Influence of weak electromagnetic fields on charged particle ISCOs

Abstract

Astrophysical black holes are often embedded into electromagnetic fields, that can usually be treated as test fields not influencing the spacetime geometry. Here we analyse the innermost stable circular orbit (ISCO) of charged particles moving around a Schwarzschild black hole in the presence of a radial electric test field and an asymptotically uniform magnetic test field. We discuss the structure of the in general four ISCO solutions for different magnitudes of the electric and the magnetic field’s strength. In particular, we find that the nonexistence of stable circular orbits of particles with equal sign of charge as the black hole for sufficiently strong electric fields can be canceled by a sufficiently strong magnetic field. In this situation, we find that ISCOs made of static particles will emerge.

A ISCO equation for the radial component r

Inserting the equation of motion (2.6) into the first condition of (3.1) and expressing the particle’s angular momentum L as well as energy E by Eqs. (3.2) and (3.3) results in the following equation for the radial component r, depending only on the magnetic field strength $\mathcal {B}$ and the electric charge $\mathcal {Q}$

$$\begin{aligned} \begin{aligned} 0=&\frac{1}{4 r^2(r-6)^2\mathcal {Q}^2}(384 \mathcal {B}^2r^3\mathcal {Q}^2+408 \mathcal {B}^4r^8-352\mathcal {B}^4r^7+96\mathcal {B}^4r^6-48\mathcal {Q}^4\\&-\,544\mathcal {B}^2r^4\mathcal {Q}^2+16\mathcal {B}^4r^{10}-144\mathcal {B}^4r^9+32\mathcal {Q}^4r-24\mathcal {B}^2r^6\mathcal {Q}^2+192\mathcal {B}^2r^5\mathcal {Q}^2\\&-\,48C\mathcal {B}r^2\mathcal {Q}^2+48rC\mathcal {B}\mathcal {Q}^2+8C\mathcal {B}r^3\mathcal {Q}^2+72\mathcal {B}^3r^5C-48r^4C\mathcal {B}^3+24r^2C\mathcal {B}\\&-\,16\mathcal {B}^3r^6C-4r^3C\mathcal {B}+36r^2-12r^3+r^4+240\mathcal {B}^2r^5-144\mathcal {B}^2r^4-168r^2\mathcal {Q}^2\\&+\,144r\mathcal {Q}^2-84\mathcal {B}^2r^6+48r^3\mathcal {Q}^2+8\mathcal {B}^2r^7-4r^4\mathcal {Q}^2), \end{aligned} \end{aligned}$$

(A.1)

with $C=\sqrt{-(r-6)r(3\mathcal {B}^2 r^4-2\mathcal {B}^2 r^3-4\mathcal {Q}^2)}$ as before. The electromagnetic field’s influence on the ISCO’s radial component r determined by $\mathcal {B}$ and $\mathcal {Q}$ can be calculated by solving Eq. (A.1) for r. Equations (3.2), (3.3) and (A.1) thus describe the charged particle along its corresponding ISCO, depending exclusively on the electromagnetic field’s properties $\mathcal {B}$ and $\mathcal {Q}$.

Considering the limiting case $\mathcal {B}=0$ from section 3.2 in turn yields the following three expressions

$$\begin{aligned}&\displaystyle 0=\frac{1}{4 r^2(r-6)^2\mathcal {Q}^2}(-48\mathcal {Q}^4+32\mathcal {Q}^4r+36r^2-12r^3+r^4-168r^2\mathcal {Q}^2+144r\mathcal {Q}^2\nonumber \\&\displaystyle +48r^3\mathcal {Q}^2-4r^4\mathcal {Q}^2), \end{aligned}$$

(A.2)

$$\begin{aligned}&\displaystyle L^2=\frac{4r\mathcal {Q}^2}{r-6}, \end{aligned}$$

(A.3)

$$\begin{aligned}&\displaystyle E=\frac{(1-2\mathcal {Q}^2)r^2+(-6+12\mathcal {Q}^2)r-12\mathcal {Q}^2}{2r(r-6)\mathcal {Q}}. \end{aligned}$$

(A.4)

In this case, the angular momentum L is of quadratic order in Eq. (2.6). The solutions $L_{\upalpha ,\upbeta }$ given by Eq. (3.2) consequently coincide, resulting in one expression $L^2$. The two solutions for the energy $E_{\upalpha ,\upbeta }$ from Eq. (3.3) coincide as well, since the distinguishing term with opposite signs $\mp 2C\mathcal {B}$ in Eq. (3.3) vanishes for $\mathcal {B}=0$, resulting in one expression for the energy E. The results presented in Fig. 1 can be obtained by solving Eq. (A.2) for r, whereas Eqs. (A.3) and (A.4) yield the corresponding square of the particle’s angular momentum and the particle’s energy, respectively.

In the limiting case $\mathcal {Q}=0$, Eq. (A.1) diverges because of a zero in the denominator of the right-hand side. It stems from having inserted the particle’s energy E given by Eq. (3.3), which diverges for $\mathcal {Q}=0$ as well. Equation (3.3) was calculated by solving $\frac{\hbox {d}^{2}\mathcal {U}}{\hbox {d}r^{2}}=0$ for E. When examining the special case $Q=0$, $\mathcal {U}(r)$ defined by Eq. (2.6) depends solely on the particle’s energy square $E^2$. It drops out when differentiating $\mathcal {U}(r)$ with respect to r. Solving $\frac{\hbox {d}^{2}\mathcal {U}}{\hbox {d}r^{2}}=0$ for E is thus not possible, making Eqs. (A.1) and (3.3) invalid in this limit.

Instead, $\mathcal {U}(r)=0$ can be solved for $E^2$ in the case $\mathcal {Q}=0$ and the result inserted into $\frac{\hbox {d}^{2}\mathcal {U}}{\hbox {d}r^{2}}=0$ to yield an analytical expression, analogous to Eq. (A.2), from which the radial component r can be calculated along an interval of $\mathcal {B}$. Since the limit of a vanishing electric charge was not the main subject of this paper and has already been treated in [22], the resulting expression will be omitted here.