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On the Fundamental Theorem of the Theory of Relativity

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Abstract

A new formulation of what may be called the “fundamental theorem of the theory of relativity” is presented and proved in (3 + 1)-space-time, based on the full classification of special transformations and the corresponding velocity addition laws. A system of axioms is introduced and discussed leading to the result, and a study is made of several variants of that system. In particular the status of the group axiom is investigated with respect to the condition of the two-way isotropy of light. Several issues which are ignored or misunderstood in the literature are emphasized.

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Notes

  1. See for instance the “Postscript” to [26] added in [27], or the “Note added in proof” of [21]. In some cases, like [4] versus [8], the ‘anticipated’ paper has come to be much more famous and cited than the ‘anticipating’ one.

  2. I remind the reader of the highly remarkable fact that of all \(\mathbb R^n\)s, \(\mathbb R^4\) is the only one to have non-diffeomorphic differential structures; indeed, there is a continuum of such structures (for an introduction to these results in differential topology, see for instance chapter I of [6]).

  3. For other purposes, which will not concern us here, one may admit nonregular curves as worldlines, e.g. with a discrete subset of points at which 3-velocity is not well-defined.

  4. In Einstein’s article [11, p. 907] one finds a statement to the effect that the special transformations must form a group, with no further elaboration or explanation.

  5. Incidentally, for systematic as well as for historical reasons, I think that room should be left in the presentation of the principle of relativity also for ‘Aristotelian’ and ‘Newtonian’ static space-times.

  6. Weyl [47, p. 179, 313] sketches a proof which applies a weaker form of the theorem to the case that the inertial worldlines belong to a “given, arbitrarily thin, cone”. Also Schwartz [36] recognizes the difficulty, but his adaptation of the theorem is not satisfactory, since it assumes, unwarrantedly in its context, that the equation of an hyperplane must be affine.

  7. “First of all, it is clear that the equations must be linear because of the properties of homogeneity which we attribute to space and time” ([11, p. 898]; italics in the original). Cf. [8, 20, 21].

  8. Details are given, for instance, in [4, p. 1519]. Other arguments assume a differentiability condition (cf. [13]).

  9. These space-times are topologically, but not metrically, equivalent to the Friedmann space-times, of well-known cosmological relevance.

  10. The obscurity of the derivation of affinity from homogeneity is pointed out, for instance, in [28, 31]. In the present context proofs based on the invariance of d’Alembert equation or on light signalling [17, 45, 48] are obviously not relevant.

  11. Of course in (12) \(B^{-1}\) stands for the inverse of the map, not the inverse of the matrix.

  12. What follows generalizes Poincaré’s argument in [35].

  13. The statement, italicized in the original, is quoted from [5, p. 100].

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Correspondence to Marco Mamone-Capria.

Appendix: Proof of Theorem 3.3

Appendix: Proof of Theorem 3.3

It is easy to verify that if \(\phi , \phi '\) are affinely equivalent cs’s, then they are also inertially equivalent.

Let us consider the converse. Suppose that every uniform worldline for \(\phi \) is also uniform for \(\phi '\). Since points at rest for \(\phi \) are physical worldlines (Axiom 3), we have for every \(\mathbf{r}_0 \in \mathbb R^3\):

$$\begin{aligned} \frac{\partial \mathbf{r}'}{\partial t}(\mathbf{r}_0 , t)=\frac{\partial t'}{\partial t} (\mathbf{r}_0, t)\mathbf{W}\end{aligned}$$
(61)

with \(\mathbf{W}\) depending at most on \(\mathbf{r}_0\). It follows that the transition function from \(\phi \) to \(\phi '\) can be re-written as

$$\begin{aligned} \left\{ \begin{array}{rcl} \mathbf{r}' &{}=&{} t' (\mathbf{r}, t)\mathbf{W}(\mathbf{r}) + \mathbf{U}(\mathbf{r}) \\ t' &{}=&{} t' (\mathbf{r}, t). \end{array}\right. \end{aligned}$$

where \(\mathbf{U}: \mathbb R^3 \rightarrow \mathbb R^3\) is a suitable differentiable function.

Claim 1: \(\mathbf{W}\) is constant.

In fact let \( \mathbf{r}(t) = \mathbf{r}_0 + (t-t_0)\mathbf{v}\) be any uniform motion for \(\phi \) with \(\mathbf{v}\) physical and \(\mathbb R\) as its domain. If we put \( f(t):= t' (\mathbf{r}_0 + (t-t_0 )\mathbf{v}, t) \) it is easy to see that f is a diffeomorphism of \(\mathbb R\) onto itself. (Let \(\mathbf{v}'\) the constant velocity of the same worldline according to \(\phi '\), and suppose that the interval \(f(\mathbb R)\) had, say, a finite supremum b; then, if \(\phi '\circ \phi ^{-1} (\mathbf{r}_0, t_0) = (\mathbf{r}'_0, t'_0)\) the subset

$$\begin{aligned} \phi \circ \phi '^{-1}\left( \{(\mathbf{r}'_0 + (t'- t'_0)\mathbf{v}')\; : \, t'\in [t'_0, b]\}\right) \end{aligned}$$

would have to be noncompact, which is absurd.) Now

$$\begin{aligned} \mathbf{v}' = \mathbf{W}+ \frac{1}{\dot{f}}(f(\mathbf{v}\cdot \nabla )\mathbf{W}+ (\mathbf{v}\cdot \nabla )\mathbf{U}) \end{aligned}$$
(62)

must not depend on t. Putting \(t=t_0\) we get that there is a function \(\mathbf{v}' = \mathbf{K}(\mathbf{r}_0, \mathbf{v})\) such that \(\mathbf{K}(\mathbf{r}(t), \mathbf{v}) \equiv \mathbf{K}(\mathbf{r}_0, \mathbf{v})\), and

$$\begin{aligned} \dot{f}(t_0)\ (\mathbf{K}(\mathbf{r}_0, \mathbf{v}) -\mathbf{W}(\mathbf{r}_0)) - f(t_0) ((\mathbf{v}\cdot \nabla )\mathbf{W})_{\mathbf{r}_0} = ((\mathbf{v}\cdot \nabla )\mathbf{U})_{\mathbf{r}_0}. \end{aligned}$$

For every fixed \(\mathbf{r}_0\) and physical velocity \(\mathbf{v}\) this is an ordinary differential equation in f(t) (after notation change from \(t_0\) to t) with vector coefficients:

$$\begin{aligned} \dot{f}\mathbf{A}+f \mathbf{B}= \mathbf{C}, \; \text{ where }\; \mathbf{A}:= \mathbf{K}- \mathbf{W}, \; \mathbf{B}:= -(\mathbf{v}\cdot \nabla )\mathbf{W}, \; \mathbf{C}:= (\mathbf{v}\cdot \nabla )\mathbf{U}. \end{aligned}$$

Now, if for all \(\mathbf{r}_0\) we have \(\mathbf{B}= {\mathbf 0}\) when we choose \(\mathbf{v}\) in three linearly independent directions, then \(\nabla \mathbf{W}=0\) and therefore \(\mathbf{W}\) is constant. But for \(\mathbf{B}\) there is no other possible value. Suppose by contradiction, that \(\mathbf{B}\ne {\mathbf 0}\) for some \(\mathbf{r}_0\) and \(\mathbf{v}\). Then f must be a solution of a (scalar) differential equation with constant coefficients of the form \(a\dot{f}+ bf = c\) with \(b\ne 0\). Now \(a\ne 0\), since otherwise f would be constant; it follows that f is of the type \( f(t) = ke^{-bt/a} + c/b \), which clearly is not onto \(\mathbb R\), no matter what the values of abc are. So claim 1 is proven.

It follows that \( \mathbf{r}' (\mathbf{r},t) = t'(\mathbf{r},t) \mathbf{W}+ \mathbf{U}(\mathbf{r}) \) with \(\mathbf{W}\) constant. Note that \(\mathbf{U}(\mathbf{r})\) cannot be constant, otherwise

$$\begin{aligned} \det \left( \frac{\partial \mathbf{r}'}{\partial \mathbf{r}}\right) = \det \left( \mathbf{W}(\nabla t')^T\right) = 0, \end{aligned}$$

since every \(3\times 3\) matrix of the form \(\mathbf{a}\mathbf{b}^T\) is singular, which would contradict Axiom 5.

Thus (62) becomes:

$$\begin{aligned} \mathbf{K}(\mathbf{r}_0, \mathbf{v}) = \mathbf{W}+ \frac{1}{\dot{f}}(\mathbf{v}\cdot \nabla )\mathbf{U}. \end{aligned}$$
(63)

By comparing the functional dependence of the various terms of this equation we get that \(\dot{f}\) must be independent of t ; since

$$\begin{aligned} \dot{f}= (\nabla t')(\mathbf{r}_0 , t)\cdot \mathbf{v}+ \frac{\partial t'}{\partial t} (\mathbf{r}_0 , t) \end{aligned}$$

this means that neither \(\displaystyle \frac{\partial t'}{\partial t}\) (put \(\mathbf{v}={\mathbf 0}\)), nor \(\nabla t'\) may depend on t. It follows that \( t' = g(\mathbf{r}) + \alpha t \) for some \(\alpha >0\) and \(g: \mathbb R^3 \rightarrow \mathbb R\) differentiable.

Claim 2: Both functions \(\mathbf{U}\) and g are affine.

Equation (63) can be re-written as

$$\begin{aligned} \mathbf{K}(\mathbf{r}_0 , \mathbf{v}) = \mathbf{W}+ (\nabla g(\mathbf{r}_0)\cdot \mathbf{v}+ \alpha )^{-1}((\mathbf{v}\cdot \nabla )\mathbf{U}) (\mathbf{r}_0). \end{aligned}$$

By substituting \(\mathbf{r}_0\) with \(\mathbf{r}(t)\) and differentiating with respect to t, we obtain zero; if we neglect the irrelevant denominator and evaluate for \(t=t_0\) we get:

$$\begin{aligned} {\mathbf 0} = \sum _{\alpha ,\beta =1}^3 \left( (\nabla g(\mathbf{r}_0)\cdot \mathbf{v}+\alpha ) \frac{\partial ^2\mathbf{U}}{\partial x^{\alpha }\partial x^{\beta }}(\mathbf{r}_0) -\frac{\partial ^2 g}{\partial x^{\alpha }\partial x^{\beta }}(\mathbf{r}_0)\frac{\partial \mathbf{U}}{\partial x^{\gamma }} (\mathbf{r}_0)v^\gamma \right) v^{\alpha }v^{\beta }. \end{aligned}$$

By re-arranging the terms we have:

$$\begin{aligned} \alpha \sum _{\alpha ,\beta =1}^3 \frac{\partial ^2\mathbf{U}}{\partial x^\alpha \partial x^\beta }v^\alpha v^\beta =-\sum _{\alpha ,\beta ,\gamma =1}^3 \left( \frac{\partial g}{\partial x^\gamma }\frac{\partial ^2\mathbf{U}}{\partial x^\alpha \partial x^\beta } - \frac{\partial ^2 g}{\partial x^\alpha \partial x^\beta }\frac{\partial \mathbf{U}}{\partial x^\gamma }\right) v^\alpha v^\beta v^\gamma . \end{aligned}$$

Now, for every fixed \(\mathbf{r}_0\) at the lefthand side there is a quadratic vector polynomial in the \(v^\alpha \), while at the righthand side there is a cubic polynomial – both homogeneous with respect to \(v^\alpha \): this is possible, for every \(\mathbf{v}\) in an open neighborhood of \({\mathbf 0}\) (Axiom 8), if and only if both polynomials vanish for every \(\mathbf{r}_0\), that is, if all their coefficients are identically zero:

$$\begin{aligned} \frac{\partial ^2 \mathbf{U}}{\partial x^\alpha \partial x^\beta } ={\mathbf 0},\; \frac{\partial ^2 g}{\partial x^\alpha \partial x^\beta }\frac{\partial \mathbf{U}}{\partial x^\gamma } ={\mathbf 0}. \end{aligned}$$
(64)

From the first equality we have that \(\mathbf{U}\) is an affine map: \(\mathbf{U}(\mathbf{r}) = X\mathbf{r}+ \mathbf{a}\), so we can write

$$\begin{aligned} \left\{ \begin{array}{rcl} \mathbf{r}' &{}=&{} (g(\mathbf{r}) + \alpha t) \mathbf{W}+ X\mathbf{r}+ \mathbf{a}, \\ t' &{}=&{} g(\mathbf{r}) + \alpha t + \ell .\end{array}\right. \end{aligned}$$

But by Axiom 5 it must be \( 0 <\det \displaystyle \left( \frac{\partial \mathbf{r}'}{\partial \mathbf{r}}\right) = \det (\mathbf{W}(\nabla g)^T + X) \) thus \(X\ne 0\). Since, as we have seen, U is nonconstant, at least one of the vectors \(\displaystyle \frac{\partial \mathbf{U}}{\partial x^\gamma }\) is nonzero; thus the second equation in (64) implies

$$\begin{aligned} \frac{\partial ^2 g}{\partial x^\alpha \partial x^\beta } \equiv 0, \end{aligned}$$

which means that also g is an affine map, and Claim 2 is proven; this ends also the proof of the theorem.

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Mamone-Capria, M. On the Fundamental Theorem of the Theory of Relativity. Found Phys 46, 1680–1712 (2016). https://doi.org/10.1007/s10701-016-0038-3

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