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Einstein Is Wrong But Einstein's Effect Is Right.

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#18 devin553344

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Posted 29 June 2020 - 06:56 AM

YES:- You are missing some thing. 

Mathematical derivatives always contain variables as it is applicable to all parameters & conditions.

In above example, I prove that in post 4 or in post 13 that in single frame analysis or in two relative frame analysis.

Even there is no force is applied in X-direction then also relativity create mathematical non-applied force in x-direction.

So, effect is applied force is different than acting force.

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In Post 13, I prove that if we do not consider that non-applied horizontal force in to the mathematical calculation then d'E=dEo/y, d'M=dMo/y

 if we do consider that non-applied horizontal force in to the mathematical calculation then only d'E=dEo, d'M=dMo

Means, non-applied mathematical horizontal force is responsible for d'E=dEo, d'M=dMo

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Means, If Bomber drop the Bomb Mo kg  moving with horizontal velocity

then for Observer on ground actual force acting is gravitational force but relativity create the mathematical horizontal force.

If we do not consider non-applied horizontal force then d'E=dEo, d'M=dMo is wrong.

This horizontal force is created due to vertical acceleration & horizontal velocity

I think I get what you saying finally. I think what you need to look at is frame dragging: https://en.wikipedia.../Frame-dragging

 

I think that is the x direction force?



#19 maheshkhati

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Posted 30 June 2020 - 05:10 AM

This is not frame dragging effect. This happen due to simple mathematics of wrong special theory of relativity. In special relativity, force is not related to acceleration but it is rate of change of moment.

This is very simple, I explain it with simple mathematics with simple event.

 

In relativity force in X-direction is mathematically calculated as

Fx  =d/dt (m ux) = dm/dt . ux + m . dux/dt = dm/dt . ux + m . ax

Let, consider fighter plane with horizontal velocity ux drop the bomb B 

& Observer is on ground

For observer :- Bomb B will move with constant horizontal velocity ux & accelerate vertically due to gravity.

So, Force applied on Bomb is gravity only in vertical direction.

 

but one mathematical acting force is created in horizontal direction i.e. Fx= dm/dt .ux + m. (0) =dm/dt .ux 

as mass of Bomb increases due to vertical acceleration.

So, applied force is different than acting force in relativity.

& even horizontal acceleration is zero but there is acting horizontal force with displacement.

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This is serious problem. It's effect is, more acting force produce more work.

So, energy produce is more than energy consume in the same frame for observer.

Detail other calculations are already given.


Edited by maheshkhati, 30 June 2020 - 05:17 AM.


#20 devin553344

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Posted 30 June 2020 - 05:46 AM

This is not frame dragging effect. This happen due to simple mathematics of wrong special theory of relativity. In special relativity, force is not related to acceleration but it is rate of change of moment.

This is very simple, I explain it with simple mathematics with simple event.

I don't see how you've proven anything wrong with relativity, you keep talking about some horizontal force. There is none? So exactly what are you trying to say? Where's this new horizontal force coming from that you speak of? Because you haven't demonstrated a new horizontal force.

 

And by the way I highly doubt relativity states that force is not related to acceleration. Again what are you saying?

 

It sounds like you don't understand math and the use of derivatives. They're nothing more than an accurate form of algebra.

 

F=ma is the same as F=dp/dt. And yes relativity must be applied here, everyone knows that, so what exactly are you doing?


Edited by devin553344, 30 June 2020 - 06:07 AM.


#21 maheshkhati

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Posted 01 July 2020 - 03:50 AM

You are referring Newtonian mathematics:-  F=ma is same to F=dp/dt in Newtonian mathematics.

In relativity :-

F is not equal to ma but F=dp/dt



#22 devin553344

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Posted 01 July 2020 - 07:43 AM

You are referring Newtonian mathematics:-  F=ma is same to F=dp/dt in Newtonian mathematics.

In relativity :-

F is not equal to ma but F=dp/dt

F=dp/dt is the accurate implementation of F=ma.



#23 maheshkhati

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Posted 02 July 2020 - 04:07 AM

No,

in relaticity

F=dp/dt =d/dt(mu) =dm/dt . u + m  . du/dt = dm/dt . u + m a

 

So, in relativity F=ma is wrong.

For Example, in x-direction

Fx=d/dt(m ux) =dm/dt . ux + m  . dux/dt = dm/dt . u + m ax

so, Fx=m ax is wrong

Total detail calculation is given in post 4


Edited by maheshkhati, 02 July 2020 - 04:10 AM.


#24 devin553344

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Posted 02 July 2020 - 05:38 AM

No,

in relaticity

F=dp/dt =d/dt(mu) =dm/dt . u + m  . du/dt = dm/dt . u + m a

 

So, in relativity F=ma is wrong.

For Example, in x-direction

Fx=d/dt(m ux) =dm/dt . ux + m  . dux/dt = dm/dt . u + m ax

so, Fx=m ax is wrong

Total detail calculation is given in post 4

F=ma is not wrong in relativity, it is still correct. You really shouldn't be saying it's wrong if you want people to think you understand the equations your presenting.

 

Why don't you read up on force, here this article covers everything you need to know: https://en.wikipedia.org/wiki/Force


Edited by devin553344, 02 July 2020 - 05:46 AM.


#25 maheshkhati

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Posted 03 July 2020 - 03:27 AM

I am pasting the content from that site

Special theory of relativity

In the Special relativity, mass and energy are equivalent (as can be seen by calculating the work required to accelerate an object). When an object's velocity increases, so does its energy and hence its mass equivalent (inertia). It thus requires more force to accelerate it the same amount than it did at a lower velocity. 

 

F=dp/dt

 

remains valid because it is a mathematical definition.But for relativistic momentum to be conserved, it must be redefined as:

 

p= mo v . (1-v2/c2)^0.5  

 

(modified because image extension can not be pasted)


Edited by maheshkhati, 03 July 2020 - 03:50 AM.


#26 devin553344

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Posted 03 July 2020 - 04:26 AM

I am pasting the content from that site

Special theory of relativity

In the Special relativity, mass and energy are equivalent (as can be seen by calculating the work required to accelerate an object). When an object's velocity increases, so does its energy and hence its mass equivalent (inertia). It thus requires more force to accelerate it the same amount than it did at a lower velocity. 

 

F=dp/dt

 

remains valid because it is a mathematical definition.But for relativistic momentum to be conserved, it must be redefined as:

 

p= mo v . (1-v2/c2)^0.5  

 

(modified because image extension can not be pasted)

Why are you copy-pasting the article, it doesn't prove your point. But it does prove mine? ie:

 

"F=dp/dt

 

remains valid because it is a mathematical definition."

 

Granted the concept of longitudinal mass is confusing. You should abandon force and go with energy, ie:

 

"E^2 = m0^2c^4+p^2c^2"

 

Or at least that's what I got from the article here: https://en.wikipedia...ngitudinal_mass

 

This is an interesting topic to me as it may relate to my theory. Thanks for the discussion.


Edited by devin553344, 03 July 2020 - 07:40 AM.


#27 maheshkhati

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Posted 06 July 2020 - 05:04 AM

You just solve the equation,

E^2 = m0^2c^4+p^2c^2

by putting value of E = y mo c^2

you get the same equation

i.e.  p= mo v . (1-v2/c2)^0.5 

so, both equations are same

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I already proved that in special  relativity, the theory is so close  that by taking one equation

like F=dp/dt 

you can prove lot of equations of relativity like transformation equations of relativity, dE=y. dEo & dM= y dMo etc

So, do not paste other equation of relativity because all equations are inter link with another or same.

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​Force is the fundamental  quantity in physics. We can apply force, do work, consume energy.

& dE=dw=F.ds , F =dp/dt is not wrong in relativity.

Physics remain same, formulas may be different.



#28 devin553344

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Posted 06 July 2020 - 05:26 AM

You just solve the equation,

E^2 = m0^2c^4+p^2c^2

by putting value of E = y mo c^2

you get the same equation

i.e.  p= mo v . (1-v2/c2)^0.5 

so, both equations are same

----------------------------------------------------------------------------------------------------------------- 

I already proved that in special  relativity, the theory is so close  that by taking one equation

like F=dp/dt 

you can prove lot of equations of relativity like transformation equations of relativity, dE=y. dEo & dM= y dMo etc

So, do not paste other equation of relativity because all equations are inter link with another or same.

--------------------------------------------------------------------------------------------------------------------

​Force is the fundamental  quantity in physics. We can apply force, do work, consume energy.

& dE=dw=F.ds , F =dp/dt is not wrong in relativity.

Physics remain same, formulas may be different.

What you're saying here does not sound correct.



#29 VictorMedvil

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Posted 06 July 2020 - 06:32 AM

What you're saying here does not sound correct.

I told you Maheshkhati is a huge crank 



#30 maheshkhati

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Posted 08 July 2020 - 01:52 AM

1) In any frame:-

Force in x-direction = Rate of change of momentum in x direction

In non-proper frame

F’x = d/dt’( y’.  mo. U’x)  where y’=(1-u’2/c2)-0.5  

 (Rate of change of momentum in x’-direction)

If this differentiation extended by proper transformation equations of frame like putting equations of  U’x, y’  & d/dt’  values then we get

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

This proves that in any frame

Force in x-direction = Rate of change of momentum in x direction

 

2) This transformation equation also indicate the same math for example

In prime frame Fx=0 , ux=0 & ax=0

Then also in non –prime frame F’x is not 0

But F’x = Fx + ( v/c2 . Fy. Uy)/(1-V .Ux/c2)   ----transformation equation in relativity.

  F’x=  v/c2 . Fy. Uy

3) Same math can be used to prove

dw’=y. dw or dE= y. dEo or dM=y. dMo

 

So, I am not wrong.

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All this is not possible by your formula F=ma    as it is wrong in relativity. or this is approximate formula

 

& In relativity Direction of acceleration & force is not same.


Edited by maheshkhati, 08 July 2020 - 02:29 AM.


#31 devin553344

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Posted 09 July 2020 - 02:56 AM

I told you Maheshkhati is a huge crank 

Crank may or may not be correct if I could figure out what he's saying. But most of his statements come across as incoherent rambling, lol.



#32 maheshkhati

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Posted 10 July 2020 - 04:01 AM

I already said that formulas may be different but physics remain same.

Even,  your said formula  E^2 = m0^2c^4+p^2c^2

Can be converted into dE=F.ds   Just differentiate above & solve little, final output is 

     dE= dP/dt . ds

& dP/dt is rate of change of momentum or force.

So, dE= F . ds

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Force is reality. We apply force & do work.

If applied force is less than acting force then energy produce will be more than energy consume.

​this is against consistency of energy 

Mathematics In post 4 is completely true. If you want to prove me wrong then you have to prove math of post 4  wrong.


Edited by maheshkhati, 10 July 2020 - 04:11 AM.


#33 devin553344

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Posted 12 July 2020 - 08:49 AM

I already said that formulas may be different but physics remain same.

Even,  your said formula  E^2 = m0^2c^4+p^2c^2

Can be converted into dE=F.ds   Just differentiate above & solve little, final output is 

     dE= dP/dt . ds

& dP/dt is rate of change of momentum or force.

So, dE= F . ds

 

First, your units appear wrong, perhaps that's what's going on here? Are you using ds to represent space?

 

Second I think you got the idea wrong, rest energy isn't part of momentum. Notice how they're represented in different areas of the equation. Separated by a plus sign.

 

 


Force is reality. We apply force & do work.

If applied force is less than acting force then energy produce will be more than energy consume.

​this is against consistency of energy 

Mathematics In post 4 is completely true. If you want to prove me wrong then you have to prove math of post 4  wrong.

Yeah and that is exactly what you didn't prove. I should be seeing an example of how two energies don't match up using values. Something you wouldn't produce for us. You just want to spit out formulas and say you're right. More incoherent rambling.

 

Lastly, no one has to prove you wrong, remember you're the one trying to prove Einstein wrong. Please keep things straight here.


Edited by devin553344, 12 July 2020 - 09:04 AM.


#34 maheshkhati

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Posted 13 July 2020 - 05:45 AM

Simple differentiation is impossible for you & you are trying to oppose me.  OK

E^2 = m0^2c^4+p^2c^2

 

Now, I do it for you

 

2 E . dE/dt = c^2 . 2. p. dp/dt

 dE/dt = c^2 (p/E). dp/dt

But ...P/E =v/c^2

So, dE/dt =v. dp/dt

dE= (v.dt) (dp/dt)

But ...(v.dt) =ds

dE= ds (dp/dt)

But dp/dt =rate of change of momentum = force

dE= F. ds

 

Remember rest energy has no contribution in doing work. It is constant ( m0^2c^4)  & it's differentiation is zero. 

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so, Mathematics in post 4 is not wrong.  you may try to do that.


Edited by maheshkhati, 13 July 2020 - 05:50 AM.




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