The signal from an emitting source moving in a Schwarzschild spacetime under the influence of a radiation field

In previous works [1, 2] we studied the motion of test particles in the gravitational background of a black hole under the influence of a superimposed radiation field. The particle–radiation interaction is assumed to take place through Thomson scattering only. Radiation exerts a drag force on the particle's motion, a well-known effect in the Newtonian regime that is often referred to as the Poynting–Robertson effect [3, 4]. We limited ourselves to the case of particles moving in the equatorial plane of the Schwarzschild and Kerr backgrounds, immersed in an (equatorial) radiation field composed of photons having the same specific angular momentum and traveling along geodesics. In [1], the simplest case was considered of a radiation field with zero angular momentum, i.e. photons moving in a purely radial direction with respect to the locally nonrotating frames, naturally associated with the family of zero angular momentum observers (ZAMOs). This analysis was extended in [2] to the case of a radiation field characterized by photons possessing all the same, non-null specific angular momentum. We found that those particles which do not escape to infinity are attracted to a single critical radius outside the horizon where they stay at rest with respect to ZAMOs; if the radiation field and the black hole have zero angular momentum, or move in a circular orbit, then a non-null angular momentum characterizes the radiation field and/or the black hole.

In this paper we consider an emitting source (hot spot) which moves in a Schwarzschild metric under the influence of a radiation field, and calculate the corresponding signal seen by an observer at infinity. In particular, we derive the flux, redshift factor and solid angle of the hot spot as a function of (coordinate) time, as well as the time-integrated image of the hot spot in the observer's sky. The results are then compared with those for a spot moving on a circular geodesic in a Schwarzschild field. This analysis will then be extended to the more complex case of a Kerr field in a forthcoming paper.

Consider a Schwarzschild spacetime, whose line element written in standard coordinates is given by

$$\begin{equation} {\rm d}s^2 = -N^2\, {\rm d}t^2 + N^{-2} {\rm d}r^2 + r^2 ({\rm d}\theta ^2 +\sin ^2 \theta\, {\rm d}\phi ^2), \end{equation} \tag{ 2.1 }$$

where N = (1 − 2M/r)^1/2 denotes the lapse function and introduces the usual orthonormal frame adapted to ZAMOs following the time lines

$$\begin{equation} n\equiv e_{\hat{t}}=N^{-1}\partial _t, \qquad e_{\hat{r}}=N\partial _r, \qquad e_{\hat{\theta }}=\frac{1}{r}\partial _\theta , \qquad e_{\hat{\phi }}=\frac{1}{r\sin \theta }\partial _\phi , \end{equation} \tag{ 2.2 }$$

where {∂_t, ∂_r, ∂_θ, ∂_ϕ} is the coordinate frame.

2.1. Circular geodesic motion

Circular geodesic motion of a test particle in the equatorial plane θ = π/2 at r = r₀ is characterized by the 4-velocity

$$\begin{equation} U=U_K=\gamma _K (n \pm \nu _K e_{\hat{\phi }}), \end{equation} \tag{ 2.3 }$$

where the Keplerian value of speed (ν_K), the associated Lorentz factor (γ_K) and angular velocity (ζ_K) are given by

$$\begin{equation} \fl \qquad \nu _K=\sqrt{\frac{M}{r_0-2M}} ,\qquad \gamma _K=\sqrt{\frac{r_0-2M}{r_0-3M}} ,\qquad \zeta _K=\sqrt{\frac{M}{r_0^3}}. \end{equation} \tag{ 2.4 }$$

The ± signs in equation (2.3) correspond to co-rotating ( + ) or counter-rotating ( − ) orbits with respect to increasing values of the azimuthal coordinate ϕ (counter-clockwise motion as seen from above).

The parametric equations of U_K are

$$\begin{equation} t_K=t_0+\Gamma _K \tau ,\qquad r_K=r_0,\qquad \theta _K=\frac{\pi }{2} ,\qquad \phi _K=\phi _0\pm \Omega _K \tau , \end{equation} \tag{ 2.5 }$$

where

$$\begin{equation} \Gamma _K=\sqrt{\frac{r_0}{r_0-3M}} , \qquad \Omega _K=\frac{1}{r_0}\sqrt{\frac{M}{r_0-3M}} . \end{equation} \tag{ 2.6 }$$

2.2. Test particle undergoing the Poynting–Robertson effect

Consider now a test particle in arbitrary motion on the equatorial plane θ = π/2, i.e. with 4-velocity and 3-velocity with respect to the ZAMOs, respectively

$$\begin{equation} \fl U=\gamma (U,n) [n+ \nu (U,n)] ,\qquad \nu (U,n)\equiv \nu ^{\hat{r}}e_{\hat{r}}+\nu ^{\hat{\phi }}e_{\hat{\phi }} = \nu \sin \alpha e_{\hat{r}}+ \nu \cos \alpha e_{\hat{\phi }}, \end{equation} \tag{ 2.7 }$$

where $\gamma (U,n)=1/\sqrt{1-||\nu (U,n)||^2}$ is the Lorentz factor and the abbreviated notation $\nu ^{\hat{a}}=\nu (U,n)^{\hat{a}}$ has been used. In a similar abbreviated notation, ν = ||ν(U, n)|| ⩾ 0 and α are the magnitude of the spatial velocity ν(U, n) and its polar angle measured clockwise from the positive ϕ direction in the r-ϕ tangent plane. Note that sin α = 0 (i.e. α = 0, π) corresponds to a pure azimuthal motion of the particle with respect to the ZAMOs, while cos α = 0 (i.e. α = ±π/2) corresponds to (outward/inward) purely radial motion with respect to the ZAMOs.

Let the particle be accelerated by a test radiation field propagating in a general direction on the equatorial plane. The corresponding equations for ν and α as derived in [2] are given by

$$\begin{eqnarray} &&\eqalign \fl \frac{{\rm d}\nu }{{\rm d}\tau } = -\frac{\sin \alpha }{\gamma }\frac{M}{r^2N} + \frac{A}{N^2r^2|\sin \beta |} [\cos (\alpha -\beta ) -\nu ][1-\nu \cos (\alpha -\beta )] , \\&&\ms \fl \frac{{\rm d}\alpha }{{\rm d}\tau } = \frac{\gamma \cos \alpha }{\nu }\frac{N}{r}\left(\nu ^2-\frac{M}{rN^2}\right) +\frac{A}{\nu } \frac{[1-\nu \cos (\alpha -\beta )] }{N^2r^2|\sin \beta |} \sin (\beta -\alpha ), \end{eqnarray} \tag{ 2.8 }$$

where τ is the proper time parameter along U, A is a positive constant for a given fixed radiation field and

$$\begin{equation} \cos \beta =\frac{b_{\rm (rad)}N}{r} , \end{equation} \tag{ 2.9 }$$

the constant b_(rad) being the photon impact parameter defined as the ratio between the conserved angular momentum and the energy associated with the rotational and time-like Killing vector fields, respectively. In the zero angular momentum limit, b_(rad) → 0, cos β → 0, then sin β → ±1, cos (α − β) → ±sin α, sin (β − α) → ±cos α and the two equations (2.8) reduce to those in [1].

Finally, we have in addition the remaining equations

$$\begin{equation} \frac{{\rm d}t}{{\rm d}\tau }= \frac{\gamma }{N} , \qquad \frac{{\rm d}r}{{\rm d}\tau }= \gamma N\nu \sin \alpha , \qquad \frac{{\rm d}\phi }{{\rm d}\tau }= \frac{\gamma }{r}\nu \cos \alpha . \end{equation} \tag{ 2.10 }$$

The system of four differential equations for ν, α, r and ϕ admits a critical solution at a radial equilibrium which corresponds to a circular orbit of constant radius r = r₀, constant speed ν = ν₀ and constant angles β = β₀, α = α₀. The constancy of the radius requires sin α₀ = 0, cos α₀ = ±1 and therefore sin (β₀ − α₀) = cos α₀sin β₀ and cos (α₀ − β₀) = cos α₀cos β₀. The force balance equation for the critical circular orbits is then

$$\begin{eqnarray} N\gamma _0^{3}\left( 1-\frac{\nu _0^2}{\nu _K^2}\right) &=& {\rm sgn}(\sin \beta _0)\frac{A}{ M } \,, \end{eqnarray} \tag{ 2.11 }$$

with

$$\begin{equation} {\pm} \nu _0 = \cos \beta _0 = \frac{ b_{\rm (rad)} N}{r_0}\quad \rightarrow \quad \gamma _0= 1/|\sin \beta _0| . \end{equation} \tag{ 2.12 }$$

In the case b_(rad) = 0 (i.e. ν₀ = 0, γ₀ = 1) and sin β₀ > 0 of purely radial outward photon motion, the previous equation reduces to the result of [1]

$$\begin{equation} \frac{A}{M} = N = \left(1-\frac{2M}{r_0}\right)^{1/2}, \end{equation} \tag{ 2.13 }$$

which requires A/M < 1 for a solution to exist.

2.3. Radiation from the spot

Let the spot to radiate isotropically in its own rest frame. Consider then a (geodesic) photon connecting the emitter world line with the observer world line, i.e.

$$\begin{equation} K=\Gamma _{\rm (ph)}\big[\partial _t +\zeta ^a_{\rm (ph)} \partial _a\big] . \end{equation} \tag{ 2.14 }$$

For a general motion we have

$$\begin{eqnarray} &&\eqalign K^t=\frac{{\rm d}t}{{\rm d}\lambda } = \frac{E}{N^2} =\Gamma _{\rm (ph)},\\&&\ms K^r=\frac{{\rm d}r}{{\rm d}\lambda }= \epsilon _r \frac{E}{r^2}\sqrt{R} =\Gamma _{\rm (ph)}\zeta ^r_{\rm (ph)},\\&&\ms K^\theta =\frac{{\rm d}\theta }{{\rm d}\lambda } = \epsilon _\theta \frac{E}{r^2}\sqrt{\Theta } =\Gamma _{\rm (ph)}\zeta ^\theta _{\rm (ph)},\\&&\ms K^\phi =\frac{{\rm d}\phi }{{\rm d}\lambda } = \frac{E}{r^2}\frac{b}{\sin ^2\theta } =\Gamma _{\rm (ph)}\zeta ^\phi _{\rm (ph)}, \end{eqnarray} \tag{ 2.15 }$$

where λ is an affine parameter, _r and _θ are sign indicators,

$$\begin{eqnarray} &&\eqalign R= r^4-r^2N^2(b^2+q^2)=r[r^3-(r-2M)(b^2+q^2)],\\&&\ms \Theta = q^2-b^2\cot ^2\theta , \end{eqnarray} \tag{ 2.16 }$$

and the following notation has been introduced:

$$\begin{equation} b=\frac{L}{E} ,\qquad q^2=\frac{{\mathcal K}}{E^2} . \end{equation} \tag{ 2.17 }$$

Here E = −K_t and L = K_ϕ are conserved Killing quantities, whereas ${\mathcal K}$ is a separation constant related to Carter's fourth constant of the motion ${\mathcal Q}$ by ${\mathcal K}={\mathcal Q}+L^2$. Note that q² can take any value.

The geodesic equations can be formally integrated by eliminating the affine parameter as follows [5]:

$$\begin{eqnarray} &&\eqalign \epsilon _r\int ^{r}\frac{{\rm d}r}{\sqrt{R(r)}}= \epsilon _\theta \int ^{\theta }\frac{{\rm d}\theta }{\sqrt{\Theta (\theta )}}, \\&&\ms t=\epsilon _r\int ^{r}\frac{r^2}{N^2\sqrt{R(r)}}\,{\rm d}r, \\&&\ms \phi =b\epsilon _\theta \int ^{\theta }\frac{{\rm d}\theta }{\sin ^2\theta \sqrt{\Theta (\theta )}} . \end{eqnarray} \tag{ 2.18 }$$

The integrals are along the path of motion.

We consider here the case of an emitter with 4-velocity U_(em) moving on the equatorial plane of a Schwarzschild spacetime. The observer with 4-velocity U_(obs) is at rest (very far from the origin) at a point not necessarily belonging to the equatorial plane. Let K be the 4-momentum of the photon connecting the two world lines. Hence we have

$$\begin{eqnarray} &&\eqalign U_{\rm (em)}=\Gamma _{\rm (em)}\big[\partial _t +\zeta ^r_{\rm (em)} \partial _r+\zeta ^\phi _{\rm (em)} \partial _\phi \big], \\&&\ms U_{\rm (obs)}=\partial _t, \\&&\ms K=\Gamma _{\rm (ph)}\big[\partial _t +\zeta ^r_{\rm (ph)} \partial _r +\zeta ^\theta _{\rm (ph)} \partial _\theta +\zeta ^\phi _{\rm (ph)} \partial _\phi\big] . \end{eqnarray} \tag{ 3.1 }$$

The energy of the photon at the emission point P_(em), as measured by U_(em), is

$$\begin{eqnarray} E_{\rm (em)}& \equiv & E(K,U_{\rm (em)}) =-K\cdot U_{\rm (em)}|_{P_{\rm (em)}}\nonumber \\ &=&- \Gamma _{\rm (ph)} \Gamma _{\rm (em)} \big[{-}N^2+N^{-2}\zeta ^r_{\rm (ph)}\zeta ^r_{\rm (em)}+r^2\zeta ^\phi _{\rm (ph)}\zeta ^\phi _{\rm (em)}\big]\big|_{P_{\rm (em)}}, \end{eqnarray} \tag{ 3.2 }$$

while the one observed at the point P_(obs) by U_(obs) is

$$\begin{equation} E_{\rm (obs)}\equiv E(K,U_{\rm (obs)}) =-K\cdot U_{\rm (obs)}|_{P_{\rm (obs)}}=-E|_{P_{\rm (obs)}} . \end{equation} \tag{ 3.3 }$$

Therefore, the ratio E_(obs)/E_(em) is

$$\begin{eqnarray} g\equiv \frac{E_{\rm (obs)}}{E_{\rm (em)}}&=& \frac{\sqrt{N^2-N^{-2}\zeta ^r_{\rm (em)}{}^2-r^2\zeta ^\phi _{\rm (em)}{}^2}}{1-N^{-4}\zeta ^r_{\rm (ph)}\zeta ^r_{\rm (em)}-b \zeta ^\phi _{\rm (em)}} \end{eqnarray} \tag{ 3.4 }$$

with all quantities evaluated at the emission point. Usually, one also introduces the 'redshift' parameter z

$$\begin{equation} z=\frac{E_{\rm (em)}-E_{\rm (obs)}}{E_{\rm (obs)}} ,\qquad g=(z+1)^{-1} . \end{equation} \tag{ 3.5 }$$

If the emitter is in a geodesic circular orbit, we have ζ^r_(em) = 0 and ζ^ϕ_(em) = ±ζ_K so that

$$\begin{equation} g=\frac{\sqrt{1- \frac{3M}{r_{\rm (em)}}}}{1\mp b \sqrt{\frac{M}{r^3_{\rm (em)}}}} . \end{equation} \tag{ 3.6 }$$

Consider a photon emitted at the points r_em, θ_em and ϕ_em at the coordinate time t_em, which reaches an observer located at r_obs, θ_obs and ϕ_obs at the coordinate time t_obs. The photon trajectories originating at the emitter must satisfy the following integral equation:

$$\begin{equation} \epsilon _r\int _{r_{\rm em}}^{r_{\rm obs}}\frac{{\rm d}r}{\sqrt{R(r)}}= \epsilon _\theta \int _{\theta _{\rm em}}^{\theta _{\rm obs}}\frac{{\rm d}\theta }{\sqrt{\Theta (\theta )}}, \end{equation} \tag{ 4.1 }$$

as from equation (2.18). The signs _r and _θ change when a turning point is reached. Turning points in r and θ are solutions of the equations R = 0 and Θ = 0, respectively. To find out which photons actually reach the observer, one thus must find those pairs (b, q²) that satisfy equation (4.1).

It is useful to introduce the new variable μ = cos θ, so that the null geodesic equations (2.18) become

$$\begin{eqnarray} &&\epsilon _r\int ^{r}\frac{{\rm d}r}{\sqrt{R(r)}}= \epsilon _\mu \int ^{\mu }\frac{{\rm d}\mu }{\sqrt{\Theta _\mu (\mu )}}, \end{eqnarray} \tag{ 4.2 }$$

$$\begin{eqnarray} &&t=\epsilon _r\int ^{r}\frac{r^2}{N^2\sqrt{R(r)}}\,{\rm d}r, \end{eqnarray} \tag{ 4.3 }$$

$$\begin{eqnarray} &&\phi =b\epsilon _\mu \int ^{\mu }\frac{1}{1-\mu ^2}\frac{{\rm d}\mu }{\sqrt{\Theta _\mu (\mu )}}, \end{eqnarray} \tag{ 4.4 }$$

where

$$\begin{equation} \fl \Theta _\mu =q^2-(q^2+b^2)\mu ^2\equiv (q^2+b^2)(\bar{\mu }^2-\mu ^2), \qquad \bar{\mu }^2=\frac{q^2}{q^2+b^2} . \end{equation} \tag{ 4.5 }$$

We will consider the case of an emitting source moving on the equatorial plane (i.e. θ_em = π/2) and a distant observer located far away from the black hole (i.e. r_obs → ∞) at the azimuthal position ϕ_obs = 0. For a photon emitted by the spot, we thus have μ_em = 0. Furthermore, for a photon crossing the equatorial plane (which is the case we are interested in), we have q² > 0 so that $\bar{\mu }^2>0$.

Consider first equation (4.2). The integral over μ is straightforward

$$\begin{equation} \int ^\mu \frac{{\rm d}\mu }{\sqrt{\Theta _\mu }}=\frac{1}{\sqrt{q^2+b^2}}\arctan \left(\frac{\mu }{\sqrt{\bar{\mu }^2-\mu ^2}}\right) . \end{equation} \tag{ 4.6 }$$

The integral over r can be worked out with the inverse Jacobian elliptic integrals. We will use the notations and conventions of [6] for the definition of the Jacobi functions

$$\begin{equation} {\rm sn}(u,k)=\sin \varphi , \qquad {\rm cn}(u,k)=\cos \varphi , \end{equation} \tag{ 4.7 }$$

where

$$\begin{equation} u=\int _0^\varphi \frac{{\rm d}\vartheta }{\sqrt{1-k^2\sin ^2\vartheta }}\ \end{equation} \tag{ 4.8 }$$

is the incomplete elliptic integral of the first kind. Let us denote the four roots of R(r) = 0 by r₁, r₂, r₃ and r₄. There are two relevant cases to be considered. In fact, since r = 0 is a root of the equation R(r) = 0, the reduced cubic equation admits either three real roots or one real root and two complex conjugate roots.

• Case A:  
  R(r) = 0 has four real roots.Let the roots be ordered so that r₁ ⩾ r₂ ⩾ r₃ ⩾ r₄, with r₃ = 0 and r₄ ⩽ 0. Physically allowed regions for photons are given by R ⩾ 0, i.e. r ⩾ r₁ (region I) and r₃ ⩽ r ⩽ r₂ (region II) with r > 2M.In region I the integral over r can be worked out by the following integration:

  $$\begin{equation} \int _{r_1}^r\frac{{\rm d}r}{\sqrt{R(r)}}=\frac{2}{\sqrt{r_1(r_2-r_4)}} {\rm sn}^{-1}(\sin \varphi _{A_I}, k_A), \end{equation} \tag{ 4.9 }$$

  where

  $$\begin{equation} \sin \varphi _{A_I}=\sqrt{\frac{(r_2-r_4)(r-r_1)}{(r_1-r_4)(r-r_2)}}, \qquad k_A=\sqrt{\frac{r_2(r_1-r_4)}{r_1(r_2-r_4)}}, \end{equation} \tag{ 4.10 }$$

  when $r_1\not=r_2$. The case of two equal roots r₁ = r₂ should be treated separately.In region II the integral over r can be worked out by the following integration:

  $$\begin{equation} \int _r^{r_2}\frac{{\rm d}r}{\sqrt{R(r)}}=\frac{2}{\sqrt{r_1(r_2-r_4)}}\, {\rm sn}^{-1}(\sin \varphi _{A_{II}}, k_A), \end{equation} \tag{ 4.11 }$$

  where

  $$\begin{equation} \sin \varphi _{A_{II}}=\sqrt{\frac{r_1(r_2-r)}{r_2(r_1-r)}}, \qquad r_1\not=r_2 . \end{equation} \tag{ 4.12 }$$
• Case B:  
  R(r) = 0 has two complex roots and two real roots.Let us assume that r₁ and r₂ are complex, with $r_1={\bar{r}_2}$, whereas r₃ = 0 and r₄ is real such that r₄ ⩽ 0. The physically allowed region for photons is given by r > 2M.The integral over r can be worked out with the following integration:

  $$\begin{equation} \int _{2M}^r\frac{{\rm d}r}{\sqrt{R(r)}}=\int _{0}^r\frac{{\rm d}r}{\sqrt{R(r)}}-\int _0^{2M}\frac{{\rm d}r}{\sqrt{R(r)}}, \end{equation} \tag{ 4.13 }$$

  with

  $$\begin{eqnarray} &&\eqalign \int _{0}^r\frac{{\rm d}r}{\sqrt{R(r)}}=\frac{1}{\sqrt{AB}}\,{\rm cn}^{-1}(\sin \varphi _{B}, k_B), \\&&\ms \int _0^{2M}\frac{{\rm d}r}{\sqrt{R(r)}}=\frac{1}{\sqrt{AB}}\,{\rm cn}^{-1}\big(\sin \varphi _{B}^H, k_B\big), \end{eqnarray} \tag{ 4.14 }$$

  where

  $$\begin{equation} A^2=u^2+v^2, \qquad B^2=(r_4-u)^2+v^2, \end{equation} \tag{ 4.15 }$$

  with u = Re(r₁) and v = Im(r₁), and

  $$\begin{equation} \sin \varphi _{B}=\frac{(A-B)r-r_4A}{(A+B)r-r_4A}, \qquad k_B=\sqrt{\frac{(A+B)^2-r_4^2}{4AB}}, \end{equation} \tag{ 4.16 }$$

  with sin φ^H_B = sin φ_B(r = 2M).

The apparent position of the image of the emitting source on the celestial sphere is represented by two impact parameters, α and β, measured on a plane centered about the observer location and perpendicular to the direction θ_obs. They are defined by [7]

$$\begin{eqnarray} \alpha &=&\lim _{r_{\rm obs}\rightarrow \infty }-r_{\rm obs}\frac{K^{\hat{\phi }}}{K^{\hat{t}}} =-\frac{b}{\sin \theta _{\rm obs}}, \nonumber \\ \beta &=&\lim _{r_{\rm obs}\rightarrow \infty }r_{\rm obs}\frac{K^{\hat{\theta }}}{K^{\hat{t}}} =\epsilon _{\theta _{\rm obs}}\sqrt{q^2-b^2\cot ^2\theta _{\rm obs}}, \end{eqnarray} \tag{ 5.1 }$$

where $K^{\hat{\alpha }}$ are the frame components of K with respect to the ZAMOs (its coordinate components are instead listed in equation (2.15)). Equivalent expressions can be obtained by decomposing the photon 4-velocity as follows:

$$\begin{equation} K=E(n)\big[e_{\hat{t}}+{\hat{\nu }}^{\hat{a}}e_{\hat{a}}\big], \qquad {\hat{\nu }}\cdot {\hat{\nu }}=1\ \end{equation} \tag{ 5.2 }$$

so that

$$\begin{eqnarray} &&\eqalign \alpha =\lim _{r_{\rm obs}\rightarrow \infty }-r_{\rm obs}{\hat{\nu }}^{\hat{\phi }}=\lim _{r_{\rm obs}\rightarrow \infty }[\vec{r}\times \hat{\nu }]^{\hat{\theta }} , \\&&\ms \beta =\lim _{r_{\rm obs}\rightarrow \infty }r_{\rm obs}{\hat{\nu }}^{\hat{\theta }}=\lim _{r_{\rm obs}\rightarrow \infty }[\vec{r}\times \hat{\nu }]^{\hat{\phi }} . \end{eqnarray} \tag{ 5.3 }$$

The line of sight to the black hole center marks the origin of the coordinates, where α = 0 = β. Now imagine a source of illumination behind the black hole whose angular size is large compared with the angular size of the black hole. As seen by the distant observer, the black hole will appear as a black region in the middle of the larger bright source. No photons with impact parameters in a certain range about α = 0 = β will reach the observer. The rim of the black hole corresponds to photon trajectories which are marginally trapped by the black hole; they spiral around many times before they reach the observer.

The image of the trajectory is thus obtained by determining all pairs (b, q²) satisfying equation (4.1) (or equivalently equation (4.2)), then substituting back into equation (5.1) in order to obtain the corresponding coordinates on the observer's plane. Alternatively, one can solve equation (5.1) for b and q², i.e.

$$\begin{equation} \fl b=-\alpha \sin \theta _{\rm obs}, \qquad q^2=\beta ^2+\alpha ^2\cos ^2\theta _{\rm obs} \quad \rightarrow \quad b^2+q^2=\alpha ^2+\beta ^2, \end{equation} \tag{ 5.4 }$$

then substituting back into equation (4.2) and solving for all allowed pairs of impact parameters (α, β).

The images of the source obtained in this way can be classified according to the number of times the photon trajectory crosses the equatorial plane between the emitting source and the observer. The trajectory of the 'direct' image does not cross the equatorial plane, while that of the 'first-order' image crosses it once, and so on.

We are interested in constructing direct images only and refer to appendix A for details.

The observed differential flux is given by [8]

$$\begin{equation} {\rm d}F_{\rm obs}=I_{\rm obs}\,{\rm d}\Omega , \end{equation} \tag{ 6.1 }$$

where dΩ is the solid angle subtended by the light source on the observer sky and I_obs is the intensity of the source integrated over its effective frequency range, i.e.

$$\begin{equation} I_{\rm obs}=g^4I_{\rm em} . \end{equation} \tag{ 6.2 }$$

The intensity I_em measured at the rest frame of the spot can be normalized as I_em = 1. Furthermore, the solid angle can be expressed in terms of the observer's plane coordinates {α, β} as

$$\begin{equation} {\rm d}\Omega =\frac{{\rm d}\alpha\, {\rm d}\beta }{r_{\rm obs}^2} . \end{equation} \tag{ 6.3 }$$

Introducing then polar coordinates on the observer's plane and switching integration over r_em and ϕ_em (see appendix B for details), the observed differential flux is expressed as

$$\begin{equation} {\rm d}F_{\rm obs}=\frac{g^4}{r_{\rm obs}^2} \frac{\cos \theta _{\rm obs}}{1-\cos ^2\phi _{\rm em}\sin ^2\theta _{\rm obs}} \left|b\frac{\partial b}{\partial r_{\rm em}}+q\frac{\partial q}{\partial r_{\rm em}}\right| {\rm d}r_{\rm em}\,{\rm d}\phi _{\rm em} . \end{equation} \tag{ 6.4 }$$

Finally, the light curve of the emitting source is constructed by introducing the time dependence of the radiation received by the distant observer, including the time delay effects. Therefore, we also need to evaluate the coordinate time interval spent by each photon to reach the observer (see appendix C).

Figures 1–5 show the apparent position of the direct image, light curve, redshift factor and solid angle for an emitting spot under the effect of both the gravitational and radiation fields. Different figures correspond to different properties of the radiation field. The distant observer is located at the polar angle θ_obs = 80° in all cases. The case of an emitting spot in circular geodesic motion on the equatorial plane of a Schwarzschild spacetime at the same initial r_em is also shown for comparison.

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In figures 1–3 the initial radius and azimuthal angle of the emitting spot are r_em(0) = 10M, ϕ_em(0) = 0; the initial velocity is that of a Keplerian circular orbit at that radius. The radiation field is radially outgoing, with different values of the luminosity parameter A/M. For small values of A/M, the emitter spirals toward the critical radius located close to the horizon, by undergoing several azimuthal cycles around the black hole (see figure 1). The trajectory is nearly circular after the first revolution so that the light curve is very close to the circular Keplerian one. The spiraling then becomes faster, and the light curve peaks occur faster and faster.

Increasing the luminosity parameter while maintaining the initial conditions fixed causes the emitting spot to drift initially to larger radii and then quickly spiral inward down to the critical radius, which, as expected, is larger in this case (see figures 2 and 3 and equation (2.13)).

Figures 4 and 5 show the effect of a nonzero angular momentum of the radiation field. In the case of figure 4 the emitting spot drifts away from the black hole before going back inward, ending in the circular equilibrium orbit in a few revolutions. This is also evident from the comparison of the corresponding light curve with the Keplerian one. In fact, the emitter spends a fairly long time far from the source before completing the first revolution, whereas in the same time interval, the Keplerian emitter orbits the source several times. Then the orbit soon becomes circular, at the smaller radius given by equation (2.11). Figure 5 refers to an emitter with the same initial conditions as in figure 4, except for the negative sign of the azimuthal velocity. In this case, the emitter initially moves clockwise around the black hole and is quickly dragged by the radiation field in the opposite direction, soon reaching the same equilibrium orbit as in figure 4.

We have calculated the signal produced by an emitting point-like source moving in the equatorial plane of a Schwarzschild spacetime under the influence of a radiation field. The latter consists of photons having the same specific angular momentum and traveling along geodesics. The interaction with the photon field leads to a friction-like drag force responsible for the so-called Poynting–Robertson effect. Previous studies have shown that, in the case of photons with zero angular momentum, i.e. propagating radially with respect to the ZAMOs, there exists an equilibrium radius representing the balance of the outward radiation force with the inward gravitational force, where the emitter can remain at rest. The location of such a critical radius depends on the luminosity parameter. If the outward photon flux possesses a nonzero angular momentum, emitting spots that do not escape end up in circular orbits. In this paper we have derived the flux, redshift factor and solid angle as a function of the (coordinate) time, as well as the time-integrated image of the spot in the observer's sky. The results are clearly different from those for an emitting spot in circular geodesic motion, as shown by numerical examples where the effect of the interaction with the radiation field has been investigated by varying both the luminosity parameter and the photon angular momentum.

The treatment and results presented here hold a potential for astrophysical applications. Matter accretion toward white dwarfs, neutron stars or black holes that emit radiation at a sizeable fraction of their Eddington luminosity (i.e. L/L_Edd ≳ 1%) will be influenced by general relativistic Poynting–Robertson-type effects. Departures from the unperturbed motion (i.e. in the absence of the radiation field) are substantial and may lead to observable phenomena. The range of astrophysical applications is vast; examples include quasi-periodic oscillations that are observed in accreting neutron stars and black holes [9, 10], thermonuclear flashes that occur on the surface of accreting neutron stars (the so-called type I bursts [11]) and the very broad Fe–Kα line profiles produced in the innermost regions of accretion disks around collapsed objects [12]. However, the treatment we have developed here is idealized in several respects and the impact of some approximations should be carefully assessed before detailed predictions for astrophysical systems are worked out. Treating the photon field as if all photons had the same angular momentum presents clear advantages for the analytical calculations presented in [1, 2], but would require some caution in an astrophysical context. For instance, the flux emitted from the surface of an accreting neutron star comprises photons emitted in virtually all directions (and thus possessing a range of different angular momenta). Moreover, in accreting black holes the motion of matter in the vicinity of the innermost stable circular orbit will be mainly affected by radiation coming from the outer disk regions, in turn involving a radiation field emitted in a range of different photon directions and emission radii. Finally, in an astrophysical environment one must also consider the impact of two key assumptions, which are intrinsic to any Poynting–Robertson-type theory, namely that matter is directly exposed to the radiation field (meaning that the optical depth to the source must be <1) and that the radiation re-emitted or scattered by matter propagates unimpeded to infinity without undergoing other interactions.

Despite these limitations the analysis presented here captures some essential features of the motion of matter in the strong field regime under the effects of an intense source of radiation. Future work will be devoted to generalizing the present treatment and addressing specific astrophysical situations in which the general relativistic version of the Poynting–Robertson effect can be relevant.

This work was partially supported through ICRANet and PRIN INAF 2008 contracts. LS and AG acknowledge the International Space Science Institute in Bern for the hospitality during part of this work was carried out. All the authors are indebted to Professor. R T Jantzen for stimulating discussions about the Poynting–Robertson effect in general relativity.

We list below for completeness the details on the construction of the direct image, the derivation of the observed energy flux and the calculation of the coordinate time interval between the emitter and observer. This is a well-known topic addressed by many authors in the literature (see, e.g., [13–17]). However, different and sophisticated techniques are used simply to show light curves as well as images, without entering the underlying analytical framework or referring to previous related works.

A.1. Constructing the direct image

The direct image results from photons which never cross the equatorial plane. As the photon reaches the observer, on the photon orbit, we have dθ/dr > 0 (i.e. dμ/dr < 0) if β > 0, and dθ/dr < 0 (i.e. dμ/dr > 0) if β < 0. Therefore, when β > 0, the photon must encounter a turning point at $\mu =\bar{\mu }$: μ starts from 0, goes up to $\bar{\mu }$ and then goes down to μ_obs (which is ${\le} \bar{\mu }$). When β < 0, the photon does not encounter a turning point at $\mu =\bar{\mu }$: μ starts from 0 and monotonically increases to μ_obs.

The total integration over μ along the path of the photon from the emitting source to the observer is thus given by

$$\begin{equation} I_\mu =\left\lbrace \begin{array}{@{}ll@{}} \left[\dsty\int _0^{\bar{\mu }}+\dsty\int _{\mu _{\rm obs}}^{\bar{\mu }}\right]\dsty\frac{{\rm d}\mu }{\sqrt{\Theta _\mu (\mu )}}&(\beta > 0)\\\ms \dsty\int _0^{\mu _{\rm obs}}\dsty\frac{{\rm d}\mu }{\sqrt{\Theta _\mu (\mu )}}=\left[\dsty\int _0^{\bar{\mu }}-\dsty\int _{\mu _{\rm obs}}^{\bar{\mu }}\right]\dsty\frac{{\rm d}\mu }{\sqrt{\Theta _\mu (\mu )}}&(\beta < 0) . \end{array} \right. \end{equation} \tag{ A.1 }$$

By using equation (4.6) we obtain

$$\begin{equation} I_\mu =\left\lbrace \begin{array}{@{}ll@{}} \dsty\frac{\pi }{\sqrt{\alpha ^2+\beta ^2}}-I_{\mu _{\rm obs}}&(\beta > 0)\\\ms I_{\mu _{\rm obs}}&(\beta < 0), \end{array} \right. \end{equation} \tag{ A.2 }$$

where

$$\begin{equation} I_{\mu _{\rm obs}}=\frac{1}{\sqrt{\alpha ^2+\beta ^2}}\arctan \left(\frac{\sqrt{\alpha ^2+\beta ^2}}{|\beta |}\cot \theta _{\rm obs}\right) . \end{equation} \tag{ A.3 }$$

Now let us consider the integration over r. Since the observer is at infinity, the photon reaching him/her must have been moving in the allowed region defined by r ⩾ r₁ when R(r) = 0 has four real roots (case A), or the allowed region defined by r > 2M when R(r) = 0 has two complex roots and two real roots (case B). There are then two possibilities for the photon during its trip: it has encountered a turning point at r = r₁, or it has not encountered any turning point in r. Define

$$\begin{equation} I_\infty \equiv \int _{r_t}^\infty \frac{{\rm d}r}{\sqrt{R(r)}}, \qquad I_{r_{\rm em}}\equiv \int _{r_t}^{r_{\rm em}}\frac{{\rm d}r}{\sqrt{R(r)}} . \end{equation} \tag{ A.4 }$$

Obviously, according to equation (4.2), a necessary and sufficient condition for the occurrence of a turning point in r on the path of the photon is that I_∞ < I_μ. Therefore, the total integration over r along the path of the photon from the emitting source to the observer is

$$\begin{equation} I_r=\left\lbrace \begin{array}{@{}ll@{}} I_\infty +I_{r_{\rm em}}&(I_\infty <I_\mu )\\\ms \dsty\int _{r_{\rm em}}^\infty \dsty\frac{{\rm d}r}{\sqrt{R(r)}}=I_\infty -I_{r_{\rm em}}&(I_\infty \ge I_\mu ) . \end{array} \right. \end{equation} \tag{ A.5 }$$

By definition, I_∞, $I_{r_{\rm em}}$ and I_r are all positive. According to equation (4.2), we must have I_r = I_μ for the orbit of a photon. The relevant cases to be considered are the following.

• Case A:  
  R(r) = 0 has four real roots.When $r_1\not=r_2$, by using equation (4.9) to evaluate integrals in equation (A.4) with r_t = r₁, we obtain

  $$\begin{eqnarray} &&\eqalign I_\infty = \frac{2}{\sqrt{r_1(r_2-r_4)}}\,{\rm sn}^{-1}\big(\sin \varphi _{A_I}^\infty , k_A\big), \\&&\ms I_{r_{\rm em}}=\frac{2}{\sqrt{r_1(r_2-r_4)}}\,{\rm sn}^{-1}\big(\sin \varphi _{A_I}^{\rm em}, k_A\big), \end{eqnarray} \tag{ A.6 }$$

  where $\sin \varphi _{A_I}^\infty =\sin \varphi _{A_I}(r\rightarrow \infty )$ and $\sin \varphi _{A_I}^{\rm em}=\sin \varphi _{A_I}(r=r_{\rm em})$. Substitute these expressions into equation (A.5); let I_r = I_μ and finally solve for r_em:

  $$\begin{equation} r_{\rm em}=\frac{r_1(r_2-r_4)-r_2(r_1-r_4){\rm sn}^2(\sin \xi _A, k_A)}{(r_2-r_4)-(r_1-r_4){\rm sn}^2(\sin \xi _A, k_A)}, \end{equation} \tag{ A.7 }$$

  where

  $$\begin{equation} \sin \xi _A=\case{1}{2}(I_\mu -I_\infty )\sqrt{r_1(r_2-r_4)} . \end{equation} \tag{ A.8 }$$

  Since sn²( − sin ξ_A, k_A) = sn²(sin ξ_A, k_A), the solution given by equation (A.7) applies whether I_μ − I_∞ is positive or negative, i.e. no matter whether there is a turning point in r or not along the path of the photon.
• Case B:  
  R(r) = 0 has two complex roots and two real roots.No turning points occur in this case. Therefore, we have

  $$\begin{equation} I_r=\int _{r_{\rm em}}^\infty \frac{{\rm d}r}{\sqrt{R(r)}}=I_\infty -I_{r_{\rm em}}, \end{equation} \tag{ A.9 }$$

  with

  $$\begin{equation} \fl \qquad I_\infty =\frac{1}{\sqrt{AB}} {\rm cn}^{-1}(\sin \varphi _{B}^\infty , k_B), \qquad I_{r_{\rm em}}=\frac{1}{\sqrt{AB}}\, {\rm cn}^{-1}\big(\sin \varphi _{B}^{\rm em}, k_B\big), \end{equation} \tag{ A.10 }$$

  by using equation (4.13) to evaluate integrals in equation (A.4); also recall equation (4.15). Solving then the equation I_r = I_μ for r_em gives

  $$\begin{equation} r_{\rm em}=\frac{r_4A[1-{\rm cn}(\sin \xi _B, k_B)]}{(A-B)-(A+B){\rm cn}(\sin \xi _B, k_B)}, \end{equation} \tag{ A.11 }$$

  where sin φ^∞_B = sin φ_B(r → ∞) and sin φ^em_B = sin φ_B(r = r_em) and

  $$\begin{equation} \sin \xi _B=(I_\mu -I_\infty )\sqrt{AB} . \end{equation} \tag{ A.12 }$$

The solid angle once expressed in terms of the observer's plane coordinates {α, β} is given by

$$\begin{equation} {\rm d}\Omega =\frac{{\rm d}\alpha\, {\rm d}\beta }{r_{\rm obs}^2} . \end{equation} \tag{ B.1 }$$

Introduce polar coordinates on the observer's plane

$$\begin{equation} \alpha =\rho \cos \psi , \qquad \beta =\rho \sin \psi . \end{equation} \tag{ B.2 }$$

The integration over the observer's plane coordinates can then be switched over r_em and ϕ_em by

$$\begin{equation} {\rm d}\alpha\, {\rm d}\beta =\rho {\rm d}\rho {\rm d}\psi =\sqrt{\alpha ^2+\beta ^2}\left|\frac{\partial (\rho ,\psi )}{\partial (r_{\rm em},\phi _{\rm em})}\right|{\rm d}r_{\rm em}\,{\rm d}\phi _{\rm em}, \end{equation} \tag{ B.3 }$$

where

$$\begin{equation} \left|\frac{\partial (\rho ,\psi )}{\partial (r_{\rm em},\phi _{\rm em})}\right| =\left|\frac{\partial \rho }{\partial r_{\rm em}}\frac{\partial \psi }{\partial \phi _{\rm em}}-\frac{\partial \rho }{\partial \phi _{\rm em}}\frac{\partial \psi }{\partial r_{\rm em}}\right|\ \end{equation} \tag{ B.4 }$$

is the Jacobian of the transformation (ρ, ψ) → (r_em, ϕ_em). Since

$$\begin{equation} \rho =\sqrt{\alpha ^2+\beta ^2}=\sqrt{b^2+q^2}\ \end{equation} \tag{ B.5 }$$

does not depend on ϕ_em, we have to evaluate only

$$\begin{equation} \frac{\partial \rho }{\partial r_{\rm em}}=\frac{1}{\rho }\left(b\frac{\partial b}{\partial r_{\rm em}}+q\frac{\partial q}{\partial r_{\rm em}}\right), \end{equation} \tag{ B.6 }$$

where the derivatives of b and q with respect to r_em are obtained simply by inverting the derivatives ∂r_em/∂q and ∂r_em/∂b which can be evaluated from equations (A.7) and (A.11).

∂ψ/∂ϕ_em instead can be evaluated from equation (4.4) governing the azimuthal motion, i.e.

$$\begin{equation} \phi =b\epsilon _\mu \int ^{\mu }\frac{1}{1-\mu ^2}\frac{{\rm d}\mu }{\sqrt{\Theta _\mu (\mu )}}, \end{equation} \tag{ B.7 }$$

taking into account that ϕ_obs = 0 and tan ψ = β/α. The integration is straightforward

$$\begin{equation} \phi =-\epsilon _\mu \arctan \left(b\frac{\mu }{\sqrt{\Theta _\mu (\mu )}}\right)\ \end{equation} \tag{ B.8 }$$

so that

$$\begin{equation} \phi _{\rm em}=\left\lbrace \begin{array}{@{}ll@{}} \pi +\arctan \left(\displaystyle \frac{\alpha }{\beta }\cos \theta _{\rm obs}\right)&(\beta > 0)\\ -\arctan \left(\displaystyle \frac{\alpha }{|\beta |}\cos \theta _{\rm obs}\right)&(\beta < 0), \end{array} \right. \end{equation} \tag{ B.9 }$$

whence

$$\begin{equation} \tan \phi _{\rm em}=\frac{\alpha }{\beta }\cos \theta _{\rm obs}=\cot \psi \cos \theta _{\rm obs} . \end{equation} \tag{ B.10 }$$

Therefore,

$$\begin{equation} \fl \sin \psi =\frac{\cos \phi _{\rm em}\cos \theta _{\rm obs}}{\sqrt{1-\cos ^2\phi _{\rm em}\sin ^2\theta _{\rm obs}}}, \qquad \cos \psi =\frac{\sin \phi _{\rm em}}{\sqrt{1-\cos ^2\phi _{\rm em}\sin ^2\theta _{\rm obs}}}, \end{equation} \tag{ B.11 }$$

implying that

$$\begin{equation} \frac{\partial \psi }{\partial \phi _{\rm em}}=\frac{\cos \theta _{\rm obs}}{1-\cos ^2\phi _{\rm em}\sin ^2\theta _{\rm obs}} . \end{equation} \tag{ B.12 }$$

Finally, the solid angle (B.1) turns out to be given by

$$\begin{equation} \fl \qquad {\rm d}\Omega =\frac{1}{r_{\rm obs}^2} \frac{\cos \theta _{\rm obs}}{1-\cos ^2\phi _{\rm em}\sin ^2\theta _{\rm obs}} \left|b\frac{\partial b}{\partial r_{\rm em}}+q\frac{\partial q}{\partial r_{\rm em}}\right| {\rm d}r_{\rm em}{\rm d}\phi _{\rm em}, \end{equation} \tag{ B.13 }$$

leading to expression (6.4) for the observed differential flux.

The light travel time between the emitter and observer is given by equation (4.3), i.e.

$$\begin{equation} t_{\rm obs}-t_{\rm em}=\epsilon _r\int ^{r}\frac{r^2}{N^2\sqrt{R(r)}} {\rm d}r\bigg \vert _{r_{\rm em}}^{r_{\rm obs}}, \end{equation} \tag{ C.1 }$$

where the integration has to be done properly. The integral can be conveniently decomposed as follows:

$$\begin{eqnarray} \fl \int ^{r}\frac{r^2}{N^2\sqrt{R(r)}}\,{\rm d}r&=&\int ^{r}\frac{r^3}{(r-2M)\sqrt{R(r)}}\,{\rm d}r\nonumber \\ \fl &=&8M^3\int ^{r}\frac{{\rm d}r}{(r-2M)\sqrt{R(r)}} +\int ^{r}\frac{r^2}{\sqrt{R(r)}}\,{\rm d}r\nonumber \\ \fl &&+2M\int ^{r}\frac{r}{\sqrt{R(r)}}\,{\rm d}r +4M^2\int ^{r}\frac{{\rm d}r}{\sqrt{R(r)}} . \end{eqnarray} \tag{ C.2 }$$

Each term can be evaluated in terms of elliptic functions, e.g., by using the table of integrals in [6].

Consider first the case A, where R(r) = 0 has four real roots. Let the roots be ordered so that r₁ > r₂ > r₃ > r₄, with r₄ < 0. Physically allowed regions for photons are given by R > 0, i.e. r > r₁ (region I) and r₂ > r > r₃ (region II). In region I the integrals entering equation (C.2) have to be worked out by the formulas nn 258.00, 258.11, 258.39 on pp 128–32 of [6]. In region II instead we refer to nn 255.00, 255.17, 255.38 on pp 116–20.

In case B the equation R(r) = 0 has two complex roots and two real roots. Let us assume that r₁ and r₂ are complex, r₃ and r₄ are real and r₃ ⩾ r₄. Then, we must have $r_1={\bar{r}_2}$, whereas r₃ ⩾ 0 and r₄ ⩽ 0. The physically allowed region for photons is given by r > r₃. The integrals entering equation (C.2) have to be worked out by the formulas nn 260.00, 260.03, 260.04 on pp 135–36 of [6].