Anamitra Palit

Important revisions/corrections have been made with the 'Extra Bit'.Relevant portion has been provided in Latex. The entire file[revised file] has been uploaded. Metric \begin{equation}c^2d\tau^2=c^2dt^2-dx^2-dy^2-dz^2 \end{equation} (1) \begin{equation}c^2=c^2\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dx}{d\tau}\right)^2-\left(\frac{dy}{d\tau}\right)^2-\left(\frac{dz}{d\tau}\right)^2\end{equation} \begin{equation}c^2=c^2{v_t}^2-{v_x}^2-{v_y}^2-{v_z}^2\end{equation}(2) We consider two proper velocities on the same manifold \begin{equation}c^2=c^2{v_{1t}}^2-{v_{1x}}^2-{v_{1y}}^2-{v_{1z}}^2\end{equation}(3.1) \begin{equation}c^2=c^2{v_{2t}}^2-{v_{2x}}^2-{v_{2y}}^2-{v_{2z}}^2\end{equation}(3.2) Adding (3.1) and (3.2) we obtain \begin{equation}2c^2=c^2\left({v_{1t}}^2+{v_{2t}}^2\right)-\left({v_{1x}}^2+{v_{2x}}^2\right)-\left({v_{1y}}^2+{v_{2y}}^2\right)-\left({v_{1z}}^2+{v_{2z}}^2\right)\end{equation} \begin{equation}2c^2=c^2\left({v_{1t}}+{v_{2t}}\right)^2-\left({v_{1x}}+{v_{2x}}\right)^2-\left({v_{1y}}+{v_{2y}}\right)^2-\left({v_{1z}}+{v_{2z}}\right)^2-2v_1. v_2\end{equation} \begin{equation}2c^2+2v_1. v_2=c^2\left({v_{1t}}+{v_{2t}}\right)^2-\left({v_{1x}}+{v_{2x}}\right)^2-\left({v_{1y}}+{v_{2y}}\right)^2-\left({v_{1z}}+{v_{2z}}\right)^2\end{equation} Since $$v1.v2\ge c^2 $$ we have \begin{equation}c^2\left({v_{1t}}+{v_{2t}}\right)^2-\left({v_{1x}}+{v_{2x}}\right)^2-\left({v_{1y}}+{v_{2y}}\right)^2-\left({v_{1z}}+{v_{2z}}\right)^2\ge 4c^2\end{equation}(4) \begin{equation}\left(v_1+v2\right ).\left(v_1+v_2\right)\ge 4c^2\end{equation} If $v_1+v_2$ is a proper velocity then \begin{equation}c^2=c^2\left(v_{1t}+v_{2t})\right)^2-\left(v_{1x}+v_{2x})\right)^2-\left(v_{1y}+v_{2y})\right)^2-\left(v_{1z}+v_{2z})\right)^2\end{equation} \begin{equation}c^2=c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2+ c^2v_{2t}^2-v_{2x}^2-v_{2y}^2-v_{2z}^2+2v_1.v_2\end{equation} \begin{equation}c^2=c^2+c^2+2v_1.v_2\end{equation} Therefore \begin{equation}v_1.v_2\le -½ c^2\end{equation} which is not true since \begin{equation}v.v=c^2\end{equation} Therefore $v_1+v_2$ is not a four vector if $v_1$ and $v_2$ are four vectors Again if $v_1-v_2$ is a four vector then \begin{equation}c^2=c^2\left(v_{1t}-v_{2t})\right)^2-\left(v_{1x}-v_{2x})\right)^2-\left(v_{1y}-v_{2y})\right)^2-\left(v_{1z}-v_{2z})\right)^2\end{equation} \begin{equation}c^2=c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2+ c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2-2v_1.v_2\end{equation} \begin{equation}c^2=c^2+c^2-2v_1.v_2\end{equation} \begin{equation} ½ c^2=v_1.v_2\end{equation} But the above formula is not a valid one. Given two infinitesimally close four velocities their difference is not a four velocity. Therefore the manifold has to be a perforated one. The manifold indeed is a mesh of worldlines and each world line is a train of proper velocity four vectors as tangents. A particle moves along a timelike path and therefore each point on it has a four velocity tangent representing the motion. The manifold is discrete and that presents difficulty with procedure like differentiation. Four352.pdf

Important revisions have been implemented in the file on Four velocity and Four momenta. The revised file has been uploaded.[pl see the section "The Extra Bit" for the revision] Four200.pdf

Time Conflict 5 does not indicate any error in the theory. The paper on Four Velocity and Four Momentum on modification indicates error in theory. "The Extra Bit " by itself is indicative of an error Four AAXYY.pdf Four50.pdf

In relation to the last post We may always choose the eight unknowns unknowns:v1_i and v2_j with each i and j=1,2,3,4 , in such a manner that the next three equations hold c^2=c^2v1_t^2-v1_x^2-v1_y^2-v1_z^2 c^2=c^2v2_t^2-v2_x^2-v2_y^2-v2_z^2+2v1.v2 and c^2=c^2(v_1+v2_t)^2_-(v1_x-v2_x)^2-(v1_y-v2_y)^2-(v1_z-v2_z)^2 They will lead to 2v1 .v2<=-c^2

I would definitely be using the Latex editor.Thanks for your suggestion. In the mean time you may consider the following c^2dtau^2=c^2dt^2-dx^2-dy^2-dz^2 c^2=c^2[dt/dtau]^2-[dx/dtau]^2-[dy/dtau]^2-[dz/dtau]^2 c^2=c^2v_t^2-v_x^2-v_y^2-v_z^2 (1) v_i are proper speeds and as such they can exceed the speed of light without hurting or violating Special Relativity For two proper velocities v1 and v2at the same point of the manifold.Since tensors are additive we have c^2=c^2(v1_t+v2_t)^2-(v1_x+v2_x)^2-(v1_y+v2_y)^2-(v1_z+v2_z)^2(2) or,c^2=c^2v1_t^2-v1_x^2-v1_y^2-v1_z^2+c^2v2_t^2-v2_x^2-v2_y^2-v2_z^2+2v1.v2 or, c^2=c^2+c^2+2v1.v2 v1.v2=-c^2 (3) By calculations we have arrived at an untenable result. Indeed v1,v2 and v=v1+v2 all satisfy(1) and hence heir existence is certified by Special relativity or even by General Relativity for that matter. An analogous result may be obtained on the General Relativity context c^2dtau^2=c^2 g_tt dt^2-g_xx dx^2-g_yy dy^2-g_zz dz^2 (4) We consider transformations g_tt dt^2=dT^2, g_xxdx^2=dX^2, g_yydy^2=dY^2,g_zzdz^2=dZ^2 (5) Local or even transformations over infinitesimally small regions would suffice. Equations (4) and (5) combined gives us the flat space time metric[mathematical form of it] c^2dtau^2=c^2 dT^2- dX^2- dY^2- dZ^2 (6) All conclusions we made earlier follow. Incidentally, there is one point to take note of:the Lorentz transformations follow from (6) in a unique manner [Reference; Steve Wienberg,Gravitation and Cosmology,Chapter 2:Special Relativity]

I would definitely be using the Latex editor.Thanks for your suggestion. In the mean time you may consider the following c^2dtau^2=c^2dt^2-dx^2-dy^2-dz^2 c^2=c^2[dt/dtau]^2-[dx/dtau]^2-[dy/dtau]^2-[dz/dtau]^2 c^2=c^2v_t^2-v_x^2-v_y^2-v_z^2 (1) v_i are proper speeds and as such they can exceed the speed of light without hurting or violating Special Relativity For two proper velocities v1 and v2at the same point of the manifold.Since tensors are additive we have c^2=c^2(v1_t+v2_t)^2-(v1_x+v2_x)^2-(v1_y+v2_y)^2-(v1_z+v2_z)^2(2) or,c^2=c^2v1_t^2-v1_x^2-v1_y^2-v1_z^2+c^2v2_t^2-v2_x^2-v2_y^2-v2_z^2+2v1.v2 or, c^2=c^2+c^2+2v1.v2 v1.v2=-c^2 (3) By calculations we have arrived at an untenable result. An analogous result may be obtained on the General Relativity context

t' would be different in the two cases (1)for (4.1) and (4.2) and(2) for (5.1),(5.2).For the first case t' is counted from the event of the origins of K and K' coinciding. For the second case t' is counted from the event of the origins of K'' and K' coinciding. That has led to an erroneous inference .This paper dos not indicate at any error in theory.

In response to the last comment made by Ghideon: Both t and t'' relate to K or to K' since they are relatively at rest.But t and t''are not identical.. t is counted from the event the origins of the reference frames K and K' meet[when the pass against each other while t'' is counted from the event when the origins of K' and K'' pass against each other.T The controversy projected through the article is hinged on the fact that the left sides of the pair (4.1) ,(4.2) and (5.1),(5.2) are x' and t' in each case. That leads to equations (6.1) and (6.2) and from there to (7) and (7') where the conflict is very much clear.

The inequalities derived in this article are correct:they follow from the reversed Cauchy inequality as suggested in an earlier posting[one may follow the derivation provided in the article] Four dot product v1.v2>=c^2 (1) [v1 and v2 represent four velocities] Four dot product p1.p1>m1 m2 c^2 (2) [p1 and p2 represent energy momentum vectors] But the final equation that causes the conflict is not correct With equation (11), v1 and v2 are ordinary speeds. In the next line v1 and v2 are proper speeds.That led to the error in the final equation ta is with (13) [Equations (1) and (2) are identical in that we may pass from one to the other. We may derive them independently as done in the article or we may derive one from the other] FourXXX.pdf

The last file with greater clarity has been uploaded again: Link to Google Drive file has also been provided. I will write in Latex at my earliest opportunity Time Conflict 5.pdf

We consider the Reversed Cauchy Inequality. For distinct two distinct vectors(t1,x1,y1,z1) and (t2,x2,y2,z2) c^2t1 t2-x1 x2-y1y2-z1 z2>Sqrt[c^2 t1^2-x1^2-y1^2-z1^2]Sqrt[c^2t2^2-x2^2-y2^2-z2^2] Link: https://en.wikipedia.org/wiki/Minkowski_space#Norm_and_reversed_Cauchy_inequality Replacing x^i by dx^i and then dividing both sides by d tau^2[tau proper time] we arrive at v1.v2>c^2[for distinct velocities](1) m1v1.m2v2>=m1 m2 c^2[m1 and m2 are rest masses of the two particles masses] Now p(i th component)=m_0 dx^i/d tau=mdx^i/dt [m_0 is rest mass;m; relativistic mass] p1.p2>m1 m2c^2(2) Formulas (1) and (2) match with what we have in the article The formula represented by(2) leads to the conflicting result [for v1 not equal to v2 v1 v2>c^2[gamma1 gamma2 -1+1/[gamma gamma2] Letting v1 tends to v2 without becoming equal we do have in the limit limit[in the absence of a discontinuity] v.v=c^2[gamma^2-1+1/gamma^2 These formulae resulting from the reversed Cauchy inequality have been gloriously overlooked by people for so many years. In fact I came to know of the reversed Cauchy inequality just yesterday.Relativity was at stake[unknowingly under the nose of people] for so many years.Quite amazing! A simpler method for arriving at contradictions/ errors in Special Relativity has been described in the uploaded article[mathematical in nature. Incidentally I can't find the code button <> which used to be there earlier Time_Conflict.pdf

I could not find the delete button for this post. Nevertheless I might repost the last one with suitable modifications

1)My paper does not claim delta_mu rho or delta_mu sigma to be tensors let alone being covariant tensors of rank two. Since the Kronecker delta, delta^a_b has the unique property of its components being invariant in all frames of reference, it may be used to represent a scalar which could be one or zero depending on whether a=b or a not equal to b (respectively).The notations delta_mu rho or delta_mu sigma in the paper have to be interpreted in such a sense. 2)The formula below: a1b1-a2b2>=Sqrt[a1^2-a2^2] Sqrt[b1^2-b2^2] with (a12^-a2^2)(a12^-a2^2)>=0 it has been deduced four dot product alpha.beta>=c^2 where alpha and beta are four velocities In the natural units c=1 Therefore, alpha.beta>=1 I did not say alpha.beta(four dot product)-|alpha||beta|>=1 or anything equivalent to it. In fact joigus writes alpha=1,beta=1 . But alpha and beta are four vectors. The objections raised by joigus do not stand to any reason. 1)Markus Hanke:" I have already shown you above that the absolute norm of a 4-velocity is always exactly equal to c. It can‘t be any different of course." The norm of a four velocity is always c. That is a fact. But the paper considered the inner product of two four velocities to obtain v1.v2>=c^2 Markus Hanke:" The mistake you made is assuming that 4-velocities can be any arbitrary 4-vector, but that isn‘t true - all 4-velocities are 4-vectors, but not all 4-vectors are automatically 4-velocities." 2)All four vectors are not four velocities. That is true and the paper has considered that. c^v1_t^2-|v1|^2=c^2 implying Sqrt[c^v1_t^2-|v1|^2]=c c^v2_t^2-|v2|^2=c^2 implying Sqrt[c^v2_t^2-|v2|^2]=c have been used in the paper. The result derived is four dot product,v1.v2>=c^2 We must not forget the equality sign while setting v1=v2. 3)Equation in point (3) v.v>=c^2[gamma^2-1+1/gamma^2] has been deduced by consistent use of mathematics on conventional theory. Its contradictory nature points to failures in conventional theory. Google dive link to paper:https://drive.google.com/file/d/1a5Sd5Vfhbg0gqERIFDcjxXmIjZXw6ulr/view?usp=sharing [For ready reckoning the google drive link has been provided;the file has also been attached Four Four 5.pdf

The mathematical framework has to be applied keeping in the mind the conditions of its application Logic is based on premises. From these premises inferences are made.By experimentation we arrive at a set of rules/laws. These are the premises. Considering them to be accurate we arrive at mathematical derivations which should hold accurately. If they do not hold accurately it means that our premises were inaccurate or incorrect.So long as the premises are not challenged mathematical derivations have to be correct.If mathematical derivations do not give us correct results we have to rethink the very foundations where the premises lie. Newtonian mechanics considers masses in slow motion in experiments. Laws were deduced accordingly.Mathematical derivations based on such laws are correct[sufficiently accurate] so long as slow moving masses are considered. Physics would follow mathematics if such conditions are maintained. Mathematical derivations based on the classical laws could be wrong if fast moving bodies are considered. Following the criterion of slow moving bodies if derivations are not physically valid then we are not to blame mathematics for it.But we are to blame incorrectness in the formulation of the laws. It does not happen like that for Newton's laws.:maintaining the criterion of slow moving bodies derivations are be correct[sufficiently accurate].Newton's laws are fine[sufficiently accurate] so long as we maintain their conditions of application.Whenever we think of a law we have to keep in mind the conditions of its application. While performing mathematical derivations involving the law concerned we have to be careful that the conditions of application of the law are taken care of.With all that if a mathematical derivation is physically untenable then we are to blame physics, in that the law itself requires revision and not mathematics.

The Kronecker delta,delta_b^a is a mixed tensor of rank two. It enjoys a unique property that its components are scalars[invariants]. They are either zero or one in their respective positions no matter what be the frame of reference.The values of these components do not change on transformation. Therefore the Kronecker delta may be used as a tensor [as it is from its transformation properties] or alternatively its components as a scalars having two possible values: zero or one. The scalar interpretation as having two values one or zero may be represented as delta_b^a:delta_b^a=1 if a=b;delta_b^a=0 if a not equal to b In this article and delta_mu,pho and delta_mu,sigma have to be interpreted with such a meaning. Link to the google drive incorporating this idea into the original file https://drive.google.com/file/d/15IZZoMEBSb59WN4SMFAX_9Y6HSQs6_BV/view?usp=sharing Mathematics is the language of physics. Mathematics follows logic. Physics since it has nothing to do with miracles also goes by logic. Therefore both the subjects should fit into each other in a coherent an a comprehensive manner. Whenever mathematics talks of some relevant logic physics cannot[should not] override it.Rather physics is expected to follow the logical conclusions ensuing mathematically. Enigma_Tensors.pdf

The mathematical formulas expressed by (3) have been applied to obtain the two formulas referred to by Ghideon. The application technique has been shown just after the results. And physics ,of course, is expected to cater to mathematics. One of the options[inequations] has to be valid. Link to google drive: https://drive.google.com/file/d/1a5Sd5Vfhbg0gqERIFDcjxXmIjZXw6ulr/view?usp=sharing

Difficulties with the theory of Tensors: The transformation of a rank two covariant tensor has been considered. Then we proceed to consider a diagonal tensor[off diagonal components are zero:A^(mu nu)=0 for mu not equal to nu] to bring out a result that all tensors should be null tensors. A link to the google drive file has been provided.A file has also been attached considering the file attachment facility that has been provided by the forum. https://drive.google.com/file/d/10z63Xidgs3m8p04_C6ZiGh-8Q6KTwPsh/view?usp=sharing Incidentally I tried the Latex with the code button.But I am not getting the correct preview. Example \begin{equation}\bar{A}^{\mu\nu}=\frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}}\frac {\partial \bar{x}^{\nu}}{\partial x^{\beta}}A^{\alpha \beta}\end {equation} \[{\ bar{A}}^{\mu\nu}=\frac{\partial {\bar{x}}^{\mu}}{\partial x^{\alpha}}\ frac{\partial {\bar{x}}^{\nu}}{\partial x^{\beta}}\] Currently for a long time I am not a frequent user of Latex thanks to the equation bar of MS Word. But I do appreciate,like many others, the application of Latex in various forums.Help is being requested from the forum regarding Latex.

Markus Hanke's Points have been clearly refuted. He could not find errors with my calculations. Something has to be wrong with the conventional theory: Some General Types of Difficulties with the theory of Tensors: The transformation of a rank two covariant tensor has been considered. Then we proceed to consider a diagonal tensor[off diagonal components are zero:A^(mu nu)=0 for mu not equal to nu] to bring out a result that all tensors should be null tensors. A link to the google drive file has been provided.A file has also been attached considering the file attachment facility that has been provided by the forum. https://drive.google.com/file/d/10z63Xidgs3m8p04_C6ZiGh-8Q6KTwPsh/view?usp=sharing Incidentally I tried the Latex with the code button.But I am not getting the correct preview. Example \begin{equation}\bar{A}^{\mu\nu}=\frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}}\frac {\partial \bar{x}^{\nu}}{\partial x^{\beta}}A^{\alpha \beta}\end {equation} \[{\ bar{A}}^{\mu\nu}=\frac{\partial {\bar{x}}^{\mu}}{\partial x^{\alpha}}\ frac{\partial {\bar{x}}^{\nu}}{\partial x^{\beta}}\] Currently for a long time I am not a frequent user of Latex thanks to the equation bar of MS Word. But I do appreciate,like many others, the application of Latex in various forums.Help is being requested from the forum regarding Latex.

The transformation of a rank two covariant tensor has been considered. Then we proceed to consider a diagonal tensor[off diagonal components are zero:A^(mu nu)=0 for mu not equal to nu] to bring out a result that all tensors should be null tensors. A link to the google drive file has been provided.A file has also been attached considering the file attachment facility that has been provided by the forum. https://drive.google.com/file/d/10z63Xidgs3m8p04_C6ZiGh-8Q6KTwPsh/view?usp=sharing Incidentally I tried the Latex with the code button.But I am not getting the correct preview. Example \begin{equation}\bar{A}^{\mu\nu}=\frac {\partial \bar{x}^{\mu}}{\partial x^{\alpha}}\frac {\partial \bar{x}^{\nu}}{\partial x^{\beta}}A^{\alpha \beta}\end {equation} \[{\ bar{A}}^{\mu\nu}=\frac {\partial {\bar{x}}^{\mu}}{\partial x^{\alpha}}\ frac{\partial {\bar{x}}^{\nu}}{\partial x^{\beta}}\] Currently for a long time I am not a frequent user of Latex thanks to the equation bar of MS Word. But I do appreciate,like many others, the application of Latex in various forums.Help is being requested from the forum regarding Latex.Thanking in advance for help provided.. Tensors_Enigma.pdf

Makus Henke has remarked; "Photons have no rest mass, so according to your expression, the inner product of photon momentum with itself is zero. Since the inner product can vanish only if the two vectors are either perpendicular, or one of the vectors has zero magnitude, that means that according to you the photon has no net momentum." Norm of the four momentum of a photon[real photon] is indeed zero. This is the mass-shell condition for the real photon.But for the spatial part of the photon ,norm of momentum is not zero[this portion in italics is a revision over the last post] Indeed, E^2-c^2|p|^2=m_0^2c^2[m_0 is the rest mass] (m_0 gamma)^2 c^2 -(m_0 gamma()^2=m_0^2c^2 The rest mass m_0 for the photon is zero . WE cannot cancel zero from the two sides. We simply write (m_0 gamma)^2 c^2 -(m_0 gamma)^2=0 For the photon m_0 gamma is of the form zero * infinity.Considering it to be a non zero finite quantity [this italized prt bis an addition to the earlier post]we have E^2=c^2|p|^2 or E=|p|c From quantum mechanics E=hc/lambda |p|c=hc/lambda or,|p|=h/lambda,a non zero value Markus Henke has remarked "Just repeating the same thing again does not make it any less wrong. " If he finds anything wrong with my paper he should point to it in a specific manner. So far I have refuted everything has has considered incorrect with my paper Quote

Thank you very much Ghideon for the information.Incidentally, can you write files of the length I have attached in that manner using the sandbox section?If the whole file is on the editor in rich text [with formulas] viewers would have an easier access to my writings Nevertheless there is something strange out here: why does the forum at all maintain the facility of attaching files if it is so suspicious about them. Links to files on the google drive, I hope, should not be considered to be of a suspicious nature.

Markus Henke: remarked "Non-sequitur. If you arrive at some kind of “discrepancy”, then that means you did something wrong, plain and simple. Tensor calculus is not a “theory”, it’s a mathematical framework that has been extensively developed, and is fully self-consistent." The important point is that the conflict exists as projected in the paper. Can he refute the contradiction shown in the paper by finding errors with my calculation? It is clear that Markus Henke has has not found any error with my calculations which,incidentally, are not of a lengthy nature. If standard theory happens to be correct everywhere, as Markus Henke believes in, he should be able to point directly to errors in my calculations. If he really such found errors he would have been vocal about them.

You do have the option of attaching files ! Nevertheless a link to my google drive has been provided: https://drive.google.com/file/d/1ncNsohI1LbqWcdZY-zvsQU8ND34gPPf6/view?usp=sharing Incidentally if you have the option of latex facilities it would be convenient for me.It would be difficult to express mathematical calculations in plain text in many situations[as with my file ]

Makus Henke has remarked; "Photons have no rest mass, so according to your expression, the inner product of photon momentum with itself is zero. Since the inner product can vanish only if the two vectors are either perpendicular, or one of the vectors has zero magnitude, that means that according to you the photon has no net momentum." Four momentum of a photon[real photon] is zero. This is the mass-shell condition for the real photon.But the spatial part of the photon momentum is not zero Indeed, E^2-c^2|p|^2=m_0^2c^2[m_0 is the rest mass] (m_0 gamma)^2 c^2 -(m_0 gamma()^2=m_0^2c^2 The rest mass m_0 for the photon is zero . WE cannot cancel zero from the two sides. We simply write (m_0 gamma)^2 c^2 -(m_0 gamma()^2=0 For the photon m_0 gamma is of the form zero * infinity. We have E^2=c^2|p|^2 or E=|p|c From quantum mechanics E=hc/lambda |p|c=hc/lambda or,|p|=h/lambda,a non zero value Markus Henke has remarked "Just repeating the same thing again does not make it any less wrong. " If he finds anything wrong with my paper he should point to it in a specific manner. So far I have refuted everything has has considered incorrect with my paper