TODO: the source light curve, its numerical contact phases and the meanings of the four plotted time labels are absent, so unique numerical stellar radii cannot be recovered. The central eclipse contact-radius relations explain what the missing measurements would determine.

If $t_1,\dots ,t_4$ denote the usual first through fourth contacts of a central eclipse, define $T_{14}=t_4-t_1$ and $T_{23}=t_3-t_2$, and let $R_L\ge R_S$ be the two radii. For a circular edge-on orbit,Taking half their sum and difference gives the radii. For short eclipses the familiar approximation is $R_L\simeq v_{\mathrm{rel}}(T_{14}+T_{23})/4$ and $R_S\simeq v_{\mathrm{rel}}(T_{14}-T_{23})/4$, with $v_{\mathrm{rel}}=2K$. At first and fourth contact the discs touch externally; at second and third the smaller disc is just internally tangent to the larger. Noncentral eclipses require the projected impact parameter as well.

Equal radial velocity amplitudes imply equal stellar masses. Under the circular, edge-on eclipsing binary interpretation, each star moves at $K=100\mathrm{km}{\mathrm{s}}^{-1}$, and the relative speed is $2K$. ThusUsing $P=5.45$ days givesMore generally, spectroscopic binary amplitudes determine $a\sin i=P(K_1+K_2)\sqrt{1-e^2}/(2\pi )$ and $(M_1+M_2){\sin }^{3}i=P(K_1+K_2{)}^3(1-e^2{)}^{3/2}/(2\pi G)$. Thus these numerical masses and separation use $e=0$, $\sin i=1$; eclipses alone imply high inclination rather than its exact value.

For two bodies of masses $M_1,M_2$, the relative Kepler orbit has semimajor axis $a$ and orbital period $P$ satisfying Kepler's third lawHere $a$ is the semimajor axis of the separation vector, not either star's individual barycentric orbit. In conventional stellar units the same relation is approximately $(P/\mathrm{yr}{)}^2=(a/\mathrm{AU}{)}^3/[(M_1+M_2)/M_{\odot }]$, with year, astronomical unit and solar mass as the units.

Substituting the electron Lorentz factor in the supplied energy-loss formula givesWithout replenishment, the electron's momentum decreases. The reference dipoles then bend it more tightly, producing a displaced orbit. In linear beam optics, $x=D_x\delta$ and $y=D_y\delta$, where $\delta =\Delta p/p\simeq -U_0/E\simeq -1.11\times 10^{-3}$ per uncompensated turn. In an ideal planar ring the horizontal dispersion is nonzero and the orbit shifts horizontally, inward where $D_x>0$. There is no corresponding vertical closed-orbit shift when $D_y=0$. Real coupling or vertical dispersion can change that idealization.

With current $I=0.500$ A, the number of circulating particles is $N=I{T}_{\mathrm{rev}}/e$, so the radiated power isThe $2\mu \mathrm{s}$ revolution time cancels when using beam current. Radio-frequency cavities restore the lost energy longitudinally, with synchronous gain $e{V}_{\mathrm{RF}}\sin {\phi }_s=U_0$ in the sine-phase convention. Their RF power system must supply at least this beam power, in addition to its own losses. Radiation damping and quantum excitation determine beam emittances; they should not be confused with the dispersive closed-orbit shift from a momentum change.

For a $5$ GeV electron, ${\gamma }_e=E/(m_e{c}^2)\simeq 9785$. With $\rho =10$ m, the stated synchrotron critical frequency formula givesAt the same total beam energy and bend radius, a proton has ${\gamma }_p=5/0.9383\simeq 5.329$, and the same supplied formula givesThe ratio is $({\gamma }_p/{\gamma }_e{)}^3\simeq (m_e/m_p{)}^3$, explaining the strong suppression for protons. The critical-energy formula is an ultrarelativistic approximation, excellent for the electron and only approximate for this moderately relativistic proton. If “5 GeV proton beam” instead denotes kinetic energy, then ${\gamma }_p=1+5/0.9383$ and the estimate is about $7.50\times 10^{-6}$ eV; the two beam-energy conventions should not be mixed.

Relativistic radiation is concentrated into an angle of order $1/\gamma$ around the instantaneous velocity. The emission direction sweeps past a tangent observer over an orbital interval of order $\rho /(\gamma c)$. Since the charge moves toward the observer, arrival times are compressed by an additional factor of order $1/{\gamma }^2$, giving the spectral scale $\omega \sim c{\gamma }^3/\rho$.

The conventional numerical factor follows from the radiation Fourier phase. Near the tangent direction, circular motion givesWith $u=c\gamma t/\rho$, this becomes $[\omega \rho /(2c{\gamma }^3)](u+u^3/3)$. The resulting Airy-type integrals are expressible through modified Bessel functions with tangent-plane argument $\xi =\omega \rho /(3c{\gamma }^3)$. The standard angle-integrated synchrotron radiation spectrum has shapeIt therefore identifiesThe beaming argument supplies the physical scaling; the spectral integral fixes the conventional factor $3/2$. Critical energy is a broad spectral scale, not a hard photon-energy cutoff.^[ref]

For a real determinant-one map, bounded generic repeated transport requires eigenvalues $e^{\pm i\mu }$, equivalently $|\mathrm{tr}M|<2$. Here both planes have $\mathrm{tr}M=2-L_1{L}_2/f^2$, soFor positive drift lengths and $f>0$, this means $f>\sqrt{L_1{L}_2}/2$. The equalities are marginal and generally produce a Jordan-block growth rather than stable motion. Moving the reference point changes the map by a similarity transformation, $M\to PM{P}^{-1}$, so the eigenvalues and full-cell phase advance are unchanged.

Write the stable map using the Courant-Snyder parameters, also called Twiss parameters:At the chosen reference point $\beta =M_{12}/\sin \mu$, $\alpha =(M_{11}-M_{22})/(2\sin \mu )$, and ${\gamma }_T=-M_{21}/\sin \mu$. Repeated traversal givesIf the full ring contains $N_c$ identical cells, its one-turn phase is $N_c\mu$ modulo $2\pi$; $n$ ring turns use $n{N}_c\mu$. The displayed $M^n$ is for $n$ repetitions of the specified cell.

Maps act on column vectors, so the first element appears at the right of the product. Starting immediately before the focusing kick, one full cell hasMultiplying givesIt has $\det M_x=1$, as required for this lossless linear beam optics. The other transverse plane experiences the opposite quadrupole signs, soThe two planes have different envelopes but the same trace and therefore the same stable phase-advance magnitude for this equal-strength doublet.

Use transverse coordinates $(x,x^{\prime })$ and a signed focusing strength $K$ in Hill's equation, $x^{\prime \prime }+Kx=0$. A drift of length $\ell$ has beam transfer matrixFor a uniform focusing quadrupole magnet with $K={\kappa }^2>0$, and a defocusing one with $K=-{\kappa }^2$,The focusing plane has oscillatory trajectories, while the defocusing plane has hyperbolic growth and decay. A magnetic quadrupole focuses one transverse plane and defocuses the other; reversing its gradient interchanges their roles.

If $|K|{\ell }^2\ll 1$ and the element's internal drift can be neglected or assigned separately, its kick may be concentrated at one position. With $1/f=|K|\ell$,Both kicks preserve position and change the slope linearly with displacement. All these ideal uncoupled maps have determinant one.

Evaluate the envelopes at their displayed arguments and set $a={\eta }^{\ast }\rho$. The common phase $e^{-i{V}_{0}t}$ cancels from every bilinear. Direct use of the Pauli matrices givesThe physical longitudinal current is $J_z=|\eta {|}^2-|\rho {|}^2$: the two density profiles move in opposite directions. Their separate transport identities give ${\partial }_t(|\eta {|}^2+|\rho {|}^2)+{\partial }_z(|\eta {|}^2-|\rho {|}^2)=0$.

For a coherent superposition, the transverse currents $J_x=2\mathrm{Re}({\eta }^{\ast }\rho )$ and $J_y=2\mathrm{Im}({\eta }^{\ast }\rho )$ generally do not vanish. They are independent of $x,y$, so they contribute no divergence in this one-dimensional construction. Omitting them would give an incorrect local four-current. Pointwise, $|\mathbf{J}{|}^2=(|\eta {|}^2+|\rho {|}^2{)}^2$, so the pure two-component Weyl current is future null wherever the density is nonzero. If only one envelope is present, the current points entirely along its propagation direction; an incoherent average can remove the transverse interference terms.

For dependence on $z$ alone and constant $V_0$, remove the common phase by $\psi =e^{-i{V}_{0}t}\tilde{\psi }$. Then ${\partial }_t\tilde{\psi }=-{\sigma }_z{\partial }_z\tilde{\psi }$. The normalized eigenvectorsobey ${\sigma }_z{u}_{\pm }=\pm u_{\pm }$. The corresponding first-order transport equations have characteristics $t-z$ and $t+z$, respectively, givingHere $\eta$ propagates toward $+z$ and $\rho$ toward $-z$, both at unit speed.

For a fixed unit vector $\hat{\mathbf{n}}$, replace $z$ by $\hat{\mathbf{n}}\cdot \mathbf{x}$ and choose orthonormal spinors satisfying $(\mathbf{\sigma }\cdot \hat{\mathbf{n}})u_{\pm }(\hat{\mathbf{n}})=\pm u_{\pm }(\hat{\mathbf{n}})$. If $\hat{\mathbf{n}}=(\sin \vartheta \cos \phi ,\sin \vartheta \sin \phi ,\cos \vartheta )$, one choice isIndependent constant phases on the spinors can be absorbed into the two arbitrary envelope functions.

Assume the scalar potential is real, so the Weyl equation has a Hermitian Hamiltonian $H=\mathbf{\sigma }\cdot \mathbf{p}+V$. Multiplying the equation and its adjoint by ${\psi }^{†}$ and $\psi$ respectively givesThus with $\rho ={\psi }^{†}\psi$ and $\mathbf{J}={\psi }^{†}\mathbf{\sigma }\psi$, the lower-index four-current $j_{\mu }=(\rho ,-\mathbf{J})$ obeys $\boxed{{\partial }^{\mu }{j}_{\mu }=0}$.

For $K=\mathbf{\sigma }\cdot \mathbf{p}$,Consequently $K$ is conserved for spatially constant $V$; a spatially uniform time-dependent scalar offset would also commute with it. For general nonuniform $V$ it is not conserved. At nonzero momentum, $(\mathbf{\sigma }\cdot \hat{\mathbf{p}}{)}^2=I$, so the helicity eigenvalues are $\boxed{+1,-1}$. In the chosen constant-potential Weyl Hamiltonian, the positive $E-V$ branch has helicity $+1$; the negative branch has $-1$. An expectation value in a superposition need not itself be an eigenvalue.