Solutions to Problem Set 7, Physics 370, Spring 2014
1
TOTAL POINTS POSSIBLE: 70 points.
1. Griffiths Problem 3.30: In Example 3.9 we derived the exact potential for a spherical shell of radius R, which carries a surface charge
σ (θ) = k cos θ.
(a) Explain, in clear English, why you might expect this charge distribution to not have a monopole moment and why it would have
a dipole moment? Do NOT rely on equations to make your case.
(b) Calculate the dipole moment of this charge distribution.
(c) Find the approximate potential, at points far from the sphere, and
3 1
compare the exact answer [V (r, θ) = kR
cos θ (r ≥ R)]. What
3ǫ0 r 2
can you conclude about the higher multipoles?
(a) (3 points possible) In this problem we consider a spherical shell
of radius R which carries a surface charge distribution σ = k cos θ.
If you attempt to visualize that charge distribution, you will see
that for every positive charge on the “northern hemisphere” of the
spherical shell (where θ = 0 → π/2), there will be a corresponding
negative charge on the “southern hemisphere,” making the total
charge on the sphere zero. This is why there is no monopole
moment, because the total charge is zero. However, there is a
systematic separation between the positive and negative charges
which seems symmetric along the z axis, therefore we would expect
a dipole moment (since the charge has a dipole distribution).
(b) (5 points possible) The dipole moment is given by
Z
p~ = ~r′ ρdτ ′ .
(1)
The x and y components of the dipole moment are zero because
of the symmetry of the charge distribution. Since all the charge
is on the shell (which has radius R), then the z-component of
Solutions to Problem Set 7, Physics 370, Spring 2014
2
the charge position can be written ~r′ · zˆ = R cos θ′ , therefore the
z-component of the dipole moment is given by
Z
Z
′
′
pz = p~ · zˆ = ~r · zˆρdτ = R cos θ′ ρdτ ′
(2)
Given this expression and the fact that ρdτ ′ = dq ′ = σda′ =
σR2 sin θ′ dθ′ dφ′ I can now manipulate this
Z π Z 2π
pz = R
σ cos θ′ R2 sin θ′ dφ′dθ′
(3a)
′
′
θ =0 φ =0
Z π
3
= 2πR
(k cos θ′ ) cos θ′ sin θ′ dθ′
(3b)
θ=′ 0
Z
π
= 2πkR3
cos2 θ′ sin θ′ dθ′
(3c)
θ=′ 0
3 ′ π
3 cos θ (3d)
= −2πkR
3 0
4
(3e)
= πkR3 ;
3
Given this z-component of the dipole moment (which is the only
non-zero component), we can write the total dipole moment as
4
~p = πkR3 zˆ
3
(4)
(c) (2 points possible) Once we have the dipole moment (from the
last part of this problem), we can compute the dipole term in the
potential of the sphere (using equation 3.99) such that
V =
rˆ · p~
p cos θ
kR3 cos θ
=
=
.
4πǫ0 r 2
4πǫ0 r 2
3ǫ0 r 2
(5)
From Equation 3.87 in the textbook we know this is the exact
potential for the sphere, so all the higher multipole moments must
be zero.
(Thanks to Dr. Craig for providing an initial draft of this solution in
LATEX form.)
Solutions to Problem Set 7, Physics 370, Spring 2014
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2. Griffiths Problem 3.46 (tweaked): A thin insulating rod, running
from z = −a to z = +a, carries the indicated line charges. In each
case, find the leading term in the multipole expansion of the potential
(a) λ = k cos πz
, where k is a constant.
2a
πz
(b) λ = k sin a , where k is a constant.
(c) λ = k cos πz
, where k is a constant.
a
For full credit you must explain clearly why the particular charge distribution leads somewhat intuitively to a given leading term in the
multipole expansion (e.g. - Why does this charge distribution look like
a monopole, dipole, etc. ?). It may be helpful to know what λ(z) means
in terms of where the charge is and to subsequently plot λ(z) versus z
in each case in order to determine what you expect!
Consider the line of charge shown below.
z
P
a
dq
θ
r’
-a
In all three of these problems, we are asked to find the potential due to
this line of charge at the point P using the multipole expansion. The
small box in the figure highlights a charge element dq = ρdτ ′ = λdz ′
a distance r ′ from the origin. The multipole expansion for this charge
element is
Z
Z
Z
1
1
1
3 cos2 θ′ 1
1
′
′
′
′
′
′ 2
V =
ρdτ + 2 r cos θ ρdτ + 3 (r )
ρdτ + · · · .
−
4πǫ0 r
r
r
2
2
(6)
Solutions to Problem Set 7, Physics 370, Spring 2014
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The expansion can also be written in the form
Z
3 cos2 θ′ 1
Q ~p · rˆ
1
1
′
′ 2
ρdτ + · · · , (7)
+ 2 + 3 (r )
−
V =
4πǫ0 r
r
r
2
2
where Q is the total charge of the distribution and p~ is its dipole moment. For each of the three charge distributions given we are supposed
to find the first nonzero term in the multipole expansion. As hinted,
the strategy for each distribution will be to first calculate Q, and if it is
zero to then calculate p~ and if that is zero then calculate the quadrupole
term.
(a) (4 points possible) The first charge distribution is λ1 = k cos(πz/2a)
and looks like
which suggests there should be a non-zero total charge. The total
charge of this distribution is
Z a
Q=
λdz ′
(8a)
−a
′
Z a
πz
=k
cos
dz ′
(8b)
2a
−a
′
Z a
πz
dz ′
(8c)
= 2k
cos
2a
0
′ a
2a
πz
(8d)
= 2k
sin
π
2a
0
4ka
.
(8e)
=
π
The approximate potential of the distribution is
V =
ka
Q
= 2 .
4πǫ0 r
π ǫ0 r
(9)
In this case, since there was an overall charge, the first-order
monopole moment will be adequate.
Solutions to Problem Set 7, Physics 370, Spring 2014
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(b) (3 points possible) The next charge distribution is λ = k sin(πz/a)
and looks like
which suggests zero total charge and a non-zero dipole moment.
Computing the total charge I find
′
Z a
πz
Q=k
sin
dz ′ = 0
(10)
a
−a
because the integrand is an odd function of z ′ and we are integrating over a symmetric interval. The dipole moment is
Z
p~ = ~r′ ρdτ ′ ;
(11)
here the x and y components are zero by symmetry so the only
(potentially) non-zero component is
Z
pz = z ′ λdz ′ .
(12)
For the charge distribution in this part of the problem this is
′
Z a
πz
′
pz =
kz sin
dz ′
(13a)
a
−a
′
Z a
πz
′
= 2k
dz ′
(13b)
z sin
a
0
′ a
′ ′
az
πz
πz
(13c)
−
cos
= 2k sin
a
π
a
0
2ka2
.
(13d)
=
π
The potential from the multipole expansion (7) is
1 ~p · rˆ ka2 cos θ
V =
=
.
(14)
4πǫ0 r 2
2π 2 ǫ0 r 2
Here the monopole term is zero, and we don’t calculate the quadrupole
because the dipole term is not zero.
6
Solutions to Problem Set 7, Physics 370, Spring 2014
(c) (3 points possible) The final charge distribution is λ = k cos(πz/a)
and looks like
which suggests zero total charge. Furthermore, if you divide this
charge distribution at z = 0, you can see that at z < 0 we have
a dipole distribution and at z > 0 we have another dipole distribution with the opposite orientation. Essentially, two opposing
dipoles separated by a small distance is a classical quadropole distribution. We’ll need to verify this. I can verify its total charge is
zero,
′ a
Z a
πz ka
′
′
= 0,
(15)
sin
Q=k
cos(πz /a)dz =
π
a −a
−a
and that its dipole moment is also zero,
′
Z a
πz
′
pz = k
z cos
dz ′ = 0
a
−a
(16)
because the integrand is an odd function of z ′ . That means we
have to evaluate the quadrupole moment
Z
k
3 cos2 θ′ 1
′ 2
V ≈
ρdτ ′
(17a)
(r )
−
4πǫ0 r 3
2
2
′
Z
πz
k (3 cos2 θ − 1) a ′ 2
dz ′ .
(17b)
(z
)
cos
=
4πǫ0 r 3
2
a
−a
The integral in (17b) is
′
′
Z a
Z a
πz
πz
′
′ 2
′ 2
dz = 2
dz ′
(z ) cos
(z ) cos
a
a
−a
0
"
′
a 3 πz ′ πz
=2
cos
+
2
π
a
a
a3
= −4 2 .
π
(18a)
πz ′
a
2
(18b)
(18c)
!
− 2 sin
πz ′
a
#a
0
Solutions to Problem Set 7, Physics 370, Spring 2014
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With this the potential of the distribution becomes
V =−
ka3 3 cos2 θ − 1
.
2π 3 ǫ0
r3
(19)
(Thanks to Dr. Craig for providing the initial draft of this solution in
LATEX form.)
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Solutions to Problem Set 7, Physics 370, Spring 2014
3. Griffiths Problem 3.53 (tweaked): In Example 3.8 we determined
the electric field outside a spherical conductor (radius R) placed in a
~ 0 using the Separation of Variables technique.
uniform electric field E
Now solve the problem now using the method of images, and check that
your results agrees with equation 3.76 which states
R3
V (r, θ) = −E0 r − 2 cos θ.
r
[Hint: Use Example 3.2, but put another charge, −q diametrically op1 2q
posite q. Let a → ∞, with 4πǫ
2 = −E0 held constant. Hint #2: It
0 a
will be useful to look at Example 3.10 and how they handle r1+ − r1−
in order to simplify this problem. But remember, in this case we are
sometimes working in the region “between” the charges instead of far
outside them, so the simplifying assumptions will be a bit different.]
(15 points possible) I take Griffith’s hint to heart and set up the image problem as shown in the figure below, with two “real” charges (at
y ± a) that generate the external electric field and the two “image”
charges in the sphere (at y ± b).
y
r4
-q
P
r3
r2
θ
+q’
-q’
r1
+q
x
{
{
{
-a
{
R
-b
b
a
(NOTE: Here +q ′ has a negative charge (whose value is given by equation 3.15, q = − Ra q, but I keep that notation to note it is the image
charge related to +q. Similarly −q ′ has a positive charge.) Now recall,
the goal in any image problem is to have the potential at the boundaries of your “image problem” match the real boundary conditions.
Given this scenario, the total potential can be written as the sum of
Solutions to Problem Set 7, Physics 370, Spring 2014
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four point charge potentials (using polar coordinates as in Example 3.2
of the book, such that θ is the angle from the x axis):
1 q
1 q′
1 q′
1 q
+
−
−
4πǫ0 1 4πǫ0 2 4πǫ0 3 4πǫ0 4
1
1
1
1
1
′
=
q
+q
−
−
4πǫ0
1
4
2
3
V (r, θ) =
(20a)
(20b)
So now we just need expressions for the separation vector magnitudes.
Exploiting the law of cosines, I can state the separation magnitudes as:
2
1
2
2
2
3
2
4
= r 2 + a2 − 2ra cos θ
= r 2 + b2 − 2rb cos θ
= r 2 + b2 − 2rb cos(π − θ)
= r 2 + a2 − 2ra cos(π − θ)
= r 2 + b2 + 2rb cos(θ)
= r 2 + a2 + 2ra cos(θ)
(21a)
(21b)
(21c)
(21d)
where for 3 and 4 I had to exploit the fact that cos(π − θ) = − cos θ.
The next step is to work toward expressions for inverses of the four
separation magnitudes in equation 20b using the multipole expansion
done in Example 3.10 as inspiration. However, keep in mind, that in
Example 3.10, Griffiths was working far outside the charge pair. I want
to build an accurate approximation between the two charges at located
~ near the origin
at +a and −a that we are using to generate a uniform E
by taking a → ∞. In essence, in this problem, I can re-write equations
21a and 21d as
h
r 2 i
r
r
2
2
2
1 − 2 cos θ +
≈ a 1 − 2 cos θ
(22a)
1 = a
a
a
a
h
r 2 i
r
r
2
2
2
1
+
2
≈
a
1
+
2
=
a
cos
θ
+
cos
θ
(22b)
4
a
a
a
2
since we are working in the region between those charges and ar ≪ ar ,
I can drop that term. Now using the binomial expansion, I can simplify
this further:
i−1/2
1
r
1
r
1h
r
1
1
1 − 2 cos θ
1 − (−2) cos θ ≈
1 + cos θ
≈
≈
a
a
a
2
a
a
a
1
(23a)
h
i
−1/2
r
1 r
r
1
1
1
1
1 + 2 cos θ
1 − 2 cos θ ≈
1 − cos θ
≈
≈
a
a
a
2 a
a
a
4
(23b)
Solutions to Problem Set 7, Physics 370, Spring 2014
10
And therefore
1
r
r
1
1 + cos θ −
1 − cos θ
a
a
a
a
1
4
1
1
2r
−
≈ 2 cos θ
a
1
4
1
−
1
≈
(24a)
(24b)
A similar approach to reworking equations 21a and 21d can be done for
equations 21c and 21b, but here we are working
in the regime far away
b 2
from those charges, where r ≫ b and r ≪ rb (this was essentially
the multipole expansion we work through in Example 3.10), and so I
can re-write those equations as:
"
2 #
b
b
b
2
2
2
1 − 2 cos θ +
≈ r 1 − 2 cos θ
(25a)
2 = r
r
r
r
"
2 #
b
b
b
2
2
2
1 + 2 cos θ +
≈ r 1 + 2 cos θ
(25b)
3 = r
r
r
r
. which we can then simplify the reciprocals of using the binomial
theorem:
−1/2
1
1
b
1
b
b
1
1
≈
1 − 2 cos θ
1 − (−2) cos θ ≈
1 + cos θ
≈
r
r
r
2
r
r
r
2
(26a)
−1/2
b
1 b
b
1
1
1
1
1 + 2 cos θ
1 − 2 cos θ ≈
1 − cos θ
≈
≈
r
r
r
2 r
r
r
3
(26b)
And therefore
1
2
−
1
3
≈
2b
cos θ.
r2
(27a)
Using equations 24b and 27a, I can write my expression for the potential
in this problem (equation 20b), as
2r
2b
1
′
q
cos θ + q
cos θ
(28)
V (r, θ) ≈
4πǫ0
a2
r2
11
Solutions to Problem Set 7, Physics 370, Spring 2014
h
Now we can apply equations (3.15) q ′ = − Ra q and (3.16) b =
of Example 3.2 in the textbook to say
2
1
2r
R
2R
V (r, θ) ≈
q
cos θ − q
cos θ
4πǫ0
a2
a
ar 2
2
2 R3
1
q r cos θ − q 2 2 cos θ
≈
4πǫ0 a2
a r
3
R
1 2q
r
−
cos θ.
≈
4πǫ0 a2
r2
R2
a
i
out
(29a)
(29b)
(29c)
In the setup of this problem, I note that I am told that in the vicinity
1 2q
of the sphere, 4πǫ
2 = −E0 , therefore, in the vicinity of the sphere, I
0 a
have:
R3
V (r, θ) ≈ −E0 r − 2 cos θ
(30)
r
which is the same result as equation (3.76) out of the textbook.
Solutions to Problem Set 7, Physics 370, Spring 2014
12
~ field there is
4. When a neutral conductor is placed in an uniform E
~ field in the conductor. When a neutral dielectric (aka
no internal E
~ field the internal E
~ field is not
insulator) is placed in an uniform E
zero, although it is typically a bit lower than the electric field outside
the dielectric.
~ field being zero in a conductor
(a) The typical explanation of the E
is that the charges move around to cancel it out. Clearly explain
how charges moving in a conductor could result in the total electric
field inside a conductor being zero. HINT: Do the charges in the
~ field? Yes or no? Explain
case of a conductor set up their own E
your answer.
(b) The charges in an dielectric are typically called “bound charges”
since they are bound to atoms and “can’t move.” Explain how
charges that “can’t move” can generate a dipole moment in re~ field. HINT: It would probably be helpsponse to an external E
ful to know what charge configuration produces a dipole moment.
(c) So, fundamentally, what is different about the response of charges
~ field in the case of a conductor versus a dielectric?
to an external E
~ field will cause any charges
(a) (4 points possible) An external E
in a conductor to feel a force. Charges in a conductor are by
definition free to move about and as such they move until a counterbalancing force is set up. That counterbalancing force is due to
~ field the free charges set up by configuring themthe internal E
selves in a dipole charge distribution. It is only when the internal
~ field produced by the charges in the conductor precisely canE
~ field, so that the total E
~ = 0, that those
cels out the external E
charges will no longer experience a force and stop moving.
(b) (4 points possible) While the charges in a dielectric are bound
to atoms, atoms themselves consist of components with different
~ field, the nucleus, which is
charges. When placed in an external E
positively charged, will experience a force in the opposite direction
as the electrons in the electron cloud around the nucleus. As such,
Solutions to Problem Set 7, Physics 370, Spring 2014
13
the nucleus and electron cloud typically shift a bit from being cocentric... the shift stops when the electric force between them is
~ field.
balancing the electric force on them due to the external E
Therefore, the electron cloud typically ends up slightly off center
of the nucleus, resulting a small dipole moment for the atom. Sum
this effect up over all the atoms in a dielectric and you can explain
the dipole moment that gets set up in the direction of the electric
field.
(c) (2 points possible) Fundamentally, the charges in a conductor
~ field to cancel
can only stop moving by setting up an internal E
~ The charges in dielectric, which are bound to
out the external E.
atoms, have the restoring force in the form of individual Coulomb’s
law forces between the electron clouds and nuclei of each atom in
the dielectric, so those charges can’t move as far, so any coun~ field they generate will not necessarily cancel out the
teracting E
~ field.
entire external E
5. Griffiths Problem 4.03 (tweaked): According to equation 4.1 (~p =
~ the induced dipole moment of an atom, p~, is proportional to the
αE),
~ This is a “rule of thumb,” not a fundamental law,
external field, E.
and it is easy to concoct exceptions — in theory. Suppose, for example, the charge density of the electron cloud were proportional to the
distance from the center, out to radius R (e.g.- ρ(r) = Ar where A
is a constant with the right units). To what power of E would p be
proportional to in that case? HINT: Assume the electron cloud is
spherically symmetric and determine a function for the electric field
given the charge distribution mentioned. This “internal” electric field
has to balance the external one when the nucleus is off center by some
amount d. Understand Example 4.1 for guidance.
(10 points possible) Start off by assuming ρ(r) = Ar in spherical coordinates. Since this is a spherically symmetric charge distribution,
we expect a spherically symmetric electric field. As such we can use a
spherical Gaussian surface centered at the origin and apply the integral
form of Gauss’ law (equation 2.44) to determine the “internal” electric
14
Solutions to Problem Set 7, Physics 370, Spring 2014
field due to this electron cloud:
I
~ · d~a = Qenc
E
ǫ0
Rr Rπ
E(4πr 2 ) =
(31a)
R2π
Ar ′ r ′2 sin θdθ′ dφ′ dr ′
r ′ =0 θ ′ =0 φ′ =0
4πA
E(4πr ) =
ǫ0
2
Zr
ǫ0
r ′3 dr ′
(31b)
(31c)
r ′ =0
πA 4
r
ǫ0
A 2
=
r
4ǫ0
E(4πr 2 ) =
Eint
(31d)
(31e)
This “internal” electric field balances the “external” electric field when
the nucleus is off-center by distance d such that:
~ ext − E
~ int = 0
E
A 2
d =0
Eext −
4ǫ0
A 2
Eext =
d
4ǫ0
(32a)
(32b)
(32c)
Now let’s rearrange equation 32c to solve for the offset distance d:
4ǫ0 Eext
rA
4ǫ0 Eext
d=
A
r
ǫ0 Eext
=2
A
d2 =
(33a)
(33b)
(33c)
Given a charge of q = e, then the induced
q dipole moment is (based on
1/2
equation 3.101) just p = qd = ed = 2e ǫ0 EAext . So p ∝ Eext , which is,
of course, different than equation 4.1.
Solutions to Problem Set 7, Physics 370, Spring 2014
15
6. Griffiths Problem 4.10 (tweaked): A sphere of radius R carries a
polarization P~ (~r) = k~r, where k is a constant and ~r is the vector from
the center.
(a) Calculate the bound charges σb and ρb .
(b) What is the meaning behind these “bound charges”? Are they
actual charges and if not, why do we bother to compute them? In
other words, of what use are they?
(c) Find the electric field inside and outside the sphere. HINT: Answering part (b) first might help you with your approach this
problem. Also, the solution to Griffiths 2.12 which was on Problem Set 3, might help.
(a) (4 points possible) The bound charges for a given polarization P~
~ · P~ .
are given by the equations 4.11, σb ≡ P~ · n
ˆ , and 4.12, ρb ≡ −∇
Its a simple matter to note that for a sphere n
ˆ = rˆ and so the
bound surface charge density is
σb = P~ · n
ˆ
h
i
= P~ · rˆ
r=R
(34a)
= [k~r · rˆ]r=R = [krˆ
r · rˆ]r=R
= kR
(34b)
(34c)
and the bound volume charge density
~ · P~
ρb = −∇
(35a)
1 ∂ 2 r Pr [using divergence of radial component in spherical coordinates]
=− 2
r ∂r
(35b)
1 ∂ 2 1 ∂
1
=− 2
r kr = − 2
kr 3 = − 2 3kr 2
(35c)
r ∂r
r ∂r
r
= −3k.
(35d)
(b) (2 points possible) The meaning of these bound charges is that
the potential produced by an object of polarization P~ is the same
Solutions to Problem Set 7, Physics 370, Spring 2014
16
as that produced by an object with volume charge density ρb =
−3k and surface charge density σb = kR. These are “real” charges
that are generating the dipole, although it may be hard to separate them from the induced dipole (you can’t easily separate the
electronic cloud from the nucleus, and when you do, it is no longer
a bound charge). So why bother with these bound charges instead
of working with the induced dipoles? Its easier to compute with
the electric field of a charge distribution than that of a bunch of
infinitesimal dipoles that contribute to the polarization.
(c) (4 points possible) For the electric field inside the sphere(r ≤ R)
with polarization P~ (~r) = k~r should be equal to that of a sphere
with a volume charge density equal to the bound volume charge
density we computed in part (a). Using Gauss’ law in problem
2.12 we found that inside the sphere,
~ = ρ rˆ
r.
E
3ǫ0
(36)
Using this solution, it is clear the electric field within the sphere
is:
−k
~ = −3k rˆ
E
r=
rˆ
r.
(37)
3ǫ0
ǫ0
For outside the sphere (r > R), symmetry states the electric field
must be radial, therefore we can use a spherical Gaussian surface
to argue the electric field should be equal to that of a sphere with
the equivalent total charge, such that
I
~ · d~a = Qenc
E
(38a)
ǫ0
ρb 43 πR3 + σb 4πR2
2
(38b)
E(4πr ) =
ǫ0
ρb 3
−3k 3
R + σb R2
R + (kR)R2
3
E= 3
=
(38c)
ǫ0 r 2
ǫ0 r 2
=0
(38d)
So the electric field outside this polarized sphere is zero!
Solutions to Problem Set 7, Physics 370, Spring 2014
17
7. Griffiths Problem 4.14 (tweaked): When you polarize a neutral
dielectric, charge moves a bit, but the total remains zero. This fact
should be reflected in the bound charges σb and ρb . Given the expressions for the bound charges from Equations 4.11 (σb ≡ P~ · n
ˆ ) and 4.12
~
~
(ρb ≡ −∇· P ), develop an expression for the total charge and show that
the total charge vanishes. What does this mean? HINT: If you want
to do this painlessly, you’ll want to exploit the divergence theorem.
(5 points possible) The total charge should be the integral over area
of the surface charge density, σb , plus the integral over volume of the
volume charge density, ρb . Using this idea and equations 4.11 and 4.12,
we can write the expression for total charge as:
I
Z
Qtot = σb da + ρb dτ
(39a)
=
=
S
I
S
I
V
P~ · n
ˆ da −
P~ · d~a −
S
Z
ZV
~ · P~ dτ
∇
~ · P~ dτ
∇
(39b)
(using d~a = daˆ
n)
(39c)
V
~ P~ states (equation 1.56)
However, the divergence theorem applied to ∇·
that:
Z
I
~
~
(∇ · P )dτ = P~ · d~a
(40)
V
S
therefore by looking at equation 39c we see the total charge must be
zero!