Măsurarea dimensiunii unei singularităţi

Dacă am raporta masa unei găuri negre (G.N.) la volumul sferei cu raza Schwarzschild am constata că densitatea respectivei G.N. este deosebit de mică. De fapt cu cât G.N. este mai masivă cu atât este mai puţin densă.

Acest fapt este doar aparent deoarece masa G.N. nu este distribuită uniform în sfera cu pricina ci concentrată într-o regiune din centrul acesteia denumită singularitate.

Masurarea diametrului unei singularităţi este imposibilă prin observare directă –acest fapt contravenind teoriei relativităţii. Asadar singurele informatii cu privire la singularităţi –şi implicit asupra densităţii reale a unei G.N.- vor trebui obtinute pe căi indirecte.

O metodă ar putea folosi o G.N. artificială, mai precis o G.N. de mişcare. În general cand vorbim despre G.N. ne referim la obiecte cosmice formate pe cale naturală, avand o masă într-atat de mare încât colapsează sub acţiunea forţei gravitaţionale. O G.N. artificială ar trebui totuşi obtinută într-un mod ceva mai complex-şi ceva mai puţin costisitor.

Astfel am putea lua o particulă încărcată electric şi am putea-o accelera cu ajutorul unui betatron ideal (a carui putere depăşeşte cu mult capacităţile actuale) până când masa de miscare a respectivei particule să depaşască masa critică.

Reamintim că prin accelerarea la viteze relativiste, masa corpurilor este substantial marită, astfel încât pentru o particula cu un volum stabilit se poate calcula viteza la care masa de miscare devine critică. Totodată şi dimensiunea particulei se micşorează pe direcţia de mişcare, ca atare transformarea se va petrece la viteza:

Odată formată această G.N. de miscare, un experimentator ar putea încerca sa îi frâneze mişcarea - şi astfel să-i reducă şi masa.

Deoarece nicio G.N. nu poate exista dacă masa critică a singularităţii nu este atinsă, am putea asista la o redresare a particulei initiale. Masurand viteza la care G.N. de mişcare a redevenit o particula normala putem lesne calcula şi masa de miscare la care redresarea a fost posibilă. Avand respectiva masă, putem cu certitudine să afirmam că stim raza singularitătii ca fiind egala cu raza Schwarzschild corespunzatoare acelei mase.

Ar mai fi de precizat ca şi în cazul în care redresarea nu se produce nici măcar când G.N. este frânată complet, tot putem extrage o informaţie despre dimensiunea singularităţii sale: aceea ca raza sa este mai mică decât raza Schwarzschild a masei de repaus a particulei initiale. În plus, deoarece G.N. este încarcată cu o sarcină electrică ea poate fi uşor reţinută şi păstrată într-o capcană de particule-desigur cu precautile de rigoare.

Dacă am putea observa totusi fenomenul de redresare atunci G.N. s-ar putea dovedi inofensiva chiar în eventualitatea în care ar fi pierduta în mod accidental. Aceasta deoarece avand dimensiuni reduse şi fiind încarcată electric ar fi imediat puternic frânată de campurile electrice ale diverselor molecule din materia pe care ar întalni-o, frânare care ar duce la redresarea sa imediată.

Aşadar, din punct de vedere teoretic o G.N. de mişcare s-ar putea dovedi edificatoare în privinţa densităţii reale a unei G.N. fără a comporta prea multe riscuri.

©Valeriu DRAGAN 2008

A new solution to Guthrie's four colour conjecture

(text only, for text&graphics click the link at the end of the article)

This problem was first formulated in October 1852 by Francis Guthrie. While he
was colouring a map, Guthrie noticed that he only needed four colours to shade the map
no matter how complicated the map was. His conjecture states that one would require
only four colours to shade any planar map so that no adjacent regions share the same
colour.
Guthrie’s conjecture stood unsolved until 1976, when two Illinois University
mathematicians, Wolfgang Haken and Kenneth Appel approached it. Their solution was
obtained using a computer program that tackled the problem.
My approach to the four colour problem is based on an analogy and a basic
topological formula.
The stages of my demonstration are:
1. Proving that if a five-vertex network exists and every vertex is connected to all
the others by at least one line, than at least two lines will intersect at least once.
2. Establishing under what restrictions a map with five regions can be judged as a
network of five vertices
(1) V+R-L=1
V=the number of vertices; R= the number of regions and L=the number of lines of a
given network (Euler formula)
We will first assume that no two lines intersect.
Let A and B be two vertices of a network and ξ1, ξ2 two lines that connect them.
According to Euler’s formula, the network only includes one region.
We will refer to the contour of the network as the continuous perimeter that
includes all of a network’s regions. Because the reunion of all regions can be thought of
as only one great region, we will restrict the use of Euler’s formula to only one region
when referring to a contour.
Let ξ3 be a new line that links A and B, we will
asume that ξ3 is exterior to the network’s contour.
Applying Euler’s formula for two vertices and three
lines we see that the new network has two regions.
However, the contour of a network can only be made-up by
two lines.
Let us take a look at the possible combinations for
the contour of the sub-regions of the network:
ξ1- ξ2; ξ1- ξ3; and ξ2- ξ3,
out of which only two are the contours for the two subregions.
We shall arbitrarily choose to eliminate one of the
combinations (as any would yield the same result).
Assuming the sub-regions were contoured by ξ1- ξ2 and ξ1- ξ3, then the network
contour could only be ξ2- ξ3, as all other combinations were already known to be only
sub-regional contours.
Notice that ξ1 is not part of the network’s contour but it is a regional contour, in
other words ξ1 belongs to the interior of the network. That is because if ξ1 belonged to
the exterior of the network it would not belong to any regional contour.
Knowing that all the sub-regions have a continuous contour (all lines are
continuous and the vertices provide continuity between them) we can state that a line
cannot belong both outisde and inside a contour which they are not part of without
intersecting it.
So if ξ1 is interior to ξ2- ξ3, then all of the points along itself, except A and B, are
interior to ξ2- ξ3.
P1)As a corollary we can state that for any network
with two vertices and three lines, there exists one and only
one line interior to the network.
Next, we analyse a network of four vertices A, B, C and D along with the lines
that connect them: AB, AC, AD, BC, BD, CD.
Analysing the network-tree formed by
the lines AB-BC-CD-DA, we notice that it
formes a continous contour, that is, at the same
time the contour of the current network.
Adding new lines to the network one by
one, we develop the following two cases:
First case:
Let BD be interior to ABCD contour.
In this case, the interior of the network will be
divided into two and only two sub-regions
ABD and DAC. We observe that C is neither
interior nor on the contour of ABD and that B
is neither interior nor on the contour of DAC.
We conclude that B is exterior to DAC and
that C is exterior to ABD.
We further assume that the line AC was,
as well, interior to the ABCD contour. Let X be
a point along AC interior to BCD (the same can
be proved if X was interior to ABD). Thus the
segment XC must belong to the BCD interior.
However in order for AC to be continuous there
must be another segment, AX, that reaches the
exterior of BCD.
If one line belongs both to the interior and the
exterior of a continuous closed perimeter (such
as BCD) then it must intersect the perimeter at
least once.
We thus reach the conclusion that AC and BD cannot
coexist in the interior of ABCD. Since BD was assumed to be
interior and AC cannot belong to the contour ABCD (as that
would mean it would intersect another line in an infinity of
points), we conclude that it must belong to the exterior of the
ABCD contour.
We observe that AB-BC and AD-DC are continuous
lines that link A and C. We also proved that AC belongs to the exterior of the network.
The entire network can thus be regarded as having three lines that connect two
vertices, the lines being: AC; AB-BC; AD-DC ; the veritces being A and C.
Applying P1) and knowing that AC belongs to the exterior of the network, we
reach the conclusion that one of the other lines must belong to the interior of the network;
because both of the remaining lines contain a vertex other than A and C.
We conclude that for this type of network there would always be a vertex that
strictlly belongs to the interior contour of the network.
Second case:
Assuming now that BD belongs to the
exterior of the ABCD contour, we observe that
BA-AD and BC-CD are continuous and link the
same vertices as BD. We apply P1) again and
reach the same conclusion: knowing that BD is
exterior, the internal line must be either BA-AD
or BC-CD. In any case, one vertex belongs strictly
to the interior of the closed contour of the
network.
Looking at the possible connection
between A and C, we assume that AC is
exterior to the network. However, knowing
that C strictly belongs to the interior of the
continuous closed contour of the network,
we end up with a contradiction: one segment
of AC must belong to the interior of the
ABD contour, although we assumed that AC
was exterior to it and that no lines intersect.
We thus reach the conclusion that
if AC exists, it must belong strictly to the
interior of the ABD network (which does
not affect the fact that C belongs strictly to
the interior of the network).
From our analysis we reached the
following conclusins: If a network is
comprised of four vertices A, B, C, D and
the lines AB, AC, AD, BC, BD, CD never
intersect each other, then there must be
one and only one vertex that strictly
belongs to the interior contour of the
network.
As a corollary, from Euler’s formula, the network must have one line that links
three points and one line that connects two points, thus the network’s contour is
comprised of all the three non-internal vertices.
Because we are analysing a network of four veritices and six lines, we must have
three and only three sub-regions, all of which have continuous and closed contours.
Assuming that C is the interior vertex of such a network, the sub-regions that
make up the network are: AB-BC-CA; BC-CD-BD and AC-CD-DA.
Analysing one of these sub-regions
(any of which can be chosen), AB-BC-CA, we
observe that D does not belong to its contour
(because then two of the lines would intersect
in an infinity of points).
Also, if we were to assume that D
belonged to the interior of AB-BC-CA, in order
for DA, DB and DC not to intersect the AB-BCCA
contour, we would have to asume that they are
interior to AB-BC-CA too. However, that would
mean that AB-BD-DA belongs to AB-BC-CA’s
interior; by hypothesis we know that AB-BC-CA
belongs to AB-BD-DA’s interior, if they are both
simultaneously correct, that would mean that ABBD-
DA and AB-BC-CA are identical. If the two
were identical, then the other two sub-regions of
the network would be non-existent, which would
break Euler’s rule.
The same technique can be used to prove
that BC-CD-BD excludes A, and that AC-CD-DA excludes B.
Thus, every sub-region inside this type of network excludes one vertex and one
vertex only.
Finally, let E be a fifth vertex added to such a network, we distinguish the
following cases:
i) E belongs to the contour of the four-point network resulting in an
intersection of at least two lines in an infinity of points. (we thus
exclude this option).
ii) E belongs to the exterior of the
network. We know that there must be a vertex that
strictly belongs to the interior of the network, which
would mean that in order to connect that vertex to E,
a continuos line shoud simultaneously belong both to
the exterior and the interior of a continuous closed
contour and thus intersecting it in at least one point
(we thus exclude this option).
iii) E belongs to the contour of an internal region thus resulting in an
intersection of at least two lines in an infinity of points. (we thus
exclude this option).
iv) E belongs to the interior of a
sub-region. We know that there must be a vertex
that strictly belongs to the exterior of the subregion,
which would mean that in order to connect
that vertex to E a continuos line should
simultaneously belong both to the exterior and the
interior of a continuous closed contour and thus
intersecting it in at least one point (we thus
exclude this option).
The conclusion we reach is that no matter
where the E vertex exists, it cannot simultaneously be connected to all the vertices of the
four-vertex network without at least two lines intersecting in at least one point other than
a vertex.
As a result of that we state:
Given a network of five vertices in which every vertex is connected to every other
vertex by continuous lines, there must be at least one point in which at least two lines
intersect.
Modelling Guthrie’s conjecture
In order to prove Guthrie’s conjecture, we must prove that no more than four
planar regions can be simultaneously connected.
Let A, B, C, D and E be planar regions that are not connected to each other (not
sharing any borders). Let AB, AC, AD, AE, BC, BD, BE, CD, CE and DE be planar
regions (called bridges) that connect only the domains that name them (i.e. AB connects
A and B etc.)
If we consider that, for instance, AB belongs to A, then A and B are connected
domains. We will further consider that each of the bridges belongs to only one region in
particular.
In order to associate the regions-bridge picture to a vertex-line network we need
to establish some minimal rules (N.B. the rules are minimal but not sufficient, following
them will not guarantee the association; however breaking them would surely invalidate
it).
Since we are by definition assuming that no two planar regions overlap, we must
impose that:
-no two bridges intersect each other
-no region intersects another bridge or region
-the border region between a bridge and its destination (i.e. the region it is not part
of) must be much smaller than the contour of the region.
-a bridge can connect only two domains
-a bridge is continuous
Thus the lines of the network must:
-be continuous
-not intersect one another
We observe that by associating the above picture to a five-vertex network we
break one of the minimal rules. We have already showed that in a five-vertex network in
which all vertices are connected to every other vertex by continuous lines, at least two
lines must intersect in at least one point.
Because of that, we conclude that we cannot simultaneously connect five distinct
planar regions.
Thus no more than four regions can be simultaneously connected, which validates
Guthrie’s four colour conjecture.

©Valeriu DRAGAN 2008 (original article from Valeriu Dragan's "Applied Geometry"© in 2006)

Applied Geometry link: http://www.lulu.com/content/553543