I Mathematically Proved That Einstein's Relativity Is Wrong

[maheshkhati]

1) Just consider that you are in revolving chair then you will find that whole world is revolving around you with different linear velocity as V= RW where W-angular velocity. Centrifugal force will act on body part of you which is angularly accelerating (with you). You will find that clock at distance R is revolving around you with velocity V=RW but no centrifugal force is acting on that clock.

              this effect is very common in world. train rider, in accelerating train from velocity 0 to V fill the acceleration & but find that he is at rest & whole world is moving in opposite direction with varying linear velocity. He sees that trees are accelerating from velocity o to V in opposite direction but find that no force is acting on that tree. 

     Means, if you put the frame of reference in accelerating frame then you will find that whole world is accelerating in opposite direction but force of acceleration will not act on substances in the world. As earth is angularly accelerating, observer on it find whole world is revolving around it with out any centrifugal force action.

2) reference of book, Element of special relativity by Dr. T. M.Karade, K.S.Adhav & Maya Bendre page no 135

      F’x = Fx – ( v/c² . Fy. Uy)/(1-V .Ux/c²) is the transformation equation for force in X-direction

So, if Fx/Fy is always equal to ( v/c² . Uy)/(1-V .Ux/c²). then F'x is zero.

Edited July 16, 2015 by maheshkhati

[Pmb]

maheshkhati - By yes did you mean that you do want me to post all the errors that I find? If so then what'd be the point? You never addressed the errors that I found so far listed in my first post.

 

You're quite wrong about things like mass and force not being observer dependent. That's quite easy to show and is shown in all special relativity textbooks. I did the same thing in my home page. See the special relativity section at http://home.comcomcast.net/~peter.m.brown/sr.htm  The derivations are there and there's no mistake in any of them save the one on inertial mass. There are a couple of typos there where I used the wrong symbol in an equation. If you follow the derivation let me know and I'll point out the typo. You can probably figure it out by context anyway.

[maheshkhati]

I try to open your http://home.comcomca....m.brown/sr.htm but fell to open. i will try to see it alter on.

Answer to previous post:- If you are sitting at the center of rotating disc then you will find that there is centrifugal force on disc's molecules in outer direction but it is stable for you because you are in same angularly accelerating  frame. You find that outer world & substances are revolving in opposite direction but the component of disc is not revolving. So, for observer on earth, all substances on earth is filling centrifugal force but stable with relative to observer on earth but at the same time find that all stars are revolving around it daily from east to west in opposite direction but centrifugal forces are not acting on it.

    This is because you are in angularly accelerating frame.  Detail in post 18

[Pmb]

I made an error when typing out the URL. Instead of ...brown/sr.htm it should have been ...brown/sr/sr.htm

 

What do you mean by "outer world & substances"? I assume that you're simply referring to all of the matter in the region of interest, plain and simple, right? What do you mean by "component of disk is not revolving"? What does "component of disk" mean?

 

How does the Earth fit into this? The Earth is also rotating and it's not clear if your coordinate system is located on the Earth and if it is then is it rotating at the same angular velocity as the Earth?

 

Please forgive my questions. The grammar is confusing me. I'm sure that with time you're grammar will improve. Mine did. :)  However, right now I don't know what "stable relative to observer on earth" means? I don't know how Earth is positioned in your description.

[Pmb]

Can't you provide a very simple derivation of how Einstein's Special Theory of Relativity (SR) implies energy is not conserved? All of those equations have been proved correct time and time again in all particle accelerators all over the world. They've been demonstrated to be correct as well has having withstood the test of time. They simply are not wrong. And that's a fact. In fact in SR energy is conserved by definition. I.e. there's a geometrical object in SR called the 4-momentum and is defined as P = (mc, p) where P is the 4-momentum (and as such it has 4 components), m is the inertial mass of the particle (aka relativistic mass), c is the speed of light and p is the particle's 3-momentum defined as mv.

 

On my personal website at http://home.comcast.net/~peter.m.brown/sr/conservation_of_mass.htm

 

I show that if mass m is conserved in one inertial frame then it's conserved in any other inertial frame. In the absence of any interaction between the particles the total energy E = mc^2 and therefore if energy is conserved in one inertial frame then it's conserved in any other inertial frame.

Edited August 15, 2015 by Pmb

[maheshkhati]

E=m C²  where m is relative mass.   So, m= y. mo  where y=(1-V²/C²)^-0.5    

  so, E= y. mo. C²

  ^{So, E = y Eo  ------- (1)}

 

Means, relative mass & energy has to increase as frame velocity V increases but we know that consumption of energy is equal to work done in that frame.

& I prove in 1St chapter that

   W = Wo/y 

(as force perpendicular to motion direction decreases & displacement (or space) in motion direction get contracted by relativity.)

hence, work done decreases as V increases.

This proves that consume energy in doing work changes as

  E = Eo/y  ------ (2)

This shows that relative consume energy decreases as frame velocity increases.  this is against relativity. 

Edited July 27, 2015 by maheshkhati

[maheshkhati]

About star velocity:- 1)What is mean by velocity?

Ans:- It is displacement per unit time. It is completely relative. If on equator of earth one train is moving from west to east with velocity V=WR where W is angular velocity of earth & R is earth radius at equator then from the spacecraft moving to another planet. You will find that train is stable at one point on earth surface & only earth is revolving. (As both are moving with same angular velocity in opposite direction) What is true earth is stable or train is stable?

        My answer is both things are true. For observer on earth, earth is stable & train is moving with very high speed V =WR is true & for observer in space, train is stable & earth is revolving is true.

It is very simple, find  displacement per unit time in your frame. that is velocity in that frame.

[maheshkhati]

For opposite angular velocity train has to move from east to west. (sorry other things are right.)

[maheshkhati]

Relativity is so wrong. It proves that there can be acceleration without force. Example is given below

In prime frame,  if Fz =0 & ratio Fx/Fy  is equal to ( v/c² . Uy)/(1-V .Ux/c²) then after transformation in S’ frame  F’x becomes F’x = 0 because

F’x = Fx – ( v/c² . Fy. Uy)/(1-V .Ux/c²)      ----transformation equation

In frame S :- Now, just consider that on magnetic substance on frictionless platform magnetic forces are acting in X-direction & in Y-direction. Magnetic force Fx is so adjusted by software program that ratio Fx/Fy is always equal to          ( v/c² . Uy)/(1-V .Ux/c²).

Then, Forces Fx (very small) & Fy in this frame will create acceleration ax & ay in direction x & y.

Observer frame S’ is moving with velocity V with relative to frame S then in frame S’ :-

There is acceleration in X' direction because ax’= ax/{r³. (1-ux. v/c²) ³ }  where r =1/(1-v²/c²) ^0.5 but there is no force in X'- direction because

               as  F’x = Fx – ( v/c² . Fy. Uy)/(1-V .Ux/c²)    &  as  Fx/Fy=( v/c² . Uy)/(1-V .Ux/c²)

         So,       F’x =0

Means, in this case in frame S’ there is acceleration in X’-direction but no force is present in X’-direction.

Means, some ghost force will accelerate substance in direction X’ in frame S’.  This is relativity.

acceleration & force are made for each other because to change state of motion, we require force. Mathematical condition where there is acceleration but no force in X- direction is very interesting condition. This is something like shoes foot print on soil without existence of any person walking on that soil.

[Pmb]

You have a severe misunderstanding of energy. Conservation of energy means that energy remains unchanged ***in a given frame of reference***. It doesn't mean that if you change the frame of reference that the energy will remain unchanged. A clear example is a particle that is at rest in an inertial frame of reference S. In S the total energy is zero. Now change to a frame S'' which is moving relative to S with speed v. in S' the total energy is the kinetic energy of the particle and has the value E = K = mv^2/2 which is greater that it was in S. But this does not mean that energy isn't conserved. By definition - conservation of energy is defined in a given frame of reference, i.e. it remains constant in that frame. It doesn't mean its constant in all frames or that it remains unchanged when you change to a new inertial frame.

 

No wonder you thought you proved SR wrong - you don't understand conservation of energy.

[maheshkhati]

I completely accept your thought. No where in my paper, I say that energy is not frame variant. I have only proved S.R. wrong & it contradiction. Read my paper completely then

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you will be surprised. I am against S.R. but I like one thought of G.R. which says that big mass changes space near to it.

I goes one step ahead & say that big mass with it gravity & other field flux not only changes the space around it but create the reference space for matter near to it.

Edited August 17, 2015 by maheshkhati

[Pmb]

I think I misread that. Let me try again. In post #23 you said that energy has to increase with frame velocity and that consumption of energy is equal to work done in that frame. Then you claim that W = Wo/y and hence work done decreases as v increases. Then you make a completely unintelligible statement "This proves that consume energy in doing work changes as E = E0/y.

 

None of that makes sense. For example; you talk about work done but you never mention what's doing the work. If you do an about of work W0 in frame S and in S' the work is W then the increase in energy in S is E0 = W0 so that whatever the energy increase in S it has to increase by the amount of work done in S'.

 

But you're expression for the work is not a general transformation of work. The amount of work done in an inertial frame S might be zero but it doesn't mean that it has to be zero in the frame S' which is in standard configuration with S. So that W = W0/y is not generally true. Right now I don't even know if that's correct to begin with.

 

In your comment about you claim that SR says that there can be acceleration without force but I know SR cold and know that it doesn't say that whatsoever.

 

Someday when I'm bored I'll read your paper like I said I would. I've been having a bad time with pain though and can't concentrate that well. Sorry.

[maheshkhati]

mathematics which proves that consumption of energy in doing work decreases as relative frame velocity increases. here, I use standard transformation equation of forces in two frames S & S' :-

Let consider that old man displaced the cart from pole A to Pole B on platform. Observers are on platform S & in train S', moving with velocity –V  then

1) When AB (displacement) parallel to the direction of train velocity.

     Then, for observer on platform:-

     so, Fy = F z = 0    , dx =d(AB)

  &  Work done W = Fx . dx ----------(1)

 

     for observer on train :-

     F’x = Fx – ( v/c² . Fy. Uy)/(1-V .Ux/c²) = Fx as Fy =0   by force transformation equation.

Here, as one meter in X- direction in S-frame is equal to 1/ γ meter in S’- frame in X-direction

     & dx' =d(AB)' = dx/ γ  where  γ = (1-V²/C²) ^–0.5

    So, W' = F'x . dx' = Fx . dx/ γ = W/ γ

  So, W' = W/ γ

    

Case 2 :- When AB perpendicular to velocity of train

    for observer on platform :-

    Fx = Fz = 0   dy = d(AB)   & dx =0

     Work done W = Fy. dy

 

    for observer in train :-

        F’y = (Fy/ γ) / (1-V .Ux/c²) = Fy/ γ    as Ux=0

       & dy'=dy as it is perpendicular to V & dx' =dx/ γ = 0

     Work done W' = F'y. dy' = (Fy/ γ) . dy = W/ γ

                      W' = W/ γ

Case 3:-Consider that old man pull the cart on platform from pole A to pole B which is not perpendicular to train velocity in straight line AB.

    Fx, Fy, dx, dy are forces & displacement on the platform in X & Y direction then

     For observer on platform:-

      Work done W = Fx.dx + Fy dy

 

    For observer in train :-

   F’x = Fx – ( v/c² . Fy. Uy)/(1-V .Ux/c²)

   F’y = (Fy/ γ) / (1-V .Ux/c²) ---------from transformation equation.

  W’ = F’x.dx’ + F’y dy’

 W’ = {Fx – ( v/c² . Fy. Uy)/(1-V .Ux/c²) } .dx’ + {(Fy/ γ) / (1-V .Ux/c²) } . dy’

Here, as one meter in X- direction in S-frame is equal to 1/ γ meter in S’- frame in X-direction

So, dx’ = dx/ γ & dy’=dy &

If m =(1-V .Ux/c²) then

W’ = (1/[m. γ]) .{Fx .dx-(Fx. V/c² . Ux. dx) - (Fy. v/c² . Uy. dx) + Fy.dy}

W’=(1/[m. γ]) .{Fx .dx(1-V .Ux/c²) + Fy . (dy - v/c² . Uy. dx) }

W’=(1/[m. γ]) .{Fx .dx(1-V .Ux/c²) + Fy . dt (Uy - v/c² . Uy. Ux) }

W’=(1/[m. γ]) .{Fx .dx.(1-V .Ux/c²) + Fy . dt .Uy. (1-V .Ux/c²) }

W’=(m/[m. γ]) .{Fx .dx+ Fy . dt .Uy}

W’=(m/[m.r]) .{Fx .dx+ Fy . dy}

W’=1/ γ.{Fx .dx+ Fy . dy} = 1/ γ . W

Or W’= W/ γ

 

This clear shows that in all cases W' = W/ γ

So, you call  generally that in all cases W' = W/y

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point 2:- Work done is not equal to energy but equal to change in energy in that frame. 

W =dE=change in energy in that frame

Means'  dE' = dE/ γ

This shows that as velocity V increases, energy consume in doing work decreases. This is against relativity.

This happens because Force perpendicular to V decreases & space get contracted in direction of V.

I only put everything in mathematical proof.

[maheshkhati]

F’x = Fx – ( v/c² . Fy. Uy)/(1-V .Ux/c²)  this is standard transformation of equation of force in X-direction

&

 ax’= ax/{r³. (1-ux. v/c²) ³ }  where r =1/(1-v²/c²) ^0.5  this is standard transformation of equation of acceleration in X-direction

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You will surprise to see when ax=0 then a'x =0  but  when Fx=0 then Fx' is not zero. This will create above problem like at some instant there is acceleration but no force in X-direction .

Edited August 19, 2015 by maheshkhati

[xyz123]

I surprised to see math in this thread because I am relativity backer. Something is wrong in your math. I will find it.

[maheshkhati]

Force without acceleration in S.R.

& acceleration without force in S.R.

& applied force is less than acting force in S.R.

STEP 1:-This problem can easily be understood by following paradox.

{Before starting this paradox, I want to put some relativity formulae’s

In any frame for force in X-direction

      Fx = d/dt( y.  mo. ux)  where y=(1-u²/c²)^{-0.5     (formula given in all standard book on S.R.)} 

After differentiation, we get

So, Fx=y. mo. ax+y³ mo. (ux/c²} (ux ax+uy ay+uz az) ------(1)}

Now, Paradox:-

On frictionless platform, object is moving with constant velocity ux in X-direction & only magnetic force is acting in Y-direction & there is acceleration in Y-direction only with velocity uy & Fz=0

If we apply eq(1) to this case then result will be

       Fx= y³ mo. (ux/c²} uy ay

  Or  Fx=Fay  as this force is form due to ‘ay’ only

Mean’s even there is no magnetic force acting on object from outside in x-direction & no ‘ax’ then also above force will act on object in +ve direction of x-axis due to ‘ay’

Important point (1):-

Mean’s applied magnetic force on object in X-direction is 0 & acting force in X-direction is

 Fx= y³ mo. (ux/c²} uy ay+0 or Fay+0=Fay

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STEP 2:-Now, Force acting in X-direction is  Fx= y³ mo. (ux/c²} uy ay or Fay

Now, after this happen, very small magnetic force of same intensity -fx = -y³ mo. (ux/c²} uy ay or  -Fay  start acting on object in direction opposite to above force (but velocity is still positive ux) & cancel that above force.

Mean’s equation (1) becomes

  0=y. mo. ax+y³ mo. (ux/c²} (ux ax+uy ay) 

Or  0 =y. mo ax. (1+ y²  {ux²/c²} ) +Fay 

(Here as  Fay= y³ mo. (ux/c²} uy ay)

Mean’s  Fay = y. mo. -ax. (1+ y². {ux²/c²} )

Mean’s there must be acceleration in –ve X-direction to fulfil above equation of S.R.

Now, see above equation carefully, it is of nature

     0= -fx + Fay

Important point (2):- Mean’s applied magnetic force on object in X-direction is -fx & acting force in X-direction is  -fx + Fay = 0 ## or 0.

Here, resultant force in X-direction is zero but there is acceleration.

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STEP3:- same things happen for +ve force in X-direction (for less than Fay or more)

Now, I am generalising above result.

Step 1&2 clearly shows that when we apply any magnetic force (Fmx) in X-direction on the object, actual force acting on object is more & that quantity is (Fmx+Fay)

Similarly,

If we apply any magnetic force (Fmy) in Y-direction on the object then actual force acting on object is more & that quantity is (Fmy+Fax)

This is completely complicated results, which says that applied force & acting forces on objects are different & more in S.R.

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STEP4:- Force does work, consume energy, gain energy & we must know that energy cannot be created. It can be transferred only:-

From above setup it must be clear that energy get transfer from magnet to object but if applied force is less than acting force then energy gain by object will be more than energy loose by the magnet. Means due to more work done by more force for same displacement, more energy get generated.

HERE, more energy (& force) is the problem.

Where this additional energy (& force) does comes from?

No answer in S.R.

(You will be surprise that I am against S.R. but not totally against G.R. )

Edited March 3, 2016 by maheshkhati

[maheshkhati]

Problem which I raised in above paradox is very serious.

For example, 

One light object is falling down under gravity in vertical Y-direction & horizontal air is pushing that object in horizontal direction.

For observer on ground

Let, fx is horizontal force applied by air on object & fy is vertical force applied by gravity on object.

Then above calculation says that

Actual force acting on objects are not fx & fy but

 Fx = fx + y³ mo. (ux/c²} uy ay =fx + Fay

& Fy = fy+ y³ mo. (uy/c²} ux ax =fx + Fax

Means, actual force acting on objects are different (more) than force applied on the object.

This is very serious output & can create very complicated problem in Special relativity.