xyzt Posted August 3, 2014 Share Posted August 3, 2014 (edited) Consider, observer is in rest train on the platform & on the platform one old man is pulling cart in perpendicular direction to length of train then in that frame of reference the work done by old man = force x displacement of cart on platform in that rest frame I think above mathematics is simple & clear. . Yes, you are demonstrating that you do not understand the Lorentz transforms. This is typical of people that claim "relativity is wrong". Now, Consider train is moving with constant velocity V then for observer in train watching the same event then due to relativity. Force (force perpendicular to train motion) decreases & displacement remain same as contraction of space is not happening in that direction. So, product of force & displacement due to force applied by old man in that frame decrease. So, for observer in train work done by old man decreases Yep, work, energy, momentum, force are not frame invariant. You haven't learned that yet? Edited August 3, 2014 by xyzt Link to comment Share on other sites More sharing options...

mahesh khati Posted August 4, 2014 Author Share Posted August 4, 2014 Force, energy, momentum, work are not frame invarient means Force, energy, momentum, work are frame varient both has same meaing & I proved that using mathematics of standard test book also refer math 3 & 4 of chapter 1 .Only problem is that force decreases with increasing velocity. To understand it is also simple refer the unit of force i.e. N or kg.m/s^{2}.. This clearly indicate that relative change in mass can not get balanced due to the relative change is time. This is not my mathematics but mathematics given in standard text book it also gives the same result. Link to comment Share on other sites More sharing options...

Strange Posted August 4, 2014 Share Posted August 4, 2014 This is not my mathematics but mathematics given in standard text book it also gives the same result. Only when you do it wrong. Are you really so arrogant to think that you have spotted a flaw in a theory that is more than 100 years old and has been analysed and rigorously tested by thousands of people? Unbelievable. Link to comment Share on other sites More sharing options...

CaptainPanic Posted August 4, 2014 Share Posted August 4, 2014 ! Moderator Note mahesh khati, please observe again the rules of our speculations forum, especially the 1st rule, which states: "Speculations must be backed up by evidence or some sort of proof. If your speculation is untestable, or you don't give us evidence (or a prediction that is testable), your thread will be moved to the Trash Can. If you expect any scientific input, you need to provide a case that science can measure." Also, I'd like to quote on of my favorite blogs: "Of course science has overturned earlier notions of how the Universe works. But sometimes, those rules are shown to be true so much and so often that when you come up with an idea that overthrows all of it, you’d better have iron-clad evidence of it." So far, it seems other members are not satisfied with your evidence. It is your task to improve this. Everybody, please also observe the other rules of the Speculations forum, especially the 2nd rule which states: "Be civil. As wrong as someone might be, there is no reason to insult them, and there's no reason to get angry if someone points out the flaws in your theory, either." Do not respond to this moderator note. Link to comment Share on other sites More sharing options...

xyzt Posted August 4, 2014 Share Posted August 4, 2014 (edited) Only problem is that force decreases with increasing velocity. And why is that a problem? Anyway, your calculations are incorrect. Here is how work transforms between frames: [math]W=dE[/math] where E represents the total energy [math]E=\frac{m_0c^2}{\sqrt{1-u^2/c^2}}[/math] wehere u is the speed of the onject of mass [math]m_0[/math] measured in frame F. So: [math]W=\frac{m_0c^2 u du}{\sqrt{1-u^2/c^2}^3}=m_0c^2 (u_xdu_x+u_ydu_y+u_zdu_z) \frac{1}{\sqrt{1-u^2/c^2}^3}[/math] Without any loss of generality, we can consider [math]u_y=u_z=0[/math] so: [math]W=m_0c^2 u_xdu_x\frac{1}{\sqrt{1-u_x^2/c^2}^3}[/math] If we want to transform the above into frame F' moving at speed V in the x direction we need to make the Lorentz transform for [math]u_x[/math]: [math]u_x=\frac{u'_x+V}{1+u'_xV/c^2}[/math] [math]du_x=\frac{(1-(u'_x/c)^2)(u'_x+V)du'_x}{(1+u'_xV/c^2)^2}[/math] [math]\frac{1}{\sqrt{1-u_x^2/c^2}}=\gamma(V)\gamma(u'_x)(1+\frac{u'_xV}{c^2})[/math] So: [math]W=m_0c^2 (u'_x+V) du'_x\gamma^3(V) \gamma(u'_x)[/math] If , on the other hand , V is in the y direction, then W transforms as: [math]W=m_0c^2 u'_xdu'_x\gamma^{-2}(V)\frac{1}{\sqrt{1+(u'_xV/c^2)^2-(u'_x/c)^2}}=W'\gamma^{-2}(V)\frac{1}{\sqrt{1+(u'_xV/c^2)^2-(u'_x/c)^2}}[/math] So, when V increases, the work W , decreases. Edited August 4, 2014 by xyzt Link to comment Share on other sites More sharing options...

CaptainPanic Posted August 4, 2014 Share Posted August 4, 2014 (edited) ! Moderator Note xyzt, when I addressed everybody regarding polite behavior, that included you too. The 2nd and 3rd sentence in your post [before you edited and improved it] add zero useful information to this post. Do not respond to this moderator note. [edit] To avoid confusion, it is noted that xyzt has edited (and improved) the above post since this mod note was posted. Edited August 5, 2014 by CaptainPanic Link to comment Share on other sites More sharing options...

Ophiolite Posted August 4, 2014 Share Posted August 4, 2014 The number of scientific theories that have turned out to be completely wrong are very few and far between. Phlogiston might be the most recent example... Prior to the development of plate tectonic theory, orogenesis - mountain building - was attributed variously to Earth contraction, Earth expansion, undation and likely a few more besides. Most remained in vogue in some quarters until the end of the 1960s, by which time the evidence and theoretical underpinning for plate tectonics was solid. Link to comment Share on other sites More sharing options...

anonymousone Posted August 4, 2014 Share Posted August 4, 2014 relativity is a theory made to detatch the masses from a logical understanding of the universe. its a real conspiracy -7 Link to comment Share on other sites More sharing options...

Bignose Posted August 4, 2014 Share Posted August 4, 2014 relativity is a theory made to detatch the masses from a logical understanding of the universe. its a real conspiracy LOL, world's worst conspiracy. Hey, let's conspire to keep the masses from "understanding the universe", but let's publish review papers laying out all the evidence that supports it: http://arxiv.org/abs/1403.7377 The greys and the illuminati are just shaking their heads in pity. 1 Link to comment Share on other sites More sharing options...

imatfaal Posted August 5, 2014 Share Posted August 5, 2014 ! Moderator Note relativity is a theory made to detatch the masses from a logical understanding of the universe. its a real conspiracy Do not derail threads with conspiracy theories. Any responses to this moderation will be hidden. Link to comment Share on other sites More sharing options...

mahesh khati Posted August 6, 2014 Author Share Posted August 6, 2014 (edited) And why is that a problem? Anyway, your calculations are incorrect. Here is how work transforms between frames: [math]W=dE[/math] where E represents the total energy [math]E=\frac{m_0c^2}{\sqrt{1-u^2/c^2}}[/math] wehere u is the speed of the onject of mass [math]m_0[/math] measured in frame F. So: [math]W=\frac{m_0c^2 u du}{\sqrt{1-u^2/c^2}^3}=m_0c^2 (u_xdu_x+u_ydu_y+u_zdu_z) \frac{1}{\sqrt{1-u^2/c^2}^3}[/math] Without any loss of generality, we can consider [math]u_y=u_z=0[/math] so: [math]W=m_0c^2 u_xdu_x\frac{1}{\sqrt{1-u_x^2/c^2}^3}[/math] If we want to transform the above into frame F' moving at speed V in the x direction we need to make the Lorentz transform for [math]u_x[/math]: [math]u_x=\frac{u'_x+V}{1+u'_xV/c^2}[/math] [math]du_x=\frac{(1-(u'_x/c)^2)(u'_x+V)du'_x}{(1+u'_xV/c^2)^2}[/math] [math]\frac{1}{\sqrt{1-u_x^2/c^2}}=\gamma(V)\gamma(u'_x)(1+\frac{u'_xV}{c^2})[/math] So: [math]W=m_0c^2 (u'_x+V) du'_x\gamma^3(V) \gamma(u'_x)[/math] If , on the other hand , V is in the y direction, then W transforms as: [math]W=m_0c^2 u'_xdu'_x\gamma^{-2}(V)\frac{1}{\sqrt{1+(u'_xV/c^2)^2-(u'_x/c)^2}}=W'\gamma^{-2}(V)\frac{1}{\sqrt{1+(u'_xV/c^2)^2-(u'_x/c)^2}}[/math] So, when V increases, the work W , decreases. Means your calculation gives the same thing i.e. as V increases Work done or consumption of energy decreases as I say. Thanks. Using formula as I use Work done in that frame= Force in that frame X displacement in that frame is also not wrong in relativity for example Work W = F X dr .....where dr is displacement is used by Einstein to prove his famous equation E = mc^{2} If you prove oppsite that work done do not decreases by increasing velovity V of train then I will be wrong. Thanks Complete theory of relativity started by considering the velocity of light in any reference frame is constant. All theory is so much interdependent on one another that this theory becomes palace of cards. Even Einstein know that very well. So, he says that several proof is not sufficient to prove me correct but one single proof is sufficient to prove me wrong. So, even if we take out one card from that palace then whole palace of card will get collapsed. So to say Einstein may be wrong in this contrain & may be true in other constrain is wrong. Relativity may be wrong or correct. Both things correct at one time is not possible. 1991, When I was in Engineer College one of my professor of physics challenge me to prove Einstein is wrong logically. I just go to black board & draw rectangle ABCD. I say consider, I have pulleys A, B, C, D at four corner of rectangle which is mounted by one string. Now, consider string has started revolving around four pulleys then what will happen by relativity. 1) As pulleys are at rest with relative to me their position can not be changed by relativity. But 2) String is moving with velocity V. So, molecular distance in string has to decrease or length of string (sides of rectangle) at any moment has to decrease with relative to me. Now, logically without changing location of pulley how can you change length of string? This is completely wrong. This is something like corner of rectangle remain as it is & length of side of rectangle decreases. I can create many such situations where we can prove that relativity is logically wrong but now we can prove it wrong mathematically.. (At the place of string you may consider flow of electrons & at the place of pulleys you may consider magnetic deflector. ) Thanks Edited August 6, 2014 by mahesh khati Link to comment Share on other sites More sharing options...

Ophiolite Posted August 6, 2014 Share Posted August 6, 2014 Means your calculation gives the same thing i.e. as V increases Work done or consumption of energy decreases as I say. Thanks. Using formula as I use Work done in that frame= Force in that frame X displacement in that frame is also not wrong in relativity for example Work W = F X dr .....where dr is displacement is used by Einstein to prove his famous equation E = mc^{2} If you prove oppsite that work done do not decreases by increasing velovity V of train then I will be wrong. Thanks Complete theory of relativity started by considering the velocity of light in any reference frame is constant. All theory is so much interdependent on one another that this theory becomes palace of cards. Even Einstein know that very well. So, he says that several proof is not sufficient to prove me correct but one single proof is sufficient to prove me wrong. So, even if we take out one card from that palace then whole palace of card will get collapsed. So to say Einstein may be wrong in this contrain & may be true in other constrain is wrong. Relativity may be wrong or correct. Both things correct at one time is not possible. 1991, When I was in Engineer College one of my professor of physics challenge me to prove Einstein is wrong logically. I just go to black board & draw rectangle ABCD. I say consider, I have pulleys A, B, C, D at four corner of rectangle which is mounted by one string. Now, consider string has started revolving around four pulleys then what will happen by relativity. 1) As pulleys are at rest with relative to me their position can not be changed by relativity. But 2) String is moving with velocity V. So, molecular distance in string has to decrease or length of string (sides of rectangle) at any moment has to decrease with relative to me. Now, logically without changing location of pulley how can you change length of string? This is completely wrong. This is something like corner of rectangle remain as it is & length of side of rectangle decreases. I can create many such situations where we can prove that relativity is logically wrong but now we can prove it wrong mathematically.. (At the place of string you may consider flow of electrons & at the place of pulleys you may consider magnetic deflector. ) Thanks Have you heard of compression and extension? Link to comment Share on other sites More sharing options...

mahesh khati Posted August 6, 2014 Author Share Posted August 6, 2014 I heard but here space contraction due to relative velocity is applicable. For example intermolecular distance in velocity direction of thread will decrease or length of thread or side will only reduce. Link to comment Share on other sites More sharing options...

Bignose Posted August 6, 2014 Share Posted August 6, 2014 I can create many such situations where we can prove that relativity is logically wrong but now we can prove it wrong mathematically.. So, I'm going to go back to my (completely ignored) request in post #9 in this thread. Using http://arxiv.org/abs/1403.7377, can you please recreate their graphs showing observations, the current predictions from GR, and the prediction made by your model? Now that you claim to be able to 'prove it [GR] wrong mathematically' it should be super easy for you to demonstrate that your model does it 'right' and since it is more right it should make even better predictions than what we have. If there are any questions about this request, please let me know. 2 Link to comment Share on other sites More sharing options...

xyzt Posted August 6, 2014 Share Posted August 6, 2014 (edited) Means your calculation gives the same thing i.e. as V increases Work done or consumption of energy decreases as I say. Thanks. Not the "same calculation". Mine is correct, yours is wrong. Using formula as I use Work done in that frame= Force in that frame X displacement in that frame is also not wrong in relativity for example Work W = F X dr .....where dr is displacement is used by Einstein to prove his famous equation E = mc^{2} If you prove oppsite that work done do not decreases by increasing velovity V of train then I will be wrong. Thanks In order to "prove relativity wrong" you need to understand relativity. You don't. Work is frame variant, you need to learn that, I just showed you the proof. Complete theory of relativity started by considering the velocity of light in any reference frame is constant. All theory is so much interdependent on one another that this theory becomes palace of cards. Even Einstein know that very well. So, he says that several proof is not sufficient to prove me correct but one single proof is sufficient to prove me wrong. So, even if we take out one card from that palace then whole palace of card will get collapsed. So to say Einstein may be wrong in this contrain & may be true in other constrain is wrong. Relativity may be wrong or correct. Both things correct at one time is not possible. The only way to disprove relativity is by constructing an experiment and RUNNING it. You are under the delusion that your INCORRECT CALCULATIONS "disprove" relativity. They don't since they only reflect your lack of knowledge. 1991, When I was in Engineer College one of my professor of physics challenge me to prove Einstein is wrong logically. I just go to black board & draw rectangle ABCD. I say consider, I have pulleys A, B, C, D at four corner of rectangle which is mounted by one string. Now, consider string has started revolving around four pulleys then what will happen by relativity. 1) As pulleys are at rest with relative to me their position can not be changed by relativity. But 2) String is moving with velocity V. So, molecular distance in string has to decrease or length of string (sides of rectangle) at any moment has to decrease with relative to me. Now, logically without changing location of pulley how can you change length of string? This is completely wrong. This is something like corner of rectangle remain as it is & length of side of rectangle decreases. I can create many such situations where we can prove that relativity is logically wrong but now we can prove it wrong mathematically.. (At the place of string you may consider flow of electrons & at the place of pulleys you may consider magnetic deflector. ) Thanks Yes, I know you can. RUN them. Come back when you have the result. For the time being, all you have is your basic misunderstandings. Edited August 6, 2014 by xyzt Link to comment Share on other sites More sharing options...

Strange Posted August 6, 2014 Share Posted August 6, 2014 (edited) All theory is so much interdependent on one another that this theory becomes palace of cards. Quite the reverse. It is like complex scaffold with all parts supporting all other parts. So, I suspect that even if you could show that a fundamental part were wrong, the rest of the structure would be fine. Of course, if you could show that a fundamental part were wrong ... But you haven't. So, he says that several proof is not sufficient to prove me correct but one single proof is sufficient to prove me wrong. Do you have any evidence, beyond your poor grasp of the theory? Edited August 6, 2014 by Strange 1 Link to comment Share on other sites More sharing options...

mahesh khati Posted August 7, 2014 Author Share Posted August 7, 2014 And why is that a problem? Anyway, your calculations are incorrect. Here is how work transforms between frames: [math]W=dE[/math] where E represents the total energy [math]E=\frac{m_0c^2}{\sqrt{1-u^2/c^2}}[/math] wehere u is the speed of the onject of mass [math]m_0[/math] measured in frame F. So: [math]W=\frac{m_0c^2 u du}{\sqrt{1-u^2/c^2}^3}=m_0c^2 (u_xdu_x+u_ydu_y+u_zdu_z) \frac{1}{\sqrt{1-u^2/c^2}^3}[/math] Without any loss of generality, we can consider [math]u_y=u_z=0[/math] so: [math]W=m_0c^2 u_xdu_x\frac{1}{\sqrt{1-u_x^2/c^2}^3}[/math] sorry please check your mathematics again. Sorry but in physics, even in relativity in any mathematical equation which we form. If we put units of all quantity in right side & left side both must be equal for example in E = m c^{2} sum of right side units & left side unit gives same result for example in E = m c^{2} N. m = kg. m^{2}/s^{2} kg , m/s^{2} . m = kg . m^{2}/s^{2} kg. m^{2}/s^{2 }=^{}kg . m^{2}/s^{2} L.H.S = R.H.S equation may be any, like famous mass equation of relativity then also this thing is true but in your expression of work this fails to do that put the units of work, C, u du you find that kg. m^{2}/s^{2 }=^{}kg . m^{4}/s^{4} means, L.H.S. is not equal to right hand side Same problem remain in your concluding expression also. 2nd point :- If your concluding expression is applied to problem of old man pulling cart in direction perpendicular to motion of train Consider man pulled cart 1 m then work done will be different If man pulled cart 10 m then work done will be more Means work is depend on displacement but your final expression do not address the displacement. It only depend on V & ux Please chech all math again. Sorry Relativities famous equation E = m c^{2} is proved by Einstein starting from expression, Work = F . dr where dr is displacement in that frame (I can give page no of that text book also) then to say using Work done = force in that frame x displacement in that frame by me is wrong. This is not correct. Mathematical proof & experimental findings are mutually related. Many times some thing is proved by mathematics & we find that it is experimently perfect after many years. So, if you want to prove me wrong mathematically then that is correct way to oppose me mathematically. Link to comment Share on other sites More sharing options...

Klaynos Posted August 7, 2014 Share Posted August 7, 2014 So, I'm going to go back to my (completely ignored) request in post #9 in this thread. Using http://arxiv.org/abs/1403.7377, can you please recreate their graphs showing observations, the current predictions from GR, and the prediction made by your model? Now that you claim to be able to 'prove it [GR] wrong mathematically' it should be super easy for you to demonstrate that your model does it 'right' and since it is more right it should make even better predictions than what we have. If there are any questions about this request, please let me know. ! Moderator Note mahesh khati, if you don't sound to the above quoted post you may be in breach of speculations rule 1 and the thread closed. I think we'd all like to see this plot. Link to comment Share on other sites More sharing options...

xyzt Posted August 7, 2014 Share Posted August 7, 2014 (edited) sorry please check your mathematics again. My calculations are correct. Yours are not. And why is that a problem? Anyway, your calculations are incorrect. Here is how work transforms between frames: [math]W=dE[/math] where E represents the total energy [math]E=\frac{m_0c^2}{\sqrt{1-u^2/c^2}}[/math] wehere u is the speed of the onject of mass [math]m_0[/math] measured in frame F. So: [math]W=\frac{m_0 u du}{\sqrt{1-u^2/c^2}^3}=m_0 (u_xdu_x+u_ydu_y+u_zdu_z) \frac{1}{\sqrt{1-u^2/c^2}^3}[/math] Without any loss of generality, we can consider [math]u_y=u_z=0[/math] so: [math]W=m_0 u_xdu_x\frac{1}{\sqrt{1-u_x^2/c^2}^3}[/math] If we want to transform the above into frame F' moving at speed V in the x direction we need to make the Lorentz transform for [math]u_x[/math]: [math]u_x=\frac{u'_x+V}{1+u'_xV/c^2}[/math] [math]du_x=\frac{(1-(u'_x/c)^2)(u'_x+V)du'_x}{(1+u'_xV/c^2)^2}[/math] [math]\frac{1}{\sqrt{1-u_x^2/c^2}}=\gamma(V)\gamma(u'_x)(1+\frac{u'_xV}{c^2})[/math] So: [math]W=m_0 (u'_x+V) du'_x\gamma^3(V) \gamma(u'_x)[/math] If , on the other hand , V is in the y direction, then W transforms as: [math]W=m_0 u'_xdu'_x\gamma^{-2}(V)\frac{1}{\sqrt{1+(u'_xV/c^2)^2-(u'_x/c)^2}}=W'\frac{\gamma^{-2}(V)\gamma^{-3}(u_x)}{\sqrt{1+(u'_xV/c^2)^2-(u'_x/c)^2}}[/math] So, when V increases, the work W , decreases. There was an extraneous [math]c^2[/math] Means work is depend on displacement but your final expression do not address the displacement. It only depend on V & ux Please chech all math again. Sorry [math]m_0 u du[/math] has dimensions of energy ([math]m_0 c^2[/math]), so you are wrong. Edited August 7, 2014 by xyzt Link to comment Share on other sites More sharing options...

Strange Posted August 7, 2014 Share Posted August 7, 2014 (edited) My calculations are correct. .... There was an extraneous [math]c^2[/math] So your calculations were correct but you just wrote them down wrong. You wouldn't let someone promoting their own theory get away with that! Edited August 7, 2014 by Strange 3 Link to comment Share on other sites More sharing options...

xyzt Posted August 7, 2014 Share Posted August 7, 2014 (edited) So your calculations were correct but you just wrote them down wrong. It was a typo, I forgot to take away the [math]c^2[/math]. Shrug. Edited August 7, 2014 by xyzt Link to comment Share on other sites More sharing options...

studiot Posted August 8, 2014 Share Posted August 8, 2014 1991, When I was in Engineer College one of my professor of physics challenge me to prove Einstein is wrong logically. I just go to black board & draw rectangle ABCD. I say consider, I have pulleys A, B, C, D at four corner of rectangle which is mounted by one string. Now, consider string has started revolving around four pulleys then what will happen by relativity. 1) As pulleys are at rest with relative to me their position can not be changed by relativity. But 2) String is moving with velocity V. So, molecular distance in string has to decrease or length of string (sides of rectangle) at any moment has to decrease with relative to me. Now, logically without changing location of pulley how can you change length of string? This is completely wrong. This is something like corner of rectangle remain as it is & length of side of rectangle decreases. I can create many such situations where we can prove that relativity is logically wrong but now we can prove it wrong mathematically.. (At the place of string you may consider flow of electrons & at the place of pulleys you may consider magnetic deflector. ) Thanks Since no one seems to have taken this statement up I offer that it is really a variation of Paul Ehrenfest's Paradox, which has been explained, in terms of relativity. http://en.wikipedia.org/wiki/Ehrenfest_paradox Link to comment Share on other sites More sharing options...

mahesh khati Posted August 11, 2014 Author Share Posted August 11, 2014 (edited) My calculations are correct. Yours are not. There was an extraneous [math]c^2[/math] [math]m_0 u du[/math] has dimensions of energy ([math]m_0 c^2[/math]), so you are wrong. I have just seen your post. I will see it in detail.afterward because civil work is going on ( on my site) but I think mathematics in the beginning is true. because if I different E = mo.c^{2} (1-u^{2}/c^{2})^{-1/2} with respect to u both side then dE/du = mo . u .(1-u^{2}/c^{2})^{-3/2} If I multiply both side by du then out put will be similar to your mathematics dE = mo . u .(1-u^{2}/c^{2})^{-3/2 }du this is your out put without C^{2 (}which was wrong). now, left side units = right side unit but this dE = change in energy per unit change in velocity (this velocity is also time dependent function) We are not interested in energy consumption for unit change of velocity in my event but in my event Displacement is fixed in both frame & force is fixed in one frame but relatively vary in second frame & as Force in stable frame = Force in moving frame . ((1-u^{2}/c^{2})^{-1/2} Work done in stable frame = work done . in moving frame . ((1-u^{2}/c^{2})^{-1/2} Energy consumption in in stable frame = Energy consumption in moving frame . ((1-u^{2}/c^{2})^{-1/2} ^{Now, if you want to apply differentiation then it will be like} ^{dW = dF . S} ^{dW/dF = S = constant in both frames} ^{means if force decreases then work done has to decrease to make S constant in different frames given in my event} ^{thank you} ^{I will post again} Edited August 11, 2014 by mahesh khati Link to comment Share on other sites More sharing options...

xyzt Posted August 11, 2014 Share Posted August 11, 2014 (edited) I have just seen your post. I will see it in detail.afterward because civil work is going on ( on my site) but I think mathematics in the beginning is true. because if I different E = mo.c^{2} (1-u^{2}/c^{2})^{-1/2} with respect to u both side then dE/du = mo . u .(1-u^{2}/c^{2})^{-3/2} If I multiply both side by du then out put will be similar to your mathematics dE = mo . u .(1-u^{2}/c^{2})^{-3/2 }du this is your out put without C^{2 (}which was wrong). now, left side units = right side unit Good, you have learned something. but this dE = change in energy per unit change in velocity (this velocity is also time dependent function) dE represents WORK. By definition. So, there is something new that you need to learn. Work done in stable frame = work done . in moving frame . ((1-u^{2}/c^{2})^{-1/2}Energy consumption in in stable frame = Energy consumption in moving frame . ((1-u^{2}/c^{2})^{-1/2} ^{This is incorrect, you are misapplying relativity in order to "prove it wrong"}. I have shown you how to do the calculations correctly, you just repeated the correct calculations at the beginning of this post, now you are reverting to your initial errors. Now, if you want to apply differentiation then it will be like ^{dW = dF . S} ^{dW/dF = S = constant in both frames} False. Relativity teaches you that length (S) is NOT "constant in both frames". Length is frame variant. You need to LEARN relativity before you try to "disprove" it. What you are really disproving is your MISCONCEPTIONS about relativity, not relativity itself. Edited August 11, 2014 by xyzt Link to comment Share on other sites More sharing options...

mahesh khati Posted August 11, 2014 Author Share Posted August 11, 2014 (edited) Good, you have learned something. Now, we do the math again your mathematics in the beginning is true. because if I differentiat E = mo.c^{2} (1-u^{2}/c^{2})^{-1/2} with respect to u both side then dE/du = mo . u .(1-u^{2}/c^{2})^{-3/2} If I multiply both side by du then out put will be similar to your mathematics dE = mo . u .(1-u^{2}/c^{2})^{-3/2 }du Now, I improvevise your math from here because their is no displacement & acceleration in your caiculation 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 As du/dt=a =acceleration & ds = u.dt =displacement dE =γ^{3}. mo. a . ds now, in relativity force parallel to direction of mation is F= γ^{3}. mo. a this is very well known to ererybody 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 of relativity also' Now, in my all event created displacement happens perpendicular to the motion of frames to avoid complecated mathematic. So, displacement remain same because space contraction due to relative motion is not happen on that direction. that is why displacement is constant. Now, it is too late 2.41 of night good night thanks & wil meet you next day. Read my mathematics in my paper. I have taken care of everythings. Edited August 11, 2014 by mahesh khati Link to comment Share on other sites More sharing options...

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