With $\tau$ the proper time of the massive test particle, the four-velocity has squared norm $-1$:The conserved quantities in part (i) giveMultiplying the normalization by $F$ and substituting gives the charged-particle radial equation in Reissner-Nordstrom spacetime:At a radial turning point $r_{\ast }>0$, the left side is zero, soThe parenthesis on the left is strictly positive. Hence $F(r_{\ast })\ge 0$. The factorization in part (b) now yieldsThere is no radial turning point between the horizons. Indeed, for $r_{-}<r<r_{+}$, the right-hand side of the radial equation is strictly positive, independently of the electric charge $e$ of the test particle. This is the charged-particle turning-point exclusion in Reissner-Nordstrom spacetime.

The inequalities are necessary, not a claim that a radial turning point exists for every $E,l,e$. At a regular horizon crossing a timelike vector cannot be tangent to a null hypersurface, so a massive test particle actually has $\dot{r}\ne 0$ there. Algebraic equality at $r_{\pm }$ would require $E=eQ/r_{\pm }$ and does not describe a regular finite-proper time radial turning point on the horizon. In the Schwarzschild spacetime limit $Q=0$, the positive-radius inner alternative disappears.

On the equatorial path, the Lagrangian per unit mass isBoth $t$ and $\varphi$ are cyclic coordinates. The Euler-Lagrange equations therefore conserve their canonical momenta:ConsequentlyHere $E$ is the total conserved energy in the specified gauge of the electromagnetic four-potential. The paper's $l$ is the angular momentum per unit mass; the physical angular momentum is $ml$. The equatorial choice is consistent with the $\theta$ Euler-Lagrange equation, since its angular force term contains $\sin \theta \cos \theta =0$ at $\theta =\pi /2$ and the electromagnetic four-potential has no angular component.

For $0<Q^2<M^2$ there are two distinct positive roots:Both roots are coordinate singularities of the static chart. In the standard extension, $r_{+}$ is the outer event horizon and $r_{-}$ is an inner Cauchy horizon. The curvature singularity is $r=0$. Between the horizons $F<0$, so the radial direction is timelike; for $0<r<r_{-}$, $F>0$ again and the central curvature singularity is timelike. These are the three radial regions of the global structure of a charged spherical black hole.^[ref]

For completeness, if $Q=0$ and $M>0$, the Schwarzschild spacetime limit has just the positive horizon $r=2M$ and a spacelike curvature singularity at $r=0$; the formal root $r_{-}=0$ of $r^{2}F$ is not a regular inner horizon. If $M=Q=0$, the metric tensor is flat and the radial origin is regular in Cartesian coordinates.

If $M^2=Q^2$ with $M>0$, thenThe curvature singularity remains at $r=0$. At $r=M$ the static chart has a coordinate singularity, removed by the Eddington-Finkelstein coordinates above. The Kretschmann scalar there equals $8/M^4$, which is finite. The coincident horizons form a degenerate Killing horizon, and its future outer branch is the event horizon of the extremal black hole. A double zero of $F$ does not remove the horizon.

If $M^2<Q^2$, thenThere is no positive root and no horizon hiding $r=0$ in the extended Reissner-Nordstrom spacetime. Its Kretschmann scalar diverges there, so $r=0$ is a naked singularity. It is a timelike curvature singularity, as the normal to nearby $r=\text{constant}$ hypersurfaces has positive squared norm $g^{rr}=F$.

Write the affine function obtained in part (iii) as $\lambda =c+2b_{\mu }{x}^{\mu }$, with constant $c$ and $b_{\mu }$. A Conformal Killing vector field with precisely this conformal Killing factor isDirect partial differentiation verifies its Conformal Killing equation. Subtracting it from any other solution leaves a Killing vector field $Y$. The formula in part (iii), now with $\lambda =0$, says ${\partial }_{\mu }{\partial }_{\nu }{Y}_{\rho }=0$. Therefore $Y$ is affine, and the Killing equation makes its linear coefficient antisymmetric after lowering an index. Every solution is consequentlyThe four constant $a^{\mu }$ generate translations, the six independent ${\omega }_{\mu \nu }$ generate Lorentz transformations, $c$ generates a dilation, and the four $b^{\mu }$ generate special conformal transformations. Their constant, linear and quadratic terms are independent; within the linear terms, the antisymmetric coefficient cannot equal a nonzero multiple of the identity. Hence the conformal Killing algebra of Minkowski spacetime hasFor $\lambda =1$, the canonical choice is the dilation generator $D=x^{\mu }{\partial }_{\mu }$. Its flow of a vector field solves $d{x}^{\mu }/ds=x^{\mu }$, givingOne may add any Killing vector field without changing $\lambda =1$, so this choice is not unique. The displayed dilation gives a global one-parameter group of diffeomorphisms. The count of fifteen concerns independent Conformal Killing vector fields; some special conformal transformations have only local flows on all of affine Minkowski spacetime.

In Cartesian coordinates on Minkowski spacetime, the Conformal Killing equation becomesDifferentiate the equations with index pairs $(\nu ,\rho )$, $(\mu ,\rho )$ and $(\mu ,\nu )$ by ${\partial }_{\mu }$, ${\partial }_{\nu }$ and ${\partial }_{\rho }$, respectively. Adding the first two and subtracting the third, Clairaut's theorem yields the flat conformal Killing integrability formulaApply ${\partial }^{\rho }$ and use ${\partial }^{\rho }{X}_{\rho }=4\lambda$. With the d'Alembert operator $\square ={\eta }^{\alpha \beta }{\partial }_{\alpha }{\partial }_{\beta }$, this givesA tensor contraction with ${\eta }^{\mu \nu }$ gives $2\square \lambda =-4\square \lambda$, so $\square \lambda =0$. Returning to the preceding equation proves the stronger statementThus the conformal Killing factor is an affine function, not an arbitrary solution of the wave equation.

Use an affine parameter $s$ on the geodesic, so its tangent vector $U$ obeys ${\nabla }_{U}U=0$. The product rule, metric compatibility, and the Conformal Killing equation giveHenceIt is conserved on every null geodesic with an affine parameter, but generally not on other geodesics. On a timelike geodesic parametrized by proper time, $g(U,U)=-1$, so the derivative is $-\lambda$. A Killing vector field, or a geodesic on which $\lambda$ vanishes, supplies the exceptional conservation. This is the conformal Killing geodesic identity.

For an arbitrary parameter with ${\nabla }_{U}U=fU$, the right-hand side acquires $f(X\cdot U)$. Thus even for a null geodesic, the simple conserved expression presupposes an affine parameter.

For a Conformal Killing vector field, the same metric compatibility calculation givesThus its Conformal Killing equation is ${\nabla }_{(\mu }{X}_{\nu )}=\lambda g_{\mu \nu }$. Taking the tensor contraction with the inverse metric, and using $g^{\mu \nu }{g}_{\mu \nu }=4$, determines the conformal Killing factor:Parentheses include the factor $1/2$ in tensor index symmetrization.

Gravitational waves also exhibit diffraction: the geometrical optics ray description fails when a propagation structure or lens delay is not large compared with the wave's wavelength or period. For a compact lens of mass $M_L$, a useful dimensionless parameter is $w=4G{M}_L{\omega }_{\mathrm{gw}}/c^3$; $w$ of order one signals potentially appreciable wave effects, with cosmological lens redshift supplying an additional factor $1+z_L$ when needed. Astrophysical gravitational lensing can therefore produce diffraction and frequency-dependent amplification. Ordinary objects are poor opaque screens for gravitational waves, so everyday slit-and-screen demonstrations are implausible despite the large wavelengths, which are about $3000$ km at $100$ Hz.

In a local inertial frame in vacuum, gravitational waves travel at $c$ in general relativity; there is no ordinary material index of refraction analogous to a transparent dielectric. Ordinary matter couples so weakly that laboratory prisms for gravitational waves are implausible. Propagation through a varying background gravitational field nevertheless changes ray directions and travel times, giving gravitational lensing and Shapiro time delay. In weak static coordinates $d{s}^2\simeq -(1+2{\Phi }_N/c^2)c^{2}d{t}^2+(1-2{\Phi }_N/c^2)d{\mathbf{x}}^2$, the null condition gives an effective coordinate index of refraction $n_{\mathrm{eff}}\simeq 1-2{\Phi }_N/c^2$. This useful description depends on coordinates and does not change the local propagation speed. Thus astronomical gravitational refraction is plausible, while a substantial everyday material effect is not.

Gravitational waves can exhibit interference. In linearized gravity, overlapping strains add, so coherent contributions can enhance or cancel the measured response. Independent astrophysical sources generally have unrelated phases and average to a stochastic background rather than stable fringes. More plausible coherent paths arise when gravitational lensing sends different parts of the same signal to the detector with different delays; fringes require sufficient overlap and coherence. The relevant quadratic radiative energy flux contains the cross terms even though the leading wave equation is linear.

Gravitational radiation admits polarization: general relativity has two transverse tensor modes, plus polarization and cross polarization, whose relative amplitudes and phase give linear, elliptical or circular gravitational-wave polarization. Their axes transform through twice a rotation of the transverse coordinate frame, unlike the vector polarization of electromagnetic radiation. Measuring these modes is physically plausible with several differently oriented detectors; a single detector generally measures only one projected combination.

Yes, for genuine precession of a nonspherical disc. Let its unit normal be $\mathbf{n}=(\sin \alpha \cos \omega t,\sin \alpha \sin \omega t,\cos \alpha )$. Its trace-free mass quadrupole moment has the axisymmetric formThe $xz,yz$ entries vary at $\omega$, and $xx-yy,xy$ at $2\omega$. Their nonzero second and third time derivatives give an oscillating far-field metric perturbation and positive average radiated power. For $\alpha =\pi /2$ the $\omega$ terms vanish but the $2\omega$ terms remain. The precise condition is $\omega \ne 0$ and $\sin \alpha \ne 0$: the literal inequality $\alpha \ne 0$ also allows $\alpha =\pi$, where the normal is fixed and no such gravitational radiation occurs. For a uniform thin disc of mass $M_d$ and radius $b$, $Q_0=-M_d{b}^2/4$, confirming that it is not zero.

No, for a uniform or otherwise axisymmetric disc in steady rotation. Although individual material elements accelerate, the mass density and mass current at every spatial point remain constant, and the mass quadrupole moment is time independent. There is no gravitational radiation. Axisymmetry of the material distribution is necessary for this conclusion: the source does not explicitly specify the disc's density. A counterexample to an unconditional answer is a disc with two equal dense spots at opposite radii; their rotating mass quadrupole moment produces gravitational waves at twice the rotation angular frequency.

No, for steady rotation of the uniform sphere. Its mass quadrupole moment is zero by spherical symmetry; its rotating mass current is stationary when the rotation axis and rate are fixed. A stationary source does not emit gravitational waves. Rotation by itself is insufficient: a time-dependent nonaxisymmetric distribution, or another changing radiative multipole moment, is required. The conclusion concerns a sphere with the stated constant density and steady motion, rather than a deformed or time-varying rotator.

No. A motionless isolated point mass produces a static gravitational field. All its mass multipole moments are time independent, so there is no outgoing gravitational radiation. In the quadrupole formula, $h_{ij}^{\mathrm{TT}}\propto d^2{Q}_{ij}^{\mathrm{TT}}/d{t}^2=0$.

Using the supplied iron parameters givesAn internal Weiss molecular field depends on temperature through $M(T)$. Its low-temperature saturation estimate uses $M_s=ng{\mu }_{B}J\simeq 1.58\times 10^6\mathrm{A}{\mathrm{m}}^{-1}$:At and above $T_C$ the zero-applied-field magnetization vanishes in this model, so this is a saturation estimate, not the field at $T_C$.

Use the paper's stipulated face-centered cubic lattice, which has twelve nearest neighbours. The saturated one-ion alignment scale isDividing the effective field among twelve equivalent neighbours gives the per-bond exchange interaction scaleTo specify the convention, write the Heisenberg model as $\mathcal{H}=-\mathcal{J}{\sum }_{⟨ij⟩}{\mathbf{J}}_i\cdot {\mathbf{J}}_j$, with dimensionless angular-momentum operators and each bond counted once. Then $g{\mu }_B{B}_e=12\mathcal{J}⟨J_z⟩$, so $\mathcal{J}=3k_B{T}_C/[12J(J+1)]\simeq 11.2\mathrm{meV}$ here. A convention with $-2{\mathcal{J}}^{\prime }$ instead has ${\mathcal{J}}^{\prime }=5.62\mathrm{meV}$. The total saturated interaction energy per ion is $-12\mathcal{J}{J}^2/2\simeq -67.4\mathrm{meV}$; the factor of one half avoids counting each bond twice.

For positive $\lambda$, a fluctuation of magnetization produces a Weiss molecular field in the same direction, reinforcing alignment. The mean-field self-consistency equation isLinearizing in the paramagnetic phase givesAt zero applied field, the right-hand side's slope at $M=0$ is $T_C/T$. Below $T_C$ this exceeds one, and the saturation of the Brillouin function allows stable nonzero solutions of opposite sign: spontaneous ferromagnetism. Above $T_C$, only the zero-field solution $M=0$ remains stable. The stated linear susceptibility is the paramagnetic branch for $T>T_C$; it must not be continued as a negative susceptibility below the transition.

The $1/x$ terms in the Brillouin function cancel. Keeping the linear terms givesConsequently,Use the dimensionless SI magnetic susceptibility $\chi =\partial M/\partial H_{\mathrm{applied}}$ and $B_{\mathrm{applied}}={\mu }_0{H}_{\mathrm{applied}}$, as in the source's molecular-field convention. The Curie law isIf susceptibility is instead defined as $\partial M/\partial B$, its coefficient is $C/{\mu }_0$ and is dimensionful. The SI convention above makes the subsequent Curie–Weiss law and $\lambda$ consistent.