hari123

Brother, I have examination papers. So, I am very busy. After 20^th 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.

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) &

Your equation

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,

See from your book

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)

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.

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/C^{2) -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

(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) }

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.