IPhO 1986 · Theory problem 2

Seismic waves and the Earth's liquid core

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  1. Early this century a model of the earth was proposed in which it was assumed to be a sphere of radius \(R\) consisting of a homogeneous isotropic solid mantle down to radius \(R_c\). The core region within radius \(R_c\) contained a liquid.
Figure 1

Figure 2.1

The velocities of longitudinal and transverse seismic waves P and S waves respectively, are constant, \(V_P\), and \(V_S\) within the mantle. In the core, longitudinal waves have a constant velocity \(V_{CP}, < V_P\), and transverse waves are not propagated.

An earthquake at E on the surface of the Earth produces seismic waves that travel through the Earth and are observed by a surface observer who can set up his seismometer at any point X on the Earth’s surface. The angular separation between E and X, \(2\theta\) given by

\[ 2\theta=\text{Angle }EOX \]

where O is the centre of the Earth.

(i) Show that the seismic waves that travel through the mantle in a straight line will arrive at X at a time \(t\) (the travel time after the earthquake), is given by

\[ t=\frac{2R\sin\theta}{v}, \qquad \text{for }\theta>\arccos\left[\frac{R_c}{R}\right], \]

where \(v=v_P\) for the P waves and \(v=v_S\) for the S waves.

(ii) For some of the positions of X such that the seismic P waves arrive to the observer after two refractions at the mantle-core interface. Draw the path of such a seismic P wave. Obtain a relation between \(\theta\) and \(i\), the angle of incidence of the seismic P wave at the mantle-core interface, for P waves.

(iii) Using the data

\(R\) \(=\) \(6370\ \mathrm{km}\)
\(R_C\) \(=\) \(3470\ \mathrm{km}\)
\(v_{CP}\) \(=\) \(10.85\ \mathrm{km\,s^{-1}}\)
\(v_S\) \(=\) \(6.31\ \mathrm{km\,s^{-1}}\)
\(v_{CP}\) \(=\) \(9.02\ \mathrm{km\,s^{-1}}\)

and the result obtained in (ii),draw a graph of \(\theta\) against \(i\). Comment on the physical consequences of the form of this graph for observers stationed at different points on the Earth's surface. Sketch the variation of the travel time taken by the P and S waves as a function of \(\theta\) for \(0\leq\theta\leq90\) degrees.

(iv) After an earthquake an observer measures the time delay between the arrival of the S wave, following the P wave, as 2 minutes 11 seconds. Deduce the angular separation of the earthquake from the observer using the data given in Section (iii).

(v) The observer in the previous measurement notices that some time after the arrival of the P and S waves there are two further recordings on the seismometer separated by a time interval of 6 minutes 37 seconds. Explain this result and verify that it is indeed associated with the angular separation determined in the previous section.

Official English text of the problem.

Answers

  • (i) The direct-wave travel time is \[ \boxed{t=\frac{2R\sin\theta}{v}}, \qquad \boxed{0\leq\theta\leq\arccos(R_c/R)}. \] Equality describes the limiting tangent ray. The printed inequality is reversed.

  • (ii) Put \(a=R_c/R\), \(b=v_{CP}/v_P\). The ray follows three straight segments, refracting towards the normal on entering the core and away from it on leaving. Its half angular sweep is \[ \boxed{\Phi(i)=\frac{\pi}{2}+i-\arcsin(a\sin i)-\arcsin(b\sin i)}. \] Ordinarily \(\theta=\Phi\); allowing rays that emerge beyond the antipode, the smaller angular separation is given by \[ \boxed{\theta=\min\{\Phi,\pi-\Phi\}}. \]

  • (iii) Direct P and S rays end at \(\theta=56.99^\circ\). Core-transmitted P rays first appear at \[ \boxed{\theta_{\min}=75.58^\circ,\qquad i_{\min}=55.02^\circ}. \] Thus the P-wave shadow is approximately \[ \boxed{113.99^\circ<2\theta<151.15^\circ}, \] while S waves have no direct or core-transmitted arrival for \(2\theta>113.99^\circ\). The refracted P travel-time branches meet at \(20.789\ \mathrm{min}\); their antipodal arrival times are \(21.733\) and \(23.397\ \mathrm{min}\). The complete graph descriptions are given below.

  • (iv) The earthquake–observer angular separation is \[ \boxed{2\theta\approx17.8^\circ}. \]

  • (v) The later signals are P and S waves reflected at the core boundary. Their calculated arrival times are \(551.3\ \mathrm{s}\) and \(947.9\ \mathrm{s}\) after the earthquake, giving \[ \boxed{\Delta t_{\mathrm{ref}}=396.65\ \mathrm{s} \approx6\ \mathrm{min}\ 37\ \mathrm{s}}. \]

Solution

We use the physically consistent reading of the numerical data: \[ v_P=10.85\ \mathrm{km\,s^{-1}},\qquad v_S=6.31\ \mathrm{km\,s^{-1}},\qquad v_{CP}=9.02\ \mathrm{km\,s^{-1}}. \] The data table in (iii) prints the label \(v_{CP}\) twice; the first value so labelled, \(10.85\ \mathrm{km\,s^{-1}}\), is \(v_P\). The inequality printed in (i) is also reversed; see part (i).

We use geometrical ray theory: rays are straight within each homogeneous region, refraction obeys Snell’s law, and reflection obeys equality of incidence and reflection angles. Diffraction, mode conversion and additional reflections at the Earth’s surface are neglected.

(i)

Let \(M\) be the midpoint of the chord \(EX\). The triangle \(OEM\) is right-angled, with \(\angle EOM=\theta\). Hence \[ EM=R\sin\theta,\qquad OM=R\cos\theta. \] The direct path length and travel time are therefore \[ EX=2R\sin\theta, \qquad \boxed{t=\frac{2R\sin\theta}{v}}. \]

The closest point of the chord to the centre is \(M\). Thus the ray avoids the core only if \[ R\cos\theta\geq R_c, \] or \[ \boxed{\theta\leq\arccos(R_c/R)}. \] Strict inequality gives a path entirely within the mantle; equality is the tangent limit. For larger \(\theta\), the chord intersects the core, so the mantle-only formula no longer describes the actual ray.

(ii)

Denote the core entry and exit points by \(A\) and \(B\). Draw the path \[ E\longrightarrow A\longrightarrow B\longrightarrow X, \] with \(EA\) and \(BX\) in the mantle and \(AB\) a chord of the core. Draw the radial normals \(OA\) and \(OB\).

Let \(r\) be the refracted angle inside the core, measured from the normal. Snell’s law gives \[ \frac{\sin i}{v_P}=\frac{\sin r}{v_{CP}}, \] so \[ r=\arcsin\left(\frac{v_{CP}}{v_P}\sin i\right). \] Since \(v_{CP}<v_P\), we have \(r<i\): the ray bends towards the normal on entering the core. By symmetry, its incidence angle at the exit is \(r\), and it emerges into the mantle at angle \(i\).

Let \(\alpha=\angle OEA\). In triangle \(EOA\), the angle at \(A\) is \(\pi-i\). The sine rule gives \[ \frac{\sin\alpha}{R_c}=\frac{\sin i}{R}, \] hence \[ \alpha=\arcsin\left(\frac{R_c}{R}\sin i\right), \qquad \angle EOA=i-\alpha. \]

The isosceles triangle \(AOB\) gives \[ \angle AOB=\pi-2r. \] The total angular sweep of the ray about \(O\) is consequently \[ 2\Phi=2(i-\alpha)+\pi-2r. \] Therefore \[ \boxed{ \Phi(i)=\frac{\pi}{2}+i -\arcsin\left(\frac{R_c}{R}\sin i\right) -\arcsin\left(\frac{v_{CP}}{v_P}\sin i\right) }. \]

For a sweep no greater than \(\pi\), \(\theta=\Phi\). If the ray emerges just beyond the antipode, its smaller source–observer angle is \(2\pi-2\Phi\), giving \[ \boxed{\theta=\min\{\Phi,\pi-\Phi\}}. \]

The requested drawing should therefore show a symmetric broken ray, with equal mantle legs, a straight core chord, and angles \(i\) and \(r<i\) marked at the entry boundary.

(iii)

The dimensionless ratios are \[ a=\frac{R_c}{R}=0.544741,\qquad b=\frac{v_{CP}}{v_P}=0.831336. \] Thus \[ \Phi(i)=\frac{\pi}{2}+i-\arcsin(a\sin i)-\arcsin(b\sin i). \]

Differentiating, with angles in radians, \[ \frac{d\Phi}{di} = 1-\frac{a\cos i}{\sqrt{1-a^2\sin^2 i}} -\frac{b\cos i}{\sqrt{1-b^2\sin^2 i}}. \] At the endpoints, \[ \Phi'(0)=1-a-b=-0.376077,\qquad \Phi'(\pi/2)=1. \] Each subtracted term decreases with \(i\), so \(\Phi'\) increases and has exactly one zero. Numerically, \[ \boxed{i_{\min}=55.017^\circ,\qquad \Phi_{\min}=\theta_{\min}=75.577^\circ}. \]

To obtain the corresponding travel-time graph, the length of either mantle leg is \[ \ell_m =\sqrt{R^2-R_c^2\sin^2 i}-R_c\cos i. \] This follows by projecting the outer and inner radii onto the mantle ray, whose perpendicular distance from \(O\) is \(R_c\sin i\).

The core chord has length \[ \ell_c=2R_c\cos r =2R_c\sqrt{1-b^2\sin^2 i}. \] Consequently, the core-transmitted P travel time is \[ \boxed{ T(i)= \frac{2}{v_P} \left(\sqrt{R^2-R_c^2\sin^2 i}-R_c\cos i\right) +\frac{2R_c}{v_{CP}}\sqrt{1-b^2\sin^2 i} }. \]

The following points specify both requested graphs:

\(i\) (degrees) Half sweep \(\Phi\) (degrees) Physical \(\theta\) (degrees) Core-transmitted P time (min)
0 90.00 90.00 21.733
20 82.74 82.74 21.504
40 77.20 77.20 21.003
55.02 75.58 75.58 20.789
70 77.84 77.84 21.171
80 82.60 82.60 22.028
89.23 90.00 90.00 23.397
90 90.76 89.24 23.538

The \(\Phi\)-versus-\(i\) curve begins at \(90^\circ\), falls smoothly to its minimum, then rises to \(90.7568^\circ\). When plotting the physical smaller angle \(\theta\), fold the last portion at \(90^\circ\): \(\theta\) reaches \(90^\circ\) at \(i=89.2322^\circ\), then decreases slightly to \(89.2432^\circ\) at grazing incidence.

The direct-wave cutoff is \[ \theta_d=\arccos(R_c/R)=56.993^\circ. \] The physical consequences are therefore:

  • For \(0\leq\theta<56.993^\circ\), direct P and S waves reach the observer.
  • For \(56.993^\circ<\theta<75.577^\circ\), neither direct nor core-transmitted P rays reach the observer. This is the P-wave shadow zone.
  • For \(\theta>56.993^\circ\), there is no direct S ray, and the liquid core supplies no transmitted S ray.
  • At \(\theta=75.577^\circ\), two refracted P paths merge. Since nearby incident rays emerge at nearly the same angular separation, ray energy is concentrated there: this is a caustic.
  • For most of \(75.577^\circ<\theta<90^\circ\), two refracted P paths give two distinct arrivals.

Using the given numbers exactly, there is a small additional feature near the antipode. For \[ 89.2432^\circ<\theta<90^\circ, \] a third refracted path reaches the observer by passing beyond the antipodal direction. At the antipode, the two oblique branches have equal travel times.

For the travel-time sketch, draw the direct branches \[ t_{P,\mathrm{dir}}=1174.19\sin\theta\ \mathrm{s}, \qquad t_{S,\mathrm{dir}}=2019.02\sin\theta\ \mathrm{s}, \] only over \(0\leq\theta\leq56.993^\circ\). Both start at the origin and increase with downward curvature; the S curve lies above the P curve. Their endpoints are \[ \begin{aligned} t_{P,\mathrm{dir}}(\theta_d)&=984.68\ \mathrm{s} =16.411\ \mathrm{min},\\ t_{S,\mathrm{dir}}(\theta_d)&=1693.16\ \mathrm{s} =28.219\ \mathrm{min}. \end{aligned} \]

The principal core-transmitted P branches begin together at \[ (\theta,T)=(75.577^\circ,20.789\ \mathrm{min}) \] and rise towards \(90^\circ\). They meet with a common tangent, since differentiation gives \[ \frac{dT}{di} =\frac{2R_c\sin i}{v_P}\frac{d\Phi}{di}, \qquad \frac{dT}{d\Phi}=\frac{2R_c\sin i}{v_P}. \] The smaller-\(i\) branch arrives earlier and flattens at the antipode, where the ray follows a diameter: \[ T(0)=\frac{2(R-R_c)}{v_P}+\frac{2R_c}{v_{CP}} =21.733\ \mathrm{min}. \] The other principal branch reaches \(23.397\ \mathrm{min}\) there. The small third branch decreases from \(23.538\ \mathrm{min}\) at \(\theta=89.243^\circ\) to \(23.397\ \mathrm{min}\) at \(90^\circ\).

For a sketch including the later reflected arrivals discussed in part (v), also draw \[ t_{v,\mathrm{ref}}(\theta) =\frac{2}{v}\sqrt{R^2+R_c^2-2RR_c\cos\theta}, \qquad 0\leq\theta\leq\theta_d. \] These curves increase from \(8.909\ \mathrm{min}\) for P and \(15.320\ \mathrm{min}\) for S at \(\theta=0\), and meet their respective direct curves tangentially at \(\theta_d\).

(iv)

For direct P and S waves the path length is the same, so their arrival-time difference is \[ \Delta t =2R\sin\theta \left(\frac{1}{v_S}-\frac{1}{v_P}\right). \] Here \[ \Delta t=2(60)+11=131\ \mathrm{s}. \] Therefore \[ \sin\theta = \frac{131} {2(6370)\left(\frac1{6.31}-\frac1{10.85}\right)} = \frac{131}{844.8239} =0.1551. \] Thus \[ \theta=8.9204^\circ, \qquad \boxed{2\theta=17.8408^\circ\approx17.8^\circ}. \] Since \(\theta<\theta_d\), the mantle-only path assumption is consistent.

The individual arrival times are \[ t_{P,\mathrm{dir}}=182.07\ \mathrm{s}, \qquad t_{S,\mathrm{dir}}=313.07\ \mathrm{s}. \]

(v)

The core boundary can reflect both P- and S-wave energy. These reflected waves take longer paths through the mantle and therefore produce later arrivals.

Let \(C\) be the reflection point on the core boundary, on the nearer radial bisector of \(\angle EOX\). The path is \[ E\longrightarrow C\longrightarrow X. \] Reflection symmetry about \(OC\) makes the incidence and reflection angles equal, and \(EC=CX\). Since \(\angle EOC=\theta\), the cosine rule gives \[ EC^2=R^2+R_c^2-2RR_c\cos\theta. \] Thus the reflected path length is \[ L_{\mathrm{ref}} =2\sqrt{R^2+R_c^2-2RR_c\cos\theta}. \]

For \(\theta=8.9204^\circ\), \[ L_{\mathrm{ref}}=5981.54\ \mathrm{km}. \] The reflected arrival times are consequently \[ t_{P,\mathrm{ref}} =\frac{5981.54}{10.85} =551.29\ \mathrm{s}, \] \[ t_{S,\mathrm{ref}} =\frac{5981.54}{6.31} =947.95\ \mathrm{s}. \] Both occur after the direct S arrival at \(313.07\ \mathrm{s}\), as required.

Their separation is \[ \begin{aligned} \Delta t_{\mathrm{ref}} &=L_{\mathrm{ref}} \left(\frac1{v_S}-\frac1{v_P}\right)\\ &=396.65\ \mathrm{s}\\ &=6\ \mathrm{min}\ 36.65\ \mathrm{s}. \end{aligned} \] Rounded to the nearest second, \[ \boxed{\Delta t_{\mathrm{ref}}=6\ \mathrm{min}\ 37\ \mathrm{s}}, \] which verifies the interpretation and the angular separation found in part (iv).

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.