Model the cart as a particle on a fixed plane curve, remaining constrained to the track. Its vertical velocity is $\dot{y}=y^{\prime }(x)\dot{x}$, so its kinetic energy, gravitational potential energy and Lagrangian areConsequently the canonical momentum conjugate to the horizontal coordinate isIt is not simply the horizontal component $m\dot{x}$ of the mechanical momentum, because a change in $x$ also moves the cart vertically. Differentiate carefully in the Euler-Lagrange equation:Subtracting yieldsThe term involving $y^{\prime \prime }$ accounts for the changing track geometry. An independent check is conservation of energy:

Smooth ground, rough cylinder contacts. Write $f$ for the tangential contact force magnitude at each upper-to-lower contact. On the top cylinder, its tangential components point upward and outward, while their horizontal components and torques cancel by symmetry. For a lower cylinder, the tangential force has an inward horizontal component $\sqrt{3}f/2$ and a downward vertical component $f/2$.

The rolling condition is now at the contact between cylinders, not at the ground. At the initial zero velocity, equality of the tangential contact-point accelerations givesThus $fR=\beta m{R}^2|\alpha |$ gives $f=2\beta mw$. The lower cylinder's outward translation equation and the top cylinder's downward equation areUsing $u=\sqrt{3}w$ gives $N=2\sqrt{3}m(1+\beta )w$ and henceFor solid cylinders this is $g/11$. Here the left lower cylinder rotates clockwise while moving left, and the right rotates counterclockwise while moving right. They may slip freely on the smooth ground, so imposing the ground-rolling relation $R|\alpha |=u$ would be incorrect. No slip at the cylinder contacts requires ${\mu }_c\ge f/N=\beta /[\sqrt{3}(1+\beta )]$ initially.

These are the two cases of the initial acceleration of three stacked cylinders. As a check, as $\beta \to 0$ the rotational contribution disappears in both formulas and they approach the smooth-contact release result $w=g/7$.

Rough ground, smooth cylinder contacts. Let $N$ be either upper-to-lower normal force, and $f_g$ the magnitude of the inward ground friction force on each lower cylinder. Pure rolling on the ground requires $|\alpha |=u/R$. The torque equation about a lower cylinder's centre isThe outward translation equation is $mu=N/2-f_g$, giving $N=2m(1+\beta )u$. The top cylinder's vertical equation is $mw=mg-\sqrt{3}N$. Using $u=\sqrt{3}w$ yieldsFor uniform solid cylinders, $\beta =1/2$ and $a_{\mathrm{top},\mathrm{down}}=g/10$. The left lower cylinder rotates counterclockwise while rolling left; the right rotates clockwise while rolling right.

The assumed no-slip ground contact is possible if its coefficient of static friction satisfies ${\mu }_g\ge \beta \sqrt{3}/(10+9\beta )$ at release. This follows from $f_g=\beta m\sqrt{3}w$ and ground normal force $N_g=mg+\sqrt{3}N/2=mw(10+9\beta )$.

During the photon's runner-frame flight, the tunnel entrance moves left by $v\Delta t_{\gamma }^{\prime }=\gamma \beta L$. It began ahead of the runner by $\gamma {\beta }^{2}L$, so at detection it lies behind byDivide this distance into the tunnel by the simultaneously measured tunnel length $L/\gamma$:The runner is stationary in $S^{\prime }$, but their location relative to the moving tunnel still has this fraction. Omitting the initial entrance gap would incorrectly count the runner's progress from an emission event at which they had not yet entered in this frame. These simultaneous positions give the tunnel photon timing in two inertial frames.

The photon starts at $x_E^{\prime }=\gamma L$ and travels left to the runner at the fixed position $x^{\prime }=0$. ThusThis distance is not the simultaneous contracted tunnel length $L/\gamma$, because the runner is not at the entrance when the photon is emitted in $S^{\prime }$. Adding the entrance-to-runner gap from part (ii) to $L/\gamma$ indeed gives $\gamma L$.

At that simultaneous entrance-clock event, $x_G=0$ and $t_G=-vL/c^2$. Its runner-frame position isThe runner is at $x^{\prime }=0$, so the entrance is still ahead by distance $\gamma {\beta }^{2}L$ at the runner-frame emission time. The simultaneously measured distance between the two tunnel ends is $x_E^{\prime }-x_G^{\prime }=\gamma L(1-{\beta }^2)=L/\gamma$, consistent with length contraction.

The entrance clock is at $x=0$. Let $G$ be the event on that clock simultaneous with emission in the runner's frame. Since $t_G^{\prime }=\gamma t_G=t_E^{\prime }$, its reading isThis is the tunnel clock's reading, not the runner-frame coordinate time $t_E^{\prime }=-\gamma vL/c^2$. It is also not a calculation of what a photograph of that distant clock would show, which would require an additional light-travel condition.

For the general Lagrangian just obtained, the Euler-Lagrange equation givesUsing the small-angle approximation $\sin \theta \simeq \theta$, this is simple harmonic motion with ${\omega }_0^2=A/I$. ThusThe endpoint mass cancels in the light-rod limit. A concrete comparison shows why the finite rod cannot silently be included: the same specified endpoint $m$ with $M_r=m$ gives $T=2\pi \sqrt{8l/(9g)}$, distinct from the negligible-rod result. This is a physical pendulum when the rod contributes, and a simple pendulum in the light-rod limit. Both expressions require small oscillation amplitude; the full sine equation remains nonlinear.

Take $\theta$ from the downward vertical. Under the conventional light-rod idealization, only the endpoint point mass contributes appreciably. Its speed is $l\dot{\theta }$, and its gravitational potential energy is $-mgl\cos \theta$. ThereforeAn additive constant in the potential energy has no effect on the Euler-Lagrange equation.

Rod-mass qualification: “thin” and “uniform” do not imply negligible mass. If the uniform rod has a finite unspecified mass $M_r$, its pivot moment of inertia is $M_r{l}^2/3$ and its centre of mass lies $l/2$ from the pivot. For a pendulum with a uniform rod and an endpoint mass, the general answer isThis formula covers either convention without inventing a rod mass. A result involving only the specified $m,l,g$ uses $M_r=0$ unless an additional relation for $M_r$ is stated.

For $m\gg M$, retain the ratio $E/m$ until a further limit is specified. Since $E\ge M$, the leading expression from a isFor an infinitely heavy target at fixed incident energy, $E/m\to 0$ and $\boxed{E^{\prime }\to E}$: the projectile reverses with negligible target recoil energy. If the incident particle is ultrarelativistic even relative to the target, $E\gg m\gg M$, the same leading expression instead gives $E^{\prime }\sim m/2$. The mass hierarchy alone does not justify replacing $E/m$ by zero.

With $M\gg m$ and $Em\gg M^2$, one also has $E\gg M^2/m\gg M$. The denominator in a is then dominated by $2Em$, and the numerator by $E{M}^2$:This residual total energy is still much larger than the incident particle's rest energy $M$, but is a small fraction of its incident energy $E$. The target acquires most of the available energy in this high-energy, heavy-projectile limit.

Launch and arrival occur on the capsule's own worldline, so its clock measures the proper time. In the ship frame its Lorentz factor is ${\gamma }_{c/s}=(1-0.9^2{)}^{-1/2}$; thereforeThis uses the station-frame distance convention stated in i. With a one-light-day distance in the ship frame instead, it becomes $\Delta {\tau }_c=\sqrt{0.19}/1.75\mathrm{day}\simeq 0.249080\mathrm{day}$. The uncertainty is a distance-frame convention in the wording, not a missing figure or an unanswered mathematical part.

Interpret the one-light-day separation $D$ as measured in the station frame at capsule launch, and retain the ship's initial inertial rest frame until reception at the station. The phrase does not explicitly name a frame; an alternative is given below. The ship-frame simultaneous separation is $D^{\prime }=D/{\gamma }_s$. The capsule moves at $0.9c$ towards a station approaching at $0.85c$, so their separation decreases at $1.75c$ in this frame. This is a separation rate, not either object's speed. Hence the elapsed ship-frame coordinate time isIt is the interval assigned using synchronized clocks in the ship frame to the launch and remote arrival. A single onboard clock does not directly mark the remote arrival event; observing its arrival signal would add a light-travel delay. If the stated one-light-day separation were instead measured in the ship frame, the answer would be $\Delta t_s=(1/1.75)\mathrm{day}=0.571429\mathrm{day}$.

In vacuum the electron-flavour survival probability depends only on $|U_{ei}{|}^2$ and the mass-squared differences. Complex conjugation of the lepton mixing matrix therefore leaves it unchanged:Thus a reactor disappearance measurement cannot search for the intrinsic CP violation phase in the standard three-neutrino model. Sensitivity to that phase calls for suitable appearance channels and, usually, comparison of neutrino and antineutrino beams. Matter can produce differences between propagation of neutrinos and antineutrinos; these are not by themselves intrinsic CP violation, and are small for the proposed short reactor baseline.

The relevant resolution condition is on the oscillation phase, not solely on which mixing angle is being measured. Since $\Delta \propto L/E$,At the same medium baseline, the fast ${\theta }_{13}$-dependent ripples have phases about $37.5$ times larger than the slow ${\theta }_{12}$ component. Resolving those ripples therefore requires a much better fractional energy resolution. At 47 km, $|{\Delta }_{23}|\simeq 59.5$; avoiding substantial phase smearing requires a resolution appreciably better than $1/59.5\simeq 1.7\%$.

However, a short-baseline ${\theta }_{13}$ experiment near 1.24 km and a ${\theta }_{12}$ experiment near 46.5 km each operate near their relevant first maximum, $\Delta \simeq \pi /2$. Their intrinsic fractional phase-resolution requirements are then comparable. The smaller ${\theta }_{13}$ disappearance amplitude demands careful relative calibration and background control, but it does not by itself impose a universally finer energy resolution. The proposal is conditional on baseline and analysis, rather than a general requirement.

A Positron signal is correct for the reactor's antineutrinos, through inverse beta decay,A true electron neutrino charged current interaction instead produces an electron, for example ${\nu }_e+n\to e^{-}+p$. This corrects the source's loose use of “neutrino” for the reactor beam.

After slowing, the Positron annihilates with an electron, generally producing two $511$ keV photons. The prompt visible energy includes both the Positron kinetic energy and annihilation energy; neglecting the small neutron recoil,A delayed neutron capture supplies a powerful coincidence tag. Detecting only two annihilation photons discards useful prompt energy and does not uniquely identify the charged current event.

The radiation length $X_0$ is the characteristic length for high-energy electron energy loss by bremsstrahlung; smaller $X_0$ means more radiative loss for the same traversed thickness. To limit that loss, minimize $x/X_0$, choosing less material or a larger $X_0$, rather than minimizing $X_0$ itself. At reactor Positron energies of a few MeV, ionization and subsequent annihilation matter more than high-energy bremsstrahlung. A sufficiently large scintillation counter measures the deposited energy, so containment of the Positron and annihilation photons, light yield and calibration are the relevant design issues.^[ref]

A fission reactor supplies predominantly electron antineutrinos, with typical energies of a few MeV. Even if they oscillate to muon antineutrinos, producing a real muon through a charged current requires roughly the muon rest energy, over $100$ MeV for a free nucleon target. For example,Thus a muon spectrometer cannot observe this appearance channel from a reactor. Measure electron antineutrino disappearance instead, most commonly through inverse beta decay, ${\bar{\nu }}_{e}p\to e^{+}n$, using the prompt Positron and delayed neutron signal.

Matched near and far particle detectors are a good design. A near particle detector, close enough that the fast oscillation phase is small, measures the unoscillated reactor spectrum and normalization. A far particle detector near the first fast maximum measures the disappearance. Their appropriately geometry-corrected ratio cancels much of the uncertain reactor flux and common cross section and detection efficiency. Relative calibration, different backgrounds, multiple reactor distances and any residual near-site oscillation still need treatment. The near site should retain enough shielding to suppress cosmic backgrounds.^[ref]

For a rate measurement of the small mixing angle ${\theta }_{13}$, choose the first fast disappearance maximum,A 47 km baseline instead lies near the slow oscillation maximum associated with ${\theta }_{12}$. Its much lower flux, approximately falling as $1/L^2$, and the many fast oscillations washed out by a reactor's broad energy spectrum make it a poor choice for this simple ${\theta }_{13}$ design. A medium-baseline spectral experiment can still constrain neutrino oscillation parameters if it has enough events and resolution; 47 km is not intrinsically useless, but it is not the optimal short-baseline rate experiment. In a, ${\theta }_{13}$ also affects the slow envelope through ${\cos }^4{\theta }_{13}$, so the sensitivity there is not identically zero.