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Take the scalar product of the Lorentz force equation with the constant magnetic field $\mathbf{B}$. Since $\mathbf{E}\cdot \mathbf{B}=0$ and $(\dot{\mathbf{x}}\times \mathbf{B})\cdot \mathbf{B}=0$,The initial velocity also obeys $\mathbf{v}\cdot \mathbf{B}=0$, so $\mathbf{x}\cdot \mathbf{B}$ remains equal to ${\mathbf{x}}_0\cdot \mathbf{B}$. The trajectory therefore lies in the planethe plane through ${\mathbf{x}}_0$ perpendicular to $\mathbf{B}$.

Let the incident and final photon momentum magnitudes beIn a nontrivial collinear collision, the photon is backscattered and the electron moves in the original photon direction, so momentum conservation gives electron momentum $P=p+p^{\prime }$. Conservation of relativistic energy givesSubstitute $P=p+p^{\prime }$, isolate the square root, and square. Cancellation leavesDividing by $p{p}^{\prime }$ and using the wavelength definitions gives the Compton scattering shiftThe other collinear possibility is the trivial forward solution ${\lambda }^{\prime }=\lambda$, in which the electron remains at rest.

A four-vector is an object whose components transform between inertial frames by a Lorentz transformation,Lorentz transformations are defined bywhere $\eta$ is the Minkowski metric. ConsequentlyThus the Minkowski norm of a four-vector is the same in every inertial frame.

Neglecting terms of order ${\Omega }^2$ and linearizing in $\theta$ givesThe second equation and the initial data implyIts contribution to the first equation is of order ${\Omega }^2$, so the retained equation isThusIntegrating the associated $y$ velocity and using $y(0)=0$ gives

The circular constraint isIts tangent in the $xz$-plane is $(\cos \theta ,0,-\sin \theta )$, while the normal force is radial. Projecting the equations from part (a) onto this tangent eliminates the normal force and givesBecause the ramp is translation-invariant along $y$, $N_y=0$; using $\dot{x}=R\dot{\theta }\cos \theta$ gives

For rest in the rotating frame, $\dot{\theta }=\dot{y}=\ddot{\theta }=\ddot{y}=0$. Assuming $\Omega \ne 0$, the second equation gives $y=0$, and the first givesOn the semicircular ramp the rest points are thereforeand, when ${\Omega }^{2}R\ge g$,

The term $-2\mathbf{\Omega }\times \dot{\mathbf{x}}$ is the Coriolis acceleration, and $-\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{x})$ is the centrifugal acceleration. With $\mathbf{\Omega }=-\Omega \hat{\mathbf{z}}$, the Cartesian components are

Part (c) givesThe half-space $x_1>0$ is a simply connected domain, so the Poincaré lemma makes this curl-free field a conservative vector field: it is the gradient of a single-valued scalar potential.

Using $\tan u+\cot u=2/\sin (2u)$,The spherical-coordinate formula from part (b) shows that a potential must satisfy $f_{\varphi }=2$ and may be independent of $r,\theta$. On $x_1>0$ the azimuthal angle has the single-valued branchHence one solution, up to an additive constant, is

Each field has only an azimuthal component and is independent of $r$ and $\varphi$. The curl in spherical coordinates therefore reduces toand its $\varphi$ component vanishes. The half-angle identitiesshow in both cases that

Apply part (a) with $g(r)=r^{-3}$. Thenaway from the origin, explicitly confirming that each resulting curl has zero divergence.

Differentiating the spherical coordinate systemgives the orthogonal decompositionwhereThus the scale factors of orthogonal coordinates are

Since the total differential satisfiescomparison of coefficients gives the gradient

Since $r=(x_j{x}_j{)}^{1/2}$, the chain rule givesFor the radial vector field $g(r)\mathbf{x}$, the product rule and Einstein summation convention give

Reflection symmetry of the cube in each coordinate plane makes every off-diagonal integralvanish because its integrand is odd in one coordinate. Permuting the three coordinate axes leaves the cube unchanged, so its three diagonal moments are equal. Thus its inertia tensor at the centre has the isotropic form

Forpart (a) and $\lambda =M{\ell }^2/6$ giveThe displayed symmetric matrix has orthogonal eigenvectorswith corresponding eigenvaluesFor a unit angular velocity, $|\mathbf{L}|=|I\mathbf{\omega }|$ is minimized and maximized along eigenvectors for the smallest and largest eigenvalues. Their ratio is

Let a change of orthonormal coordinates be represented by a rotation matrix $R$. Since both angular momentum and angular velocity are vectors,Using $L_i=I_{ij}{\omega }_j$ in both frames givesfor every vector $\mathbf{\omega }$. Multiplication by $R^{-1}=R^T$ yieldswhich is exactly the transformation law for a rank-two tensor. Thus $I_{ij}$ is a rank-two tensor.

Put $\mathbf{r}=\mathbf{x}-\mathbf{a}$. The vector triple product givesThereforeAt the centre of mass,Expanding the first formula, the terms linear in $x_i$ vanish because the centre of mass is the origin, while $\rho {\int }_{B}dV=M$. Hence the parallel axis theorem in tensor form is