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• 1. Photon Size and Weight
  1.2 Viscosity of Light and Photons per Cubical Metre
  1.3 New Photon Modell and its Prove

• 1.2.1 Light Particles per cubical metre and ' Ratio ' - Approach 1
  1.2.1.1 Light Particles per cubical metre and ' Ratio ' - Approach 1 . 1
• 1.2.2 Light Particles per cubical metre and ' Ratio ' - Approach 2
• 1.2.3 Light Particles per cubical metre and ' Ratio ' - Approach 3
• 1.2.4 Light Particles per cubical metre and ' Ratio ' - Approach 4


• 1. Photon Size and Weight

' Photon Size 2020 . Graphical Solution '













' Classic Approach '

This ' Classic Approach ' results cubical metres for expanded Light Speed .
By using the cubical root of mass delta Sun as the value for temperature delta ,
we might have a look at other Stars like ' R 136 a1 ' or ' Sirius A ' and ' Rigel ' etc .












' Right click on the graphic in the web browser firefox to enlarge '



Using Earth weight we are going to calculate with the temperature delta
' Approach 2 ' ( see below ) delivers
The temperature delta of the light emitting zone should result values exactly ...

We just might to have found a solution for calculating with the
higher temperature of White Dwarfs beside same mass like our Sun !
For example , if we have a look at the White Dwarf Sirius b :
Temperature Delta of Sirius b and Sun is 4 ;
mass is the same approximately .
Temperature = Energy = E = 4 times more than our Sun ;
E = m c c = 4
c = Spin Speed of the Super - Quantums around the Photon Core = 2
because c = speed E = 4 and root of 4 ( = c ) = 2
resulting higher Frequency of the wave ( and shorter wavelenght , too ) but
Amplitude = 2 * Radius of the Photon = Diametre
E = Diametre or Amplitude squared
E = 4 * Energy of our Sun
root of 4 = 2
resulting double diametre and half of the wavelenght for E = 4
which should to be equal to the original wavelength ...

Using the higher mass - but lower temperature - of Red Supergiants
we might have seen , that energy is less what might mean
a lower spin-speed inheriting a longer wavelength , resulting :
Red Light !


To precise the work written above with a delta pi :

The Star Sirius B owns a mass almost equal to our Sun
but it is about 4 times hotter .
We regard temperature as Energy E ( = 4 )
E = m c c with c = spin speed , m is very small
and
E = Amplitude squared = ( Photon ) diametre squared
( 4 * Energy = double spin speed )
spin speed ( or 1 spin ) = diametre * pi
E = spin speed squared
E = diametre squared
root of E = spin speed ( double spin speed in case of E = 4 )
root of E = diametre
root of E * diametre * pi = 1 Spin
( because 1 spin would be a squared diametre ( diametre increases ) * pi )
( diametre squared ) is not equal to ( root of E * diametre * pi ) = diametre squared * pi

We will see the delta pi
showing diametre squared with increasing energy but the spinned distance is diametre squared * pi
what results a shorter wavelenght
( despite a larger diametre , because of the pi times larger spinned distance )
which is seen as blue light - we even might talk about a ' Blue Shift ' - the opposite of Red Shift ...

' The part responsible for speeding photons up to light speed , while breaking gravitational influence '













' Assuming energy to increase with mass ... '


































1.2 Viscosity of Light and Photons per Cubical Metre


Knowing Light Weight per Cubical Metre enables
the calculation of Viscosity of Light and Photons per Cubical Metre ...












1.3 New Photon Modell and its prove


' Photon Spin 2020 . Assuming a new photon modell . '

The Prove for Photon Spin












It has been a long way to get up to here :

1.2.1 Light Particles per cubical metre and ' Ratio ' - Approach 1

Assuming an average 'Light Particle' as a cube with lenght '1',
( 1 per Force ) to be ( 1 cubical metre per flexible weight ) times ( 1 second per flexible way ),
there should be about (( number times 10 to the power of number ) to the power of 3 )
Light particles per 1 cubical metre .

It is a rough calculation related to the units meter, second and kilogramm ,
using given Light Speed ( 3 times 10 to the power of 8 metres per second ) ,
Absolute Density of Ratio '1' as Ratio '2', Mass of Sun ( 2 times 10 to the power of 30 kilogramms ) ,
Ratio '1' as 'Lightsize per Lightweight', and Ratio '3'

If a Light Particle would have a weight of 0 Kilogramms,
it would move with a Speed of ' 1 ' within a scale from 0 to 1 .
And each mass should have its relative weight .

Note 1:
Using the term 'Mass' as kilogramm with one kilogramm as a unit to measure Gravitation
and 'Density of Mass' as 1 cubical metre per a variable size of kilogramm .

Note 2:
This calculation has led to an assumed number which is rounded roughly.
Shown by the graphics below :)

Note 3:
Be the cause
a Sun Light Particle needs exactly that impulse to leave gravitation of Sun,
similar to air particles to be bound to gravitation of Earth .

Note 4:
' Impulse making Particles slowing down to the power of two or even to the ' multiplication of Π ' in general '


1.2.1.1 Light Particles per cubical metre and ' Ratio ' - Approach 1 . 1

Approach 1.1 resulting 1 per ( 8 times 10 to the power of 90 ) kilogramms
per 1 cubical metre of Light
and an average Size of about 1 per ( 10 to the power of 40 ) ‚a' lenght of a cube
( the cubical root of about 1 per ( 10 to the power of 120 ) metres )




1.2.2 Light Particles per cubical metre and ' Ratio ' - Approach 2

The whole Approach 2 has led our idea to calculate with :
( Sun weight ) times ( light speed - from Sun - to the power of 3 ) is equal to
( Earth weight ) times ( ( fleeing Velocity of Earth times ( temperature of our Sun ) ) to the power of 3 ) .
What results an average temperature of a value about a million degrees celsius on our Sun , heat meaning particle movement ...

A new idea ( and it should to be the right one :) leads us to use ( 1 per Sun weight ) times ( light speed squared ) is equal to
( 1 per Earth Weight ) times ( fleeing velocity , vertically , of Earth squared ) times temperature of our Sun
because E = Temperature = m c c and an object would have to overcome more force to leave a celestial body vertically ,
our result is something about 400 kilometres per second of an object to speed up to for leaving
gravitational pull of earth vertically , using Sun Corona with a temperature of something between one and two million Kelvin ...




Light Density Calculation

Calculating with Gravitational Force, for light you might use a = 1 because close to sun

I found some ' magical ' 1 metre per 1 second what might be the fleeing Speed of our Sun
from the core of our Milky Way

Using 1 light-kilogramm per one light-metre , which is the very first idea from 2016
and cubical value of kilogramms of an object effecting one cubical metre of another object

Calculating with density of Jupiter would show
that Jupiter might leave Sun orbit about 2 cm a year, using a = 1 .

What is slower than the fleeing speed of Earth from Sun
despite of a smaller density
what would end up in an assumption of planets , that are attracted
by core of Milky Way rather by a star , if once have left gravitational influence of this star .




Note 5:
To get 'f' you might use ' 1 per force = density of mass times (*) density of speed ,
( (1/1/x per 1 kg) * 1 / (3 * 10 to the power of 8) ) '
by using x as light weight and density of average speed of sunlight
instead of calculating with using gravitational weight of sun (in kilogramms)
times speed how shown before by the upper graphics , what was I thought that days -
resulting to calculate with ( density of ) force , ( 1 / f ) and 'ratio 1' ( 1 / 1/x ) .


Note 6:
Or you just go ahead with the good old Isaac Newton
for ' force ' so you will use ' f = maa '
within the calculations described above ,
simply set his ' f ' in the formula with the squared root ( yellow highlighted ) .
Meaning light weight for mass ( 1/x ) , using its speed squared
as ' f ' in order to go with density of force ' f ' = density of mass times density of speed .

Note 7:
You may use speed ( of light ) squared , it is allowed ,
because you might think about unfolding an ' a lenght ' squared area
to get the point pressure a force is acting on by its speed .

This result has led my calculation to something
close to a trillionst part of a kilogramm ( 1 per squared speed )
for the weight of one cubical metre of light ,
using the calculation mentioned by note 6 and described above .
And if you will look through a microscope some day ,
seeing the little craters by ' light particle impacts '
you can check for the size if right or no ;) .



Note 8:
The closest approach to get the weight of one cubical metre of light I have found , is using just
1 cubical metre per Gravitational Pull of Sun in kilogramm seconds ( 2 times 10 to the power of 30 ) =
1 / 1/x ( 1 per light weight , or let us just say ' size per size per weight ' , what would result weight ) times
Speed in metres per second ( of Sunlight, 3 times 10 to the power of 8 ) to the power of 3 ,
see the graphic below .

Note 9:
The final result now for calculating the weight of one cubical metre of Light
should be to use
1 per ( Gravitational Pull of Sun ( 2 times 10 to the power of 30 ) in kg (*) times 1 second )
= equal to
( 1 per ( 1 / (1/x) ) ) as light density (*) times
cubical value of speed ( of Sunlight 2,7 times 10 to the power of 25 cubical meters per second ) .
What keeps gravitation which should to be measured in kilogramms in balance and
regarding unfolding an 'a' length cube
to get the point pressure a force is acting on by its speed , and

be the cause

a Sun Light Particle needs exactly that impulse to overcome gravitation of Sun,
similar to air particles to be bound to gravitational force of Earth .
A stone moving off earth with 3 meters per second
should have a ' gravitational weight ' of 1 per ( 3,6 times 10 to the power of 25 ) kilogramms
and its impulse ( ! should there be a need for an impulse ? )
to keep calculating with gravitational balance , no matter its size .

( see the graphic with the dark sun , shown below )

The Gravitational Force or Pull of Sun is more powerful the closer to Sun ,
and Light particles might react ( or ' being reflected ' ) faster
from Earth because the Earth has less of gravitational force ( measurement in kilogramms ) than sun .

It's the 21th of November 2018 today and I will continue from time
to time with thinking past about the offered ways to calculate ,
but I strongly think that it is done now :) ? !



Note 10 ( old ) :
' Number ay caramba , the way I pray to stay ' -
' Die Zahl zur Qual , der Weg der Steg '
The result should be the ( 5,4 times ten to the power of 55 )th part
of a one kilogramm for light weight per cubical metre
*** and *** the size of an average light particle, taken by 'ratio 1' :) ,
might unlikely (?) be the cubical root of 1 per ( 2,916 times 10 to the power of 111 ) metres
' a lenght ' of a cube .
Because of using ' ratio 1 ' to be 1 ( light size ) per ( 1 / light weight ) -
and ratio super emotio ;)-


A size of 1 cubical metre of light ,
it would have the 1 per ( 5,4 times 10 to the power of 55 ) part of a one kilogramm of weight .

Or let's just stay with a one kilogramm of light should measure about
5,4 times 10 to the power of 55 cubical meters ;)-

' Größe zu Gewicht wie ( 5,4 mal 10 hoch 55 ) zu 1 '
' Size per Weight is equal to ( 5,4 times 10 to the power of 55 ) per 1 '




Relations : Ratio '1' , and Ratio 'B' ( 1 per a variable size )

Note 11 ( new ) :
' Number ay caramba , the way I pray to stay ' -
' Die Zahl zur Qual , der Weg der Steg '
The result should be the cubical root of ( 1,6 times ten to the power of 167 )th part
of a one kilogramm for light weight per cubical metre
*** and *** the size of an average light particle, taken by 'ratio 1' :) ,
might unlikely (?) be ' the cubical root of the cubical root ' of
1 per ( 2,479 times 10 to the power of 334 ) ' a lenght ' metres of a cube .
Because of using ' ratio 1 ' to be 1 ( light size ) per ( 1 / light weight ) -
and ratio super emotio ;)-



An interesting thing I found , is that using Moon or Earth weight and their orbiting speed
returns results which are very close to the approach of using the calculation mentioned by Approach 4 .

Calculating with weight of Earth ( ~ partially ) and Sun ( partially, too ) and
density of Moon should result in speed Moon away from Earth.
Use Sun weight in kilogramms , because the Earth and Moon both are attracted by Sun .
Using Earth weight and density of Moon shows , that
our Moon would move about 2,5 m away from Earth during a year .
Calculating with 1 per kgs of Sun results 0,03745 metres a year of Moon leaving Earth orbit ,
which is very close to a current result of measurement .
Using 1 per Earth weight in kgs is an ' ideal sphere experiment ' with Earth and Moon isolated .
The result is regardless of other forces within our solar system or universe
and doesn't take distance and the by going loss of gravitational force into consideration .
To get a clear result you should make use of the variable 'a' ;)-

What I found too , is that ' speed squared ' per mass ( in kg ) of a moving object
might be of a fixed aspect ratio : 1 per ~lightSpeed squared ;
e.g. speed of Moon around Earth 'squared' per its weight should be about 1 per lightSpeed 'squared' ,
works with Sun and Earth also , approximately .

A later idea went me to use orbiting speed to the power of four
per weight what should result the fixed aspect ratio
of something about 1 per ten to the power of 10 .
I have tested on moon , earth and sun , and it seems
to work out without many doubts - or just limited ! ;)
Try using core of milky way , its moving speed within space
to the power of four per its weight .
Using kilogramms and metres per second .

Note :
I have been looking for a good ' all in one formula '
to describe and calculate objects within our universe ,
and some further ideas are in development ...



' Impulse making particles slowing down to the power of two '
In true , it should be the relation of a quarter circle instead of being squared , if so and if ...
And gravitational force should become less by ' a graph of a quarter circle ' , too


A trial to prove Approach 2

Prove 1 ( within the formula )

According to the ratio used by Approach 2 , an object within space
with a weight of 5 kilogramms would require another object
to be of a density of 0,025 kilogramms per cubical metre ,
in order this another object to move away with a speed of
2 metres per second , what results 8 by its cubical value .

Weight is relative , and the 5 kilogramm space object
might to be in relation of earth ( or earth kilogramms )
what would result the calculation
' ( 0,025 kg per 6 times 10 to the power of 24 , earth weight )
times five kilogramms '
to receive the actual weight in relation to earth
of the 0,025 kg per 1 cubical metre object ,
which is the same density
required to
leave earth at a speed of 2 metres per second
using the calculation with earth weight .

This result actually is closer to the density of vacuum
( using an assumed value )
regarding the 5 kg object to be made of compressed density .
And according to Archimedes Law , things just arrange to
same ( average ) density .

The ratio used by Approach 2 returns Weight per Second
( while one Second can be very relative , too ) .
Surely , the pyramides have been sunken ' a bit '
since the first corner stone was set ? ;)-
And Jupiter might become an ' Earth II ' one day
when it finally has compressed by gravitational force :)



Prove 2 ( external prove )

Think a stone with 1 kilogramm of Weight ,
( and to be about 1 cubical metre of size or just infinitely small ) ,
it would need an impulse or force
higher than 6 times 10 to the power of 24 ( weight of Earth )
times (*) cubical speed
to overcome gravitational force of Earth .

1 kg ^= Earth Weight times cubical Speed

For light, it might be the same :

1 / x / 1 times ( 1 second per ( 1 kg times 1 cubical metre ) , ( light weight ) )
^=
1 second per ( Sun Weight times cubical Light Speed )
      ( in order to leave gravitational force of Sun )

what results 1 cubical metre of Light to be about
1 per ( 5,4 times 10 to the power of 55 ) kg of weight per second

The temperature of our Sun is about 10 to the power of 6 degrees Celsius
( a value rounded very roughly and used by an assumed average )

That means the impulse of Sun light to be that times stronger
than on Earth because you can use an average temperature of about
1 degree Celsius on our Earth .
For , let's just say shooting a little stone
( or light ball ) from Earth ,
we would need an impulse of 1 per one million because
our Earth is that times lighter than our Sun .
As Earth is something of about a million times lighter than Sun ,
and about a million times colder ,
light would need a 1 per one million impulse to leave Earth
which could be provided by electricity ,
and if our Earth would be of the same temperature like Sun ,
light would leave at a million times faster speed , assuming .

Another thing , which would be in harmony with Prove 1 too ,
is to use Earth weight in kilogramms instead of Sun Weight ,
to calculate Light Weight per cubical metre .
Because if we regard Prove 1 , we would need to divide
the Light Weight of 1 per 5,4 times 10 to the power 55 kilogramms
per 1 cubical metre
by earth weight firstly ,
and then to multiply with Sun Weight ,
because we think and measure Sun Weight in relation to
earth weight and Earth Kilogramms
( like shown by the graphic above and by Prove 1 ) .
Regarding a 1 kilogramm stone on Earth might weight
something of about a million kilogramms on Sun be the Cause
Sun is a million times heavier than Earth , too .

That would result a Weight of something about
1 per ( 1,6 times ten to the power of 50 ) kilogramms of Light Weight
in relation to Earth , using the Ratio and
what would be a million times heavier than
calculating with total Sun Weight ,
and our Sun is about a million times hotter .



We think the 1 kg stone to leave Earth again :

1 kg from Earth ^= 1 per ( 5,4 times 10 to the power of 55 ) kg light Weight
per second , leaving Sun
Impulse on Sun for Light is 10 to the power of 6 times stronger than on Earth ,
surely we are talking about heat , meaning Particle Movement

The 1 kg stone might need an impulse of 6 times ten to the power of 24 ( earth weight )
times
a speed of x to the power of 3 to leave Earth , which should be equal to the light impulse ,
( 5,4 times 10 to the power of 55 ) and we might regard the temperature delta of earth and sun which is something about 10 to the power of 6
in comparison to weight delta which should to be the same approximately or even exactly :)-

What should result a speed of something about
10 kilometres per second this 1 kg stone needs to be speeded up
in order to leave earth ;
which is a result very close to the speed Isaac Newton
has proven .
Claimed as common ( and assuming ) light to act with same speed more or less ,
but to mind the weight of the space object it leaves , like e.g. earth or sun ,
the calculation should provide a solid value to deal and to work with
if using it for e.g. the calculation a rocket needs to be speeded up to leave our earth .



You are using cubical value of Speed in metres to the power of 3
because of
calculating with e.g. Sun weight to be compressed within 1 cubical metre .

We might see the calculation below ...





Using Ratio '1' meaning 1 for light size as a unit , this are 'y' meters .

Approach 2 resulting 1 per (5,4 times 10 to the power of 55 ) kilogramms per
1 cubical metre of Light
and an average Size of about 1 per ( 10 to the power of 37 ) ‚a' lenght of a cube
( the cubical root of about 1 per ( 10 to the power of 111 ) metres )

Graphical Solution of the Calculation

Light Particle Attraction by Gravitational Force of e.g. Moon , Earth Satellite ;
An average white Light Particle sent from Earth by an impulse should be slower
and of shorter distance than sent from Moon with the same impulse .
Light should obey physics like any other object .
Just think about a Light Particle to be the same size like Earth or Moon in relation to infinity :)

The upper graphic would show a nice way to calculate the weight
of an assumed average white Light Particle
by knowing force and all other characteristics of a ' stone ' or ' iron ' ball
shot from the near area of e.g. Moon or Earth etc. while comparing that data
by measured Light Particle attraction of a ' light ray ' attracted by an object like Moon etc .
like the experiment of the thrown Paper Jet , see below ;)-

A short story ... I have been living near a train station .
While I was watching the tree behind a locomotive and its smoke stack
the tree seemed to flicker and to blur behind the chimney .
I suddenly had the idea that the little light molecules
have been thrown out of their natural passes
and to be pushed upward a little bit ,
surely because of the heat meaning particle movement .
Then I started to think about twice such a things :
Might this be an answer to the question of
light molecules reacting like domino particles either
or just behaving like little iron balls , once shot off
reaching their targets in a straight way .
An interesting quiz , which seems to remain a riddle to me up to today .

But keeping that little secret , it told me another one :
Throwing a paper jet above a heating source ,
we all might know speed , weight and hence force , too , of this experiment .
So , if measurement of light molecule distraction above a heating source
like that one above the smoke pipe of the locomotive would be possible
by a strong objective ; we would know speed of light as well and
comparing that data with the thrown paper jet and its force and
degree of distraction , it would be an easy thing
to calculate the weight of a single light particle .

Short words , long mathematics ;)-

Paper Jet weight = 1 / 1000 kilogramms
Paper Jet speed = 4 metres per 1 second
Paper Jet Force = mass times way per time = 1 per 250 kilogramm metres
Paper Jet Distraction by the Heat Source = 1 metre upward

1 metre upward = 1 / 250 kgm


Light Particle weight = x kilogramms
Light Particle speed = 3 times ten to power of eight metres per one second
Light Particle Force = 3 times ten to power of eight metres times x kilogramms
Light Particle distraction by the Heat Source = y metres ; known

1 metre upward of the paper Jet = 'z' times upward of the Light Particle , 'z' times y metres


1 per 'z' = Light Speed times 'x' kilogramms

Nono end ;)-

1.2.3 Light Particles per cubical metre and ' Ratio ' - Approach 3

A Light Particle to be same size and weight like Moon and
to be blown up to 1 kilogramm of weight and 1 metre of size

If a Light Particle would ' weight ' 1 kilogramm , and
our Moon would weight 6 times 10 to the power of 22 kilogramms ,
may it to be a Light Particle the
1 per ( 6 times 10 to the power of 22 ) th part of a one kilogramm firstly ,
and then the ( 9 times 10 to the power of 17 ) th part of that .
Considering force of a Light Particle and force of Moon to be the same ,
and cubical value of gravitational force , kilogramms to the power of 3 ,
acting per cubical metre by cubical speed .



Assuming a Constant of Gravitational Force , meaning each object
to have the same energy or to be influenced by , within our universe .
Resulting the calculation of the weight of an average light molecule
to be estimated about 1 per ( 10 to power of 120 ) kilogramms ,

and about 1 per ( 10 to the power of 37 ) metres of size
regarding our Moon measures about 10 times 10 to the power of 9 cubical metres
which is that times bigger than one cubical metre a Light Particle is set to .
What would provide a value some how in harmony with
the rework of the very first approach from the year 2016
and with the second approach of august 2018 , too .


Approach 3 resulting 1 per ( 10 times 10 to the power of 120 ) kilogramms
per 1 cubical metre of Light
and an average Size of about 1 per ( 10 to the power of 37 ) ‚a' lenght of a cube
( the cubical root of 1 per ( 10 to the power of 111 ) metres again )

Multiplication of the weight with Moon Size
is because of the Light Particle being blown up to all three parametres
Weight , Size and Speed
in order to achieve a balance

And using this calculation with Earth and Moon
it gives something about 4 times ten to the power of three
kilogramms per cubical metre of moon density .
Just use the Division Delta of both Earth and Moon fleeing Speed .
Division Delta means to reduce the ratio of Moon speed per Earth speed
which should to be ( 0,038 metres per year ) per ( 0,044 metres per year )
by using 0,38 per 0,44 ( just get rid of additional potences, meaning
erasing 10 to the power of -9 ( speed per second ) on both top and bottom
of the ratio to get a pure ratio , which is what we call ' Division Delta ' ) .

The whole Approach 3 is based on Delta Force of Space Objects
doesn ' t exist in relation to infinity .
And some idea about 2*2*2 = 8 and 3*3*3 = 27 ,
with a delta of either 1 or 19 ,
and too, because of assuming the cubical value of Speed acting
per cubical metre by cubical value of kilogramms like mentioned above .





1.2.4 Light Particles per cubical metre and ' Ratio ' - Approach 4 ( Mathematical error ? )

Note 1
1 / f = 1 / x * 1/average Lightspeed
isolate f to :
f = 1 per ( 1 / x * 1/average Lightspeed )
and use this 'f' (force or impulse) now
within the calculation described below ,
simply set this 'f' in the formula
with the squared root , see about that above .



Calculation using both formulas in combination ,

1 / force is equal to 1 cubical metre per flexible weight ( not density of mass )
times speed ( 1 second per flexible way )
as force of movement

and

1 / force is equal to squared root of ( ( 1 cubical metre per x kilogramm ( as density of mass ) )
times speed ( 1 second per flexible way )
as gravitational force

and setting values of force equal to each other

what ended up in a mathematical error ( ? ) to receive a ratio

This has led my idea to distinguish between two main kinds of energy :
Gravitational Force beside ( and resulting in ) Force of Movement .
You might use speed squared , too .
A candy I can still show .
At the end , it is the relativity of weight which is causing clutter :)-
as well as the first three approaches result at least similar values .
It might be the cause light speed is used by its average , too ,
which could be very relative instead .



Air and Light !

Vacuum Density Ballooning

' Actually , a rocket drive might to be compared with a heated Vacuum ,
may be at an elephant's level '

Dedication


This piece of research work is dedicated especially to The Royal Swedish Academy Of Sciences







and to my friends and parents




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