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At instantaneous height $z$, the image charge is a distance $2z$ below the real charge, giving force $m\ddot{z}=-K/z^2$ with $K=q^2/(16\pi {\epsilon }_0)$. Multiplying by $\dot{z}$ and using rest at $z=d$ givesThe induced-conductor potential energy is $-K/z$, not the full energy of a pair of independently movable charges. On the falling branch, integrate the reciprocal speed:The last integral is $\pi /2$ by $s={\sin }^2\theta$. Hence the fall time of a charge toward a grounded plane in the electrostatic Newtonian model isFor $q=0$ there is no force and the particle remains at rest.

The charge and its proposed image charge give the exterior electric potentialAt $z=0$ the terms cancel, so $\Phi =0$ everywhere on the plane and its tangential derivatives vanish. Thus the image construction satisfies the earthed-conductor boundary conditions. It has the correct point-charge singularity, is harmonic elsewhere in $z>0$, and decays at infinity; electrostatic uniqueness identifies it with the exterior solution.

In electrostatic equilibrium, the electric field inside a perfect conductor is zero. Apply the Faraday's law $\oint \mathbf{E}\cdot d\mathbf{l}=0$ to a thin rectangular loop crossing the surface, with its long sides tangential and its short sides shrinking to zero. It yields continuity of the tangential electric field. Since its interior value is zero, $E_x(0+)=E_y(0+)=0$. Earthing fixes the electric potential at the plane to zero. For completeness, a thin Gauss law pillbox gives $E_z(0+)=\sigma /{\epsilon }_0$; the normal component can be nonzero because of induced surface charge. These are the electrostatic boundary conditions at a conductor.

Take $f(0)=0$ and $f(x)=1/x$ for $x\ne 0$. Its graph is $\{(x,y):xy=1\}\cup \{(0,0)\}$, a union of two closed sets, since multiplication is continuous. But $f(1/n)=n$ does not tend to $f(0)$, so the function is not continuous.

If $f$ were not continuous at $x$, there would be $\epsilon >0$ and $x_n\to x$ with $d_Y(f(x_n),f(x))\ge \epsilon$. A compact metric space is sequentially compact, so some subsequence $f(x_{n_k})$ converges to $y$. The closed graph then contains its product limit $(x,y)$, forcing $y=f(x)$, contrary to the lower bound. Thus $f$ is continuous, the compact-codomain form of the closed graph theorem for compact spaces.

Suppose $(x_n,f(x_n))$ converges to $(x,y)$. Continuity gives $f(x_n)\to f(x)$, and uniqueness of limits in a metric space gives $y=f(x)$. Hence every convergent sequence in the graph has its limit in the graph, proving ${\Gamma }_f$ is closed.

For finite sets $X_1,\dots ,X_N$, the inclusion-exclusion principle statesTo prove it, fix an element belonging to exactly $s\ge 1$ of these sets. At level $r$ it is counted in precisely $\left(\genfrac{}{}{0ex}{}{s}{r}\right)$ intersections, so its total multiplicity on the right isby the binomial theorem. An element in no set contributes zero. Summing these equal multiplicities over the finite union proves inclusion-exclusion. This also explains why pairwise subtraction alone is insufficient: elements in triple and higher intersections must receive the alternating corrections.

Choose a right inverse $s:A\to B$ of the surjection $f$, so $f(s(a))=a$, and defineFor each $a$, the element $b=s(a)$ then witnesses both $f(b)=a$ and $g(b)=h(a)$. As in part (a), arbitrary simultaneous choices use the axiom of choice. If $A$ is empty, surjectivity and the existence of $f:B\to A$ force $B$ to be empty, and the unique empty function $h:A\to C$ already meets the requirement.

Suppose $g$ is constant on each fiber of a function of $f$. Since every fiber is nonempty, $h(a)$ must be its common $g$-value: any witness $b$ in that fiber gives that value. This defines exactly one $h$ and yields $g=h\circ f$, an instance of factorization through a surjection.

Conversely, suppose a fiber contains $b,b^{\prime }$ with $f(b)=f(b^{\prime })=a_0$ but $g(b)\ne g(b^{\prime })$. Starting with any admissible $h$, define $h_1,h_2$ to agree with it away from $a_0$, but put $h_1(a_0)=g(b)$ and $h_2(a_0)=g(b^{\prime })$. Both remain admissible, witnessed at $a_0$ by $b$ and $b^{\prime }$ respectively, yet they are distinct. HenceThe distinction is important: the original existence property asks only for one witness in each fiber. Only when $g$ is fiberwise constant does that property force $g=h\circ f$ for every $b$.

A left inverse of $f$ is a function $\ell :B\to A$ satisfying $\ell \circ f={\mathrm{id}}_A$. If such $\ell$ exists and $f(a)=f(a^{\prime })$, applying $\ell$ gives $a=a^{\prime }$, so $f$ is an injection. Conversely, suppose $f$ is an injection and fix $a_0\in A$, possible because $A$ is nonempty. DefineThe first value is unique by injectivity, and $\ell (f(a))=a$ for every $a$. ThusNo choice from a family of nonempty sets is needed here: each preimage in the first case is unique, and the second case uses one fixed element.

A right inverse is a function $r:B\to A$ with $f\circ r={\mathrm{id}}_B$. Its existence makes every $b$ the image of $r(b)$, proving that $f$ is a surjection. Conversely, if $f$ is a surjection, every fiber of a function $f^{-1}(\{b\})$ is nonempty. Using the usual axiom of choice for arbitrary sets, choose $r(b)\in f^{-1}(\{b\})$ for each $b$. Then $f(r(b))=b$, givingThe converse in this second equivalence is the right-inverse characterization of the axiom of choice; it is not an unrestricted choice-free theorem about arbitrary sets.

With the Sun fixed, its gravitational force on the planet is a central force. In the fixed orbital plane, the transverse acceleration in polar coordinates therefore vanishes:During $dt$, the position vector sweeps an infinitesimal sector of area $dA=\frac{1}{2}{r}^{2}d\theta$. Hence the areal velocity isIntegrating over any two equal time intervals gives equal swept areas, proving Kepler's second law. Here $L$ is the signed component of angular momentum normal to the orbital plane. The argument uses conservation of angular momentum; an inverse-square dependence is not needed for this particular law, only a central force.

The particle's angular momentum about the force centre is $\mathbf{L}=m\mathbf{r}\times \dot{\mathbf{r}}$. A central force exerts zero torque about that centre, soBy conservation of angular momentum, $\mathbf{L}$ is a constant vector. If it is nonzero, the identity $\mathbf{r}\cdot \mathbf{L}=0$ places every point of the orbit in the fixed plane through the centre perpendicular to $\mathbf{L}$:If $\mathbf{L}=0$, position and velocity are parallel. Away from the force centre, ${\dot{\mathbf{e}}}_r=(\dot{\mathbf{r}}-\dot{r}{\mathbf{e}}_r)/r=0$, so the motion is radial along a fixed line, which lies in any plane containing that line. Thus the zero-angular-momentum case also gives planar motion. As usual, these assertions apply while the force law supplies a regular orbit; continuation through a singular force centre is a separate issue.

A central force acts along the line joining the particle to a fixed centre. Taking that centre as the origin, its usual position-dependent form isPositive $f$ gives an outward force and negative $f$ an inward one. For a conventional central force the magnitude depends only on distance from the centre, not on direction. The angular-momentum conclusions below require only that the force remain radial about that fixed centre.

The integer $137$ is prime: none of the primes $2,3,5,7,11$ below $\sqrt{137}$ divides it. Since $43\not\equiv 0(\mathrm{mod}137)$, Fermat's little theorem gives $43^{136}\equiv 1$. Therefore $43^{135}$ is the modular inverse of $43$. The extended Euclidean algorithm yieldsand back-substitution gives $1=51(43)-16(137)$. ConsequentlyIndeed, $43\cdot 51=2193=16\cdot 137+1$, directly checking the required modular inverse.

Eliminate $x$ by twice the first modular congruence minus three times the second. This givesThe modular inverse of $3$ is $16$, because $3\cdot 16=48\equiv 1$. Hence $y\equiv 16\cdot 13\equiv 20$. Substitution into the second modular congruence gives $6x\equiv 14-7(20)\equiv 15$, and $6^{-1}\equiv 8$, so $x\equiv 8\cdot 15\equiv 26$. ThusIn particular, $(x,y)=(26,20)$ works: $9(26)+12(20)=474=10(47)+4$ and $6(26)+7(20)=296=6(47)+14$. The elimination is reversible because $3$ and $6$ have modular inverses, so the displayed classes contain all solutions.

Place the four fixed charges at $(sa,ta)$ with $s,t\in \{-1,1\}$, and write the moving particle's position as $(x,y)$. Introduce its mass $m>0$, which is needed for an oscillation frequency but is not named in the question. Let $C=Qq/(4\pi {ϵ}_0)>0$. By Coulomb's law and superposition, the moving particle's potential energy isThe electrostatic potential itself is $V/q$. Put $r^2=x^2+y^2$ and ${\delta }_{st}=-(sx+ty)/a+r^2/(2a^2)$. Near the centre, a Taylor expansion givesOn summing over the four corners, linear terms cancel. To quadratic order, $\sum {\delta }_{st}=2r^2/a^2$ and $\sum {\delta }_{st}^2=4r^2/a^2+O(r^4/a^4)$, since $\sum st=0$. Reflection symmetry removes all odd-degree terms. HencewhereThus the centre has zero planar force and a positive-definite planar Hessian matrix $\kappa I_2$. It is a strict local minimum of the constrained potential energy, so it is a stable equilibrium. More explicitly, conservation of the total energy confines sufficiently small-energy motions within a small neighbourhood of this isolated local minimum.

Keeping only the quadratic potential, the two coordinates obey independent small oscillations:Both planar modes have the same angular frequency, and the ordinary frequency isThe planar constraint is essential to this constrained electrostatic equilibrium. Along the perpendicular axis, $V(0,0,z)=4C/(2a^2+z^2{)}^{1/2}=V(0,0)-C{z}^2/(\sqrt{2}{a}^3)+O(z^4)$, so the same stationary point is unstable to unconstrained normal displacements. That also checks the cancellation of the three-dimensional Hessian trace required by Laplace's equation away from charges; it does not contradict the positive-definite restriction to the given plane.

With no electric field, the Lorentz force isFor nonrelativistic motion, the work-energy theorem givesConsequently $\boxed{K(t)=K(0)}$. The reason is geometric: magnetic forces do no work because they are perpendicular to the instantaneous velocity. They can change its direction but not its magnitude. The same energy conclusion holds relativistically, using $dE/dt=\mathbf{F}\cdot \mathbf{v}$ and subtracting the constant rest energy to obtain kinetic energy.

Let the incoming projectile momentum be ${\mathbf{p}}_1$. The stationary target has four-momentum $(m_{2}c,0)$, so the total incoming norm isHere the relativistic energy-momentum relation eliminated $|{\mathbf{p}}_1{|}^2$. Write $W$ for the total energy in the centre-of-momentum frame; then $W^2=c^2{P}_{\mathrm{in}}^2$. If $M_f={\sum }_i{\mu }_i$, the relativistic collision mass threshold gives $W\ge M_f{c}^2$. Squaring and rearranging with $m_2>0$ yields the fixed-target production thresholdThe actual incident total energy also obeys $E_1\ge m_1{c}^2$. Thus a negative algebraic reaction bound would not allow a negative-energy projectile.

For a photon, $m_1=0$ and $E_1=h\nu$, where $h$ is the Planck constant. The necessary bound becomesIn the usual massive multi-product channel with $M_f>m_2$, conservation alone permits the threshold configuration in which all products are at rest relative to one another in the centre-of-momentum frame. Boosting that configuration supplies the required laboratory momentum. Hence the photon fixed-target production threshold isFor a multi-product channel with $M_f\le m_2$, there is no positive mass threshold: the lower limit of allowed positive frequencies is zero. The compact expression for that lower limit is $\boxed{{\nu }_{\inf }=\max \{0,M_f^2-m_2^2\}{c}^2/(2m_{2}h)}$; it need not correspond to a zero-frequency incident photon.

For completeness, if the only product is one particle of mass ${\mu }_1$, its invariant mass must match exactly. A positive incident frequency is possible only when ${\mu }_1>m_2$, and is then fixed at $\nu =({\mu }_1^2-m_2^2)c^2/(2m_{2}h)$; frequencies above that value do not produce the same single particle while conserving four-momentum. If a multi-product channel includes massless particles, an all-products-at-rest configuration is impossible for those particles with nonzero energy, so the nominal massive threshold may be only an infimum. These qualifications distinguish the requested necessary inequalities from unrestricted claims of sufficiency.

The two incoming particles have equal energies and opposite momenta, so their total four-momentum in this frame isIt is already a centre-of-momentum frame. By four-momentum conservation, the final spatial momenta sum to zero and their energies sum to $2E$. Each product has energySumming gives the necessary relativistic collision mass thresholdFor massive products the equality threshold has all of them at rest in this frame. Above it, multiple products can carry balancing momenta and relative kinetic energy. The displayed condition is necessary, as requested; a single final particle must instead have exactly the incoming invariant mass, and additional physical conservation laws could further restrict a real reaction.

For $|\mathbf{v}|<c$, define the Lorentz factor $\gamma =(1-|\mathbf{v}{|}^2/c^2{)}^{-1/2}$. A particle of rest mass $m$ hasThese expressions give the relativistic energy-momentum relation directly:so $\boxed{E^2=c^2\mathbf{p}\cdot \mathbf{p}+m^2{c}^4}$.

The four-momentum iswhere $U^{\mu }$ is the four-velocity, the proper-time derivative of $(ct,\mathbf{x})$. It is a four-vector, so its Minkowski metric norm is invariant under Lorentz transformations. Choose metric signature $(+,-,-,-)$; in the particle's rest frame, $P^{\mu }=(mc,0)$ and $P_{\mu }{P}^{\mu }=m^2{c}^2$. In any frame the same norm is $E^2/c^2-|\mathbf{p}{|}^2$. Equating the two yields the same relativistic energy-momentum relation. With the opposite metric convention both norm expressions change sign, leaving the energy relation unchanged.

Again assume independent die scores. Their sum has the discrete convolution of the two score distributions, soAll terms are nonnegative. The arithmetic-geometric mean inequality applied to the two extreme-score contributions givesIf all eleven probabilities were equal, normalization would make each $1/11$, but the inequality would require $1/11\ge 2/11$. This contradiction proves the sum cannot be uniform on $\{2,\dots ,12\}$, whatever the individual die probabilities.