= Solution
Because $s$ is <arc length>, $|d\mathbf x/ds|=1$. Thus $\mathbf y=\mathbf n/\alpha$, where $\mathbf n$ is the unit tangent to the <elastic ray>. In an <isotropic> medium this is also the <wavefront> normal. Consequently \b[$\mathbf y$ is the vector <slowness>], of magnitude $1/\alpha$; locally $\mathbf y=\nabla T=\mathbf k/\omega$, with $T$ the <travel time> and $\mathbf k$ the <wavevector>.
For depth-dependent <wave speed> $\alpha(z)$, the ray equation gives $dy_x/ds=dy_y/ds=0$. The horizontal component of $d\mathbf x/ds$ therefore has a fixed direction, and the <elastic ray> lies in the vertical plane determined by that direction and its initial point. A vertical <elastic ray> is the limiting case. If $i$ is the angle to the vertical, its conserved horizontal <slowness> gives
$$
\boxed{\frac{\sin i}{\alpha(z)}=p_{\mathrm{flat}}.}
$$
At a horizontal interface, matching the tangential <wavevector> at fixed <frequency> gives the same invariant on both sides: this is <Snell's law>.
For radial <wave speed> $\alpha(r)$, take the <cross product> of position with <slowness>. The ray equations give
$$
\frac{d}{ds}(\mathbf x\times\mathbf y)
=\alpha\mathbf y\times\mathbf y+\mathbf x\times\nabla(1/\alpha)=0,
$$
since the second vector is radial. A nonzero constant $\mathbf L=\mathbf x\times\mathbf y$ is perpendicular to every position on the <elastic ray>, so the ray lies in a fixed plane through the centre. Radial rays obey the same conclusion with a nonunique plane. The <spherical elastic ray invariant> is therefore
$$
\boxed{p=|\mathbf L|=\frac{r\sin i}{\alpha(r)},}
$$
where $i$ is measured from the radial direction. Across a spherical interface the radius is common and tangential <slowness> is continuous, so this form of <Snell's law> survives an interface discontinuity.
In a homogeneous sphere a <P wave> follows the straight chord between the two surface points. If their principal angular separation is $0\leq\Delta\leq\pi$, its length is $2R\sin(\Delta/2)$. For constant <P wave> speed $c_P$,
$$
\boxed{T_D(\Delta)=\frac{2R}{c_P}\sin\frac\Delta2.}
$$
Now put $a=R'$ and denote the mantle and core <P wave> speeds by $c_m>c_c$. Equal incidence and reflection angles put the core-boundary reflection point on the angular bisector. Each mantle leg has length $L$ given by $L^2=R^2+a^2-2Ra\cos(\Delta/2)$. Hence the <core-reflected travel time> is
$$
\boxed{T_R(\Delta)=\frac{2}{c_m}\sqrt{R^2+a^2-2Ra\cos(\Delta/2)},\qquad
0\leq\Delta\leq\Delta_g=2\arccos(a/R).}
$$
The stated domain is essential. At the reflection point the incoming line approaches from the mantle only if $R\cos(\Delta/2)\geq a$. Otherwise it has already crossed the core boundary. The mantle-only direct chord also exists only for $\Delta\leq\Delta_g$, with $T_D=2R\sin(\Delta/2)/c_m$. In this interval,
$$
L^2-R^2\sin^2(\Delta/2)=[R\cos(\Delta/2)-a]^2\geq0,
$$
so the reflected arrival is later. Its intercept is $2(R-a)/c_m$, whereas the direct intercept is zero. At grazing incidence the branches meet at $T=2\sqrt{R^2-a^2}/c_m$, and both have slope $dT/d\Delta=a/c_m$. A continuation of the homogeneous-sphere direct formula beyond this endpoint is not a mantle-only arrival in the modified model.
The core-transmitted <elastic rays> consist of three straight legs. Their impact distances in mantle and core are $b_m=pc_m$ and $b_c=pc_c$, respectively. At core entry,
$$
\frac{\sin i_m}{c_m}=\frac{\sin i_c}{c_c}=\frac pa,
\qquad \sin i_R=\frac{pc_m}{R},\qquad 0\leq p\leq a/c_m.
$$
Since $c_c<c_m$, the <elastic ray> bends towards the normal on entry and away from it on exit. The mantle leg subtends $i_m-i_R$ at the centre, and the core chord subtends $\pi-2i_c$. Thus the total angular sweep along the transmitted path is $2(i_m-i_R)+\pi-2i_c$. This sweep need not equal the principal separation if it exceeds $\pi$; the diagram uses a path with sweep below $\pi$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/iii/paper-54-earth-rays.png]
{title=Direct and core-reflected travel-time branches and refraction through a slower spherical core}
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