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. ------------------------------------------------------------------------------------------------------------------------------------ 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. -------------------------------------------------------------------------------------------------------------------------------------------------- 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. ------------------------------------------------------------------------------------------- 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. ---------------------------------------------------------------------------------------------------------------------------------------------------- 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. ----------------------------------------------------------------------------------------------------------------------------------- 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 ----------------------------------------------------------------------------------------------------------------------------------------- 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 Important point (2):- Mean’s applied magnetic force on object in X-direction is -fx & acting force in X-direction is -fx + Fay = 0 ## or 0. Here, resultant force in X-direction is zero but there is acceleration. ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ 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) This is completely complicated results, which says that applied force & acting forces on objects are different & more in S.R. ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- 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 ----------------------------------------------------------------------------------------------------------------------------------------- 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 Important point (2):- Mean’s applied magnetic force on object in X-direction is -fx & acting force in X-direction is -fx + Fay = 0 ## or 0. Here, resultant force in X-direction is zero but there is acceleration. ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ 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) This is completely complicated results, which says that applied force & acting forces on objects are different & more in S.R. ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- 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 ---------------------------------------------------------------------------------------------------------------------------------------------------------- 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.

-------------------------------------------------------------------------------------------------------------------------------------------------- , so: -------------------------------------------------------------------------------------------------------------------------------------------------- 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: -------------------------------------------------- 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: ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------- 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 --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- 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

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.

-------------------------------------------------------------------------------------------------------------------------------------------------------------- --------------------------------------------------------------------------------------------------------------------------------------------------------------- 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. -------------------------------------------------------------------------------------------------------------------------- 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/{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 -------------------------------------------------------------------------------------------- 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. --------------------------------------------------------------------------------------------- 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. ------------------------------------------------------------------------------------------------------- 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

------------------------------------------------------------------------------------------------------------------------------------- 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 ------------------------------------------------------------------------------------------------------------------------------------------------ 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. ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- 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.

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 C2 (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