Distribution of static equilibrium points in static electric field produced by fixed electrons

The Journal of China Universities of Posts and Telecommunications August 2014, 21(4): 92–95 www.sciencedirect.com/science/journal/10058885

http://jcupt.xsw.bupt.cn

Distribution of static equilibrium points in static electric field produced by fixed electrons GUO Feng ( ) School of Science, China University Mining & Technology(Beijing), Beijing 100083, China

Abstract Fix n electrons on the disk | z | ≤R . They generate a static electric field. Let a be an electron located in the disk | z | ≤R sin π n . It shows that the closed disk with center a and radius R contains at least one static equilibrium point. Keywords fixed electrons, static electric field, static equilibrium points

1

Introduction

In this article, the basic notations and definitions was referred to Refs. [1–2]. Many infinitely electrons are fixed in a plane (electron may be a unit charged wire perpendicular to the plane). Then they generate a static electric field. In 1995, Qiao [3] studied the following problem: whether the field infinitely has many static electric equilibrium points. If we place a free electric charge at a static electric equilibrium point, the charge cannot make displacement, the resultant force that the charge receive is 0, and the electric field intensity at the point is 0. What’s more, the potential gradient at the point is also 0. Obviously, the equilibrium point is important for the field. Considering electric plane with direction as the complex plane，we assume the locations of electrons in the complex plane are respectively {ω j }∞j =1 . Then the logarithmic

potential

∞

∑ log | ω − ω

j

|

of

the

field

is

given

by

and the gradient of this potential is

j =1

∞

1

∑ω −ω j =1

∞ ),

. If j

∞

∑ (1 | ω |) < ∞ j

(the electrons extend to

j =1

∞

∑1 (ω − ω ) j

is a meromorphic function and its

j =1

Received date:

Corresponding author: GUO Feng, E-mail: [email protected] DOI: 10.1016/S1005-8885(14)60321-6

zeros are static electric equilibrium points. In the filed above, Qiao proved that it has infinitely many static electric equilibrium points; especially, when all the electrons lie on a straight line, then equilibrium points of the electric field lie on this straight line and there is one and only one equilibrium point between any two adjacent points of {ω j }∞j =1 . In details, it is: Theorem A Let {ω j }∞j =1 be a sequence of distinct complex

numbers

satisfying

∞

∑1 | ω j =1

∞

∑1 (ω − ω ) j

j

| < ∞.

Then

has infinitely many zeros.

j =1

Similarly, for static electric field produced by many finitely fixed electrons, what’s the distribution of static equilibrium points is also significant. Fix n( n≥2 ) electrons in a disk whose radius is R, they can generate a static electric field, suppose the electric potential of every electric is 0 and the electric potential of ∞ is ∞ , it is natural to study such a problem: take every electron as center to make a closed disk with radius l such that the disk contains at least one static equilibrium point, what’s the connection of l and R ? Let D(a; R) = {z :| z − a | ≤R} be the closed disk with center a and radius R, the problem above can be refined as: without loss of generality we assume R = 1 . Let all the locations of fixed electrons are zν (ν = 1, 2,..., n) , the

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GUO Feng, et al. / Distribution of static equilibrium points in static electric field produced by fixed electrons

93

n

electric potential is given by p(z) = ∏ ( z − zν ) , then the ν =1

locations of static equilibrium point are the zeros of p ′( z ) . For l and R, Sendov raised a conjecture [4]: n

p(z) = ∏ ( z − zν ); n≥2

(1)

ν =1

If all zeros of a polynomial are all in the unit closed disk D(0;1) . Then, for zν (ν = 1, 2,...n) , the closed disk D( zν ;1) contains at least one zero of p ′( z ) . The conjecture was published in Ref. [4], it attracted attention of many mathematicians. For example, the conjecture is true when p(0) = 0 [5], and when the degree of p ( z ) is at most eight [6-9], but it has not been solved completely. Here we introduce some important results. Let a be a zero of p( z ) , M J Miller [7] proved that if | a | is close to 1 enough. Then the conclusion is true for

a . That is Ref. [10]: Theorem B Let p( z ) be a polynomial with degree n + 1 , all its zeros belong to D(0;1) . Let β be a zero and | p |β be the shortest distance between zero of p′( z ) and β . Then there exists a sequence {K } , K n > 0 , such that lim K n = 1/ 3 and when β is closed to 1 ∞ n n =1

x →∞

enough, | p |β ≤1 − K n (1 − β ) . For any zero zν of p( z ) , we can make a proper rotation for ^ such that zν eiθ ∈ \ , it is easy to know that D( zν ;1) contains at least one zero of p′( z ) , equivalent to that D( zν eiθ ;1) contains at least one zero of p '( zeiθ ) . The hypothesis that zero close to 1 enough in Theorem B is not necessary, it has proved that when zero zν satisfies | zν | close to 1, D( zν ;1) contain at least one zero of p′( z ) . The proof of Miller require | β |> 1 − (1/ n ) here we

Fig. 1

The result of Miller

M J Miller [11] has studied the supremum of | p |β in Theorem B and have: Theorem B* Define S (n, β ) to be the set of complex polynomials of degree n ≥ 2 with all roots in the unit disk and at least one root at β . For a polynomial p( z ) , define | p |β to be the distance between β and the

closest root of the derivative p( z ) . Define: rn ( β ) = sup{| p |β : p ( z ) ∈ S (n, β )}

(2)

Then when | β | is close to 1 enough, 3 (1− | β |) + O(1− | β |) 2 (3) 10 Dégot, Jérôme has studied Sendov’s conjecture for high degree polynomials in 2014 [12]: Theorem C Let p ( z ) be a polynomial with all its zeros in the unit closed disk D(0;1) , and a be a zero of p( z ) , there exists an integer N such that the disk | ζ − a| rn ( β )≤1 −

≤1 contains a critical point of p( z ) when the degree of p( z ) is larger than N. Here we give a figure to show the result, which can be seen in Fig. 2.

give a simple figure of polynomial with 6 degrees for showing the result of Miller’s [7]: In Fig. 1, the radius of disk OR is 1 and the radius of disk OK is 5/6. The point A is a zero of polynomial p( z ) in annulus (undertint domain), the location of A is zν , the closed disk AB with center A and radius 1 − K 6 (1− | zν |) contains at least one zero of p ′( z ) .

Fig. 2

The result of Dégot, Jérôme

94

The Journal of China Universities of Posts and Telecommunications

In Fig. 2, let a be any zero of polynomial p( z ) , the point A is the location of a, all the other black points are the zeros of p ( z ) , there exists an integer N such that if the number of zeros of p( z ) is larger than N, then the closed disk AB contains at least one zero of p′( z )

2

Main result and it’s proof

2014

arg{ p0 ( z ′) − p0 (a )} − arg{ p0 ( z ′′) − p0 (a)} = n

n

j =1

j =1

∑ arg( z ′ − a) − ∑ arg( z ′′ − a)≤nθ < 2π

(4)

where θ is the angle subtended at a by the tangent from a to the circle | z |= sin π / n . Then the segment z ′, z ′′ cannot be mapped into a closed curve, and therefore p0 ( z ) is univalent in D(0;sin π / n).

The authors prove here that when the zeros close to the origin enough, they will have the same property. Let a be a zero of p( z ) in Sendov’s conjecture, when a belong to closed disk D(0;sin π / n) , it is proven that D(a;1) contains at least one zero of p ′( z ) . Theorem 1 Let p ( z ) be a polynomial with degree n , all the zeros of p ( z ) are in D(0;1) . Assume a be a zero of p( z ) with | a | ≤ sin π / n , then D(a;1) contains at least one zero of p ′( z ). We give a figure of polynomial with degree 6 to show the result:

Fig. 3

The result of Theorem 1

In Fig. 3, the radius of disk OR is 1, the radius of OK is sin π / 6 = 1/ 2 . Let the point A be a zero of p( z ) in the closed disk OK (gray disc), then the closed disk AB with radius 1 (dotted-line disk) contains at least one zero of p′( z ) Let p( z ) be a polynomial with degree n which satisfies p ′( z ) ≠ 0 when z ∈ D(0; R ) , then p( z ) is univalent in D(0;sin π / n). Lemma 1

The proof of Lemma 1 can be referred to Refs. [13–15]. Here we give an example of the polynomial p0 ( z ) = ( z − a ) n ( | a |> 1) to show the result: For any two different points z ′, z ′′ ' ∈ D(0;sin π / n)

Fig. 4

The result of Lemma 1

In Fig. 4, the point A is the location of zero a of polynomial p0 ( z ) , α = z ′Az ′′ is the angle from any two different points z ′, z ′′ to the point A and the ∠θ = MAN is the possibly widest angle of α , because sin θ / 2 = OM / OA = [( sin π / n ) / OA] < sin π / 2, then θ < 2π / n . Proof of Theorem 1: If the conclusion is not true, then there exists a zero a ∈ {zν | ν = 1, 2,..., n}, | a | ≤ sin π / n , such that p′( z ) have no zero in D(a;1) . Let Q( z ) = p ( z + a) . Then when z ∈ D(0;1) we have Q(0) = 0 and Q′(0) ≠ 0 . By Lemma 1, Q( z ) is univalent in D(0;sin π / n) . Let

z0 be a point with | z0 |=| a | and | Q( z0 ) |=

min | Q( z ) | . | z | =|a|

Because Q′(0) ≠ 0 when z ∈ D(0;1) and we get: | Q( z0 ) |= min | Q( z ) | (5) | z |=|a|

In Eq. (5), the segment [0,| Q( z0 ) |] has a lift curve γ : z = z (t ) (0≤t≤1) in D(0;sin π / n). It is easy to see that arg Q( z (t )) (0≤t≤1) is constant, and then: 1

| Q( z0 ) |= ∫ | Q( z (t )) || z ′(t ) | dt 0

We can write:

(6)

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GUO Feng, et al. / Distribution of static equilibrium points in static electric field produced by fixed electrons n −1

Q′( z ) = n∏ ( z − ω j ) (n≥2)

(7)

j =1

Because Q′( z ) ≠ 0 in D(0;1) , we have | ω j |> 1 . Thus when | z | ≤1 , we can have the equations as follows: | z − ω j | ≥ | ω j | − | z |> 1− | z | (8) | Q ′( z ) |> n(1− | z |) n −1 Therefore, when | z |=| a | , we have:

(9)

1

| Q( z ) | ≥ | Q( z0 ) |> n ∫ n(1− | z (t ) |) n −1 | z ′(t ) | dt≥ 0

1

n ∫ n(1− | z (t ) |) 0

n −1

d | z (t ) | = 1 − (1− | a |) n

(10)

Further, when | z − a |=| a |, | p ( z ) |> 1 − (1− | a |) n It is obvious that 0 ∈ {z :| z − a |=| a |} , and then we get: | Q(− a) |=| p (0) |=| z1 ⋅⋅⋅ zn |> 1 − (1− | a |) n

(11)

By Eq. (11), we have: (1− | a |) n + | z1 ⋅⋅⋅ zn |> 1, and note that | a | ≥ | z1 ⋅⋅⋅ zn | , then (1− | a |) n + | a | ≥(1− | a |) n + | z1 ⋅⋅⋅ zn |> 1

(12)

However, we see: (1− | a |) n < 1− | a |< 1 , when 1≥ | a |

(13)

therefore (1− | a |) + | a |< 1 , which Eq. (12) and the proof n

is completed.

3

Conclusions Fix n electrons in the disk | z | ≤R . Then they generate

static electric field. Let a be an electron in the closed concentric disk with radius | z | ≤R sin π / n , then the

closed disk

D ( a; R )

95

contains at least one static

equilibrium point. It describes the distribution of static equilibrium points in the static electric field produced by finite fixed electrons.

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