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mahesh khati

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About mahesh khati

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  1. 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.
  2. 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
  3. 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.
  4. 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.
  5. 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.
  6. 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.
  7. 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.)
  8. 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).
  9. 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?
  10. but in above paradox, I have not consider uy=0. (& same calculation is true in Y-direction for same conditions.)
  11. 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?
  12. Means, we can not differentiate in S' frame.
  13. 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
  14. 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
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