• Author / Uploaded
• batpatman

Citation preview

8 July 2009 Physics of the Earth, F D Stacey and P M Davis Solutions to problems in Appendix J 1.1

(a) Consider a mass element dm at (x,y,z) in a thin spherical shell of radius r, centred at the coordinate origin, so that r2 = (x2 + y2 + z2). The moments of inertia of dm about the x, y and z axes are dI x = ( y 2 + z 2 )dm; dI y = ( x 2 + z 2 )dm; dI z = ( x 2 + y 2 )dm; 2 2 2 2 so that dI x + dI y + dI z = 2( x + y + z )dm = 2 r dm The same relationship applies to all mass elements and therefore to 2 their total, I x + I y + I z = 2mr But by symmetry I x = I y = I z so that for a spherical shell

I = (2 / 3)mr 2 (b) Now let this be an element of a uniform sphere, so that dm = 4 πr 2ρdr and 2 8π R 8π I = ∫ r 2dm = ρ ∫ r 4dr = ρR 5 . 3 3 0 15 3 Substituting M = ( 4 3) πR ρ we have the result for a uniform

sphere:

(c)

I = ( 2 5 ) MR 2

Let ρ = ρ0 ( R / r ), where ρ0 is the surface density (at R) R

R

M = ∫ 4 πr ρdr = 4 π ∫ r 2ρ0 ( R / r )dr 2

0

0

= 2πR ρ0 3

R

2 I = ∫ r 2 dm 3 0 where dm = 4πr2 ρdr = 4πRρ0rdr is the mass of a spherical shell of radius r. R

8 2 I = πRρ0 ∫ r 3dr = πρ0 R 5 3 3 0

I /MR2 = 1/3, compared with 0.3307 for the Earth.

1

1.2

(a) The total mass and moment of inertia are obtained as a sum of the values for a sphere of radius R and density ρ and a superimposed sphere of radius R/2 and density (f - 1) ρ 3

π 4 4 ⎛R⎞ M = πR 3ρ + π ⎜ ⎟ ( f − 1) ρ = R 3ρ ( 7 + f ) 3 3 ⎝2⎠ 6 3 ⎤ ⎛ R ⎞2 π 5 2 ⎡4 2 ⎡4 ⎛ R ⎞ ⎤ I = ⎢ πR 3ρ ⎥ R 2 + ⎢ π ⎜ ⎟ ( f − 1) ρ ⎥ ⎜ ⎟ = R ρ ( 31 + f ) 5 ⎣3 5 ⎢⎣ 3 ⎝ 2 ⎠ ⎦ ⎥⎦ ⎝ 2 ⎠ 60 ( π 60 )( 31 + f ) = 31 + f I = 2 MR ( π 6 )( 7 + f ) 10 ( 7 + f ) 2 For I MR = 0.3307, f = 3.403 (Earth) 0.365 2.06 (Mars) 0.391 1.25 (Moon) (b) Let the core have radius kR; then 4 4 4 M = πR 3ρ + π(kR)3 × 2ρ = πR 3ρ (1 + 2k 3 ) 3 3 3 8π 5 8π 8π 5 5 I= R ρ + ( kR ) × 2ρ = R ρ (1 + 2k 5 ) 15 15 15 5 2 (1 + 2k ) I = MR 2 5 (1 + 2k 3 )

(Earth) For I MR 2 =0.3307, k = 0.5480 0.365 0.385 (Mars) 0.391 0.230 (Moon) The algebra admits alternative, very high, values of k that are incompatible with plausible densities. 1.3

As a simple approximation, assume each layer to have uniform density, equal to the average of values at the top and bottom of the layer. Then, starting from the outside we can obtain the increments in mean density for successive layers. This allows us to use the method of solving Problem 1.2, by taking the sum of moments of inertia for a series of uniform spheres with densities Δρ , as listed r rS ρ ρ Δρ 0 0.04 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.95 1.0

160000 141000 89000 41000 13300 3600 1000 350 80 18 2 0.4 0

150500 115000 65000 27150 8450 2300 675 215 49 10 1.2 0.2

35500 50000 37850 18700 6150 1625 460 166 39 8.8 1.0 0.2

2

5 8π 5 5 5 rS ⎡⎣(1.0 ) × 0.2 + ( 0.95) × 1.0 + ( 0.9 ) × 8.8 15 5 +.... + ( 0.04 ) × 35500 ⎤ ⎦ 5 8π = ( 6.96 ×108 ) × 254 = 7.0 ×1046 kg m 2 15 The mass of the model is also an overestimate 4 3 3 3 M = πrS3 ⎡ (1.0 ) × 0.2 + ( 0.95) × 1.0 + .......... + ( 0.04 ) × 35500⎤ ⎣ ⎦ 3 4 8 3 30 = π ( 6.96 ×10 ) ×1640 = 2.3 ×10 kg 3 Thus, if we reduce the moment of inertia estimate by the factor needed to bring the mass of the model into line with the observed mass, we have I sun ≈ 6.0 ×1046 kg m 2 For a uniform sphere of the same mass and radius I 0 = 0.4 MrS 2 = 3.85 × 1047 kg m 2

Thus I =

I 5.7 ×1046 = = 0.148 Using Allen’s value, I 0 3.85 ×1047 1.4

The Doppler shift is seen by a stationary observer. To an observer rotating with the Sun the radiation would not appear Doppler shifted. Thus the radiation exerts no retarding torque on the Sun. The angular momentum lost by radiation of energy dE is the angular momentum of a spherical shell of radius R and mass dE/c2 (c is the speed of light). Thus the rate of loss of angular momentum is 2 daS dt = R 2 dE dt . ω c 2 3 2 dE dt = 4πrE S = 3.825 ×1026 W , where rE is the radius of the −2 Earth’s orbit and S = 1360W m is the solar constant. Thus daS / dt = 3.94 ×1021 kg m 2s −2 compared with the present angular momentum aS = I ω = 5.7 ×1046 × 2.87 ×10−6 = 1.6 ×1041 kg m 2s −1 1 daS = 2.4 × 10−20 s −1 = 7.6 × 10−13 /year aS dt This would not be a significant effect, even in the life-time of the Sun. Corresponding slowing of rotation depends upon the radial movement of mass to compensate for the mass lost from the surface. If the radius of gyration, k = I M , is maintained constant then the moment of inertia is proportional to the mass and its rate of change is given by 1 dI 1 dM 1 dE . = = = 2.14 × 10−21 s −1 2 I dt M dt Mc dt 2 and since aS = Mk ω , 3

1 dω 1 das 1 dM = − = 2.2 × 10−20 s −1 = 6.9 × 10−13 year −1 ω dt as dt M dt 1.5

The cross-sectional area intercepting radiation is πr , so energy 2 2 absorbed is πr S . Area radiating is 4πr , so energy lost is 4πr 2σT 4 . Equating these 1 T = ( S 4σ ) 4 = 278K (5o C)

1.6

In this case the same area (A) receives and radiates energy so 1 AS = AσT 4 ; T = ( S σ ) 4 = 393K (120o C)

1.7

Power/unit area at the solar surface is P = S (rEarth orbit / RSun ) 2 = σTS 4

2

1

2 TS = ⎡ S ( rE / RS ) / σ ⎤ = 5770K ⎣ ⎦

1.8

4

Power/unit area received by planet ∝ ( rorbit )

−2

Power radiated per unit area ∝ T 4 Therefore T ∝ 1 rorbit 1.9

Luminosities of the planets are proportional to the squares of their radii and inversely proportional to squares of distances from the Sun. Both quantities are listed, relative to Earth values, in Table 1.1. The order is Planet Jupiter Venus Earth Mercury Saturn Mars Moon Uranus Neptune Pluto

Luminosity, relative to Earth, ( Rplanet Rorbit )

2

4.45 1.726 1.000 0.979 0.918 0.122 0.074 0.043 0.016 2 × 10−5

1.10 If we assume the flux of dust particles towards the Sun to be in equilibrium with their production in the asteroidal belt , then they have the same size distribution. However, the radial speed of their motion towards the Sun, obtained by differentiating Eq. (1.25), is −1 −1 proportional to d and therefore to m 3 , where d is particle diameter and m is mass. The number density in space is the ratio flux/speed, and is therefore −1 dN ′ ∝ m − n dm m 3 = m −( n −1 3)dm

4

The Earth samples this distribution. 1.11 On free fall from infinite distance the meteoroids gain kinetic energy equal to the gravitational potential energy lost, so the speed at the Earth is v, given by 1 2 GM Sun GM Earth v = + 2 rorbit rEarth 2 This is compounded with the Earth’s orbital speed, ( GM Sun rorbit ) which is perpendicular to the free fall into the Sun. Since the Earth’s gravity gives the minor component of v above and so deflection by the Earth is small, we can add the orbital velocity in quadrature also. Thus 1

1

vTotal

⎛ 3GM Sun 2GM Earth ⎞ −1 =⎜ + ⎟ = 52787m s rEarth ⎠ ⎝ rorbit 2

1

⎛ 2GM Earth ⎞ −1 ⎟ = 10883m s Escape speed from the Earth is vescape = ⎜ r ⎝ Earth ⎠ which is about 20% of the speed of escape from the solar system when the Sun’s gravity is also included.

2.1

2.2

2

m + 0.1mchondrites Core mass = 0.326 = irons Earth mass mirons + mchondrites whence mirons = 0.251 mTotal Let r = core radius / total radius Then, ρ = r 3ρcore + (1 − r 3 ) ρmantle = 5280kg m −3 Assuming uncompressed densities (low pressure phases), ρcore = 7800kg m −3 , ρmantle = 3300kg m −3 , r = ⎡⎣( ρ − ρm ) / ( ρc − ρm ) ⎤⎦ 3 = 0.76 1

2.3

The simple supposition that the isotopes with the smallest (or most negative) mass excesses would be the most abundant is not generally correct. The table gives some clues to the reason for this. First, we note that 1H and 3He are the only isotopes with fewer neutrons than protons. This is true for the whole periodic table and not just for the elements selected in Table 2.1. Generally the proportion of neutrons increases with atomic number and 40Ca is the heaviest isotope for which the number of protons is even equal to the number of neutrons. The reason is that the mutual electrostatic repulsion of the protons reduces the binding energy, so that the total binding energy is increased by diluting the protons with neutrons. This effect increases with atomic number and all the heavy elements are neutron-rich. As noted in Section 2.1, isotopes with atomic masses that are multiples of 4 are favoured and, of those in the table, all except 56Fe are multiples of 4He, reflecting the strong binding of the 4He structure. In comparing these nuclei with one another we have a valid indication of the general 5

variation of binding energy with nuclear size, but we need caution in comparing them with the less abundant isotopes of the same elements, all of which have extra neutrons. The neutron mass (1.008665u, where u is the atomic mass unit, as listed in Table A1) exceeds the sum of the proton and electron masses (1.007276u), so that the added neutrons increase the nuclear mass excess without any consideration of nuclear binding. In spite of this, the binding energies apparent from the mass excesses appear to favour many of the heavier, more neutron-rich isotopes, reinforcing the argument that dilution of the neutrons reduces the repulsive energy. This does not very effectively answer the question, why is there not a more obvious correlation between binding energy and abundance? There is no simple answer. The relative abundances are consequences of multiple processes of nuclear synthesis, as illustrated by the fact that isotopic ratios in some of the pre-solar system grains found in carbonaceous chondrites are quite different from the bulk of the solar system (Section 4.5). Abundances of isotopes reflect the processes that produced them, with binding energy introducing only a statistical bias. 3.1 (a) If the density is ρ, then at radius r, accumulated mass m = ( 4π 3) r 3ρ , the gravitational potential energy released by adding a shell of mass dm = 4πr 2ρdr is Gmdm 16 2 2 4 dE = = π Gρ r dr r 3 Integrating from r = 0 to R and substituting M = ( 4π 3) R 3ρ , E = ( 3 / 5 ) GM 2 / R (b) See Eq. (4.1). 3.2

If the cosmic ray flux is φ, the rates of production of S,R are dS dt = φσ S P, dR dt = φσ R P , where P is the concentration of parents, from which they are produced. For uniform φ over exposure time t, S = φσ S Pt , but if t is long compared with the half-life of R, then an equilibrium concentration is established, such that dR dt = φσ R P − λR = 0 so R = φσ R P / λ

Thus S R = ( σ S σ R ) λt Eq.(1.8) follows and Eq.(1.9) assumes an effective production rate of S proportional to ( σ S + σ R ) .

3.3

−λt Given the number remaining at time t, N = N 0 e , the number decaying in the interval t to (t + dt ) is dN * = −dN = λN 0 e −λt dt The average life is therefore

6

t =

3.4

1 N0

∫

∞

t =0

(a) by Eq.(3.7),

∞

tdN * = λ ∫ te −λt dt = 0

d ( 40 Ar ) d ( K) 40

1 λ

= 0.105 ( eλ K t − 1) = 0.098 ± 0.005

9 whence tAr = (1.20 ± 0.05 ) × 10 years.

By Eq.(3.12),

d ( 87 Sr

d ( 87 Rb

86

Sr )

86

Sr )

(

)

= eλ Rbt − 1 = 0.0215 ± 0.0007

9 whence tSr = (1.50 ± 0.05 ) × 10 years.

By Eq.(3.15),

d ( 207 Pb d ( 206 Pb

204 204

Pb )

Pb )

=

U eλ235t − 1 = 0.090 ±0.005 238 U eλ238t − 1 235

tPb = (1.43 ± 0.10 ) ×109 years. whence (b) Disparities are well outside the formal errors of individual ‘ages’ and the Ar age is lowest, implying an argon loss which (with the small error) is consistent for all of the minerals used. This suggests a 9 reasonably complete argon loss at about 1.2 × 10 years rather than partial loss later. The Sr and Pb ages are not clearly different, but consistent with a slight U-Pb diffusion. In this case it must be noted that the error of the Pb age is greatest, implying variability of diffusion. We must therefore take the Sr age as the most reliable estimate of the formation age and treat the Ar age as evidence of a 9 metamorphic event at about 1.2 ×10 years, although argon diffusion without appreciable reheating cannot be ruled out.

3.5

As noted in Section 4.3, the Pb-Pb isochron gives an age τ = ( 4.54 ± 0.03) × 109 years for the meteorites. The Rb-Sr isochron

(

)

gives eλRb τ − 1 = 0.0664 (see Eq.(4.4). Thus λ Rb = 4.1

ln(1.0664) year −1 = (1.416 ± 0.009) ×10−11 year −1 9 (4.54 ± 0.03) × 10

The rate of production of 129 I during the interval τ was 127 I /τ where 127 I represents the total production of that isotope. The net rate of production of 129 I during this interval is therefore 127 d 129 I I = − λ 129 I ( ) dt τ so that

∫

( I ) τ d ( 129 I ) 129

127

τ

= ∫ λdt

I 129 − I λτ ⎛ 129 I ⎞ 1 −λτ which gives ⎜ 127 ⎟ = (1 − e ) ⎝ I ⎠ τ λτ 0

0

7

129

The Xe eventually produced in the meteorite corresponds to the 129 I remaining at time t after cessation of synthesis, i.e. 129 Xe ⎛ 129 I ⎞ e −λt = = (1 − e−λτ ) ⎜ ⎟ 129 I ⎝ 127 I ⎠ τ+t λτ Xe e −λt = (1 − 1 + λτ − ..) = e−λt 127 I λτ

129

−1 For τ 2 J2×Mα = 5.8×1014 kg, which corresponds to 580 km3 of water. This is much more than the capacity of any man-made lake and would represent a level change of about 1.5 m on the largest natural lake, the Caspian Sea, or 7 m on Lake Superior.

8.1

With respect to the Earth’s axis the second term in Eq.(8.1) has an asymmetry because r is measured from an axis through the combined centre of mass of the Earth and Moon. This asymmetry is cancelled −1 by that in the first term, as becomes apparent when ( R′ ) is substituted by Eq.(8.3), leaving the geocentric rotational term (the third term in Eq.8.8). Note that the cos ψ term formally disappears by the substitution of Eq.(8.5).

8.2

The Earth is deformed tidally to a prolate ellipsoidal form in response to potential W2 (Eq.8.9), ψ being the angle to the EarthMoon axis. The additional potential due to the Earth itself is then that arising from its ellipticity, as may be expressed by the second term of Eq.6.13, where we now put φ = (π / 2 − ψ ) , being the angle to the plane normal to the Earth-Moon axis, so that (C-A) is negative. Then k2 is defined as the ratio of the angular dependence (second term) in V to the excitation (W2), so that 3

C − A⎛ R ⎞ M C − A⎛ R ⎞ k2 = − ⎟ =− ⎜ ⎟ 2 ⎜ ma ⎝ r ⎠ m Ma 2 ⎝ r ⎠

3

17

The density is assumed to be uniform so that ( A − C ) Ma 2 is very simply related to the semi-axes, a and c (noting that c > a in the present situation), because ⎛ a2 + c2 ⎞ C = 0.4 Ma 2 and A = 0.4M ⎜ ⎟ ⎝ 2 ⎠ Then

⎛ c2 ⎞ A−C ⎤ ⎛c ⎞ ⎡ ⎛c ⎞ = 0.2 ⎜ 2 − 1⎟ ≈ 0.4 ⎜ − 1⎟ , ⎢for ⎜ − 1⎟ > h. cos θ does not vanish in this way for the infinite sphere. Note also an explanation in terms of Gauss’s theorem, as in electrostatics. The gravitational field lines of a thin isolated plane sheet are normal to the sheet and are equally distributed on either side of it, whereas a spherical shell exerts no gravitational field inside itself and all of the field lines from the 1

23

surface distribution of mass emerge from one side of it (the outside) making the field there twice as strong. 9.3 (a) In the approximation of a spherically symmetric Earth, surface gravity is g = GM/R2, independently of the radial density distribution, and M = (4π/3)R3 ρ . g is well measured, so if ρ is determined then G is fixed, or vice-versa. (b) At depth z (radius r = R − z) gravity is due only to the material inside r, mass m(r), that is g = Gm(r)/r2 The material between r and R is a spherical shell outside the point of observation and does not contribute. If the density of any layer is ρ then the gravity gradient within it is dg G dm 2Gm(r ) = − dr r 2 dr r3 G 2g 2g = 3 4πr 2ρ − = 4πGρ − r r r This is zero if 2πGρ = g/r = Gm(r)/r3 = G(4π/3) ρ r3/r3= (4π / 3)Gρ that is ρ = (2/3) ρ Note that the free air gradient just above the surface, where ρ→0, is −(8π/3)G ρ = −2g/R = −3.086×10-6 s-1 = −0.3086 mGal m-1 taken as negative here because g decreases with r. (c) If ρ = ρ /2, then we can rewrite the gradient as dg/dr = 2πG ρ − 2g/r = 2πG ρ − (8π/3)G ρ = −(2π/3)G ρ = 0.07715 mGal m-1, a quarter of the free air gradient. (d) If ρ = 1025 kg m-3, the density term in the gradient is 4πGρ = 8.60×10-7 s-2 = 0.0860 mGal m-1 and the gradient becomes 0.2226 mGal m-1 (e) The density of air at sea level is 1.23 kg m-3, so 4πGρ = 1×10-4 mGal m-1, 0.03% of the free air gradient. This is the difference between the free air and free vacuum gradients.

10.1 From Appendix D

K=

2μ(1 + ν) 3(1 − 2ν)

so that ⎤ 4 2μ(1 + ν) 4 2 ⎡ (1 + ν) K+ μ= + μ = μ⎢ + 2⎥ 3 3(1 − 2ν) 3 3 ⎣ (1 − 2ν) ⎦ 1 − ν ⎛ ⎞ = 2μ. ⎜ 1 − 2ν ⎟ ⎝ ⎠ 2

Thus

K + 43 μ ⎛ VP ⎞ 2(1 − ν) =⎜ ⎟ = μ ⎝ VS ⎠ (1 − 2ν)

24

⎡⎛ V ⎞ 2 ⎤ ⎡⎛ V ⎞ 2 ⎤ P P From which ν = ⎢⎜ ⎟ − 2 ⎥ / 2 ⎢⎜ ⎟ − 1⎥ V V ⎢⎣⎝ S ⎠ ⎥⎦ ⎢⎣⎝ S ⎠ ⎥⎦

10.2 The simpler case is the calculation of V⊥ , for which we need the modulus χ for the combined layers, for compression perpendicular to the layers. If stress σ is applied, with lateral constraint preventing dimensional changes in the plane of the layers, then the strains are σ χ1 and σ χ 2 . The strain for the 1 combined layers is 2 σ (1 χ1 + 1 χ 2 ) = σ χ so that χ = 2χ1χ 2 / ( χ1 + χ 2 )

and therefore 1

2 2 ⎛ χ ⎞ 2 ⎡ 2χ1χ 2 ⎤ ⎡ 2 ( χ1 ρ ) . ( χ 2 ρ ) ⎤ V⊥ = ⎜ ⎟ = ⎢ ⎥ =⎢ ⎥ ⎝ ρ ⎠ ⎣ ρ ( χ1 + χ 2 ) ⎦ ⎣ χ1 ρ + χ 2 ρ ⎦ 2V1V2 = 1 2 (V1 + V22 ) 2 1

1

To calculate V|| , consider Eqs.(10.4) for each material. (x-y) is the plane of layering, with propagation in the x direction and z perpendicular to the layers. There are six equations relating the strain components ε x1 , ε y1 , ε z1 , ε x 2 , ε y 2 , ε z 2 , to the normal stresses σ x1 , σ y1 , σ z1 , σ x 2 , σ y 2 , and σ z 2 . The following conditions simplify the equations: 1. Strains must be the same in both media in the plane of layering, so ε x1 = ε x 2 = ε x and ε y1 = ε y 2 . 2. The total strain in y and z directions is zero, so ε y1 = ε y 2 = 0 and ε z1 + ε z 2 = 0 . 3. In the z direction stresses are equal in the two media, σ z1 = σ z 2 = σ z With these constraints the six equations are ε x = ⎡⎣σ x1 − ν ( σ y1 + σ z ) ⎤⎦ / E1 0 = ⎡⎣ σ y1 − ν ( σ x1 + σ z ) ⎤⎦ / E1

ε z1 = ⎡⎣σ z − ν ( σ x1 + σ y1 ) ⎤⎦ / E1

ε x = ⎡⎣σ x 2 − ν ( σ y 2 + σ z ) ⎤⎦ / E2 0 = ⎡⎣σ y 2 − ν ( σ x 2 + σ z ) ⎤⎦ / E2

−ε z1 = ⎡⎣σ z − ν ( σ x 2 + σ y 2 ) ⎤⎦ / E2

(1) (2) (3) (4) (5) (6)

From these equations we must calculate the modulus for strain in the x direction, that is χ|| = σ x ε x = ( σ x1 + σ x 2 ) / 2ε x .

25

First, use (2) and (5) to substitute for σ y1 and σ y 2 in the other four equations. Then add (3) and (6) to eliminate ε z1 . This gives ε x = ⎡⎣σ x1 (1 − ν 2 ) − σ z ( ν 2 + ν ) ⎤⎦ / E1 ε x = ⎡⎣σ x 2 (1 − ν 2 ) − σ z ( ν 2 + ν ) ⎤⎦ / E2 σ z = ⎡⎣ ν (1 − ν ) ⎤⎦ .[ E2σ x1 + E1σ x 2 ] / ( E1 + E2 )

(1’) (4’) (3’)+(6’)

Substituting for σ z in (1’) and (4’), we have ⎡ ν 2 (1 + ν) E2 σ x1 + E1σ x 2 ⎤ ε x = ⎢σ x1 (1 − ν 2 ) − . (1”) ⎥ / E1 E1 + E2 ⎦ (1 − ν) ⎣ ⎡ ν 2 (1 + ν) E2 σ x1 + E1σ x 2 ⎤ ε x = ⎢σ x 2 (1 − ν 2 ) − . (4”) ⎥ / E2 E1 + E2 ⎦ (1 − ν) ⎣ These are simultaneous equations that can be solved for σ x1 and σx2 .

At this point it is convenient to substitute χ1 and χ 2 for E1 and E2, using the relation (Appendix D) E = χ (1 + ν )(1 − 2ν ) (1 − ν ) Then (1”) and (4”) become ⎡ χ ⎤ 2 ε x (1 − 2ν )( χ1 + χ 2 ) = σ x1 ⎢(1 − ν ) + (1 − 2ν ) 2 ⎥ − ν 2 σ x 2 χ1 ⎦ ⎣ ⎡ χ1 ⎤ 2 2 = σ x 2 ⎢(1 − ν ) + (1 − 2ν ) ⎥ − ν σ x1 χ2 ⎦ ⎣

These give

χ|| =

σ x1 + σ x 2 2ε x

⎡ 2 ⎛ χ1 χ 2 ⎞ ⎤ 2 ⎢ 2ν + 2 (1 − ν ) + (1 − 2ν ) ⎜ + ⎟ ⎥ (1 − 2ν )( χ1 + χ 2 ) 1 ⎝ χ 2 χ1 ⎠ ⎦ = ⎣ 2 ⎡ χ1 ⎤ ⎡ χ2 ⎤ 4 2 2 ⎢(1 − ν ) + (1 − 2ν ) χ ⎥ . ⎢(1 − ν ) + (1 − 2ν ) χ ⎥ − ν 2⎦ ⎣ 1 ⎦ ⎣

which simplifies to ⎡ ⎤ ⎛ χ1 χ 2 ⎞ 2 ⎢(1 − 2ν ) ⎜ 2 + + ⎟ + 4ν ⎥ χ 2 χ1 ⎠ ( χ + χ2 ) . ⎣ ⎝ ⎦ χ|| = 1 2 ⎛ χ1 χ 2 ⎞ 2 (1 − ν ) ⎜2+ + ⎟ χ 2 χ1 ⎠ ⎝ ⎡ ⎛ χ1 χ 2 ⎞ ⎤ ⎢ ⎜2− − ⎟⎥ 2 χ 2 χ1 ⎠ ⎥ χ1 + χ 2 ⎢ ν 1+ .⎝ 2 = 2 ⎢ (1 − ν) ⎛ χ1 χ 2 ⎞ ⎥ ⎢ ⎜2+ + ⎟⎥ χ 2 χ1 ⎠ ⎥⎦ ⎝ ⎣⎢

26

χ + χ2 = 1 2

2 ⎡ ( χ1 − χ2 ) ⎤ ν2 1 − . ⎢ 2 ⎥ 2 ⎣⎢ (1 − ν) ( χ1 + χ 2 ) ⎦⎥

2 Substituting VP ρ for each χ 12

2 ⎧ 2 2 ⎡ VP12 − VP22 ) ⎤ ⎫⎪ ( ν2 ⎪VP1 + VP2 ⎢ ⎥⎬ α|| = ⎨ 1− . 2 2 2 2⎥ ⎢ 2 (1 − ν ) ⎪⎩ (VP1 + VP2 ) ⎦ ⎪⎭ ⎣ For slight anisotropy we can write it as 12 VP|| 1 ⎡ 2 ν2 2 2 2 2 2⎤ −1 = (VP1 + VP2 ) − (1 − ν)2 (VP1 − VP2 ) ⎥ − 1 2VP1VP2 ⎢⎣ VP ⊥ ⎦ For ν = 0.27, VP1 VP2 = 1.164 gives 1% anisotropy VP1 VP2 = 1.402 gives 5% anisotropy.

10.3 (a) By Eq.(10.23) ( π / QP ) T = 6s where T = 20.25min = 1215s is the trans-Earth travel time for a PKP wave (see Fig. 17.8). Thus QP = 636 . (b) If the core Q is assumed to be infinite, Eq.(10.23) still applies but QP is the mantle value and T is the travel time in the mantle only. This is equal to the PcP travel time at Δ = 0°, T = 8.5min. = 510s. Thus QP (mantle) = 267. (c) Taking QS ≈ QP / 2 (Eq.10.24), QS (mantle) ≈ 133 . (d) S waves do not penetrate the core. If we consider SKS waves with no attenuation in the core then Eq.(10.23) applies with T = 15.6 min = 936s for ScS waves at Δ = 0° and QS ≈ 133 as above, giving d ln A / df ≈ −22 10.4 If the total width of a hysteresis loop, measured on the strain axis, is Δε and the peak-to-peak stress is 2σ (amplitude σ of the stress cycle), then the area of an elliptical loop is ΔE = ( π 4 ) Δε.2σ = ( π 2 ) Δε.σ . The peak elastic energy, for strain ε is (1 2 ) ε.σ So that

ΔE ( π2 ) Δε.σ πΔε 2π = 1 = = E ε Q ( 2 ) ε.σ −8

Thus Δε = 2ε / Q = 1.0 ×10 To measure this to 10% requires a strain resolution of 10-9.

27

11.1 We determine the vertical strain to obtain displacement which will have a maximum value at the upper surface. Let z be measured downwards from the top of the block. Taking compression as negative, Eq. (11.34) gives the vertical stress as σ zz = −ρgz Because the forces at the sides are zero, σ xx = σ yy = 0 Hooke's law Eq.(11.1) gives σ zz = λ (exx + eyy + ezz ) + 2μezz The strains are related by Eq.(11.11) ezz = −νexx = −νeyy The vertical stress becomes σ zz = λ (1 − 2ν)ezz + 2μezz = (λ − 2λν + 2μ)ezz −ρgz = (λ − 2λν + 2μ)

∂w ∂z

Integrating ρg z2 w = A− (λ − 2λν + 2μ) 2 where A is the constant of integration. Applying boundary conditions w = 0, z = L; ρg L2 A= (λ − 2λν + 2μ) 2 ⎧ L2 z 2 ⎫ ρg ⎨ − ⎬ (λ − 2λν + 2μ) ⎩ 2 2 ⎭ if λ =μ w=

ρg L2 = 3 × 103 ×10 × 108 /(5 × 3 × 1010 ) (5μ) = 20 m Note that the shear stress implied by this deformation would exceed the yield stress of rock. A mountain with these dimensions would collapse. wmax =

11.2 For a block of material as described in problem 11.1 the vertical stress is σ zz = ρgz (now taken positive as we are considering friction). The normal stress, σ n , and shear stress, τ, on a plane dipping at angle

28

α are obtained from Eq.(11.26, assuming σ zz =σ 22 ) as σ n = ρgz cos 2 α τ=(1/2)σ 22 sin 2α = (1/ 2)ρgz sin 2α From Eq. (11.36) slipping occurs when τ ≥ S0 + μ f σ n (1/ 2)ρgz sin 2α ≥ S0 + μ f ρgz cos 2 α ρgz sin α cos α ≥ S0 + μ f ρgz cos 2 α

S0 (1) ρgz cos 2 α The angle is steeper at low elevations. Note that if we consider a mass m of area A, on a dipping plane and resolve forces rather than stresses, we obtain tan α ≥ μ f + S0 A / mg cos α tan α ≥ μ f +

Since m = ρAz cos α, we also obtain Eq.(1) tan α ≥ μ f + S0 /(ρgz cos 2 α)

11.3 (a) To find the constant term we equate the radial stress to the pressure at r=R, A σ rr (r = R) = 3 = P; R 3 A = PR , so PR 3 PR 3 PR 3 r r , σ ( ) = − , σ ( ) = − θθ φφ r3 2r 3 2r 3 To find the displacements we integrate the strain field, noting that σ rr (r ) =

29

σ rr + σφφ + σθθ = 0 ⇒ e = err + eφφ + eθθ = 0 Considering the x-direction, Hooke's law, Eq.(11.6) gives σ xx = λe + 2μ

P R3 ∂u ∂u = 2μ = σ rr (θ = 0, φ = 0) = 0 3 ∂x ∂x x

∂u 1 P0 R 3 = ∂x 2μ x 3 ux = A −

P0 R 3 , u x = 0, x → ∞, ⇒ A=0, 4μx 2

P0 R 3 4μx 2 From symmetry

ux = −

ur =

P0 R 3 , u φ = uθ = 0 4μr 2

(b) A uniform pressure at infinity corresponds to σ rr = σθθ = σφφ = P In order to satisfy the boundary condition of zero radial stress at the cavity we superpose the negative solution from part (a) with that of a uniform P throughout the medium PR 3 r3 R3 σθθ = P(1 + 3 ) 2r R3 σφφ = P (1 + 3 ) 2r σ rr = P −

11.4 Cracks in a hydrofracture experiment open against the least principal stress (compressions positive). The transition from horizontal to vertical fracture will occur when the least principal stress equals the overburden pressure. Above this depth we infer that the least principal stress is the overburden pressure because the cracks are horizontal.

30

σ1 is larger than overburden pressure at depths < 1 km σ3 =ρgz and σ 2 >σ3 . We use the condition that at the transition depth the least and intermediate principal stresses are equal to one another and to the overburden pressure to find σ 2 σ 2 = σ3 =ρgz=3 × 103 × 9.8 × 103 = 29.5 MPa Then at 2 km if the only change is the overburden pressure it then becomes the intermediate principal stress (σ 2 ). The principal stresses at 2km depth become, [σ1 , σ 2 , σ3 ] = [100,59, 29.5] MPa

11.5 (a) From Eq.(11.53) Pb = 3σ3 − σ1 =40 MPa; σ3 = Pc = 25 MPa σ1 = 3Pc − Pb = 35 MPa σ3 = ρgh=3 × 103 ×10 × 103 = 30MPa σ=[35,30,25] MPa, [North-South, Vertical, East-West] (b) Strike slip earthquakes with maximum and minimum principal stresses horizontal. (c) The angle the slip plane makes with σ1 is given by π 1 1 − θ = tan -1 (1/ μ f ) = tan -1 (1/ 0.6) = 29.50. 2 2 2 The dip angle will be 90 degrees and the strike angle 29.5 degrees Eq.(11.39) as δ = to north.

11.6

(a) The radial and tangential stresses for a cylindrical pressurized magma chamber are AP AP σ rr = 2 0 ; σθθ = − 2 0 ; σφφ = 0 r r The quilibrium equations equations are given by Eq.(11.49) as

∂σrr 1 ∂σ rθ σrr − σθθ + + =0 r ∂θ r ∂r ∂σrθ 1 ∂σθθ σ + + 2 rθ = 0 r ∂θ r ∂r since σ rθ = 0; ∂σrr σ rr − σθθ + ∂r r

AP0 AP0 + 2 2 2 AP0 r r =0 =− 3 + r r

31

(b) Since σ rr ( r = R ) = P0 =

AP0 R2

A = R2.

(c) To obtain displacement we integrate the strain field noting, P0 R 2 P0 R 2 + − ( )+0=0 r2 r2 so e = err + eθθ + eφφ = 0 σ rr + σθθ + σφφ =

σ rr = λe + 2μ

∂u ∂u = 2μ ∂r ∂r

∂u P0 R 2 = 2 r ∂r P R2 u = C − 0 , where C is a constant. As r → ∞, u = 0, so C = 0. 2μr 2μ

P0 R 2 2μr for P0 negative, u decays as 1/ r.

u=−

11.7 (a) We substitute in the equilibrium eq.(11.58) derivatives of σ zz =

2 Xz 2 y 2 Xy 3 2 Xzy 2 ; σ = ; σ = yy zy πr 4 πr 4 πr 4

i.e. ∂σ zz ∂σ zy 4 Xzy 8 Xz 3 y 4 Xzy 8 Xz 2 y 2 + = − + − ∂z ∂y πr 4 πr 6 πr 4 πr 6 8 Xzy 8 Xzy ( z 2 + y 2 ) − =0 πr 4 πr 6 ∂σ yy ∂σ zy 6 Xy 2 8 Xy 4 2 Xy 2 8 Xz 2 y 2 + = − + − ∂y ∂z πr 4 πr 6 πr 4 πr 6

=

8 Xy 2 8 Xy 2 ( z 2 + y 2 ) = − =0 πr 4 πr 6 (b) At the free surface tractions are generated by normal and shear stresses, 2 Xz 2 y 2 Xy 3 2 Xzy 2 σ zz = ; σ yy = ; σ zy = ; πr 4 πr 4 πr 4 At z = 0, σ zz = σ zy = 0, i.e., surface tractions Eq(11.19) are zero.

32

(c) To find the net force we integrate the surface tractions around a cylindrical surface surrounding the line of forces. Consider cylindrical coordinates with θ measured from y. 2 X cos θ sin 2 θ 2 X cos3 θ 2 X cos 2 θ sin θ ; σ yy = ; σ zy = πr πr πr Traction on any area is (Eq.11.18) σijn j , where n are direction cosines.

σ zz =

T2 = σ 23 n3 + σ22 n2 = σ yy n y + σ zy nz For an area rdθ on the edge of a cylinder direction cosines are [n z ,n y ]= [sinθ, cosθ], π/2

F=Ty =

∫ 2T rdθ y

0

π/2

=

∫ 2(σ

yy

n y + σzy nz ) rdθ

0

= = =

4X π

π/2

4X π

π/2

4X π

π/2

∫ (cos

4

θ + cos2 θ sin 2 θ)dθ

4

θ + cos2 θ(1 − cos2 θ))dθ

0

∫ (cos 0

∫ cos

2

θdθ

0

4 X θ sin 2θ [ + ] 2π 2 4 =X So F=X =

11.8

Stresses for a normal line load at the surface of a half space are given by 2 Xz 3 2 Xzy 2 2 Xz 2 y σ zz = ; σ = ; σ = yy xy πr 4 πr 4 πr 4 As in Problem 11.8, consider cylindrical coordinates with θ measured from z. 2 X cos3 θ 2 X cos2 θ sin θ 2 X cos2 θ sin θ ; σ yy = ; σ zy = πr πr πr Traction on any area is σijn j Eq.(11.18), where n are direction cosines. σ zz =

T3 = σ33n3 + σ32n2 = σzz nz + σzy n y For an area rdθ on the edge of a cylinder surrounding the line load, direction cosines are [n z ,n y ]= [cosθ, sinθ],

33

π/ 2

F = Tz =

∫

π/2

2Tz rd θ =

0

= = =

4X π

π/2

4X π

π/2

4X π

π/2

∫ 2(σ

n + σ zy n y )rdθ

zz z

0

∫ (cos θ+cos θsin θ)dθ 4

2

2

0

∫ (cos

4

θ + cos 2 θ(1 − cos 2 θ))dθ

0

∫ cos 0

2

θd θ =

4 X θ sin 2θ [ + ] π 2 4

=X So F=X

11.9 (a) Applying the first of Eqs. (10.4), we have εx = 0 and σz =0, so that σx = νσy. Substituting this in the second of Eqs. (10.4), with σz =0 as before, εy = σy(1 − ν2)/E. The required modulus is E* = σy / εy = E/(1 − ν2). (b) We first allow the strip to expand freely in width, with strain e0, and then apply line forces to its edges, such that the strain is reduced to the value, e, that would be obtained by heating it while welded to the half space. Then the stress in the strip is σyy = E*(e0 − e). This requires a force per unit length F = hE*(e0 − e). Equal and opposite forces are applied to the surface of the half space and, using the equation from Problem 11.7, at distance W from each of these forces the stress is σyy = 2F/πW, so that at the mid-point of the strip (at distance W from each of the two line forces), the half space stress is 4F/πW. The corresponding strain is e = (4F/πW)/ E*. Equating the forces, F, in the strip and the half space, (πW/4)e = h(e0 − e), so that 4h e0/(πW + 4h) ≈ (4h/πW)e0 The possibility that uneven diurnal (or annual) heating of surface rocks generates stresses that penetrate to kilometre depths can be considered as an explanation of the paradoxical observation of a diurnal variation in earthquake frequency in the vicinity of Kilauea volcano, Hawaii (Rydelek, Davis and Koyanangi, J. Geophys.Res., 93,B5, 4401-4411, 1988). A variation at tidal frequency is not seen and this means that, to be an effective earthquake trigger, thermal stress would need to exceed the tidal stress, ~ 25 kPa. We can imagine a surface strip of rock exposed to strong sunlight, with surroundings insulated by soil cover, to have a ‘free’ thermal strain e0 ≈ 2×10-4 (α = 10-5 K-1 (linear coefficient), ΔT = 20 K). Its effective thickness is h = √(2η/ω) ≈ 0.19 m, that is the skin depth for a thermal wave of diurnal frequency, ω ≈ 7.3×10-5 s-1, in rock of thermal diffusivity η = 1.3×10-6 m2s-1. The necessary width of the strip, W, is

34

the depth of seismic activity, ~ 3000 m, so that the deep strain is e ≈ (4h/πW)e0 = 1.6×10-8 and if E* ≈ 80 GPa, the thermal stress is 1.3 kPa, less than 10% of the tidal stress, making it an unpromising earthquake trigger. 12.1 Using subscripts I, P, H for the Indian plate, the Pacific plate and the Hawaii hot-spot, Eq.(12.1) is H ωI = H ωP + P ωI A straightforward way to add these vectors is to resolve them into components in directions x (through 0°N, 0°E), y (through 0°N, 90E) and z (90°N). Then for each vector ω with a pole at (φ, λ) the components are x = ω cos φ cos λ; y = ω cos φ sin λ; z = ω sin φ Corresponding components are added so that xHI = xHP + xPI , etc The magnitude of H ωI is 2 2 2 ωI = ( xHI + + yHI + zHI )

12

H

and its direction is given by sin φHI = zHI / H ωI tan λ HI = yHI / xHI with the convention that -90° 52 km. This thickness is well inside the boundary layer for transfer of core heat to the mantle and cannot imply a static boundary layer, but demonstrates the inevitability of mantle convection. 13.5 Eq. (13.13) gives plume radius as a ¼ power of viscosity. Thus, a factor 100 change in viscosity only changes the radius by a factor

41

(100)1/4 = 3.16. If the viscosity is 1020 Pa s, then the radius is 66 km or if viscosity is 1016 Pa s, the radius becomes 6.6 km.

13.6 We derive Eq.(13.20) σ zz − σ xx = (1/ 2)ρs ge using Eq.(13.25) D

( σ zz − σ xx ) = Δσ = e

1 zΔρ( z ) gdz D ∫0

1 1 Δσ = ∫ z (0 − ρ S ) gdz + zc 0 zc

zc

∫ z(ρ

f

− ρ S ) gdz

e

=−

1 1 gρ S e 2 + g (ρ f − ρ S )( zc 2 − e 2 ) 2 zc 2 zc

=−

1 1 1 1 1 gρ S e 2 + gρ f z c 2 − gρ f e 2 − gρ S z c 2 + gρ S e 2 2 zc 2 zc 2 zc 2 zc 2 zc

1 1 1 gρ f z c − gρ f e 2 − g ρ S z c 2 2 zc 2 1 1 = g (ρ f − ρ S ) zc − gρ f e 2 2 2 zc

=

(1)

Isostasy=> ρ S zc = ρ f ( zc − e); ρf =

ρ S zc ( zc − e )

(2)

Substitute (2) in (1) ρ z 1 ρ z 1 g S c e2 ( σ zz − σ xx ) = g ( S c − ρ S ) zc − 2 zc ( zc − e ) 2 ( zc − e ) ⎡ z 2 c − z 2 c + ezc − e 2 ⎤ 1 ⎡ z 2c ⎤ 1 zc ( zc − e ) e2 = gρ S ⎢ = gρ S ⎢ − − ⎥ ⎥ 2 ( zc − e ) ⎣ ⎦ 2 ⎣ ( zc − e ) ( zc − e ) ( zc − e ) ⎦ 1 = gρ S e 2

13.7 (a) Substitution in Eq.(13.43) with φ = atan(0.85), β = 6o , α = 2o , λ w = 0.7 , μ w = μb = 0.85 , λ b = 1 − {(α + β)[1 + (1 − λ w ) / ( csc φ sec(2α + 2β) − 1)] − β}/ μ b ,

gives λ b = 0.88 . (b) The effective friction for the low Andes is μb (1 − λ b ) = μ b (1 − 0.88) = 0.12μb . Double the friction for the high Andes means that μb (1 − λ b ) = 0.24μ b , λ b = 0.76. Substitution back into Eq.(13.43) rearranging terms and solving for λ w = 0.5 .

42

These values are consistent with the basal shear zone of the high Andes being drier λ b = 0.70 compared with λ b = 0.88 for the low Andes and the overlying wedge material also being drier, λ w = 0.70 compared with λ w = 0.50 , by a similar amount. 14.1 At distance r from the axis of the dislocation the strain energy 1 density is με2 = μS 2 / 8π2 r 2 . Thus the energy per unit length of an 2 annulus between r and (r+dr) is dE = (μS 2 / 8π 2 r 2 ) 2πrdr and so the total energy between radii r1 and r2 is E1,2 =

μS 2 4π

∫

r2

r1

dr μS 2 ⎛ r2 ⎞ = ln ⎜ ⎟ r 4π ⎝ r1 ⎠

Thus we cannot consider either r1 → 0 or r2 → ∞ . The difficulty with the lower limit arises from the fact that the equations refer to a simple dislocation, that is one in which there is a sudden change in displacement on the axis, and not a more realistic graded dislocation. r1 = 0 implies infinite stress and energy density on the axis. Thus r1 cannot be less than the distance at which the failure stress (με)max is reached. The upper limit problem is typical of the difficulties that arise when two parameters are allowed to become infinite. Thus by assuming that the dislocation is infinitely long, we mean longer than any other dimension including r. With this assumption we cannot also allow r to approach infinity. The strain falls off more rapidly than by Eq.(14.1) at distances comparable to the dislocation length. An approximate solution is therefore to put r2 equal to the (finite) length. 14.2 Eliminating MS from Eqs. (14.34) and (14.36) a log10 E = 1.5log10 + 4.8 + f ( Δ, h ) T Considering earthquakes all at the same depth and distance 1.5 E ∝ (a / T ) 2 But strain, ε ∝ a / T and E ∝ ε × τ whereτ is the duration of a wave passing a stationary point and is proportional to the total length of the wave train. Thus −0.5 τ ∝ (a / T ) This appears to say that, as the amplitude (or magnitude) increases, so the wavetrain becomes shorter and is clearly not correct. To see 0.3 why, consider the body wave case, for which τ ∝ ( a / T ) . This is

reasonable because the surface observations of a, T are representative of the spherical spreading wave-front. But for surface waves the energy is spread over different depths, depending on T. Although a/ T measures the surface strain in a surface wave and (a/

43

T)2 gives the energy density, we really need the integration of the energy flux with depth. Thus the energy density argument cannot be applied directly to surface wave energies. 14.3 We can write the displacement y in a propagating wave t x⎞ ⎛ as y = a sin ⎜ 2π − 2 π ⎟ λ⎠ ⎝ T λ = α T where ⎛ dy 2π t x ⎞ Strain ε = = − a. cos ⎜ 2π − 2π ⎟ dx VPT VPT ⎠ ⎝ T so that ε max =

2πa VPT

1 2 χε = 2 π2 χa 2 / VP 2T 2 2 This is the total wave energy/unit vol which oscillates between kinetic energy and strain energy. The volume through which wave 2 energy is distributed is 2πR ΔR where ΔR = ατ (assuming a surface earthquake that radiates into a hemisphere). Thus total energy E = 2πR 2VP τ.2 π 2χa 2 / VP 2T 2 s

Energy/unit vol =

But

χ = VP 2ρ , so

E = 4 π3VPρR 2a 2 τ / T 2 14.4 Differentiating Eq. (14.53) and substituting dv/dk in Eq.(16.49), we have −1/ 2 k ⎡g g ⎤ ⎡ g ⎤ u = v + ⎢ tanh(kh) ⎥ ⎢ − 2 tanh(kh) + h sec h 2 (kh) ⎥ k 2 ⎣k ⎦ ⎣ k ⎦ 1 = v + ⎡⎣ −v 2 + gh(1 − tanh 2 (kh) ⎤⎦ 2v v gh ⎡⎣1 − tanh 2 (kh) ⎤⎦ = + 2 2v This general expression can be written in alternative ways by substituting for one of the parameters by Eq. (14.53), but is convenient for demonstrating the two limits. If (kh) is very large, that is λ> h, as for long waves, such as tsunamis, then tanh2 (kh) → (kh) 2, which is negligible compared with unity and the second term reduces to gh/2v. But, by Eq. (14.53), in this case v → (gh)1/2, and the second term becomes v/2, so that u = v. 14.5 (a) Eq.(14.8) gives the displacement in the i-direction from a unit point force in the k-direction as

44

1 ⎡ δik 1 ∂ 2r ⎤ − 4 πμ ⎣⎢ r 4(1 − ν) ∂xi ∂xk ⎥⎦ Expanding this expression 1 ⎡1 1 1 ⎡1 1 1 1 x2 ⎤ ∂x ⎤ G11 = − = − + ⎢ ⎥ 4πμ ⎣⎢ r 4(1 − ν) ∂xr ⎦⎥ 4πμ ⎣ r 4(1 − ν) r 4(1 − ν) r 3 ⎦ Gik =

1 1 ⎡ (3 − 4ν) x 2 ⎤ = + 3⎥ 4πμ 4(1 − ν) ⎢⎣ r r ⎦ λ 2λ + 2μ 4λ 6λ + 6μ − 4λ (λ + 3μ) (3 − 4ν) = 3 − = = 2λ + 2μ 2λ + 2μ (λ + μ ) Recalling (Table D1) that ν =

1 1 ⎡ (λ + 3μ) x 2 ⎤ G11 = + 4πμ 4(1 − ν) ⎢⎣ r (λ + μ) r 3 ⎥⎦ 1 1 yx G12 = 4πμ 4(1 − ν) r 3 1 1 xz G13 = 4πμ 4(1 − ν) r 3 (b) The solution for displacement from an earthquake source is equivalent to that of a double couple as given by Eq.(14.10). For the ∂ first couple, note the derivative w.r.t. the source point, x’ , ∂x ' , is ∂ ∂ negative with respect to a field point, x, i.e., ∂x' = − ∂x . Using the results from part (a) ⎡ (λ + 3μ) x 2 ⎤ ⎡ y (λ + 3μ) 3 yx 2 ⎤ + 3 ⎥ ; G11,2 = B ⎢ 3 + 5 ⎥; G11 = B ⎢ r ⎦ ⎣ r (λ + μ) r ⎦ ⎣ r (λ + μ) yx x 3xy 2 G21 = B 3 ; G21,2 = − B 3 + B 5 r r r xz 3 yxz G31 = B 3 ; G31,2 = B 5 r r For the second couple yx y 3 yx 2 ; G = − B + B 12,1 r3 r3 r5 ⎡ (λ + 3μ) y 2 ⎤ ⎡ x (λ + 3μ) 3xy 2 ⎤ G22 = B ⎢ + 3 ⎥ ; G22,1 = B ⎢ 3 + 5 ⎥; r ⎦ ⎣ r (λ + μ) r ⎦ ⎣ r (λ + μ) G12 = B

G32 = B

xz 3 yxz ; G31,2 = B 5 3 r r

45

Adding individual couples to make a double couple located at z = c ux = G12,1 + G11,2 = − B

⎡ y (λ + 3μ) 3 yx 2 ⎤ y 3 yx 2 + B 5 + B⎢ 3 + 5 ⎥ 3 r r r ⎦ ⎣ r (λ + μ)

⎡ y y (λ + 3μ) 6 x 2 y ⎤ = B ⎢− 3 + 3 + 5 ⎥ r (λ + μ) r ⎦ ⎣ r Let x = 0, λ = μ y y 1 y ux = B 3 = B 2 = 2 3/ 2 2 r (y + c ) 12πμ ( y + c 2 )3/ 2

(c) To find the displacement from an earthquake source (Eq. 14.10), multiply the double couple displacements given in part (b) by the total moment M 0 = μbS to obtain μbS y bS y ux = = 2 2 3/ 2 2 12πμ ( y + c ) 12 π ( y + c 2 )3/ 2 (d ) The moment magnitude relation is given by Eq.(14.38) as M W = (2 / 3) log10 M 0 − 6.07 M 0 = 103/ 2( M W + 6.07) For an equidimensional fault, length is related to moment and stress drop though Eq.(15.23) 1/ 3

1/ 3

⎛ 103/ 2( M W + 6.07) ⎞ ⎛M ⎞ l =⎜ 0 ⎟ =⎜ ⎟ Δσ ⎝ Δσ ⎠ ⎝ ⎠ = 7.515 km. b= =

σ l μ

3 × 106 7515 = 0.751 m 3 ×1010

(e) x km -5 -4 -3 -2 -1 0 1 2 3 4 5

ux (metres) -0.0214 -0.0249 -0.0270 -0.0252 -0.0161 0 0.0161 0.0252 0.0270 0.0249 0.0214 46

14.6 It is convenient to break the integral in Eq.(14.27) into three regimes (Figures 1 and 2). We compare the pulse received from the end of the fault ( r = L / 2 ) with that received from the beginning of the fault ( r = − L / 2 ). For rupture towards the end of the fault, the integral in Eq.(14.21) becomes L/2 M0 P(t ) = I [t − r / V − ( L / 2 + y ) / VR + ycosψ / V )]dy LT − L∫/ 2

L/VR+T VR

L/VR

t l1

t’ l2

L/VS+T

VS

L/VS T

y

0

L

Figure 1. Time space development of the slipping Haskell patch. The thick lines denote the start (lower) and end (upper) of the region that is slipping. The thin lines describe the propagation of the radiation. Points l1 and l2 denote the beginning and end of the slipping patch. Radiation from all points along the diagonal line between l1 and l2 arrive simultaneously at time t. At the beginning of rupture l1 increases at the rupture speed while l2 remains at zero until the patch grows to its final size. Similarly at the end of the rupture, when l1 reaches the end, l2 continues until the patch size decreases to zero.

Referring to Figure (1) for a pulse received from the end of the fault,

M P(t ) = 0 LT

L/2

∫ I [t − r / V − ( L / 2 + y ) / V

R

+ ycosψ / V )]dy

−L/ 2

M0 (l1 (t ) − l2 (t )) LT For the starting phase, L/V < t < L / V + T

P(t ) = M0 LT

L/2

∫ I [t − L / V

R

+ y (1/ V − 1/ VR )]dy

−L/ 2

47

t ' = l1 / VR = t - L / V + l1 / V ; l1 = P (t ) = =

t − L /V 1/ VR − 1/ V

M 0 t − L /V T L / VR − L / V

M 0 t − L /V T X For the intermediate phase, L / VS + T < t < L / VR

t ' = T + l2 / VR = t − L / V + l2 / V ; l2 = l1 − l2 =

t − L /V −T 1/ VR − 1/ V

T 1/ VR − 1/ V

M0 M T = 0 T L / VR − L / V X For the ending phase, L / VR < t < T + L / VR P (t )=

l1 = L l1 − l2 = =

l2 =

t − L /V − T 1/ VR − 1/ V

L / VR − L / V − t + L / V + T 1/ VR − 1/ V

L / VR − t + T 1/ VR − 1/ V

P (t ) =

M 0 L / VR − t + T M 0 L / VR − t + T = T L / VR − L / V T X

For a pulse received from the beginning of the fault For the starting phase, 0 < t < T t ' = t − l1 / V = l1 / VR , l2 = 0 l1 =

t 1/ VR + 1/ V

For the intermediate phase, T < t