−10 )
(32)
≈ 2.1 × 10−10
where the vector form is written in spherical coordinates. The magnitude of the acceleration is written because the next step is to solve for the electron velocity in
its orbits.
In the case of centripetal acceleration and circular orbits, the tangential velocity of the particle is found using
a = v 2 /r. The distance r changes as the electron moves
closer to the proton. As r decreases along this path the
initial velocity must be the lowest velocity throughout
the trip. This initial velocity is found using,
v2
r
(34)
This is another instance in which the energy lost to radiative processes is negligible compared to other loss
mechanisms.
=
q2
4πo me r2
s
v
=
[3.] Problem 11.14 from Griffiths
Consider the Bohr model of the hydrogen atom. In this
model the ground state electron travels around the proton in a circular orbit of radius ro = 5 × 10−11 m, due to
the Coulomb attraction between the particles. Classical
electrodynamics requires that this accelerating electron
emit radiation and therefore continuously lose energy.
As it loses energy this electron will move inward and
eventually impact the proton. Show that the Larmor
formula is valid for most of the electron’s inward motion (i.e. v c). Calculate the typical lifetime of such
an atom if the electron’s orbits can always be considered
circular.
Solution
=
where me is the electron’s mass, q is the electron’s
~ is the electric field due only to the procharge, and E
ton. Recall that the force experienced by the electron
does not include its own fields in this problem.
The electric field of the proton is known, but it’s also
given as Eq. 2.10 if you wish to review. The acceleration
is,
q2
r̂
4πo me r2
~a
=
−
a
=
q2
4πo me r2
(36)
(37)
r
1
πo me r
(39)
(40)
All of the parameters in (40) are known numerically. The
initial (and lowest) velocity of this electron occurs at its
initial position where r = ro ,
vo =
1.6 × 10−19
2
≈
To show that the Larmor formula is valid I will solve
for the initial velocity of the electron in functional form
and then determine how this changes as the electron approaches the proton. This is accomplished by looking at
the forces involved,
~
F~ = me~a = −q E
(35)
q
2
q2
4πo me r
(38)
s
π(8.85 ×
2.2 × 106
1
× 10−31 )(5 × 10−11 )
(41)
10−12 )(9.11
(42)
This is a large velocity compared to everyday experience, but vo /c ≈ 0.007, which satisfies the small velocity
condition. As the electron approaches
the proton this
√
velocity increases by a factor of 1/ r. This factor causes
the velocity to slowly increase and we may therefore
claim that the electron’s velocity is less than the speed
of light for most of the trip. Of course, once the electron gets incredibly close to the proton the distance goes
to zero and the velocity goes to infinity. We ignore this
situation here.
The lifetime of the electron in this configuration is given
by the time necessary for the electron to radiate away
all of its energy. The total energy of the electron, Ee , is
its kinetic energy plus its potential energy. The kinetic
energy of the electron is given by (17) replacing the initial velocity there with the instantaneous velocity here.
The potential energy is P.E.= −qV , where V is the scalar
potential field due to the proton and the negative sign is
included because of the electron charge. The scalar potential is known for a single charged particle, giving the
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following expression for the energy,
1
q
Ee =
me v 2 − q
2
4πo r
(43)
Simplify using (39),
Ee
Now we equate (48) and (53) so that we can solve for
the time it takes for the electron to impact the proton.
In (48) we have an expression for the amount of energy
radiated away as a function of the acceleration (with the
radial dependence coming from the acceleration term).
In (53) we have the same quantity, but have derived it
from energy principles directly since we know the initial
energy and can quantify its rate of change as a function
of time and radial position.
=
me
q2
q2
−
2 4πo me r 4πo r
(44)
=
q2
4πo r
(45)
µo q 6
3
96π 2o cm2e r4
=
−
q 2 dr
8πo r2 dt
(54)
=
−
(46)
dt
=
−
96π 3 2o cm2e r4
q2
·
dr
2
8πo r
µo q 6
(55)
=
−
12π 2 o cm2e r2
dr
µo q 4
(56)
Z
0
me
−1
2me
q2
8πo r
This is the total amount of energy that must be lost to
radiation before the electron impacts the proton.
The energy per unit time radiated away is given by (5),
and we have solved for the acceleration in (37).
2
q2
µo q 2
(47)
P =
6πc 4πo me r2
µo q 6
=
3
96π 2o cm2e r4
(48)
Z
dt
ro
∆E
∆t
=
(49)
where E represents energy and not electric field.
t = −
dE
dt
t
(50)
where the negative sign is placed because we know
the energy is decreasing in the system but the radiated
power must be positive.
Using our expression for the total energy that must be
lost as given in (46), and noting that r = r(t), we have,
dEe
d
q2
= −
−
(51)
−
dt
dt
8πo r
=
q2
8πo
1
− 2
r
dr
dt
q
dr
2
8πo r dt
Z
0
r2 dr
(58)
ro
=
−
0
12π 2 o cm2e r3
µo q 4
3 ro
(59)
=
−
4π 2 o cm2e
−ro3
4
µo q
(60)
=
4π 2 o cm2e ro3
µo q 4
(61)
t
(62)
=
4π 2 (8.85 × 10−12 )(3 × 108 )(9.11 × 10−31 )2 (5 × 10−11 )3
(4π × 10−7 )(1.6 × 10−19 )4
(63)
(52)
2
−
12π 2 o cm2e
µo q 4
All of these values are known,
≈
=
(57)
To be more mathematically rigorous we may require
that the left side of (58) be t − to , but since we can always
adjust the initial time to be to = 0 it will not matter.
As we reduce the ∆t to an infinitesimal time period we
have,
P =−
12π 2 o cm2e r2
dr
µo q 4
In (57) the limits of the r integral reflect the fact that the
electron begins at the ground state radius and moves
into the proton.
To be able to compare these energy terms we take advantage of what power represents. Since power is energy per unit time it can be written as,
P
−
=
(53)
1.3 × 10−11 s
(64)
If this truly explained hydrogen, then it would be an incredibly short lived atom.
5
[4.] Problem 11.15 from Griffiths
Example 11.3 provides an expression for the power radiated as a function of angle θ, given as Eq. 11.74,
Regarding example 11.3 (page 463), find the angle, θmax ,
at which the maximum radiation emission occurs. Show
that,
r
1−β
∼
θmax =
for v ≈ c
(65)
2
Calculate the intensity of the radiation at θmax for the ultrarelativistic case (i.e. where v ≈ c), but write this value
in proportion to the same quantity for an instantaneous
velocity v = 0. State the ratio in terms of γ. Reference
figure 11.14.
Solution
The parameters in the problem statement are from page
463 of Griffiths,
β≡
v
c
γ≡p
1
1 − v 2 /c2
(66)
dP
dΩ
=
µo q 2 a2
sin2 θ
16π 2 c (1 − β cos θ)5
(67)
To find the angle at which the maximum radiation occurs we take the relevant derivative of (67) and set it
equal to zero. We already know that the minimum in
radiation (equal to zero) occurs at θ = 0. So although
this method can only show us where extrema occur, unless the solution is θ = 0 we can safely assume we have
found a maximum. If you are still unsure, then you can
take the second derivative of (67) at the value of θ found
from setting the first derivative to zero and if this result
is negative (positive) then the point represents a maximum (minimum).
Finding θmax ,
0
=
d µo q 2 a2
sin2 θ
dθ 16π 2 c (1 − β cos θ)5
=
µo q 2 a2 d
16π 2 c dθ
=
µo q 2 a2 (1 − β cos θ)5 (2 sin θ cos θ) − sin2 θ(5(1 − β cos θ)4 (−β)(− sin θ))
16π 2 c
(1 − β cos θ)10
(70)
=
(1 − β cos θ)5 (2 sin θ cos θ) − sin2 θ(5(1 − β cos θ)4 (−β)(− sin θ))
(1 − β cos θ)10
(71)
=
(1 − β cos θ)5 (2 sin θ cos θ) − sin2 θ(5(1 − β cos θ)4 (−β)(− sin θ))
(72)
=
(1 − β cos θ)4 (1 − β cos θ)(2 sin θ cos θ) − 5β sin3 θ
(73)
=
sin θ (1 − β cos θ)(2 cos θ) − 5β sin2 θ
(74)
=
2 cos θ − 2β cos2 θ − 5β(1 − cos2 θ)
(75)
=
2 cos θ − 2β cos2 θ − 5β + 5β cos2 θ
(76)
=
3β cos2 θ + 2 cos θ − 5β
(77)
sin2 θ
(1 − β cos θ)5
(68)
(69)
6
This is the form of a quadratic equation and may be
solved as such. I have not labeled this angle as θmax ,
but by the method taken we are finding that value.
dure in asymptotic analysis.
cos θmax =
cos θ
=
−2 ±
p
4 − 4(3β)(−5β)
6β
=
=
−1 ±
p
1+
3β
15β 2
p
1 + 15β 2
3β
(84)
−1 +
p
1 + 15(1 − )2
3(1 − )
(85)
(78)
=
p
−2 ± 2 1 + 15β 2
6β
−1 +
(79)
=
hp
i
1
1 + 15(1 − 2 + 2 ) − 1
3(1 − )
(86)
=
√
1
1 + 15 − 30 − 1
3(1 − )
(87)
(80)
We choose the sign in (80) based on information about
θmax provided in example 11.3. From figure 11.14 and a
statement on the bottom of page 463 we know that the
radiation is concentrated through a narrow cone near
the forward direction. Figure 11.14 is turned sideways,
but if we look at the location of θmax it is evident that
concentrating the radiation in a narrow region near the z
axis means that θmax < 45◦ . This requires that cos θ > 1.
Meeting this condition requires that we take the plus
sign in (80).
" r
#
1
30
4 1+ −1
=
3(1 − )
16
(88)
" #
1/2
15
1
4 1+ −1
=
3(1 − )
8
(89)
Recall the binomial expansion, which will be applied
multiple times in the steps to follow,
Finally,
(1 + x)n
p
1 + 15β 2
cos θ =
3β
"
#
p
2
−1
+
1
+
15β
θmax = cos−1
3β
−1 +
1 + nx
(90)
(81)
for small x.
(82)
Applying a binomial expansion to the square root term
in (89) provides,
For the next part of the problem it is more direct to use
the cosine form of θmax as written in (81). Since we are
in the ultrarelativistic case β ≡ v/c ≈ 1, but slightly less
than one since the velocity is still less than the speed of
light. When the velocity is very close to the speed of
light we know that β is so close to unity that it may be
written as,
β ≈1−
≈
(83)
where is an expansion parameter and is very small.
The rest of this problem concerns rewriting terms and
solving for this expansion parameter.
To begin, replace β with its expansion in (81). When
writing out the terms we should neglect anything that
is of order 2 or higher. Since 1, any term containing an 2 is so small it should be ignored. This process is
known as expanding to order one in , a common proce-
cos θmax
1 15
1
=
4 1+
−1
3(1 − )
2 8
(91)
=
1
15
4+ −1
3(1 − )
4
(92)
=
1
5
1− 1−
4
(93)
We can solve further by using the binomial expansion
on the 1/(1 − ) term.
1
1−
=
(1 − )−1
(94)
≈
1+
(95)
7
5
cos θmax ≈ (1 + ) 1 − 4
(96)
5
5
≈ 1 − + − 2
4
4
(97)
1
≈1− 4
(98)
We invoke another expansion of the cosine function for
comparison with (98).
cos x
=
cos θmax
∞
X
(−1)n x2n
(2n)!
n=0
The direction at which the maximum power is radiated
occurs at θ = 90◦ . This is known because the maximum
must occur when the sine term is equal to unity. The
maximum power radiated in this zero velocity instant
is,
max
dP
µo q 2 a2
=
(108)
dΩ v=0
16π 2 c
The other radiated power term is (67) evaluated at the
already determined θmax .
(99)
=
1−
x4
x2
+
− ...
2!
4!
(100)
≈
1−
2
θmax
2
(101)
dP
dΩ
max
µo q 2 a2
sin2 θmax
2
16π c (1 − β cos θmax )5
=
θ=θmax
The desired ratio is,
dP max
dΩ θ=θmax
dP max
dΩ v=0
=
sin2 θmax
µo q 2 a2
16π 2 c (1−β cos θmax )5
µo q 2 a2
16π 2 c
(110)
=
sin2 θmax
(1 − β cos θmax )5
(111)
4
since keeping the θmax
term seems unnecessary.
Equate (98) and (101),
2
θmax
2
1
1− 4
=
1−
2
=
2
θmax
(102)
(103)
(109)
Since this is evaluated at θmax we know that β can be
expanded as in (83). The plan is to write (111) in terms
of and then relate this result to γ as requested. The
numerator of (111) can be rewritten using yet another
trigonometric property; sin x ≈ x for small x. We know
that θmax is small in this case because we solved in terms
of β ≈ 1. As such, we know from (105) that the angle is
small. The denominator is rewritten using (98).
2
r
2
=
(104)
θmax
Using (83) to rewrite the above,
r
1−β
θmax ≈
2
=
(105)
In the next part of the problem we want to write the radiated power in the direction of θmax as a fraction of the radiated power in the equivalent maximum direction for
a particle of instantaneous velocity equal to zero. Notice
that the particle is still accelerating, as it must in order to
radiate at all, but that we are concerned with the instant
at which its velocity reaches zero.
Begin by writing (67) for v = 0, which is the same as
setting β = 0.
dP
dΩ
2 2
=
=
(112)
2
θmax
(1 − (1 − )(1 − 4 ))5
(113)
p 2
=
2
(1 − (1 −
4
−+
2 5
4 ))
(114)
Once again, drop the 2 term.
p 2
2
(1 − (1 −
4
−+
2
4
))5
=
2
5 5
( 4 )
=
1
2
=
0.16384−4
2
µo q a
sin θ
16π 2 c (1 − (0) cos θ)5
µo q 2 a2
sin2 θ
16π 2 c
sin2 θmax
(sin θmax )
≈
5
(1 − β cos θmax )
(1 − β(1 − 4 ))5
5
4
−4
5
(115)
(116)
(106)
(107)
(117)
where inverse powers of small numbers are large and
therefore kept.
8
It is necessary to express in terms of γ.
γ
=
arrive at a solution of,
1
S floor
(118)
p
1 − β2
=
µo q 2 d2 ω 4 R2 h
5/2
32π 2 c (R2 + h2 )
(126)
=
1
p
1 − (1 − )2
(119)
(b) Using your result from (a), calculate the energy per
unit time across the entire (infinite) floor. Comment on
this value and describe whether it makes sense.
=
1
p
1 − (1 − 2 + 2 )
(120)
≈
1
√
2
(121)
(c) The spring/charge system is losing energy through
this radiation. Solve for the time τ at which the oscillation amplitude has been reduced to d/ exp, assuming
that the energy loss per cycle is very small.
Solution
≈
1
2γ 2
(122)
Our final answer is given in (117), and we can now write
this in terms of γ,
dP max
dΩ θ=θmax
dP max
dΩ v=0
= 0.16384
1
2γ 2
−4
= 0.16384(24 )γ 8
≈ 2.6γ 8
(123)
(a) From Eq. 9.63 we know that the intensity of EM
waves (which compose radiation) is equivalent to the
time averaged value of their Poynting flux. The hint reminds us to make sure we only solve for the intensity
that actually strikes the floor. This is the amount of energy flux that passes through the floor. Defining a geometry in which the floor is at a constant z = 0 value
means that the area vector of the floor is given by −ẑ.
The intensity striking the floor is,
(124)
(125)
There are probably more equations in this solution
where it is appropriate to have an “approximately
equal” sign instead of an “equal to” sign, but in all the
steps above the approximations are so good that there
is little error. You can use the equal signs in this case,
ultrarelativistic, without large error.
hSi floor
Griffiths’ Hints: Assume d λ h. The calculation only concerns the average energy per unit time
that strikes the floor. Neglect radiative damping and
(127)
The oscillating charge is essentially an electric dipole.
The intensity of the radiation from such a system is
given by Eq. 11.21,
[5.] Problem 11.21 from Griffiths
(a) Solve for the radiation intensity that strikes the
floor as a function or R, the distance away from the
floor point located directly beneath the charge. Solve
for the value of R at which the radiation intensity is a
maximum.
~ · ẑ
hSi
where I am taking the absolute value of this expression
because we already know the radiated energy is leaving
the oscillating charge.
~
hSi
A particle (mass = m and charge = q) is attached to
a spring and suspended from a ceiling. Reference figure 11.19 in Griffiths. The equilibrium position of this
spring-mass system is a distance h above the floor. At
time t = 0 the mass is pulled downward a distance d
and then released to begin oscillatory motion.
=
=
µo p2o ω 4 sin2 θ
r̂
32π 2 cr2
(128)
The dipole moment amplitude is po = qd because d is
the maximum displacement from the origin. The origin
is set as the equilibrium position of the charge. The angularp
frequency of a spring/mass system is known to be
ω = k/m. Inserting this information into the expression for intensity gives,
r !4
2
µ
sin
θ
k
o
2
~
hSi
=
r̂
(129)
(qd)
32π 2 cr2
m
=
µo q 2 d2 k 2 sin2 θ
r̂
32π 2 cm2 r2
(130)
The R dependence is contained in the r and sin θ terms.
The following image shows the charge (solid black circle) suspended from the spring (coil). The charge is
placed at the center of a spherical coordinate system.
9
which an extrema occurs and this will be the location of
maximum intensity.
0=
=
d
hSi floor
dR
d
dR
=
(139)
µo q 2 d2 k 2 R2 h
2
32π cm2 (h2 + R2 )5/2
µo q 2 d2 k 2 h
32π 2 cm2
(140)
d
R2
2
dR (h + R2 )5/2
(141)
From this drawing we see that,
sin θ
=
R
r
r
=
p
(131)
h2 + R 2
(132)
=
(h2 + R2 )5/2 (2R) − R2 (5/2)(h2 + R2 )3/2 (2R)
(h2 + R2 )5
(142)
= 2R(h2 + R2 )5/2 − 5R3 (h2 + R2 )3/2
(143)
= (h2 + R2 )3/2 2R(h2 + R2 ) − 5R3
(144)
Returning to (130) with this information,
2 2 2
~ = µo q d k
hSi
32π 2 cm2
=
R
r
µo q 2 d2 k 2 R2
32π 2 cm2
2 √
1
√
2
h + R2
1
h2 + R 2
2
2 (133)
r̂
= R 2h2 + 2R2 − 5R2
1
h2 + R 2
r̂ (134)
=
µo q 2 d2 k 2 R2
2
32π cm2 (h2 + R2 )2
(135)
r̂ · ẑ
µo q 2 d2 k 2 R2
cos θ
32π 2 cm2 (h2 + R2 )2
R=h
=
(146)
2
3
(147)
The radial distance to the position of maximum intensity
is slightly shorter than the distance of the particle from
the floor.
(136)
(137)
(b) The total energy per unit time (the power) across the
floor is found by integrating the expression in (138) over
the floor. This is written as,
Z
µo q 2 d2 k 2 R2 h
32π 2 cm2 (h2 + R2 )5/2
(145)
= 2h2 − 3R2
r
The intensity striking the floor is the z component of this
value,
=
µo q 2 d2 k 2 R2
=
r̂
2
32π cm2 (h2 + R2 )2
hSi floor
Pfloor
=
~ floor · dA
~
hSi
(148)
(138)
Z
where I have used the drawing to determine that cos θ =
h/r and included the value of r as given in (132). This
answer agrees with the solution provided by Griffiths if
we replaced the k and m factors with their equivalent
representation in terms of ω. It is good form to always
solve problems in terms of the specific variables given.
To solve for the value of R at which this intensity is a
maximum, set the R derivative of (138) equal to zero.
One of the solutions should be R = 0, which we already
know to be a minimum. Find the other value of R for
∞
Z
2π
hSiẑ · R dR dφẑ
=
0
Z
∞
Z
=
0
(149)
0
0
2π
µo q 2 d2 k 2 R2 h
(R
2
32π cm2 (h2 + R2 )5/2
dR dφ)
(150)
=
µo hq 2 d2 k 2
(2π)
32cπ 2 m2
Z
0
∞
(h2
R3
dR
+ R2 )5/2
(151)
10
The remaining integral is found in a table of integrals,
Z
a
1
x3
dx = 2
− √
2
5/2
2
3/2
2
(a + cx )
3c (a + cx )
c a + cx2
(152)
Considering our specific integral,
Z ∞
R3
dR
(153)
2
2 5/2
0 (h + R )
=
h2
1
−√
2
3(h + R2 )3/2
h2 + R 2
=
0−0−
1
h2
−√
3(h2 )3/2
h2
∞
(154)
determine when it reaches the specified value. One expression that includes the oscillation amplitude is the
potential energy of the spring/charge system,
U
=
1 2
kz
2
(162)
where z = d in the first parts of this problem and z =
z(t) now.
The time derivative of this energy is,
0
dU
dt
=
k d 2
z
2 dt
(163)
(155)
h2
1
+
3h3
h
=
−
=
1
1
−
h 3h
(157)
=
2
3h
(158)
(156)
The power radiated by the dipole is also an expression
for the energy change per unit time. Note, however, that
the radiation emitted by the dipole carries energy away
to both the ceiling and the floor. Using (161) with the k
and m dependence,
P
dU
dt
=
=
µo q 2 k 2 z 2
12πcm2
(164)
and the d term has been replaced with the time dependent z.
This result is now applied to (151),
Pfloor
=
=
µo hq 2 d2 k 2
16cπm2
µo q 2 d2 k 2
24cπm2
2
3h
Since the spring system is losing energy we can write,
(159)
k d 2
z
2 dt
(160)
Consider Eq. 11.22, the total power radiated by an electric dipole,
hP i
=
µo p2o ω 4
12πc
=
−
µo q 2 k 2 z 2
12πcm2
(165)
which leads to a familiar differential equation and solution,
d 2
z
dt
=
−
µo kq 2 2
z
6πcm2
(166)
d(z 2 )
z2
=
−
µo kq 2
dt
6πcm2
(167)
=
−
µo kq 2
t + z0
6πcm2
(168)
z2
=
µo kq 2
d2 exp −
t
6πcm2
(169)
z
=
µo kq 2
t
d exp −
12πcm2
(170)
(161)
where Griffiths correctly applies the angle brackets to
indicate that this is a time average. All of the work in
this solution should have included these angle brackets
as well. We knew from the beginning that these were
time averages, however, so I left off the brackets.
If we rewrite (160) in terms of the variables given in
(161), replacing the ω and dipole moment amplitude
dependencies, then we see that the power striking the
floor is equal to half of the total energy radiated by this
dipole. We integrated out to infinity so any radiation
that does not reach the floor must hit the ceiling.
(c) The oscillation amplitude is given by d, the distance
by which the charge is initially displaced. We’ll need to
solve for this amplitude as a function of time and then
ln z 2
where the initial amplitude, zo = d, is given in the problem statement. The amplitude is equal to d/e when the
exponential argument above is equal to −1. This occurs
11
when,
moment of this particle/image pair is,
z
p~ = q d~
d
e
=
(176)
(171)
(177)
= q(2~z)
2
−
µo kq
t
12πcm2
t
=
−1
=
12πcm2
µo kq 2
(172)
(173)
where my t is the τ that we have been asked to find.
Notice that this time decreases with increasing value of
the charge q. The radiated power increase with charge
so this is a sensible result.
[6.] Problem 11.25 from Griffiths
A charged particle (mass m and charge q) is a distance
z from a conducting surface. As this particle moves toward (or away) from the conducting surface it emits radiation. This process may be treated as an interaction
between the particle and its image due to the conducting surface. This pair forms an effective dipole, the moment of which changes in time. Find the total radiated
power as a function of the particle’s height z above the
conducting surface.
=
1
6m2 z 4
µo cq
4π
2
(174)
The problem statement gives the method by which to
solve it. Treat this situation as an electric dipole where
the separation distance is changing in time. The general
expression for the power radiated by an electric dipole
is,
∼
=
µo p̈2
6πc
For the term in (175),
(179)
p̈ = 2qz̈
It remains to solve for z̈. This is done by examining the
force on the particle due to its image. Considering magnitudes only,
F
=
ma
=
m
d2
z
dt2
=
11.60
(175)
where p is the dipole moment and the double dot represents a double time derivative.
This may be treated as a one dimensional problem along
the line that connects the charged particle to its image.
This particle will accelerate as it approaches the conducting surface due to the greater amount of induced
surface charge that will exert a force on it. The dipole
mz̈
(180)
This force is due to the electric influence of the image.
The primed field is that due to the image particle, which
must have charge −q (review ch. 3 in the text if you do
not believe this).
3
Solution
P
where in the final step we keep the magnitude of the
dipole moment and I indicated the time dependence
contained therein. The factor of 2 comes from the separation between the particle and its image.
F = qE 0
Griffiths Hint: The solution is,
P (z)
(178)
p(t) = 2qz(t)
(181)
= q
−q
4πo r2
= −
q2
4πo (2z)2
(183)
= −
q2
16πo z 2
(184)
(182)
Equating (180) and (184),
mz̈ = −
z̈ = −
q2
16πo z 2
(185)
q2
16mπo z 2
(186)
Solving for the second time derivative of the dipole moment provides,
q2
p̈ = 2q −
(187)
16mπo z 2
= −
q3
8mπo z 2
(188)
12
tionship µo o = 1/c2 .
Back to (175),
µo
P =
6πc
µo
=
6πc
=
q3
−
8mπo z 2
2
q6
64m2 π 2 2o z 4
µo q 6
384cm2 π 3 2o z 4
(189)
1
= c4 µ2o
2o
(190)
P =
(191)
=
This does not look like the solution Griffiths provided.
His solution, (174), has no o dependence so it is probably best to remove this in our solution noting the rela-
µo q 6
(c4 µ2o )
384cm2 π 3 z 4
µ3o q 6 c3
384m2 π 3 z 4
3
µo cq 2
1
=
6m2 z 4
4π
(192)
(193)
(194)
(195)
Our solution is equivalent to that provided by Griffiths.