# Use of incorrect transformation equation

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When I refer post

http://www.scienceforums.net/topic/84740-relativity-is-wrong/

I was surprised to see calculation

F’x = Fx –[(V/C^2 ) (Fy.Uy)]/[(1-V Ux/C^2)].

As Fx =0

F’x = –[(V/C^2 ) (Fy.Uy ]/[(1-V Ux/C^2)].

This type of transformation is not correct.

This is only transformation equation which transform Fx in S frame to F’x in S’ frame. So, when Fx =0 then F’x =0. because

F’x = Fx – n Fy here n=[(V/C^2 ) (Fy.Uy)]/[(1-V Ux/C^2)]

1)in non relativity expreassion

Fy/Fx =(m ay)/(m ax) = ay/ax

Fy= (ay/ax) Fx

2) In relativity

Now, Fy = dPy/dt = (dPy/dUy ) . (dUy/dt) = ay . (dPy/dVy )

So, Fy/Fx = (ay/ax) . {(C^2-Ux^2)/(C^2-Uy^2) }

If Q ={(C^2-Ux^2)/(C^2-Uy^2) } then

Fy = (ay/ax) Q . Fx

So, F’x = Fx – n Fy this equation can be written as

F’x = Fx – n. (ay/ax) Q . Fx

F’x = Fx {1- n. (ay/ax) Q}

So, If Fx = 0 then F’x = 0

This happens because Fy/Fx can be express as function of ay/ax.

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For those who are also searching I think this refers to post#53 in the linked thread.

hari123 welcome to SF, I look forward to some interesting posts, but please give a full reference in future to help others.

Can you confirm I have identified the correct post?

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!

Moderator Note

That thread is locked. As such we will not be discussing any of mahesh khati's claims. If this was a response, please provide a link to the specific post in question (click on the post number in the upper right of the post, and copy/paste the link)

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When I refer post

http://www.scienceforums.net/topic/84740-relativity-is-wrong/

I was surprised to see calculation

Well, it is textbook physics.

F’x = Fx –[(V/C^2 ) (Fy.Uy)]/[(1-V Ux/C^2)].

As Fx =0

F’x = –[(V/C^2 ) (Fy.Uy ]/[(1-V Ux/C^2)].

This type of transformation is not correct.

This is only transformation equation which transform Fx in S frame to F’x in S’ frame. So, when Fx =0 then F’x =0.

You are making the same exact kind of errors as Mahesh Khati. Force is not frame invariant, so $F_x=0$ in frame S does not mean $F'_x=0$ in frame S'. Crack open any book on relativity and you will see how forces transform. The formula in post 53 in the thread is correct.

So, Fy/Fx = (ay/ax) . {(C^2-Ux^2)/(C^2-Uy^2) }

Wrong. I will let you figure out where you made your error. I will give you a hint: $F_x=\frac{d}{dt} \frac{m_0u_x}{\sqrt{1-(u/c)^2}}$ and $F_y=\frac{d}{dt} \frac{m_0u_y}{\sqrt{1-(u/c)^2}}$

Calculate the derivatives, the first time you did it, you messed up.

Edited by xyzt
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Well? Are you going to admit to your errors or are you going to disappear?

Edited by xyzt
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(dr)^2 = (dx)^2 + (dy)^2

[dr/dt]^2 = [dx/dt]^2 + [dy/dt]^2

U^2 = Ux^2 + Uy^2

2 U dU/dUx =2. Ux

dU/dUx = Ux/U ............1

Px = (mo/(1-U^2/C^2)) . Ux as per your expression

As U^2 = Ux^2 + Uy^2

dPx/dUx ={1-U^2/C^2}^-3/2 . mo . (1-Uy^2/C^2)

If you want complete above calculation I will give you.

but Fx = dPx/dt =(dPx/dVx ) . (dVx/dt) =(dPx/dVx ) ax

={1-U^2/C^2}^-3/2 . mo . (1-Uy^2/C^2) . ax

similar for Fy

This shows that Fy/Fx = (ay/ax) . {(C^2-Ux^2)/(C^2-Uy^2) }

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Well, it is textbook physics.

You are making the same exact kind of errors as Mahesh Khati. Force is not frame invariant, so $F_x=0$ in frame S does not mean $F'_x=0$ in frame S'. Crack open any book on relativity and you will see how forces transform. The formula in post 53 in the thread is correct.

Wrong. I will let you figure out where you made your error. I will give you a hint: $F_x=\frac{d}{dt} \frac{m_0u_x}{\sqrt{1-(u/c)^2}}$ and $F_y=\frac{d}{dt} \frac{m_0u_y}{\sqrt{1-(u/c)^2}}$

Calculate the derivatives, the first time you did it, you messed up.

I gave you a hint but you continued to make elementary mistakes, I'll give you another hint:

$F_x=\frac{d}{dt} \frac{m_0u_x}{\sqrt{1-(u/c)^2}}=m_0 \frac{du_x}{dt}\frac{1-u^2/c^2+u_xu/c^2}{\sqrt{1-u^2/c^2}^3}$

$F_y=\frac{d}{dt} \frac{m_0u_y}{\sqrt{1-(u/c)^2}}=m_0 \frac{du_y}{dt}\frac{1-u^2/c^2+u_yu/c^2}{\sqrt{1-u^2/c^2}^3}$

So $F_x/F_y=?$

You need to learn how to do differentiation correctly. Until you do that, you will continue to make gross errors in your "physics". I put "physics" in quotes because what you are doing is not physics, it is the same fringe stuff Mahesh Khati was doing, combining basic errors.

Or you can always go learn how it is done properly in the textbooks, Here is the link I gave you earlier, I suggest that you study it.

Edited by xyzt
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where am I wrong ?

(dr)2 = (dx)2 + (dy)2
[dr/dt]2 = [dx/dt]2 + [dy/dt]2
U2 = Ux2 + Uy2 As Uy2 is not a function of Ux

2U dU/dx =2Ux

dU/dUx = Ux/U
This is wrong

Or

Px = (mo/(1-U2/C2) -1/2) . Ux as per your expression
dPx/dUx = -1/2 . mo. (1-U2/C2)-3/2 . (-2U/C2) . (dU/dUx) . Ux + (mo/(1-U2/C2) -1/2)

put dU/dUx =Ux/U

dPx/dUx = mo. (1-U2/C2)-3/2 . (Ux2/C2) + (mo/(1-U2/C2) -1/2)

As U2 = Ux2 + Uy2
dPx/dUx = (1-U2/C2)-3/2 . mo . (1-Uy2/C2)

but Fx = dPx/dt =(dPx/dUx ) . (dUx/dt) =(dPx/dUx ) ax by chain theo. of differ.

={1-U2/C2}3/2 . mo . (1-Uy2/C2) . ax
similar for Fy

This shows that Fy/Fx = (ay/ax) . {(C2-Ux2)/(C2-Uy2) }

This is wrong

Edited by hari123
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where am I wrong ?

but Fx = dPx/dt =(dPx/dUx ) . (dUx/dt) =(dPx/dUx ) ax by chain theo. of differ.

={1-U2/C2}3/2 . mo . (1-Uy2/C2) . ax

similar for Fy

This shows that Fy/Fx = (ay/ax) . {(C2-Ux2)/(C2-Uy2) }

This is wrong

You will have to figure where you went wrong by yourself, I narrowed it down for you. Have fun! Do you go to the same school as Mahesh Khati?

I gave you a hint but you continued to make elementary mistakes, I'll give you another hint:

$F_x=\frac{d}{dt} \frac{m_0u_x}{\sqrt{1-(u/c)^2}}=m_0 \frac{du_x}{dt}\frac{1-u^2/c^2+u_xu/c^2}{\sqrt{1-u^2/c^2}^3}$

$F_y=\frac{d}{dt} \frac{m_0u_y}{\sqrt{1-(u/c)^2}}=m_0 \frac{du_y}{dt}\frac{1-u^2/c^2+u_yu/c^2}{\sqrt{1-u^2/c^2}^3}$

So $F_x/F_y=?$

You need to learn how to do differentiation correctly. Until you do that, you will continue to make gross errors in your "physics". I put "physics" in quotes because what you are doing is not physics, it is the same fringe stuff Mahesh Khati was doing, combining basic errors.

Or you can always go learn how it is done properly in the textbooks, Here is the link I gave you earlier, I suggest that you study it.

What happens when you calculate:

$\frac{d}{du_y} \frac{u_y}{\sqrt{1-(u/c)^2}}=\frac{1-u^2/c^2+u_yu/c^2}{\sqrt{1-u^2/c^2}^3}$

If you do it correctly, you get the above. If you do it wrong, as you keep doing it, you get what you are getting.

Edited by xyzt
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This your equation is only true when dU/dUy = 1,

as small change in U i.e.dU can not be equivalent to small change in Uy i.e. dUy. This is wrong.

Physics is not to create equation but more than that.

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This your equation is only true when dU/dUy = 1,

as small change in U i.e.dU can not be equivalent to small change in Uy i.e. dUy. This is wrong.

So, you do not know how to calculate derivatives. I gave you some hints, didn't seem to register with you.

You must be related to Mahesh Khati. Or you are going to the same school, where they teach relativity denying based on bad calculus.

Physics is not to create equation but more than that.

I agree. It must be difficult for you to do any physics when you are unable to calculate a basic derivative. It is even more disturbing when you cannot do that after being shown how to do the calculations correctly.

Let's try a different way to show what you are doing wrong.

So, Fy/Fx = (ay/ax) . {(C^2-Ux^2)/(C^2-Uy^2) }

If Q ={(C^2-Ux^2)/(C^2-Uy^2) } then

Fy = (ay/ax) Q . Fx

Let's admit that for a minute.

So, F’x = Fx – n Fy

Nope, this doesn't follow, you are simply making up nonsense.The correct derivation is:

$F'_x=\frac{d}{dt'} \frac{m_0u'_x}{\sqrt{1-(u'/c)^2}}$

Edited by xyzt
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So, you do not know how to calculate derivatives. I gave you some hints, didn't seem to register with you.

You must be related to Mahesh Khati. Or you are going to the same school, where they teach relativity denying based on bad calculus.

I agree. It must be difficult for you to do any physics when you are unable to calculate a basic derivative. It is even more disturbing when you cannot do that after being shown how to do the calculations correctly.

Let's try a different way to show what you are doing wrong.

Let's admit that for a minute.

Nope, this doesn't follow, you are simply making up nonsense.The correct derivation is:

$F'_x=\frac{d}{dt'} \frac{m_0u'_x}{\sqrt{1-(u'/c)^2}}$

So, nothing?

$u'_x=\frac{u_x+V}{1+u_xV/c^2}$

$t'=\gamma(V)(t+Vx/c^2)$

You will also need:

$\frac{dt'}{dt}=\gamma(V)$

and

$\frac{1}{\sqrt{1-u'^2/c^2}}=\frac{1+u_xV/c^2}{\sqrt{1-u^2/c^2}\sqrt{1-V^2/c^2}}$

$\frac{u'_x}{\sqrt{1-(u'/c)^2}}=\frac{u_x+V}{\sqrt{1-u^2/c^2}\sqrt{1-V^2/c^2}}$

All you have left to do is to calculate:

$\frac{d}{dt'}\frac{u_x+V}{\sqrt{1-u^2/c^2}\sqrt{1-V^2/c^2}}=\gamma \frac{dt}{dt'}\frac{d}{dt}\frac{u_x+V}{\sqrt{1-u^2/c^2}}=\frac{d}{dt}\frac{u_x+V}{\sqrt{1-u^2/c^2}}=\frac{d}{dt}\frac{u_x}{\sqrt{1-u^2/c^2}}+V\frac{d}{dt}\frac{1}{\sqrt{1-u^2/c^2}}$

The first term (multiplied with $m_0$) gives $F_x$, the second term gives out the symmetric terms dependent on $F_y,F_z$

Do you think you could do that without introducing new mistakes?

Edited by xyzt
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So, nothing?

$u'_x=\frac{u_x+V}{1+u_xV/c^2}$

$t'=\gamma(V)(t+Vx/c^2)$

You will also need:

$\frac{dt'}{dt}=\gamma(V)$

and

$\frac{1}{\sqrt{1-u'^2/c^2}}=\frac{1+u_xV/c^2}{\sqrt{1-u^2/c^2}\sqrt{1-V^2/c^2}}$

$\frac{u'_x}{\sqrt{1-(u'/c)^2}}=\frac{u_x+V}{\sqrt{1-u^2/c^2}\sqrt{1-V^2/c^2}}$

All you have left to do is to calculate:

$\frac{d}{dt'}\frac{u_x+V}{\sqrt{1-u^2/c^2}\sqrt{1-V^2/c^2}}=\gamma \frac{dt}{dt'}\frac{d}{dt}\frac{u_x+V}{\sqrt{1-u^2/c^2}}=\frac{d}{dt}\frac{u_x+V}{\sqrt{1-u^2/c^2}}=\frac{d}{dt}\frac{u_x}{\sqrt{1-u^2/c^2}}+V\frac{d}{dt}\frac{1}{\sqrt{1-u^2/c^2}}$

The first term (multiplied with $m_0$) gives $F_x$, the second term gives out the symmetric terms dependent on $F_y,F_z$

Do you think you could do that without introducing new mistakes?

I made a mistake earlier (I took the partial derivative instead of the total one):

$\frac{dt'}{dt}=\gamma(V)(1+V/c^2 \frac{dx}{dt})=\gamma(1+Vu_x/c^2)$

$F'_x=\frac{m_0}{1+Vu_x/c^2}( \frac{d}{dt}\frac{u_x}{\sqrt{1-u^2/c^2}}+V\frac{d}{dt}\frac{1}{\sqrt{1-u^2/c^2}})=\frac{1}{1+Vu_x/c^2}(F_x+m_0V\frac{d}{dt}\frac{1}{\sqrt{1-u^2/c^2}})$

Edited by xyzt
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Don’t make simple things complicated; we have already formula for F’x which express in terms of Fx & Fy. Only problem is Fx & Fy are related by which relation.

You say my derivative is wrong &

Now, I derivate this in detail

d/dUy [(1-U2/C2) -1/2 . UY ] = (1-U2/C2) -1/2 + {[ d/dU (1-U2/C2) -1/2] ] . [dU/dUy] . UY …….. by chain rule of deri

=(1-U2/C2) -1/2+ { [-1/2 . (1-U2/C2)-3/2 . (-2U/C2)] . [dU/dUy] . UY}

=(1-U2/C2)-3/2 . [ 1-U2/C2 +(U Uy/C2)( dU/dUy) ]

Now, you have multiple choices

1)Put dU/dUy = 1 (which is wrong) &

2)Let’s put something else & create new equation. Put, dU/dUy = (dU/dt) / (dUy/dt) = a/ay

d/dUy [(1-U2/C2) -1/2 . UY ] =(1-U2/C2)-3/2 . [ 1-U2/C2 +(U Uy/C2) . (a/ay) ]

hence, Fy = mo . dUy/dt . (1-U2/C2)-3/2 . [ 1-U2/C2 +(U Uy/C2) . (a/ay) ]

Fy = mo . ay . (1-U2/C2)-3/2 . [ 1-U2/C2 +(U Uy/C2) . (a/ay) ]

Hence,

Fy/Fx =(ay/ax) . [{1-U2/C2 +(U Uy/C2) . (a/ay) } / {1-U2/C2 +(U Ux/C2) . (a/ax) } ]

This shows that whenever Fy, Fx are in existence they are related until Fx =0 & your transformation eqn can be express as Fx. f(ax,ay...)

This shows that your transformation equation is in existence until Fx =0 & when Fx =0 it is not required

Now,

dE/c2 represent mass dm.

mass create force when there is change of state of motion.

If I am moving with constant velocity in the train. This motion will not create any force in the train direction or parallel to it in the world. (Even you add some mass some where)

Edited by hari123
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hari post#14

Has someone written a book ?

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dE/c2 represent mass dm.

mass create force when there is change of state of motion.

By your style, you must be Mahesh Khati. Why do you need two different accounts? Anyways:

$\frac{dE}{dt}=\vec{F}\vec{u}=F_xu_x+F_yu_u+F_zu_z$

so:

$F'_x=\frac{F_x}{1-Vu_x/c^2}-\frac{V}{c^2}\frac{F_xu_x+F_yu_u+F_zu_z}{1-Vu_x/c^2}=F_x-\frac{V}{c^2}\frac{F_yu_u+F_zu_z}{1-Vu_x/c^2}$

If you replace V with -V you get:

$F'_x=F_x+\frac{V}{c^2}\frac{F_yu_u+F_zu_z}{1+Vu_x/c^2}$

(for a different derivation, see below)

If I am moving with constant velocity in the train. This motion will not create any force in the train direction or parallel to it in the world. (Even you add some mass some where)

You aren't "moving with constant velocity". The motion is accelerated, remember?

I made a mistake earlier (I took the partial derivative instead of the total one):

$\frac{dt'}{dt}=\gamma(V)(1+V/c^2 \frac{dx}{dt})=\gamma(1+Vu_x/c^2)$

$F'_x=\frac{m_0}{1+Vu_x/c^2}( \frac{d}{dt}\frac{u_x}{\sqrt{1-u^2/c^2}}+V\frac{d}{dt}\frac{1}{\sqrt{1-u^2/c^2}})=\frac{1}{1+Vu_x/c^2}(F_x+m_0V\frac{d}{dt}\frac{1}{\sqrt{1-u^2/c^2}})$

So, you do not want to finish the calculations, after all the hints I gave you?

$F'_x=\frac{1}{1+Vu_x/c^2}(F_x+m_0V\frac{d}{dt}\frac{1}{\sqrt{1-u^2/c^2}})=\frac{1}{1+Vu_x/c^2}(F_x+\frac{V}{c^2}(F_xu_x+F_yu_y+F_zu_z))=F_x+\frac{V}{c^2}\frac{F_yu_y+F_zu_z}{1+Vu_x/c^2}$

Has someone written a book ?

You can find the correct derivation in any book. I gave "hari" a reference. This is what he's talking about.

Heck, I just got tired of waiting for him to realize his mistakes and I finished the proof for him.

Edited by xyzt
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Brother, I have examination papers. So, I am very busy. After 20th oct, we will discuss this in detail. I have face book account also I can give it to you for reference.

I am not against your transformation equation given in http://www.sciencebits.com/node/176 but it works when there is force in X – direction in S frame & there is acceleration in that direction in S- frame. I try to explain you in detail.

In book given in above link. Transformation force in X- direction

Step before giving final transformation equation

Here, -V/c2 (dE/dt) = -V (dm/dt) as dm = dE/c2 This is like rate of change of momentum in that direction

This mass (or energy) is not created but transformed from one system to another & both systems are at the same state of motion when Fx = 0 & no acceleration in X-direction.

So, force is not created in S’ frame.

This is difficult for you to understand. I explain it by one similar event

1) There are two fighter planes of same mass ‘m’ moving with constant acceleration ‘a’ due to same thrust created by both the engines in same direction parallel to one another.

Now, one plane ‘A’ start refueling other plane ‘B’ by rate dm/dt

Now, what will happen?

Plane ‘B’ will become more massive & start de-accelerating (Opposing force to Fx)

This is effect of -V/c2 (dE/dt) or -V.(dm/dt) in your formula.

2)Now, engines are shut down then what will happen.

Acceleration in X-direction = 0 & both planes are moving with constant velocity in X-direction.

,dm/dt or fuel transfer will not create any –ve effect on motion of two plane & any new force will not created in X-direction.

Because both systems are at the same state of motion in X-direction.

This clearly tells you that there is force in X-direction only when Fx is present.

I want to put one formula given by relativity

This formula clear that when there is force then there is acceleration in that direction.

Edited by hari123
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Brother, I have examination papers. So, I am very busy. After 20th oct, we will discuss this in detail. I have face book account also I can give it to you for reference.

Not interested, I am not your "brother", I am not friends with cranks who deny mainstream science.

I am not against your transformation equation given in

http://www.sciencebits.com/node/176 but it works when there is force in X – direction in S frame & there is acceleration in that direction in S- frame.

False, the formula is derived for non-zero components of the acceleration in x, y AND z directions.

I want to put one formula given by relativity

This formula clear that when there is force then there is acceleration in that direction.

The above teaches you that $\vec{F}=(F_{parallel},F_{perpendicular})$

where:

$F_{parallel}=\gamma^3 m_0a_{parallel}$

$F_{perpendicular}=\gamma m_0a_{perpendicular}$

So, there is force in BOTH parallel direction (as you claim) AND in the perpendicular direction (as you keep denying). That is consistent with the formulas explained to you, since the formulas contain $\frac{du_x}{dt}$ AND $\frac{du_y}{dt}$ AND $\frac{du_z}{dt}$

Edited by xyzt
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You are both wrong & doing math in wrong way. I was just spectacle from many days but as discussion is going on about my thread. I put one post if allow

1)For Hari, You are wrong from some steps in your math of post 14. I am pasting your math below

Now,I derivate this in detail

d/dUy [(1-U2/C2) -1/2 . UY ] = (1-U2/C2) -1/2 + {[ d/dU (1-U2/C2) -1/2] ] . [dU/dUy] . UY …….. by chain rule of deri

=(1-U2/C2) -1/2+ { [-1/2 . (1-U2/C2)-3/2 . (-2U/C2)] . [dU/dUy] . UY}

=(1-U2/C2)-3/2 . [ 1-U2/C2 +(U Uy/C2)( dU/dUy) ]

Your are wrong from this step , now. i do it in my way

let, consider U = K Uy

then dU = K dUy

& dU/dUy = U/Uy

here, K = constant & if needed for some one can be consider as 1/cos(a) where a is angle between U & Uy

d/dUy [(1-U2/C2) -1/2 . UY ] =(1-U2/C2)-3/2 . [ 1-U2/C2 +(U Uy/C2) . (U/Uy) ] =(1-U2/C2)-3/2 . [ 1-U2/C2 +(U 2/C2)

put this in Fy = mo . dUy/dt . (d/dUy [(1-U2/C2) -1/2 . UY ]

hence, Fy = mo . dUy/dt . (1-U2/C2)-3/2 . [ 1-U2/C2 +(U 2/C2) ]

Fy = mo . ay . (1-U2/C2)-3/2 .

& similarly Fx = mo . ax . (1-U2/C2)-3/2 .

similarly F =(Fx2+Fy2)1/2 =mo . (1-U2/C2)-3/2 .(ax 2+ay2)1/2 =mo . (1-U2/C2)-3/2 .a

similarly Fy/Fx = ay/ax

Now,

transformation equation becomes

F’x = Fx {1- n. (ay/ax) }

where n=[(V/C^2 ) (Uy)]/[(1-V Ux/C^2)]

& F'x =0 when Fx=0

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

2)For XYZT, You are also wrong

& Now, differ similarly to above way result will be

F'x = mo. (dU'x/dt') . d/dU'x [u'x . (1-U'2/C2)-1/2 ].

F'x = mo . a'x . (1-U' 2/C2)-3/2 .

we know that a'x = (ax/y3 ) .(1-ux.V/c2 )-3 (transformation of acceleration )

This shows that when ax =0 i.e. acceleration in train cabin in X - direction is zero then a'x =0

& altimately F'x = 0

Thanks

Edited by mahesh khati
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2)For XYZT, You are also wrong

Rubbish, this is the DEFINITION of force, in case you don't recognize it. You have been doing this for quite a while. Do you and "hari123" go to the same school? Because you both have the same crank ideas.

Edited by xyzt
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Means, we can not differentiate in S' frame.

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Means, we can not differentiate in S' frame.

Rubbish, you can differentiate in any frame.

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