Null geodesics in the C-metric with cosmological constant

As mentioned above, the geodesics can only exist in the domains of $x$ where $X\geq 0$. This condition immediately requires $Q\geq 0$, and that $Q=0$ is only possible for geodesics of zero angular momentum.

Let us denote the roots by $\{x_{*},x_{1},x_{2}\}$ with the ordering $x_{*}\leq x_{1}\leq x_{2}$ whenever they are real. Assuming for the moment that this is the case, we see from Eq. (19) that $X$ is obtained from $G$ by multiplication with $Q$, then shifted down by $\Phi^{2}$ units, as sketched in Fig. 1. From the figure we infer that the relative positions of the roots of $G$ and $X$ are

and that geodesics exist in the domain

for which $X\geq 0$, this domain is marked blue in Fig. 1. The other domain $x\leq x_{*}$ is outside (4) and is therefore irrelevant.

The reality of the roots are determined by the discriminant

The function $X$ will have real and distinct roots if $\Delta(X)>0$, or

where $\Omega_{\pm}$ is as defined in Eq. (3). If $\Phi^{2}>-\Omega_{-}Q$, $x_{1}$ and $x_{2}$ become a complex conjugate pair, while if $\Phi^{2}<-\Omega_{+}Q$, the complex conjugate pair is $x_{*}$ and $x_{1}$. As we have previously noted, $\Omega_{+}$ is positive for $0<mA<\frac{1}{2}$. Therefore this latter case is physically irrelevant.

With the Descartes rule of signs, we can further pin down the specific regions occupied by the roots. Using this rule, we see that if $Q-\Phi^{2}>0$, $X$ has one positive root, and if $Q-\Phi^{2}>0$, $X$ has two positive roots. Furthermore, since $-\Omega_{-}$ is always greater than $1$ for $0<mA<\frac{1}{2}$, then $Q-\Phi^{2}$ crosses from negative to positive before reaching $\Phi^{2}=-\Omega_{-}Q$, where the discriminant vanishes. In other words, $x_{1}$ crosses from negative to positive before coalescing with $x_{1}$ in the positive domain.

From these considerations, we may organise the $x$-motion of the geodesics into five cases:

Case A, $\Phi^{2}=0$. These are geodesics with zero angular momentum in the $z$-component. In this case, $x_{1}=-1$ and $x_{2}=1$ and the photon can access the full Lorentzian domain $-1<x<1$. By Eq. (18b), $\phi$ is constant for these geodesics. These are polar orbits that intersects with the north or south axis of the C-metric (and possibly colliding with the cosmic string/strut).

Case B, $0<\Phi^{2}<Q$. In this case the motion is restricted to $x_{1}\leq x\leq x_{2}$ which is a proper subset of the Lorentzian region (4). The root $x_{1}$ is negative while $x_{2}$ is positive, so this domain contains the ‘equatorial plane’ $x=0$. The distinguishing feature of this case is that the geodesics oscillates in the polar direction between $x_{1}$ and $x_{2}$, and crosses the ‘equatorial plane’ $x=0$ in between. Increasing the angular momentum narrows the range between $x_{1}$ and $x_{2}$. As $\Phi^{2}$ is increased towards $Q$, the root $x_{1}$ approaches zero.

Case C, $\Phi^{2}=Q$. In this case, the constant term of $X$ vanishes and $0$ is one of its roots. By continuity from the previous case, we identify this root to be $x_{1}=0$. The other two roots can be written explicitly as

In this case, the photon’s domain of polar oscillation is bounded at one side by the ‘equatorial plane’ as measured in coordinates $x=\cos\theta=0$, and the ‘southern hemisphere’ is no longer accessible to the photons.

Case D, $Q<\Phi^{2}<-\Omega_{-}Q$. By the Descartes rule of signs, there are now at least two positive roots. This is by continuity from the previous cases as $x_{1}$ crosses the past the point $x=0$ and now becomes a positive root. Therefore the roots in this case are ordered by

In this case, the domain of polar oscillation is now strictly to the north of the ’equatorial plane’. Further increasing $\Phi^{2}$ towards $Q\Omega_{-}$ continues to narrow the domain as $x_{1}$ approaches $x_{2}$.

Case E, $\Phi^{2}=-\Omega_{-}Q$. Here the discriminant $\Delta(X)$ vanishes and the two roots $x_{1}$ and $x_{2}$ coalesce into a degenerate root given by

Since this is a degenerate root, we have $X(x_{\mathrm{pr}})=X^{\prime}(x_{\mathrm{pr}})=0$. By Eq. (18c), we see that this trajectory corresponds to a constant $x=x_{\mathrm{pc}}>0$ fixed at a polar angle located north of the ‘equator’. This is the case referred to by Ref. [27] as the photon cone. The other root is $x_{*}=\frac{-\left(1+\sqrt{1+12m^{2}A^{2}}\right)}{6mA}$, which is still located beyond the physically relevant range of interest.

For values of $\Phi^{2}>-\Omega_{-}Q$ the two roots $x_{1}$ and $x_{2}$ become complex and $X$ is negative in the domain $-1\leq x\leq 1$. No geodesics can exist in this case. The cases described above can be summarised in Fig. 2, showing the root structure of $X$ as $\Phi^{2}$ varies from $0$ to $-\Omega_{-}Q$.

Let us denote the roots of $Y$ by $\{y_{1},y_{2},y_{*}\}$ with the ordering $y_{1}\leq y_{2}\leq y_{*}$ whenever they are real. The constant term of $Y$ is always positive. The Descartes rule of sign tells us that there is only one positive root, which would be the largest, $y_{*}$.

We note that Eq. (18d) requires $E^{2}$ to be non-zero. So in the following we consider $E^{2}>0$. Assuming for the moment that all roots of $F$ and $Y$ are real, we notice that $Y$ is obtained from $-F$ by multiplication of $Q$, then shifted upwards by $E^{2}$ units. The existence of geodesics requires $Y\geq 0$. Recalling that the domain for our static Lorentzian patch is where $F>0$, the relevant domain for null geodesics are marked blue in Fig. 3. From this graph we infer the roots are ordered by

From Fig. 3, we can see that as $E^{2}$ increases, the graph of $Y$ is raised further up and eventually $y_{1}$ coalesces with $y_{2}$, becoming complex. The precise value of $E^{2}$ where this happens can be determined from the discriminant of $Y$,

The polynomial $Y$ has real and distinct roots for $\Delta(Y)>0$, or

where $\Omega_{\pm}$ is the same as in Eq. (3). As $E^{2}$ approaches the upper bound, $y_{1}$ approaches $y_{2}$, whereas if $E^{2}$ approaches the lower bound, $y_{2}$ approaches $y_{*}$. Now, in the AdS ($\ell^{2}>0$) case the lower bound consists of a positive and negative term, and one may wonder whether this gives a positive non-zero lower bound for $E^{2}$. However, we can show that this lower bound is always negative as long as $F$ has distinct real roots. Recalling Eq. (2), this condition requires $\ell^{2}>\frac{54m^{2}A^{2}\Omega_{+}}{2A^{2}\left(1-4m^{2}A^{2}\right)^{2}}$. Upon rearranging, we have

Using the relation $\Omega_{+}\Omega_{-}=-\frac{\left(1-4m^{2}A^{2}\right)^{2}}{27m^{2}A^{2}}$, one directly computes

Therefore this lower bound could not be reached by a positive $E^{2}$.

Next we organise the different cases as we increase $E^{2}$ continuously:

Case I, $0<E^{2}<\left(\Omega_{+}+\frac{1}{A^{2}\ell^{2}}\right)Q$. This is the case depicted in Fig. 3, where $Y$ has distinct roots. A photon may propagate in one of the two possible disconnected domains

In other words, there is a ‘potential barrier’ at $y_{1}<y<y_{2}$ which prevents photons in $a<y\leq y_{1}$ to access $y_{1}\leq y<b$, and vice versa. As $E^{2}$ is increased, the potential barrier narrows and $y_{1}$ and $y_{2}$ approach each other.

Case II, $E^{2}=\left(\Omega_{+}+\frac{1}{A^{2}\ell^{2}}\right)Q$. Here $y_{1}$ and $y_{2}$ coalesce into a degenerate root given by

In this case, $Y(y_{\mathrm{ps}})=Y^{\prime}(y_{\mathrm{ps}})=0$, and Eq. (18d) is solved by $y$ being constant at $y_{\mathrm{ps}}$. This is the aspherical photon surface of the C-metric. This terminology was introduced by Gibbons and Warnick [29] where they have shown that the geometry of constant-$y$ surfaces is that of a deformed sphere. This is also the analogue to the photon spheres around the Kerr black hole [30].

Case III, $E>\left(\Omega_{+}+\frac{1}{A^{2}\ell^{2}}\right)Q$. In this case $y_{1}$ and $y_{2}$ become complex conjugate pairs. Then $F$ is always positive in the Lorentzian region $a<y<b$ and the potential barrier no longer exists. So photons from the neighbourhood of the black-hole horizon have access to the acceleration horizon, and vice versa.

The cases described above are summarised in Fig. 4, showing the different cases as $E^{2}$ varies from $E^{2}>0$.

In a metric of mass parameter $m$, acceleration parameter $A$ and (A)dS parameter $\ell^{2}$, a photon trajectory can take either of the following cases classified by A, B, C, or D based on its angular momentum, and cases I, II, or III based on its energy. We have seen that the root structure of $X$ depends on $\Phi^{2}/Q$, whereas the root structure of $Y$ depends on $E^{2}/Q$. It is then convenient to define dimensionless quantities

The type of motion can now be parametrised in $(\eta,\xi)$-space subdivided into 12 domains which we denote AI, AII, AIII, BI, …, EIII. For example, a photon trajectory with angular momentum in the range $1<\eta<1$ (Case B) and with energy $\xi=\Omega_{+}+\frac{1}{A^{2}\ell^{2}}$ (Case I), will be called Case BII. For another example, the circular photon orbits of constant $x$ and $y$ belongs to EII, where $\left(\eta,\xi\right)=\left(-\Omega_{-},\Omega_{+}+\frac{1}{A^{2}\ell^{2}}\right)$.

The respective domains for each case are sketched in Fig. 5. The cases AI, AIII, BII, CI, CIII, DII, EI, and EIII occupy one-dimensional lines, while AII, CII, and EII are points. The aspherical photon surfaces are represented by the blue horizontal line, starting from the part where $0\leq\eta<1$ is the case AII, for which the domain of polar oscillation crosses the ‘equatorial plane’ $x=0$. For $1<\eta<-\Omega_{-}$ on this line is where the surface lies completely in the ‘northern hemisphere’ $x>0$ and never crosses the ‘equatorial plane’. The critical case between the two is the point BII, where the southernmost part of the polar oscillation just touches the equatorial plane. The vertical red line represents $\ell=-\Omega_{-}$ where the geodesics have constant $x$, containing the cases DI, DII, and DIII. The photon surface and constant-$x$ lines intersect at the point $(-\Omega_{-},\Omega_{+}+\frac{1}{A^{2}\ell^{2}})$, corresponding to the circular photon orbit where both $x$ and $y$ are constant.

Let us now demonstrate some examples of the various cases. The trajectories are obtained by either by solving Eq. (13) or using the analytical solutions of Sec. 4, giving parametric equations for $x$ and $y$. We can convert to more familiar Boyer–Lindquist-type coordinates by the transformation

In the Ricci-flat case, these coordinates are those used in Refs. [18, 20, 21]. We further transform to Cartesian-like coordinates by the transformation

For the moment, we assume the azimuthal coordinate has periodicity $\phi\in[0,2\pi)$, leaving both the cosmic strut and string present. As mentioned in Sec. 2 and in the literature, we can remove one of them by appropriately redefining the periodicity of $\phi$. For concreteness, the following examples in this section take place in an AdS C-metric with parameters

For these values, we have³³3The symbol ‘$\simeq$’ indicates that numerical values are being displayed up to five significant figures.

We start with an example from Case BI shown in Fig. 6, with $(\eta,\xi)=(0.9,1.1)$. In this case, geodesics can propagate in two disconnected domains separated by a potential barrier $y_{1}<y<y_{2}$. The blue curves in Fig. 6 represent a photon starting just before the acceleration horizon at $y=-0.6$ towards the black hole, encountering the potential barrier at $y_{1}$ before going off into the acceleration horizon. The red curve represents a photon starting just outside the black hole horizon at $y=-2.6$ heading outwards. It encounters the potential barrier at $y_{2}$ and subsequently falls into the black hole.

For the next example let us consider photon cones where $\eta=-\Omega_{-}$. Here the geodesics are confined to $x_{\mathrm{pc}}=\mathrm{constant}$. In Fig. 7 we show explicitly the cases EI and EIII. For the case EI shown in Fig. 7(a), the blue curve shows the photon initially at a position just before the acceleration horizon, $y=-0.6$, heading towards the black hole until it encounters the potential barrier $y_{1}$, before heading away and falling beyond the acceleration horizon. The red curve is a trajectory starting just outside the black-hole horizon, $y=-2.6$ heading outwards before encountering the potential barrier at $y_{2}$, upon which it turns back and falls into the black hole. The case EIII example is shown in Fig. 7(b), showing the photon initially at $y=-0.6$ taking a trajectory that eventually falls into the black-hole horizon. At all times the $x$ coordinate is constant at $x=x_{\mathrm{pc}}$.

Perhaps the more interesting case would be the aspherical photon surface with $\xi=\Omega_{+}+\frac{1}{A^{2}\ell^{2}}$ and $y$ is constant at $y_{\mathrm{ps}}$. The examples are depicted in Fig. 8. In Fig. 8(a), we show a geodesic from case BII for $\eta=0.9$. Here the geodesics oscillates in the polar direction that contains the ‘equatorial plane’. Increasing $\eta$ to $1.02$, the domain of the $x$-oscillation narrows. This value of $\eta$ corresponds to case DII shown in Fig. 8(b), where $x_{2}$ is positive. So the photon oscillates in a domain contained in the ‘northern hemisphere’ and does not cross the ‘equatorial plane’. Further increasing $\eta$ to the value $-\Omega_{-}$, we now have $x$ also being constant at $x_{\mathrm{pc}}$. This is now a circular photon ring of Case EII, shown in Fig. 8(c).

Here, it is convenient to express $X$ in terms of its roots. Here we shall assume the roots are distinct (corresponding to cases A–D), since there are no physically relevant geodesics for when $X$ has complex roots, and the case with degenerate roots simply corresponds to the photon cones of constant $x$. Therefore we write

Choosing the branch and the initial conditions will determine the specific forms of the solutions. When a geodesic encounters a turning point where $X=0$, care must be taken in correctly choosing the correct branch for its subsequent motion. Essentially one must always choose the branch such that future-directed geodesics remain future-directed, and similarly for the past-directed geodesics. Future-directed geodesics are those where $\lambda$ increases throughout the motion, since by Eq. (40a) lead to $t$ increasing as well, and the opposite holds for past-directed geodesics.

Firstly, suppose we take the initial conditions $x(0)=x_{2}$. In this case we choose the lower sign of Eq. (40c) so that $\lambda$ increases as the $x$-coordinate of the photon decreases away from $x_{2}$. Then, Eq. (40c) can be integrated as

This integral can be evaluated exactly (see, e.g., [32]). The result is⁴⁴4Note that $\lambda$ lies outside the square root.

where $\mathrm{F}(\varphi,k)$ is the elliptic integral of the first kind. Inverting to express $x$ as a function of $\lambda$, we get

where $\mathrm{sn}(\psi,q)$ is the Jacobi elliptic function of the first kind.

On the other hand, if we take the initial conditions $x(0)=x_{1}$, we choose the upper sign of Eq. (40c) so that $\lambda$ increases as the $x$-coordinate of the photon increases away from the southern boundary $x_{1}$. In this case the integral is

Inverting this to solve for $x$, we have

Similar to the solutions for $x(\lambda)$ the specific forms of the solutions depend on the initial conditions and the choice of branch. As before we shall choose the branches that correspond to future-directed geodesics. Here we shall consider Cases I and III, since Case II is simply the photon surface where $y$ is constant.

We first consider Case I, where the roots of $Y$ are real and distinct. Hence we write

In this case the geodesics can either exist in $y>y_{2}$ or $y<y_{1}$. In the former case, we take the initial conditions $y(0)=y_{2}$. We choose the upper branch of the square root so that $\lambda$ increases as the photon increases in $y$ away from $y_{2}$ (and heads towards the acceleration horizon). Then Eq. (40d) can be integrated as

We invert this to solve for $y$,

On the other hand, for photons in the domain $y<y_{1}$, we shall take $y(0)=y_{1}$ and choose the lower branch of the square root so that $\lambda$ increases as the photon’s $y$-coordinate decreases away from $y_{1}$ (and heads towards the black hole). Here the integral is