mahesh khati

My child was young. He was very sensitive for cold. Even small change in weather causes cough & throat infection. Even antibiotic gives very temporary relief to him. I think about this problem & do some different thing different.

I give medicine & side by side, I mix 1/4 tablespoon of turmeric in bowl of water.

I give 1 tablespoon of that very dilute turmeric water to my child after every 1/2 an hour for a complete day. So that his throat will be yellow from inside always.

I find that this give him relief early.

Sometime even medicine is not require for mild infection.

& today his immunity for cough is improved.

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I consider that in this treatment. I keep the receptor in the throat engage with turmeric (Which is a medicinal powder present in Kitchen) & for virus receptor in throat is not available.  So, to multiply the virus, additional space in the throat is not available. This turmeric has very different medicinal smell. That may go directly to lunge & again engage receptors there. So, our body get more time to create antibodies for virus.  

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I think in covid-19 treatment. Above treatment is given side by side then patient will get relief very early. This is simple, harmless, economical, scientific & very supportive. Today, I give some Ayervidic syrup & this turmeric water & many time I do not require any other medicine. 

Here, we are parking the turmeric in the receptor of throat before virus will get park on it. This is simple but effective side by side treatment.

I want your comment please. 

I have paper on relativity, space gun & on other topics on my web url removed. Please visit & give your suggestion. That may be opposite to my thought then also it is welcome.

einstein.pdf

After nearly one month, there is no solution to above problem in S.R. It is really difficult because I have not change any calculation in S.R. Only proves that there are two forces in S.R.

One is applied & other is actual acting force on object & that force is more than applied force.

I was smelling this problem from last 3 years because in S.R. force (in X-direction) is not depend on change of state of motion (acceleration) in that direction but depends on change of momentum. If some ball is falling & if I am moving toward it horizontally then also

fx= y³ mo. (ux/c²} uy ay ----------- as ax=0

this force will act on the ball in horizontal direction.

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this problem can be complicated as we want

for example man is pulling cart in any direction on platform with force f let, fx & fy are there components in X & Y direction

Then calculations given in post 1 says that actual forces acting on cart, in x & y directions are

Fx = fx +y³ mo. (ux/c²} uy ay & Fy = fy + y³ mo. (uy/c²} ux ax

& this can be complicated further

Now, I have to stop.

I like this post Mr Mordred & also think about this solution previously but I have problem. In nature we not apply Fx but we get Fx

for example.

On platform old man applied force perpendicular to velocity of train on cart & it accelerate in Y-direction only (acceleration & force has same direction on platform)

let, train velocity is -ux

Then for observer on train :-

then cart will moves with constant velocity 'ux' in x-direction, it accelerate with some acceleration 'ay' in Y-direction only with velocity 'uy'.

Now, if I apply above equation for this frame

Fx = d/dt( y. mo. ux) where y=(1-u²/c²)^-0.5

then, output will be

Fx= y³ mo. (ux/c²} uy ay ----------- as ax=0

here, I have not applied any force in X-direction but constant velocity in X-direction creates this force.

As velocity of train is more this force is more & when it is zero, force is zero.

Above are not the general formula for force in X-directions but conditional, Now, I am giving general formula's

these are the general formula in S.R. for forces in X, Y & Z-direction given in web site https://en.wikisource.org/wiki/The_Direction_of_Force_and_Acceleration

&

in book element of special relativity (Academic book) by Dr T M Karade, Dr K.S. Adhav & Dr Maya S. Bendre page no 135 gives the same formulas

So,

in S.R., in any frame for force in X-direction

Fx = d/dt( y. mo. ux) where y=(1-u²/c²)^-0.5 is true & then simple differentiation

So, after differentiation

Fx= y. mo. (dux/dt) + y³. mo. {ux/c²}. (u . du/dt)

Fx= y. mo.ax + y³. mo. {ux/c²}. (u . a) -----(1)

We know, u²=ux²+uy²+uz²

So, 2 u. du/dt =2 ux (dux/dt) +2 uy (duy/dt) + 2 uz (duz/dt)

u. a = ux ax +uy ay + uz az --------(2)

from (1) & (2)-------- (you can directly differentiate without this two step also)

So, Fx=y. mo. ax+y³ mo. (ux/c²} (ux ax+uy ay+uz az)

So, calculation given in post 1 is true.

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Mr Ematfaal, above general equation can easily converted in to your conditional equation by just two or three steps.

please, pointed out where I am wrong.

This is very simple derivative but if you find something wrong in it.

Please,inform me, I will be very much thankful to you.

in S.R., in any frame for force in X-direction

Fx = d/dt( y. mo. ux) where y=(1-u²/c²)^-0.5

So, after differentiation

Fx= y. mo. (dux/dt) + y³. mo. {ux/c²}. (u . du/dt)

Fx= y. mo.ax + y³. mo. {ux/c²}. (u . a) -----(1)

We know, u²=ux²+uy²+uz²

So, 2 u. du/dt =2 ux (dux/dt) +2 uy (duy/dt) + 2 uz (duz/dt)

u. a = ux ax +uy ay + uz az --------(2)

from (1) & (2)-------- (you can directly differentiate without this two step also)

So, Fx=y. mo. ax+y³ mo. (ux/c²} (ux ax+uy ay+uz az)

So, calculation given in post is true (please find out if anything wrong in it.)

I can give detail calculation:-

In S.R., in any frame for force in X-direction

Fx = d/dt( y. mo. ux) where y=(1-u²/c²)^-0.5

So, after differentiation

Fx= y. mo. ax + y³. mo. {ux/c²}. (u . a) -----(1)

We know, u²=ux²+uy²+uz²

So, u. a = ux ax +uy ay + uz az --------(2)

from (1) & (2)-------- (you can directly differentiate without this two step also)

So, Fx=y. mo. ax+y³ mo. (ux/c²} (ux ax+uy ay+uz az)

So, calculation given in post is true.

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Point 1:-

Problem which I raised from above paradox is very serious.

For example,

One light object is falling down under gravity in vertical Y-direction & horizontal air is pushing that object in horizontal direction.

For observer on ground

Let, fx is horizontal force applied by air on object & fy is vertical force applied by gravity on object.

Then above calculation says that

Actual force acting on objects are not fx & fy but

Fx = fx + y³ mo. (ux/c²} uy ay =fx + Fay

& Fy = fy+ y³ mo. (uy/c²} ux ax =fx + Fax

Means, actual force acting on objects are different (more) than force applied on the object.

This is very serious output & can create very complicated problem in relativity. (if not solve).

Ok, I again re-write the paradox with above improvement

STEP 1:-This problem can easily be understood by following paradox.

{Before starting this paradox, I want to put some relativity formula's

In any frame for force in X-direction

Fx = d/dt( y. mo. ux) where y=(1-u²/c²)^-0.5

After differentiation, we get

So, Fx=y. mo. ax+y³ mo. (ux/c²} (ux ax+uy ay+uz az) ------(1)}

Now, Paradox:-

On friction-less platform, object is moving with constant velocity ux in X-direction & only magnetic force is acting in Y-direction & there is acceleration in Y-direction only & uy is not zero & Fz=0

If we apply eq(1) to this case then result will be

Fx= y³ mo. (ux/c²} uy ay

Or Fx=Fay as this force is form due to ‘ay (& uy)’

Mean’s even there is no magnetic force acting on object from outside in x-direction & no ‘ax’ then also above force will act on object in +ve direction of x-axis due to ‘ay ’

Important point (1):-

Mean’s applied magnetic force on object in X-direction is 0 & acting force in X-direction is

Fx= y³ mo. (ux/c²} uy ay+0 or Fay+0=Fay

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STEP 2:-Now, Force acting in X-direction is Fx= y³ mo. (ux/c²} uy ay or Fay

Now, after this happen, very small magnetic force of same intensity -fx = -y³ mo. (ux/c²} uy ay or -Fay start acting on object in direction opposite to above force (but velocity is still positive ux & cancel that above force.

Mean’s equation (1) becomes

0=y. mo. ax+y³ mo. (ux/c²} (ux ax+uy ay)

Or 0 =y. mo ax. (1+ y² {ux²/c²} ) +Fay

(Here as Fay= y³ mo. (ux/c²} uy ay)

Mean’s Fay = y. mo. -ax. (1+ y². {ux²/c²} )

Mean’s there must be acceleration in –ve X-direction to fulfill above equation of S.R.

Now, see above equation carefully, it is of nature

0= -fx + Fay

STEP3:- same things happen for +ve force in X-direction (for less than Fay or more)

Now, I am generalizing above result.

Step 1&2 clearly shows that when we apply any magnetic force (Fmx) in X-direction on the object, actual force acting on object is more & that quantity is (Fmx+Fay)

Similarly,

If we apply any magnetic force (Fmy) in Y-direction on the object then actual force acting on object is more & that quantity is (Fmy+Fax)

STEP4:- Force does work, consume energy, gain energy & we must know that energy cannot be created. It can be transferred only:-

From above setup it must be clear that energy get transfer from magnet to object but if applied magnetic force is less than acting force then energy gain by object will be more than energy loose by the magnet. Means due to more work done by more force for same displacement, more energy get generated.

HERE, more energy & force is the problem.

Where this additional energy (& force) does comes from?

but in above paradox, I have not consider uy=0.

(& same calculation is true in Y-direction for same conditions.)

Force without acceleration in S.R.

& acceleration without force in S.R.

& applied force is less than acting force in S.R.

STEP 1:-This problem can easily be understood by following paradox.

{Before starting this paradox, I want to put some relativity formulae’s

In any frame for force in X-direction

Fx = d/dt( y. mo. ux) where y=(1-u²/c²)^-0.5

After differentiation, we get

So, Fx=y. mo. ax+y³ mo. (ux/c²} (ux ax+uy ay+az az) ------(1)}

Now, Paradox:-

On frictionless platform, object is moving with constant velocity ux in X-direction & only magnetic force is acting in Y-direction & there is acceleration in Y-direction only & Fz=0

If we apply eq(1) to this case then result will be

Fx= y³ mo. (ux/c²} uy ay

Or Fx=Fay as this force is form due to ‘ay’ only

Mean’s even there is no magnetic force acting on object from outside in x-direction & no ‘ax’ then also above force will act on object in +ve direction of x-axis due to ‘ay’

Important point (1):-

Mean’s applied magnetic force on object in X-direction is 0 & acting force in X-direction is

Fx= y³ mo. (ux/c²} uy ay+0 or Fay+0=Fay

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STEP 2:-Now, Force acting in X-direction is Fx= y³ mo. (ux/c²} uy ay or Fay

Now, after this happen, very small magnetic force of same intensity -fx = -y³ mo. (ux/c²} uy ay or -Fay start acting on object in direction opposite to above force (but velocity is still positive ux) & cancel that above force.

Mean’s equation (1) becomes

0=y. mo. ax+y³ mo. (ux/c²} (ux ax+uy ay)

Or 0 =y. mo ax. (1+ y² {ux²/c²} ) +Fay

(Here as Fay= y³ mo. (ux/c²} uy ay)

Mean’s Fay = y. mo. -ax. (1+ y². {ux²/c²} )

Mean’s there must be acceleration in –ve X-direction to fulfil above equation of S.R.

Now, see above equation carefully, it is of nature

0= -fx + Fay

STEP3:- same things happen for +ve force in X-direction (for less than Fay or more)

Now, I am generalising above result.

Step 1&2 clearly shows that when we apply any magnetic force (Fmx) in X-direction on the object, actual force acting on object is more & that quantity is (Fmx+Fay)

Similarly,

If we apply any magnetic force (Fmy) in Y-direction on the object then actual force acting on object is more & that quantity is (Fmy+Fax)

STEP4:- Force does work, consume energy, gain energy & we must know that energy cannot be created. It can be transferred only:-

From above setup it must be clear that energy get transfer from magnet to object but if applied magnetic force is less than acting force then energy gain by object will be more than energy loose by the magnet. Means due to more work done by more force for same displacement, more energy get generated.

HERE, more energy & force is the problem.

Where this additional energy (& force) does comes from?

Means, we can not differentiate in S' frame.

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

Now your above equation becomes

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

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2)For XYZT, You are also wrong

you formula in your post

& 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

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

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

what is your displacement?.

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

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

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.

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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

So, in your case also

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

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

, 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)

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=F⋅v, 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

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)

So, In your case

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.

So, your calculation is wrong

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.

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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

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/{γP^3P. (1-ux. v/cP^2P)P^3P } where γ =1/(1-vP^2P/cP^2P)P^0.5

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,

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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.

Good, you have learned something.

 

 

Now, we do the math again

your mathematics in the beginning is true.

because if I differentiat E = mo.c² (1-u²/c²)^-1/2 with respect to u both side then

dE/du = mo . u .(1-u²/c²)^-3/2

If I multiply both side by du then out put will be similar to your mathematics

dE = mo . u .(1-u²/c²)^-3/2 du

Now, I improvevise your math from here because their is no displacement & acceleration in your caiculation

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

As du/dt=a =acceleration & ds = u.dt =displacement

dE =γ³. mo. a . ds

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

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 of relativity also'

 

 

Now, in my all event created displacement happens perpendicular to the motion of frames to avoid complecated mathematic. So, displacement remain same because space contraction due to relative motion is not happen on that direction.

that is why displacement is constant.

Now, it is too late 2.41 of night good night

thanks & wil meet you next day.

 

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

 

 

 

 

 

My calculations are correct. Yours are not.

There was an extraneous [math]c^2[/math]

[math]m_0 u du[/math] has dimensions of energy ([math]m_0 c^2[/math]), so you are wrong.

I have just seen your post. I will see it in detail.afterward because civil work is going on ( on my site) but I think

mathematics in the beginning is true.

because if I different E = mo.c² (1-u²/c²)^-1/2 with respect to u both side then

dE/du = mo . u .(1-u²/c²)^-3/2

If I multiply both side by du then out put will be similar to your mathematics

dE = mo . u .(1-u²/c²)^-3/2 du

this is your out put without C^{2 (}which was wrong).

now, left side units = right side unit

but this dE = change in energy per unit change in velocity (this velocity is also time dependent function)

We are not interested in energy consumption for unit change of velocity in my event but in my event

Displacement is fixed in both frame & force is fixed in one frame but relatively vary in second frame &

as Force in stable frame = Force in moving frame . ((1-u²/c²)^-1/2

Work done in stable frame = work done . in moving frame . ((1-u²/c²)^-1/2

Energy consumption in in stable frame = Energy consumption in moving frame . ((1-u²/c²)^-1/2

^{Now, if you want to apply differentiation then it will be like}

^{dW = dF . S}

^{dW/dF = S = constant in both frames}

^{means if force decreases then work done has to decrease to make S constant in different frames given in my event}

^{thank you}

^{I will post again}