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maheshkhati

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maheshkhati last won the day on May 10

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  1. I am very sure, if Einstein develop general theory of relativity first & then see toward local theory of relativity. He will develop it in my way. Big object not only bend the space but create the space. Vacuum is different than space. Particle is local vibrating charge spike in a space as given in last chapter OR world is the interaction of big charge with local charges. Many scientist knows that Einstein special theory of relativity is wrong & they try to prove it wrong by experiment but relativity wins because time & mass are the properties of space. They do not remain same. So
  2. https://youtu.be/W9yWv5dqSKk This video beautifully explain the particle, wave duality. This also proves that this is only possible when particle (Droplet) moving in some medium then only duality is possible. I consider the space as given in 2nd chapter of my paper https://vixra.org/pdf/2006.0033v1.pdf act as medium through which particle travel. Last chapter explain how electromagnetic vibration happen in particle level.
  3. In relativity force in X-direction is mathematically calculated as Fx =d/dt (m ux) = dm/dt . ux + m . du/dt = dm/dt . ux + m . ax Let, consider fighter plane with horizontal velocity ux drop the bomb B & Observer is on ground For observer :- Bomb B will move with constant horizontal velocity ux & accelerate vertically due to gravity. So, Force applied on Bomb is gravity only in vertical direction. but one mathematical acting force is created in horizontal direction i.e. Fx= dm/dt .ux + m. (0) =dm/dt .ux as mass of Bomb increases due to vertical acceleration. So, applied force is dif
  4. Mathematics in post 4 indicate that step 1 proves that when applied force in x-direction is 0 then force acting in x-direction Fx= 0+y3 mo. (ux/c2} uy ay= y3 mo. (ux/c2} uy ay ---------------------------------------------------------------------------------------------------------------------------------------- STEP 2 proves that applied force in x-direction is -Fx = -y3 mo. (ux/c2} uy ay then force acting in x-direction Fx= -y3 mo. (ux/c2} uy ay +y3 mo. (ux/c2} uy ay =0 ------------------------------------------------------------------------------------------------------------------------
  5. Thanks, you understand something. You are true. Relativity requires more force in the x direction since it increases momentum but mathematics in post 4 proves that the additional force is not the applied force but mathematical force. & That additional force is not responsible for acceleration in that direction. That is the problem. OR mathematics in post 4 proves that more rate of change of moment happen than applied force in that direction due to combination of velocity in that direction & acceleration perpendicular to that direction.
  6. Just consider that one man is at location (x,y) is pulling Box on friction less platform with rope with force F whose component in x-direction & y-direction are Fx & Fy respe. Calculation of relativity math in Post 4 suggest that actual force acting on Box is more i.e. Fx+Fmy & Fy+Fmx. Due to more force , more work done happen. So, less consumption of energy create more work or more energy in relativity. This is against consistency of energy in that frame.
  7. Simple differentiation is impossible for you & you are trying to oppose me. OK E^2 = m0^2c^4+p^2c^2 Now, I do it for you 2 E . dE/dt = c^2 . 2. p. dp/dt dE/dt = c^2 (p/E). dp/dt But ...P/E =v/c^2 So, dE/dt =v. dp/dt dE= (v.dt) (dp/dt) But ...(v.dt) =ds dE= ds (dp/dt) But dp/dt =rate of change of momentum = force dE= F. ds Remember rest energy has no contribution in doing work. It is constant ( m0^2c^4) & it's differentiation is zero. ---------------------------------------------------------------------------------------------------------------------------------- so, Mathemati
  8. I already said that formulas may be different but physics remain same. Even, your said formula E^2 = m0^2c^4+p^2c^2 Can be converted into dE=F.ds Just differentiate above & solve little, final output is dE= dP/dt . ds & dP/dt is rate of change of momentum or force. So, dE= F . ds --------------------------------------------------------------------------------------------------------------------- Force is reality. We apply force & do work. If applied force is less than acting force then energy produce will be more than energy consume. ​this is against consistency of energy
  9. 1) In any frame:- Force in x-direction = Rate of change of momentum in x direction In non-proper frame F’x = d/dt’( y’. mo. U’x) where y’=(1-u’2/c2)-0.5 (Rate of change of momentum in x’-direction) If this differentiation extended by proper transformation equations of frame like putting equations of U’x, y’ & d/dt’ values then we get F’x = Fx + ( v/c2 . Fy. Uy)/(1-V .Ux/c2) ----transformation equation in relativity. This proves that in any frame Force in x-direction = Rate of change of momentum in x direction 2) This transformation equation also indicate the same math for exam
  10. You just solve the equation, E^2 = m0^2c^4+p^2c^2 by putting value of E = y mo c^2 you get the same equation i.e. p= mo v . (1-v2/c2)^0.5 so, both equations are same ----------------------------------------------------------------------------------------------------------------- I already proved that in special relativity, the theory is so close that by taking one equation like F=dp/dt you can prove lot of equations of relativity like transformation equations of relativity, dE=y. dEo & dM= y dMo etc So, do not paste other equation of relativity because all equations are inter link wi
  11. I am pasting the content from that site Special theory of relativity In the Special relativity, mass and energy are equivalent (as can be seen by calculating the work required to accelerate an object). When an object's velocity increases, so does its energy and hence its mass equivalent (inertia). It thus requires more force to accelerate it the same amount than it did at a lower velocity. F=dp/dt remains valid because it is a mathematical definition.But for relativistic momentum to be conserved, it must be redefined as: p= mo v . (1-v2/c2)^0.5 (modified because image extension can
  12. No, in relaticity F=dp/dt =d/dt(mu) =dm/dt . u + m . du/dt = dm/dt . u + m a So, in relativity F=ma is wrong. For Example, in x-direction Fx=d/dt(m ux) =dm/dt . ux + m . dux/dt = dm/dt . u + m ax so, Fx=m ax is wrong Total detail calculation is given in post 4
  13. You are referring Newtonian mathematics:- F=ma is same to F=dp/dt in Newtonian mathematics. In relativity :- F is not equal to ma but F=dp/dt
  14. This is not frame dragging effect. This happen due to simple mathematics of wrong special theory of relativity. In special relativity, force is not related to acceleration but it is rate of change of moment. This is very simple, I explain it with simple mathematics with simple event. In relativity force in X-direction is mathematically calculated as Fx =d/dt (m ux) = dm/dt . ux + m . dux/dt = dm/dt . ux + m . ax Let, consider fighter plane with horizontal velocity ux drop the bomb B & Observer is on ground For observer :- Bomb B will move with constant horizontal velocity ux & acce
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