Neither statement is well defined. The equivalence of vectors is a statement about the transformations they are subject to. If one is discussing 4-vectors and Minkowski spacetime, the transformations are Lorentz transformations, so two vectors are equivalent, if they can be mapped to one another by a proper Lorentz transformation. 

Tachyons describe instabilities, that's why they can't describe stable matter. Since dark matter does have ordinary gravitational interactions, it cannot be described by tachyons-it would have decayed. 

Dear Stam,

Your viewpoint looks rather strange. You suggest to define equivalence of  two vectors via the transformation of  coordinates. The coordinate system is simply a method of the geometry description. Mathematicians tend to give geometrical definitions in terms of  geometry itself, but not in  terms of  methods of  the geometry description.

The geometrical definition is provided by specifying the (properties of the) transformations, not the coordinates, of course-that's what was done above. Transformations realize mappings of vectors, as elements of vector spaces, to vectors, independently of any particular coordinate system. So the statement that the suggestion was via transformation of coordinates is incorrect, since that is not what is written in the first answer.

And as is known, proper Lorentz transformations map time-like vectors to time-like vectors, light-like vectors to light-like vectors and space-like vectors to space-like vectors.

Dear Stam,

You have not answered my question, how is defined equivalence of two vectors. For timelike vectors both definitions coincide. For spacelike vectors they are different. Let us consider three spacelike vectors with coordinates AB={0,0,0,z} and CD={5,3,4,z}, EF={5,4,3,z} in the same coordinate system. Are vectors AB and CD equivalent? Are vectors AB and EF equivalent? Are vectors CD and EF equivalent?

Two 4-vectors could be considered equivalent, if they can be related by a Lorentz transformation, since, in that case, they have the same norm. The invariants of the transformations define equivalence.So the norm of CD would be 25-9-16-z^2=-z^2 and the norm of AB is, similarly, -z^2, which means that they can be related by a Lorentz transformation. Similarly for EF. All these vectors are space-like, with the same norm, therefore there exist Lorentz transformations that map one to another and, since the composition of Lorentz transformations is a Lorentz transformation, there isn't any invariant way to distinguish them, if that's all that's known about them. These statements,  of course, are valid in Minkowski spacetime.

Dear Stam,

I am sorry, but you do not answer my question again. As far as I have understood, you suppose that two vectors are equivalent (equal), if the vectors have equal lengths (norm). Unfortunately, equality of lengths is insufficient for equality of vectors even in the proper  Euclidean geometry. Vectors are to be in parallel besides. It means that (AB.CD)=|AB| |CD|, where (AB.CD)is  the scalar product of vectors This relation in the geometry of Minkowski is also invariant with respect to any Lorentz transformation.

As to vectors AB, CD, and EF in my previous post, this relation is fulfilled for vectors AB=CD and AB=EF, but it is not fulfilled for vectors CD and EF, because

(CD.EF)=25-12-12-z2=1-z2, whereas |CD||EF|=-z2. It means that the equality relation, defined by (AB.CD)=|AB| |CD| and |AB|=|CD|  is intransitive. It means, that the geometry of Minkowski equipped by this definition is not a logical construction, because in any logical construction the equivalence relation is transitive.

I  am repeating my question: What of two definitions of equivalence is valid?

The statement that the vectors have the same 4-norm implied that they are related by Lorentz transformations. These transformations preserve the lorentzian scalar product, also. The first relation implies that the vectors are equivalent, the second doesn't, since the scalar product is always invariant, independent of its value, whereas the point of the first relation isn't that the norms are invariant, which they are but that they're equal.

Dear Stam,

You do not answer my question. You write about transformations. What relation have transformations to definition of of two vectors equivalence? If you disagree with two given definitions, then give your own definition of two vectors equivalence. But give it.

AB, CD and EF have the same 4-norm. That *means* that there exist Lorentz transformations such that AB = L1.CD and CD = L2.EF, which means that AB = (L1.L2).EF = L3.EF, since Lorentz transformations form a group. That's why they may be considered  equivalent. Their scalar products are irrelevant for the discussion. If they didn't have the same 4-norm, this would mean that there can't exist Lorentz transformations that relate them. After that it's a question what one wants to do with such vectors. it doesn't make sense to ask about equivalence as such. 

Your first proposition looks more acceptable to me both from physics and math point of view.Mathematician does the work for application but not for mathematics .There are many math formulas still are not useful.I suggest  you to take a physical problem and apply .Then you will really understand which one you accept ?.

                                                                                         B.Rath

Dear Stam,

It is very pity, that you cannot give a definition of two vectors equality. You write about transformations, which have no relation to this definition. It follows from your post that two vectors are equal, if their norms are equal. This definition is wrong in the proper Euclidean geometry and for timelike vectors in geometry of Minkowski, because in these cases both my definitions coincide and are valid.

 Dear Biswanath,

The problem is such, that using the first (coordinate) definition we obtain the geometry of Minkowski as a logical construction and smooth world lines of tachyons. Using the second (coordinatless) definition, the geometry of Minkowski cannot be a logical construction, because the equality relation is intransitive for spacelike vectors. Besides, world lines of tachyons wobbles with infinite amplitude, and a single tachyon cannot be detected because of this wobbling.

Equality and equivalence are two different things, so it's useful to keep them separate. It does not follow from anything I've written that two vectors are equal, if their norms are equal. Two vectors, that are equal, have equal norms. The point is that it's useful, in many cases,  to consider equivalence  relations, *other* than equality.

Regarding tachyons: it does not follow from anything written here that worldlines of tachyons ``wobble''. The reason tachyons can't be detected doesn't have anything to do with any wobbling-it has to do with the fact that tachyons describe the instability of an unstable equilibrium, which *means* that, they are absent, as excitations, at stable equilibria, since otherwise these equilibria would be unstable. This is well known.

Dear Stam, 

Please give definition of equality and definition of equivalence. It is useless to discuss without these dfinitions. As to tachyons. I do not know anything about unstable equilibrium. It is nongeometrical concept. But we speak about geometries. Tachyon by definition is a particle, whose world line is spacelike. Any element of the tachyon world line is spacelike. Tachyon world line wobbles. For spacelike vector P0P1 there are many vectors P1P2, P1P'2, P1P''2,... which are equal to P0P1 , but they are not equal between themselves.This indefinite direction of adjacent element P1P2 with respect to element P0P1 of the tachyon world line is a reason of wobbling. This multivariance of two vectors equality depends on definition of the vector equivalence.

Vectors A and B are equal, iff for any C

(A,C)=(B,C).

Dear Mikhail,

Your definition of equality contains infinite number of equations. Not all equations are independent. Choosing four coordinate vectors as vectors C, we obtain the first condition of equality in the form ABk=CDk. From practical viewpoint your definition is inconvenient because of  infinite number of relations in your definition. The second definition contains only two equation. This fact is of most interest. Besides, this definition can be used only for continuous geometries.

Once more: both definitions are, individually,  ``valid''-by definition. They are, also, consistent, in the sense that there do exist vectors, that are satisfy one, if they satisfy the other. They aren't equivalent, since there do exist vectors that satisfy one, but not the other. So the only issue is, in what ways they're ``useful'' and that depends on the applications one has in mind. However the consistent description of tachyons doesn't rely on the equivalence of the two definitions. 

Dear Stam,

You are a past master in going around the discussed problem. You write many words and nothing to the point. Discussion with you is of no interest.

 Personal comments on a discussion thread are irrelevant and inappropriate. What's only relevant is to present arguments on the content of statements. The statements about the vectors have been dealt with, as have some of the statements about tachyons. The statement that tachyons can be discovered by the gravitation of a tachyon gas is, however, incorrect, since identifying a contribution with that of unstable excitations is problematic. Additionally, the statement that tachyons can describe dark matter is wrong, because dark matter satisfies energy conditions different from tachyons. Finally, the statements about the equality or equivalence of 4-vectors are completely irrelevant to any properties of tachyons, beyond the well-known fact that tachyons are described by space-like worldlines.

Dear Stam,

As I have understood from your post, you consider wobbling of the tachyon world lines, as some unstable excitations of the tachyon world lines and state that such situation is impossible. In reality, the wobbling is a natural state of the spacelike world line, and this fact is connected with the definition of two vectors equality. Any world line (timelike and spacelike) is a set of connected vectors … S0S1,S1S2,S2S3,… These vectors have the same length. If  it is a world line of a free particle, the adjacent vectors are in parallel, and S0S1=S1S2,…. If there is only one vector S!S2, which is equal to S0S1, then the world line is smooth in the limit, when lengths of vectors vanish. If there are many vectors S1S2 which are equal to vector S0S1, then position of vector S1S2 with respect to adjacent vector S0S1 is random, and the world line wobbles.

True definition of two vectors equality is the second one (by means of two equations). According this definition the equal timelike equal  vector is determined univocal. However, the spicelike vector is determined randomly. Thus, spacelike world line wobbles.

We know about tachyons only, that their world lines are spacelike.  Tachyons have not been detected experimentally. It means, that either tachyons are absent, or a single tachyon cannot be detected because its properties. If world line of tachyon wobbles, a single tachyon cannot be detected.

Now about the second definition of two vectors equality. Being  applied to spacelike vectors, it is intransitive. It means that  geometry of Minkowski  is not a logical construction, because the equivalence relation is transitive in any logical construction. Mathematical community suppose, that any generalized geometry is a logical construction, because it is constructed by means of the Euclidean method, when all statements of geometry are deduced from a system of axioms. However, if the Euclidean geometry GE has been constructed, a generalized geometry can be constructed as a deformation of GE. All statement of GE are expressed via Euclidean  distance function \rhoE (metric) or via Euclidean world function \sigmaE=\rhoE2/2. Substituting Euclidean world function \sigmaE in all geometrical relations of GE by world function \sigma of generalized geometry  G, one obtains all geometrical relations of  G. The deformation method is simpler and more effective, than the Euclidean method, which needs proofs of numerous theorems. The second definition of two vectors equality is an attribute of the deformation method of the geometry construction, whereas the first one is an attribute of the Euclidean method of the geometry construction.

Why the scientific community does not accept construction of a geometry by means of a deformation of GE, I do not know. This fact associates with the Galiley  phenomenon, when concept of inertia, suggested by Galiley, has been not accepted by scientific community during almost a century. It was accepted only after Newton introduced his first law of mechanics.

Thus, difference between two definitions has fundamental character. It seems to me, that one does not accept new fundamental concepts, because these new fundamental concepts were not teached in the school.

Mikhil G  Ivanov has the same has the same bent of mind that of mine .I suggest again

make graphical plot of your idea and discuss .

B Rath

Hyperbolic geometry, which is the geometry of spacetime, is quite different from Euclidian geometry. The errors come from failing to take this into account.

Dear Stam,

World function of a geometry G determines geometry G completely. World function determines even dimension of a geometry, although many researchers cannot believe this. If we substitute Euclidean world function \sigmaE in all relations of GE by world function \sigma of a deformed geometry G, we obtain all relations of G. Of course, world functions of Euclidean geometry and  of pseudo-Euclidean geometry are different. But formal method of deformation is better, than any mental considerations about this difference. There is no mistake. Deformation distroys logical connection between the statements of Euclidean geometry GE. It is reasonable, because deformation is not a logical operation. 

To establish equivalence of two "things" we require a third "thing" e. g. a ruler or another vector c or a transformation or a world geometry.

Can this be remedied?