Some Interesting Maths

Theorem 1

Two equal squares cannot be converted into a bigger and a smaller square.

Proof: Let us assume everything is made up of unit tiles. One square + a figure = a bigger square, only if the figure can be converted into a L shape which engulfs the smaller square from two sides. The L shape must be able to form the outer boundary of the square. In a square each layer of the boundary contains an odd number that is 2 more than the previous number. For example 1+3= 4 , 1+3+5 = 9 etc. To engulf a^(2) the boundary must be of b^(2) + 2ab. Since the boundary has to be taken out from a equal square leaving another square ( c^2) the boundary must be of d^(2) + 2cd . So, b^(2)+2ab= a^(2) - c^(2)

Let c= (a-x), then we have

b^(2)+2ab= a^(2) - (a-x)^(2)

b^(2) +2ab= 2ax - x^(2)

let x = b.f where f is a rational number. Then we have,

b^(2) +2ab - f(2ab) +f^(2)(b^2) = 0

Now solving for x using the quadratic equation formula.

[2ab + - √{(2ab)^(2) -8a(b^3) -4(b^(4))}]/2(b^2)

or, [2ab + - b√{(2a)^(2) -8ab -4(b^(2))}]/2(b^2)

or, [a + - √{(a)^(2) -2ab -(b^(2))}]/(b)

Now for f to be rational (a)^(2) -2ab -(b^(2)) must be a perfect square. Let us assume (a)^(2) -2ab -(b^(2))= (a-b-x)^(2) where x is a rational number such that b+x is less than a. We have,

(a)^(2) -2ab -(b^(2))= (a-b-x)^(2)

or, -(b^(2)) = b^(2) + x^(2) +2bx - 2a(b+x)

or, -(b^(2))= (b+x)^(2) -2a(b+x)

or, -(b^(2))= (b+x)(b+x-2a)

or, -(b^(2))/(b+x) = b+ x -2a

or, 2a = b+ x + -(b^(2))/(b+x)

now b+x < a  and -(b^(2))/(b+x)<b<a so in no possible way can  2a be equal to b+ x + -(b^(2))/(b+x). Hence x is not rational and therefore f is also not rational and our initial assumption that 2 equal squares can be turned into 2 unequal squares was wrong.

Poetry

Poetry

chess

I make moves without seeing, what's the need to see. Emotions are betrayed by the eyes, what's the need to say.

flower

Everything looks beautiful in youth, who can beat time. All flowers blossom in the day, which one survives the night.

light

I may not have the appearance, but I do have foresight. I may not know colors, but I do know light.

Everyone wants gains but no pains,I think differently. Even when offered pain, I just convert it into a gain.

chess heart

I used to consider myself a chess player, but the heart turned out to be that of a lover. Whom I used to consider my savior turned out to be my killer.

cry

Harsh times make me cry, but I don't cheat. I may not know how to win but I don't accept defeat.

victory

People drink to overcome shyness, I prefer to remain shy. Every heart has got its own desires, even I get a high.

A new and very simple technique to approximate pi

Pi

Let the radius of a circle be R. We take a quarter of that circle.quarter circleNow a right angled triangle ABC can be drawn starting from center of the circle with AB (its height) extending along the left edge of the quarter circle, it’s base BC starting from B and touching the circumference of the quarter circle at C and AC being the hypotenuse of the circle also joins center of the circle to its circumference, in other words it is the radius of the circle

pi

 

 

 

 

When AB = radius of the circle then BC will be zero as AC is always equal to the radius of the circle, when AB is ‘h’ less than R then BC will be equal to √((R)^2-(R-h)^2) ,or √(2Rh-h^2). Now if this quarter circle is converted into a rectangle with length R, it’s breath(b) will be the average of  √(2Rh-h^2) as h increases from 0 to R . Area of the quarter circle may be written as R*b, we know it’s also equal to pi(R)^2/4 ,therefore b=piR/4 is the average of  √(2Rh-h^2) .The area of the quarter circle is also equal to the integral of  √(2Rh-h^2)dh as h moves from 0 to R. Now if we take the integral of (2Rh-h^2)dh instead of √(2Rh-h^2)dh we will get R(R^2)-(R^3)/3 which is the integral of squares of side  √(2Rh-h^2) as h moves from 0 to R, so can we approximate the average square to have a side of b? We will check that later, for now let us do it, the average square has an area of b^2(dh) so the total area of the squares is b^2(number of squares)dh, or b^2dh(R/dh)=R(b^2) but the area of the figure is also = R(h^2)-(h^3)/3,

Therefore we have:-

R(b^2)= R(R^2)-(R^3)/3

We know that b=piR/4

R(piR/4)^2=2(R^3)/3

Or, (pi^2)/16 =2/3

Or, (pi^2)=32/3

This gives us the value of pi to be 3.266, so what did we do wrong..well we approximated the value of average square to be b^2. b^2 is the square of integral of '√(2Rh-h^2)dh divided by R', while average square is the integral of (2Rh-h^2)dh divided by R so we are substracting the square of integral of (√x/R) from integral of (x/R) where x=2Rh-h^2. So what will be the difference? The integral of x/r is (x)^2/2R(dx/dt) if dx/dt is constant, if it isn’t it will turn into an infinite series and we don’t want that so instead of taking the value of dx/dt at h=R we take the average value of dx/dt. Dx/dt = d(2Rh-h^2) or, 2R-2h which changes from 2R to 0 , so the average value of dx/dt will be R. Therefore the integral of x/R is is (x)^2/2(R)^2 and the integral of √(x)/R is 2x√(x)/3R(dx/dt) or 2x√(x)/3(R^2) and that squared is 4(x^3)/9(R^4) ,so we have :- (x^2)/2(R^2)-4(x)^3/9(R^4) ,the value of x is (2Rh-h^2) at h=R it becomes (R)^2, so we have:-

((R^2)^2)/2(R)^2 - 4((R^2)^3)/9(R)^4

=(R^2)(1/2 – 4/9)

=(R^2)(1/18)

This (multiplied by dh) is the difference in the area of average square from b^2, therefore the difference in area of the whole figure will be R*(R^2)(1/18) or, (R^3)/18.

So we have  :-  R(b^2) +(R^3)/18= R(R^2)-(R^3)/3

Or, R(piR/4)^2 + (R^3)/18= R(R^2) - (R^3)/3

Or, (pi)^2/16 = 2/3 – 1/18

Or, (pi)^2 =11*16/18

Or, (pi)^2 =176/18

Or, (pi)^2=9.7777777778

Or, pi = 3.127

So what did we do wrong now, well we integrated x^2 and √x for dx instead of dh but that was a fine approximation as you can see.

A Theory for Finding the Best Move in Chess at Any Given Position

Theory 1

Any player's chances of winning in chess can be approximated to two major factors, the amount of squares he controls and the amount of time he has. Amount of squares controlled is simply the number of squares in which he can bring one of his pieces but  the opponent cannot because of impossibility of the feat or because of obvious loss of material and the amount of time can be said to be the amount of difference between moves required by the opponent to accomplish his best attack and the moves required by the player to thwart that attack. For example opponent is threatening to kill your bishop in 3 moves with that being his best attack and it can be defended in 1 move then you have time of 2 moves to accomplish your own plans. Now your score at any given moment is (number of squares you control)*(amount of time you have) - (" " opponent controls)*(" " time opponent has). Your goal is to make a move that increases your score by maximum amount. 1 pawn has a value of 3 squares and can be at most times traded for control of 3 extra squares.

Theory 2

Okay so theory 1 a very basic theory but it had some flaws, it only counts the total squares not the squares of individual pieces, so some pieces may be hyperactive others under active which is not a desirable situation, so a better approach would be to take into account ∆S/2S. Where ∆S is the increase in number of squares available to a particular piece and S the number of squares it initially had (if ∆S is much more than 1 please take the average value of S instead of its initial value in the denominator). If one move increases the squares of more than 1 piece add them up separately. Another flaw in theory 1 was that time can vary absurdly, so we have no standard value of a square and no correlation between the value of a piece and value of a square, if I say 3 squares is worth 1 pawn, you may say when time doubles -'now any 1.5 squares is worth a pawn' or any 1 square is worth a pawn if time has tripled which is absurd. Let us take another approach. 1 threat may require 1 move to be solved, in that case opponent solves that threat instead of increasing S, so a threat that requires 1 move to be solved, or in other words wasting your opponent's 1 move is worth average ∆S/2S, but then in one move opponent may also cause some material damage. Average game lasts 40 moves and excluding the king there are 39 points on the board. But then average game may not be played till checkmate or without mistakes so let's take a conservative figure, say 100 moves. Then 39 points require 100 moves to be extinguished so 1 move is worth 0.4 pawns + average ∆S/2s gained (which can be approximated to 1/4). So a threat that requires 1 move to be solved is worth 2/3 pawns provided you do also increase the average number of squares while giving the threat. A more rigorous approach will be to calculate the decrease in advantage provided by opponent's best move(∆bm). For example opponent was threatening to kill your pawn, you in turn threatened opponent's queen, now the opponent saves his queen instead of killing your pawn so ∆bm is 1. So the score change can be approximated to be 'forced win +  ∆S/2s + ∆bm '(or "∆S/2s +2(opp's move wasted)/3" if you want to calculate lesser). Now will the move with highest score increase be the best move? It may be but there is a big catch, 1 best move may be better than a number of good moves. So a move that does not increase or even decreases the score but prepares for a second move that will increase the score by a higher amount than 2 alternate best moves combined is probably better than the alternate moves, so you must look for combination of moves that increase the score by maximum amount but that again would increase our work.

Theory 2++

Forced win + ∆S/2S + ∆bm(or 2∆(opp's moves)/3) , is a good theory and if practiced rigorously would easily bring your rating to 1800. But it's not the end of of the story, it explains the importance of mobility, tempo, and piece activity but fails to incorporate 'piece coordination'-a very important principle of chess. It basically means your pieces must defend each other, attack together (i.e. if one piece becomes a martyr while vanquishing a greater foe his teammate must revenge him) and work together towards decreasing the possibilities and eventually annihilating a piece. Let's go step by step more the defense the better. Each piece with 1 extra defender than the number of attackers is worth 0.2 points more. For 2 extra defenders 0.282, and for 3 extra defenders 2root(3) and so on(also don't forget to take the value of the 1/2S of the square where you are defending a piece). Next pieces working together to attack the opponent accomplish more. Now if number of attackers is more than the number of defenders, you get a win so we won't account for that. Now if 2 pieces are working together to reduce 2 square available to an opponent's pieces the total value of squares reduced is 2*root(2+1)/2S where S is the initial number of squares that piece had. Also don't forget to add 2*root(2+1)/2S1 and 2*root(2+1)/2S2  for two of your own pieces. The general algorithm is (no. of sqs changed)*root(no. of pieces involved)/2(total no. of sqs). So the total score change in a move is Forced win + ∆Sroot(pieces involved)/2S + ∆bm(or 2∆(opp's moves)/3) +0.2∆root(extra defense), your goal is to find a combination of moves that increase the score by maximum amount.

Why Don't We Use Pythagoras Theorem for Finding Tangential Velocity or Centripetal Acceleration

If we use pythagoras theorem for finding tangential velocity or centripetal acceleration that is, we

√(R^2 + (vsinx)^2) - R

or,(√(R^2 + (vsinx)^2) - R )* (√(R^2 + (vsinx)^2) + R)/(√(R^2 + (vsinx)^2) + R)

or, (R^2 + (vsinx)^2) - R^2/(√(R^2 + (vsinx)^2) + R)

or,(vsinx)^2/(√(R^2 + (vsinx)^2) + R)

or, approximately : (vsinx)^2/2R

But we know centripetal force is (vsinx)^2/R , so what did we do wrong over here? Is there some law which states we must use differentiation and not pythagoras theorem? Well there is a problem in taking what we got as centripetal acceleration. How do we know it's acceleration? Sure tangential velocity is acceleration but then how do we know it's velocity and not distance travelled? For time =1 unit we got displacement or tangential velocity whatever it is to be (vsinx)^2/2R. Now let's put t=2 seconds and find the total of this change, if the result gets multiplied by 2 then it's proportional to t and we can say it's tangential velocity but if it gets multiplied by 4 then it's proportional to t^2 and we may say it's distance traveled and in case it's neither proportional to t nor t^2 we may say we did some mistake somewhere. At t=1 total change was (vsinx)^2/2R and cosx increased by v(sinx)^2/R and sinx changed by an extremely small amount and can still be approximated as 1. After t=1 and before t=2 change will be vcosx +(√(R^2 + (vsinx)^2) - R ) = (vsinx)^2/2R +(vsinx)^2/R . Total change = change at t=1 and change at t=2 or,  =(vsinx)^2/2R + (vsinx)^2/2R +(vsinx)^2/R =4(vsinx)^2/2R

So as you can see what we get from pythagoras theorem is proportional to t^2 and hence is displacement not velocity.

Displacement(S)= u^2 + a(t^2)/2

at t=1, S= a(t^2)/2

= Acceleration/2

Therefore Acceleration = (vsinx)^2/R

And if we want displacement to be zero we must ensure acceleration is zero. Hence gravitational acceleration will be equal to (vsinx)^2/R for our familiar circular or elliptical orbit.

New, very powerful meditation techniques

Buddha

 

Benefits of Meditation:

  1. You will  have a calm, clear and stress free mind.
  2. With as little as half hour practice daily, your concentration power will increase, you will perform better and take less time to do so in tasks requiring high concentration. Results will begin to show in 8 weeks.
  3. You will feel more motivated.
  4. Your thoughts will become much faster . Each thought will have more emotion (which can be a problem, so consult your doctor before starting the practice). So why will the thoughts become faster and more emotional? Its because you are eliminating a lot of thoughts during this meditation but you cannot eliminate all of them, the fittest at this game will manage to survive and more of a kind of thought you have, more the brain will adapt to produce that kind of thought so after a period of time you will have much faster and emotional thoughts. Its plain evolution in action.

Best technique

Write down each and every thought you have, write them in short. Write only the names of persons and things you think about not what they are doing or it will become too complicated. Describe your thoughts only in nouns not verbs, and write only the first 2 letters of each word.

Technique 1

Repeat every image that comes into your mind. Keep focusing on the last image that came in your mind till the next one comes. Make sure that no image goes unnoticed or unrepeated.

 

Technique 2

You can do this technique anytime , anywhere, in any pose. All you have do is keep repeating the word 'blank'. Now as soon as a thought enters your mind you have to say  'Not blank' instead of 'blank', while the number of blanks can be as high as your repeating frequency(the higher the better), the number of 'Not blanks' should be equal to the number of thoughts you had, lesser the time gap between a thought coming and your saying Not blank (mentally please, both blank and Not blank) the better.

Technique 3

Don't let your mind trick you into thinking meditation

Keep on repeating the word killed until you have some thing very important to think. And when you have something very important to think which you will have coz of your mind tricking you, start repeating 'not killed' till the urgency of that thought has died down. This technique is much easier to do than technique 1 but gives almost same benefits.

UPDATE: technique 3 is very difficult to do and technique 4 is not that powerful so what do we do, well not to worry I have got another few techniques  for making your life stress free in the short run and making your thoughts faster and stronger in the long run (through survival of fittest, duh).

Technique 4

mantra mindfulness meditation

Keep repeating any word of your choice and at the same time be aware of your surroundings, by aware I mean aware of all sights and sounds that arise in the environment surrounding you. Now whenever a thought arises which distracts you from the present moment, make a note of it using your index finger as if counting 1, now you don't need to actually count, that would make the technique very difficult to do. All you need to do is raise your index finger whenever a thought arises and don't forget the mantra, keep on repeating it no matter what (till your meditation practice doesn't end, of course), and also try to be aware of surroundings while noting your thoughts, if you find this difficult to do you may skip the noting with index finger part but the mantra and the surroundings part are essential. As you can see this is a mixture of 2-3 meditations, traditional meditations by themselves are not powerful by themselves to bring any notable benefits or to even stop the thought train at least. These combination are much more beneficial and powerful, try it if you don't believe me.

Technique 5

Breathing mantra meditation

Firs step in this meditation is to become aware of your breathing, then slow it down a bit, in such a way that there is a pause between breath in and breath out. Now as you are breathing in start repeating the word 'breath in' and keep on repeating till the breath is released, and as soon as you start releasing the breath start repeating 'breath out' and keep on repeating till you again start breathing in. Now you have to keep on repeating well during the pause as well, breath in or breath out depending upon whether your breath is inside or not. This technique combines mantra and mindfulness in most powerful way i know of till now.In traditional  TM practice practitioner soon becomes mindless and starts repeating the mantra like a robot while the mind wanders, this BM meditation forces you to be aware and hence rise above thoughts

Technique 6

Counting meditation

Keep on counting non stop Start from 1, and do not repeat after reaching a certain number. The faster you count the better. This way you will know how efficient your session was. Ideally you should reach at least 1500 for a 30 minute session. This meditation will apart from providing all the benefits of meditation also increase your working memory. The important thing to keep in mind is that counting should be continuous, there should be no pauses in between.

By Vaibhav Jain

Universe is not dark: A new theory of gravitation and other things

Update: this theory is wrong please don't waste your time reading it, I will upload the correct theory of everything soon

THEORY OF ALMOST EVERYTHING

NEW CALCULUS

UNIVERSE WITHOUT DARK MATTER OR DARK ENERGY

GRAVITATION WITHOUT ANY CONSTANTS

UNIFICATION OF GRAVITY AND ELECTROMAGNETISM

SCORE THEORY

BY VAIBHAV JAIN

 

Chapter 1: New calculus

Chapter 2: Score

Chapter 3: Gravitation

Chapter 4: Photons

Chapter 5: Momentum and Kinetic energy

Chapter 6: E.M.F.

 

 

 

 

 

1

NEW CALCULUS

A time interval is always continuous, no matter how small, there can be no smallest possible time interval as any you take will be made up of smaller components. There can only be one smallest possible time interval that is zero. Same is true for any continuously changing variable.

While using differentiation for finding or expressing the laws of the universe we always need the ‘first cause’ and the effect and for finding the initial cause of an event we always need ‘the smallest possible time interval an extremely small one won’t do.

Suppose we are differentiating a variable x with respect to time, we first need to take the change in x in the smallest possible time interval, which has to be zero, meanwhile in an extremely small time interval x becomes x + Δx but we ignore the Δx and take dx to be zero(as dt is zero). In this way if we keep on ignoring Δx the difference between actual value of x ie x + Δx , and book value of x  keeps growing ,now suppose x + Δx is multiplied by another variable (or even constant) the difference between x + Δx and x increases more rapidly. Suppose x is a constantly changing variable then at the beginning when x’s value was Δx, here comes our first rule ,none of our chosen variables(namely distance, velocity, cos a, sin a) can ever be zero, so when x was Δx and increased to 2Δx, then we ignored 100% of increase in x, now when x changes from 2Δx to 3Δx we again ignore Δx ,ie an increase of 50% of x ,when x is 3Δx and changes to 4Δx we ignore 33.33 % of x ,when x is x and increases to x+ Δx ,we ignore an increase of (Δx/x)*100% of x, in this way we ignore 100% +50% +33.33% +25%... approximately lnx*100% of x . So while the book value is x the actual value is x(1+lnx) ,the book value is such because we have taken dx instead of Δx at all intervals ,but then why take percentage ,why not just add up and conclude that we have taken 0 instead of x. That is because when x is multiplied with other numbers the difference also grows larger at each step ,but what when something is added to x, then the difference doesn’t grow larger ,well I will explain that case later.

To differentiate a variable we will first have to convert the book value of the variable into actual value

x = x(1+lnx)

Now dx = 0 but d(lnx) is not = 0  ,You may say to differentiate lnx we again need to convert it into actual value, and it will be an endless cycle but over here we need the ‘change in x’ in the smallest possible interval ‘the percentage of x we are ignoring’ can vary and we are under no constrain to ‘ignore the smallest possible percentage of x’ at each step in other words Δx can vary and Δx does not need to be measured in the smallest possible time interval ,a reasonably small one will do.

d(x(1+lnx))/dt= 0(1+lnx) + xd(1+lnx)

=0 +x*(dx/dt)/x

But the x above is x at book value(xb) and x at denominator is x at actual value

=(dx/dt)/(1+lnx)

This is the differential of x. Over here dx/dt is actually Δx/ Δt where Δx and Δt are extremely small(but not zero). We can rewrite dx/dt as                     Δx/ Δt(1+lnx). Now before taking ln of x we must convert physical quantities into their natural units ,we cannot take just any unit. Natural unit of velocity is c, that of distance is 1.2*10^(-25)meters. This is necessary as we won’t be using any constants except c (and h once to derive natural units of distance, apart from that only measured masses and distances between things)

Although we refer to x(1+lnx) as actual value ,for calculations it will always have to be written as x(1+lnx) and never as only x and the distinction between actual value and book value will come into play only when we differentiate a continuously changing variable in other words x will become x(1+lnx) only before we differentiate a continuously changing variable x.

The differential of 1/x will be

1/x= (1/x)b (1+ln(1/x))

=(1/x)b (1-lnx)

=d((1/x)b(1-lnx))/dt

=(1/x)bΔ(1-lnx)/Δt

=(1/x)b*(- Δx/x)                     ,over here Δx is short for Δx/Δt

=- (Δx(1+ln(1/x))/x)b*(1/x)b

=- Δx(1-lnx)/x*x                    as xb is written simply as x while x as x(1+lnx)

Now if we want to differentiate xy

xy=(xy)b (1+ln(xy))

=xbyb (1+lnx+lny)

= xbyb Δ(1+lnx+lny)/Δt

= xbyb(Δx/x + Δy/y)

= Δxy/(1+lnx) + Δyx/(1+lny)

Now you may ask why not take xy as one variable and get the differential as Δ(xy)/(1+ln(xy)),because, here comes the second rule-

You must separarate the ln of a term into ln of independent variables before differentiating, an independent variable may be defined as one which does not neccecarily change when some other variable changes and whose slope at a point(ie an extremely small space) is the least curved.

Now if you have to differentiate √x

√x  = √xb (1+ln√x ) ,now since √x is not an indepent variable we must convert the ln of it into ln of the independent variable

= √xb (1+(lnx)/2)

d√xb(1+(lnx)/2)/dt

= √xbΔ(1+(lnx)/2)/Δt

= √xb (Δx/2x)

= Δx/2√xb(1+lnx)

= Δx/2√x(1+lnx)

If a variable x is written as x/2+x/2 and then differentiated

The differential will be (Δx/2)/(1+lnx) + (Δx/2)/(1+lnx) it will not become (Δx/2)/(1+ln(x/2)) + (Δx/2)/(1+ln(x/2))

Suppose we can write x as y+z ,now if we differentiate x and supposing y and z are independent ,the differential will be Δy/(1+lny) + Δz/(1+lnz) the fact that z has been added to y makes no difference to lny but if x were=y*z and now even if y and z are independent ,the differential will be z*Δy/(1+lny) + y*Δz/(1+lnz) and as you can see lny is part denominator of a much larger number as along with the differential of y it(lny) has also been multiplied with z ,so in multiplication the distinction between book value an actual value grows while in addition it does not and that is why while finding the actual value of a continuously changing variable we took percentage at each step.

The differential has changed but the integral will remain the same as traditional calculus because during integration we just sum up extremely small or even smallest possible changes.

∫ Δxdx/(1+lnx) =x/(1+lnx) -1/(1+lnx)^2 …

 

 

 

   2

SCORE

The universe is infinite in space and continuously evolving. In this dynamic universe many systems exist, each of them faces 2 risks ,firstly due to circumstances it might change so much that the new system has no resemblance whatsoever to the previous system and the previous system may become history or in other words die. Second risk a system faces is that in this infinite universe it may get so far away from other systems that it may have no effect on any other system in other words it may get lost and isolated in space. Let us assume this universe is made up of empty spaces and tiny particles, then any system can be defined as the number of particles and the spaces between them. The best system will be the one that is changing the least (at any given period of time) and is closest to other systems. Another factor to consider will be that if 2 particles separated by 10000 meter start moving away at 1 cm per second ,we will call that change as slow while if 2 particles separated by 0.00001 cm start moving away at 1 cm per second we will call that change as fast.

Let’s go step by step

A system that changes the least is the best

1-(vcosa)^2/2  where v stands for velocity of a particleand cosa for the cosine of direction of same particle with respect to some other particle.

A system that is closest to other to other systems is the best.

(1/r)-(vcosa)^2/2 where r stands for distance from 1 other particle

Considering the third factor

(1/r)-((vcosa)^2/2r)

Now the vcosa term will be multiplied by the mass of the particle ,the relativistic mass .For now we will assume all particles have equal and unit mass.

(1/r)-((vcosa)^2/2r(√(1-v^2/c^2))

This is the score of a particle with respect to one other particle. Total score of a particle is the sum of it’s score with respect to all particles.

= (1/r)-((vcosa)^2/2r(√(1-v^2/c^2)) + (1/r2)-((v2cosb)^2/2r2(√(1-v2^2/c^2))  + (1/r3)-((v3cosc)^2/2r3(√(1-v3^2/c^2))…

Where v2 ,r2 ,cosb are velocity distance and direction with respect to a second particle and v3,r3,cosc with respect to a third particle. Each 1/r term will be multiplied by respective mass of the corresponding particle.

If most of other particles are concentrated at a far away point, score of a particle can be approximated as

Ng/Rg –Ng(vcosa)^2/2Rg(y) where Ng stands for mass of other particles in the galaxy/galaxy cluster and Rg your distance from the point where other particles are concentrated ,v is your velocity with respect to that point and cosa the cosine of your angle from that point and y the relativistic term. But this equation must be written in natural units. If v refers to velocity in meter per second ,velocity in natural units become v/c.

The equation now becomes

Ng/Rg –Ng(vcosa)^2/2(c)^2Rg(y) where c stands for speed of light in vaccum.

This: Ng/Rg –Ng(vcosa)^2/2(c)^2Rg(y)=score of a particle

Score of all particles summed up is the score of the universe and the score of the universe is always conserved. The score cannot increase to infinity, who would want to live in a black hole anyway. In fact the score of the universe cannot increase (or decrease) at all, in this universe you can improve only at the cost of others that’s why its evolving, only the fit survive.

The universe is effectively divided into galaxy, galaxy clusters and super clusters with most mass being centered at the center of galaxy, galaxy cluster and super cluster (ignoring the dark matter which trust me doesn’t exist). The distance between two super clusters is huge and score from other super clusters will get too small for particles in one super cluster and can be effectively ignored. While analyzing the motion of a collection of particles that move together (a body) we will have to ignore the score of constituent particles from each other because no score disturbance takes place over there and even if there is relative motion between the particles the body’s movement is independent of it.

More than 99% of the mass of a galaxy is concentrated at the center of the galaxy (again if we assume dark matter does not exist). And most of the mass of galaxy super cluster exists at the center of the super cluster (at least in the case of Virgo super cluster if not always). For particles which are not too close to a large body the 1/r + 1/r2 +1/r3 .. part of its score can be approximated as (Ng/Rg +Ngs/Rgs) where Ng is mass of particles in the central region of the galaxy ang Rg, the chosen particles distance from it and Ngs mass of particles in the central region of galaxy super cluster and Rgs the chosen particle’s distance from it. If unlike local cluster which is a poor galaxy cluster if a galaxy cluster is proportionately massive enough we will also have to take the number of particles in galaxy cluster center in addition to galaxy center and galaxy super cluster center.

Now let us start by analyzing the motion of a star around the center of its galaxy. For sake of simplicity let’s assume it’s the sole galaxy in the universe (no, no cluster etc) and almost all its mass is concentrated in the center. Then the score of the star will be

nNg/Rg –nNg(vcosa)^2/2Rg(c^2)(y) where v is the velocity with respect to the center of galaxy, cosa the cosine of angle it makes from the center of galaxy and n the number of particles in the star. Unless some other particle compensates for change in score of a particle at the exact same time the score a particle remains constant. The score of the star which is sum of scores of constituent particles will also be constant.

3

GRAVITATION

Therefore we have

nNg/Rg –nNg(vcosa)^2/2Rg(c^2)(y)=constant

Differentiating both sides with respect to time (we no longer need to write further derived quantities in natural units except the units with ln in front of them)

  • -nNg(1-lnRg)vcosa/Rg^2 –nNgvA(cosa)^2/Rg(c^2)(y)(1+lnv) + nNgv^3cosa(sina)^2/(Rg^2)(c^2)(y)(1+lncosa) + nNg(vcosa)^3(1-lnRg)/(Rg^2) +nNg(vcosa)^2(Δy/Δt)(1-lny)/2Rg(c^2)(y^2) = 0 ,where A stands for acceleration
  • nNgvcosa(-(Acosa/1+lnv) +((vsina)^2/Rg(1+lncosa))+((vcosa)^2(1-lnRg)/2Rg+(vcosa(Δy/Δt)(1-lny)/y)/Rg(c^2)y= nNg(1-lnRg)vcosa/Rg^2 (solving this we will et acceleration in 3*10^8 meters per second square ,if we want it in meter per second square we must multiply L.H.S. by c ,we get
  • nNgvcosa(-(Acosa/1+lnv) +((vsina)^2/Rg(1+lncosa))+((vcosa)^2(1-lnRg)/2Rg+(vcosa(Δy/Δt)(1-lny)/y))/Rg(c)y= nNg(1-lnRg)vcosa/Rg^2
  • -Acosa/(1+lnv)+ (vsina)^2/Rg(1+lncosa)+ (vcosa)^2(1-lnRg)/2Rg+ vcosa(Δy/Δt)(1-lny)/y= c(1-lnRg)(y)/Rg
  • Now let us assume that our chosen galaxy is an average spiral galaxy. Scientists expect it to be between 0.02 and 0.08, and here I make my first prediction an average spiral galaxy will have an eccentricity of 0.05. Average cosa will be equal to (e^2)/π for ellipses close to a circle, or, (0.05)^2/3.14

Or, 0.0008 apprx.

Now ln(cosa)=ln(0.0008) = -7.2 apprx.

Now considering the star is at a distance of 3*(10^20) meters away from the center of the galaxy (which is the distance between Sun and the center of milky way) ln(Rg) will become

ln(3*10^20 *1.2*10^25) ,=ln(3.6*10^35)=105 apprx.

Now, -Acosa/(1+lnv)+(vsina)^2/Rg(1+lncosa)+(vcosa)^2(1-lnRg)/2Rg =c(y)(1-lnRg)/Rg  ,ignoring the Δy term as it would be very small ,for now we can also ignore (vcosa)^2 term as cosa is very small.

Or, -Acosa/(1+lnv) +(vsina)^2/Rg(1-7.2) = c(1-105)/Rg ,ignoring y as it’s very close to 1

Or,     -Acosa/(1+lnv) =c(-104)/Rg –(vsina)^2/-6.2Rg

For acceleration to be zero

104c/Rg=(vsina)^2/6.2Rg

Since sina is soo close to 1 we may ignore it, then we have

v^2=6.2*104c

Or, v^2=632c

Or, v=√632c

Or, v=4.354 *10^5 meters per second

This is the velocity at which stars will revolve around the center of an isolated galaxy provided it’s an average spiral one.

Now to analyze our equation,

  • -Acosa/(1+lnv)+ (vsina)^2/Rg(1+lncosa)+ (vcosa)^2(1-lnRg)/2Rg+ vcosa(Δy/Δt)(1-lny)/y= c(1-lnRg)(y)/Rg

vcosa(Δy/Δt)(1-lny)/y can be simplified to                              (v^2)Acosa(1-ln(c^2-v^2))/2(c^2-v^2)

For   v^2<<c^2 we may ignore this term.

  • -Acosa/(1+lnv)+ (vsina)^2/Rg(1+lncosa)+ (vcosa)^2(1-lnRg)/2Rg+ = c(1-lnRg)(y)/Rg

Now when cosa>>sina we may ignore (vsina)^2/Rg(1+lncosa)

  • -Acosa/(1+lnv)+ (vcosa)^2(1-lnRg)/2Rg = c(1-lnRg)(y)/Rg
  • -Acosa/(1+lnv)= - (vcosa)^2(1-lnRg)/2Rg+ c(1-lnRg)(y)/Rg

 

For (vcosa)^2/2 >c acceleration will be in opposite direction or away from the center of galaxy, the critical velocity for outward acceleration of an object from a galaxy will be √2c or 24.5 km/s directly away from the center of galaxy. If cosa were 1/3 or a body were moving at an angle of 70.53 degrees away from center of galaxy ,the critical velocity would become 73.5 km/s and this is very much plausible if we assume a body generally moves only around the center of galaxy and gets a cosine velocity only due to chance (in spiral galaxy even the more).

If Acosa had been = -c/Rg +(vcosa)^2/2Rg then the velocity would increase proportionately with Rg or v/Rg=k, because of (1+lnv/c)(1-lnRg) velocity increase will be slightly greater or  v/Rg=k+x, where x will increase very slowly.

Now, acceleration is dependent on (vcosa)^2/Rg ,and cosa approaches 1 as a body moves away from a reference frame so if cosa were less than 1 initially as the body started accelerating away from the center of galaxy cosa kept on increasing and so did acceleration and this increase will be much greater than the increase due to (1-lnRg)(1+lnv/c).

Now when A and cosa are very close to zero.

 

(1+lnv/c)(vsina)^2/Rg(1+lncosa) =c(y)(1-lnRg)(1+lnv/c)/Rg

Or, (vsina)^2=c(y)(1-lnRg)lncosa

Now this lncosa component is very important, because when cosa changes vsina changes by a huge amount due to this lncosa component and this is what happens in elliptical galaxies and star over there do not have a fixed velocity because lncosa keeps on changing. Now, before we come to the case of a planet revolving around a star in this galaxy let me remind you, differential of x is Δx/Δt(1+lnx) where x is an independent variable, if x is of the form a+b+c then differential will be Δa/Δt(1+ln(a+b+c))+ Δb/Δt(1+ln(a+b+c))+ Δc/Δt(1+ln(a+b+c)) provided a,b and c are not independent of each other. In case of a body moving with velocity v1 ,its velocity is same with respect to all bodies at rest or with respect to all bodies moving at velocity v2 however it’s velocity with respect to bodies at rest is different from the one with respect to bodies moving at v2. Therefore velocity with respect to one group of inertial objects (all objects moving at one velocity) is one variable, velocity with respect to another group of inertial objects is another variable. For Rg, for changing your distance from one object while while keeping it constant with respect to another, you must move in a curved path and that is not allowed as it will result in acceleration which will cause distance change, so when you change your distance with respect to one object, you change  your distance from every object.

Now let us come to the case of a planet revolving around our star.

Suppose the planet moves towards the center of galaxy with velocity v1cosa and then it also starts moving towards the star with velocity v2 and the star makes an angle a+b from the direction of motion of the planet then the velocity towards the center of galaxy also increases or decreases and angle from the center of galaxy to the direction of motion changes from a to c.

We have

N(v2)^2/2(y)R + Ng((v1+v2cos(a+b))cosc)^2/2(y)Rg= Nc/R +Ngc/Rg

Now, N(v2)^2/2(y)R is very small compared to Ng((v1+v2cos(a+b))cosc)^2/2(y)Rg ,even for a planet as close as mercury to a star as heavy as the sun N/R is 100 times smaller than Ng/Rg so we may ignore N(v2)^2/2(y)R in L.H.S. we have,

Ng((v1+v2cos(a+b))cosc)^2/2(y)Rg= Nc/R +Ngc/Rg

Now v1+v2cos(a+b) can be written as V and we may ignore y

we have,

Ng(Vcosc)^2/2Rg =Nc/R + Ngc/Rg

Differentiating both sides we get,

NgVΔVcosc^2/Rg(1+lnV/c)-NgV^3sinc^2cosc/Rg(1+lncosc)=                          -N(1-lnRg)Vcos(a+c)/R^2 –Ngc(1-lnRg)Vcosc/Rg^2

We ignored some terms which have already been explained earlier

Or, ΔVcosc/(1+lnV/c) –V^2sinc^2/Rg(1+lncosc)) =                        -cRgN(1-lnRg)cos(a+c)/NgR^2cosc – c(1-lnRg)/Rg

Now acceleration due to the star is

ΔVcosc/(1+lnV/c) = -cRgN(1-lnRg)cos(a+c)/NgR^2cosc

V was (v2cos(a+b)+v1)

Δ(v2cos(a+b)+v1)cosc=-cRgN(1-lnRg)cos(a+c)(1+lnV/c)/NgR^2cosc

(A2cos(a+b)+A1-v2^2(sin(a+b))^2/R)cosc=-cRgN(1-lnRg)cos(a+c)(1+lnV/c)/NgR^2cosc

Or, Acosc - v2^2(sin(a+b)^2)cosc/R=-cRgN(1-lnRg)cos(a+c)(1+lnV/c)/NgR^2cosc

The cos(a+c)/cosc in the R.H.S. is because we are measuring the direction with respect to the center of the galaxy.

The equation can be simplified into

A - v2^2(sin(a+b)^2)/R=-cRgN(1-lnRg)(1+lnV/c)/NgR^2

Solve this and you will get double the gravitational acceleration from any object, double because we have taken an isolated galaxy. Now what if the galaxy isn’t an isolated galaxy but part of a galaxy cluster and super cluster. Let’s take Milky Way for instance, the closest large galaxy to it is the Andromeda galaxy but it will have a small effect as Ng/Rg due to it will be more than 1 magnitude lesser than Ng/Rg due to the center of Milky Way, however the virgo cluster situated at the center of our supercluster consisting of 2500 galaxies situated at a distance of 6.5*10^23 meters (2.75*10^3 times the distance between earth and center of galaxy) will exert considerable (almost 0.91% of Ng/Rg due to milky way) influence, as milky way is part of a weak cluster we may ignore the local cluster. The acceleration of sun towards the center of milky way will be obtained by solving

1.9Ng(vcosa)^2/2c(y)Rg=Ng/Rg

Or, vsina =435.4/1.9 km/s

=229.16 km/s

Now solving Rgc(1-lnRg)(1+lnv)/1.9Ng you will get the exact value of gravitational constant. (1+lnv/c) =ln229000/3*10^8 +1=-7.2+1=-6.2.the plane of solar system is mostly perpendicular(at an angle of more than 60 degrees) to the plane of the galaxy, so the velocity of earth around the galaxy won’t exceed 232 km/s ,therefore (1+lnv/c) will be -6.2 for earth also.

Now,

 

 

 

1.9Ng(vcosa)^2/2√(c^2-v^2)Rg=Ng/Rg+N/R

Differentiating v^2/2√(c^2-v^2) we get,

vA/√(c^2-v^2)(1+lnv) + (v^3)A(1-ln(c^2-v^2))/2√((c^2-v^2))(c^2-v^2)

Over here while measuring ln(1+lnv) and ln(c^2-v^2) we must not forget c is one unit.

=-vA/√(c^2-v^2)6.2 + vAv^2(1-ln(c^2-v^2))/√(c^2-v^2)2(c^2-v^2)

= -(vA/√(c^2-v^2)6.2)(1-3.1v^2/c^2-v^2)

This -3.1v^2/c^2-v^2 will give the precession of planets like mercury. It will be slightly greater than relativistic values.(about 1 in 30 parts)

 

4

PHOTON

Till now we assumed all particles have equal and unit mass but this is not true for photons whose mass approaches 0, let m be the mass of a particle then its score is

Ng/Rg –mNg(vcosa)^2/2Rg(c^2)(y)

A photon’s velocity approaches c so y approaches 0, so does m so they cancel out.

Ng/Rg –Ng(vcosa)^2/2Rg(c^2)

A photon has another component in its score called frequency which is something like the number of particles. Frequency can only change discreetly and therefore It’s differential won’t be divided by 1+lnf. But frequency change will make the score uneven as it will happen in an instant while other variables are changing continuously. Over here comes the concept of wave, score is unbalanced inside a wave, suppose there is a frequency change of 10 units in one go, then the photon will enter a wave till continuous change of other variables has fulfilled the score gap and then only the next frequency change will happen.

Score of a photon –

fNg/Rg - fNg(ccosa)^2/2Rgc^2

Now we want acceleration equivalent to be in m/s^2 and angle changes in radians so we multiply ccosa term with c.

fNg/Rg  - fNg(ccosa)^2/2Rgc

or, fNg/Rg (1-c(cosa)^2/2)

The score of a photon is not balanced or zero, that is, it provides different amount of accelerations (or decelerations) to particles by being fired at different directions with different frequency. While being created the photon gives away it’s unbalanced score so the change in the remaining score has to be zero to keep total universe’s score change zero

Differentiating the score of a photon we get

df/dt. Ng/Rg (1-c(cosa)^2/2) –fNgccosa(1-c(cosa)^2/2)(1-lnRg)/Rg^2 +fNg/Rg (-A(cosa)^2/2) +Ng c^2cosa(sina)^2/(1+lncosa)Rg^2=0

df/dt. Ng/Rg (1-c(cosa)^2/2) +Ng c^2cosa(sina)^2/(1+lncosa)Rg^2–fNg(ccosa)^2(1-lnRg)/2Rg^2-fNg/Rg(A(cosa)^2/2) = fNg(ccosa)(1-lnRg)/Rg^2

Now for the moment let us ignore acceleration term and fNg(ccosa)^2(1-lnRg)/2Rg^2.

Then we have

df/dt. Ng/Rg (1-c(cosa)^2/2) +Ng c^2cosa(sina)^2/(1+lncosa)Rg^2= fNg(ccosa)(1-lnRg)/Rg^2

Now whenever cosa is naturally changing, the change will be reduced to compensate for right hand side term which you may call gravitation. When cosa isn’t changing the frequency will change to compensate for R.H.S. term.

Case 1 : A photon directly moving towards the center of galaxy.

df/dt. Ng/Rg (1-c(cosa)^2/2)= fNg(ccosa)(1-lnRg)/Rg^2

df/dt=fccosa(1-lnRg)/Rg(1-c(cosa)^2/2)

let us approximate the denominator as -c(cosa)^2/2 then,

df/dt=-2fccosa(1-lnRg)/Rgc(cosa)^2

df/dt =-2f(1-lnRg)/Rgcosa

For a small Δh corresponding frequency change will be

Δf = -2fΔh(1-lnRg)/ccosaRgcosa

Case 2: Photon is moving directly towards a massive body.

df/dt. Ng/Rg (1-c(cosa)^2/2)=2fN(1-lnRg)ccosb/R^2

df/dt=2fNccosb(1-lnRg)Rg/Ng(1-c(cosa)^2/2)R^2

let’s us approximate 1-c(cosa)^2 as –c(cosa)^2 and cosb =1

df/dt=-2fN(1-lnRg)Rg/Ngc(cosa)^2(R)^2

we know G=-Rgc(1-lnRg)6.2/2Ng

df/dt=4fNG/6.2(ccosa)^2(R)^2

=2fNG/3.1(ccosa)^2(R)^2

This is the change in frequency due to gravity.

Case 4: A photon moves towards a star grazes it’s surface then passes away.

In this case cosa is naturally changing and total cosa change will decrease such that there will seem to be a deflection, there will be no frequency change or gravitational redshift in this case as deflection will compensate for the required score.

fNgccosaΔ(cosa)/Δt(1+lncosa)Rg =fNccosb(1-lnRg)/R^2

Δ(cosa)/Δt(1+lncosa)=Ncosb(1-lnRg)Rg/Ngcosa(R)^2

Now, when N/R is much smaller than Ng/Rg no significant change will occur , significant changes will occur only when N/R is comparable to Ng/Rg, there will come a time when N/R will be very close or even greater than Ng/Rg from center of galaxy, we may write total Ng/Rg as (Ng/Rg + N/R). To further simplify our calculations we will take R/N instead of Rg/Ng and multiply the whole by N/R*Rg/Ng.

Δ(cosa)/Δt(1+lncosa)=Ncosb(1-lnRg)R/Ncosa(R)^2 * N/R * Rg/Ng

Δ(cosa)/Δt= (1+lncosa)Ncosb(1-lnRg)R/Ncosa(R)^2 * N/R * Rg/Ng

Since total Ng/Rg= N/R + Ng/Rg

Rg/Ng= RRg/NRg+NgR

Δ(cosa)/Δt= (1+lncosa)Ncosb(1-lnRg)R/Ncosa(R)^2 * N/R * RRg/NRg+NgR

Δ(cosa)/Δt= (1+lncosa)cosb(1-lnRg)/cosaR * NRg/NRg+NgR

NRg=6*10^50 For N = mass of our Sun

NgR=2*10^50 for R=6.95*10^8

Now let us integrate the instantaneous change in cosa, we get approximately

Δ(cosa)= Δ ln(R)(1-lnRg)(1+lncosa)cosb/c(cosa)^2 * NRg/NRg+NgR

Now let us assume N/R = total Ng/Rg we will later multiply the value we get by the fraction N/R was of total Ng/Rg during the whole journey ,then cosa becomes cosb

Δ(cosa)= Δ ln(R)(1-lnRg)(1+lncosb)cosb/c(cosb)^2

This is the value we get if photon moves towards the center of galaxy (with cosb being the angle it makes with the center of galaxy)

Now suppose the photon started travelling from hades group of stars (as was used in famous experiment of 1919) 10^18 meters away from the Sun and grazes the Sun with a radius of 7*10^8 meters and reaches earth 10^13 meters away. ΔR=10^18 - 7*10^8 + 10^13 meters, now we must take Δ ln(R) in solar units.

Δ ln(R)=ln(1.428*10^4)-ln(1) –(ln(1) –ln(1.428*10^9))

Δ ln(R)=ln(1.428*10^4)+ln(1.428*10^9)

Δln(R)=ln(2*10^13)

=30.5

(1-lnRg)=104

Cos b changes from -1 to 1 so (1+lncosb)=1

Δ(cosa)= Δ ln(R)(1-lnRg)(1+lncosb)cosb/c(cosb)^2

Δ(cosa)=30.5*104/3*10^8

=2.15 arc seconds

This would be the deflection if the photon grazes the center of galaxy travelling equivalent distance

For our Sun the deflection would be

2.15*∫dx , (where x = NRg/NgR+NRg)

Now x changes from a very minuscule quantity  to 6*10^50/1.4*10^50+6*10^50 at cosb=0 and back to a very small quantity at cosb=1,(mass of center of galaxy is 2*10^41 as its measured to be 10^42 including dark matter which does not exist)

We have

2.15*(6/7.4 )

=1.743 Arc seconds

As you can see from the equation this should be a somewhat longer range effect than Einstein thought, meaning that for R>7*10^8 this deflection should decrease lesser than he thought.

Now coming to the other 2 terms in photon’s score

fNgACosa/Rg=-fNg(ccosa)^2(1-lnRg)/Rg^2

Now photon’s velocity is always c, how can it accelerate, well it doesn’t. Look at it this way, for a particular score change if you could accelerate a particle to velocity v, at twice that much distance from the center of galaxy for same amount of score change you can accelerate that same particle to more than 2v. Time is approximately equal to Rg/Ng(y), such that velocity of light is always constant. As photon moves away from center of galaxy its velocity decreases due to time  dilation,because of Acosa= (ccosa)^2(1-lnRg)/Rg, the velocity remains c.

5

MOMENTUM AND KINETIC ENERGY

A photon increases the number of particles in the universe, therefore when a photon is released all the particles in the universe will experience a score deccrease of f(ccosa)^2/2cR, there will be a total score decrease of Ngf(ccosa)^2/2cRg , a photon carries an unbalanced score of -Ngf((ccosa)^2 -2c)/2cRg, (which is immediately given away at the time of creation )  so the net change in score of the universe will be –fNg(ccosa)^2/2cRg from the creation of a photon. So a photon release should always be accompanied by a net velocity decrease in the universe and photon acceptance by an increase.

-(vsina)^2/Rg(1+lncosa) - (vcosa)^2(1-lnRg)/Rg +Acosa/(1+lnv) =                   -c(1-lnRg)/Rg

vsina term and acceleration term will always sum up to the c term unless Acosa changes due to a significant change in Ng/Rg ,if vcosa changes due to orbital perbutations, vsina or unbalanced acceleration will increase.

-(vsina)^2/Rg(1+lncosa)  +Acosa/(1+lnv) =  -c(1-lnRg)/Rg  unless Acosa changes due to change in Ng/Rg. - (vcosa)^2(1-lnRg)/Rg ensures that vcosa/time or Ngvcosa/Rg(y) remain constant (a little more than constant due to new calculus) .

Ngvcosa/Rg(y) -(vsina)^2/Rg(1+lncosa) +Acosa/(1+lnv)=-c(1-lnRg)/Rg +k

Ngvcosa/Rg(y)=k

For n number of particles nNgvcosa/Rg(y)=nk

Now if this body of n particles collides with another body of n2 particles moving at v2 velocity

n2Ngv2cosa/Rg(y)=n2k2

Now change in nk must be equal and opposite to change in n2k2 else score will change,

Δ nNgvcosa/Rg(y)= Δn2Ngv2cosa/Rg(y)

Δ nvcosa /(y)= Δn2v2cosa/(y)

This gives us the conservation of momentum. For the change in kinetic energy photons will be released.

6

EMF

Whenever a charged particle spots another charged particle it releases a photon inside itself which is again accepted inside itself, this way no extra particle is added into the universe but the unbalanced score of the photon will have to be accounted for and it will be by acceleration towards or away from the other charged particle.

Unbalanced score of a photon = -fNg(c(cosa)^2-2)/2Rg=Ng/cRRg or –1/cR where R is our chosen particle’s distance from the other charged particle .

f(c(cosa)^2-2)/2=1/cR or -1/cR

-fNg(c(cosa)^2-2)/2Rg will cause a (vcosa)^2 decrease of f(c*c(cosa)^2-2)/2

Δ(vcosa)^2 should be = f((ccosa)^2-2)/2

But v is a continuously changing variable because of gravity so the discrete release of photons will not affect (1+lnv) but velocity is a continuously changing variable and any change in it should always be divided by (1+lnvₒ) where vₒ is the velocity due to continuous acceleration such as gravitation.

Δ(vcosa)^2/(1+lnvₒ) = f((ccosa)^2-2)/2

Δ(vcosa)^2/(1+lnvₒ) = 1/R

Now vcosa will have a discrete change and it will enter a wave where R will change continuously till the score gap is filled.

Δ(vcosa)^2/Δt(1+lnvₒ) =Δ(1/R)/Δt

2vcosaAcosa/(1+lnvₒ) = (1-lnRg)vcosa/R^2

Acosa=(1-lnRg)(1+lnvₒ)/2(R^2)

For R=1meter the acceleration of a electron towards a positron will be

Acosa=-104*6.2/2

=-322.4 m/s^2

This is very close to the experimental value of acceleration of electron towards positron. Now a proton’s acceleration towards an electron, in a proton only one particle does the attraction probably a positron inside it, the rest of the particles composing the proton are ordinary particles.

Proton’s acceleration towards electron:

nΔ(vcosa)^2/Δt(1+lnvₒ)=Δ(1/R)Δt

Where n is the number of particles in the proton into their respective mass, which is equal to mass of proton/mass of electron =1837

Acosa =322.4/1837 m/s^2 for R = 1

Because acceleration from photons is instantaneous and R change continuous there will be a component of vsina in R change. R after a given time t will be R+vcosa(t)+√(R^2+(vsina^2)*t) –R

=R + vcosa(t) + (vsina^2)*t/2R

nAv(cosa)^2=-(1-lnRg)(vcosa)/R^2 - (1-lnRg)(vsina^2)/R^3

nAcosa=-(1-lnRg)/R^2 - (1-lnRg)v(tana)(sina)/R^3

The second term in R.H.S. is for magnetism.

 

 

The End