# Relativity is wrong

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Read my mathematics in my paper. I have taken care of everythings.

I read your "paper", it is the same mistakes you are posting in the forum.

So, dE = F . ds

I.e. Work done = force in that frame X displacement in that frame

Yet, you fail to transform $ds$ when passing from one frame to the other. You cling to your incorrect claim that $ds$ is frame-invariant. SR teaches you that it ISN'T.

now, in relativity force parallel to direction of mation is F= γ3. mo. a this is very well known to ererybody

Err, wrong again, the general formula is:

$\mathbf{F} = \gamma(\mathbf{v})^3 m_0 \, \mathbf{a}_\parallel + \gamma(\mathbf{v}) m_0 \, \mathbf{a}_\perp$

Only when the component of the acceleration perpendicular on the velocity $\mathbf{v}$ is zero does the formula reduce to what you wrote. In general case, it doesn't.

I am going to tell you one last time : you CANNOT USE SR in order to "disprove" SR.

Edited by xyzt

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dE = mo . u .(1-u2/c2)-3/2 du this is last step taken from your calculation

Now, I improve vise your math from here because their is no displacement & acceleration in your calculation

dE = mo . u .(1-u2/c2)-3/2 du =dE = mo . u .(1-u2/c2)-3/2 du/dt . dt =γ3. mo. du/dt . u.dt =γ3. mo. a . ds

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Now, in under line step taken from your calculation

u is velocity in X-direction & du is change of velocity in X - direction

Then acceleration must be in X direction it can not be perpendicular to u

or perpendicular component of acceleration is zero. (at this particular step)

So,

In this particular case F= γ3. mo. a &

dE =γ3. mo. a . ds &

So, dE = F . ds

I.e. Work done = force in that frame X displacement in that frame

This formula is true & can be derived from your calculation.

This is not generalize formula but used in this particular case.

Now, length contraction

If one observer is moving with velocity u with relative to second observer where u is relative velocity in X- direction then

dx = dx'/γ , dy=dy' & dz = dz'

i.e. length is contracted in x =direction only not in Y or Z direction.

In my all event of my paper displacement happen in Y direction (event created specially in this way to reduce the complication)

So, in my all events in my paper displacement =ds remain same in all frames.& not very.

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In your post you use formula

here E represents the total energy where u is the speed of the object of mass measured in frame F.

E=γ mo.c2

E =γ Eo

means, energy also changes similar to mass changes in relativity or mass & energy are same but we see it differently

Now, I create circular reference in relativity. displacement was consider perpendicular to relative motion of frames.

This changes relative mass, relative acceleration, force, work done & energy

& now, check the relativity energy relation &I find that

SR fails here to form circular reference

In in all calculation, I use very event specific mathematics deals to that event only.

​Mathematics 1 is very general mathematics to get the filling of slow down of event in frame motion by comparing accelerating falling ball to tied ball in rail cabin

Mathematics 2 is given to give general filling of how mass increases & acceleration decreases more than mass.

Mathematics 3 & 4 is given to show that standard text book mathematics gives the same out put as my mathematics

because mathematics remain same.

Edited by mahesh khati
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If, in frame S you have* $dW=F_ydy$, then in frame S', moving in the x direction with speed $V$, you will have: $F'_y=F_y \sqrt{1-V^2/c^2}=\frac{F_y}{\gamma}$

$F'_x=\frac{u_yV}{c^2}F_y$

$dy'=dy$

$dx'=\gamma(dx+Vdt)=\gamma V dt$

so

$dW'=F'_ydy'+F'_xdx'=F_y(dy \sqrt{1-V^2/c^2}+\gamma V dt \frac{u_yV}{c^2})=F_y dy(\frac{1}{\gamma}+\gamma \frac{V^2}{c^2})=F_ydy \gamma=\gamma dW > dW$

So? Work, is frame variant, as explained multiple times to you.

What I see that having failed to prove your case for "work", you graduated to your other "disproofs" of SR. I have already explained to you that you cannot disprove SR by using SR, all you can do is to demonstrate your inability to understand and learn SR. Have fun!

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* This is not how mechanical work is described in relativity. I explained that to you in a previous post but it fell on deaf years.

Edited by xyzt
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Mahesh - just in case you think it is just XYZT saying this I will repeat; Special Relativity is mathematically internally consistent. It is a very simple and self-contained theory which allows us to say that as long as the postulates are held to be true then the theory must follow. There is simply no room for a mathematically correct situation within SR which leads to an internal contradiction; the only way to show SR is flawed/incomplete is to move outside its frame of application or to provide experimental evidence that contradicts a postulate.

If you have a thought experiment that is within the realm of application and postulate of SR yet still contradicts SR then you have made a mistake.

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When ever there is force, there is acceleration

Or there is acceleration then there is force, because force is rate of change of momentum per unit time in that frame in that direction

but it is very easy to prove that in my specific case that acceleration in X-direction in both frames are absent

Now, go to frame transformation equation of acceleration

.ax’= ax/{γP3P. (1-ux. v/cP2P)P3P } where γ =1/(1-vP2P/cP2P)P0.5

Means, if ax =0 then a'x=0

or acceleration in X-direction in both frames are absent.

Then force in X-direction must be absent.in both frames.

(In vector mathematics, effect of any vector may be acceleration in perpendicular direction is zero)

Or d'w = F'y. d'y only possible work in my specific case

Or d'w = Fy/y dy

Or d'w = (1/γ) F'y . dy =.(1/γ) dw

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Now,

dx'/dt' = V as y dt =d't

i.e. whole event setup is moving with constant velocity V in direction X or not accelerating

here, this displacement is only due to inertia of whole substance in that direction X.

Not due to force or acceleration in that direction.

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If ball fall from height dy or ds in train cabin & if I put stick of height ds near to that fall it remains vertical of same length in both frames

moving with constant velocity due to inertia with train velocity. for man on platform.

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If your equation is very general equation then it must be true for all values of uy

let's consider that ball start falling now

then uy at that instant will be uy =0

then your final equation become (for small dy)

d'w=(1/γ) F'y . dy =.(1/γ) dw

Edited by mahesh khati
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When ever there is force, there is acceleration

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Then force in X-direction must be absent.in both frames.

(In vector mathematics, effect of any vector may be acceleration in perpendicular direction is zero)

Relativity teaches you that your claim (like most , if not all of your other claims) is false. You keep piling false claims on top of false claims.

$F_x=0$

$u_x=0$

$F'_x=F_x+\frac{u_yV}{c^2+\frac{u_xV}{c^2}}F_y=0+\frac{u_yV}{c^2}F_y=\frac{u_yV}{c^2}F_y$

See how easy is to disprove your falsities?

The guy is a mechanical engineer. I wonder how he got his degree.

Now,

Yep

dx'/dt' = V as y dt =d't

Nope, I'll let you figure out your mistakes by yourself on this one.

then your final equation become (for small dy)

d'w=(1/γ) F'y . dy =.(1/γ) dw

No, it doesn't, $dW'=\gamma F_y dy=\gamma dW$. I'll let you figure out your errors by yourself. You can't do physics and you demonstrated that you can't do basic math either.

Means, if ax =0 then a'x=0

or acceleration in X-direction in both frames are absent.

Then force in X-direction must be absent.in both frames.

Ahh, I missed this glaring error as well. You see, force in SR is NOT $\vec{F}=m \vec{a}$, it is defined as $\vec{F}=\frac{d}{dt}(\gamma(u) m \vec{u})$. Try to do the Lorentz transform , you will see your errors. Or , most likely, you will add more errors on top of the existent errors.

Edited by xyzt
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Thanks, I realy like your this calculation but in work, there is two parts force & displacement

work = force x displacement

displacement happen due to force then only work is done

Here,consider old man is pulling cart on platform perpendicular to train velocity & observer is in moving train

Then Y-displacemet d'y or dy is forced diaplacement happen due to force applied by old man &

displacement in train direction or x-direction is not forced displacement

but happens automatically due to relative velocity of the frames which is V

i.e. even there is no event then also relative displacement will be above displacement due to V velocity of train

So, this is not any the forced displacement in X- direction happens due to action of force but happens automatically

So, work in X -direction will be absent because forced displacement is absent in that direction

Be kind to that old man, in train frame he is appling some force in X-direction & displaceing that cart with train velocity in X-direction is wrong.

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In above event if I apply

this is your equation using acceleration

In this event acceleration is perpendicular to V only

So, Total Force in train frame = y mo (acceleration in perpendicular direction in frame of train) -----eq 1

but by relativity transformation

(here, Total force is perpendicular to train velocity)

(acceleration in train frame) = 1/ y2 ( acceleration in platform frame).

put this in euation (1) becomes

total force in train frame F =1/y mo (acceleration in platform frame done by old man)

total force in train frame F =1/y force in platform frame

As force displacement perpendicular to train motion is dy & constant

(dy) x Total force in train frame F =1/y force in platform frame x (dy)

Work done in train frame = 1/y . work done in platform frame

Edited by mahesh khati
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Have you ever considered the remote possibly that it is you that is wrong, rather than the hundreds of thousands of physicists who preceded you?

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Have you ever considered the remote possibly that it is you that is wrong, rather than the hundreds of thousands of physicists who preceded you?

He knows he's wrong, he will just never admit it.

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Thanks, I realy like your this calculation but in work, there is two parts force & displacement

work = force x displacement

displacement happen due to force then only work is done

...which is exactly what I explained to you in an earlier post:

$dW'=F'_ydy'+F'_xdx'=F_y(dy \sqrt{1-V^2/c^2}+\gamma V dt \frac{u_yV}{c^2})=F_y dy(\frac{1}{\gamma}+\gamma \frac{V^2}{c^2})=F_ydy \gamma=\gamma dW > dW$

There is a $F'_ydy'$ component and a $F'_xdx'$ component.

You keep denying the $F'_xdx'$ component because you don't know relativity.

In above event if I apply

this is your equation using acceleration

In this event acceleration is perpendicular to V only

....in frame S ONLY. In frame S', the acceleration is no longer perpendicular on the speed V, there is also a parallel component that shows up. This is the cause of the component $F'_x \ne 0$ that shows up in frame S'.

Edited by xyzt
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This is the transformation equation

So, if Fx =0 & Vx =0 & Fz =0

Then

F'x = - (V. Uy/c2) Fy

Now, direction of this force is opposite to relative displacement of frames as you consider

As it opposes displacement of object(This is because dE/dt increase or my. Vy increases)

this force will not displace the particle or ball or cart in the direction of relative frame motion but it will oppose the relative displacement.

It acts in opposite direction to displacement d'x or try to reduce that displacement.

point 2 :- displacement which you consider in above your calculation is inertial displacement

This displacement happens automatically due to relative motion of the frames.

This displacement happen even event is not happened.

& force F'x acts opposite to this displacement due to increase in energy objects due to uy velocity.

Edited by mahesh khati
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This is the transformation equation

So, if Fx =0 & Vx =0 & Fz =0

Then

F'x = - (V. Uy/c2) Fy

Good, you are starting to learn. Took you a few days to reproduce the formula that I have shown you a few days ago. You got the sign wrong because you have the first formula wrong:

$F'_x=F_x+\frac{u_yV}{c^2+\frac{u_xV}{c^2}}F_y+\frac{u_zV}{c^2+\frac{u_xV}{c^2}}F_z$

point 2 :- displacement which you consider in above your calculation is inertial displacement

This displacement happens automatically due to relative motion of the frames.

This displacement happen even event is not happened.

& force F'x acts opposite to this displacement due to increase in energy objects due to uy velocity.

I'll have Ranch with the above. You keep making the mistakes, I keep correcting them.

The only "mistakes" that you have found in SR are your own.

Edited by xyzt
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He knows he's wrong, he will just never admit it.

!

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Let us develop the transformation of force from a system S where a particle is moving at a speed v to any system S' which is moving relative to system S at a speed V.

Calculating the transformation is very similar to the transformation of velocities. We begin with the momentum transformation (and not the coordinate transformation as we did for the velocities). We have:

P'x =γ (Px - B/c . E) , P'y = Py , P'z = P'z

with β and γ defined using the coordinate system velocity V:

y =(1-V2/c2)-1/2 & B= V/c

Using chain rule, we can write

F'y= dP'y/dt' = (dP'y/dt) / (dt'/dt) = (dPy/dt) / (dt'/dt)

However, we have from the Lorentz transformation for the time, that

t' = y (t-Bx/c)

dt'/dt = y {1-(B/c).(dx/dt)} = y (1-V.ux/c2)

Thus,

F'y = Fy/ { y (1-V.ux/c2)}

For the x component, we have (again, using the Lorentz transformation) that:

F'x=dP'x/dt' = (dP'x/dt) / (dt'/dt)

= {y (dPx/dt - B/c .dE/dt) } / {y (1-V.ux/c2) }

We have seen, however, from the definition of force, that dE/dt=Fv, and thus

F'x ={Fx - B/c . (Fx.ux + Fy. uy + Fz uz) } / {(1-V.ux/c2) }

As B =V/c

F'x = Fx - {v/c2. ( Fy. uy + Fz uz) } / {(1-V.ux/c2) }

To summarize, the complete transformation is:

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Now, above complete calculation which you can get from any other book indicate that

if Fx =0 & Vx =0 & Fz =0

Then

F'x = - (V. Uy/c2) Fy

& direction of force F'x is opposite to displacement.

Now, you post previously that

....in frame S ONLY. In frame S', the acceleration is no longer perpendicular on the speed V, there is also a parallel component that shows up. This is the cause of the component that shows up in frame S'.

You are true
If such component is their in frame S' in -ve direction then object will accelerate in opposite direction & will try to decrease inertial displacement in that direction.

This is decrease in displacement in that frame will not add the energy of object in that frame but it decreases energy level of object in that frame due to work done
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Now,
I want to give math 1 & math 2
In math 1, I will give mathematics used in my paper & in math 2, I will used your mathematics to prove that both gives same result
Math 1 :- Math in my paper
Let, consider same event ball is falling in train cabin & observer is on the platform
this event happens perpendicular to X- direction or V
So, displacement in X -direction in event is absent in train cabin
So, dx = 0
If this displacement is transform in to platform frame then displacement will be
d'x = (1/y) dx = 0
in above event by relativity, displacement d'x will be absent.
This can be also obtained by transformation equation of acceleration that
acceleration will be there in frame S' only when there is acceleration in the frame S in X -direction
This proves that acceleration & displacement d'x is absent in frame of platform if it is absent in frame of train rider
This proves that in both frame complete event is perpendicular to V
Now, you can apply equation
So, in this particular case
Fy'=y mo a'
other steps are similar given in the paper & finally we get that
work done in platform frame=(1/ y) . F dy = (1/ y) W < W
Now math 2 :- I consider your math is true
there is acceleration & displacement in X -direction in frame S' (platform) (-------accepting for some time)
From above calculation it becomes quite clear that force in X-direction will displace object in opposite direction.
This will decrease the inertial displacement of object which is ( y V dt) happens due to frame motion
let this decrease in displacement as d'x then
Total work done by object in frame S' = (1/ y) . Fy. dy - F'x . d'x
or this horizontal work done by the object in reverse direction decreases energy level of the object again
So, Total work done = (1/ y) .W -- F'x . d'x < W
--------------------------------------------You will say that I will be happy (with this result due to decrease in the energy again) but I am not, Math 1 & math 2 must be similar if relativity math is true
Now, what happens this small decrease of energy is compensated as fallows
here decrease in the horizontal inertial displacement, d'x= ( y V dt) decreases the velocity V, this increases (1/ y) & ultimately
(1/ y) . Fy. dy increases
And as both decrease in energy F'x . d'x is compensated by increase in the energy in Y -direction
Total work done remain same = (1/ y) .W < W
Now, it becomes quite clear that even if I have not consider x- displacement then also work done = (1/ y) .W & even if I consider x-displacement then also work done is (1/ y) .W. then why particle will displace in X- direction. When energy level of the object will not changed
This is why in this case also X=displacement is absent & event will happen in Y direction only.
Now, you can apply
where event is perpendicular to V only & there is acceleration when there is force as I applied
Edited by mahesh khati
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To summarize, the complete transformation is:

You are true
If such component is their in frame S' in -ve direction then object will accelerate in opposite direction & will try to decrease inertial displacement in that direction.

This is decrease in displacement in that frame will not add the energy of object in that frame but it decreases energy level of object in that frame due to work done

The transformation of force with the minus sign is derived from the Lorentz transform:

$x'=\gamma(x-Vt)$, so:

$dx'=\gamma(dx-Vdt)=-\gamma V dt$

Now try calculating $dW'=F'_xdx'+F'_ydy'$. If you do it correctly you should be getting $dW'=\gamma dW$.

So, once again, all you are finding is your mistakes in understanding SR.

Edited by xyzt
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, so:

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You are again mixing displacement between frames & displacement with in a frame. So, it must be address now.

I am in train moving with velocity V, I will see that trees are traveling in opposite direction with velocity yVdt this is not the displacement by force.

SMALL EXAMPLE (TO SOLVE THIS PROBLEM)

In laboratary I have big frictionless table. On that table one iron block is moving in X-direction

with velocity V.

Then Displacement of that iron block in X-direction will be y V dt.for me

Case 1:-

Now. one magnate is held in such a way that that force is applied in direction of motion of iron block

Then displacement of iron block will not be yV dt but (yV dt+dx) ....... where dx is additional displacement by magnetic force

& work done = (Magnetic force) X dx only (here displacement is not yV dt or yV dt +dx both are wrongs)

Case 2 :-

Now. one magnate is held in such a way that that force is applied in opposite direction to motion of iron block

Then displacement of iron block will not be yV dt but (yV dt - dx) ....... where dx is displacement by magnetic force in opposite dierction

& work done = (Magnetic force) X dx only

Displacement can not be yV dt which is only inertial displacement. & you must consider displacement by force some thing different than yV dt

For other things, I will post later

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, so:

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This is L- transformation equation of co-ordinates transformation

&

this is differential form of that only

&

If you want L-transformation of displacement in frames then it is only

dS'x = dSx. (1/y)

or, d'x =dx . (1/y)

Edited by mahesh khati
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, so:

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You are again mixing displacement between frames & displacement with in a frame. So, it must be address now.

No, I am not. I am simply trying to teach you relativity. But you seem unwilling and unable to learn.

You would much rather cling to your misinterpretations of SR than learn SR.

If you want L-transformation of displacement in frames then it is only

dS'x = dSx. (1/y)

or, d'x =dx . (1/y)

Not when $dx=0$.

One gets $dx'=dx/\gamma$ when one imposes $dt'=0$ in $dx'=\gamma(dx-Vdt)$ and $dt'=\gamma(dt-Vdx/c^2)=0$. This is NOT the case here.

The case here is $dx=0$ and basic math teaches you that $dx'=-\gamma V dt$ as I tried to teach you. Sorry I failed, there is nothing more that I can do for you.

Edited by xyzt
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, so:

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You are again mixing displacement between frames & displacement with in a frame. So, it must be address now.

Nope $dx'$ is the displacement in the frame where I also calculated $dy',F'_x,F'_y,dW'$.

You have to come to grips with the fact that you have no clue when it comes to SR.

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You are missing basic concept how to calculate length in displacement in frame

you first mark the beginning point in the frame or cabin & then after displacement. You mark the finishing point

& after event you measure distance between these two point & that length is displacement.

(You can not measure displacement at the beginning or at the midway of the event)

Means, to measure distance, you must be on same time co-ordinate in that frame i.e. dt =0

(We can not measure any length partially. It may be length of object or displacement. It must be measure fully in same time co-ordinate t )

Now, other calculation is given everywhere how to calculate (X2-X1) = y (x'2-x'1)

& dx =y dx'

dx' = dx/y

Now, your displacement dx'=yV dt in one frame becomes

dx=y (yV dt) in rest frame but this is wrong &

dx=dx'=0

This is the right way to calculate displacement in relativity.when you are comparing quantities in frames

Now, I want to address the bigger issue

In the event of ball falling & observer is on the platform

every body accept that work done in Y - direction (vertical) is

dW'y=(1/y).fy. dy

​ Now, only controversy is about work done in horizontal or X-direction

I am sure that this work done is zero because d'x=dx=0

but in this case

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1st equation shows that force Fy create de-accelerating force or retarding force, (-ve sign is very important here)

In our case

Fy.Vy+Fz.dz =Fy. dy/dt + 0 =dWy/dt

This calculation shows that work done or energy increase in vertical Y- displacement de-accelerate or retard the object in X-direction

This is because object becomes relatively heavy due to energy accumulation.due to dy displacement

This retarding force will be in action when some external force is acting in X- direction & displace the object in that direction.

If this external displacing force is not their in x-direction then this retarding force will not be their because object can not displace our self & loose its energy by moving in opposite direction.

(Or work done by the object only when external force act on it )

& as in our event there is no external force which act on object & displace it in X=direction this retarding force will not be their.

So, there is no work done in X- direction.......... or object will not loose its energy our self

So, total work done in our event =(1/y).fy. dy =(1/y) dW<dW only

& there is no horizontal (X) relative displacement in both frames as external displacing force for object is not present in this event.in X-direction.

Edited by mahesh khati
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"you first mark the beginning point in the frame or cabin & then after displacement. You mark the finishing point

& after event you measure distance between these two point instantly. & that length is displacement."

I may be mistaken but I think that, according to SR, "instantly" doesn't have a defined meaning for events separated in space.

So, at best, you have ignored one of the axioms of SR, in order to prove it false.

That's not logical.

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If I want to measuer displacement in that frame then I have to marked begining of displacement (physically) & end of that displacement. in that frame & at the end of the event. in that time co-ordinate t i.e.dt =0, in that frameI have to find out displacement length. I have not said about any simultenious event. If I speak some word instantenous & it sound like that then I am sorry

2nd point:- We can not measure any length partially. It may be length of object or displacement. It must be measure fully in same time co-ordinate t. When we read derivation of length contraction.in relativity similar expression is consider. This is already given in relativity

I want to know if there is very small force Fx is present in rest frame S & that displace the object dx in that rest frame in +ve X direction then what will be displacement of that object in S' frame in our event

I thing it will be

dx' =(1/y). dx

Edited by mahesh khati
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So, at best, you have ignored one of the axioms of SR, in order to prove it false.

That's not logical.

It is called "beating the strawman to death". We should let him do it, I corrected all his errors but to no avail.

What I explained to you. Not what you continue to erroneously claim.

Edited by xyzt
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Sorry for any wrong word I use but

Point 1 :- Displacement in any event is the distance measure between finished point of the object in the event to start point of object in the event measure at the end of the event.

For example. If ball falls in train cabin from tied thread which is cut suddenly. This ball is under horizontal force of attraction also. Now, ball will not fall vertically but this fall will tilted towards horizontal attractive force & falls down on floor.

We can not measure this displacement in between the happening of the event. It must be measure at the end of the event. This displacement is the distance measure between point where ball finally fall on the floor & point of relies of the ball measure at the end of the event.

Means, in rail cabin two points are fixed ( after the event), point where ball is relies ed from thread & point where it is fall on the floor.

For observer on platform as length of train contracted due to relativity. These two points in the cabin also come closure & horizontal distance between these two points will decrease.

so, for observer on platform

dx'=(1/y) dx

This is very simple. At the end of the event displacement is just distance & it obeys the low of contraction of relativity.

in my event , horizontal force is absent

So, ball will fall vertically in cabin

so, dx = 0 & dx' =0

Point 2:- If cart of mass m is pulled by force Fx then cart will accelerate.

Due to acceleration & continuous change in the state of motion, opposite force is developed by the cart

Force = m ax

Now, consider that I put up some additional mass (one form of energy) dm in the cart suddenly then additional retarding force will get develop

Now, opposing force = (m+dm). ax

Now, for same acceleration we will require some additional force. or cart will retard or de-accelerate.

but this all is happen only when force is acting on cart & it is accelerating

If cart is at rest & no force is acting then after adding some additional mass(one form of energy) will not create any additional force.

Same thing happen in our case. Detail is given in post 69

when on object (ball,cart or particle) force Fx is act in X- direction then only

F'x = Fx - {v/c2. ( Fy. uy )} / {(1-V.ux/c2) }

because Fy.Vy =Fy. dy/dt =dWy/dt =dE/dy

This {v/c2. ( Fy. uy )} / {(1-V.ux/c2) } or {v/c2. (dE/dt} } / {(1-V.ux/c2) } is not a physical force.

This is retarding effect developed due to additional energy accumulated due to motion in y-direction

This retarding effect is only there until object will accelerate in +ve X- direction due to external force Fx

If Fx is not there then there will not be any horizontal displacement or any retarding force because

Anybody can not accelerate our self & loose its energy until some external force will act on it.

So, if Fx in not present in rest frame & if there is no acceleration in X-direction in rest frame then in S' frame

opposing force (retarding force) will not come into action because these are internally created forces.

& object will not create force our self & get displace in that direction & loose our energy.

Work is done only when external force will act on object & it will get displaced in that direction.

Point 3 :- Means, math 1 which I have done in post 64 is true

There is no relative displacement & work in X-direction

So, work done = (1/y) Fy.dy =(1/y) W

Edited by mahesh khati
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I am done correcting your errors. Bye!

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!

Moderator Note

And with that we're closed.

mahesh khati, do not open a new thread on the subject. You have gotten a tremendous amount of constructive feedback at the expense of others' time and effort. Use it.

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