mahesh khati Posted February 20, 2016 Posted February 20, 2016 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?
mahesh khati Posted February 22, 2016 Author Posted February 22, 2016 (edited) but in above paradox, I have not consider uy=0. (& same calculation is true in Y-direction for same conditions.) Edited February 22, 2016 by mahesh khati
swansont Posted February 22, 2016 Posted February 22, 2016 but in above paradox, I have not consider uy=0. Yes, you did. "object is moving with constant velocity ux"
mahesh khati Posted February 25, 2016 Author Posted February 25, 2016 (edited) 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? Edited February 25, 2016 by mahesh khati
swansont Posted February 25, 2016 Posted February 25, 2016 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 How do you achieve a magnetic force in the y direction only? Anyway, I suspect your mistake is trying to break down the force into components before taking the derivative.
mahesh khati Posted February 29, 2016 Author Posted February 29, 2016 (edited) 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). Edited February 29, 2016 by mahesh khati
swansont Posted February 29, 2016 Posted February 29, 2016 I suspect your mistake is trying to break down the force into components before taking the derivative.
Strange Posted February 29, 2016 Posted February 29, 2016 HERE, more energy & force is the problem. Where this additional energy (& force) does comes from? Clearly, you have made an error somewhere. Are you asking for help in finding it?
mahesh khati Posted March 2, 2016 Author Posted March 2, 2016 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.)
Strange Posted March 2, 2016 Posted March 2, 2016 So, calculation given in post is true Obviously not because you get the wrong answer.
Strange Posted March 2, 2016 Posted March 2, 2016 please, pointed out where I am wrong. Why? Whenever people explain your mistakes you either ignore them or insist that you are right and the rest of the world is wrong (as in this thread). It is a waste of time trying to help you.
swansont Posted March 2, 2016 Posted March 2, 2016 please, pointed out where I am wrong. I have done this twice. Here's the third time: I suspect your mistake is trying to break down the force into components before taking the derivative. IOW, find the net force by taking the derivative of the momentum, and then break it down into components. 1
imatfaal Posted March 2, 2016 Posted March 2, 2016 The relativistic expression relating force and acceleration for a particle with constant non-zero rest mass moving in the direction is: From wiki
mahesh khati Posted March 3, 2016 Author Posted March 3, 2016 (edited) 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. Edited March 3, 2016 by mahesh khati
Mordred Posted March 3, 2016 Posted March 3, 2016 (edited) You posted the solutions page, is there something you don't understand by this statement? "Hence in order to accelerate a body in a given direction, we may apply any force in the desired direction, but must at the same time apply at right angles another force whose magnitude is given by equation. 5. https://en.wikisourc...nd_Acceleration Edited March 3, 2016 by Mordred 1
mahesh khati Posted March 4, 2016 Author Posted March 4, 2016 (edited) 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. Edited March 4, 2016 by mahesh khati
Mordred Posted March 4, 2016 Posted March 4, 2016 OK well I did some digging I beleive this page has the equations your looking for. https://en.wikisource.org/wiki/Some_Transformation_Equations
mahesh khati Posted March 5, 2016 Author Posted March 5, 2016 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.
mahesh khati Posted March 15, 2016 Author Posted March 15, 2016 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.
Strange Posted March 15, 2016 Posted March 15, 2016 After nearly one month, there is no solution to above problem in S.R. More accurately, after nearly one month you still haven't understood your error.
Phi for All Posted March 15, 2016 Posted March 15, 2016 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. ! Moderator Note This is completely unacceptable. You were shown where your calculations may be wrong, and have done nothing about them, yet you insist you're right and no one has offered a solution to your problem. We can't help fight ignorance when you purposely ignore learning anything. Thread closed due to insufficient rigor.
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