I Mathematically Proved That Einstein's Relativity Is Wrong

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#1 maheshkhati

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Posted 02 June 2015 - 03:35 AM

On my web site www.maheshkhati.com , In my paper on relativity, I mathematically proved that Einstein's theory of relativity is wrong . I mathematically proved that A) CONSUMPTION OF ENERGY IN TWO DIFFERENT FRAMES PROVES EINSTEIN IS WRONG INCONSISTANCY OF FORCES IN TWO DIFFERENT FRAMES PROVES RELATIVITY WRONG C) INCONSISTANCY OF MASS IN RELATIVITY D) RELATIVE VELOCITY WILL BE MORE THAN VELOCITY OF LIGHT. E) ABSOLUTE INERTIAL REFERENCE FRAME IS A RELATIVISTIC CONCEPT F) IN TWO REFERENCE FRAMES, WHICH ARE MOVING WITH CONSTANT RELATIVE VELOCITIES, ONE FRAME IS SPECIAL

#2 pgrmdave

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Posted 02 June 2015 - 09:40 AM

No, you didn't.

#3 Buffy

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Posted 02 June 2015 - 09:44 AM

Cool! Would you care to summarize your argument here? You'll generally find that people are skeptical of folks who consistently misspell words like "inconsistency" though, so you've got your work cut out for you.

Hard to trust honesty of inconsistent person,

Buffy

#4 CraigD

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Posted 02 June 2015 - 06:35 PM

Welcome to hypography, Maheshkhati.

You’ve clearly put a lot of thought and effort into your paper “Einstein’s Theory of Relativity Is Wrong & What Is the World Made Up Of?” (PDF).

Rather than summarizing it as Buffy suggests, let’s review it, starting at the beginning.

On page 4 and 5, you present a paradox that seems to arise from the mass dilation feature of Special Relativity. There’s a small but interesting literature of paradoxes of this kind (here are a few of the more well-know ones).

I think yours has an obvious mistake: you assume that the equation for force due to gravity for a body at rest relative the Earth, also called weight, is
$F_g = m g$
where $m$ is the mass of the body being weighed, and $g$ is a local constant, the acceleration of gravity.

Your mistake is that g is not an elementary constant, but a function of a quantity that isn’t constant according to SR:
$g = \frac{M G}{r^2}$
where $M$ is the mass of the Earth, $G$ the gravitational constant, and $r$ the distance between the center of the Earth and the body.

So you should have used
$F_g = \frac{m M G}{r^2}$

According to SR, as observed by the person on the moving train, A,
$F_g = \frac{\ ( m \gamma ) M G}{r^2}$
where $\gamma = \frac{1}{\sqrt{1 - \left( \frac{v}{c}\right)^2 } }$,
where $v$ is the speed of the train and $c$ is the speed of light.

As observed by the persons at rest relative to Earth watching the train, B,
$F_g = \frac{m ( M \gamma ) G}{r^2}$

So both A and B agree that the weight of the hanging ball, and that it snapped its string and crushed the unfortunate cockroach. No paradox occurs.
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#5 maheshkhati

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Posted 03 June 2015 - 03:07 AM

Let, consider that old man is pulling cart on railway platform perpendicular to train velocity V.

For observer on platform old man apply force Fo on cart & displace the cart up to small distance dy

Then, Work done for man on platform Wo = Fo . dy

Now, For train rider:-

Force applied by old man for the train rider will be  F = Fo/r       where r =(1-v2/c2)-1/2

(Refer :- first chapter of 1St paper on www.maheshkhati.com where detail calculation for such case is given. Which is based on standard text book of relativity)

Work done by old man for train rider will be W = Force x displacement =(Fo/r) x dy

(As dy is perpendicular to train velocity it remain same as space contraction is not happen in that direction)

So,   W = Wo/r

We know energy get consumed in doing work & is equal to work done. Let, Eo & E are energy consumed in doing work for man on platform & for train rider then

E = Eo/r ---------------result (1)

This is against relativity. Because in relativity. Energy has to increase not to decrease &

E= r. Eo

is true.

This all happen because force perpendicular to relative frame motion decreases by increasing velocity (of the observer in the frame) but displacement remain same.

Why do we get such result against relativity?

Edited by maheshkhati, 03 June 2015 - 03:19 AM.

#6 maheshkhati

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Posted 04 June 2015 - 03:56 AM

I think, You have also proved Einstein wrong because by transformation of force in relativity in your case also force Fy get transform as

F'y = Fy/r (here Fy may be any force i.e. due to gravity or any. Transformation formula of forces in relativity transform similarly)

So, F'g = Fg/r

(Refer :- first chapter of 1St paper on www.maheshkhati.com where detail calculation for such case is given. Which is based on standard text book of relativity)

but you proved that F'g = Fg

γγ Then you are also proved that Einstein is wrong

#7 Pmb

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Posted 28 June 2015 - 01:25 PM

I looked through your paper and found that it was filled with misconceptions in all cases. If you'd like I'd be more than happy to print out the entire paper, read it carefully and then write a post or series of posts (if there is a word limit for individual posts) to explain where you went wrong.

Here's one as an example: On page 20 you argue "But R does not have any limit because it is distance. So if R > C then V > C."

Surely you must have been aware that all these incredibly brilliant physicists who make relativity their life's work didn't let this one slip by them, do you? Your error is in your assumption that the radius of a rotating disk can be arbitrarily large. In fact it can't be. Such a disk is made of atoms. There are no atoms which can move at the speed of light so there is no atom in the disk which can be moving at a speed greater than or equal to the speed of light. This implies that the radius of any disk is limited by its angular velocity.  The faster a disk rotates the more mass each atom gains and therefore the greater the mass of the disk. If the angular velocity w is greater than w = c/R then the mass of those atoms becomes infinite which means it can't rotate that fast.

Your comments which follow based on that are therefore wrong. Nothing can move at or faster than the speed of light other than massless elementary particles.

So I'll read your paper and get back to you. Let me know if you'd like me to post all the errors that are in the paper.

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#8 maheshkhati

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Posted 03 July 2015 - 02:58 AM

1) Yes, if relativity is true then this should not be happen but for the man on revolving earth all stars on equator are moving from east to west daily with velocity more than light because nearest star of earth is at distance of 4 light years away i.e. R = 4 x 8 x 108 x 60 x 60 x 24 x 365

V = R . W -------------------(1)

Here, W = 2.pai  / T = 2 x  3.14 / 24 x 60 x 60 = 7.26 x 10-5  radians per second is angular velocity of earth.

Therefore, for nearest star.

V = 7.326 x 1012 m/s > C

This equation clearly indicate that even nearest star is moving with velocity 9000 times more, than velocity of light in reference frame which fixed on revolving earth.

This clearly shows that, the stars which you see in the sky above equator moving from East to West having linear velocity more than C relative to man on revolving earth.

Thanks for taking interest in my paper.

2) My thoughts is clear & simple. Only kinematics and observer can not create & change the things(like mass, force etc) . It happens due interaction of fields.

Edited by maheshkhati, 03 July 2015 - 03:03 AM.

#9 CraigD

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Posted 03 July 2015 - 03:06 PM

1) Yes, if relativity is true then this should not be happen but for the man on revolving earth all stars on equator are moving from east to west daily with velocity more than light ...

This is incorrect.

Relativity, both Galilean and Einsteinean, state that the laws of physics are the same in every inertial frame of reference. They do not state that they are the same in rotating frames of reference.

If the universe were as you describe it, maheshkhati, bodies far from the Earth would not only have velocities greater than c, they would experience a centrifugal pseudoforce away from the Earth. The Moon, for example, which is about $r$=384400000 m from the Earth, so would have a speed of about $\frac{2 \, \pi \, r}{1 \,\mbox{day}} \,\dot=\, 27950 \,\mbox{m/s}$, and would experience a pseudoacceleration of about $\frac{v^2}{r} \,\dot=\, 2 \,\mbox{m/s/s}$. This is greater than its surface gravity, so bodies on the far side of the moon would fly off of it, which we can see is not happening.

There’s a more profound reason why relativity can’t hold for rotating frames of reference. If that were the case, the universe would have a center about which the rest of it rotates – in your example, the Earth.

Another way to understand why your argument is wrong is to consider it on a smaller scale using common sense intuition. A person can easily rotate at about a rate of 0.25 rotations/s. If one were to do this in the center of a 200 m-wide parking lot, the cars at the edge of it would be moving at a speed of about 157 m/s, and experience an outward centrifugal force of about 247 m/s/s, about 25 times the Earth’s surface gravity, much greater than the traction of any car’s tires on pavement can withstand. We know, and can easily show, however, that standing in the middle of a parking lot and turning in pace doesn’t cause cars to be thrown out of the parking lot!

2) My thoughts is clear & simple. ...

Having clear, simple thoughts does not assure that your thoughts are correct.

#10 maheshkhati

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Posted 06 July 2015 - 03:21 AM

Same thing was told by me when I put up this argument. They say that this is not a inertial frame of reference due to angular acceleration. When I try to find out perfect inertial frame in the world. I do not get any frame.

Refer chapter E :- ABSOLUTE INERTIAL REFERENCE FRAME IS A RELATIVISTIC CONCEPT in my paper

​Every frame in the world is accelerating. Inertial filling is the local effect or relativistic effect. Even line drawn by you on earth surface which is perfect linear is not linear when you see it from space.

I only prove that if i put reference frame on earth with Y is vertical axis then star on that axis will not remain there but get displace with velocity V which is more than light. this is only to prove that velocity more than light exist in the world. This is not something out of world or you can see such things from your necked eyes every day.

Main point:- chapter A, B, C are more important for me because I used relativity & proved it wrong.

Now, I have math which proves that displacement may be any direction in frame S then also dE' = dE/r in frame S' moving with velocity V

Means, as observer velocity increases consume energy in doing work decreases. I will put this math on this thread.

Thanks

#11 maheshkhati

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Posted 06 July 2015 - 05:04 AM

As both system (earth & star) are not attached by gravity directly this centrifugal pseudo force will not come in to existence.

#12 maheshkhati

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Posted 09 July 2015 - 05:37 AM

mathematics which proves that consumption of energy in doing work decreases as relative frame velocity increases:-

Let consider that old man displaced the cart from pole A to Pole B on platform. Observers are on platform & in train 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

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.

#13 maheshkhati

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Posted 09 July 2015 - 05:50 AM

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.

#14 A-wal

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Posted 09 July 2015 - 11:46 AM

The fact that there's no such thing as a true inertial frame isn't the point. It's an idealised frame of reference to show the relationships at different relative velocities only. If you only take those effects into account then SR describes them, but nothing else. So what's the problem?

Angular velocity isn't covered by SR so it's meaningless to use that in any attempt to refute it.

I don't know how you got to no force being present. Acceleration is not frame dependant. The amount of acceleration varries, so the mass of an object varries with it but if an object accelerates then it accelerates in every frame of reference.

#15 maheshkhati

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Posted 13 July 2015 - 02:51 AM

1) Yes, angular velocity is not cover in SR but it is present every where & due to that we got linear velocity. Only I want to show that in earth frame, stars have velocity more than light velocity.

2) Inertial condition is local effect because true inertial frame is not present in the world.

3) Transformation formulas are

for acceleration in X direction :-ax’= ax/{r3. (1-ux. v/c2) 3 }  where r =1/(1-v2/c2) 0.5  means as ax =0 then ax' =0

but for transformation of forces in x -direction :- F’x = Fx – ( v/c2 . Fy. Uy)/(1-V .Ux/c2) ---(1)   means as Fx =0 then Fx' = – ( v/c2 . Fy. Uy)/(1-V .Ux/c2)  not zero

In our case in post 13 :-

This will create the problem because  if Fx/Fy is always equal to ( v/c2 . Uy)/(1-V .Ux/c2).as given in post 13 then Fx =( v/c2 . Fy. Uy)/(1-V .Ux/c2)  & if I put this in transformation equation of relativity (1) then

F'x =0

Means, there is Fx, so ax & ax'  but F'x =0

Means, in frame S' there is acceleration a'x but no force in X'- direction. Means' ghost force will accelerate the object in S' frame in X'-direction.

Edited by maheshkhati, 13 July 2015 - 02:59 AM.

#16 CraigD

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Posted 13 July 2015 - 01:12 PM

Only I want to show that in earth frame, stars have velocity more than light velocity.

I think most interested readers understand what you want to show, but also understand that what you are trying to show is simply misguided and wrong.

I tried to explain this to you in with an appeal to intuition and common sense this post, but as you haven’t abandoned your argument, gather I failed. Here’s then, is an explanation of why you cannot, for a given angular velocity of a body (in your example, a star) around a specified axis (in your example, the axis of rotation of the Earth) alone, draw any conclusion about the linear velocity of the body:

Angular velocity is not relative. That is, it is possible to determine by measurement, whether the angular velocity of a the star is due to the body having been accelerated, or because the body being used to define the Earth has been accelerated.

If the star was accelerated, and its radial distance from the Earth is not changing, it experience a specific centripetal acceleration toward the Earth, which could be measured by a centrifugal pseudo force.

If the Earth was accelerated, and the radial distance of parts of it are not changing (that is, it’s not flying apart), parts of it (every where on its surface except the polls) will experience a specific centripetal acceleration toward its center, which can be measured by a centrifugal pseudo force.

If a centripetal acceleration is measured on Earth, none will exist on the star. If no centripetal acceleration is measured on Earth, one would be on the star.

Without measuring one of these centripetal accelerations, we cannot determine whether the angular velocity of the star is due to the Earth having been accelerated, or the star.

This measurement has been done, many times, finding the expected centripetal acceleration on Earth.

Therefore, we know that the angular velocity of a star around the Earth’s axis of rotation is due to the parts of the Earth having been accelerated, not the star. In other words, we know that the Earth is rotating, rather than that the Earth is stationary, and the universe rotating around it.

Had this measurement not shown the expected angular velocity of points on the Earth’s equator, we would have to conclude that many stars were moving faster than the speed of light, and theories that predict this is impossible, such as Special Relativity would be false. The measurement did show that the Earth, not the universe, is rotating around the Earth’s axis, however, so no stars are moving faster than the speed of light, and theories that predict that the relative speed of bodies cannot be greater than the speed of light may be true.

#17 A-wal

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Posted 14 July 2015 - 10:25 AM

Rather than looking at it as no object being able to reach the speed of light relative to any other object a more general rule is that the distance between any two objects can never change as fast as the speed of light from the frame of reference of those objects. When an object is spinning the distance between it and other objects isn't changing at all.

Transformation formulas are

for acceleration in X direction :-ax’= ax/{r3. (1-ux. v/c2) 3 }  where r =1/(1-v2/c2) 0.5  means as ax =0 then ax' =0

but for transformation of forces in x -direction :- F’x = Fx – ( v/c2 . Fy. Uy)/(1-V .Ux/c2) ---(1)   means as Fx =0 then Fx' = – ( v/c2 . Fy. Uy)/(1-V .Ux/c2)  not zero

In our case in post 13 :-

This will create the problem because  if Fx/Fy is always equal to ( v/c2 . Uy)/(1-V .Ux/c2).as given in post 13 then Fx =( v/c2 . Fy. Uy)/(1-V .Ux/c2)  & if I put this in transformation equation of relativity (1) then

F'x =0

Means, there is Fx, so ax & ax'  but F'x =0

Means, in frame S' there is acceleration a'x but no force in X'- direction. Means' ghost force will accelerate the object in S' frame in X'-direction.

I don't understand. In order for an object to accelerate it needs the application of a force. The amount of acceleration that the object undergoes is frame dependant so the mass of the object is also frame dependant, E=mc^2, but the energy isn't frame dependant. You can't affect the force that's applied to the object by changing frames.

Had this measurement not shown the expected angular velocity of points on the Earth’s equator, we would have to conclude that many stars were moving faster than the speed of light, and theories that predict this is impossible, such as Special Relativity would be false. The measurement did show that the Earth, not the universe, is rotating around the Earth’s axis, however, so no stars are moving faster than the speed of light, and theories that predict that the relative speed of bodies cannot be greater than the speed of light may be true.

Except that apparently it's fine for objects to move away from each other faster than the speed of light if it's due to the expansion of the universe because it's the space in between that's changing faster the the speed of light, not the objects themselves. WTF? It's the same thing! That's one of the reasons I think the big bang is utter BS, but that's a different discussion.

Edited by A-wal, 14 July 2015 - 10:33 AM.