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SQUARE 2 MAGAZINE
Number 24. April 2011.
We remind you that news about what you are doing should be sent to the Editor,
David Penman, email address [email protected] You can access all copies of
the magazine at our website http://www.essex.ac.uk/maths/dept/square2/index.shtm.
Earnshaw’s theorem.
Suppose I have a number of fixed charges at various points in space. Can I use these
charges to hold some other charge fixed at some other point? (We are talking here, to
begin with an idealised model, where the charges have no masses so there are no
gravitational or other forces between them: just the usual forces repulsion between
two positively charged, or two negatively charged, particles, or of attraction between
two oppositely charged particles).
The answer is, rather obviously, yes: indeed suppose we put four negative charged
particles, with equal negative charges on each of them, at the four vertices of a square.
Then rather clearly a positively charged particle at the centre of the square will
experience (add vectors!) overall no force so should remain where it is.
However the question becomes rather harder if we ask whether we can keep a point
charge in stable equilibrium this way. To get the idea of this, reflect again on the
square example. If the positively charged particle is deflected even slightly in one
direction, it will of course start being pulled towards some of the charges: in other
words, its equilibrium is not stable. Can one find another configuration where there is
a charge in stable equilibrium?
The answer to this question is “No”, and the answer was found by Samuel Earnshaw,
a Sheffield clergyman in 1842. Earnshaw had been born in Sheffield and schooled
there and in nearby Doncaster. He subsequently studied mathematics at St. John’s
College Cambridge where he was Senior Wrangler (i.e. best student in mathematics in
the year in which he graduated) in 1831. This would normally have made him moreor-less automatically a fellow at St. John’s eventually, but in 1832 he married his
childhood sweetheart Ann Wall and of course in those days the fellows had to be
unmarried. He then worked for several years as “coach” for candidates for the
Cambridge examinations, apparently very successfully, and was in 1846 appointed to
the parish church St. Michael Cambridge. However he had health problems in 1847
and moved back to Sheffield where he was Chaplain on Queen Mary’s foundation
from 1847-1888 when he died in Earnscliffe, a suburb of Sheffield. In addition to his
many writings on mathematics and physics, in his role as a minister he also wrote a
number of theological treatises. No photograph of him appears to be extant, though a
photograph of a bust of him can be seen at
http://www.chem.yale.edu/~chem125/levitron/Earnshaw.html
It is not hopelessly difficult to get an idea of why Earnshaw’s theorem might be true.
Suppose we could, with a collection of fixed charges, keep another in stable
equilibrium. Then there must be some small ball around the charge being kept fixed,
not containing any of the other charges, where all the overall force lines are pointing
towards the point it is being kept fixed at. (Otherwise it could “get away” along the
force line going out of the ball, so the equilibrium would not be stable). That is saying
that the force lines point in towards the point, i.e. that it is a sink.
Recall that the divergence of a vector in component form,
each is a (well-behaved) function of , , , is
,
,
where
•
It is a measure, crudely speaking, of the extent to which a point is a sink or source. It
is not zero when the point is a sink or a source. So if our putative point of stable
equilibrium existed, it would have to be a sink
However it is a known fact about electrostatics that, letting
denote the force
vector at the position , is divergenceless, i.e. •
0. This is one of
Maxwell’s equations. (Strictly speaking, the equation in question is usually written as
•
/ where is the electric field and is the (continuous) charge density.
Here the electric field is the force felt by a unit charge, but of course that is a scalar
multiple of the force F, and in the situation we are discussing there is no continuous
charge, so the two equations are equivalent).
This contradiction now completes the proof.
Somewhat more mathematically, we have that the force vector is a gradient of some
potential function:
for a suitable function . This means
, ,
: this fact follows from another of Maxwell’s equations, namely
that the curl of is the rate of change of the magnetic field induced by the electric
field with time. (The curl of a vector is a measure of the amount of rotation going on
, ,
at the point – hence its name. The formula for the curl of a vector
is
(again choosing a co-ordinate system)
,
It is easy to see that if
,
,
,
,
,
then
just using
symmetry of mixed partial derivatives. The converse direction (that
implies that
is harder and does involve some assumptions on the domain
one is operating in, e.g. that it is convex (the line segment between any two points,
both in the domain, is entirely contained in the domain).
In our case there is no time-dependence and so the right-hand side is zero, and it is a
standard theorem that, in a “simply connected region” (roughly, one where there no
holes: the ball around a point is such a region) having curl zero implies that
This means that •
0, i.e. that satisfies Laplace’s equation ∆
0.
.
It is a standard exercise in courses on Laplace’s equation that nonconstant functions
which satisfy it cannot have maxima or minima, just saddle points (i.e. unstable
equilibria), inside the open connected set with compact closure. This is clearly
equivalent to Earnshaw’s theorem. To see roughly why it is true, we illustrate the
argument in two dimensions, where the equation is
0
At a maximum, we must obviously have
0 and
0,
0.
Since we also have
0 we must have
0,
0. To show that
actually this case cannot arise other than for constant functions, let
| | where
0: that is, we perturb a little (which we can only
do if we are in the interior). Then
∆
∆
∆| |
0
4 .
But also ∆ must be 0 at its maxima (by the same argument as for . Thus
cannot have its maximum in the interior . Thus since is continuous on the compact
set
we have it attains a maximum somewhere or other: by the previous
argument, this somewhere must be on the boundary
we
, say at Then, for
have the obvious string of inequalities
|
Letting
0 we get the result:
max
|
max
max | |
.
We leave as an exercise the fact that this “maximum principle” implies uniqueness,
i.e. that there is only one solution of Laplace’s equation satisfying the boundary
condition of being zero on
.
Earnshaw’s theorem actually applies more generally. It also applies to magnetic fields
(by more use of Maxwell’s equations) and to gravity, which also leads to a potential
satisfying Laplace’s equation. And also, of course, by superposition, to combinations
of forces of these types. It turns out that for an irrotational vector field (i.e. one whose
curl is the zero vector) in 3-dimensional space, having divergence zero is essentially
the property of having an inverse square law, which perhaps goes some way towards
explaining the fact that there are many inverse square laws in nature. To give some
rather informal idea of this argument, we note that if a vector has curl equal to the
zero vector, then (in a reasonable domain) it is the grad of some function : thus, if
its divergence is also zero, we have that ∆
0. Now since the force should be
emerging in a spherically symmetric fashion, the obvious thing to try is to write
Laplace’s equation in spherical polar coordinates, and think that the force should
depend only on the distance from the source rather than the angles it makes with
various axes. The derivation of Laplace’s equation in (any) polar coordinates is a truly
horrible exercise which generations of mathematics undergraduates have sweated
over, dropping signs (and worse) to get wholly duff answers: we will merely crib,
sorry, give the answer, which is that (remembering we are assuming there is no
dependence on the angles: the general formula is much nastier)
∆
.
Thus solutions have the property that
for a suitable constant . Now we do the
comparatively mild exercise of writing the grad of a function in polar coordinates.
Again, assuming no dependence on angles, we get that the force
is
times a unit
vector in the direction of the current position, and so we indeed get to an inverse
square law. This perhaps gives some insight into the fact that inverse square laws are
widespread in nature. Not all forces in nature obey inverse square laws – for
example, the van der Waals force in particle physics (a very weak force between
molecules compared with, say, normal chemical bonds) is roughly proportional to the
inverse sixth power of the distance between the two molecules. [Note also that in
dimensions, a vector field has curl 0) if and only if (up to technicalities) it obeys an
inverse
1
th power law.]
Of course our analysis above has not been terribly rigorous, and there are “loopholes”.
For example, the theorem would appear to rule out the possibility of “magnetic
levitation” - that is, keeping a body in a certain position using only magnetic forces.
However there are certain substances which, when put into a magnetic field, produce
a magnetic field opposite to the external field – this property is called diamagnetism.
(It goes back to the 18th century observation that the rare elements bismuth and
antimony were repelled by magnetic fields.) In this situation, one can levitate an
object using fixed magnets only. And indeed frogs and various other objects, e.g.
superconductors, have been made to magnetically levitate in this way. Other
“loopholes” include having the fixed magnet spinning.
(The magnetic levitation idea is a potentially useful one in transport, in that if a train
or similar can be magnetically held above tracks, and pushed along them in some
way, it might in the long run be able to move much faster than any wheel-bases
system. Indeed Japanese scientists have got an experimental MAGLEV train to run at
speeds of up to about 580 kilometres per hour. Some older readers may dimly recall
the maglev transit between the terminal and railway station at Birmingham airport in
the early 1990s. However there are still serious problems with the technology in
practice).
Earnshaw’s theorem did create difficulties when people first tried to investigate the
structure of the atom, as atoms do appear to stay put much of the time. J. J. Thomson
was apparently in some sense led to his “plum-pudding” model of the atom, where the
the (negatively charged) electrons of the atom scattered around inside the (positively
charged) nucleus by the difficulties arising from Earnshaw’s theorem. Of course our
current model of the atom is more or less opposite to that, with the electrons outside
the nucleus: but of course the point is that nowadays the electrons are orbiting the
nucleus, so we do not have fixed charges any more.
Problems Corner
Remember our problems from last time:
Problem 1. Choose two points uniformly at random on (not inside!) a circle of radius
1. Show that the expected distance between them is 4/ .
Solution. Without loss of generality (translate!) the circle is
1 and
(rotate!) one of the points is 1,0 . Let the other be cos , sin
: then the distance
squared is cos
1
sin
2 2 cos
4 sin
. (Who forgot the
double angle formula cos 2
1 2 sin
?). Thus the actual distance is
2 sin
. We thus want the average of this quantity, for uniformly distributed on
0,2 . Since of course a random variable uniformly distributed on , has density
1/
we get
2 sin /2
4cos /2
2
1
2
4
1
2
4
4
2
as required.
Problem 2. Recall that a Mersenne prime is a prime number of the form 2
1
where is itself a prime. Show that a Mersenne prime cannot have the property that
2
1 (mod ). (See the article on the ABC conjecture last time for more on
this).
Solution. We follow the hint given with the problem (which, alas contained a
possibly very confusing typo – sorry). Let
2
1. By the Little Fermat theorem,
for any prime number we have 2
1
and so | 2
1 and thus
| 2
2 , i.e. |
1 . Put
1
. Of course
0.
1 / 2
1 mod - the division sign was missing from
We now consider 2
1 (mod ),
the original hint, apologies. Now if we have that 2
2
1
2
1
0 mod
2
1
On the other hand, the left-hand side of the last equation is also
1 2
2
2
.
But also of course 2
1 (mod ) and so 2
1 (mod ) for each 1
2
2
(mod ).
Thus we get is 1 2
Thus we have that
this won’t do since
0 (mod ), so | and thus
(being prime) is at least 2.
The ABC conjecture is still emphatically unsolved.
. Thus
1
1.
and
New Problems.
As usual, Problems 1 and 2 are rather standard. A (valid....) solution to the unsolved
problem would be a very different kettle of fish....
Problem 1. Let , , … be a sequence of elements in the integers modulo .
Show that there is some subsequence of consecutive terms
,
,…
whose
sum is zero.
Problem 2. Let be any finite group. Show that any sequence , … . | | |
at least | | | | 1
1 elements of contains a subsequence ,
…
elements whose product is the identity of the group.
|
| |
of
of | |
Unsolved Problem The unsolved problem this time is a question that may well have
occurred to many people: it is not at all clear how one would go about proving it.
Recall Euclid’s proof that there are infinitely many prime numbers: let (looking for a
contradiction) , , … be a putative exhaustive list of the prime numbers. Then
…
1 has a prime factor, but that factor cannot be any of , , …
as if it was we would get some
dividing 1, contradiction.
The Euclid-Mullin sequence
2 and that then for
is defined by saying that
∏
any
2, is the smallest prime factor of 1
. The idea is that, by the last
paragraph, this will generate a new prime number at each stage, so (as there are
infinitely many primes) we really will get an infinite sequence. It is (rather obviously)
a cumbersome way of generating new primes.
Check that the first few terms of this are 2,3,7,43,13,53... However the numbers
is the product of a 68 digit prime
involved rapidly get out of control: for example,
and an 112 digit prime. The next few have obvious small factors (note that the
sequence is not monotonic, as was shown much earlier), but I believe the value of
is still unknown.
Problem. Do all prime numbers eventually appear in the sequence?
No-one really knows how to tackle this problem. I think the smallest prime whose
being in the sequence is still not known is 31.
One can of course consider various variants, e.g. starting with a different prime. For
example if we start with an (odd) Sophie Germain prime (one for which 2
1 is
also a prime) then of course the sequence begins , 2, 2
1, … However in general
the pattern is very hard to predict: one can get runs of very small primes followed by
an absolutely enormous one, and generally a big new idea seems needed.
Here is a dodgy heuristic, taken essentially from http://wwwdidactique.imag.fr/preuve/Newsletter/000910Theme/000910ThemeUKillust.html for
why the answer to the problem might be “yes”. Suppose the sequence did not include
all of the prime numbers. Let be the smallest prime number not in the sequence.
Beyond a certain , all prime numbers less than will be included in the numbers
, … , . If
, pretend (dishonestly) that the number
…,
1 is
sufficiently random that it has probability 1/ of being a multiple of and thus has
probability 1
1/ of not being a multiple of : in the Euclid-Mullin set-up, is
thus the probability that
is different from . The probability that none of
,…,
is equal to is (under a further bogus independence assumption) thus
1
, which tends to 0 as
tends to infinity. This argument is riddled with dodgy
assumptions, but it is at least suggestive.
One can of course ask all sorts of variant questions, e.g. start at a different prime from
2...... Might this help???
Unfunny Joke for 2011.
Why did the two vectors start an internet-based company?
Because they thought they had a good dot product.