devin553344 8 Posted June 29, 2020 Report Share Posted June 29, 2020 YES:- You are missing some thing. Mathematical derivatives always contain variables as it is applicable to all parameters & conditions.In above example, I prove that in post 4 or in post 13 that in single frame analysis or in two relative frame analysis.Even there is no force is applied in X-direction then also relativity create mathematical non-applied force in x-direction.So, effect is applied force is different than acting force.------------------------------------------------------------------------------------------------------------------------------------------------------------------------------In Post 13, I prove that if we do not consider that non-applied horizontal force in to the mathematical calculation then d'E=dEo/y, d'M=dMo/y if we do consider that non-applied horizontal force in to the mathematical calculation then only d'E=y dEo, d'M=y dMoMeans, non-applied mathematical horizontal force is responsible for d'E=y dEo, d'M=y dMo-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Means, If Bomber drop the Bomb Mo kg moving with horizontal velocity then for Observer on ground actual force acting is gravitational force but relativity create the mathematical horizontal force.If we do not consider non-applied horizontal force then d'E=y dEo, d'M=y dMo is wrong.This horizontal force is created due to vertical acceleration & horizontal velocityI think I get what you saying finally. I think what you need to look at is frame dragging: https://en.wikipedia.org/wiki/Frame-dragging I think that is the x direction force? Quote Link to post Share on other sites

maheshkhati 5 Posted June 30, 2020 Author Report Share Posted June 30, 2020 (edited) 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 asFx =d/dt (m ux) = dm/dt . ux + m . dux/dt = dm/dt . ux + m . axLet, consider fighter plane with horizontal velocity ux drop the bomb B & Observer is on groundFor 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 different than acting force in relativity.& even horizontal acceleration is zero but there is acting horizontal force with displacement.-----------------------------------------------------------------------------------------This is serious problem. It's effect is, more acting force produce more work.So, energy produce is more than energy consume in the same frame for observer.Detail other calculations are already given. Edited June 30, 2020 by maheshkhati Quote Link to post Share on other sites

devin553344 8 Posted June 30, 2020 Report Share Posted June 30, 2020 (edited) 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.I don't see how you've proven anything wrong with relativity, you keep talking about some horizontal force. There is none? So exactly what are you trying to say? Where's this new horizontal force coming from that you speak of? Because you haven't demonstrated a new horizontal force. And by the way I highly doubt relativity states that force is not related to acceleration. Again what are you saying? It sounds like you don't understand math and the use of derivatives. They're nothing more than an accurate form of algebra. F=ma is the same as F=dp/dt. And yes relativity must be applied here, everyone knows that, so what exactly are you doing? Edited June 30, 2020 by devin553344 Quote Link to post Share on other sites

maheshkhati 5 Posted July 1, 2020 Author Report Share Posted July 1, 2020 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 Quote Link to post Share on other sites

devin553344 8 Posted July 1, 2020 Report Share Posted July 1, 2020 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/dtF=dp/dt is the accurate implementation of F=ma. Quote Link to post Share on other sites

maheshkhati 5 Posted July 2, 2020 Author Report Share Posted July 2, 2020 (edited) No,in relaticityF=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-directionFx=d/dt(m ux) =dm/dt . ux + m . dux/dt = dm/dt . u + m axso, Fx=m ax is wrongTotal detail calculation is given in post 4 Edited July 2, 2020 by maheshkhati Quote Link to post Share on other sites

devin553344 8 Posted July 2, 2020 Report Share Posted July 2, 2020 (edited) No,in relaticityF=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-directionFx=d/dt(m ux) =dm/dt . ux + m . dux/dt = dm/dt . u + m axso, Fx=m ax is wrongTotal detail calculation is given in post 4F=ma is not wrong in relativity, it is still correct. You really shouldn't be saying it's wrong if you want people to think you understand the equations your presenting. Why don't you read up on force, here this article covers everything you need to know: https://en.wikipedia.org/wiki/Force Edited July 2, 2020 by devin553344 Quote Link to post Share on other sites

maheshkhati 5 Posted July 3, 2020 Author Report Share Posted July 3, 2020 (edited) I am pasting the content from that siteSpecial theory of relativityIn 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 not be pasted) Edited July 3, 2020 by maheshkhati Quote Link to post Share on other sites

devin553344 8 Posted July 3, 2020 Report Share Posted July 3, 2020 (edited) I am pasting the content from that siteSpecial theory of relativityIn 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 not be pasted)Why are you copy-pasting the article, it doesn't prove your point. But it does prove mine? ie: "F=dp/dt remains valid because it is a mathematical definition." Granted the concept of longitudinal mass is confusing. You should abandon force and go with energy, ie: "E^2 = m0^2c^4+p^2c^2" Or at least that's what I got from the article here: https://en.wikipedia.org/wiki/Mass_in_special_relativity#Transverse_and_longitudinal_mass This is an interesting topic to me as it may relate to my theory. Thanks for the discussion. Edited July 3, 2020 by devin553344 Quote Link to post Share on other sites

maheshkhati 5 Posted July 6, 2020 Author Report Share Posted July 6, 2020 You just solve the equation,E^2 = m0^2c^4+p^2c^2by putting value of E = y mo c^2you get the same equationi.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 equationlike F=dp/dt you can prove lot of equations of relativity like transformation equations of relativity, dE=y. dEo & dM= y dMo etcSo, do not paste other equation of relativity because all equations are inter link with another or same.--------------------------------------------------------------------------------------------------------------------Force is the fundamental quantity in physics. We can apply force, do work, consume energy.& dE=dw=F.ds , F =dp/dt is not wrong in relativity.Physics remain same, formulas may be different. Quote Link to post Share on other sites

devin553344 8 Posted July 6, 2020 Report Share Posted July 6, 2020 You just solve the equation,E^2 = m0^2c^4+p^2c^2by putting value of E = y mo c^2you get the same equationi.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 equationlike F=dp/dt you can prove lot of equations of relativity like transformation equations of relativity, dE=y. dEo & dM= y dMo etcSo, do not paste other equation of relativity because all equations are inter link with another or same.--------------------------------------------------------------------------------------------------------------------Force is the fundamental quantity in physics. We can apply force, do work, consume energy.& dE=dw=F.ds , F =dp/dt is not wrong in relativity.Physics remain same, formulas may be different.What you're saying here does not sound correct. Quote Link to post Share on other sites

VictorMedvil 349 Posted July 6, 2020 Report Share Posted July 6, 2020 What you're saying here does not sound correct.I told you Maheshkhati is a huge crank Quote Link to post Share on other sites

maheshkhati 5 Posted July 8, 2020 Author Report Share Posted July 8, 2020 (edited) 1) In any frame:-Force in x-direction = Rate of change of momentum in x directionIn non-proper frameF’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 getF’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 exampleIn prime frame Fx=0 , ux=0 & ax=0Then also in non –prime frame F’x is not 0But F’x = Fx + ( v/c2 . Fy. Uy)/(1-V .Ux/c2) ----transformation equation in relativity. F’x= v/c2 . Fy. Uy3) Same math can be used to provedw’=y. dw or dE= y. dEo or dM=y. dMo So, I am not wrong.------------------------------------------------------------------------------------------------All this is not possible by your formula F=ma as it is wrong in relativity. or this is approximate formula & In relativity Direction of acceleration & force is not same. Edited July 8, 2020 by maheshkhati Quote Link to post Share on other sites

devin553344 8 Posted July 9, 2020 Report Share Posted July 9, 2020 I told you Maheshkhati is a huge crank Crank may or may not be correct if I could figure out what he's saying. But most of his statements come across as incoherent rambling, lol. Quote Link to post Share on other sites

maheshkhati 5 Posted July 10, 2020 Author Report Share Posted July 10, 2020 (edited) I already said that formulas may be different but physics remain same.Even, your said formula E^2 = m0^2c^4+p^2c^2Can 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 Mathematics In post 4 is completely true. If you want to prove me wrong then you have to prove math of post 4 wrong. Edited July 10, 2020 by maheshkhati Quote Link to post Share on other sites

devin553344 8 Posted July 12, 2020 Report Share Posted July 12, 2020 (edited) I already said that formulas may be different but physics remain same.Even, your said formula E^2 = m0^2c^4+p^2c^2Can 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 First, your units appear wrong, perhaps that's what's going on here? Are you using ds to represent space? Second I think you got the idea wrong, rest energy isn't part of momentum. Notice how they're represented in different areas of the equation. Separated by a plus sign. 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 Mathematics In post 4 is completely true. If you want to prove me wrong then you have to prove math of post 4 wrong.Yeah and that is exactly what you didn't prove. I should be seeing an example of how two energies don't match up using values. Something you wouldn't produce for us. You just want to spit out formulas and say you're right. More incoherent rambling. Lastly, no one has to prove you wrong, remember you're the one trying to prove Einstein wrong. Please keep things straight here. Edited July 12, 2020 by devin553344 Quote Link to post Share on other sites

maheshkhati 5 Posted July 13, 2020 Author Report Share Posted July 13, 2020 (edited) Simple differentiation is impossible for you & you are trying to oppose me. OKE^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/dtBut ...P/E =v/c^2So, dE/dt =v. dp/dtdE= (v.dt) (dp/dt)But ...(v.dt) =dsdE= ds (dp/dt)But dp/dt =rate of change of momentum = forcedE= F. ds Remember rest energy has no contribution in doing work. It is constant ( m0^2c^4) & it's differentiation is zero. ----------------------------------------------------------------------------------------------------------------------------------so, Mathematics in post 4 is not wrong. you may try to do that. Edited July 13, 2020 by maheshkhati Quote Link to post Share on other sites

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