PRL-Bench

astro

Polarization Eigenmode Mixing and Mode Conversion Caused by the Competition Between Strongly Magnetized Plasma and QED Vacuum Polarization

We study **radiative transfer and X-ray polarization in the atmospheres of compact objects with ultra-strong magnetic fields**, focusing on the mixing of the two polarization eigenmodes and the energy-dependent mode conversion caused by the competition between **magnetized plasma dispersion** and **QED vacuum-polarization birefringence**. Normalized notation: $B_{12}\equiv B/(10^{12}\,\mathrm{G})$, $B_{14}\equiv B/(10^{14}\,\mathrm{G})$, $E_1\equiv E/(1\,\mathrm{keV})$, and $Y_e$ is the electron fraction ($0<Y_e\le 1$). The QED critical magnetic field is $$ B_Q\equiv \frac{m_e^2c^3}{e\hbar}= …

1. (5pt) Characteristic density of the strongest-mixing layer $\rho_\star(B,E,Y_e)$ - (R1) The constant coefficient $C_\rho$ (keep 4 significant digits). - (R2) Rewrite the same formula using $B_{14}$ as the normalized variable, and give the new leading coefficient (keep 4 significant digits). - (R3) The leading asymptotic form of $f(B)$ in the two limits: - 1) $B\ll B_Q$; - 2) $B\gg B_Q$ (the asymptotic expression must include the numerical constant, e.g. of the form $(B/\kappa B_Q)^{1/2}$, where $\kappa$ must also be given numerically). - (R4) Take $f(B)=1$, …

1. (5pt) Answer: - (1pt) R1 = $C_\rho=9.640\times10^{-5}$ - (1pt) R2 = $$\rho_\star = 0.9640\,Y_e^{-1}B_{14}^{2}E_1^{2}f(B)^{-2}\ \mathrm{g\,cm^{-3}}$$ - (1pt) R3 = $$f(B)\simeq\begin{cases} 1, & B\ll B_Q,\\ \left(\dfrac{B}{5B_Q}\right)^{1/2}, & B\gg B_Q. \end{cases}$$ - (1pt) R4 = $3.07018\ \mathrm{g\,cm^{-3}}$ - (1pt) R5 = $29.3134\ \mathrm{g\,cm^{-3}}$ 2. (2pt) Answer: - (1pt) R6 = $E_{\rm ad}=2.520\bigl[f(B)\tan\theta_{kB}|1-u_i|\bigr]^{2/3}\left(\frac{1\mathrm{cm}}{H_\rho}\right)\ \mathrm{keV}$ - (1pt) R7 = $2.10770\ \mathrm{keV}$ 3. (3pt) …

1. For the characteristic density of the strongest-mixing layer, the required results are: - $C_\rho=9.640\times10^{-5}$. - Rewriting the normalization from $B_{12}$ to $B_{14}$ gives $$ \rho_\star = 0.9640\,Y_e^{-1}B_{14}^{2}E_1^{2}f(B)^{-2}\ \mathrm{g\,cm^{-3}}. $$ - The asymptotic behavior of $f(B)$ is $$ f(B)\simeq\begin{cases} 1, & B\ll B_Q,\\ \left(\dfrac{B}{5B_Q}\right)^{1/2}, & B\gg B_Q. \end{cases} $$ - For **Set-A**, with $f(B)=1$, $$ B=7.53\times10^{13}\,\mathrm{G},\quad E=2.37\,\mathrm{keV},\quad Y_e=1, $$ …

astro

Invariant-space singular mapping at the Schwarzschild ISCO and exact cancellation of environmental dephasing

In extreme mass-ratio inspirals (EMRIs), weak non-geodesic forces generated by a collisionless environment (e.g. dynamical friction from gravitational scattering) drive secular drifts of the orbital invariants. Observable waveform systematics are not controlled by the local force components themselves, but by how they map into the invariant space and, through a two-timescale construction, into the accumulated gravitational-wave (GW) phase. In the Schwarzschild strong field this mapping is singular at the innermost stable circular orbit (ISCO), so demonstrating a *finite* gauge-invariant enviro …

1. Reduce the two tangential-drag invariant-flux relations to explicit closed-form functions of \(R/M\) only: \[ \mathcal A(R)\equiv \frac{1}{a^3}\frac{\dot\varepsilon_e}{\varepsilon},\qquad \mathcal B(R)\equiv \frac{1}{a^3}\frac{\dot l_{z,e}}{l_z}. \] Your final expressions must be real for \(R>3M\) and contain no remaining \(\varepsilon\) or \(l_z\). Also determine whether \(\mathcal A(R)\mathcal B(R)=1-3M/R\) holds identically (Yes/No). 2. Define the invariant-space mapping slopes \[ \Gamma(R)\equiv\left.\frac{\partial\ln\Omega}{\partial\ln\varepsilon}\right|_{l_z},\qq …

1. Answer: (1pt) \[ \boxed{\mathcal A(R)=\sqrt{\frac{M(R-3M)}{R(R-2M)}}\,,\qquad \mathcal B(R)=\sqrt{\frac{(R-3M)(R-2M)}{MR}}} \] and \(\boxed{\mathcal A\mathcal B=1-\frac{3M}{R}:\ \text{Yes}}\). Rubrics: - (1pt) Shows the product identity holds *exactly* as an algebraic consequence of the circular-orbit relations (not as a numerical check). 2. Answer: (2pt) \[ \boxed{\Gamma(R)=-\frac{3(R-2M)}{R-6M}},\qquad \boxed{\Upsilon(R)=0}, \qquad \boxed{\Gamma\text{ diverges at }R=6M:\ \text{Yes}}. \] Rubrics: - (1pt) Computes \(\Gamma\) via an explicit *fixed-\(l_z\)* direction in invariant spac …

1. Using \(l_z^2=MR^2/(R-3M)\), \[ \left(\frac{l_z}{R}\right)^2=\frac{M}{R-3M},\quad 1+\left(\frac{l_z}{R}\right)^2=\frac{R-2M}{R-3M}, \] and \[ \left(\frac{R}{l_z}\right)^2=\frac{R-3M}{M},\quad 1+\left(\frac{R}{l_z}\right)^2=\frac{R-2M}{M}. \] Also \(\sqrt{1-3M/R}=\sqrt{(R-3M)/R}\). Substituting into the definitions gives \[ \mathcal A(R)=\sqrt{\frac{M(R-3M)}{R(R-2M)}},\qquad \mathcal B(R)=\sqrt{\frac{(R-3M)(R-2M)}{MR}}. \] Their product is \[ \mathcal A\mathcal B=\frac{R-3M}{R}=1-\frac{3M}{R}, \] hence the identity holds (Yes). 2. The key is that \((\varepsilon,l_z)\) are independent coordi …