On Four Velocity and Four Momentum

[Anamitra Palit]

6 hours ago, joigus said:

In fact, a Kronecker delta that is twice covariant or twice contravariant is not an isotropic tensor either. It must be 1-covariant 1-contravariant. Also arbitrary tensor products of 1-covariant 1-contravariant Kronecker deltas is an isotropic tensor. Arbitrary products are not. That's because contravariant indices transform with the inverse matrix with respect to covariant ones (that's why they're called "contra"). If you multiply twice by the same matrix you don't get back to Kronecker deltas.

You must go carefully through all these checks in order not to make elementary mistakes. It's a natural rite of passage.

The literature is full of mistakes of this nature. Not in the really prestigious books, of course.

No. Mathematical physics is expected to cater to physics. Mathematics doesn't need any imput from physics.

Mathematical physics is expected to be self-consistent, and further, it is expected to comply with what we measure in the laboratory.

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

Edited December 22, 2020 by Anamitra Palit

[Ghideon]

16 minutes ago, Anamitra Palit said:

Whenever mathematics talks of some relevant logic physics cannot[should not] override it.Rather physics is expected to follow the logical conclusions ensuing mathematically.

There are plenty of examples in physics where a certain mathematical framework works for some cases and not for some others. Newtonian physics vs special relativity is one such example. The physics does not have to follow the logical conclusions of the mathematics developed by Newton. 

[Anamitra Palit]

5 hours ago, Ghideon said:

There are plenty of examples in physics where a certain mathematical framework works for some cases and not for some others. Newtonian physics vs special relativity is one such example. The physics does not have to follow the logical conclusions of the mathematics developed by Newton. 

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.

[Ghideon]

7 hours ago, Anamitra Palit said:

The mathematical framework has to be applied keeping in the mind the conditions of its application 

I asked you to show that for the inequalities you used. Please provide an answer.

Here is a link to the unanswered question: scienceforums.net/.../comment=1163549

 

Edited December 23, 2020 by Ghideon

[Markus Hanke]

On 12/22/2020 at 6:55 AM, Anamitra Palit said:

Markus Hanke's Points have been clearly refuted. He could not find errors with my calculations.

You haven‘t even addressed them yet, let alone refuted anything. 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 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. The spatial and temporal part of a 4-vector need to be related in a specific way (as I showed above), or else it isn‘t a 4-velocity. When you take this into account, you end up with an equality, not “>=“. The same is of course true for 4-momenta, since \(p^{\mu}=mu^{\mu}\). 

The algebraic expressions you derived on the RHS of (1) and (2) are correct, but the “>=“ inequality is not.

Point (3) is trivially wrong, since the norm of a 4-vector is an invariant, and hence cannot functionally depend \(\gamma\).

On 12/22/2020 at 6:55 AM, Anamitra Palit said:

Some General Types of Difficulties with the theory of Tensors:

You are just copy-pasting the same stuff over and over again, without addressing anything that is being said to you. 

Edited December 23, 2020 by Markus Hanke

[joigus]

15 hours ago, Anamitra Palit said:

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.

That's what I said. Are you repeating what I said?

Only the property is not unique. Any tensor product of 1-covariant 1-contravariant tensor also has that property. And same with epsilon tensors (in that case the components are 1, 0, and -1).

The Kronecker delta \( \left. \delta^{\mu} \right._{\nu}\) is an isotropic tensor. The Kronecker deltas \( \delta^{\mu\nu}\), \(\delta_{\mu\nu} \) are not.

On the other hand, from your document (ineq. 1), it does not follow, as you say, that,

\[ \alpha\beta -\left| \boldsymbol{\alpha} \right| \left| \boldsymbol{\beta} \right| \geq 1 \]

(expressed in a lighter notation). Your ineq. 1), e.g., would be,

\[ \alpha\beta -\left| \boldsymbol{\alpha} \right| \left| \boldsymbol{\beta} \right| \geq \sqrt{ \alpha^2 - \left|  \boldsymbol{\alpha} \right|^2 } \sqrt{\beta^2 - \left| \boldsymbol{\beta} \right|^2 } \]

The above expression does not follow, as this counterexample shows:

Pick \( \alpha = 1 = \left| \boldsymbol{\alpha} \right| \); \( \beta = 1 = \left| \boldsymbol{\beta} \right| \), but,

\[ \alpha\beta -\left| \boldsymbol{\alpha} \right| \left| \boldsymbol{\beta} \right| =0 < 1\]

So you're wrong here.

Other mistakes have been pointed out to you repeatedly. Time to go back to a relativity book and do the exercises.

 

Edited December 23, 2020 by joigus
improving LateX

[Anamitra Palit]

17 hours ago, joigus said:

That's what I said. Are you repeating what I said?

Only the property is not unique. Any tensor product of 1-covariant 1-contravariant tensor also has that property. And same with epsilon tensors (in that case the components are 1, 0, and -1).

The Kronecker delta δμν is an isotropic tensor. The Kronecker deltas δμν , δμν are not.

On the other hand, from your document (ineq. 1), it does not follow, as you say, that,

 

αβ−|α||β|≥1

 

(expressed in a lighter notation). Your ineq. 1), e.g., would be,

 

αβ−|α||β|≥α2−|α|2−−−−−−−−√β2−|β|2−−−−−−−√

 

The above expression does not follow, as this counterexample shows:

Pick α=1=|α| ; β=1=|β| , but,

 

αβ−|α||β|=0<1

 

So you're wrong here.

Other mistakes have been pointed out to you repeatedly. Time to go back to a relativity book and do the exercises.

 

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.

 

18 hours ago, Markus Hanke said:

You haven‘t even addressed them yet, let alone refuted anything. 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 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. The spatial and temporal part of a 4-vector need to be related in a specific way (as I showed above), or else it isn‘t a 4-velocity. When you take this into account, you end up with an equality, not “>=“. The same is of course true for 4-momenta, since pμ=muμ . 

The algebraic expressions you derived on the RHS of (1) and (2) are correct, but the “>=“ inequality is not.

Point (3) is trivially wrong, since the norm of a 4-vector is an invariant, and hence cannot functionally depend γ .

You are just copy-pasting the same stuff over and over again, without addressing anything that is being said to you. 

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

Edited December 24, 2020 by Anamitra Palit

[joigus]

4 hours ago, Anamitra Palit said:

In fact joigus writes alpha=1,beta=1 . But alpha and beta are four vectors.

No. There is a time component alpha and the corresponding Euclidean norm of the 3-vector alpha (that's why I wrote it in boldface). Same for beta. IOW, there are four numbers whose value I've chosen to be 1, producing a very clear counterexample of your assertion.

It's very common notation in relativity.

\[ \left( \alpha^{\mu} \right) = \left( \alpha, \boldsymbol{\alpha} \right) \]

Do you understand the notation now?

The question about constraints for physical 4-vectors, that Markus, Ghideon and I have told you about, still stands.

As I said, Anamitra Palit must go back to elementary relativity books.

Edited December 24, 2020 by joigus

[Anamitra Palit]

1 hour ago, joigus said:

No. There is a time component alpha and the corresponding Euclidean norm of the 3-vector alpha (that's why I wrote it in boldface). Same for beta. IOW, there are four numbers whose value I've chosen to be 1, producing a very clear counterexample of your assertion.

It's very common notation in relativity.

 

(αμ)=(α,α)

 

Do you understand the notation now?

The question about constraints for physical 4-vectors, that Markus, Ghideon and I have told you about, still stands.

As I said, Anamitra Palit must go back to elementary relativity books.

Goigus:

"IOW, there are four numbers whose value I've chosen to be 1, producing a very clear counterexample of your assertion."

1. Time component of velocity,v_t=1,other components:v_x=v_y=v_z=1[c=1,natural units]

Therefore norm-square of v for each velocity =1^2-1^2-1^2-1^2=-2 that is norm^2  of velocity 4-vector=-2c^2 if we are not using the natural units. The norm^2 of each velocity four vector should be c^2 [signature of metric(+,-,-,-)]

In so far as the example of Giogus. The norm of v is not correct.

2. If Goigus means there  for each vector  v_t=1 and |v|=1 then norm square,v.v for each velocity=1^2-1^2=0: [norm alpha]^2=0;[norm beta]^2=0

Again this will not hold.

Goigus should reconsider his views.There is uncompromisable reason why I don't require to go back to the texts on relativity for understanding Goigus.

 

 

Edited December 24, 2020 by Anamitra Palit

[Markus Hanke]

5 hours ago, Anamitra Palit said:

The result derived is four dot product,v1.v2>=c^2

The result is manifestly wrong, as shown by the maths above, irrespective of how you might have arrived at it.

5 hours ago, Anamitra Palit said:

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

Again, see the maths above. The inner product of 4-velocities cannot logically exceed c^2.

5 hours ago, Anamitra Palit said:

Its contradictory nature points to failures in conventional theory.

Again, the inner product is an invariant, so it cannot depend on \(\gamma\). It is trivially easy to formally prove that. Also, if that weren’t so, Lorentz invariance could not hold in the real world - but of course it does, as shown by experiment and observation. I don’t know what you are actually trying to show here, but whatever it is, it is physically and mathematically meaningless.

I think it is fairly obvious by now that you aren’t actually prepared to listen to any of the feedback given to you here. You are just repeating the same things again and again, and ignore the fact that you have already been shown wrong. If you were actually interested in learning something, you would listen to what has been said, and use it to improve your own understanding of the subject matter. It is not enough to just blindly do algebraic manipulations - you need to also understand the geometric meaning of these objects, and how they are related to one another, and how that fits into overall physics. 

[joigus]

6 hours ago, Anamitra Palit said:

Goigus should reconsider his views.There is uncompromisable reason why I don't require to go back to the texts on relativity for understanding Goigus.

Quote

 

Proof: First we consider the relations

1) \( c^{2}v_{1t}v_{2t}-\left|\overrightarrow{v_{1}}\right|\left|\overrightarrow{v_{2}}\right|\geq\sqrt{c^{2}v_{1t}^{2}-\left|\overrightarrow{v_{1}}\right|^{2}}\sqrt{c^{2}v_{12t}^{2}-\left|\overrightarrow{v_{2}}\right|^{2}} \)

2) \( c^{2}v_{1t}v_{2t}-\left|\overrightarrow{v_{1}}\right|\left|\overrightarrow{v_{2}}\right|\leq -\sqrt{c^{2}v_{1t}^{2}-\left|\overrightarrow{v_{1}}\right|^{2}}\sqrt{c^{2}v_{12t}^{2}-\left|\overrightarrow{v_{2}}\right|^{2}} \)

The above two relations have been written from (1) where we have considered the following, \( a_1 = cv_{1t} \) ,  \( a_2 = \left| \overrightarrow{v_1} \right| \), \( a_1 = cv_{2t} \) , \( b_2 = \left| \overrightarrow{v_2} \right| \) [suffix ’t’ denotes the time component] But \( c^2v_{1t}^2 - v_{1t}^2v_{2t}^2 = c^2 \); \( c^2 v_{12t}^2 - v_{1t}^2v_{2t}^2 = c^2 \) 

The first inequality gives us

\[ c^2v_{1t}v_{2t} - \left| \overrightarrow{v_1} \right| \left| \overrightarrow{v_2} \right| \]

 

"Makus Henke" is Markus Hanke, "Goigus" is Joigus, and as to \( v_{12t} \) nobody knows what it is.

It can be nothing other than a typo, you clearly meant either \( v_{1t} \) or \( v_{2t} \) (otherwise you're summoning something here which is neither \( a_1 \), \( a_2 \), \( b_1 \) or \( b_2 \), the four numbers of your previous (correct) mathematical lemma.

Once you either correct this typo, or tell us what \( v_{12t} \) is, I'm afraid nothing's going to save the incorrect proof, for reasons very much already explained.

Anamitra Palit should reconsider his writing before Joigus (at least) considers his views any further.

Edited December 24, 2020 by joigus

[Anamitra Palit]

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

Edited January 9, 2021 by Anamitra Palit

[Anamitra Palit]

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

Edited January 10, 2021 by Anamitra Palit

[Anamitra Palit]

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

[joigus]

On 12/24/2020 at 3:23 AM, Anamitra Palit said:

In fact joigus writes alpha=1,beta=1 . But alpha and beta are four vectors.

No. I meant the Euclidean norm of both vectors \( \boldsymbol{\alpha} \) and \( \boldsymbol{\beta} \). That's why I wrote them with boldface characters.

If you read carefully, I did introduce this qualification:

On 12/24/2020 at 7:24 AM, joigus said:

It's very common notation in relativity.

\[ \left(\alpha^{\mu}\right)=\left(\alpha,\boldsymbol{\alpha}\right) \]

So I meant different things by plain \( \alpha \) and \( \left| \boldsymbol{\alpha} \right| \). I should have been more explicit --and standard-- though, and written,

\[ \left(\alpha^{\mu}\right)=\left(\alpha^{0},\boldsymbol{\alpha}\right) \]

I'm sorry about that. Nevertheless, the argument stands.

Edited January 10, 2021 by joigus

[Ghideon]

Question:

In your attached PDF-document there is this section:

 

Why is the time in frame K" t and not t"? Or in other words why have you equation (3.1) written as

[math]x'=\gamma(x''-a-vt)[/math] 

instead of this:

[math]x'=\gamma(x''-a-vt'')[/math]

 

 

5 hours ago, Anamitra Palit said:

Incidentally I can't find the code button <> which used to be there earlier

Please note that you can use latex within math-sections on the forum:

[math]x'=\gamma(x''-a-vt'')[/math]

I guess that is what @joigus uses as well.

Edited January 10, 2021 by Ghideon
format

[joigus]

1 hour ago, Ghideon said:

I guess that is what @joigus uses as well.

I use the backslash + square brackets for maths sections, and the backslash + round brackets for inline maths. I don't know how to escape LateX to show the code. Perhaps a "code" tag...

\\[ \\] \\( \\)

Simple backslash is a common escape tag. But it doesn't work here.

[Anamitra Palit]

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

Edited January 11, 2021 by Anamitra Palit

[Anamitra Palit]

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.

Edited January 11, 2021 by Anamitra Palit

[Ghideon]

1 hour ago, Anamitra Palit said:

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

Thanks for your comment. I agree that t and t'' are not identical and that was the point. You did not answer the question as far as I can tell. Why do you use t instead of t''? What happens to (3.1) when you use an expression for t" as you did for x"?

In other words: Your PDF has: 
What is the resulting expression for t" and why is that not needed?

Edited January 11, 2021 by Ghideon

[Anamitra Palit]

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.

[Ghideon]

17 minutes ago, Anamitra Palit said:

This paper dos not indicate at any error in theory.

(Emphasis mine)

By "This paper" you mean the attachment named "Time Conflict 5.pdf" ? 

[joigus]

@Anamitra Palit,

Maybe you can try some LateX editor, like,

https://www.google.com/search?q=wysiwyg+latex+editor+online&oq=wysiwyg+LateX+editor+online

and then nest your equations with braces, like this:

\[ \textrm{ \[ <LateX code here> \] } \]

or with round brackets for inline maths.

This would considerably improve the communication of mathematical arguments.

Edited January 11, 2021 by joigus

[Anamitra Palit]

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 

[Anamitra Palit]

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]

Edited January 12, 2021 by Anamitra Palit