xyzt Posted August 11, 2014 Share Posted August 11, 2014 (edited) Read my mathematics in my paper. I have taken care of everythings. I read your "paper", it is the same mistakes you are posting in the forum. So, dE = F . ds I.e. Work done = force in that frame X displacement in that frame Yet, you fail to transform [math]ds[/math] when passing from one frame to the other. You cling to your incorrect claim that [math]ds[/math] is frame-invariant. SR teaches you that it ISN'T. now, in relativity force parallel to direction of mation is F= γ^{3}. mo. a this is very well known to ererybody Err, wrong again, the general formula is: [math]\mathbf{F} = \gamma(\mathbf{v})^3 m_0 \, \mathbf{a}_\parallel + \gamma(\mathbf{v}) m_0 \, \mathbf{a}_\perp [/math] Only when the component of the acceleration perpendicular on the velocity [math]\mathbf{v}[/math] is zero does the formula reduce to what you wrote. In general case, it doesn't. I am going to tell you one last time : you CANNOT USE SR in order to "disprove" SR. Edited August 11, 2014 by xyzt Link to comment Share on other sites More sharing options...

mahesh khati Posted August 13, 2014 Author Share Posted August 13, 2014 (edited) ------------------------------------------------------------------------------------------------------------------------------------- dE = mo . u .(1-u^{2}/c^{2})^{-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-u^{2}/c^{2})^{-3/2 }du =dE = mo . u .(1-u^{2}/c^{2})^{-3/2 }du/dt . dt =γ^{3}. mo. du/dt . u.dt =γ^{3}. 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= γ^{3}. mo. a & dE =γ^{3}. 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.c^{2} E =γ Eo means, energy also changes similar to mass changes in relativity or mass & energy are same but we see it differently Now, I create circular reference in relativity. displacement was consider perpendicular to relative motion of frames. This changes relative mass, relative acceleration, force, work done & energy & now, check the relativity energy relation &I find that SR fails here to form circular reference In in all calculation, I use very event specific mathematics deals to that event only. Mathematics 1 is very general mathematics to get the filling of slow down of event in frame motion by comparing accelerating falling ball to tied ball in rail cabin Mathematics 2 is given to give general filling of how mass increases & acceleration decreases more than mass. Mathematics 3 & 4 is given to show that standard text book mathematics gives the same out put as my mathematics because mathematics remain same. Edited August 13, 2014 by mahesh khati Link to comment Share on other sites More sharing options...

xyzt Posted August 13, 2014 Share Posted August 13, 2014 (edited) If, in frame S you have* [math]dW=F_ydy[/math], then in frame S', moving in the x direction with speed [math]V[/math], you will have: [math]F'_y=F_y \sqrt{1-V^2/c^2}=\frac{F_y}{\gamma}[/math] [math]F'_x=\frac{u_yV}{c^2}F_y[/math] [math]dy'=dy[/math] [math]dx'=\gamma(dx+Vdt)=\gamma V dt[/math] so [math]dW'=F'_ydy'+F'_xdx'=F_y(dy \sqrt{1-V^2/c^2}+\gamma V dt \frac{u_yV}{c^2})=F_y dy(\frac{1}{\gamma}+\gamma \frac{V^2}{c^2})=F_ydy \gamma=\gamma dW > dW [/math] So? Work, is frame variant, as explained multiple times to you. What I see that having failed to prove your case for "work", you graduated to your other "disproofs" of SR. I have already explained to you that you cannot disprove SR by using SR, all you can do is to demonstrate your inability to understand and learn SR. Have fun! ----------------------------------------------------------------------------------------------------------------------------------------- * This is not how mechanical work is described in relativity. I explained that to you in a previous post but it fell on deaf years. Edited August 13, 2014 by xyzt Link to comment Share on other sites More sharing options...

imatfaal Posted August 13, 2014 Share Posted August 13, 2014 Mahesh - just in case you think it is just XYZT saying this I will repeat; Special Relativity is mathematically internally consistent. It is a very simple and self-contained theory which allows us to say that as long as the postulates are held to be true then the theory must follow. There is simply no room for a mathematically correct situation within SR which leads to an internal contradiction; the only way to show SR is flawed/incomplete is to move outside its frame of application or to provide experimental evidence that contradicts a postulate. If you have a thought experiment that is within the realm of application and postulate of SR yet still contradicts SR then you have made a mistake. 3 Link to comment Share on other sites More sharing options...

mahesh khati Posted August 16, 2014 Author Share Posted August 16, 2014 (edited) 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^{3}^{P}. (1-ux. v/cP^{2}^{P})P^{3}^{P} } where γ =1/(1-vP^{2}^{P}/cP^{2}^{P})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 -------------------------------------------------------------------------------------------- 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 Edited August 16, 2014 by mahesh khati Link to comment Share on other sites More sharing options...

studiot Posted August 16, 2014 Share Posted August 16, 2014 When ever there is force, there is acceleration What about equilibrium? Link to comment Share on other sites More sharing options...

xyzt Posted August 16, 2014 Share Posted August 16, 2014 (edited) Then force in X-direction must be absent.in both frames. (In vector mathematics, effect of any vector may be acceleration in perpendicular direction is zero) Relativity teaches you that your claim (like most , if not all of your other claims) is false. You keep piling false claims on top of false claims. [math]F_x=0[/math] [math]u_x=0[/math] [math]F'_x=F_x+\frac{u_yV}{c^2+\frac{u_xV}{c^2}}F_y=0+\frac{u_yV}{c^2}F_y=\frac{u_yV}{c^2}F_y[/math] See how easy is to disprove your falsities? What about equilibrium? The guy is a mechanical engineer. I wonder how he got his degree. Now, Yep dx'/dt' = V as y dt =d't Nope, I'll let you figure out your mistakes by yourself on this one. then your final equation become (for small dy) d'w=(1/γ) F'y . dy =.(1/γ) dw No, it doesn't, [math]dW'=\gamma F_y dy=\gamma dW[/math]. I'll let you figure out your errors by yourself. You can't do physics and you demonstrated that you can't do basic math either. Means, if ax =0 then a'x=0 or acceleration in X-direction in both frames are absent. Then force in X-direction must be absent.in both frames. Ahh, I missed this glaring error as well. You see, force in SR is NOT [math]\vec{F}=m \vec{a}[/math], it is defined as [math]\vec{F}=\frac{d}{dt}(\gamma(u) m \vec{u})[/math]. Try to do the Lorentz transform , you will see your errors. Or , most likely, you will add more errors on top of the existent errors. Edited August 16, 2014 by xyzt Link to comment Share on other sites More sharing options...

mahesh khati Posted August 18, 2014 Author Share Posted August 18, 2014 (edited) -------------------------------------------------------------------------------------------------------------------------------------------------------------- --------------------------------------------------------------------------------------------------------------------------------------------------------------- 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 Edited August 18, 2014 by mahesh khati Link to comment Share on other sites More sharing options...

Ophiolite Posted August 18, 2014 Share Posted August 18, 2014 Have you ever considered the remote possibly that it is you that is wrong, rather than the hundreds of thousands of physicists who preceded you? Link to comment Share on other sites More sharing options...

xyzt Posted August 18, 2014 Share Posted August 18, 2014 (edited) Have you ever considered the remote possibly that it is you that is wrong, rather than the hundreds of thousands of physicists who preceded you? He knows he's wrong, he will just never admit it. -------------------------------------------------------------------------------------------------------------------------------------------------------------- --------------------------------------------------------------------------------------------------------------------------------------------------------------- Thanks, I realy like your this calculation but in work, there is two parts force & displacement work = force x displacement displacement happen due to force then only work is done ...which is exactly what I explained to you in an earlier post: [math]dW'=F'_ydy'+F'_xdx'=F_y(dy \sqrt{1-V^2/c^2}+\gamma V dt \frac{u_yV}{c^2})=F_y dy(\frac{1}{\gamma}+\gamma \frac{V^2}{c^2})=F_ydy \gamma=\gamma dW > dW [/math] There is a [math]F'_ydy'[/math] component and a [math]F'_xdx'[/math] component. You keep denying the [math]F'_xdx'[/math] component because you don't know relativity. In above event if I apply this is your equation using acceleration In this event acceleration is perpendicular to V only ....in frame S ONLY. In frame S', the acceleration is no longer perpendicular on the speed V, there is also a parallel component that shows up. This is the cause of the component [math]F'_x \ne 0[/math] that shows up in frame S'. Edited August 18, 2014 by xyzt Link to comment Share on other sites More sharing options...

mahesh khati Posted August 18, 2014 Author Share Posted August 18, 2014 (edited) This is the transformation equation So, if Fx =0 & Vx =0 & Fz =0 Then F'x = - (V. Uy/c^{2}) 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. Edited August 18, 2014 by mahesh khati Link to comment Share on other sites More sharing options...

xyzt Posted August 18, 2014 Share Posted August 18, 2014 (edited) This is the transformation equation So, if Fx =0 & Vx =0 & Fz =0 Then F'x = - (V. Uy/c^{2}) Fy Good, you are starting to learn. Took you a few days to reproduce the formula that I have shown you a few days ago. You got the sign wrong because you have the first formula wrong: [math]F'_x=F_x+\frac{u_yV}{c^2+\frac{u_xV}{c^2}}F_y+\frac{u_zV}{c^2+\frac{u_xV}{c^2}}F_z[/math] point 2 :- displacement which you consider in above your calculation is inertial displacement This displacement happens automatically due to relative motion of the frames. This displacement happen even event is not happened. & force F'x acts opposite to this displacement due to increase in energy objects due to uy velocity. I'll have Ranch with the above. You keep making the mistakes, I keep correcting them. The only "mistakes" that you have found in SR are your own. Edited August 18, 2014 by xyzt Link to comment Share on other sites More sharing options...

CaptainPanic Posted August 20, 2014 Share Posted August 20, 2014 He knows he's wrong, he will just never admit it. ! Moderator Note In my role as moderator, I would like to advise people who think that a discussion is going round in circles (or that the topic has otherwise been depleted) to stop participating. Do not continue a thread just to have the last word, or to "win" a discussion. If someone is here to learn, they will read your valuable posts again. If the other participants are here to "win" a discussion, then any reply from your side will not help anyway. Please note that I am not suggesting that this discussion is done, and this is not an official warning to anyone. I am not moderating the contents here. Anyone is free to continue. It is just the above sentence (in the quote) that triggered me to make this friendly comment. Please stay on topic, which means you shouldn't respond to this moderator tip. Link to comment Share on other sites More sharing options...

mahesh khati Posted August 20, 2014 Author Share Posted August 20, 2014 (edited) 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-V^{2}/c^{2})^{-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/c^{2}) Thus, F'y = Fy/ { y (1-V.ux/c^{2})} 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/c^{2}) } 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/c^{2}) 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 Edited August 20, 2014 by mahesh khati Link to comment Share on other sites More sharing options...

xyzt Posted August 20, 2014 Share Posted August 20, 2014 (edited) To summarize, the complete transformation is: You are true If such component is their in frame S' in -ve direction then object will accelerate in opposite direction & will try to decrease inertial displacement in that direction. This is decrease in displacement in that frame will not add the energy of object in that frame but it decreases energy level of object in that frame due to work done The transformation of force with the minus sign is derived from the Lorentz transform: [math]x'=\gamma(x-Vt)[/math], so: [math]dx'=\gamma(dx-Vdt)=-\gamma V dt[/math] Now try calculating [math]dW'=F'_xdx'+F'_ydy'[/math]. If you do it correctly you should be getting [math]dW'=\gamma dW[/math]. So, once again, all you are finding is your mistakes in understanding SR. Edited August 20, 2014 by xyzt Link to comment Share on other sites More sharing options...

mahesh khati Posted August 22, 2014 Author Share Posted August 22, 2014 (edited) -------------------------------------------------------------------------------------------------------------------------------------------------- , 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) Edited August 22, 2014 by mahesh khati Link to comment Share on other sites More sharing options...

xyzt Posted August 22, 2014 Share Posted August 22, 2014 (edited) -------------------------------------------------------------------------------------------------------------------------------------------------- , so: -------------------------------------------------------------------------------------------------------------------------------------------------- You are again mixing displacement between frames & displacement with in a frame. So, it must be address now. No, I am not. I am simply trying to teach you relativity. But you seem unwilling and unable to learn. You would much rather cling to your misinterpretations of SR than learn SR. If you want L-transformation of displacement in frames then it is only dS'x = dSx. (1/y) or, d'x =dx . (1/y) Not when [math]dx=0[/math]. One gets [math]dx'=dx/\gamma[/math] when one imposes [math]dt'=0[/math] in [math]dx'=\gamma(dx-Vdt)[/math] and [math]dt'=\gamma(dt-Vdx/c^2)=0[/math]. This is NOT the case here. The case here is [math]dx=0[/math] and basic math teaches you that [math]dx'=-\gamma V dt[/math] as I tried to teach you. Sorry I failed, there is nothing more that I can do for you. Edited August 22, 2014 by xyzt Link to comment Share on other sites More sharing options...

xyzt Posted August 23, 2014 Share Posted August 23, 2014 -------------------------------------------------------------------------------------------------------------------------------------------------- , so: -------------------------------------------------------------------------------------------------------------------------------------------------- You are again mixing displacement between frames & displacement with in a frame. So, it must be address now. Nope [math]dx'[/math] is the displacement in the frame where I also calculated [math]dy',F'_x,F'_y,dW'[/math]. You have to come to grips with the fact that you have no clue when it comes to SR. Link to comment Share on other sites More sharing options...

mahesh khati Posted August 25, 2014 Author Share Posted August 25, 2014 (edited) 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. Edited August 25, 2014 by mahesh khati Link to comment Share on other sites More sharing options...

John Cuthber Posted August 25, 2014 Share Posted August 25, 2014 "you first mark the beginning point in the frame or cabin & then after displacement. You mark the finishing point & after event you measure distance between these two point instantly. & that length is displacement." I may be mistaken but I think that, according to SR, "instantly" doesn't have a defined meaning for events separated in space. So, at best, you have ignored one of the axioms of SR, in order to prove it false. That's not logical. 1 Link to comment Share on other sites More sharing options...

mahesh khati Posted August 25, 2014 Author Share Posted August 25, 2014 (edited) 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?. Edited August 25, 2014 by mahesh khati Link to comment Share on other sites More sharing options...

xyzt Posted August 25, 2014 Share Posted August 25, 2014 (edited) So, at best, you have ignored one of the axioms of SR, in order to prove it false. That's not logical. It is called "beating the strawman to death". We should let him do it, I corrected all his errors but to no avail. what is your displacement?. What I explained to you. Not what you continue to erroneously claim. Edited August 25, 2014 by xyzt Link to comment Share on other sites More sharing options...

mahesh khati Posted August 27, 2014 Author Share Posted August 27, 2014 (edited) Sorry for any wrong word I use but Point 1 :- Displacement in any event is the distance measure between finished point of the object in the event to start point of object in the event measure at the end of the event. For example. If ball falls in train cabin from tied thread which is cut suddenly. This ball is under horizontal force of attraction also. Now, ball will not fall vertically but this fall will tilted towards horizontal attractive force & falls down on floor. We can not measure this displacement in between the happening of the event. It must be measure at the end of the event. This displacement is the distance measure between point where ball finally fall on the floor & point of relies of the ball measure at the end of the event. Means, in rail cabin two points are fixed ( after the event), point where ball is relies ed from thread & point where it is fall on the floor. For observer on platform as length of train contracted due to relativity. These two points in the cabin also come closure & horizontal distance between these two points will decrease. so, for observer on platform dx'=(1/y) dx This is very simple. At the end of the event displacement is just distance & it obeys the low of contraction of relativity. in my event , horizontal force is absent So, ball will fall vertically in cabin so, dx = 0 & dx' =0 Point 2:- If cart of mass m is pulled by force Fx then cart will accelerate. Due to acceleration & continuous change in the state of motion, opposite force is developed by the cart Force = m ax Now, consider that I put up some additional mass (one form of energy) dm in the cart suddenly then additional retarding force will get develop Now, opposing force = (m+dm). ax Now, for same acceleration we will require some additional force. or cart will retard or de-accelerate. but this all is happen only when force is acting on cart & it is accelerating If cart is at rest & no force is acting then after adding some additional mass(one form of energy) will not create any additional force. Same thing happen in our case. Detail is given in post 69 when on object (ball,cart or particle) force Fx is act in X- direction then only F'x = Fx - {v/c2. ( Fy. uy )} / {(1-V.ux/c2) } because Fy.Vy =Fy. dy/dt =dWy/dt =dE/dy This {v/c2. ( Fy. uy )} / {(1-V.ux/c2) } or {v/c2. (dE/dt} } / {(1-V.ux/c2) } is not a physical force. This is retarding effect developed due to additional energy accumulated due to motion in y-direction This retarding effect is only there until object will accelerate in +ve X- direction due to external force Fx If Fx is not there then there will not be any horizontal displacement or any retarding force because Anybody can not accelerate our self & loose its energy until some external force will act on it. So, if Fx in not present in rest frame & if there is no acceleration in X-direction in rest frame then in S' frame opposing force (retarding force) will not come into action because these are internally created forces. & object will not create force our self & get displace in that direction & loose our energy. Work is done only when external force will act on object & it will get displaced in that direction. Point 3 :- Means, math 1 which I have done in post 64 is true There is no relative displacement & work in X-direction So, work done = (1/y) Fy.dy =(1/y) W Edited August 27, 2014 by mahesh khati Link to comment Share on other sites More sharing options...

xyzt Posted August 27, 2014 Share Posted August 27, 2014 I am done correcting your errors. Bye! 1 Link to comment Share on other sites More sharing options...

swansont Posted August 27, 2014 Share Posted August 27, 2014 ! Moderator Note And with that we're closed.mahesh khati, do not open a new thread on the subject. You have gotten a tremendous amount of constructive feedback at the expense of others' time and effort. Use it. 2 Link to comment Share on other sites More sharing options...

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