The Lorentz invariant cone $\mathcal{T}^+$

Assuming $m = 0$, namely disregarding the internal gauge transformations, if we want to satisfy the equity and the minimum time principles (see Section 2.4), a nonvanishing element of $\mathcal{T}^+$ must have $b^0 > 0$, since all the physical operations ``take some time''. It follows that $\mathcal{T}^+$ is a closed cone with a nonempty interior and we call it the feasibility cone. We say that a theory of this kind has a modified geometry. In the presentation of this argument we follow refs. [7,11,14].

In order to determine the structure of $\mathcal{T}^+$, we have to assume some symmetry property and a natural requirement is that, as in the normal theories, it is symmetric under the proper orthochronous Lorentz group $\mathcal{L} = SO^{\uparrow}(1, 3)$. More precisely, we write an element of $\mathcal{T}^+$ in the form (3.3) and we require the invariance under the Lorentz transformations

$$b^i \to \Lambda^i{}_k b^k, \qquad b^{[ik]} \to \Lambda^i{}_j \Lambda^k{}_l b^{[jl]}.$$ (3.7)

If $\mathcal{T}^+$ contains an element with coordinates $(b^0, \mathbf{b}, \mathbf{b}', \mathbf{b}'')$, it contains also the element with co-ordinates $(b^0, \mathbf{0}, \mathbf{0}, \mathbf{0})$, which belongs to the convex hull of a finite set of points obtained from the given point by means of suitable rotations. The dilatation invariant set of the possible values of $b^0$ cannot be reduced to $0$, because $\mathcal{T}^+$ generates $\mathcal{T}$ and cannot be the whole real line because $\mathcal{T}^+$ is a cone. It follows that it is the half line $b^0 \leq 0$ or the half line $b^0 \geq 0$. For physical reasons, we choose the second possibility. In order to avoid that, after a Lorentz transformation, $b^0$ becomes negative, we have to assume that $\mathcal{T}^+$ is contained in the wedge defined by eq. (3.5).

We consider an arbitrary element of the interior of $\mathcal{T}^+$ and we simplify its coordinates by means of a suitable Lorentz transformation. Since the four-vector $b$ is timelike, we can obtain $\mathbf{b} = \mathbf{0}$. Then, by means of a rotation, we can cancel the third components of $\mathbf{b}'$ and $\mathbf{b}''$. In conclusion, we obtain

$$b^0 > 0, \qquad b^1 = b^2 = b^3 = b^{[21]} = b^{[03]} = 0.$$ (3.8)

We indicate by $b^{\alpha}(\zeta)$ the coordinates obtained from $b^{\alpha}$ by means of a Lorentz boost with rapidity $\zeta$ along the third space axis and we define the quantities

$$b^{\alpha}_{\pm} = 2 \lim_{\zeta \to \pm \infty} \exp(- \vert \zeta \vert) b^{\alpha}(\zeta).$$ (3.9)

Since $\mathcal{T}^+$ is Lorentz and dilatation invariant and it is also closed, it contains the points with co-ordinates $b^{\alpha}_{\pm}$. One can easily see that the only nonvanishing limits are

$$b^0_{\pm} = b^0, \quad b^{[01]}_{\pm} = b^{[01]} \pm b^{[31]}, \quad b^{[32]}_{\pm} = b^{[32]} \pm b^{[02]},$$

$$b^3_{\pm} = \pm b^0, \quad b^{[02]}_{\pm} = b^{[02]} \pm b^{[32]}, \quad b^{[13]}_{\pm} = b^{[13]} \pm b^{[10]}$$ (3.10)

and it follows that

$$b^{\alpha} = 2^{-1} (b^{\alpha}_+ + b^{\alpha}_-).$$ (3.11)

We have shown that the element considered is the sum of two elements of $\mathcal{T}^+$ that satisfy the conditions

$$\Vert\mathbf{b}\Vert = b^0 > 0, \qquad \Vert\mathbf{b}'\Vert = \Vert\mathbf{b}''\Vert = a b^0,$$

$$\mathbf{b}' \cdot \mathbf{b}'' = 0, \qquad \mathbf{b}' \times \mathbf{b}'' = a^2 b^0 \mathbf{b}.$$ (3.12)

These conditions are equivalent to the Lorentz invariant conditions

$$b^0 > 0, \qquad b^k b_k = 0 , \qquad b^{[ik]} b_{[ik]} = 0,$$

$$\epsilon_{ikjl} b^{ik} b^{jl} = 0, \qquad b^{ji}b_{jk} = a^2 b^i b_k \qquad a \geq 0$$ (3.13)

and, for any given $a \geq 0$, define a set $\mathcal{T}_a \subset \mathcal{T}$ invariant with respect to the proper orthochronous Lorentz group $\mathcal{L}$. It follows that the decomposition found above in a special case is also possible for an arbitrary element of the interior of $\mathcal{T}^+$.

One can easily see that $\mathcal{L}$ acts transitively on the sets $\mathcal{T}_a$, which are either contained in $\mathcal{T}^+$ or do not intersect it. The first possibility is realized if $a$ belongs to a closed convex set containing the point $a = 0$ and bounded, otherwise for fixed $b^0$ arbitrarily large values of $\mathbf{b}'$ and $\mathbf{b}''$ would be permitted and $\mathcal{T}^+$ would not be a cone. In other words, we must have $0 \leq a \leq \ell^{-1}$, where $\ell$ is a positive fundamental length. One can easily see that if this inequality is satisfied, every element of $\mathcal{T}_a$ can be written as the sum of two elements of $\mathcal{T}_{1/\ell}$.

In conclusion, we have proven that all the elements of the interior of $\mathcal{T}^+$ can be written as the sum of two elements of $\mathcal{T}_a$ with $a \leq 1/\ell$ or of four elements of $\mathcal{T}_{1/\ell}$. Since $\mathcal{T}^+$ is the closure of its interior, it is easy to show that all its elements have the same decomposition. Note that, if $B \in \mathcal{T}_{1/\ell}$, $(b^0)^{-1} \mathbf{b}$, $\ell (b^0)^{-1} \mathbf{b}'$ and $\ell (b^0)^{-1}\mathbf{b}''$ form a left-handed triad of normalized orthogonal vectors.

From the decomposition of an element of $\mathcal{T}^+$ into elements of $\mathcal{T}_{1/\ell}$ and eq. (3.12), we obtain immediately the inequalities

$$\Vert\mathbf{b}\Vert \leq b^0, \qquad \Vert\mathbf{b}'\Vert ... ...^{-1} b^0, \qquad \Vert\mathbf{b}''\Vert \leq \ell^{-1} b^0,$$ (3.14)

which can be interpreted as limitations to the velocity, the angular velocity and the acceleration of a moving frame. The relevance of these inequalities for the motion of a particle connected to the frame are discussed in Chapter 8. It easily follows, using the properties of the norm, that an element $B \in \mathcal{T}_{1/\ell}$ can be decomposed as the sum of elements of $\mathcal{T}^+$ only if all of them are proportional to $B$. This means that $\mathcal{T}_{1/\ell}$ is the set of all the extremal elements of the cone $\mathcal{T}^+$.