Relativistic Engineering

8/8/2019 Relativistic Engineering

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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:

._

_

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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

._

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:_2

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constnalgravitatioG

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Fonacceleratinalgravitatio

G

dcx

ctxmeaningG

dt

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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

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0

2

20

0

2

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comparing

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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


54/97

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

surface.65


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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

seen at Chambord documents. 74


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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

perimeter at least once. 84


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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

Documents

Relativistic Engineering