User:Ageary/Theory of seismic waves

Problems in Exploration Seismology and their Solutions

Series	Geophysical References Series
Title	Problems in Exploration Seismology and their Solutions
Author	Lloyd P. Geldart and Robert E. Sheriff
DOI	http://dx.doi.org/10.1190/1.9781560801733
ISBN	ISBN 9781560801153
Store	SEG Online Store

2.1 The basic elastic constants

2.1a Show that when the only nonzero applied stress is ${\sigma }_{ii}$, Hooke’s law requires that the normal strains ${\varepsilon }_{yy}={\varepsilon }_{zz}$, and that Poisson’s ratio ${\sigma }$, defined as ${\sigma }=-{\varepsilon }_{yy}/{\varepsilon }_{xx}=-{\varepsilon }_{zz}/{\varepsilon }_{xx}$, satisfies the equation

${\begin{aligned}{\sigma ={\frac {\lambda }{{2}\left({\lambda }+{\mu }\right)}}}.\end{aligned}}$

(2.1a)

Background

Stress is force/unit area and is denoted by ${\mathrm {\sigma } }_{xy}$, etc., where a force in the $x$-direction acts upon a surface perpendicular to the $y$-axis. The stresses ${\mathrm {\sigma } }_{xx}$ and ${\mathrm {\sigma } }_{xy}$ are, respectively, a normal stress and a shearing stress.

Stresses produce strains (changes in size and/or shape). If the stresses cause a point $P\left(x,\;y,\;z\right)$ to have displacements $\left(u,\;v,\;w\right)$ along the coordinate axes, the basic strains are given by derivatives of these displacements as follows:

${\begin{aligned}{Normal}\ {strains}:\ \,\ {\mathrm {\varepsilon } }_{xx}={\frac {{\mathrm {\partial } }u}{{\mathrm {\partial } }x}},{\quad }{\mathrm {\varepsilon } }_{yy}={\frac {{\mathrm {\partial } }v}{{\mathrm {\partial } }y}},{\quad }{\mathrm {\varepsilon } }_{zz}={\frac {{\mathrm {\partial } }w}{{\mathrm {\partial } }z}}.\end{aligned}}$

(2.1b)

${\begin{aligned}{Shearing}\ {strains}:\ {\mathrm {\varepsilon } }_{xy}={\mathrm {\varepsilon } }_{yx}={\frac {{\mathrm {\partial } }v}{{\mathrm {\partial } }x}}+{\frac {{\mathrm {\partial } }u}{{\mathrm {\partial } }y}},{\quad }{\hbox{and so on for }}{\mathrm {\varepsilon } }_{yz}{\hbox{ and }}{\mathrm {\varepsilon } }_{zx}.\end{aligned}}$

(2.1c)

The vector displacement ${\mathrm {\zeta } }$ is

${\begin{aligned}{\mathrm {\zeta } }=ui+vj+wk,\end{aligned}}$

(2.1d)

where ${\textbf {i}},{\textbf {j}},{\textbf {k}}$ are unit vectors in the ${\textit {x}}-,{\textit {y}}-,{\textit {z}}$-directions (see Sheriff and Geldart, 1995, problem 15.3). The dilatation $\Delta$ is the change in volume per unit volume, i.e.,

${\begin{aligned}\Delta \approx \left(1+{\mathrm {\varepsilon } }_{xx}\right)\left(1+{\mathrm {\varepsilon } }_{yy}\right)\left(1+{\mathrm {\varepsilon } }_{zz}\right)-1\approx {\mathrm {\varepsilon } }_{xx}+{\mathrm {\varepsilon } }_{yy}+{\mathrm {\varepsilon } }_{zz}={\nabla }\cdot {\mathrm {\zeta } }.\end{aligned}}$

(2.1e)

A pressure ${\mathcal {P}}$ produces a decrease in volume, the proportionality constant being the bulk modulus $k$:

${\begin{aligned}k=-{\mathcal {P}}/\Delta .\end{aligned}}$

(2.1f)

Sometimes the compressibility, $1/k$, is used instead of $k$.

In addition to creating strains, stresses cause rotation of the medium, the vector rotation ${\theta }$ being equal to

${\begin{aligned}{\theta }={\mathrm {\theta } }_{x}i+{\mathrm {\theta } }_{y}j+{\mathrm {\theta } }_{z}k={\frac {1}{2}}{\nabla }\times {\mathrm {\zeta } },\end{aligned}}$

(2.1g)

where ${\mathrm {\theta } }_{x}=\left(1/2\right)\left({\mathrm {\partial } }w/{\mathrm {\partial } }y-{\mathrm {\partial } }v/{\mathrm {\partial } }z\right)$, ${\mathrm {\theta } }_{y}=\left(1/2\right)\left({\mathrm {\partial } }u/{\mathrm {\partial } }z-{\mathrm {\partial } }w/{\mathrm {\partial } }x\right)$, ${\mathrm {\theta } }_{z}=\left(1/2\right)\left({\mathrm {\partial } }v/{\mathrm {\partial } }x-{\mathrm {\partial } }u/{\mathrm {\partial } }y\right).$

For small strains and an isotropic medium (where properties are the same regard-less of the direction of measurement), Hooke’s law relates the stresses to the strains as follows:

${\begin{aligned}{\mathrm {\sigma } }_{ii}={\mathrm {\lambda } }{\mathrm {\Delta } }+2{\mu }{\mathrm {\varepsilon } }_{ii},{\qquad }i=x,\;y,\;z;\end{aligned}}$

(2.1h)

${\begin{aligned}{\mathrm {\sigma } }_{ij}={\mathrm {\mu } }{\mathrm {\varepsilon } }_{ij},{\qquad }{\qquad }{\quad }i\neq j.\end{aligned}}$

(2.1i)

where ${\mathrm {\lambda } }$ and ${\mathrm {\mu } }$ are Lamé’s constants (${\mathrm {\mu } }$ is usually called the modulus of rigidity or the shear modulus).

Solution

Subtracting equation (2.1h) for $i=y$ from the same equation for $i=z$ gives ${\mathrm {\varepsilon } }_{yy}={\mathrm {\varepsilon } }_{zz}$.

Dividing equation (2.1h) for $i=y$ by ${\mathrm {\varepsilon } }_{xx}$ gives

${\begin{aligned}0={\mathrm {\lambda } }\left(1+2{\frac {{\mathrm {\varepsilon } }_{yy}}{{\mathrm {\varepsilon } }_{xx}}}\right)+2{\mathrm {\mu } }{\frac {{\mathrm {\varepsilon } }_{yy}}{{\mathrm {\varepsilon } }_{xx}}}={\mathrm {\lambda } }-2\left({\mathrm {\lambda } }+{\mathrm {\mu } }\right)\sigma ,\end{aligned}}$

so

${\begin{aligned}{\mathrm {\sigma } }={\mathrm {\lambda } }/2\left({\mathrm {\lambda } }+{\mathrm {\mu } }\right).\end{aligned}}$

2.1b Show that Young’s modulus $E$, defined as ${\mathrm {\sigma } _{xx}/{\mathrm {\varepsilon } }_{xx}}$, is given by the equation

${\begin{aligned}{E={\frac {{\mu }\left({3}{\lambda }+{2}{\mu }\right)}{\left({\lambda }+{\mu }\right)}}.}\end{aligned}}$

(2.1j)

Solution

Adding the three equations (2.1h) for $i=x$, $y$, and $z$, and recalling that ${\mathrm {\sigma } }_{yy}=0={\mathrm {\sigma } }_{zz}$, we get

${\begin{aligned}{\mathrm {\sigma } }_{xx}=\left[3{\mathrm {\lambda } }{\mathrm {\Delta } }+2{\mathrm {\mu } }\left({\mathrm {\varepsilon } }_{xx}+{\mathrm {\varepsilon } }_{yy}+{\mathrm {\varepsilon } }_{zz}\right)\right]=\left(3{\mathrm {\lambda } }+2{\mathrm {\mu } }\right)\Delta .\end{aligned}}$

Dividing both sides by ${\mathrm {\varepsilon } }_{xx}$ gives

${\begin{aligned}E={\mathrm {\sigma } }_{xx}/{\mathrm {\varepsilon } }_{xx}=\left(3{\mathrm {\lambda } }+2{\mathrm {\mu } }\right)\left(1+2{\mathrm {\varepsilon } }_{yy}/{\mathrm {\varepsilon } }_{xx}\right)=\left(3{\mathrm {\lambda } }+2{\mathrm {\mu } }\right)\left(1-2{\mathrm {\sigma } }\right).\end{aligned}}$

Using equation (2.1a) we get

${\begin{aligned}E={\frac {{\mathrm {\mu } }\left(3{\mathrm {\lambda } }+2{\mathrm {\mu } }\right)}{\left({\mathrm {\lambda } }+{\mathrm {\mu } }\right)}}.\end{aligned}}$

2.1c A pressure $\mathbf {\mathcal {P}}$ is equivalent to stresses ${\sigma }_{xx}={\sigma }_{yy}={\sigma }_{zz}=-{\mathcal {P}}$. Derive the following result for the bulk modulus $k$:

${\begin{aligned}k{\mathbf {=} \left({\lambda }\mathbf {+{\frac {2}{3}}} {\mu }\right)}.\end{aligned}}$

(2.1k)

Solution

Since ${\mathcal {P}}=-{\mathrm {\sigma } }_{xx}=-{\mathrm {\sigma } }_{yy}=-{\mathrm {\sigma } }_{zz}$, we add equation (2.1h) for each of the three values $i=x,y,z$, obtaining $-3{\mathcal {P}}=\left[3{\mathrm {\lambda } }{\mathrm {\Delta } }+2{\mathrm {\mu } }\left({\mathrm {\varepsilon } }_{xx}+{\mathrm {\varepsilon } }_{yy}+{\mathrm {\varepsilon } }_{zz}\right)\right]=\left(3{\mathrm {\lambda } }+2{\mathrm {\mu } }\right)\Delta$, so from equation (2.1f),

${\begin{aligned}k=-{\mathcal {P}}/{\mathrm {\Delta } }=\left({\mathrm {\lambda } }+{\frac {2}{3}}{\mathrm {\mu } }\right).\end{aligned}}$

2.2 Interrelationships among elastic constants

The entries in Table 2.2a express, for isotropic media, the quantities at the heads of the columns in terms of the pairs of elastic constants or velocities at the left ends of the rows. The first three entries in the ninth row are equations (2.1j), (2.1a), and (2.1k) and the next two entries in the same row are formulas for the P- and S- wave velocities ${\alpha }$ and ${\beta }$ (see problem 2.5). Starting from these five relations, derive the other relations in the table.

Background

For isotropic media, any two of the elastic constants can be considered as independent and the others can be expressed in terms of these two. The P- and S-wave velocities, ${\mathrm {\alpha } }$ and ${\mathrm {\beta } }$, given by the equations

${\begin{aligned}{\mathrm {\rho } }{\mathrm {\alpha } }^{2}=\left({\mathrm {\lambda } }+2{\mathrm {\mu } }\right),{\quad }{\mathrm {\rho } }\beta ^{2}=\mu \end{aligned}}$

(2.2a)

[see Sheriff and Geldart, 1995, equations (2.28) and (2.29)] can also be expressed in terms of any two elastic constants (plus the density ${\mathrm {\rho } }$).

Solution

Denoting the equations by row and column (as for matrix elements) and using a comma instead of a period, we use equations (9,1) to (9,3) to derive the equations that do not involve ${\mathrm {\alpha } }$ and ${\mathrm {\beta } }$, then we use equations (9,6) and (9,7) (see equation (2.2a)) to derive the rest. From equation (9,1),

${\begin{aligned}{{\mathrm {\lambda } }\left(E-3{\mathrm {\mu } }\right)=2{\mathrm {\mu } }^{2}-{\mathrm {\mu } }E,}\\{{\mathrm {\lambda } }={\mathrm {\mu } }\left(E-2{\mathrm {\mu } }\right)/\left(3{\mathrm {\mu } }-E\right).}\end{aligned}}$

(3,5)

From equation (9,2),

${\begin{aligned}{{\mathrm {\lambda } }\left(2{\mathrm {\sigma } }-1\right)+2{\mathrm {\mu } }{\mathrm {\sigma } }=0,}\\{{\mathrm {\lambda } }=2{\mathrm {\mu } }{\mathrm {\sigma } }/\left(1-2{\mathrm {\sigma } }\right).}\end{aligned}}$

(5,5)

Solving equation (9,2) for ${\mathrm {\mu } }$, we have $2{\mathrm {\mu } }{\mathrm {\sigma } }={\mathrm {\lambda } }\left(1-2{\mathrm {\sigma } }\right)$,

${\begin{aligned}{\mathrm {\mu } }={\mathrm {\lambda } }\left(1-2{\mathrm {\sigma } }\right)/2{\mathrm {\sigma } }.\end{aligned}}$

(6,4)

From equation (9,3),

${\begin{aligned}{\mathrm {\lambda } }=k-{\frac {2}{3}}{\mathrm {\mu } }.\end{aligned}}$

(7,5)

Equating ${\mathrm {\mu } }$ from equations (9,2) and (9,3) [or from equations (6,4) and (7,5)] gives

${\begin{aligned}{\mathrm {\mu } }={\mathrm {\lambda } }\left(1-2{\mathrm {\sigma } }\right)/2{\mathrm {\sigma } }={\frac {3}{2}}\left(k-{\mathrm {\lambda } }\right),\end{aligned}}$

(8,4)

thus

${\begin{aligned}k={\mathrm {\lambda } }\left(1-2{\mathrm {\sigma } }\right)/3{\mathrm {\sigma } }+{\mathrm {\lambda } }={\mathrm {\lambda } }\left(1+{\mathrm {\sigma } }\right)/3\sigma ,\end{aligned}}$

(6,3)

and

${\begin{aligned}{\mathrm {\lambda } }={\frac {3{\mathrm {\sigma } }k}{\left(1+{\mathrm {\sigma } }\right)}}.\end{aligned}}$

(4,5)

Solving equation (4,5) for ${\mathrm {\sigma } }$ gives

${\begin{aligned}{\mathrm {\sigma } }={\frac {\mathrm {\lambda } }{3k-{\mathrm {\lambda } }}}.\end{aligned}}$

(8,2)

Substituting equation (4,5) into equation (8,4), we get,

${\begin{aligned}{\mathrm {\mu } }={\frac {3}{2}}\left(k-{\frac {3{\mathrm {\sigma } }k}{\left(1+{\mathrm {\sigma } }\right)}}\right)={\frac {3k\left(1-2{\mathrm {\sigma } }\right)}{2\left(1+{\mathrm {\sigma } }\right)}}.\end{aligned}}$

(4,4)

We use equation (7,5) to eliminate ${\mathrm {\lambda } }$ from equation (6,3):

${\begin{aligned}k=\left(k-{\frac {2}{3}}{\mathrm {\mu } }\right)\left({\frac {1+{\mathrm {\sigma } }}{3{\mathrm {\sigma } }}}\right)={\frac {k\left(1+{\mathrm {\sigma } }\right)}{3{\mathrm {\sigma } }}}-{\frac {2{\mathrm {\mu } }\left(1+{\mathrm {\sigma } }\right)}{9{\mathrm {\sigma } }}},\end{aligned}}$

that is,

${\begin{aligned}k\left({\frac {1+{\mathrm {\sigma } }}{3{\mathrm {\sigma } }}}-1\right)={\frac {2{\mathrm {\mu } }\left(1+{\mathrm {\sigma } }\right)}{9{\mathrm {\sigma } }}},\end{aligned}}$

so

${\begin{aligned}k={\frac {2{\mathrm {\mu } }\left(1+{\mathrm {\sigma } }\right)}{3\left(1-2{\mathrm {\sigma } }\right)}}.\end{aligned}}$

(5,3)

Solving for ${\mathrm {\sigma } }$, we get

${\begin{aligned}{\mathrm {\sigma } }={\frac {\left(3k-2{\mathrm {\mu } }\right)}{2\left(3k+{\mathrm {\mu } }\right)}}.\end{aligned}}$

(7,2)

Figure 2.2a)  Relations between elastic constants and velocities for isotropic media.

We use equation (8,4) to express ${\mathrm {\mu } }$ in equation (9,1) in terms of $\left(k,\;{\mathrm {\lambda } }\right)$. Thus

${\begin{aligned}E={\frac {{\frac {3}{2}}\left(k-{\mathrm {\lambda } }\right)\left[3{\mathrm {\lambda } }+3\left(k-{\mathrm {\lambda } }\right)\right]}{{\mathrm {\lambda } }+{\frac {3}{2}}\left(k-{\mathrm {\lambda } }\right)}}={\frac {9k\left(k-{\mathrm {\lambda } }\right)}{\left(3k-{\mathrm {\lambda } }\right)}}.\end{aligned}}$

(8,1)

Solving equation (8,1) for ${\mathrm {\lambda } }$ gives

${\begin{aligned}{\mathrm {\lambda } }=3k\left(3k-E\right)/\left(9k-E\right).\end{aligned}}$

(2,5)

We now eliminate ${\mathrm {\lambda } }$ from equation (9,1) using equation (7,5)

${\begin{aligned}{E={\frac {\mu \left({3k-2\mu +2\mu }\right)}{\left({k-2\mu /3+\mu }\right)}}}\\{=9k\mu /\left({3k+\mu }\right).}\end{aligned}}$

(7,1)

Solving equation (7,1) for $k$ and ${\mathrm {\mu } }$ gives

${\begin{aligned}k={\mathrm {\mu } }E/3\left(3{\mathrm {\mu } }-E\right),\end{aligned}}$

(3,3)

${\begin{aligned}{\mathrm {\mu } }=3kE/\left(9k-E\right).\end{aligned}}$

(2,4)

Next we use equation (6,4) to eliminate ${\mathrm {\mu } }$ from equation (9,1):

${\begin{aligned}E={\mathrm {\mu } }\left({\frac {6{\mathrm {\mu } }{\mathrm {\sigma } }+2{\mathrm {\mu } }-4{\mathrm {\mu } }{\mathrm {\sigma } }}{2{\mathrm {\mu } }{\mathrm {\sigma } }+{\mathrm {\mu } }-2{\mathrm {\mu } }{\mathrm {\sigma } }}}\right)={\mathrm {\lambda } }\left(1-2{\mathrm {\sigma } }\right)\left(1+{\mathrm {\sigma } }\right)/{\mathrm {\sigma } }.\end{aligned}}$

(6,1)

Using equations (9,1) and (5,5) we get

${\begin{aligned}{E={\mathrm {\mu } }\left(6{\mathrm {\mu } }{\mathrm {\sigma } }\right)/\left(2{\mathrm {\mu } }+{\mathrm {\mu } }-2{\mathrm {\mu } }{\mathrm {\sigma } }\right)}\\{{\quad }=2{\mathrm {\mu } }^{2}\left(1+{\mathrm {\sigma } }\right)/{\mathrm {\mu } }=2{\mathrm {\mu } }\left(1+{\mathrm {\sigma } }\right).}\end{aligned}}$

(5,1)

Solving equation (5,1) for ${\mathrm {\sigma } }$ and ${\mathrm {\mu } }$ gives

${\begin{aligned}{\mathrm {\sigma } }=E/2{\mathrm {\mu } }-1=\left(E-2{\mathrm {\mu } }\right)/2\mu ,\end{aligned}}$

(3,2)

${\begin{aligned}{\mathrm {\mu } }=E/2\left(1+{\mathrm {\sigma } }\right).\end{aligned}}$

(1,4)

Using equations (4,4) and (4,5) to replace $\left({\mathrm {\lambda } },\;{\mathrm {\mu } }\right)$ in equation (9,1) by $\left({\mathrm {\sigma } },\;k\right)$ gives

${\begin{aligned}E={\frac {3k\left(1-2{\mathrm {\sigma } }\right)}{2\left(1+{\mathrm {\sigma } }\right)}}\times {\frac {\left(9{\mathrm {\sigma } }k+3k-6{\mathrm {\sigma } }k\right)}{\left[3{\mathrm {\sigma } }k+\left(3k-6{\mathrm {\sigma } }k\right)/2\right]}}=3k\left(1-2{\mathrm {\sigma } }\right).\end{aligned}}$

(4,1)

Solving equation (4,1) for $k$ and ${\mathrm {\sigma } }$, we get

${\begin{aligned}k=E/3\left(1-2{\mathrm {\sigma } }\right),\end{aligned}}$

(1,3)

${\begin{aligned}{\mathrm {\sigma } }=\left(3k-E\right)/6k.\end{aligned}}$

(2,2)

The last equation (of this group) can be obtained by substituting equation (1,3) into equation (4,5):

${\begin{aligned}{\mathrm {\lambda } }={\frac {3{\mathrm {\sigma } }k}{\left(1+{\mathrm {\sigma } }\right)}}={\frac {3{\mathrm {\sigma } }E}{3\left(1-2{\mathrm {\sigma } }\right)\left(1+{\mathrm {\sigma } }\right)}}={\frac {{\mathrm {\sigma } }E}{\left(1+{\mathrm {\sigma } }\right)\left(1-2{\mathrm {\sigma } }\right)}}.\end{aligned}}$

(1,5)

Equations (10,1) to (10,3) express $E$, ${\mathrm {\sigma } }$, and $k$ in terms of the P- and S-wave velocities, ${\mathrm {\alpha } }$ and ${\mathrm {\beta } }$ (see equation 2.2a). To introduce ${\mathrm {\alpha } }$ and ${\mathrm {\beta } }$, we write equations (10,1) to (10,3) in terms of $\left({\mathrm {\lambda } }+2{\mathrm {\mu } }\right)$ and ${\mathrm {\mu } }$. Thus,

${\begin{aligned}{E={\mathrm {\mu } }\left({\frac {3{\mathrm {\lambda } }+2{\mathrm {\mu } }}{{\mathrm {\lambda } }+{\mathrm {\mu } }}}\right)={\mathrm {\mu } }\left[{\frac {3\left({\mathrm {\lambda } }+2{\mathrm {\mu } }\right)-4{\mathrm {\mu } }}{\left({\mathrm {\lambda } }+2{\mathrm {\mu } }\right)-{\mathrm {\mu } }}}\right]}\\{{\qquad }{\quad }={\frac {{\mathrm {\rho } }\beta ^{2}\left(3{\mathrm {\rho } }{\mathrm {\alpha } }^{2}-4{\mathrm {\rho } }\beta ^{2}\right)}{{\mathrm {\rho } }\left({\mathrm {\alpha } }^{2}-\beta ^{2}\right)}}={\frac {{\mathrm {\rho } }\beta ^{2}\left(3{\mathrm {\alpha } }^{2}-4\beta ^{2}\right)}{{\mathrm {\alpha } }^{2}-\beta ^{2}}}},\end{aligned}}$

(10,1)

${\begin{aligned}{\mathrm {\sigma } }={\frac {\mathrm {\lambda } }{2\left({\mathrm {\lambda } }+{\mathrm {\mu } }\right)}}={\frac {\left({\mathrm {\lambda } }+2{\mathrm {\mu } }\right)-2{\mathrm {\mu } }}{2\left({\mathrm {\lambda } }+2{\mathrm {\mu } }\right)-2{\mathrm {\mu } }}}={\frac {{\mathrm {\alpha } }^{2}-2\beta ^{2}}{2\left({\mathrm {\alpha } }^{2}-\beta ^{2}\right)}},\end{aligned}}$

(10,2)

${\begin{aligned}k={\mathrm {\lambda } }+{\frac {2{\mathrm {\mu } }}{3}}=\left({\mathrm {\lambda } }+2{\mathrm {\mu } }\right)-{\frac {4}{3}}{\mathrm {\mu } }={\mathrm {\rho } }\left({\mathrm {\alpha } }^{2}-{\frac {4}{3}}\beta ^{2}\right).\end{aligned}}$

(10,3)

Equation (10,4) is the second equation in equation (2.2a). To get equation (10,5), we write

${\begin{aligned}{\mathrm {\lambda } }=\left({\mathrm {\lambda } }+2{\mathrm {\mu } }\right)-2{\mathrm {\mu } }={\mathrm {\rho } }\left({\mathrm {\alpha } }^{2}-2\beta ^{2}\right).\end{aligned}}$

(10,5)

To verify column 6, we start with equation (9,6) and express ${\mathrm {\lambda } }$ and ${\mathrm {\mu } }$ in terms of the required pair of constants. Thus, using equations (1,4) and (1,5) we get

${\begin{aligned}{{\mathrm {\rho } }{\mathrm {\alpha } }^{2}=\left({\mathrm {\lambda } }+2{\mathrm {\mu } }\right)={\frac {E{\mathrm {\sigma } }}{\left(1+{\mathrm {\sigma } }\right)\left(1-2{\mathrm {\sigma } }\right)}}+{\frac {2E}{2\left(1+{\mathrm {\sigma } }\right)}}}\\{{\qquad }=\left({\frac {E}{1+{\mathrm {\sigma } }}}\right)\left({\frac {\mathrm {\sigma } }{1-2{\mathrm {\sigma } }}}+1\right)={\frac {E\left(1-{\mathrm {\sigma } }\right)}{\left(1+{\mathrm {\sigma } }\right)\left(1-2{\mathrm {\sigma } }\right)}}}.\end{aligned}}$

(1,6)

Following the same procedure, using equations (2,4) and (2,5), we get

${\begin{aligned}{{\mathrm {\rho } }{\mathrm {\alpha } }^{2}=3k\left({\frac {3k-E}{9k-E}}\right)+\left({\frac {6kE}{9k-E}}\right)}\\{{\qquad }=\left({\frac {3k}{9k-E}}\right)\left(3k-E+2E\right)={\frac {3k\left(3k+E\right)}{9k-E}}.}\end{aligned}}$

(2,6)

In the same way, we get the following results:

${\begin{aligned}{{\mathrm {\rho } }{\mathrm {\alpha } }^{2}={\mathrm {\mu } }\left({\frac {E-2{\mathrm {\mu } }}{3{\mathrm {\mu } }-E}}\right)+2{\mathrm {\mu } }=\left({\frac {\mathrm {\mu } }{3{\mathrm {\mu } }-E}}\right)\left(E-2{\mathrm {\mu } }+6{\mathrm {\mu } }-2E\right)}\\{{\qquad }=\left({\frac {\mathrm {\mu } }{3{\mathrm {\mu } }-E}}\right)\left(4{\mathrm {\mu } }-E\right),}\end{aligned}}$

(3,6)

${\begin{aligned}{{\mathrm {\rho } }{\mathrm {\alpha } }^{2}=\left({\frac {3k{\mathrm {\sigma } }}{1+{\mathrm {\sigma } }}}\right)+3k\left({\frac {1-2{\mathrm {\sigma } }}{1+{\mathrm {\sigma } }}}\right)=\left({\frac {3k}{1+{\mathrm {\sigma } }}}\right)\left({\mathrm {\sigma } }+1-2{\mathrm {\sigma } }\right)}\\{{\qquad }={\frac {3k\left(1-{\mathrm {\sigma } }\right)}{\left(1+{\mathrm {\sigma } }\right)}},}\end{aligned}}$

(4,6)

${\begin{aligned}{\mathrm {\rho } }{\mathrm {\alpha } }^{2}=\left({\frac {2{\mathrm {\mu } }{\mathrm {\sigma } }}{1-2{\mathrm {\sigma } }}}\right)+2{\mathrm {\mu } }=2{\mathrm {\mu } }\left({\frac {\mathrm {\sigma } }{1-2{\mathrm {\sigma } }}}+1\right)=2{\mathrm {\mu } }\left({\frac {1-{\mathrm {\sigma } }}{1-2{\mathrm {\sigma } }}}\right),\end{aligned}}$

(5,6)

${\begin{aligned}{\mathrm {\rho } }{\mathrm {\alpha } }^{2}={\mathrm {\lambda } }+{\frac {{\mathrm {\lambda } }\left(1-2{\mathrm {\sigma } }\right)}{\mathrm {\sigma } }}={\frac {\mathrm {\lambda } }{\mathrm {\sigma } }}\left(1-{\mathrm {\sigma } }\right),\end{aligned}}$

(6,6)

${\begin{aligned}{\mathrm {\rho } }{\mathrm {\alpha } }^{2}=\left(k-{\frac {2}{3}}{\mathrm {\mu } }\right)+2{\mathrm {\mu } }=k+{\frac {4}{3}}{\mathrm {\mu } },\end{aligned}}$

(7,6)

${\begin{aligned}{\mathrm {\rho } }{\mathrm {\alpha } }^{2}={\mathrm {\lambda } }+3\left(k-{\mathrm {\lambda } }\right)=3k-2\lambda .\end{aligned}}$

(8,6)

Column 7 is merely the square root of column 4 after dividing by ${\mathrm {\rho } }$. Column 8 is obtained by dividing column 7 by column 6.

2.3 Magnitude of disturbance from a seismic source

2.3a Firing an air gun in water creates a pressure transient a small distance away from the air gun with peak pressure ${\mathcal {P}}$ of 5 atmospheres ($5.0\times 10^{5}$ Pa). If the compressibility of water is $4.5\times 10^{-10}$/Pa, what is the peak energy density?

Background

Air guns (see problem 7.7) suddenly inject a bubble of high‐pressure air into the water to generate a seismic wave.

Stresses acting upon a medium cause energy to be stored as strain energy, because the stresses are present while the medium is being displaced. Strain energy density (energy/unit volume) $E$ is equal to [see Sheriff and Geldart, 1995, equation (2.22)]

${\begin{aligned}{E={\frac {1}{2}}\sum \limits _{i}\sum \limits _{j}{\mathrm {\sigma } }_{ij}{\mathrm {\varepsilon } }_{ij}={\frac {1}{2}}{\mathrm {\lambda } }{\mathrm {\Delta } }^{2}+{\mathrm {\mu } }\left({\mathrm {\varepsilon } }_{xx}^{2}+{\mathrm {\varepsilon } }_{yy}^{2}+{\mathrm {\varepsilon } }_{zz}^{2}\right)}\\{{\qquad }{\qquad }{\qquad }{\qquad }+{\frac {1}{2}}{\mathrm {\mu } }\left({\mathrm {\varepsilon } }_{xy}^{2}+{\mathrm {\varepsilon } }_{yz}^{2}+{\mathrm {\varepsilon } }_{zx}^{2}\right).}\end{aligned}}$

(2.3a)

Solution

From problem 2.1c, we see that $k=1/\left(4.5\times 10^{-10}\right)=2.2\times 10^{9}$ Pa. Also, ${\mathrm {\mu } }=0$ for water, so $E={\frac {1}{2}}{\mathrm {\lambda } }{\mathrm {\Delta } }^{2}$. From equation (7,5) of Table 2.2a we find that ${\mathrm {\lambda } }=k$ when ${\mathrm {\mu } }=0$. Also, $k=-{\mathcal {P}}/{\mathrm {\Delta } }$ (see equation (2.1f)), so

${\begin{aligned}\Delta =\left|{\cal {P}}/k\right|=5.0\times 10^{5}/2.2\times 10^{9}=2.3\times 10^{-4}.\end{aligned}}$

Using equation (2.3a) we find that

${\begin{aligned}E={\frac {1}{2}}\left(2.2\times 10^{9}\times 2.3^{2}\times 10^{-8}\right)=58\,\,\,\mathrm {J/m} ^{3}.\end{aligned}}$

[The dimensions of $E$ are the same as those of stress, since strains are dimensionless. Thus, stress units are ${\rm {N/m}}^{2}={\rm {Nm/m}}^{3}={\rm {J/m}}^{3}$.]

2.3b If the same wave is generated in rock with ${\mathrm {\lambda } }={\mathrm {\mu } }=3.0\times 10^{10}$ Pa, what is the peak energy density? Assume a symmetrical $P$-wave with ${\mathrm {\varepsilon } }_{xx}={\mathrm {\varepsilon } }_{yy}={\mathrm {\varepsilon } }_{zz},{\mathrm {\varepsilon } }_{ij}=0$ for $i\neq j$.

Solution

We have ${\mathrm {\Delta } }={\mathrm {\varepsilon } }_{xx}+{\mathrm {\varepsilon } }_{yy}+{\mathrm {\varepsilon } }_{zz}=3{\mathrm {\varepsilon } }_{xx}$, ${\mathrm {\varepsilon } }_{ij}=0,i\neq j$, so equation (2.3a) becomes

${\begin{aligned}E={\frac {1}{2}}{\mathrm {\lambda } }{\mathrm {\Delta } }^{2}+3{\mathrm {\mu } }{\mathrm {\varepsilon } }_{xx}^{2}={\mathrm {\lambda } }{\mathrm {\Delta } }^{2}\left({\frac {1}{2}}+{\frac {1}{3}}\right)={\frac {5}{6}}{\mathrm {\lambda } }({\mathcal {P}}/k)^{2}.\end{aligned}}$

Equation (9,3) in Table 2.2a gives $k={\mathrm {\lambda } }+{\frac {2}{3}}{\mathrm {\mu } }={\frac {5}{3}}{\mathrm {\lambda } }$ so that

${\begin{aligned}E={\frac {5}{6}}{\mathrm {\lambda } }{\mathcal {P}}^{2}\left/\right.\left({\frac {5}{3}}{\mathrm {\lambda } }\right)^{2}=0.3\times (5.0\times 10^{5})^{2}/3\times 10^{10}=2.5\ \mathrm {J/m} ^{3}.\end{aligned}}$

2.4 Magnitudes of elastic constants

To illustrate the relationships and magnitudes of the elastic constants, complete Table 2.4a.

[Note that these values apply to specific specimens; the elastic constants for rocks range considerably, especially as porosity and pressure change.]

Solution

We use the row-column notation to designate equations from Table 2.2a.

Water: Since water is a fluid we know that ${\mathrm {\mu } }=0=\beta$. Equation (4,1) shows that $E=0$ also. From equation (4,5) we get ${\mathrm {\lambda } }=k=2.1\times 10^{9}$ Pa.

Table 2.4a. Magnitudes of elastic constants and velocities.

Constant	Water	Stiff mud	Shale	Sandstone	Limestone	Granite
$E\,\,\,({\times }10^{9}{\hbox{Pa}})$				16	54	50
$k\,\,\,({\times }10^{9}{\hbox{Pa}})$	2.1
${\mathrm {\mu } }\,\,\,({\times }10^{9}{\hbox{Pa}})$
${\mathrm {\lambda } }\,\,\,({\times }10^{9}{\hbox{Pa}})$
${\mathrm {\sigma } }$	0.50	0.43	0.38	0.34	0.25	0.20
${\mathrm {\rho } }\,\,\,({\hbox{g/cm}}^{3})$	1.0	1.5	1.8	1.9	2.5	2.7
${\alpha }\,\,\,({\hbox{km/s}})$	1.5	1.6	3.2
${\beta }\,\,\,({\hbox{km/s}})$

Stiff mud: Because ${\mathrm {\sigma } }<0.5$, stiff mud is equivalent to a solid, hence ${\mathrm {\mu } }\neq 0$. From ${\mathrm {\rho } }$ and ${\mathrm {\alpha } }$ we get $\left({\mathrm {\lambda } }+2{\mathrm {\mu } }\right)$ using equation (9,6), while equation (9,2) expresses ${\mathrm {\sigma } }$ in terms of ${\mathrm {\lambda } }$ and ${\mathrm {\mu } }$, thus enabling us to find both ${\mathrm {\lambda } }$ and ${\mathrm {\mu } }$.

We have:

${\begin{aligned}{{\rho }{{\alpha }^{2}}=\left({{\lambda }+2{\mu }}\right)=\left({1.5\;{\rm {g/c}}{{\rm {m}}^{3}}}\right){\times }{{(1.6{\times }{{10}^{5}}\;{\rm {cm/s}})}^{2}}}\\{{\qquad }=3.8{\times }{{10}^{10}}\;{\rm {dynes/c}}{{\rm {m}}^{2}}=3.8{\times }{{10}^{9}}\;{\rm {Pa}},}\\{{\quad }{\sigma }=0.43={\lambda }/2\left({{\lambda }+{\mu }}\right),\;{\rm {i}}.{\rm {e}}.,\;0.86{\mu }=0.14{\lambda }.}\end{aligned}}$

Solving the two equations, we get ${\mathrm {\lambda } }=2.9{\times }10^{9}$ Pa, ${\mathrm {\mu } }=0.47{\times }10^{9}$ Pa. Using equation (6,1),

${\begin{aligned}{E={\mathrm {\lambda } }{\frac {\left(1+{\mathrm {\sigma } }\right)\left(1-2{\mathrm {\sigma } }\right)}{\mathrm {\sigma } }}=2.9{\times }10^{9}{\times }1.43{\times }0.14/0.43}\\{{\qquad }{\qquad }{\qquad }{\qquad }{\quad }=1.4{\times }10^{9}\ \mathrm {Pa} .}\end{aligned}}$

Equation (6,3) gives

${\begin{aligned}k={\mathrm {\lambda } }{\frac {\left(1+{\mathrm {\sigma } }\right)}{3{\mathrm {\sigma } }}}=2.9{\times }10^{9}{\times }1.43/1.29=3.2{\times }10^{9}\ \mathrm {Pa} .\end{aligned}}$

Finally, to get ${\mathrm {\beta } }$ we note that ${\mathrm {\lambda } }$, ${\mathrm {\mu } }$, and $k$ are in $N/m^{2}$, and so we must express ${\mathrm {\rho } }$ in appropriate units of $kg/m^{3}$, i.e., ${\mathrm {\rho } }=1.5{\times }10^{3}$ $kg/m^{3}$. We now have

$\beta =({\mathrm {\mu } }/{\mathrm {\rho } })^{1/2}=\left(0.47{\times }10^{9}/1.5{\times }10^{3}\right)=0.56\ \mathrm {\hbox{km/s}} .$

Shale. As with stiff mud, we have been given ${\mathrm {\rho } }$, $\alpha$, and ${\mathrm {\sigma } }$, so that again ${\mathrm {\rho } }$ and ${\mathrm {\alpha } }$ give us $\left({\mathrm {\lambda } }+2{\mathrm {\mu } }\right)$ and ${\mathrm {\sigma } }$ gives us ${\mathrm {\lambda } }/2\left({\mathrm {\lambda } }+{\mathrm {\mu } }\right)$ so that we can solve these equations for ${\mathrm {\lambda } }$ and ${\mathrm {\mu } }$, then find $k$ and $E$ using equations (9,3) and (9,1). Thus,

${\begin{aligned}\left(\lambda +2\mu \right)=\rho \alpha ^{2}=\left(1.8\ \mathrm {g/cm} ^{3}\right)\times (3.2\times 10^{5}\ \mathrm {cm/s} )^{2}=18.4\times 10^{9}\ \mathrm {Pa} ,\end{aligned}}$

${\begin{aligned}\sigma =0.38=\lambda /2\left(\lambda +\mu \right),\quad \mathrm {i.e.} ,0.76\mu =0.24\lambda .\end{aligned}}$

Solving the two equations gives ${\mathrm {\lambda } }=11{\times }10^{9}$ Pa, ${\mathrm {\mu } }=3.6{\times }10^{9}$ Pa.

From equation (9,3), $k={\mathrm {\lambda } }+{\frac {2}{3}}{\mathrm {\mu } }=14{\times }10^{9}$ Pa, and equation (9,1) gives

${\begin{aligned}E={\mathrm {\mu } }{\frac {\left(3{\mathrm {\lambda } }+2{\mathrm {\mu } }\right)}{\left({\mathrm {\lambda } }+{\mathrm {\mu } }\right)}}=9.9{\times }10^{9}\ \mathrm {Pa} .\end{aligned}}$

Finally,

${\begin{aligned}\beta =({\mathrm {\mu } }/{\mathrm {\rho } })^{1/2}=(3.6{\times }10^{9}/1.8{\times }10^{3})^{1/2}=1.4\ \mathrm {\hbox{km/s}} .\end{aligned}}$

Sandstone: We are given the elastic constants $E$ and ${\mathrm {\sigma } }$ (plus ${\mathrm {\rho } }$), so we get $k$, ${\mathrm {\mu } }$, ${\mathrm {\lambda } }$, $\alpha$, ${\mathrm {\beta } }$ using equations (1,3) to (1,7) in Table 2.2a. Thus

${\begin{aligned}{k=16{\times }10^{9}/3{\times }0.32=17{\times }10^{9}\ \mathrm {Pa} ,}\\{{\mathrm {\mu } }=16{\times }10^{9}/2{\times }1.34=6.0{\times }10^{9}\ \mathrm {Pa} ,}\\{{\mathrm {\lambda } }=16{\times }10^{9}{\times }0.34/1.34{\times }0.32=13{\times }10^{9}\ \mathrm {Pa} ,}\\{{\mathrm {\alpha } }=[16{\times }10^{9}{\times }0.66/1.34{\times }0.32{\times }1.9{\times }10^{3}]^{1/2}=3.6\ \mathrm {\hbox{km/s}} ,}\\{{\mathrm {\beta } }=[16{\times }10^{9}/2{\times }1.34{\times }1.9{\times }10^{3}]^{1/2}=1.8\ \mathrm {\hbox{km/s}} .}\end{aligned}}$

[We could also have obtained ${\mathrm {\alpha } }$ and ${\mathrm {\beta } }$ by using equations (9,6) and (9,7).]

Limestone: We solve in the same way as with sandstone since we are given the same constants:

${\begin{aligned}{k=54{\times }10^{9}/3{\times }0.50=36{\times }10^{9}\ \mathrm {Pa} ,}\\{{\mathrm {\mu } }=54{\times }10^{9}/2{\times }1.25=22{\times }10^{9}\ \mathrm {Pa} ,}\\{{\mathrm {\lambda } }=54{\times }10^{9}{\times }0.25/1.25{\times }0.50=22{\times }10^{9}\ \mathrm {Pa} ,}\\{{\mathrm {\alpha } }=[54{\times }10^{9}{\times }0.75/1.25{\times }0.50{\times }2.5{\times }10^{3}]^{1/2}=5.1\ \mathrm {\hbox{km/s}} ,}\\{{\mathrm {\beta } }=[54{\times }10^{9}/2{\times }1.25{\times }2.5{\times }10^{3}]^{1/2}=2.9\ \mathrm {\hbox{km/s}} .}\end{aligned}}$

Granite: Again the solution is the same as for sandstone.

${\begin{aligned}{k=50{\times }10^{9}/3{\times }0.60=28{\times }10^{9}\ \mathrm {Pa} ,}\\{{\mathrm {\mu } }=50{\times }10^{9}/2{\times }1.2=21{\times }10^{9}\ \mathrm {Pa} ,}\\{{\mathrm {\lambda } }=50{\times }10^{9}{\times }0.20/1.2{\times }0.60=14{\times }10^{9}\ \mathrm {Pa} ,}\\{{\mathrm {\alpha } }=[50{\times }10^{9}{\times }0.80/1.2{\times }0.60{\times }2.7{\times }10^{3}]^{1/2}=4.5\ \mathrm {\hbox{km/s}} ,}\\{{\mathrm {\beta } }=[50{\times }10^{9}/2{\times }1.2{\times }2.7{\times }10^{3}]^{1/2}=2.8\ \mathrm {\hbox{km/s}} .}\end{aligned}}$

Table 2.4b summarizes the results.

Table 2.4b Magnitudes of elastic constants.

Constant	Water	Stiffmud	Shale	Sandstone	Limestone	Granite
$E\,\,\,({\times }10^{9}{\hbox{Pa}})$	0	1.4	9.9	16	54	50
$K\,\,\,({\times }10^{9}{\hbox{Pa}})$	2.1	3.2	13	17	36	28
${\mathrm {\mu } }\,\,\,({\times }10^{9}{\hbox{Pa}})$	0	0.47	3.6	6.0	22	21
${\mathrm {\lambda } }\,\,\,({\times }10^{9}{\hbox{Pa}})$	2.1	2.9	11	13	22	14
${\mathrm {\sigma } }$	0.50	0.43	0.38	0.34	0.25	0.20
${\mathrm {\mu } }\,\,\,({\hbox{g/cm}}^{3})$	1.0	1.5	1.8	1.9	2.5	2.7
${\mathrm {\alpha } }\,\,\,({\hbox{km/s}})$	1.5	1.6	3.2	3.6	5.1	4.5
${\mathrm {\beta } }\,\,\,({\hbox{km/s}})$	0	0.56	1.4	1.8	2.9	2.8

2.5 General solutions of the wave equation

2.5a Verify that $\psi =f\left(x-Vt\right)$ and $\psi =g\left(x+Vt\right)$ are solutions of the wave equation (2.5b).

Background

When unbalanced stresses act upon a medium, the strains are propagated throughout the medium according to the general wave equation

${\begin{aligned}{\nabla }^{2}\psi ={\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }x^{2}}}+{\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }y^{2}}}+{\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }z^{2}}}={\frac {1}{V^{2}}}{\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }t^{2}}},\end{aligned}}$

(2.5a)

$\psi$ being a disturbance such as a compression or rotation. $\psi$ is propagated with velocity $V$ (see Sheriff and Geldart, 1995, Section 2.2). The disturbance is the result of unbalanced normal stresses, shearing stresses, or a combination of both. When normal stresses create the wave, the result is a volume change and $\psi$ is the dilitation [see equation (2.1e)], and we get the P-wave equation, $V$ becoming the P-wave velocity ${\mathrm {\alpha } }$. Shearing stresses create rotation in the medium and $\psi$ is one of the components of the rotation given by equation (2.lg) ; the result is an S-wave traveling with velocity ${\mathrm {\beta } }$. Various expressions for ${\mathrm {\alpha } }$ and ${\mathrm {\beta } }$ in isotropic media are given in Table 2.2a.

In one dimension the wave equation (2.5a) reduces to

${\begin{aligned}{\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }x^{2}}}={\frac {1}{V^{2}}}{\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }t^{2}}}.\end{aligned}}$

(2.5b)

Solution

We use subscripts to denote partial derivatives and primes to denote derivatives with respect to the argument of the function. Then, writing $\zeta =\left(x-Vt\right)$, we have

${\begin{aligned}{{{\mathrm {\psi } }_{x}}={\frac {\partial {\mathrm {\psi } }}{\partial x}}=\left({\frac {df}{d{\mathrm {\zeta } }}}\right)\left({\frac {\partial {\mathrm {\zeta } }}{\partial x}}\right)={\frac {df}{d{\mathrm {\zeta } }}}=f',}\\{{{\mathrm {\psi } }_{xx}}=\left({\frac {df'}{d{\mathrm {\zeta } }}}\right)\left({\frac {\partial {\mathrm {\zeta } }}{\partial x}}\right)=f'',}\\{{{\mathrm {\psi } }_{t}}=\left({\frac {df}{d{\mathrm {\zeta } }}}\right)\left({\frac {\partial {\mathrm {\zeta } }}{\partial t}}\right)=-Vf',}\\{{{\mathrm {\psi } }_{tt}}=-V\left({\frac {df'}{d{\mathrm {\zeta } }}}\right)\left({\frac {\partial {\mathrm {\zeta } }}{\partial t}}\right)={V^{2}}f''}\end{aligned}}$

Substituting in equation (2.5b), we get the identity $f''=f''$ so $f\left(x-Vt\right)$ is a solution. We get the same result when $\psi =g\left(x+Vt\right)$. A sum of solutions is also a solution, so $f\left(x-Vt\right)+g\left(x+Vt\right)$ is a solution.

2.5b Verify that ${\psi =f\left(\ell x+my+nz-Vt\right)+g\left(\ell x+my+nz+Vt\right)}$ is a solution of equation (2.5a), where ${\left(\ell ,{\rm {\;}}m,\;n\right)}$ are direction cosines.

Solution

Let $\left(\ell x\;+my+nz-Vt\right)=\zeta ,\left(\ell x\;+my+nz+Vt\right)=\xi$. We now must show that $\psi =f\left(\zeta \right)+g\left(\xi \right)$ is a solution of equation (2.5a). Proceeding as before, we have

${\begin{aligned}{{\mathrm {\psi } }_{x}=\left({\frac {df}{d\zeta }}\right)\left({\frac {{\mathrm {\partial } }\zeta }{{\mathrm {\partial } }x}}\right)+\left({\frac {dg}{d\xi }}\right)\left({\frac {{\mathrm {\partial } }\xi }{{\mathrm {\partial } }x}}\right)=\ell (f'+g'),}\\{{\mathrm {\psi } }_{xx}=\ell ^{2}(f''+g'').}\end{aligned}}$

In the same way we get

${\begin{aligned}\psi _{yy}=m^{2}(f''+g''),\;{\quad }\psi _{zz}=n^{2}(f''+g'').\end{aligned}}$

But $\left(\ell ^{2}+m^{2}+n^{2}\right)=1$ (see Sheriff and Geldart, 1995, problem 15.9a), so $\left(\psi _{xx}+\psi _{yy}+\psi _{zz}\right)=(f''+g'')$.

Following the same procedure we find that $\left(1/V^{2}\right)\psi _{tt}=(f''+g'')$ thus verifying that $\psi =f\left(\ell x+my+nz-Vt\right)+g\left(\ell x+my+nz+Vt\right)$ is a solution of equation (2.5a).

2.5c Show that

${\begin{aligned}\psi \left(r,\;t\right)=\left(1/r\right)f\left(r-Vt\right)+\left(1/r\right)g\left(r+Vt\right)\end{aligned}}$

is a solution of the wave equation in spherical coordinates (see problem 2.6b) when the wave motion is independent of ${\theta }$ and $\phi$:

${\begin{aligned}{\frac {1}{V^{2}}}{\frac {{\mathrm {\partial } }^{2}{\mathrm {\psi } }}{{\mathrm {\partial } }t^{2}}}={\frac {1}{r^{2}}}\left[{\frac {\mathrm {\partial } }{{\mathrm {\partial } }r}}\left(r^{2}{\frac {{\mathrm {\partial } }{\mathrm {\psi } }}{{\mathrm {\partial } }r}}\right)\right].\end{aligned}}$

(2.5c)

Solution

The wave equation in spherical coordinates is given in problem 2.6b. When we drop the derivatives with respect to ${\mathrm {\theta } }$ and ${\mathrm {\phi } }$, the equation reduces to equation (2.5c). Writing ${\mathrm {\zeta } }=(r-Vt)$, we proceed as in part (a). Starting with the right-hand side, we ignore $g\left(r+Vt\right)$ for the time being and obtain

${\begin{aligned}{\mathrm {\psi } }_{r}=-\left(1/r^{2}\right)f\left({\mathrm {\zeta } }\right)+\left(1/r\right)f'\left({\mathrm {\zeta } }\right),\\r^{2}{\mathrm {\psi } }_{r}=-f\left({\mathrm {\zeta } }\right)+rf'\left({\mathrm {\zeta } }\right),\\{\frac {\mathrm {\partial } }{{\mathrm {\partial } }r}}\left(r^{2}{\mathrm {\psi } }_{r}\right)=-f'\left({\mathrm {\zeta } }\right)+f'\left({\mathrm {\zeta } }\right)+rf''\left({\mathrm {\zeta } }\right)=rf''\left({\mathrm {\zeta } }\right),\\\left(1/r^{2}\right){\frac {\mathrm {\partial } }{{\mathrm {\partial } }r}}\left(r^{2}{\mathrm {\psi } }_{r}\right)=\left(1/r\right)f''\left({\mathrm {\zeta } }\right),\\\psi _{t}=-\left(V/r\right)f';{\quad }{\mathrm {\phi } }_{tt}=\left(V^{2}/r\right)f''.\end{aligned}}$

Substitution in equation (2.5c) shows that $\left(1/r\right)f\left({\mathrm {\zeta } }\right)$ is a solution. In the same way we can show that $\left(1/r\right)g\left({\mathrm {\xi } }\right)$ is also a solution, hence the sum is a solution.

2.6 Wave equation in cylindrical and spherical coordinates

2.6a Show that the wave equation (2.5a) can be written in cylindrical coordinates (see Figure 2.6a as

${\begin{aligned}{\frac {\partial ^{2}\psi }{\partial r^{2}}}+{\frac {1}{r}}{\frac {\partial \psi }{\partial r}}+{\frac {1}{r^{2}}}{\frac {\partial ^{2}\psi }{\partial \theta ^{2}}}+{\frac {\partial ^{2}\psi }{\partial z^{2}}}={\frac {1}{V^{2}}}{\frac {\partial ^{2}\psi }{\partial t^{2}}}.\end{aligned}}$

(2.6a)

Solution

Figure 2.6a.  Cylindrical coordinates.

We shall solve by direct substitution. We have $x=r\cos {\mathrm {\theta } }$, $y=r\sin {\mathrm {\theta } }$, $z=z$, and $r^{2}=x^{2}+y^{2}$, ${\mathrm {\theta } }=\tan ^{-1}\left(y/x\right)$. The following solution is lengthy, so we use subscripts to denote partial derivatives and write

${\begin{aligned}a=\sin {\mathrm {\theta } }=y/r,{\mathrm {\quad } }b=\cos {\mathrm {\theta } }=x/r.\end{aligned}}$

We shall require the derivatives:

${\begin{aligned}{\mathrm {\partial } }r/{\mathrm {\partial } }x=r_{x}=x/r=\cos {\mathrm {\theta } }=b,\\{\mathrm {\partial } }r/{\mathrm {\partial } }y=r_{y}=y/r=\sin {\mathrm {\theta } }=a;\\{\mathrm {\partial } }\theta /{\mathrm {\partial } }x=\theta _{x}=\left[{\frac {\mathrm {\partial } }{{\mathrm {\partial } }x}}\left(y/x\right)\right]/[1+\left(y/x\right)^{2}]=-{\frac {y}{x^{2}}}\left({\frac {1}{1+({\frac {y}{x}})^{2}}}\right)\\=-\left(\sin \theta \right)/r=-a/r,\\{\mathrm {\partial } }\theta /{\mathrm {\partial } }y=\theta _{y}=\left[{\frac {\mathrm {\partial } }{{\mathrm {\partial } }y}}\left(y/x\right)\right]/[1+\left(y/x\right)^{2}]={\frac {1}{x}}\left({\frac {1}{1+({\frac {y}{x}})^{2}}}\right)\\=\left(\cos \theta \right)/r=b/r.\end{aligned}}$

To replace ${\mathrm {\psi } }_{xx}$ and ${\mathrm {\psi } }_{yy}$ with derivatives with respect to $r$ and ${\theta }$, we write:

${\begin{aligned}\psi _{x}&=\psi _{r}r_{x}+\psi _{\theta }\theta _{x}=\psi _{r}b-\psi _{\theta }a/r,\\\psi _{y}&=\psi _{r}r_{y}+\psi _{\theta }\theta _{y}=\psi _{r}a+\psi _{\theta }b/r.\end{aligned}}$

Then,

${\begin{aligned}{\mathrm {\psi } }_{xx}={\frac {\mathrm {\partial } }{{\mathrm {\partial } }r}}\left({\mathrm {\psi } }_{r}b-{\mathrm {\psi } }_{\mathrm {\mathrm {\theta } } }a/r\right)r_{x}+{\frac {\mathrm {\partial } }{{\mathrm {\partial } }{\mathrm {\theta } }}}\left({\mathrm {\psi } }_{r}b-{\mathrm {\psi } }_{\mathrm {\theta } }a/r\right)\theta _{x}\\=\left({\mathrm {\psi } }_{rr}b-{\mathrm {\psi } }_{r{\mathrm {\theta } }}a/r+{\mathrm {\psi } }_{\mathrm {\mathrm {\theta } } }a/r^{2}\right)b\\{\quad }+\left({\mathrm {\psi } }_{r{\mathrm {\theta } }}b-{\mathrm {\psi } }_{r}a-{\mathrm {\psi } }_{\theta \theta }a/r-{\mathrm {\psi } }_{\mathrm {\theta } }b/r\right)\left(-a/r\right)\\=[{\mathrm {\psi } }_{rr}b^{2}-{\mathrm {\psi } }_{r{\mathrm {\theta } }}\left(2ab/r\right)+{\mathrm {\psi } }_{\mathrm {\theta } }\left(2ab/r^{2}\right)+{\mathrm {\psi } }_{r}\left(a^{2}/r\right)+{\mathrm {\psi } }_{\theta \theta }\left(a/r)^{2}\right],\\{\mathrm {\psi } }_{yy}={\frac {\mathrm {\partial } }{{\mathrm {\partial } }r}}\left({\mathrm {\psi } }_{r}a+{\mathrm {\psi } }_{\mathrm {\theta } }b/r\right)r_{y}+{\frac {\mathrm {\partial } }{{\mathrm {\partial } }{\mathrm {\theta } }}}\left({\mathrm {\psi } }_{r}a+{\mathrm {\psi } }_{\mathrm {\theta } }b/r\right)\theta _{y}\\=\left({\mathrm {\psi } }_{rr}a+{\mathrm {\psi } }_{r{\mathrm {\theta } }}b/r-{\mathrm {\psi } }_{\mathrm {\mathrm {\theta } } },b/r^{2}\right)a\\{\quad }+\left({\mathrm {\psi } }_{r{\mathrm {\theta } }}a+\psi _{r}b+\psi _{\theta \theta }b/r-\psi _{\mathrm {\theta } }a/r\right)\left(b/r\right)\\=\left({\mathrm {\psi } }_{rr}a^{2}+\psi _{r{\mathrm {\theta } }}2ab/r-{\mathrm {\psi } }_{\mathrm {\theta } }2ab/r^{2}\right)+\left({\mathrm {\psi } }_{r}b^{2}/r+{\mathrm {\psi } }_{\theta \theta }b^{2}r^{2}\right).\end{aligned}}$

Thus

${\begin{aligned}{\nabla }^{2}{\mathrm {\psi } }={\mathrm {\psi } }_{rr}\left(a^{2}+b^{2}\right)+{\mathrm {\psi } }_{r}\left(a^{2}+b^{2}\right)/r+{\mathrm {\psi } }_{\theta \theta }\left(a^{2}+b^{2}\right)/r^{2}+{\mathrm {\psi } }_{zz},\end{aligned}}$

so

${\frac {{\mathrm {\partial } }^{2}{\mathrm {\psi } }}{{\mathrm {\partial } }r^{2}}}+{\frac {1}{r}}{\frac {{\mathrm {\partial } }{\mathrm {\psi } }}{{\mathrm {\partial } }r}}+{\frac {1}{r^{2}}}{\frac {{\mathrm {\partial } }^{2}{\mathrm {\psi } }}{{\mathrm {\partial } }{\mathrm {\theta } }^{2}}}+{\frac {{\mathrm {\partial } }^{2}{\mathrm {\psi } }}{{\mathrm {\partial } }z^{2}}}={\frac {1}{V^{2}}}{\frac {{\mathrm {\partial } }^{2}{\mathrm {\psi } }}{{\mathrm {\partial } }t^{2}}}.$

2.6b Transform the wave equation into spherical coordinates (see Figure 2.6b), showing that it becomes

${\begin{aligned}{\frac {1}{r^{2}}}\left[{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial \psi }{\partial r}}\right)+{\frac {1}{\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial \psi }{\partial \theta }}\right)+{\frac {1}{{\sin }^{2}\theta }}{\frac {\partial ^{2}\psi }{\partial \phi ^{2}}}\right]={\frac {1}{V^{2}}}{\frac {\partial ^{2}\psi }{\partial t^{2}}}.\end{aligned}}$

(2.6b)

Solution

Spherical coordinates $\left(r,\;\theta ,\;\phi \right)$ and rectangular coordinates are related as follows (see Figure 2.6b):

${\begin{aligned}x=r\sin \theta \cos \phi ,\\y=r\sin \theta \sin \phi ,\\z=r\cos \theta ,\\r=(x^{2}+y^{2}+z^{2})^{1/2},\\\theta ={\rm {cos}}^{-1}\left(z/r\right),\\\phi =\tan ^{-1}\left(y/x\right).\end{aligned}}$

We continue to use subscripts to denote derivatives and letters to denote sines and cosines:

${\begin{aligned}{\begin{array}{l}a=\sin \theta ,\\m=\sin \phi ,\\a_{\mathrm {\theta } }=b,\\m_{\phi }=n,\end{array}}{\quad }{\begin{array}{l}b=\cos \theta ,\\n=\cos \phi ,\\b_{\mathrm {\theta } }=-a,\\n_{\phi }=-m.\end{array}}\end{aligned}}$

The derivatives of $r$, ${\theta }$, and $\phi$ now become:

${\begin{aligned}r_{x}&=x/r=an,\\r_{y}&=y/r=am,\\r_{z}&=z/r=b;\\\theta _{x}&={\frac {\partial }{\partial x}}[{\rm {cos}}^{-1}\left(z/r\right)]=-{\frac {\partial }{\partial r}}\left(z/r\right)[1-(z/r)^{2}]^{-1/2}={\frac {Z}{r^{2}}}r_{x}\left(1/a\right)=bn/r,\\\theta _{y}&=-{\frac {\partial }{\partial y}}\left(z/r\right)[1-(z/r)^{2}]^{-1/2}=-{\frac {Z}{r^{2}}}r_{y}\left(1/a\right)=bm/r,\\\theta _{z}&=-{\frac {\partial }{\partial z}}\left(z/r\right)[1-(z/r)^{2}]^{-1/2}=-\left(1/r-z/r^{2}r_{z}\right)[1-(z/r)^{2}]^{-1/2}\\&=\left(-1/r\right)\left(1-zb/r\right)\left(1/a\right)=\left(-1/r\right)\left(1-b^{2}\right)\left(1/a\right)=-a/r,\\\phi _{x}&={\frac {\partial }{\partial x}}[\tan ^{-1}\left(y/x\right)]={\frac {\partial }{\partial x}}\left(y/x\right)[1+(y/x)^{2}]^{-1}=-\left(y/x^{2}\right)[1+(y/x)^{2}]^{-1}\\&=-\left(y/x\right)\left(1/x\right)[1+(y/x)^{2}]^{-1}=-\left(m/n\right)\left(1/ran\right)(1/n^{2})^{-1}=-m/ar,\\\phi _{y}&={\frac {\partial }{\partial y}}\left(y/x\right)[1+(y/x)^{2}]^{-1}=\left(1/x\right)[1+(y/x)^{2}]^{-1}=\left(1/ran\right)(1/n^{2})^{-1}=n/ar,\\\phi _{z}&=0\ {{\text{(because }}\phi {\text{ is not a function of }}z)}.\end{aligned}}$

Figure 2.6b  Spherical coordinates.

Summarizing these results, we have

${\begin{aligned}{\begin{array}{l}r_{x}=an,\\\theta _{x}=bn/r,\\\phi _{x}=-m/ar,\end{array}}\quad {\begin{array}{l}r_{y}=am,\\\theta _{y}=bm/r,\\\phi _{y}=n/ar,\end{array}}\quad {\begin{array}{l}r_{z}=b,\\\theta _{z}=-a/r,\\\phi _{z}=0.\end{array}}\end{aligned}}$

We now calculate the derivatives $\phi _{xx}$, etc.:

${\begin{aligned}\psi _{x}&=\psi _{r}r_{x}+\psi _{\theta }\theta _{x}+\psi _{\phi }\phi _{x}\\&=\psi _{r}an+\psi _{\theta }bn/r-\psi _{\phi }m/ar;\\\psi _{xx}&={\frac {\partial }{\partial r}}\left(\psi _{r}an+\psi _{\theta }bn/r-\psi _{\phi }m/ar\right)\left(an\right)\\&\quad +{\frac {\partial }{\partial \theta }}\left(\psi _{r}an+\psi _{\theta }bn/r-\psi _{\phi }m/ar\right)\left(bn/r\right)\\&\quad +{\frac {\partial }{\partial \theta }}\left(\psi _{r}an+\psi _{\theta }bn/r-\psi _{\phi }m/ar\right)\left(-m/ar\right),\\&=\left(\psi _{rr}an+\psi _{r\theta }bn/r-\psi _{\theta }bn/r^{2}-\psi _{\gamma \phi }m/ar+\psi _{\phi }m/ar^{2}\right)an\\&\quad +(\psi _{r\theta }an+\psi _{r}bn+\psi _{\theta \theta }bn/r-\psi _{\theta }an/r-\psi _{\theta \phi }m/ar\\&\quad +\psi _{\phi }mb/a^{2}r)\left(bn/r\right)+(\psi _{\gamma \phi }an-\psi _{r}am+\psi _{\theta \phi }bn/r\\&\quad -\psi _{\theta }bm/r-\psi _{\phi \phi }m/ar-\psi _{\phi }n/ar)\left(-m/ar\right)\\&=\psi _{rr}a^{2}n^{2}+\psi _{r\theta }\left(2abn^{2}/r\right)-\psi _{\gamma \phi }\left(2mn/r\right)+\psi _{r}\left({\frac {b^{2}n^{2}+m^{2}}{r}}\right)\\&\quad +\psi _{\theta \theta }\;\left({\frac {bn}{r}}\right)^{2}+\psi _{\theta \phi }\left({\frac {b}{a}}\right)\left({\frac {2mn}{r}}\right)+\psi _{\theta }\;\left({\frac {bm^{2}}{ar^{2}}}-{\frac {2abn}{r^{2}}}\right)\\&\quad +\psi _{\phi \phi }\left({\frac {m}{ar}}\right)^{2}+\psi _{\phi }\left({\frac {2mn}{a^{2}r^{2}}}\right);\\\psi _{y}&=\psi _{r}r_{y}+\psi _{\theta }\theta _{y}+\psi _{\phi }\phi _{y}\\&=\psi _{r}am+\psi _{\theta }bm/r+\psi _{\phi }n/ar;\\\psi _{yy}&={\frac {\partial }{\partial r}}\left(\psi _{r}am+\psi _{\theta }bm/r+\psi _{\phi }n/ar\right)\left(am\right)\\&\quad +{\frac {\partial }{\partial \theta }}\left(\psi _{r}am+\psi _{\theta }bm/r+\psi _{\phi }n/ar\right)\left(bm/r\right)\\&\quad +{\frac {\partial }{\partial \phi }}\left(\psi _{r}am+\psi _{\theta }bm/r+\psi _{\phi }n/ar\right)\left(n/ar\right)\\&=\left(\psi _{rr}am+\psi _{r\theta }bm/r-\psi _{\theta }bm/r^{2}+\psi _{\gamma \phi }n/ar-\psi _{\phi }n/ar^{2}\right)am\\&\quad +(\psi _{r\theta }am+\psi _{r}bm+\psi _{\theta \theta }bm/r-\psi _{\theta }am/r+\psi _{\theta \phi }n/ar\\&\quad -\psi _{\phi }bn/a^{2}r)\left(bm/r\right)\\&\quad +(\psi _{\gamma \phi }am+\psi _{r}an+\psi _{\theta \phi }bm/r+\psi _{\theta }bn/r+\psi _{\phi \phi }n/ar\\&\quad -\psi _{\phi }m/ar)\left(n/ar\right)\\&=\psi _{rr}a^{2}m^{2}+\psi _{r\theta }\;\left({\frac {2abm}{r}}\right)+\psi _{\gamma \phi }\left({\frac {2mn}{r}}\right)+\psi _{r}\left({\frac {1-a^{2}m^{2}}{r}}\right)\\&\quad +\psi _{\theta \theta }\;\left({\frac {bm}{r}}\right)^{2}+\psi _{\theta \phi }\left({\frac {b}{a}}\right)\left({\frac {2mn}{r^{2}}}\right)+\psi _{\theta }\;\left(-{\frac {2abm^{2}}{r}}+{\frac {bn^{2}}{ar^{2}}}\right)\\&\quad +\psi _{\phi \phi }\left({\frac {n}{ar}}\right)^{2_{-\psi }}\phi \left({\frac {mn}{r^{2}}}\right)\left(1+{\frac {b^{2}}{a^{2}}}+{\frac {1}{a^{2}}}\right);\\\psi _{z}&=\psi _{r}r_{z}+\psi _{\theta }\theta _{z}+\psi _{\phi }\phi _{z}=\psi _{r}b-\psi _{\theta }a/r;\\\psi _{zz}&=\left(\psi _{rr}b-\psi _{r\theta }a/r+\psi _{\theta }a/r^{2}\right)b\\&\quad +\left(\psi _{r\theta }b-\psi _{r}a-\psi _{\theta \theta }a/r-\psi _{\theta }b/r\right)\left(-a/r\right)\\&=[\psi _{rr}b^{2}-\psi _{r\theta }\left(2ab/r\right)+\psi _{r}\left(a^{2}/r\right)+\psi _{\theta \theta }\left(a/r)^{2}+\psi _{\theta }\left(2ab/r^{2}\right)\right].\end{aligned}}$

Adding the three derivatives, we get

${\begin{aligned}\nabla ^{2}\psi &=\psi _{rr}\left(a^{2}m^{2}+a^{2}n^{2}+b^{2}\right)+\psi _{r\theta }\;\left[{\frac {2ab\left(m^{2}+n^{2}\right)}{r}}-{\frac {2ab}{r}}\right]\\&\quad +\psi _{r}\left[{\frac {\left(b^{2}n^{2}+m^{2}\right)+\left(1-a^{2}m^{2}\right)+a^{2}}{r}}\right]\\&\quad +\psi _{\theta \theta }\;\left({\frac {b^{2}n^{2}+b^{2}m^{2}+a^{2}}{r^{2}}}\right)\\&\quad +\psi _{\theta }\;\left[{\frac {\left(bm^{2}/a-2abm^{2}\right)+\left(-2abm^{2}+bn^{2}/a+2ab\right)}{r}}\right]\\&\quad +\psi _{\phi \phi }\left[(-m/ar)^{2}+\left(n/ar\right)^{2}\right]\\&=\psi _{rr}+\left({\frac {2}{r}}\right)\psi _{r}+\left({\frac {1}{r^{2}}}\right)\psi _{\theta \theta }\;+\left({\frac {\cot \theta }{r^{2}}}\right)\psi _{\theta }\;+\left({\frac {1}{a^{2}r^{2}}}\right)\psi _{\phi \phi }.\end{aligned}}$

Substituting the values of $a$, $b$, $m$, and $n$, we get for the wave equation

${\begin{aligned}{\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }r^{2}}}+{\frac {2}{r}}{\frac {{\mathrm {\partial } }\psi }{{\mathrm {\partial } }r}}+{\frac {1}{r^{2}}}{\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }\theta ^{2}}}+\left({\frac {{\rm {\;cot\;}}\theta }{r^{2}}}\right){\frac {{\mathrm {\partial } }\psi }{{\mathrm {\partial } }\theta }}+\left({\frac {1}{r^{2}{\rm {sin}}^{2}\theta }}\right){\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }\phi ^{2}}}={\frac {1}{V^{2}}}{\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }t^{2}}}.\end{aligned}}$

This is often written in the more compact form

${\begin{aligned}{\frac {1}{r^{2}}}\left[{\frac {\mathrm {\partial } }{{\mathrm {\partial } }r}}\left(r^{2}{\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }r}}\right)+{\frac {1}{\sin \theta }}{\frac {\mathrm {\partial } }{{\mathrm {\partial } }\theta }}\left(\sin \theta {\frac {{\mathrm {\partial } }\psi }{{\mathrm {\partial } }\theta }}\right)+{\frac {1}{{\rm {sin}}^{2}\theta }}{\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }\phi ^{2}}}\right]={\frac {1}{V^{2}}}{\frac {{\mathrm {\partial } }^{2}\psi }{{\mathrm {\partial } }t^{2}}}.\end{aligned}}$

2.7 Sum of waves of different frequencies; Group velocity

2.7a A pulse composed of two frequencies, $\omega _{0}\pm {\mathrm {\Delta } }\omega$, can be represented by factors involving the sum and difference of the two frequencies. If the two components have the same amplitudes, we can write

${\begin{aligned}A\cos \left(\kappa _{1}x-\omega _{1}t\right),{\quad }A\cos \left(\kappa _{2}x-\omega _{2}t\right),\end{aligned}}$

where $\omega _{1}=\omega _{0}+\Delta \omega ,\omega _{2}=\omega _{0}-\Delta \omega ,k_{0}=2\pi /\lambda _{0}=\omega _{0}/V$,

${\begin{aligned}\kappa _{1}&\approx \kappa _{0}+\Delta \kappa \approx \left(\omega _{0}+\Delta \omega \right)/V,\ \mathrm {and} \\\kappa _{2}&\approx \kappa _{0}-\Delta \kappa \approx \left(\omega _{0}-\Delta \omega \right)/V.\end{aligned}}$

Show that the composite wave is given approximately by the expression

${\begin{aligned}B\cos(\kappa _{0}x-\omega _{0}t),\end{aligned}}$

where $B=2A\cos \left\{{\mathrm {\Delta } }\kappa \left[x-\left({\mathrm {\Delta } }\omega /{\mathrm {\Delta } }\kappa \right)t\right]\right\}$.

Background

When different frequency components in a pulse have different phase velocities $V$ (the velocity with which a given frequency travels), the pulse changes shape as it moves along. The group velocity $U$ is the velocity with which the envelope of the pulse travels.

The envelope of a pulse comprises two mirror-image curves that are tangent to the waveform at the peaks and troughs, and therefore define the general shape of the pulse.

Solution

Adding the two components and using the identity

${\begin{aligned}\cos \theta +\cos \phi =2\cos \left[{\frac {1}{2}}\left(\theta +\phi \right)\right]\cos \left[{\frac {1}{2}}\left(\theta -\phi \right)\right],\end{aligned}}$

we get for the composite wave

${\begin{aligned}2A\cos \left\{{\mathrm {\Delta } }\kappa \left[x-\left({\mathrm {\Delta } }\omega /{\mathrm {\Delta } }\kappa \right)t\right]\right\}\cos \left(\kappa _{0}x-\omega _{0}t\right)=B\cos \left(\kappa _{0}x-\omega _{0}t\right).\end{aligned}}$

2.7b Why do we regard $B$ as the amplitude? Show that the envelope of the pulse is the graph of $B$ plus its reflection in the $x$-axis.

Solution

The solution in (a) can be written $\psi =B\psi _{0}$. We regard $B$ as the amplitude for two reasons: (1) $B$ repeats every time that $x$ increases by $2\pi /{\mathrm {\Delta } }\kappa$ or each time that $t$ increase by $2\pi /{\mathrm {\Delta } }\omega$. But ${\mathrm {\Delta } }\kappa \ll \kappa _{0}$, ${\mathrm {\Delta } }\omega \ll \omega _{0}\}$, so $B$ must repeat more slowly than $\psi _{0}$. (2) Each time that $\psi _{0}$ attains its limiting values, $\pm 1$, $\psi$ has the value $\pm B$ and therefore never exceeds $\left|B\right|$; thus the curves of $+\left|B\right|$ and $-\left|B\right|$ pass through the maxima and minima of $\psi _{0}$ and therefore constitute the envelope.

2.7c Show that the envelope moves with the group velocity $U$ where

${\begin{aligned}U={\frac {{\mathrm {\Delta } }\omega }{{\mathrm {\Delta } }\kappa }}\approx {\frac {\mathrm {d} \omega }{\mathrm {d} \kappa }}\approx V-{\mathrm {\lambda } }{\frac {\mathrm {d} V}{\mathrm {d} {\mathrm {\lambda } }}}=V+\omega {\frac {\mathrm {d} V}{\mathrm {d} \omega }}\end{aligned}}$

(2.7a)

(see Figure 2.7a).

Solution

If we consider the quantity $B$ as a wave superimposed on the primary wavelet $\cos \left(\kappa _{0}x-\omega _{0}t\right)$, comparison with the basic wave type $f(x-Vt)$ of problem 2.5a shows that $\left[x-\left({\mathrm {\Delta } }\omega /{\mathrm {\Delta } }\kappa \right)t\right]$ takes the place of $(x-Vt)$, i.e., $\left({\mathrm {\Delta } }\omega /{\mathrm {\Delta } }\kappa \right)$ is the velocity $U$ with which the envelope travels. In the limit, $U$ is given by

$\begin{array}{r}U=\mathrm{d}\mathrm{\&\#x}\end{array}$