# The Finsler Spacetime Condition for (α,β)-Metrics and Their Isometries

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## Abstract

**:**

## 1. Introduction

- 1.
- The necessary and sufficient conditions for an $(\alpha ,\beta )$-metric to define a Finsler spacetime structure;
- 2.
- Determining the isometries of general $(\alpha ,\beta )$-metrics.

## 2. Preliminaries

- 1.
- Positive two-homogeneity: $L(x,\alpha \dot{x})={\alpha}^{2}L(x,\dot{x}),\phantom{\rule{0.166667em}{0ex}}\forall \alpha >0,$ $\forall (x,\dot{x})\in \mathcal{A}.$
- 2.
- At any $\left(x,\dot{x}\right)\in \mathcal{A}$ and in one (and then, in any) local chart around $(x,\dot{x}),$ the Hessian:$${g}_{ij}={\displaystyle \frac{1}{2}}{\displaystyle \frac{{\partial}^{2}L}{\partial {\dot{x}}^{i}\partial {\dot{x}}^{j}}}={\displaystyle \frac{1}{2}}{L}_{\xb7i\xb7j}$$

**Definition**

**1**

- 3.
- There exists a connected conic subbundle $\mathcal{T}\subset \mathcal{A}$ with connected fibers ${\mathcal{T}}_{x}=\mathcal{T}\cap {T}_{x}M,$ $x\in M,$ such that on each ${\mathcal{T}}_{x}:$$L>0,$g has a Lorentzian signature $(+,-,-,-)$ and L can be continuously extended as 0 to the boundary $\partial {\mathcal{T}}_{x}.$

- On a Finsler spacetime, there exists the following important subsets of $TM$:

- The conic subbundle $\mathcal{A}$ where L is defined, smooth, and contains nondegenerate Hessians is called the set of admissible vectors; we will typically understand $\mathcal{A}$ as the maximal subset of $\stackrel{\circ}{TM}$ with these properties.
- The conic subbundle $\mathcal{T},$ where the signature of g and the sign of L agree, will be interpreted as the set of future-pointing time-like vectors.

**Note.**From the above definition, it follows that all the fibers ${\mathcal{T}}_{x}$ of $\mathcal{T}$ are actually convex cones, see [39].

## 3. Spacetime Conditions for $\left(\mathbf{\alpha},\mathbf{\beta}\right)$-Metrics

**Assumption**

**1.**

**Remark**

**1.**

- 1.
- Using the above assumption ${\mathcal{T}}_{x}\cap {\mathcal{T}}_{x}^{a}\ne \u2300$ and ${L}_{|{\mathcal{T}}_{x}}>0$, it follows that:$$\forall \dot{x}\in {\mathcal{T}}_{x}:\phantom{\rule{4pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}A>0,\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\mathsf{\Psi}>0;$$Indeed, according to the mentioned assumption, for any $x\in M$, there exists at least one vector y in the intersection ${\mathcal{T}}_{x}\cap {\mathcal{T}}_{x}^{a}$ which thus must satisfy $A\left(y\right)>0,\mathsf{\Psi}\left(y\right)>0$. As both a A and Ψ are assumed to be smooth (and hence, continuous) on ${\mathcal{T}}_{x}$, in order to change sign, they should pass through 0. But the vanishing of either A or Ψ entails $L=0$, which is in contradiction with the positiveness axiom for L inside $\mathcal{T}$; therefore, (8) must hold throughout ${\mathcal{T}}_{x}$.
- 2.
- On the boundary $\partial {\mathcal{T}}_{x},$ L can be continuously prolonged as to satisfy:$$\phantom{\rule{3.33333pt}{0ex}}A=0\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}or\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\mathsf{\Psi}=0.$$

**Lemma**

**1 (**

**The domain of**

**$s$).**

**Proof.**

**Remark**

**2 (sharpness (or non-sharpness) of the bounds for**$s$). To precisely establish the interval I, we first need the critical points of the function $s=s\left(\dot{x}\right)$. These turn out to be situated:

- In the plane $B=0$; in this case, the corresponding critical value is $s=0$.
- On the ray directed by $\dot{x}=\tilde{b}$ emanating from the origin; this yields the critical value $s=\u2329b,b\u232a$.

- Taking into account the above Lemma, we note that the corresponding critical values are—if attained inside ${\mathcal{T}}_{x}$—minimal values for s. This way, we find:

- 1.
- The lower bound $s={s}_{0}$ is attained for $\dot{x}\in {\mathcal{T}}_{x}$ in two situations:
- i.
- When $b\in {\mathcal{T}}_{x}$ is L-time-like (which implies that, in particular, it is also a-time-like, meaning that $\u2329b,b\u232a>0$) and $\dot{x}$ is collinear to $\tilde{b}$;
- ii.
- When the critical hyperplane $B=0$ for s intersects ${\mathcal{T}}_{x}$ (a necessary condition for this is that b is a-space-like).

- 2.
- For the upper bound, we have two possibilities:
- i.
- If the boundary $\partial {\mathcal{T}}_{x}$ contains points where $A=0,B\ne 0,$ (as in Figure 1a,c,d), then, approaching these points, we will have $s\to \infty ;$ therefore, in this case, $s\in [{s}_{0},\infty ),$ where the upper bound is sharp.
- ii.
- If A does not vanish anywhere on $\partial {\mathcal{T}}_{x}$ (situation (b) in Figure 1, when $\partial {\mathcal{T}}_{x}$ consists only of points where $\mathsf{\Psi}=0,$ $A>0$), then—since obviously we cannot have $A=0$inside${\mathcal{T}}_{x}$ either—it follows that s has a finite supremum on ${\mathcal{T}}_{x}$, in other words, we can safely write $s\in [{s}_{0},{s}_{1})$ for some finite value ${s}_{1}>0.$

**Lemma**

**2.**

**Proof.**

- Step 1.
- There exists at least one $\dot{x}$ on the boundary $\partial {\mathcal{T}}_{x}$ such that $B\left(\dot{x}\right)\ne 0:$

- -
- We first note that ℓ has no common points with $H.$ Indeed, for any $\lambda \in \mathbb{R},$ we have $B(u-\lambda v)=B\left(u\right)-\lambda B\left(v\right)=B\left(u\right)\ne 0.$
- -
- Second, $\ell \cap {\overline{\mathcal{T}}}_{x}$ is a non-empty, connected set. Indeed, ℓ contains an interior point of ${\mathcal{T}}_{x},$ which is u. Moreover, since ${\mathcal{T}}_{x}$ is convex, this means that its closure—which is also necessarily convex—must intersect ℓ by a segment, a half-line, or the whole ℓ. In any case, $\ell \cap {\overline{\mathcal{T}}}_{x}$ is connected.
- -
- Third, we show that the set ${\overline{\mathcal{T}}}_{x}\cap \ell $ intersects the cone ${\mathcal{T}}_{x}^{a}:A=0.$ To this aim, let us build the function $f:\mathbb{R}\to \mathbb{R},$$$f\left(\lambda \right):=A\left(u-\lambda v\right)=\u2329u,u\u232a-2\lambda \u2329u,v\u232a+{\lambda}^{2}\u2329v,v\u232a.$$This function always has at least one root, ${\lambda}_{0},$ as follows. (i) If $\u2329v,v\u232a=0$, i.e., $v\in \partial {\mathcal{T}}_{x}^{a}$, then, since u is a-time-like, it cannot be a-orthogonal to $v,$ which means $\u2329u,v\u232a\ne 0$. Hence, in this case f is of first degree in $\lambda $ and thus has one zero. (ii) If $\u2329v,v\u232a>0$, i.e., $v\in {\mathcal{T}}_{x}^{a},$ then f is quadratic with a halved discriminant $\Delta ={\u2329u,v\u232a}^{2}-\u2329u,u\u232a\u2329v,v\u232a\ge 0$ by virtue of the reverse Cauchy–Schwarz inequality for a. Hence, again, f has real roots.

- -
- Finally, taking into account the connectedness of ${\overline{\mathcal{T}}}_{x}\cap \ell $, we find that, moving away from u along the line $\ell ,$ we stay in ${\mathcal{T}}_{x}$ and, at some point, we must hit the boundary $\partial {\mathcal{T}}_{x}.$ If, in a worst case scenario, we do not hit first a point where $\mathsf{\Psi}=0$, then we must anyway reach a root ${\lambda}_{0}$ of f, which will thus give a boundary point $\dot{x}$ for ${\mathcal{T}}_{x}\subset {\mathcal{T}}_{x}^{a}.$ Moreover, since $\ell \cap H=\u2300,$ at this point we always have $B\left(\dot{x}\right)\ne 0.$

- Step 2.
- There exists a subset ${I}_{0}\subset I$ on which $\mathsf{\Psi}-s{\mathsf{\Psi}}^{\prime}>0$:

- Step 3.
- $\mathsf{\Psi}-s{\mathsf{\Psi}}^{\prime}\ne 0$ must have a constant sign on the entire I:

**Lemma**

**3.**

- (i)
- ${g}_{\dot{x}}$ has $(+,-,-,-)$ signature and is negative definite on the ${g}_{\dot{x}}$-orthogonal complement of $\dot{x};$
- (ii)
- $det{g}_{\dot{x}}<0.$

**Proof.**

**Proposition**

**1.**

**Theorem**

**1 (**

**The spacetime conditions**

**)**

- (i)
- $A>0,\mathsf{\Psi}>0$ on ${\mathcal{T}}_{x}$ and $\underset{\dot{x}\to \partial {\mathcal{T}}_{x}}{lim}\left(A\mathsf{\Psi}\right)=0.$
- (ii)
- For all values of s corresponding to vectors $\dot{x}\in {\mathcal{T}}_{x}:$$$\mathsf{\Psi}-s{\mathsf{\Psi}}^{\prime}>0,\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\left(s-\u2329b,b\u232a\right){\displaystyle \frac{d}{ds}}ln\sigma >-1.$$

**Proof.**

- $\leftarrow :$ Assume now that there exists a conic subbundle $\mathcal{T}\subset \mathcal{A}$ with connected fibers and obeying conditions $\left(i\right)$ and $\left(ii\right).$ Then, in order to prove that $(M,L)$ is a Finsler spacetime, it is sufficient to show that ${g}_{\dot{x}}$ has $\left(+,-,-,-\right)$ signature for all $\dot{x}\in {\mathcal{T}}_{x}$, which, using Lemma 3, is the same as $det{g}_{\dot{x}}<0.$ With this aim, we note that, using $\mathsf{\Psi}>0,$ $\mathsf{\Psi}-s{\mathsf{\Psi}}^{\prime}>0$, this reduces to:$$\frac{d}{ds}}\left[(s-\u2329b,b\u232a)\sigma \right]>0.$$

**Remark**

**3.**

- 1.
- Since $s-\u2329b,b\u232a\ge 0,$ then any strictly positive and monotonically increasing function σ will obey our condition, as, in this case, we will have the stronger inequality: $\left(s-\u2329b,b\u232a\right){\displaystyle \frac{d}{ds}}ln\sigma \ge 0.$
- 2.

## 4. Examples

#### 4.1. Lorentzian Metrics $L=\kappa A$

#### 4.2. Randers Metrics $L=\u03f5{(\sqrt{A}+B)}^{2},\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\u03f5=sign(\sqrt{A}+B)$

- The cone ${\mathcal{T}}_{x}$ is a connected component of the conic set $\sqrt{A}+B>0$, more precisely, the connected component of the cone$$({a}_{ij}-{b}_{i}{b}_{j}){\dot{x}}^{i}{\dot{x}}^{j}>0$$
- On the boundary $\partial {\mathcal{T}}_{x},$ we must have $L=0$, which implies:$$B=-\sqrt{A}\le 0.$$This leads to the even more interesting conclusion below.
- At any point $x\in M$ of a Randers spacetime, the whole cone ${\mathcal{T}}_{x}$ lies in the half-space $B<0$:$$B\left(\dot{x}\right)<0,\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{1.em}{0ex}}\forall \dot{x}\in {\mathcal{T}}_{x}.$$In particular, the hyperplane $B=0$ cannot intersect the interior of ${\mathcal{T}}_{x}$.To justify the above, let us first show that $B=0$ cannot occur inside ${\mathcal{T}}_{x}$. Indeed, if this were the case, then, fixing an arbitrary $u\in {\mathcal{T}}_{x}\cap \{B=0\}$, the fact that ${\mathcal{T}}_{x}$ is open in the topology of ${T}_{x}M\simeq {\mathbb{R}}^{4}$ guarantees that there must be an entire Euclidean ball centered at u which remains in ${\mathcal{T}}_{x}$. However, since u is in the hyperplane $B=0$, half of this ball will be contained in the hyperspace $B>0$; in other words, the intersection $\mathcal{C}:={\mathcal{T}}_{x}\cap \{B>0\}$ is non-empty. By its definition, $\mathcal{C}$ is an open, conic, and convex set and, moreover, by continuity, on the boundary $\partial \mathcal{C}$, we have $B\ge 0$. On the other hand, proceeding as in the first step of the proof of Lemma 2, we find that there must exist a point $\dot{x}\in \partial \mathcal{C}$ where $B\ne 0$. Such a point is, thus, a point of $\partial {\mathcal{T}}_{x}$, where, in addition, $B>0$, which is in contradiction with the above remark. Therefore, $B=0$ cannot happen in the interior of ${\mathcal{T}}_{x}$ (but only possibly on its boundary). An immediate consequence of this is that B must have a constant nonzero sign on ${\mathcal{T}}_{x}$; moreover, this sign is negative.

**Proposition**

**2.**

**Proof.**

- Furthermore, according to Theorem 1, on its future-pointing time-like cones, we must have $\mathsf{\Psi}-s{\mathsf{\Psi}}^{\prime}=1-\sqrt{s}>0$, which is equivalent to $s<1$. Using Lemma 1, this gives$$\u2329b,b\u232a\le s<1.$$

- To prove the first inequality $\u2329b,b\u232a\ge 0$, let us assume that this is not the case, i.e., $\u2329b,b\u232a<0$. We will show that, in this case, the hyperplane $B=0$ must intersect ${\mathcal{T}}_{x}$. To this aim, fix an arbitrary point $x\in M$ and an a-orthonormal basis $\{{e}_{0},{e}_{1},{e}_{2},{e}_{3}\}$, with ${e}_{0}$ being a-time-like and ${e}_{1}$ being collinear to $\tilde{b}$. It follows that $\tilde{b}=\lambda {e}_{1}$ for some $\lambda \in {\mathbb{R}}^{*}$. Accordingly, $A=0$ becomes equivalent to ${\left({\dot{x}}^{0}\right)}^{2}-{\left({\dot{x}}^{1}\right)}^{2}-{\left({\dot{x}}^{2}\right)}^{2}-{\left({\dot{x}}^{3}\right)}^{2}=0$ and the hyperplane $B=0$ is described by the equation ${\dot{x}}^{1}=0$. Taking into account (18), it follows that the intersection between this hyperplane and ${\mathcal{T}}_{x}$ is the region of the three-dimensional cone:$${\left({\dot{x}}^{0}\right)}^{2}-{\left({\dot{x}}^{2}\right)}^{2}-{\left({\dot{x}}^{3}\right)}^{2}>0,$$

- The functions A and $\mathsf{\Psi}$ are, by definition, positive inside $\mathcal{T}$ and, when approaching the boundary of $\mathcal{T}$ (that is, for $s\to 1$), we find that $\mathsf{\Psi}\to 0$ and hence, $L\to 0$. The smoothness of $\mathsf{\Psi}$ follows as the value $s=0$ (which would correspond to $B=0$) cannot appear inside ${\mathcal{T}}_{x}$. This happens since b is, by hypothesis, non-space-like with respect to a; that is, $B=0$ cannot occur within ${\mathcal{T}}_{x}^{a}$ and accordingly cannot occur in ${\mathcal{T}}_{x}$.
- The conditions in (ii) of Theorem 1 follow immediately using (21).

- Thus, $L=A\mathsf{\Psi}$ defines a Finsler spacetime structure. □

**Remark**

**4.**

- 1.
- The result above extends those in [42] by taking into consideration the case when b is a-space-like and actually proving that this never defines a Randers spacetime.
- 2.
- A very recent article1, [21], proposed a modification of the standard Randers metric. With our above notations and signature convention $(+,-,-,-)$, on the set of interest $A>0$, this reads:$$L=\u03f5(\sqrt{A}-{\left|B\right|)}^{2}.$$This is a very interesting extension, as it removes the request that B should be strictly negative inside ${\mathcal{T}}_{x}$. Thus, for modified Randers metrics (24), the restriction $\u2329b,b\u232a\ge 0$ (which was motivated precisely by the fact that once crossing the hyperplane $B=0$, B changes sign) is no longer necessary. Moreover, on regions where $L>0$ and $A>0$ (in which the cones ${\mathcal{T}}_{x}$ must be contained), we have $\mathsf{\Psi}={(1-\sqrt{s})}^{2}$ just as above (with the only difference that, in the definition of s, we no longer have $\left|B\right|=-B$). Thus, a completely similar reasoning shows that (24) leads to a well-defined spacetime structure if and only if $\u2329b,b\u232a<1$. This is consistent with the result in [21].

#### 4.3. Bogoslovsky–Kropina Metrics $L={A}^{1-q}{B}^{2q}$

**Proposition**

**3.**

- (i)
- $\u2329b,b\u232a>0$ and $q\in [-1,1).$ In this case, the future-pointing cones of L coincide with those of A.
- (ii)
- $\u2329b,b\u232a=0$ and $q\in (-1,1).$ In this case, the future-pointing cones of L coincide with those of A.
- (iii)
- $\u2329b,b\u232a<0$ and $q\in (0,1)$. In this case, the future-pointing cones of L are obtained by intersecting the future-pointing cones of A with the half-space $B>0$.

**Proof.**

- $\to :$ Assume that L defines a spacetime structure and let us first identify the future-pointing cones ${\mathcal{T}}_{x}$ of L, together with the domain of the definition of $\mathsf{\Psi}$.

- 1.
- If b is non-space-like with respect to a, then $B=0$ cannot happen inside ${\mathcal{T}}_{x}^{a}$ (it can, in the worst case, when b is a-light-like, happen on its boundary). In this case, we thus have ${\mathcal{T}}_{x}={\mathcal{T}}_{x}^{a}$; moreover, the reverse Cauchy–Schwarz inequality tells us that $s\ge \u2329b,b\u232a$, with equality for $\dot{x}\sim \tilde{b}$. In other words, the minimum value ${s}_{0}=\u2329b,b\u232a$ is always attained on the closure of ${\mathcal{T}}_{x}$.
- 2.
- If b is a-space-like, then points with $B=0$ will always exist inside ${\mathcal{T}}_{x}^{a}$. We note that, in order to have a finite limit for $L={A}^{1-q}{B}^{q}$ as we approach the hyperplane $B=0$, we must necessarily have:$$q>0.$$In this case, the cone ${\mathcal{T}}_{x}$ will be the region of the cone ${\mathcal{T}}_{x}^{a}$ situated in the half-space $B>0$ and s will tend to zero as we approach the boundary points with $B=0$.
- 3.
- On the other hand, as shown above, the boundary of each cone ${\mathcal{T}}_{x}$ of any $(\alpha ,\beta )$-metric spacetime must contain at least one point where $B\ne 0$. In our case, this means that there will always be at least one boundary point satisfying $A=0$, i.e., we necessarily have in ${\mathcal{T}}_{x}$, values $s\to \infty $.

- Briefly: in any case, $\mathsf{\Psi}$ is defined on the entire interval:$$s\in ({s}_{0},\infty ),\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{s}_{0}:=max\{\u2329b,b\u232a,0\}.$$

- Having identified the function $\mathsf{\Psi}$, we are now ready to rewrite, in our case, the conditions in Theorem 1.

- Since, inside the above-identified cones ${\mathcal{T}}_{x}$, we obviously have $A>0$ and $\mathsf{\Psi}>0$, where $\mathsf{\Psi}$ is smooth, in order to check these conditions, we must impose that when approaching any point of the boundary ${\mathcal{T}}_{x}$, we should have $L={A}^{1-q}{B}^{q}\to 0$. This immediately implies:$$q<1.$$
- The first inequality $\mathsf{\Psi}-s{\mathsf{\Psi}}^{\prime}>0$ becomes ${s}^{q}(1-q)>0$, i.e., it is equivalent to the same inequality $q<1$.

- Noting that $\sigma :={s}^{q}{(q-1)}^{2},$ the second inequality (16) reads $(s-\u2329b,b\u232a){\displaystyle \frac{q}{s}}>-1$, which, taking into account that $s>0$, is equivalent to$$\left(q+1\right)s>\u2329b,b\u232aq.$$

- If $\u2329b,b\u232a>0$, then at the lower bound ${s}_{0}=\u2329b,b\u232a>0$, the (non-strict) inequality (29) is identically satisfied. Imposing it for $s\to \infty $, we find:$$q\ge -1,$$
- If $\u2329b,b\u232a=0,$ then for ${s}_{0}=0$, the (non-strict) inequality (29) is identically satisfied. Imposing it for $s\to \infty $, we find:$$q\ge -1.$$Yet, we note that $q=-1$ cannot actually happen, as it would lead to equality in (29) for all s, which proves the necessity of condition (ii) in our statement.
- If $\u2329b,b\u232a<0$, then for $s\to \infty $, we find $q>-1$, whereas the lower bound ${s}_{0}=0$ gives $q\ge 0$, proving the necessity of condition (iii).

- ←: Assuming one of the situations (i)–(iii) happens, then the conditions in Theorem 1 are immediately satisfied on the said cones by $\mathsf{\Psi}:({s}_{0},\infty )\to \mathbb{R},\phantom{\rule{4pt}{0ex}}\mathsf{\Psi}\left(s\right)={s}^{q}$. Hence, L defines a spacetime structure. □

#### 4.4. Generalized m-Kropina (VGR) Metrics

**Proposition**

**4.**

- (i)
- $\u2329b,b\u232a\ge 0\phantom{\rule{4pt}{0ex}}\phantom{\rule{4.pt}{0ex}}\mathrm{and}\phantom{\rule{4.pt}{0ex}}\phantom{\rule{4pt}{0ex}}p\in (-1,1]$.In this case, the future-pointing time-like cones of L are the connected components of the cones $kA+m{B}^{2}>0$ contained in the future-pointing time-like cones ${\mathcal{T}}_{x}^{a}$.
- (ii)
- $\u2329b,b\u232a<0\phantom{\rule{4pt}{0ex}}\phantom{\rule{4.pt}{0ex}}\mathrm{and}\phantom{\rule{4.pt}{0ex}}\phantom{\rule{4pt}{0ex}}p\in (-1,0)$.In this case, the future-pointing time-like cones of L are regions of the the cones $kA+m{B}^{2}>0$ lying both in the future-pointing time-like cones ${\mathcal{T}}_{x}^{a}$ and in the half-space $B>0$.

**Proof.**

- 1.
- Assume $\u2329b,b\u232a\ge 0$. Then, $B=0$ cannot happen inside the future-pointing time-like cones of a, and hence cannot occur in the Finslerian ones ${\mathcal{T}}_{x}$. Thus, in this case, each of the cones ${\mathcal{T}}_{x}$ is the connected component of the convex cone:$$kA+m{B}^{2}>0$$$$(k{a}_{ij}+m{b}_{i}{b}_{j}){\tilde{b}}^{i}{\tilde{b}}^{j}=\u2329b,b\u232a(k+m\u2329b,b\u232a)\ge 0,$$At this lower bound, inequality (36) reduces to:$$\u2329b,b\u232a(k+m\u2329b,b\u232a)\ge 0$$
- 2.
- Suppose $\u2329b,b\u232a<0$. Then, the hyperplane $B=0$ has common points with the cone $kA+m{B}^{2}>0$. However, points with $B=0$ are either singular points for L (if $p>0$), or null cone points if $p<0$. Obviously, the only viable situation is the second one, namely:$$p\in (-1,0).$$Knowing this, evaluation of (36) at the lower bound of the interval for s, which is in this case $s=0$, gives:$$\u2329b,b\u232akp\ge 0$$The future-pointing time-like cones of L are then the intersections of the cone $kA+m{B}^{2}>0$ with the future-pointing cones ${\mathcal{T}}_{x}^{a}$ and the half-space $B>0$.

- $\leftarrow :$ Assuming $k>0$ and $m<0$, then the matrix $(k{a}_{ij}+m{b}_{i}{b}_{j})$ has a Lorentzian signature $(+,-,-,-)$, meaning that the set $kA+m{B}^{2}>0$ is a convex cone, which is, moreover, contained in the cone $A>0$. Then, assuming one of the situations (i) or (ii) holds, one can immediately check the conditions of Theorem (1) for the specified ${\mathcal{T}}_{x}$ (note that statement (ii) in the theorem is equivalent to (34) and (36)). □

#### 4.5. Exponential Metrics $L=A{e}^{P\left(s\right)}$

**Proposition**

**5.**

- If this is the case, then the future-pointing time-like cones of L coincide with those of the Lorentzian metric A.

**Proof.**

**Example**

**1.**

## 5. Isometries of $(\mathbf{\alpha},\mathbf{\beta})$-Metric Spacetimes

**Particular case: trivial symmetries.**Assume L is non-Riemannian, that is, ${\mathsf{\Psi}}^{\prime}\left(s\right)\ne 0$ at least on some interval ${I}_{0}\subset I.$ If $\xi $ is a Killing vector field of $L,$ then

**Nontrivial symmetries.**In the following, let us explore which $\left(\alpha ,\beta \right)$-metrics admit nontrivial symmetries. We find the following result, which is valid for all pseudo-Finsler $\left(\alpha ,\beta \right)$ metrics, independently of their signature.

**Theorem**

**2.**

**Proof.**

- 1.
- ${\lambda}_{1},{\mu}_{1}=0.$ In this case, $\frac{s{\mathsf{\Psi}}^{\prime}-\mathsf{\Psi}}{2{\mathsf{\Psi}}^{\prime}}}={\displaystyle \frac{{\lambda}_{2}}{{\mu}_{2}}}=:\kappa \left(x\right)$ only. Integration of this equation gives:$$\mathsf{\Psi}=c\left(s-2\kappa \right),\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}c=c\left(x\right),$$
- 2.
- ${\lambda}_{1}=0,{\mu}_{1}\ne 0.$ In this situation, the first equality (48) gives, after simplification by B,$$\frac{{\xi}^{C}\left(B\right)}{{\xi}^{C}\left(A\right)}}={\displaystyle \frac{{\lambda}_{2}B}{{\mu}_{1}A+{\mu}_{2}{B}^{2}}}.$$We note that, on the right hand side, on the one hand, we must have ${\lambda}_{2}\ne 0$ (otherwise we only get trivial symmetries ${\xi}^{C}\left(B\right)=0$) and, on the other hand, the numerator and the denominator must be given by relatively prime polynomials; in the contrary case, A would admit B as a factor, which is not possible, since its $\dot{x}$-Hessian $\left(2{a}_{ij}\right)$ is nondegenerate. However, then B must divide ${\xi}^{C}\left(B\right)$, that is,$${\xi}^{C}\left(B\right)=\kappa {\lambda}_{2}B,$$$${\xi}^{C}\left(A\right)=\kappa \left({\mu}_{1}A+{\mu}_{2}{B}^{2}\right).$$The functions $\mathsf{\Psi}$ corresponding to such symmetries are then obtained from (49), which in our case reads$$s-\frac{\mathsf{\Psi}}{{\mathsf{\Psi}}^{\prime}}=\frac{2{\lambda}_{2}s}{{\mu}_{1}+{\mu}_{2}s}$$$$\mathsf{\Psi}=cexp\left(\int {\displaystyle \frac{\left({\mu}_{1}+s{\mu}_{2}\right)ds}{{\mu}_{2}{s}^{2}+\left({\mu}_{1}-2{\lambda}_{2}\right)s}}\right).$$
- 3.
- ${\lambda}_{1}\ne 0,{\mu}_{1}=0.$ This gives $\frac{B{\xi}^{C}\left(B\right)}{{\xi}^{C}\left(A\right)}}={\displaystyle \frac{{\lambda}_{1}A+{\lambda}_{2}{B}^{2}}{{\mu}_{2}{B}^{2}}$ and, equivalently,$${B}^{3}{\mu}_{2}{\xi}^{C}\left(B\right)={\xi}^{C}\left(A\right)\left({\lambda}_{1}A+{\lambda}_{2}{B}^{2}\right).$$This implies that B must divide ${\lambda}_{1}A+{\lambda}_{2}{B}^{2}$, since ${\xi}^{C}\left(A\right)$ is of degree two and hence it cannot “swallow” more than a factor of ${B}^{2}$ of the ${B}^{3}$ from the left hand side. This in turn implies that B divides $A,$ thus leading to a degenerate $\dot{x}$-Hessian for A, which is impossible.
- 4.
- ${\lambda}_{1},{\mu}_{1}\ne 0.$ In this case, we have:$$\frac{B{\xi}^{C}\left(B\right)}{{\xi}^{C}\left(A\right)}}={\displaystyle \frac{{\lambda}_{1}A+{\lambda}_{2}{B}^{2}}{{\mu}_{1}A+{\mu}_{2}{B}^{2}}}.$$The ratio on the right hand side is either irreducible or a function of x only. This is derived as follows. Assuming that it can be simplified by a first degree factor, then both ${\lambda}_{1}A+{\lambda}_{2}{B}^{2}$ and ${\mu}_{1}A+{\mu}_{2}{B}^{2}$ must be decomposable; in particular, they must have degenerate $\dot{x}$-Hessians, i.e.,$$det\left({\lambda}_{1}{a}_{ij}+{\lambda}_{2}{b}_{i}{b}_{j}\right)=det\left({\mu}_{1}{a}_{ij}+{\mu}_{2}{b}_{i}{b}_{j}\right)=0.$$Using Lemma A1 (see Appendix A), that is ${\lambda}_{1}^{4}deta\left(1+{\displaystyle \frac{{\lambda}_{2}}{{\lambda}_{1}}}\u2329b,b\u232a\right)=$${\mu}_{1}^{4}deta\left(1+{\displaystyle \frac{{\mu}_{2}}{{\mu}_{1}}}\u2329b,b\u232a\right)=0,$ leads to$$\frac{{\lambda}_{2}}{{\lambda}_{1}}}={\displaystyle \frac{{\mu}_{2}}{{\mu}_{1}}}.$$The latter actually means that the ratio $\frac{{\lambda}_{1}A+{\lambda}_{2}{B}^{2}}{{\mu}_{1}A+{\mu}_{2}{B}^{2}}}={\displaystyle \frac{{\lambda}_{1}}{{\mu}_{1}}$ depends on x only (i.e., it is actually simplified by a second degree factor, not a first degree one, as assumed).Thus, we only have two possibilities:
- (a)
- $\frac{{\lambda}_{1}A+{\lambda}_{2}{B}^{2}}{{\mu}_{1}A+{\mu}_{2}{B}^{2}}}=\kappa \left(x\right)$, that is, $\frac{s{\mathsf{\Psi}}^{\prime}-\mathsf{\Psi}}{2{\mathsf{\Psi}}^{\prime}}}=\kappa $, which, by a similar reasoning to Case 1, entails that L is actually pseudo-Riemannian.
- (b)
- $\frac{{\lambda}_{1}A+{\lambda}_{2}{B}^{2}}{{\mu}_{1}A+{\mu}_{2}{B}^{2}}$ is irreducible. In this case, from (53), we find that B must either divide ${\xi}^{C}\left(A\right)$ (but, then, $\frac{{\lambda}_{1}A+{\lambda}_{2}{B}^{2}}{{\mu}_{1}A+{\mu}_{2}{B}^{2}}}={\displaystyle \frac{B{\xi}^{C}\left(B\right)}{{\xi}^{C}\left(A\right)}$ would equal a ratio of first degree polynomials, which contradicts the irreducibility assumption) or it must divide ${\lambda}_{1}A+{\lambda}_{2}{B}^{2}$, which, taking into account that A and B are always relatively prime, leads to ${\lambda}_{1}=0,$ in contradiction with the hypothesis ${\lambda}_{1}\ne 0.$Therefore, there are no properly Finslerian functions L with ${\lambda}_{1},{\mu}_{1}\ne 0.$

**Example**

**2.**

**Nontrivial symmetries of a Bogoslovsky–Kropina spacetime metric.**Consider, on $M={\mathbb{R}}^{4},$ a conformal deformation a of the Minkowski metric $\eta =diag(1,-1,-1,-1)$ and a light-like 1-form b, as follows:

## 6. Conclusions

- 1.
- 2.
- To calculate and understand the precise physical interpretation of the various non-Riemannian notions such as Berwald or Landsberg curvature (see, e.g., [47]), with a special focus on spacetimes with $(\alpha ,\beta )$-metrics. We conjecture that these non-Riemannian quantities could account for at least a part of the observed dark energy phenomenology.
- 3.
- In light of the above, it will be important to construct some non-trivial Ricci flat Finsler $(\alpha ,\beta )$-metric spacetimes with some nonzero non-Riemannian quantities. Some interesting Ricci flat Finsler spacetimes have been constructed, although they are not in an $(\alpha ,\beta )$-form, see, e.g., [48].
- 4.
- To construct the most general spatially spherically symmetric and the most general cosmologically symmetric spacetimes of $(\alpha ,\beta )$-metric type, whose underlying Lorentzian metrics do not possess such symmetries.
- 5.
- To completely classify $(\alpha ,\beta )$-metrics which lead to well-defined Finsler spacetimes, for which the underlying pseudo-Riemannian metric a has a non-Lorentzian signature.

## Author Contributions

## Funding

## Acknowledgments

## Conflicts of Interest

## Appendix A. Calculation of $det\left({\mathit{g}}_{\dot{\mathit{x}}}\right)$

**Lemma**

**A1**

- 1.
- $det({Q}_{\mathit{ij}}+\delta {C}_{i}{C}_{j})=det\left({Q}_{\mathit{ij}}\right)\left(1+\delta {C}^{k}{C}_{k}\right)$ for all $n$-vectors $C=\left({C}_{i}\right)\in {\mathbb{C}}^{n}$, $\delta \in \mathbb{R},$ where the indices of ${C}_{i}$ have been raised by means of the inverse $\left({Q}^{\mathit{ij}}\right).$
- 2.
- If $1+\delta {C}^{k}{C}_{k}\ne 0,$ the inverse of the matrix $\left({Q}_{\mathit{ij}}+\delta {C}_{i}{C}_{j}\right)$ exists and has the entries:$${Q}^{\mathit{ij}}-{\displaystyle \frac{\delta {C}^{i}{C}^{j}}{1+\delta {C}^{k}{C}_{k}}}.$$

**Note.**Some other versions of this result, e.g., [49], require the matrix $Q$ to be symmetric. Yet, as stated in [45], this restriction (though it applies to all the examples in this article), is not necessary, so we preferred to present the results in full generality.

**Corollary**

**A1.**

**Proof.**

- To calculate the blocks appearing in (A4), we use ${A}_{\xb7i}=2{a}_{\mathit{ik}}{\dot{x}}^{k}$, which yields: ${a}^{\mathit{ij}}{A}_{\xb7i}=2{\dot{x}}^{j},{a}^{\mathit{ij}}{A}_{\xb7i}{b}_{j}=2B,{s}_{\xb7i}=\frac{2}{A}\left({\mathrm{Bb}}_{i}-s{\tilde{\dot{x}}}_{i}\right)$ (where ${\tilde{\dot{x}}}_{i}={a}_{\mathit{ij}}{\dot{x}}^{j}$) and finally:$${B}^{k}{B}_{k}=\u2329b,b\u232a,\phantom{\rule{2.em}{0ex}}{C}^{k}{C}_{k}={\displaystyle \frac{4s}{A}}\left(\u2329b,b\u232a-s\right),\phantom{\rule{2.em}{0ex}}{\left({B}^{k}{C}_{k}\right)}^{2}={\displaystyle \frac{4s}{A}}{\left(\u2329b,b\u232a-s\right)}^{2}.$$
- On subsets where $1+\delta {b}^{k}{b}_{k}\ne 0$, we can apply the corollary, which gives, after a brief computation:$$det\left({g}_{\mathit{ij}}\right)={(\mathsf{\Psi}-s{\mathsf{\Psi}}^{\prime})}^{n-2}det\left({a}_{\mathit{ij}}\right)\left[\mathsf{\Psi}\left(\mathsf{\Psi}-s{\mathsf{\Psi}}^{\prime}\right)+\left(\u2329b,b\u232a-s\right)\left(\mathsf{\Psi}{\mathsf{\Psi}}^{\prime}+2s\mathsf{\Psi}{\mathsf{\Psi}}^{\prime \prime}-s{{\mathsf{\Psi}}^{\prime}}^{2}\right)\right].$$The square bracket can be rewritten as:$$\frac{{\mathsf{\Psi}}^{2}}{\left(\mathsf{\Psi}-s{\mathsf{\Psi}}^{\prime}\right)}}{\displaystyle \frac{\partial}{\partial s}}\left(\left(s-\u2329b,b\u232a\right){\displaystyle \frac{{\left(\mathsf{\Psi}-s{\mathsf{\Psi}}^{\prime}\right)}^{2}}{\mathsf{\Psi}}}\right),$$
- The result can then be prolonged by continuity also at points where $1+\delta {b}^{k}{b}_{k}=0$. To see this, we note that this equality cannot happen on any entire interval ${I}_{0}\subset I$, as this would entail that, for $s\in {I}_{0}$,$$1+{\displaystyle \frac{{\mathsf{\Psi}}^{\prime}\u2329b,b\u232a}{\mathsf{\Psi}-s{\mathsf{\Psi}}^{\prime}}}=0\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{2.em}{0ex}}\iff \phantom{\rule{2.em}{0ex}}{\displaystyle \frac{{\mathsf{\Psi}}^{\prime}}{\mathsf{\Psi}}}={\displaystyle \frac{1}{s-\u2329b,b\u232a}}.$$That is, $\mathsf{\Psi}=\kappa \left(s-\u2329b,b\u232a\right),$ where $\kappa =\kappa \left(x\right)$ only. However, on the (open) subset ${s}^{-1}\left({I}_{0}\right)\subset \mathcal{A}$, we would then have $L=A\mathsf{\Psi}=\kappa \left({B}^{2}-\u2329b,b\u232aA\right),$ which has degenerate Hessian ${g}_{\mathit{ij}}=\kappa \left({b}_{i}{b}_{j}-\u2329b,b\u232a{a}_{\mathit{ij}}\right)$ and hence does not represent a pseudo-Finsler function.

## Notes

1 | This article appeared during the production process of the present paper. |

2 |

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**Figure 1.**The relative positions of the spacetime cones. (

**a**) $\partial {\mathcal{T}}_{x}:\{A=0\}$, (

**b**) $\partial {\mathcal{T}}_{x}:\{\mathsf{\Psi}=0\}$, (

**c**) $\partial {\mathcal{T}}_{x}:\{A=0\}\mathrm{or}\{\mathsf{\Psi}=0\}$, (

**d**) $\partial {\mathcal{T}}_{x}\cap \partial {\mathcal{T}}_{x}^{a}=\ell $.

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Voicu, N.; Friedl-Szász, A.; Popovici-Popescu, E.; Pfeifer, C.
The Finsler Spacetime Condition for (*α*,*β*)-Metrics and Their Isometries. *Universe* **2023**, *9*, 198.
https://doi.org/10.3390/universe9040198

**AMA Style**

Voicu N, Friedl-Szász A, Popovici-Popescu E, Pfeifer C.
The Finsler Spacetime Condition for (*α*,*β*)-Metrics and Their Isometries. *Universe*. 2023; 9(4):198.
https://doi.org/10.3390/universe9040198

**Chicago/Turabian Style**

Voicu, Nicoleta, Annamária Friedl-Szász, Elena Popovici-Popescu, and Christian Pfeifer.
2023. "The Finsler Spacetime Condition for (*α*,*β*)-Metrics and Their Isometries" *Universe* 9, no. 4: 198.
https://doi.org/10.3390/universe9040198