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Relativistic Engineering
Valeriu Drgan
Bucharest 2010
ISBN:978-973-0-09404-6
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SummaryForeword
Relativistic propulsion
Positron-electron propulsion
Photonic vs. material momentum
Classic vs. Relativity
Lavoisier and the relativistic gravitational paradox
Interfering light wavesOptical and Infra Red second order perpetuum mobile
Another abatement from Clausiusspostulate
The isothermal Engine
Molecular Mach number
Macroscopic magnetic monopoles
On light and darkness
The mass of the kinetic energy Vs. the motion mass increase
Measuring the radius of a black hole singularityA photonic black hole
A Heisenbergian Black Hole
The maximum gravitational acceleration
Faster than the light?
Stretching the event horizon
A matter of tempo
The alibi principleApparent faster than light travelling
The apparent temporal inertial force
Hyper time concept, the P.C.-game analogy
The door-bell solution
Information dissipation
Geometrical thinking
The Mobius strip, Klein bottle and J surface
Unconventional number systems
Guthries conjecture proof
Reference
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Foreword
The truly important things
are not the ones nobodyhas ever done before but
rather the ones that
everybody should havedone.
A good physicist will look
for rules and a good
engineer will look for the
exceptions.
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Relativistic propulsion
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Positron-electron
propulsion The main aspect of this propulsion system is
that it doesnt require a work fluid (or mass).
Another very important aspect is that if
accelerated beyond a critical velocity, theparticles we create are more efficient at givingus thrust than photons would be.
Basically, the e+e- propulsor uses the pairformation effect encountered when a highenergy photon enters the field of a heavy atom.The energy of the photon is converted into apair of matter and anti-matter particles. In ourcase we shall discuss the electron-positronformation.
After their formation we might capture thoseparticles and guide them to a betatron where wecould accelerate them at relativistic speeds andeject them out into space trough magneticnozzles.
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The schematics:
Magnetic nozzles
Betatrons
Magnetic guides
High
energyphoton
Massive
atom
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Photonic vs. corpuscular
momentum
Following on the positron-electron propulsion system I sat
out to see how it wouldcompare to a simpler photon-momentum engine (basically alaser)
Comparing the two momentafor the same energyrequirement bearing in mindthe fact that we have to
generate the electron andpositron. Meaning that even ifthey were materialand thus
better for propulsion, they still
were expensive energy-wise.
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The mathwork
2
2
0
1c
v
vmpe
2
2
2
02
0_
12c
v
vmcmE etotal
2
2
220
12c
v
vc
c
m
c
h
c
hp
2
2
22
2
2
12
1?
1c
v
vcc
c
v
v
c
v
c
vcv
21?
2
2
2
c
v
c
v
v
c
2
1?12
2
c
v
v
c
2
1?1
2
25.02?0:___ 24 aaareacheventuallywethen
1:
v
caif
With the only positive real root:
a=1.1483805939739538 so:v>0.8708c
Momentum: Corresponding energy:
comparison:
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Comments
The surprising result we
obtained should scare eventhe bravest physicist!
It would appear that, with
the right light source as
our power supply, a
spaceship could navigate
upstream using the
above propulsor
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Classic vs. Relativity
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Lavoisier and the
relativistic
gravitational paradox
One of the properties of the theory ofrelativity is that, at low velocities, the
relativistic results can be approximated byNewtonian mechanics. However, at highvelocities, relativity looks radically differentfrom classical mechanics, a common examplebeing the decrease in the acceleration of a bodythat is acted upon by a constant force. Because
of the relativistic mass increase, the force canno longer provide the same initial accelerationwhich fades to zero as the speed of the object inquestion approaches the speed of light.
There is however a force available to us thatcan provide a paradoxical result: gravity.
Gravitational forces depend on the mass of theobjects they act upon and thus the gravitationalacceleration is no longer decreased as the bodyreaches relativistic velocities.
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The experiment:
Lets imagine the following experiment: a
body of mass m is situated at an altitude h in a
gravitational field thus having a potential energy , if
we consider as the equivalent constant gravitational
acceleration we can write :
If we drop the body, the terminal velocity will
be given, in Newtonian mechanics:
However this is not all. Because the
gravitational acceleration is immune to relativistic
effects, the same terminal velocity will be obtained
by using the theory of relativity.
Because of this, the kinetic energy calculated
at the moment of impact is slightly higher than the
initial potential energy, i.e. the kinetic energy is
higher than the energy required to lift the body at theconsidered altitude.
ghv 2
ghmE 0
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The mathwork
2
2
0
1c
v
mm
hgv
_
2
2
2
2
0
12c
v
vmEkinetic
2d
mGg terra
1
1
1
2
20
c
vmm
2_
2
_0
2
_
_
0
2
_
2
0
2
1
21
1
:____
1
21
1
:
1
21
1
:______
chg
c
hg
mE
totalthetoitbringing
c
hg
hgmEE
also
c
hg
cmmcE
masssurplusthetoingcorrespondenergyThe
na lgravitatio
potentialkinetic
surp
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Comments and
interpretations
The comment being that it
would seen that gravitygenerates mass under certain
conditions.
It is also important toacknowledge that gravity
alone cannot (by this line of
thought) generate mass. Onewould have to use any other
force to lift (slowly!) the
body at the required altitude.
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Salisbury screens
The upper layer is semi-transparent and splits theincident beam into two :the first partial beam is
reflected by it and the second partial beam is reflected
by the ground plane below. The thickness of the
screen determines the wavelength of optimum
operation i.e. t=/4.Because of this, the operation range is rather limited
and hence the decreased efficiency with angle of
incidence and frequency spectrum.20
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Optical and Infra Red
second order perpetuum
mobile Rudolf Clausius, a nineteen century
physicist postulated that on its own, heatcannot naturally flow from a colder bodyto a hotter one. This is true if we only lookat convection and conduction. Withradiation, a third way of transferring heat,things get a little more complicated.
Using the standard radiation model, we can
affirm that all bodies radiate heat, nomatter how hot or cold they are. By this itmeans that, if we were to construct adevice such as the following ones, wecould-in fact- heat a hotter body using acolder one.
Another thing: an infrared diode woulddirectly transform thermal radiation intoelectricity (and from there to mechanicalwork) without the need of a cold fountain!
Which denies the Caratheodoryformulation ofClausiusspostulate.
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The Infra-Red device
Adiabatic
wall
hotspot
Heat sink forthe
concentrator
Concave radiator with
focal point in the
hotspot
By Kirchoffs radiation laws, because the
small surface has to emit as much energy as it is receiving
from the concave radiator, the temperature of it has to be
higher at equilibrium and thus we have ourselves a
natural temperature gradient.
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The optical device
The basic idea is the same only this time
its in the optical spectrum.
The number of photons focused by the lens into the
focal point is greater than that of the photons let in by
the window located there. Hence we have an optical
hotspot that is easy and quite cheap to
manufacture. To put it simply, more energy comes infrom the lens than the lens receives from inside
trough the window.
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Another abatement
from Clausiuss
postulateGravitational segregation on temperature levels
due
to the Archimedes force is another fine example of
how a temperature gradient can spontaneously
appear and provide us with a means for extracting
energy from a certain room without the need of a
cold source. The way things work is:
first we isolate a chamber filled with gas.
Then we place that chamber inside a
gravitational (or centrifugal) field. Due to theArchimedes force, the hotter-less dense-gas
will rise and the colder denser gas will fall. If
the gravitational field is intense enough, the
gradient of temperature will be grater and
will permit for a conventional thermodynamic
machine to work with the two sources of
heat: the cold and the hot one.24
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The schematics of a gravitational
second order perpetuum mobile
(by definition)
Gravitational
field
Hot source
cold source
Kalinacycle
Thermo-
dynamic
machine
Q in
Q out
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The isothermal Engine
The approximate expression of theconclusions of the Joule-Thomsonexperiment is that the internal energy U ofa gas is independent of the volume it fillsand only depends upon the temperature. Ifthis is so, then the following machine willwork with the perfect efficiency of 1.
The device uses the ideal gas law:
To vary the volume within the chamber weonly change the number of gas molecules(controlling vaporization) at constant
temperature. After the gas has reached themaximum expansion, we isothermally coolthe thing, condensing the vapors back intoliquid and the cycle could start all overagain.
We should also have to account for the latentheats of vaporization and condensation!
RTpV
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The working of an
isothermal engine
In the first stage the boiling fluid vaporizes
generating moles of steam which push the piston
up; The second stage, the isothermal cooling beginsand the steam condenses into the initial fluid ; Note
that at all times the temperature of both liquid and
vapor is constant- this should not work according to
the second law of thermodynamics27
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Molecular Mach
number Considering the perfect gas model, one
can calculate the so called thermal
velocity (i.e. the mean statistic speed atwitch a molecule will move).
One could also calculate the speed of
sound in that gas.
Upon calculating the Mach number ofa molecule we find out that thetemperature factors out of the equation,leaving us with a constant (which isdifferent for each real gas or gas
mixture)
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The mathwork:
._
_
_
:___
constBolzmannk
tcoefficienadiabatic
etemperaturT
massmolecularm
m
RTcsoundofspeedthe
m
kTvvelocitythermal
3:_
m
kTm
kT
c
vnumberMachmolecular
3
:__
Tk
Tk
c
v
3
/3
c
v
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Macroscopic magnetic
monopoles It may seem silly but this may
be the next best thing to a true
magnetic monopole. If the gapswere filled by better magnetunits then the field around thesphere created should look
mono-polar.
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On light and darkness
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Light wave
interference Another amusing
gedankenexperiment involves using
two parallel coherent beams of light. If we consider the gravitational
attraction between them then we couldcalculate the time (and distance atwitch) they will superimpose and
interfere (either partially cancelling orsumming up)
._'
__
:__2
constsPlanckh
lightofspeedc
frequency
c
hmphotonofmass
._
._
:_2
distinitiald
constnalgravitatioG
d
mG
m
Fonacceleratinalgravitatio
G
dcx
ctxmeaningG
dt
d
mG
dteconvergencuntiltime
3
3
2
:
:__
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The mass of the kinetic
energy Vs. the motion
mass increase As theory goes, each material
body travelling at a relativistic
speed, will have a mass increasegiven by the Lorentz equation.
We also observe the relativistickinetic energy has a
corresponding mass (seeEinsteins )
One might assume that thiskinetic energy mass is equal to themass increase given byLorenzbut not really.
The mass corresponding to thekinetic energy is smaller than themass increase..
2mcE
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The mathwork:
4
2
2
2
2
20
2
22
2
0
2
20
0
2
22
2
0
2
2
2
2
0
8/
8?
:__
11
1?
12
:
1
1
1?
12
:
1
1
1
12
:
12
:__
cv
and
a
endthein
aa
a
c
vaif
c
vm
c
vc
vm
comparing
cv
mm
mmm
c
vc
vmmtherefore
mcE
c
v
vmEenergykineticThe
k
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Measuring the radius
of a black hole
singularity If we were to calculate the density of a
black hole in a traditional way, i.e. mass over
volume we would find a counter-intuitiveresult: that the more massive a black hole is, the
less dense it is. This off course is not a physical
truth because, as theory goes, the mass of such
a body is not uniformly distributed inside the
sphere of the respective Schwarzschild radiusbut rather concentrated within a singularity.
By means of pure observation, we cannot
assess the size-hence volume- of the said
singularity, simply because everything we canmeasure, even theoretically, is outside the event
horizon of the black hole. Nevertheless, there is
still hope for indirect observationor data
gathering.
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The gedankenexperiment this paper proposes
involves the creation of an artificial black hole of a
certain kind: a kinematic black hole. This kinematic
black hole (KBH) is based on the relativistic principleof mass increase as a body is accelerated at near the
speed of light. It is therefore possible for an object to
be accelerated at such a speed that its motion mass
exceeds the critical mass required for creating a black
hole for its respective size.
(1)
Equation (1) is a direct derivative of the
Schwarzschild equation for determining the radius ofthe event horizon.
Using the Lorenz equation that gives the mass
of an object moving at a certain speed we can derive
the following relation that shows at which speed a
spherical object can be transformed into a KBH.
(2)
The above equation also factors in the relative
length deformation associated with relativistic
speeds, i.e. the radius of the spherical particle will
contract in the direction of motion.
G
cRm
particle
critical2
2
2
0
021cR
mGcv
particle
transform
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Closing the deal..
Once such a KBH is formed anexperimenter can then try decelerating it and
thus reducing the total mass of it. Due to thefact that a black hole cannot exist without itscritical mass, we might witness the reversion ofthe KBH back into the original particle that weused in the beginning of the experiment. All weneed to do at this stage is to measure the
velocity at which this reversion has occurredand calculate the motion mass at that time.Knowing the value of the instantaneous masswill immediately give us the information weneeded about the radius of the singularity asgiven by the equation (3).
(3)
Equation (3) expresses the fact that the radiusof the KBH singularity cannot be lower than theSchwarzschild radius corresponding to the
motion mass of the particle at the speed ofwhich it reverted.
2ildSchwarzsch2
cmGR motion
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Final remarks
The above experiment dwelledon the assumption that the KBH
will eventually revert to theoriginal particle at one point.However it is impossible to becertain of such a fact especiallysince some theorists predict thatsingularities are infinitesimal. Theonly thing that we can be certainof after the above experiment hasbeen carried out is that, if not
even when fully stopped the KBHdoes not revert the radius of itssingularity will definitely besmaller than the Schwarzschildradius of its original particlemass.
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A photonic black hole
Weve seen how a kinematic
black hole can be created byaccelerating a particle
beyond a critical speed.
The nextgedankenexperiment tries to
figure out if a photon-of all
things-can become a blackhole due to its energy
density at a certain
frequency.
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The math
2
2
2
2
2
...
2
:
_
2
_
chG
rearanging
c
h
c
G
then
Rif
cGmR
massphotonm
where
c
h
c
hm
hildschwartzsc
hildschwartzsc
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A Heisenbergian Black
Hole Even more abstract is the notion of a
black hole generated by the uncertainty
regarding the momentum of a particlewithin an ever smaller space.
Heisenbergs principle states thatthere is a balance between the
precision wth which we can measurethe momentum and position of anygiven particle.
There is a law governing thisbalance but as a rule of thumb, themore precise you are at determiningthe spatial boundaries of a particle, theless you know about the magnitude ofits momentumand this gave me anidea
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The math behind the
principle
3
2
2
2
2
:________
2
...
2
:
_
2
:__
4
:
4
:_'
c
hGL
lenghtscalePlanktheequalsphotonaforwhich
vc
hGx
rearangingv
p
c
Gx
then
xRif
c
Gm
R
radiushildSchwartzscthe
x
hp
hence
hmvx
principlesHeisenberg
p
hildschwartzsc
hildschwartzsc
vc
hGx
vc
hGx
toleading
c
Gmx
volumespherehildSchwartzsctheequalsvolumecubetheif
2
2
3
2
3
3
4
2
:_
2
3
4
:_______
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The maximum
gravitational acceleration
It may be a simplistic approach but if we
were to calculate the gravitational
acceleration taken at the surface of a blackhole we would obtain a maximum.
That calculated valuewhich is a
function of mass- is the greatest
gravitational acceleration available for therespective mass and hence the title of this
exercise.
2
2
c
GmR hildschwartzsc
Gm
cg
R
mGg
distinitiald
constnalgravitatioG
d
mG
m
Fgonacceleratinalgravitatio
4
._
._
:_
4
max
2max
22
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Stretching the event
horizon Weve seen previously how the gravitational
acceleration measured on the event horizon (of
a spherical non rotational black hole) can becalculated. It would appear that it is inversely
proportionate to the mass.
This is in agreement with the fact that the
density of black holes gets lower and lower as
they get more and more massive. Two questions could be asked:
Could a black hole be so heavy that its density be as
low as common objects? And
Could a the gravitational pull of a black hole be so low
on the event horizon that we could actually physicallygo back and forth?
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Faster than light?
This is a purely speculative exercise but it
is nevertheless one that arises from the fact
that a gravitational force is immune to theincrease of mass of the object its
accelerating.
The question is this: If we had a particle
accelerated at a certain initial speed, then
let it fall freely within a gravitational field
(see the Lavoisier paradox), then could that
particle reach the speed of light as a result
of the gravitational acceleration before
hitting the ground (or in this case the event
horizon)?
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Further discussions When designing this gedankenexperiment, we cannot work
under the assumption that the mass of the falling object is
negligible in regard to the mass of the black hole.
This is because, as the velocity increases towards the speed
of light so will the mass, at one point our object will
inherently become just as heavy as the black hole itself.
Another problem with this is that the mathematical modelwill have to account for the increase in the gravitational
acceleration, not only as the result of the approach to the
impact point but also because the mass of the body on which
were falling will increase as it is itself accelerated.
Yet another problem will arise if we account for the
transformation of our object into a black hole. This is goingto happen because of the mass increase with velocity.
In the end, our experiment will reach another obstacle
which will probably end it altogether..
Because of the mass increase, the two black holes will
expand their radii. Meaning that, as we approach the speed of
light the masses will increase approaching infinity. Becauseof that, their Schwartzschild radii will approach infinity. So
no matter how far apart we assume they are, in the end the
speed of light will Not be acheved outside any of the event
horizons.
So, to answer the question will any of the objects reach the
speed of light? , maybe but as they tend to do so, theduration (due to the gravitational time dilation) will be
infinite-in the end we wont have either infinite mass nor
faster than light objectsnot like this anyway.
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A matter of tempo
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The alibi principle Time travelling has fascinated me for as long
as I can remember. However there is also thematter of the grandfather paradox, should atime traveler kill his grandfather (be it by
accident) before his father was born he wouldeffectively erase himself from historyhencenot being able to time travel in the first place.Its a classic.
In this paragraph Id like to explain a morebenign aspect of time travel: assuming that all
the particles of the Universe existed in one formor another and can only exist in one place at atime. Assume now that one would go back intime for a couple of decades or so, the particlesthat are in his body at the time he sat to go backin time did in fact exist 30 years ago in theexact same state (assuming hes not made ofnuclear decaying materials). Having said that,we reach an impossible situation:
How can one (or more) particle be bothmaking part of our time traveler's body and atthe same time be part of their original,respective bodies?
In law there is this thing called an alibi, so Iguess, if all assumptions are correct grandpawill be safe.
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Apparent faster than
light travelling The theory of relativity teaches that one
cannot travel faster than light. This much weknow. But is it possible to make a trip such that
the passengers will feel they travelled fasterthan light?
It is possible to engineer such a trip. Due tothe Lorentz time dilation, the passengers willalways feel their trip was shorter than the ones
waiting for them at the destination (whichseems to be valid even at non-relativisticvelocities).
The question now is just how fast should ourstarship go in order for the passengers to feelone second has passed for each light-secondthey travel.
Without spoiling the excitement, the speed inquestion will be v=0.707c
Feasable..
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The calculations
behind it all:
2
1
c
vtt
ship
Terraship
ndestinatioreachtolightfornecessarytime
v
ct
ship
ship
______
1
2
2
1
astronava
ship
t
cv
travelt
if:
2cvship
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Taking things a step
further: If such a thing can be done with a ship, why
not with a colony all together?
For instance say you had to travel to a distant
place but want to be in the same timeframe as
the people at your destination and those at your
departure. All you need to do in order to by-
pass the twin paradox is to travel at 0,707c andto make sure that both the departure and the
arrival colonies are also dilating time (be it by
gravity or by relativistic velocity) the same way
you do.
We therefore pose the question: if everyone
feels like they traveled faster than light whos to
say they didnt ? In this case the distance they
covered can be considered to have shrunk, but
this is going in a more philosophical direction
than intended in this paper.
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The apparent temporal
inertial force It sounds gibberish but its
actually a neat little conceptIve devised :
If we had a string attachedto a body which was in a
different gravitational field(more or less intense) than,by pulling the string wellget a supplementary force
due to the fact that theacceleration in that timereference frame flowsslower or faster.
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The principle of the
temporal inertial force
Gravitational
field intensity
Perceived acceleration due to gravitational
time dilation 53
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The difference between
philosophers and logiciansis that logicians cover their
reasoning with the words
assuming that
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Hypertime concept,
the P.C.-game analogy Speculating on the time traveling
concepts, a couple of thoughts came up:
The concept of hyper-time, seeks todescribe the way time itself dilates andcontracts.
Also causality might not be necessarilyinfringed by time travelers should theycome from outside the studied system-thecomputer game analogy:
Say you are playing a game, at one pointyou click save, make a mistake and goback to the saved file. The game appearsthe same for the characters inside but it isnot the same in your frame of reference.
The same thing can happen with a timetraveler not-tied up by the causality of theactions of the group hes studying.
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The door-bell solution
Should causality be infrindged, we mightwithness a door-bell situation. The classicaldoorbell has a magnetic switch that, whenclosed by the button will pull the tongue on
the bell and also break the circuit. Because thetongue is elastic, it will go back and re-closethe circuit only to begin the process again.
This non-stationary situation in which thecircuit cannot be stable either way could benatures way to deal with the grandfather
paradox. In one stage the traveler exists only togo back and extinguish himself and then-sincethere is no killer he could exist oncemore.
It is my strong belief that at one point oranother most of the information fades into thebig picture and so, if given time, the two
paralel non-stationary, paradoxal, situationswill eventually re-merge into one. And it worksfor everything from flipping coins to timetravelling.
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Time travelling
the lottery safety rule If you went back in time
youd know which numbersto pickor would you?Some processes in natureare truly random (in the
sense that they are notcertain to repeat should thesituation be recreatedexactly) off course this
doesn't happen to lotteryextractions but it does applyto quantum processes wherethe Heisenberg principle is
extensively applicable.
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Door-bell solution for
the grandfatherparadox in hypertime
Reality splits at the beginning of theparadoxal situation creating twoparallel and oscillating realities untilthe two (or more) realities merge dueto the information dissipation
hypertimeline
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Information
dissipation This is a topic many might not like to accept,
mainly because were used to the butterfly effect
which does make a lot of sense. The problem with it
is that its not air-tight. Say you were playing dice
and need a certain number to win (all other numberswill make you lose), you roll and the number doesnt
match so you lose. Its almost certain that the number
rolled will not be remembered by anyone and so the
influence of it being a 4 or a 2 will be negligible.
An interesting thing about the butterfly effect is
they way it was discovered-on a simulated reality
model which in those days was highlyexperimental
(for lack of a better word)
We also have to keep in mind an insect
Langtons ant, its a cellular autonoma designed to
make a left turn if it steppes on a black square or a
right if its on a white square while changing the
color of the square it previously inhabited. Sounds
simple right? The mystery with this ant is that, nomatter how complex the map it is placed on is, it
will always enter into a loop, generating its own
pattern which is the same every time.
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Geometrical thinking
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The Mobius strip, Klein
bottle and J surface
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The Mobius strip When it comes to recreational topology, one of the
most famous models is that created by AugustFerdinand Mbius.
The strip that bears his name is the first in a serioesof special topological surfaces. The property that itholds is having only one face and one edge asopposed to a sheet of paper which has one edge buttwo faces or to a sphere which also has two faces butno edge. Until this model was invented, in 1858, itwas believed that the minimum number of faces areal physical body could have was two.
We can manufacture a mobius strip by glueing theedges of a ribbon after twisting them 180 degrees.Following a path on the surface of this strip wellrealise that the distance were required to walkuntil we reach the starting point is twice the lenght of
the initial ribbon.This is because by twisting theribbon, weve joined the two faces and so both willhave to be walked before reaching the starting point.This yields an engineering application for conveior
belts which will have a double lifespan due to thedoubling of the active surface.
Another interesting fact about mobius strips is thatif we were to try and make two at the same time(holding two ribbons) well only end up with just one
bigger strip instead of two ... Same goes for cuttingone in half (along its lenght offcourse).
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A Mobius strip
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The Klein bottle Assuming we had two mobius strips one being the
mirror immage of the other, we could join them on
their edges to form a Klein bottle.
The Klein bottle is a topological surface with one
face and no edge. The man who designed it was Felix
Klein in 1882. His initial name for it was KleinscheFlche-the Klein surface but, as always a trivial
typo made it into what we have today: Kleinsche
Flasche-the Klein bottle.
Looking at its surface, we might think that its
intersecting at one point and forming an edge. That
would be wrong because the Klein bottle is a 4D
surface which we can only project into our 3D
geometric world. In other words what youll see in
this book is a 2D projection of a 3D projection of a
4D body.
Another way of looking at the edge is bycomparison to the Mobius strip: if we draw it in 2D,
the edge willat one point-overlap and apear as if it
would intersect itself. Luckally for us, we have
enough geometrical dimensions in Nature.
We do have another parametrisation for the Kleinbottle called the figure 8. Its generated by railing
an 8 figure along a twisted loop-to be honest this is
the simplest way you can explain visually a Klein
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Classical Klein bottle
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Figure 8 Klein bottle
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The J surface
The final topological surface Ill bediscussing is the J surface. Itsprobabllythe simplest of them all and it can becreated in a variety of ways.
First way is to section a torus and join theupper edge to the lower edge like in thefollowing graphic.
What we have obtained is a body withone surface and one edge that istopologically different from the Mobiusstrip.
It can also be cut into two mirroredMobius stripsbut is also distinct from theKlein bottle, since the Klein bottle has noedge and the J surface has one.
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A quick way to obtain
a J surface
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The type I J surface
There are three ways to generate aJ surface, all being perfectlyequivalent:
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Type II J surface
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Type III J surface
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Unconventional
number systems The role of number systems in the development of
human civilization is undoubtedly significant. These
systems reflect firstly a concept and even the degree
of abstractisation a civilization possesses.
In genere, a number system is defined by a series of
conventions consisting of digits-we may actually find
numbers amongst them- as morphological entities
and a base in witch to express those numbers.
Surprising or not, the first primitive number system
had the base 1. Even if drawing lines on a wall to
count didnt really made everyone aware of that fact.
By expressing numbers in 1 base, the primitive mancould represent quite easelly a bigger or lesser
number (even compare two numbers). However at
high enough values, such processess became teadious
and probablly resulted in violent outbursts-some of
which can still be seen amongst more evolved men
who operate in the 2-base system.
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The next step
1=| 2=||
19=|||||||||||||||||||
et c.
The Etruscans, have invented a more useful tool (which
was later adapted by the Romans) : the digit that marked
every fifth bar |. This notation meant to bring some rhythm
which in terms help the usage of the 1 base system can be
considered a first step towards a more evolved number
system.
19= |||| |||| |||| ||||
It didnt take too long until the clever people found the four
bars preceding the to be useless and hence made theprocess even better:
19= ||||
The idea caught and before too soon, new notations began
appearing . Similar to the previous system, the Etruscans
noted X for .
They also had different digits for larger numbers like 50 or
100=C.
The usage refined the writing to below three units to be
added or subtracted from the number. Similarly, Latin
writings displayed numbers by a series of additions and
subtractions-which can still be observed in the French
language today .
In later years, this plurality of digits became a victim of its
own success, masons reached record number of digits,
basically every number up to 100 has its own character as
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Expressing nothing
A popular misconception is that the numberzero was imported from the Arabs in the
XII-th century by Fibonacci. However there are
a lot of such concepts indigenous to Europe thatpredate Fibonacci.
For instance, the Greek astronomers used asystem (that predated the Etruscans) whos rolewas even more complex than that of the Mayanzerowitch was a gap filler really. According tosome authors, the symbol 0 is of Greek originand the name is of Arab legacy shafira=empty.
Around 500 A.D. the dobrogean monk
Dionysius Exiguus notes nulla as the result ofa calendar computation-because the romansystem had no zero.
This short introduction could not be endedwithout mentioning the similarities to the
Sanskrit names of numbers of their Latinpronunciation : eka, dvau, trayas, catvaras,panca,shats, sapta, ashtsau, nava, dasa.
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Unconventional bases
The writing of numbers in different basismight prove a bit tricky especially becauseof the rules for division.
One of the fathers of the number systems,Bergman, is the first to analyze writingnumbers in an irrational base and it is 41years later, in 1998 that Knuth takes intoconsideration a transcendent base.
These approaches require a more clearspelling of numbers so there should besome notations:
n = number
b = base
a = base power coefficient j = sum ordinal
For instance to write a number in the 10base we give b=10 and write it as a sum ofpowers-for decimal places we shall use
negative powers.
x
j
j
j ban1
1
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Unusual properties
One base that Knuth analyzed was . The property ofthis base is that it allows the writing of the same numberin different forms
E.g.:1010 = 10100,010010101011 = 10100,0101
These isotopes can be encountered in other bases too.
Other properties of such as 2- -1=0 , allow thewriting of integral numbers.
In 2003, Allouche and Shallit introduce the nega-binaryand nega-decimal systems. For the nega-decimal system,
the generalized expression is this:
So that the first 9 numbers are the same but the tenth is
190-10. It may seem a lucky choice for the primitive man to use
the positive 1 base, because the negative 1 base can onlyexpress 0 and 1.
Until now weve seen bases belonging to R or at leastcomparable to real numbers. But what if we used the
imaginary unit?
x
j
j
jan1
1)10(
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Imagination
Because i is not comparable toany Real number, we can nolonger use the extended sum.
Not only that, but we can nolonger know what kind of digits
we could use. Because the baseof the numbering system is acomplex number, why notconsider the digits as
complex numbers as well? Thisis useful since the digits aremerely the coefficients ofvarious powers of the base.
If
x
j
j
j ian1
1)32(Caj
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Being un-elegant
Although it is not a usual practice to write a digit that is higherthan the base, this is an un-elegant solution to a practical problem,and it works:
|1710|1110|2410|= 24100+11101+17102=1834
This necessity is due to the fact that we cannot know for surehow the digits are in respect to the base.
Since were using complex numbers for both the basis and thedigits, we might use complex powers aswell.
Previouslly weve shown how a positive power corresponds to adigit and a negative to a decimal (i.e. to the left and to the right ofthe decimal point) but where would we place a rational power?
Obviouslly a power of order 0.5 or -2.4 is somewhere in an
intermediate position but expressable by conventional summing.
Where j is rational an a and b are complex
It is arguablly difficult to work with the 3,5 th digit or the sqrt(3)decimal but the purpouse here is to have a generalisation.
The next forseable step would be to have even the power
expressed as a complex number but such a number system is ofpure abstract interest right now.
x
j
j
j ian1
1)32(
x
uvi
uviuvi zyiqpin
1
1)()(
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Guthries conjecture
proof This problem was first formulated in October
1852 by Francis Guthrie. While he wascolouring a map, Guthrie noticed that he only
needed four colours to shade the map no matterhow complicated the map was. His conjecturestates that one would require only four coloursto shade any planar map so that no adjacentregions share the same colour.
Guthries conjecture stood unsolved until
1976, when two Illinois Universitymathematicians, Wolfgang Haken and KennethAppel approached it. Their solution wasobtained using a computer program that tackledthe problem via brute-force.
My approach to the four colour problem is
based on an analogy and a basic topologicalformula.
The purpose of this demonstration is not toprove that all maps can be coloured by fourcolours but rather to show that, on a planesurface there is no way to design such a map
that would require more than four.
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The proof
1. Proving that if a five-vertex network exists and everyvertex 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 fiveregions 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 (Eulerformula)
We will first assume that no two lines intersect.
Let A and B be two vertices of a network and 1, 2two lines that connect them.
According to Eulers 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 networksregions. Because the reunion of all regions can bethought of as only one great region, we will restrict theuse of Eulers formula to only one region when referringto a contour.
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Let 3 be a new line that links A and B, we will asumethat 3 is exterior to the networks contour.
Applying Eulers formula for two vertices and threelines 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:
1- 2; 1- 3; and 2- 3, out of which only two are thecontours for the two sub-regions.
We shall arbitrarily choose to eliminate one of thecombinations (as any would yield the same result).
Assuming the sub-regionswere contoured by 1- 2and 1- 3, then the networkcontour could only be 2-3, as all other combinationswere already known to beonly sub-regional contours.
Notice that 1 is not part ofthe networks contour but itis a regional contour, in
other words 1 belongs tothe interior of the network.
That is because if 1belonged to the exterior ofthe network it would notbelong to any regionalcontour.
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Assuming the sub-regions were contoured by 1- 2
and 1- 3, then the network contour could only be
2- 3, as all other combinations were already known
to be only sub-regional contours.
Notice that 1 is not part of the networks contour
but it is a regional contour, in other words 1 belongs
to the interior of the network. That is because if 1
belonged to the exterior of the network it would not
belong to any regional contour.
Knowing that all the sub-regions have a continuouscontour (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 1 is interior to 2- 3, then all of the points
along itself, except A and B, are interior to 2- 3.
P1) As a corollary we
can state that for anynetwork
with two vertices and
three lines, there
exists one and only
one line interior to
the network.
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Next, we analyse a network of four verticesA, B, C and D along with the lines that connectthem: AB, AC, AD, BC, BD, CD.
Analysing the network-tree formed by thelines AB-BC-CD-DA, we notice that it formes acontinous contour, that is, at the same time thecontour 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 thenetwork will be divided intotwo and only two sub-regionsABD and DAC. We observethat C is neither interior nor onthe contour of ABD and that Bis neither interior nor on thecontour of DAC.
We conclude that B is exterior toDAC and that C is exterior to ABD.We further assume that the line AC
was, as well, interior to the ABCDcontour. Let X be a point along ACinterior 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 becontinuous there must be another
segment, AX, that reaches theexterior of BCD.If one line belongs both to the
interior and the exterior of acontinuous closed perimeter (such asBCD) then it must intersect the
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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 thatconnect 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 0exterior of the ABCD contour,
we observe that BA-AD and BC-CD are continuous and link the same
verticesas 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.
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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 Eulers formula, the network must haveone line that links three points and one line that connects two
points, thus the networks 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 whichhave 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).
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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-CAs interior; by hypothesis we know that AB-BC-CA
belongs to AB-BD-DAs 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 wouldbreak Eulers 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).
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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 shouldsimultaneously 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 twolines 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, theremust be at least one point in which at least two lines
intersect.
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Modeling Guthries
conjecture 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 planarregions 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 continuous92
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Modeling Guthries
conjectureThus the lines of the network must:
-be continuous
-not intersect one another We observe that by associating the below
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 distinctplanar regions.
Thus no more than four regions can be
simultaneously connected, which validates
Guthries fourcolour conjecture.
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The topological
model
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Reference:
[1] Biography of Physics, by GeorgeGamow (1961) Harper & Row
[2] One, Two, Three...Infinity, byGeorge Gamow (1947), Viking Press
[3] Entertaining Mathematical Puzzles,by Martin Gardner (1986) Dover
[4]de Bonos thinking course by
Edward de Bono (1995) BBC books [5]De ce tac civilizaiile extraterestre
by Dan Farca (1983) Editura Albatros
[6]mathworld.wolfram.com
[7]Pi by Darren Aronofsky(1998)Artizan entertainment
[8]Fermats enigma by Simon Singh(1997) Anchor books
[9]Jurnal filozofic by Constantin Noica(2002) Editura Humanitas
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Before the end
This book took less than 24hours to compile and almost 10
years to develop, even if latelyIve been thinking more aboutengineering than theoreticalphysics or geometry, they stilland always be- the purer part ofmy mind.
Many thanks to everyone who,at one point or another, inspiredme to all these ideas. Things may
change but I hope the past neverill b d it