x
Fig. 7.11. Deriving Eq. (99).
As this sketch shows, the elementary area ratios participating in Eq. (96) may be readily expressed via the derivatives of these functions: dAx/ dA = sin = dy/ dl , dAy/ dA = cos = – dx/ dl, so that we may write
dy
dx
y
x
.
0
(7.97)
x
dl
y
dl border
Introducing, instead of , a new function ( x,y), defined by its derivatives as
1
1
x
,
y
,
(7.98)
x
2
y
y
2
x
we may rewrite Eq. (97) as
dy dx
d
2
2
,
0
(7.99)
border
y dl
x dl
dl
border
so that the function has to be constant at each border of the cross-section.
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In particular, for a singly-connected cross-section, limited to just one continuous border line (as in Fig. 11), this constant is arbitrary, because according to Eqs. (98), its choice does not affect the longitudinal deformation function ( x,y) and hence the deformation as a whole. Now let us use the definition (98) of ( x, y) to calculate the 2D Laplace operator of this function: 2
2
2
1
1
x
y
.
1
(7.100)
x, y
2
2
x
y
2
x
y 2
y
y
This is a 2D Poisson equation (frequently met, for example, in electrostatics), but with a very simple, constant right-hand side. Plugging Eqs. (98) into Eqs. (95), and those into Eq. (88), we may express the torque z, and hence the torsional rigidity C, via the same function:
C for
C
z
2
x
y
.
dxdy
arbitrary
(7.101a)
cross-
x
y
A
section
Sometimes, it is easier to use this result in either of its two different forms. The first of them may be readily obtained from Eq. (101a) using the integration by parts:
C 2
dy xd
dx yd 2
dy x
dx
dx y
dy
border
border
(7.101b)
4 dxdy
dxdy,
border
A
A
while the proof of one more form,
2
C 4
,
(7.101c)
x, y
dxdy
A
is left for the reader’s exercise. Thus, if we need to know the rod’s rigidity alone, it is sufficient to calculate the function ( x, y) from Eq. (100) with the boundary condition border = const, and then plug it into any of Eqs. (101). Only if we are also curious about the longitudinal deformation (93) of the cross-section, we may continue by using Eq. (98) to find the function ( x,y) describing this deformation.
Let us see how does this recipe work for the two examples discussed above. For the round cross-section of radius R, both the Poisson equation (100) and the boundary condition, = const at x 2 + y 2 =
R 2, are evidently satisfied by the following axially-symmetric function:
1
( 2
2
x y )
.
const
(7.102)
4
For this case, Eq. (101a) yields
1 2 1 2
C 4 x y dxdy
2
2
x y 2
d r,
(7.103)
2
2
A
A
i.e. the same result (89) that we had for = 0. Indeed, plugging Eq. (102) into Eqs. (98), we see that in this case / x = / y = 0, so that ( x,y) = const, i.e. the cross-section is not bent. (As was discussed in Sec. 1, a uniform translation dqz = = const does not constitute a deformation.) Now, turning to a rod with a narrow rectangular cross-section A = w t with t << w, we may use this strong inequality to solve the Poisson equation (100) approximately, neglecting the second Chapter 7
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derivative of along the wider dimension (say, y). The remaining 1D differential equation d 2/ d 2 x = –1, with boundary conditions x=+ t/2 = x =- t/2, has an obvious solution: = – x 2/2 + const. Plugging this expression into any form of Eq. (101), we get the following (correct!) result for the torsional rigidity: 1
3
C
wt
.
(7.104)
3
Now let us have a look at the cross-section bending law (93) for this particular case. Using Eqs.
(98), we get
x 2
x,
y 2
.
y
(7.105)
y
x
x
y
Integrating these differential equations over the cross-section, and taking the integration constant (again, not contributing to the deformation) for zero, we get a beautifully simple result:
xy,
q
i.e.
xy
.
(7.106)
z
It means that the longitudinal deformation of the rod has a “propeller bending” form: while the regions near the opposite corners (on the same diagonal) of the cross-section bend toward one direction of the z-
axis, the corners on the other diagonal bend in the opposite direction. (This qualitative conclusion remains valid for rectangular cross-sections with any “aspect ratio” t/ w.) For rods with several surfaces, i.e. with cross-sections limited by several boundaries (say, hollow pipes), finding the function ( x, y) requires a bit more care, and Eq. (103b) has to be modified, because the function may be equal to a different constant at each boundary. Let me leave the calculation of the torsional rigidity for this case for the reader’s exercise.
7.7. 3D acoustic waves
Now moving from the statics to dynamics, we may start with Eq. (24), which may be transformed into the vector form exactly as this was done for the static case at the beginning of Sec. 4.
Comparing Eqs. (24) and (52), we immediately see that the result may be represented as
2
Elastic
q
E
E
continuum:
2
q
( q) f ( ,t
r ) .
(7.107)
2
dynamics
t
2 1
( )
2 1
( 1
)( 2 )
Let us use this general equation for the analysis of the perhaps most important type of time-dependent deformations: acoustic waves. First, let us consider the simplest case of a virtually infinite, uniform elastic medium, with no external forces: f = 0. In this case, due to the linearity and homogeneity of the equation of motion, and taking clues from the analysis of the simple 1D model (see Fig. 6.4a) in Secs. 6.3-6.5,32 we may look for a particular time-dependent solution in the form of a sinusoidal, linearly-polarized, plane traveling wave
Plane,
sinusoidal
ik r
t
q r
( , t)
Re
wave
e
a
,
(7.108)
32 Note though that Eq. (107) is more complex than the simple wave equation (6.40).
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where a is the constant complex amplitude of a wave (now a vector!), and k is the wave vector, whose magnitude is equal to the wave number k. The direction of these two vectors should be clearly distinguished: while a determines the wave’s polarization, i.e. the direction of particle displacements, the vector k is directed along the spatial gradient of the full phase of the wave Ψ k r t arg a ,
(7.109)
i.e. along the direction of the wave front propagation.
The importance of the angle between these two vectors may be readily seen from the following simple calculation. Let us point the z-axis of an (inertial) reference frame along the direction of vector k, and the x-axis in such direction that the vector q, and hence a, lie within the { x, z} plane. In this case, all variables may change only along the z-axis, i.e. = n z(/ z), and the amplitude vector may be represented as the sum of just two Cartesian components:
a a n a n .
(7.110)
x
x
z
z
Let us first consider a longitudinal wave, with the particle motion along the wave direction: ax =
0, az = a. Then the vector q in Eq. (107) describing this wave, has only one ( z-) component, so that q
= qz/ z and (q) = n z(2q/ z 2), and the Laplace operator gives the same expression: 2q =
n z(2q/ z 2). As a result, Eq. (107), with f = 0, is reduced to a 1D wave equation 2
2
2
q
E
E
q
E 1
( )
q
z
z
z
,
(7.111)
2
2
2
t
1
(
2 )
1
(
2 1
)( 2 ) z
1
( 1
)( 2 ) z
similar to Eq. (6.40). As we already know from Sec. 6.4, this equation is indeed satisfied with the solution (108), provided that and k obey a linear dispersion relation, = v l k, with the following longitudinal wave velocity:
E
K
2
1
(
)
(4 / )
3
Longitudinal
v
.
(7.112)
l
waves:
1
( 1
)( 2 )
velocity
The last expression allows for a simple interpretation. Let us consider a static experiment, similar to the tensile test experiment shown in Fig. 6, but with a sample much wider than l in both directions perpendicular to the force. Then the lateral contraction is impossible ( sxx = syy = 0), and we can calculate the only finite stress element, zz, directly from Eq. (34) with Tr (s) = szz:
1
1
4
2 s s 3 K s K s .
(7.113)
zz
zz
3 zz
3 zz
3
zz
We see that the numerator in Eq. (112) is nothing more than the static elastic modulus for such a uniaxial deformation, and it is recalculated into the velocity exactly as the spring constant in the 1D
waves considered in Secs. 6.3-6.4 – cf. Eq. (6.42).
Formula (114) becomes especially simple in fluids, where = 0, and the wave velocity is described by the well-known expression
1/ 2
K
Longitudinal
v
.
(7.114) waves:
l
velocity
in fluids
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Note, however, that for gases, with their high compressibility and temperature sensitivity, the value of K
participating in this formula may differ, at high frequencies, from that given by Eq. (40), because fast compressions/extensions of gas are usually adiabatic rather than isothermal. This difference is noticeable in Table 1, one of whose columns lists the values of vl for representative materials.
Now let us consider an opposite case of transverse waves with ax = a, az = 0. In such a wave, the displacement vector is perpendicular to n z, so that q = 0, and the second term on the right-hand side of Eq. (107) vanishes. On the contrary, the Laplace operator acting on such vector still gives the same nonzero contribution, 2q = nz(2q/ z 2), to Eq. (107) so that the equation yields 2
2
q
x
E
q
x
,
(7.115)
2
t
2 1
( )
2
z
and we again get the linear dispersion relation, = v t k, but with a different velocity: 33
Transverse
waves:
2
E
v
.
(7.116)
t
velocity
1
(
2 )
We see that the speed of the transverse waves depends exclusively on the shear modulus of the medium.34 This is also very natural: in such waves, the particle displacements q = n xq are perpendicular to the elastic forces dF = nz dF, so that the only one element xz of the stress tensor is involved. Also, the strain tensor sjj’ has no diagonal elements, Tr (s) = 0, so that is the only elastic modulus actively participating in Hooke’s law (32). In particular, fluids cannot carry transverse waves at all (formally, their velocity (116) vanishes), because they do not resist shear deformations. For all other materials, the longitudinal waves are faster than the transverse ones.35 Indeed, for all known natural materials’
Poisson’s ratio is positive so that the velocity ratio that follows from Eqs. (112) and (116), 1/ 2
v
2 2
l
,
(7.117)
v
1 2
t
is above 2 1.4. For the most popular construction materials, with 0.3, Poisson’s ratio is about 2 –
see Table 1.
Let me emphasize again that for both the longitudinal and the transverse waves, the dispersion relation between the wave number and frequency is linear: = vk. As was already discussed in Chapter 6, in this case of acoustic waves (or just “sound”), the phase and group velocities are equal, and waves of more complex form, consisting of several (or many) Fourier components of the type (108), preserve 33 Just as in Chapter 6, let me emphasize that the wave velocities we are discussing in this section and Sec. 8
below have nothing to do with particle velocities q/ t. For example, in the transverse wave we are discussing now, v t is the velocity in the z-direction, while the particles of the medium move across it, in the x-direction. Also, v l and v t do not depend on the wave amplitudes, while the particle velocities are proportional to them.
34 Because of that, one can frequently meet the term shear waves. Note also that in contrast to the transverse waves in the simple 1D model analyzed in Chapter 6 (see Fig. 6.4a), those in a 3D continuum do not need a pre-stretch tension T. We will return to the effect of tension in the next section.
35 Because of this difference between v l and v t, in geophysics, the longitudinal waves are known as P-waves (with the letter P standing for “primary”) because they arrive at the detection site, say from an earthquake, first – before the transverse waves, called the S-waves, with S standing for “secondary”.
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their form during propagation. This means that both Eqs. (111) and (115) are satisfied by solutions of the type (6.41):
z
q ( z, t)
,
(7.118)
f t
v
where the functions f describe the propagating waveforms. (However, if the initial wave is a mixture, of the type (110), of the longitudinal and transverse components, then these components, propagating with different velocities, will “run from each other”.) As one may infer from the analysis of a periodic system model in Chapter 6, the wave dispersion becomes essential at very high ( hypersound) frequencies where the wave number k becomes close to the reciprocal distance d between the particles of the medium (e.g., atoms or molecules), and hence the approximation of the medium as a continuum, used through this chapter, becomes invalid.
As we already know from Chapter 6, besides the velocity, the waves of each type are characterized by one more important parameter, the wave impedance Z – for acoustic waves frequently called the acoustic impedance of the medium. Generalizing Eq. (6.46) to the 3D case, we may define the impedance as the ratio of the force per unit area (i.e. the corresponding element of the stress tensor) exerted by the wave, and the particles’ velocity. For the longitudinal waves,
s
q
/ z
Z
zz
zz
zz
zz
z
.
(7.119)
l
q
/ t
s
q
/ t
s
q
/ t
z
zz
z
zz
z
Plugging in Eqs. (108), (112), and (113), we get
Longitudinal
Z K
(7.120)
l
(
4 / )
3 1/2 ,
waves:
impedance
in a clear analogy with the first of Eqs. (6.48). Similarly, for the transverse waves, the appropriately modified definition, Z t xz/( qx/ z), yields Transverse
Z
(7.121) waves:
t
1/2.
impedance
Just like in the 1D models studied in Chapter 6, one role of the wave impedance is to scale the power P carried by the wave. For plane 3D waves in infinite media, with their infinite wave front area, it is more appropriate to speak about the power density, i.e. power p = d P/ dA per unit area of the front, and characterize it by not only its magnitude,
dF q
p
,
(7.122)
dA t
but also the direction of the energy propagation, that (for a plane acoustic wave in an isotropic medium) coincides with the direction of the wave vector: p pnk. Using the definition (18) of the stress tensor, the Cartesian components of this Umov vector 36 may be expressed as
36 Named after N. A. Umov, who introduced this concept in 1874 – ten years before a similar notion for electromagnetic waves (see, e.g., EM Sec. 6.4) was suggested by J. Poynting. In a dissipation-free elastic medium, the Umov vector obeys the continuity equation ( v 2/2 + u)/ t + p = 0, with u given by Eq. (52), which expresses the conservation of the total (kinetic plus potential) energy of the elastic deformation.
Chapter 7
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q
p
.
(7.123)
j
j'
jj'
j'
t
Returning to plane waves propagating along axis z, and acting exactly like in Sec. 6., for both the longitudinal and transverse waves we again arrive at Eq. (6.49), but for p rather than P (due to a different definition of the wave impedance – per unit area rather than per particle chain). For the sinusoidal waves of the type (108), it yields
2
Z
*
p
aa ,
(7.124)
z
2
with Z being the corresponding impedance – either Z l or Z t.
Just as in the 1D case, one more important effect, in which the notion of impedance is crucial, is the partial wave reflection from at an interface between two media. The two boundary conditions, necessary for the analysis of the reflection, may be obtained from the continuity of the vectors q and dF.
(The former condition is evident, while the latter one may be obtained by applying the 2nd Newton law to any infinitesimal volume dV = dAdz, where the segment dz straddles the interface.) Let us start from the simplest case of the normal incidence on a plane interface between two uniform media, each with its own elastic moduli and mass density. Due to the symmetry of the system, it is obvious that the longitudinal/transverse incident wave may only excite similarly polarized reflected and transferred waves. As a result, we may literally repeat the calculations of Sec. 6.4, again arriving at the fundamental relations (6.55) and (6.56), with the replacement of Z and Z’ with the corresponding values of either Z l (120) or Z t (121). Thus, at the normal incidence, the wave reflection is determined solely by the acoustic impedances of the media, while the sound velocities are not involved.
The situation, however, becomes more complicated at a nonzero incidence angle (i) (Fig. 12), where the transmitted wave is generally also refracted, i.e. propagates under a different angle, ’ (i), beyond the interface. Moreover, at (i) 0 the directions of particle motion (vector q) and of the stress forces (vector dF) in the incident wave are neither exactly parallel nor exactly perpendicular to the interface, and thus this wave may serve as an actuator for the reflected and refracted waves of both polarizations – see Fig. 12, drawn for the particular case when the incident wave is transverse. The corresponding four angles, (r)
(r)
t , t , ’ l, ’ t, may be readily related to t(i) by the “kinematic” condition that the incident wave, as well as the reflected and refracted waves of both types, must have the same spatial distribution along the interface plane, i.e. for the interface particles participating in all five waves. According to Eq. (108), the necessary boundary condition is the equality of the tangential components (in Fig. 12, kx), of all five wave vectors:
r
r
'
'
'
'
i
k sin
k sin k sin k sin k k sin .
(7.125)
t
t
l
l
l
l
t
t
x
t
t
Since the acoustic wave vector magnitudes k, at fixed frequency , are inversely proportional to the corresponding wave velocities, we immediately get the following relations:
r
'
'
i
sin
sin
sin
sin
r
i
l
l
t
t
,
,
(7.126)
t
t
v
v'
v'
v
l
l
t
t
so that generally all four angles are different. (This is of course an analog of the well-known Snell law in optics – where, however, only transverse waves are possible.) These relations show that, just like in optics, the direction of a wave propagating into a medium with lower velocity is closer to the normal (in Chapter 7
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Essential Graduate Physics
CM: Classical Mechanics
Fig. 12, to the z-axis). In particular, this means that if v’ > v, the acoustic waves, at larger angles of incidence, may exhibit the effect of total internal reflection, so well known from optics37, when the refracted wave vanishes. In addition, Eqs. (126) show that in acoustics, the reflected longitudinal wave, with velocity v l > vt, may vanish at sufficiently large angles of the transverse wave incidence.
z
r
l
i
k
r
k
t
i
r
t
t
t
r
v , v
k
l
t
l
'
'
v , v
x
l
t
kx
Fig. 7.12. Deriving the “kinematic”
'
t
'
conditions (126) of the acoustic wave
kl
'
reflection and refraction (for the case of
l
a transverse incident wave).
'
kt
All these facts automatically follow from general expressions for amplitudes of the reflected and refracted waves via the amplitude of the incident wave. These relations are straightforward to derive (again, from the continuity of the vectors q and dF), but since they are much bulkier than those in the electromagnetic wave theory (where they are called the Fresnel formulas 38), I would not have time/space to spell out and discuss them. Let me only note that, in contrast to the case of normal incidence, these relations involve eight media parameters: the impedances Z, Z’, and the velocities v, v’
on both sides of the interface, and for both the longitudinal and transverse waves.
There are other interface effects as well. Within certain frequency ranges, interfaces and surfaces of elastic solids may sustain so-called surface acoustic waves (SAW), in particular, the Rayleigh waves and the Love waves.39 The main feature that distinguishes such waves from their bulk (longitudinal and transverse) counterparts discussed above, is that the displacement amplitudes are largest at the interface and decay exponentially into the bulk of both adjacent media, so that the waves cannot be plane in the usual sense of being independent of two Cartesian coordinates.
For an analysis of such waves, it is important that in a uniform medium, even non-plane elastic waves may be always separated into independent longitudinal and transverse components. Indeed, it is straightforward (and hence left for the reader) to prove that Eq. (107) may be satisfied by a vector sum q(r, t) = ql(r, t) + qt(r, t), with the former component having zero curl (ql = 0) and propagating with the velocity (112), and the latter component having zero divergence (qt = 0) and propagating with the velocity (116). The plane waves q ln z and q tnx analyzed above certainly fall into these two categories, but in more general waves, there may be no clear association between the longitudinal and transverse components and their polarization.
This is true, in particular, in the Rayleigh waves, where the particle displacement vector q may be represented as the sum ql + qt, each of the vectors having more than one Cartesian component. In 37 See, e.g., EM Sec. 7.5.
38 Their discussion may be also found in EM Sec. 7.5.
39 Named, respectively, after Lord Rayleigh (born J. Strutt, 1842-1919) who has theoretically predicted the very existence of surface acoustic waves, and A. Love (1863-1940).
Chapter 7
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contrast to the bulk waves, the longitudinal and transverse components are coupled via their interaction with the interface, and as a result, propagate with a single velocity v R. A straightforward analysis of the Rayleigh waves on the surface of an elastic solid (i.e. its interface with free space) yields the following equation for v R:
4
2
2
2
v
v
v
1
R
1
R
1
R
.
(7.127)
2 2
2
2
v
v
v
t
t
l
According to this formula, and Eqs. (112) and (116), for realistic materials with the Poisson index between 0 and ½, the Rayleigh waves are slightly (by 4 to 13%) slower than the bulk transverse waves –
and hence substantially slower than the bulk longitudinal waves.
In the simplest case a “1D-plane” Rayleigh wave, independent of one Cartesian coordinate, the net vector q has just two Cartesian components (each contributed by ql and qt): one parallel to the propagation direction and hence to the interface, and another one normal to it. As a result, the trajectory of each particle in the wave is an ellipse in the plane normal to the interface. In contrast, the Love waves are purely transverse, with q oriented parallel to the interface. However, the interaction of these waves with the interface reduces their velocity v L in comparison with that ( v t) of the bulk transverse waves, keeping it within the narrow interval between v t and v R:
v v v v .
(7.128)
R
L
t
l
The practical importance of surface acoustic waves is that their amplitude decays very slowly with distance r from their point-like source: a 1/ r 1/2, while any bulk waves decay much faster, as a
1/ r. (Indeed, in the latter case the power P a 2, emitted by such source, is distributed over a spherical surface area proportional to r 2, while in the former case all the power goes into a thin surface circle whose length scales as r.) At least two areas of applications of the surface acoustic waves have to be mentioned: geophysics (for the earthquake detection and the Earth crust seismology), and electronics (for signal processing, with a focus on frequency filtering). Unfortunately, I cannot dwell on these interesting topics and I have to refer the reader to special literature.40
7.8. Elastic waves in restricted geometries
From what was discussed at the end of the last section, it should be pretty clear that generally, the propagation of acoustic waves in elastic bodies of finite size is rather complicated. There is, however, one important limit in which several important simple results may be readily obtained. This is the limit of (relatively) low frequencies, where the corresponding wavelength is much larger than at least one dimension of the system.
Let us consider, for example, various waves that may propagate along thin rods, in this case
“thin” meaning that the characteristic size a of the rod’s cross-section is much smaller than not only the length of the rod but also the wavelength = 2/ k. In this case, there is a considerable range z of distances along the rod,
a z ,
(7.129)
40 See, for example, K. Aki and P. Richards, Quantitative Seismology, 2nd ed., University Science Books, 2002
and D. Morgan, Surface Acoustic Waves, 2nd ed., Academic Press, 2007.
Chapter 7
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in that we can neglect the material’s inertia, and apply the results of our earlier static analyses. For example, for a longitudinal wave of stress, which is essentially a wave of periodic tensile extensions and compressions of the rod, within the range (129) we may use the static relation (42):
Es .
(7.130)
zz
zz
In this simple case, it is easier to use the general equation of elastic dynamics not in its vector form (107), but rather in the precursor, Cartesian-component form (25), with fj = 0. For the plane waves of stress, propagating along the z-axis, only one component (with j’ z) of the sum on the right-hand side of that equation is not equal to zero, and it is reduced to
2
q
j
jz
.
(7.131)
2 2
t
z
In our current case of longitudinal waves, all components of the stress tensor but zz are equal to zero.
With zz from Eq. (130), and using the definition szz = qz/ z, Eq. (131) is reduced to a simple 1D wave equation,
2
2
q
q
z
E
z ,
(7.132)
2 2
2
t
z
which shows that the velocity of such tensile waves is
1/ 2
E
Tensile
v .
(7.133) waves:
velocity
Comparing this result with Eq. (112), we see that the tensile wave velocity, for any realistic material with a positive Poisson’s ratio, is lower than the velocity v l of longitudinal waves in the bulk of the same material. The reason for this difference is simple: in thin rods, the cross-section is free to oscillate (e.g., shrink in the longitudinal extension phase of the passing wave),41 so that the effective force resisting the longitudinal deformation is smaller than in a border-free space. Since (as it is clearly visible from the wave equation), the scale of the force determines that of v 2, this difference translates into slower waves in rods. Of course, as the wave frequency is increased to ka ~ 1, there is a (rather complicated and cross-section-depending) crossover from Eq. (133) to Eq. (112).
Proceeding to transverse waves on rods, let us first have a look at long bending waves for which the condition (129) is satisfied, so that the vector q = n xqx (with the x-axis being the bending direction –
see Fig. 8) is virtually constant in the whole cross-section. In this case, the only element of the stress tensor contributing to the net transverse force Fx is xz, so that the integral of Eq. (131) over the cross-section is
2
q
F
x
A
x ,
with
2
F d r.
(7.134)
x
2
t
xz
z
A
Now, if Eq. (129) is satisfied, we again may use the static local relations (75)-(77), with all derivatives d/ dz duly replaced with their partial form / z, to express the force Fx via the bending deformation qx.
Plugging these relations into each other one by one, we arrive at a rather unusual differential equation 41 For this reason, the tensile waves can be called longitudinal only in a limited sense: while the stress wave is purely longitudinal: xx = yy = 0, the strain wave is not: sxx = syy = – szz 0, i.e. q(r, t) n zqz.
Chapter 7
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2
4
q
q
A
x EI
x .
(7.135)
2
y
4
t
z
Looking for its solution in the form of a sinusoidal wave (108), we get the following dispersion relation:42
Bending
EI
waves:
2
y
4
k .
(7.136)
dispersion
A
relation
Such relation means that the bending waves are not acoustic at any frequency, and cannot be characterized by a single velocity that would be valid for all wave numbers k, i.e. for all spatial Fourier components of a waveform. According to our discussion in Sec. 6.3, such strongly dispersive systems cannot pass non-sinusoidal waveforms too far without changing their waveform rather considerably.
This situation changes, however, if the rod is pre-stretched with a tension force T – just as in the discrete 1D model that was analyzed in Sec. 6.3. The calculation of the effect of this force is essentially similar; let us repeat it for the continuous case, for a minute neglecting the bending stress – see Fig. 13.
T ( z dz)
( z dz)
( z)
y
y
- T ( z)
Fig. 7.13. Additional forces in a thin rod
q
q dq
x
x
x
(“string”), due to the background tension T.
z
z dz
Still sticking to the limit of small angles , the additional vertical component d T x of the net force acting on a small rod fragment of length dz is T x( z – dz) – T x( z) = T y( z + dz) – T y( z) T ( y/ z) dz, so that Fx/ z = T ( y/ z). With the geometric relation (77) in its partial-derivative form qx/ z = y, this additional term becomes T (2 qx/ z 2) . Now adding it to the right-hand side of Eq. (135), we get the following dispersion relation
2
1
4
2
EI k k .
(7.137)
A
y
T
Since the product A in the denominator of this expression is just the rod’s mass per unit length (which was denoted in Chapter 6), at low k (and hence low frequencies), this expression is reduced to the linear dispersion law, with the velocity given by Eq. (6.43):
1/ 2
T
v
.
(7.138)
A
So Eq. (137) describes a smooth crossover from the “guitar-string” acoustic waves to the highly dispersive bending waves (136).
42 Note that since the “moment of inertia” Iy, defined by Eq. (70), may depend on the bending direction (unless the cross-section is sufficiently symmetric), the dispersion relation (136) may give different results for different directions of the bending wave polarization.
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Now let us consider another type of transverse waves in thin rods – the so-called torsional waves, which are essentially the dynamic propagation of the torsional deformation discussed in Sec. 6.
The easiest way to describe these waves, again within the limits (129), is to write the equation of rotation of a small segment dz of the rod about the z-axis, passing through the “center of mass” of its cross-section, under the difference of torques = n z z applied on its ends – see Fig. 10: 2
z
I
dz
d ,
(7.139)
z
2
z
t
where Iz is the “moment of inertia” defined by Eq. (91), which now, after its multiplication by dz, i.e.
by the mass per unit area, has turned into the genuine moment of inertia of a dz-thick slice of the rod.
Dividing both sides of Eq. (139) by dz, and using the static local relation (84), z = C = C( z/ z), we get the following differential equation
2
2
I
z C
z .
(7.140)
z
2
2
t
z
Just as Eqs. (111), (115), and (132), this equation describes an acoustic (dispersion-free) wave, which propagates with the following frequency-independent velocity
1/ 2
C
Torsional
v
.
(7.141) waves:
I z
velocity
As we have seen in Sec. 6, for rods with axially-symmetric cross-sections, the torsional rigidity C is described by the simple relation (89), C = Iz, so that Eq. (141) is reduced to Eq. (116) for the transverse waves in infinite media. The reason for this similarity is straightforward: in a torsional wave, particles oscillate along small arcs (Fig. 14a), so that if the rod’s cross-section is round, its surface is stress-free, and does not perturb or modify the motion in any way, and hence does not affect the transverse velocity.
(a)
(b)
Fig. 7.14. Particle trajectories in two
different transverse waves with the same
velocity: (a) torsional waves in a thin
round rod and (b) circularly-polarized
waves in an infinite (or very broad)
sample.
This fact raises an interesting issue of the relation between the torsional and circularly-polarized waves. Indeed, in Sec. 7, I have not emphasized enough that Eq. (116) is valid for a transverse wave polarized in any direction perpendicular to the wave vector k (in our notation, directed along the z-axis).
In particular, this means that such waves are doubly-degenerate: any isotropic elastic continuum can carry simultaneously two non-interacting transverse waves propagating in the same direction with the same velocity (116), with two mutually perpendicular linear polarizations (directions of the vector a), Chapter 7
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for example, directed along the x- and y- axes.43 If both waves are sinusoidal (108), with the same frequency, each point of the medium participates in two simultaneous sinusoidal motions within the [ x, y] plane:
q Re
i( kz t)
a e
A cos ,
Ψ
q Re
i( kz t)
a e
A
(7.142)
x
x
x
y
y
y
cos Ψ ,
where kz – t + x, and y – x. Basic geometry tells us that the trajectory of such a motion on the [ x, y] plane is an ellipse (Fig. 15), so that such waves are called elliptically polarized. The most important particular cases of such polarization are:
(i) = 0 or : a linearly-polarized wave, with the displacement vector a is directed at angle =
tan-1( Ay/ Ax) to the x-axis; and
(ii) = /2 and Ax = Ay: two possible circularly-polarized waves, with the right or left polarization, respectively.44
Ay
q( t)
qy
Fig. 7.15. The trajectory of a particle in an
0
qx
elliptically polarized transverse wave, within
A
the plane perpendicular to the direction of
x
wave propagation.
Now comparing the trajectories of particles in the torsional wave in a thin round rod (or pipe) and the circularly-polarized wave in a broad sample (Fig. 14), we see that, despite the same wave propagation velocity, these transverse waves are rather different. In the former case (Fig. 14a) each particle moves back and forth along an arc, with the arc’s length different for different particles (and vanishing at the rod’s center), so that the waves are not plane. On the other hand, in a circularly polarized wave, all particles move along similar, circular trajectories, so that such a wave is plane.
To conclude this chapter, let me briefly mention the opposite limit when the size of the body, from whose boundary the waves are completely reflected,45 is much larger than the wavelength. In this case, the waves propagate almost as in an infinite 3D continuum (which was analyzed in Sec. 7), and the most important new effect is the finite number of wave modes in the body. Repeating the 1D analysis at the end of Sec. 6.5, for each dimension of a 3D cuboid of volume V = l 1 l 2 l 3, and taking into account that the numbers kn in each of the three dimensions are independent, we get the following generalization of 43 As was discussed in Sec. 6.3, this is also true in the simple 1D model shown in Fig. 6.4a.
44 The circularly polarized waves play an important role in quantum mechanics, where they may be most naturally quantized, with their elementary excitations (in the case of mechanical waves we are discussing, called phonons) having either positive or negative angular momentum Lz = .
45 For acoustic waves, such a condition is easy to implement. Indeed, from Sec. 7 we already know that the strong inequality of the wave impedances Z is sufficient for such reflection. The numbers in Table 1 show that, for example, the impedance of a longitudinal wave in a typical metal (say, steel) is almost two orders of magnitude higher than that in air, ensuring their virtually full reflection from the surface.
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Eq. (6.75) for the number N of different traveling waves with wave vectors within a relatively small volume d 3 k of the wave vector space:
V
1
3D
3
3
3
dN g
density
d k
d k k ,
(7.143)
2
,
1
for
3
V
of modes
where k >>> 1/ l 1,2,3 is the center of this volume, and g is the number of different possible wave modes with the same wave vector k. For the mechanical waves analyzed above, with one longitudinal mode, and two transverse modes with different polarizations, g = 3.
Note that since the derivation of Eqs. (6.75) and (143) does not use other properties of the waves (in particular, their dispersion relations), this mode counting rule is ubiquitous in physics, being valid, in particular, for electromagnetic waves (where g = 2) and quantum “de Broglie waves” (i.e.
wavefunctions), whose degeneracy factor g is usually determined by the particle’s spin.46
7.9. Exercise problems
7.1. Derive Eqs. (16).
Hint: Besides the definition of the cylindrical coordinates and basic calculus, you may like to use Eq. (4.7) with d = ( d)n z.
7.2. A uniform thin sheet of an isotropic, elastic material, of
thickness t and area A >> t 2, is compressed by two plane, parallel, broad, rigid surfaces – see the figure on the right. Assuming that there is no
t
slippage between the sheet and the surfaces, calculate the relative
compression (- t/ t) as a function of the compressing force. Compare the A
result with that for the tensile stress calculated in Sec. 3.
7.3. Two opposite edges of a thin, very wide sheet of an isotropic, elastic material have been clamped in two rigid, plane, parallel walls that are pulled apart with force F, along the sheet’s length l.
Find the relative extension l/ l of the sheet in the direction of the force and its relative compression t/ t in the perpendicular direction, and compare the results with Eqs. (45)-(46) for the tensile stress and the solution of the previous problem.
R
t << R
7.4. Calculate radial extension R of a thin, long, round cylindrical
pipe, due to its rotation with a constant angular velocity about its
symmetry axis (see the figure on the right), in terms of the elastic moduli E
and . The external pressure both inside and outside the pipe is negligible.
7.5.* A static force F is exerted on an inner point of a uniform,
isotropic elastic body. Calculate the spatial distribution of the deformation created by the force, assuming that far from the point of its application and the points we are interested in, the body’s position is kept fixed.
46 See, e.g., EM Secs. 7.8 and QM Sec. 1.7.
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y
t
7.6. A long, uniform rail with the cross-section shown in the figure
on the right, is being bent with the same (small) torque twice: first within
the xz- plane and then within the yz-plane. Assuming that t << l, find the l x
ratio of the rail bending deformations in these two cases.
l
7.7. Two thin rods of the same length and mass are
made of the same elastic, isotropic material. The cross-section
of one of them is a circle, while the other one is an equilateral
triangle – see the figure on the right. Which of the rods is
stiffer for bending along its length? Quantify the relation. Does
the result depend on the bending plane orientation?
7.8. A thin, elastic, uniform, initially straight beam is placed on
x
two point supports at the same height – see the figure on the right. .
z
z
0
0
Calculate the support placements that:
a
a
0
z
(i) ensure that the beam ends are horizontal, and
(ii) minimize the largest deviation of the beam from the
g
horizontal baseline.
Hint: For Task (ii), an approximate answer (with an accuracy better than 1%) is acceptable.
7.9. Calculate the largest longitudinal compression force T
T
T
that may be withstood by a thin, straight, elastic rod without i
( )
bucking (see the figure on the right) for two shown cases:
l
(i) the rod’s ends are clamped, and
T
T
ii
( )
(ii) the rod is free to turn about the support points.
7.10. An elastic, light, thin poll with a square cross-section of area A = a a, had been firmly dug into the ground in the vertical position, sticking out by height h >> a. What largest compact mass M may be placed straight on the top of the poll without the stability loss?
7.11. Calculate the potential energy of a small and slowly changing, but otherwise arbitrary bending deformation of a uniform, elastic, initially straight rod. Can the result be used to derive the dispersion relation (136)?
7.12. Calculate the torsional rigidity of a thin, uniform rod whose cross-section is an ellipse with semi-axes a and b.
7.13. Calculate the potential energy of a small but otherwise arbitrary torsional deformation z( z) of a uniform, straight, elastic rod.
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d
N
R
7.14. Calculate the spring constant dF/ dl of a F
F
coil made of a uniform, elastic wire with a circular cross-
section of diameter d, wound as a dense round spiral of N
>> 1 turns of radius R >> d – see the figure on the right.
l
F
R
7.15. The coil described in the previous problem is now used as what is
sometimes called the torsion spring – see the figure on the right. Find the d
corresponding spring constant d/ d, where is the torque of the external forces F
0
relative to the center of the coil (point 0).
F
7.16. Use Eqs. (99) and (100) to recast Eq. (101b) for the torsional rigidity C of a thin rod into the form given by Eq. (101c).
7.17.* Generalize Eq. (101b) to the case of rods with more than one cross-section’s boundary.
Use the result to calculate the torsional rigidity of a thin round pipe, and compare it with Eq. (91).
7.18. Prove that in a uniform medium, any (not necessarily plane) elastic wave may be decomposed into a longitudinal wave with ql = 0 and a transverse wave with qt = 0, and find the equations satisfied by these functions.
7.19.* Use the wave equations derived in the solution of the previous problem and the semi-quantitative description of the Rayleigh surface waves given in Sec. 7 of the lecture notes, to calculate the structure of the waves and derive Eq. (127).
7.20.* Calculate the modes and frequencies of free radial oscillations of a sphere with radius R, made of a uniform elastic material.
7.21. A long steel wire has a circular cross-section with a 3-mm diameter, and is pre-stretched with a constant force of 10 N. Which of the longitudinal and transverse waves with frequency 1 kHz has the largest group velocity in the wire? Accept the following parameters for the steel (see Table 1): E =
170 GPa, = 0.30, = 7.8 g/cm3.
7.22. Define and calculate the wave impedances for (i) tensile and (ii)
P
torsional waves in a thin rod, appropriate in the long-wave limit. Use the
results to calculate the fraction of each wave’s power P reflected from a firm
connection of a long rod with a round cross-section to a similar rod, but with a
twice smaller diameter – see the figure on the right.
P
7.23. Calculate the fundamental frequency of small transverse standing waves on a free and uniform thin rod, and the position of displacement nodes in this mode.
Hint: Numerical solution of the final transcendent equations is acceptable.
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Chapter 8. Fluid Mechanics
This chapter describes the basic notions of fluid mechanics, discusses a few core problems of statics and dynamics of ideal and viscous fluids, and gives a very brief review of such a complicated phenomenon as turbulence. In addition, the viscous fluid flow discussion is used as a platform for an elementary introduction to numerical methods of the partial differential equation solution – whose importance extends well beyond this particular field.
8.1. Hydrostatics
The mechanics of fluids (defined as the materials that cannot keep their geometric form on their own, and include both liquids and gases) is both more simple and more complex than that of the elastic solids, with the simplifications mostly in statics.1 Indeed, fluids, by definition, cannot resist static shear deformations. There are two ways to express this fact. First, we can formally take the shear modulus , describing this resistance, to equal zero. Then the Hooke’s law (7.32) shows that the stress tensor is diagonal:
.
(8.1)
jj'
jj
jj'
Alternatively, the same conclusion may be reached just by looking at the stress tensor definition (7.19) and/or Fig. 7.3, and saying that in the absence of shear stress, the elementary interface dF has to be normal to the area element dA, i.e. parallel to the vector dA.
Moreover, in fluids at equilibrium, all three diagonal elements jj of the stress tensor have to be equal at each point. To prove that, it is sufficient to single out (mentally rather than physically), from a static fluid, a small volume in the shape of a right prism, with mutually perpendicular faces normal to the two directions we are interested in – in Fig. 1, along the x- and y-axes.
dA
dA ( dA cos)
xx
x
xx
cos
y
Fig. 8.1. Proving the pressure isotropy.
x
dA
The prism is in equilibrium if each Cartesian component of the vector of the total force exerted on all its faces equals zero. For the x-component this balance may be expressed as xxdAx – ( dA)cos
= 0. However, from the geometry (Fig. 1), dAx = dA cos, so that the above expression yields = xx.
A similar argument for the y-component gives = yy so that xx = yy. Changing the orientation of the prism, we can get such equalities for any pair of diagonal elements of the stress tensor, jj, so that all three of them have to be equal.
1 It is often called hydrostatics because water has always been the most important liquid for the human race and hence for science and engineering.
© K. Likharev
CM: Classical Mechanics
This common diagonal element of the stress matrix is usually denoted as (- P), because in the vast majority of cases, the parameter P, called pressure, is positive. Thus we arrive at the key relation (which was already mentioned in Sec. 7.2):
.
(8.2)
Pressure
jj'
P
jj'
In the absence of bulk forces, pressure should be constant through the volume of fluid, due to the translational symmetry. Let us see how this result is affected by bulk forces. With the simple stress tensor (2), the general condition of equilibrium of a continuous medium, expressed by Eq. (7.25) with the left-hand side equal to zero, becomes
P f 0 ,
(8.3)
j
rj
and may be re-written in the following convenient vector form:
P f 0.
(8.4)
In the simplest case of a heavy fluid with mass density , in a uniform gravity field f = g, the equation of equilibrium becomes,
P g
0,
(8.5)
with only one nonzero component – near the Earth’s surface, the vertical one. If, in addition, the fluid may be considered incompressible, with its density constant,2 this equation may be readily integrated over the vertical coordinate (say, y) to give the so-called Pascal equation:3
P gy
const,
(8.6)
Pascal
equation
where the direction of the y-axis is taken opposite to that of vector g.
Two manifestations of this key equation are well known. The first one is the fact that in interconnected vessels filled with a fluid, its pressure is equal at all points at the same height ( y), regardless of the vessel shape, provided that the fluid is in equilibrium.4 In particular, if a heavy liquid has an open surface, then in equilibrium, it has to be horizontal – at least, not too close to the retaining walls (see Sec. 2).
The second manifestation of Eq. (6) is the buoyant force Fb exerted by a liquid on a (possibly, partly) submerged body, i.e. the vector sum of the elementary pressure forces dF = PdA exerted on all elementary areas dA of the submerged part of the body’s surface – see Fig. 2. According to Eq. (6), with the constant equal to zero (corresponding to zero pressure at the liquid’s surface taken for y = 0, see Fig.
2a), the vertical component of this elementary force is
2 As was discussed in Sec. 7.3 in the context of Table 7.1, this is an excellent approximation, for example, for human-scale experiments with water.
3 The equation, and the SI unit of pressure 1 Pa 1N/m2, are named after Blaise Pascal (1623-1662) who has not only pioneered hydrostatics, but also invented the first mechanical calculator, and made several other important contributions to mathematics – and to Christian philosophy!
4 This simple fact opens wide opportunities for the engineering field of hydraulics, in particular enabling a very simple and efficient way to magnify forces, using interconnected hydraulic cylinders of different diameters.
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dF dF cos dA
P cos gy
cos dA
gydA .
(8.7)
y
h
where dAh = cos dA is the horizontal footprint (say, dxdz) of the elementary area dA. Now integrating this relation over all the surface, we get the total vertical buoyant force:5
Archimedes
principle
F g
y dA gV ,
b
(8.8)
h
S
where V is the volume of the submerged part of the body’s volume, while is the liquid’s density, so that by magnitude, F b equals the weight of the liquid which would fill the submerged volume.
(a)
(b)
y
dF
dA
after
0
a
V
before
b
dA
A
Fig. 8.2. Calculating
g
g
the buoyant force.
This well-known Archimedes principle may be proved even more simply using the following argument: the liquid’s pressure forces, and hence the resulting buoyant force, cannot depend on what is inside the body’s volume. Hence F b would be the same if we filled the volume V in question with a liquid similar to the surrounding one. But in this case, the liquid should be still in equilibrium even if the surface is completely flexible, so that both forces acting on its inner part, the buoyant force Fb and the inner liquid’s weight mg = Vg, have to be equal and opposite, thus proving Eq. (8) again.
Despite the simplicity of the Archimedes principle, its erroneous formulations, such as “The buoyant force’s magnitude is equal to the weight of the displaced liquid” [WRONG!] creep from one undergraduate textbook to another, leading to application errors. A typical example is shown in Fig. 2b, where a solid vertical cylinder with the base area A is pressed into a liquid inside a container of comparable size, pushing the liquid’s level up by distance a. The correct answer for the buoyant force, following from Eq. (8), is
F gV
,
(8.9a)
b
gA a b
because the volume V of the submerged part of the cylinder is evidently A( a + b). But the wrong formulation cited above, using the term displaced liquid, would give a different answer: F gV
gAb . [WRONG!]
(8.9b)
b
displaced
(The latter result is correct only asymptotically, in the limit b/ a .) Another frequent error in hydrostatics concerns the angular stability of a freely floating body –
the problem of vital importance for the boat/ship design. It is sometimes claimed that the body is stable 5 The force is vertical, because the horizontal components of the elementary forces dF exerted on opposite elementary areas dA, at the same height y, cancel.
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only if the so-called buoyancy center, the effective point of buoyant force application (in Fig, 3, point B),6 is above the center of mass (C) of the floating body. However, as Fig. 3 shows, this is unnecessary; indeed in the shown case, point B remains below point C, even at a small tilt. Still, in this case, the torque created by the pair of forces Fb and mg tries to return the body to the equilibrium position, which is therefore stable. As Fig. 3 shows, the actual condition of the angular stability may be expressed as the requirement for point M (in shipbuilding, called the metacenter of the ship’s hull) to be above the ship’s center of mass C.7
Fb
M
C
B
Fig. 8.3. Angular stability of a
floating body.
g
m
To conclude this section, let me note that the integration of Eq. (4) may be more complex in the case if the bulk forces f depend on position,8 and/or if the fluid is substantially compressible. In the latter case, Eq. (4) has to be solved together with the medium-specific equation of state = ( P) describing its compressibility law – whose example is given by Eq. (7.38) for ideal gases: mN/ V =
mP/ k B T, where m is the mass of one gas molecule.
8.2. Surface tension effects
Besides the bulk (volume-distributed) forces, one more possible source of pressure is surface tension. This effect results from the difference between the potential energy of atomic interactions on the interface between two different fluids and that in their bulks, and thus may be described by an additional potential energy
U A
,
(8.10) Surface
i
tension
where A is the interface area, and is called the surface tension constant – or just the “surface tension”.
For a stable interface of any two fluids, is always positive.9 For surfaces of typical liquids (or their interfaces with air), at room temperature, the surface tension equals a few 10-2 J/m2,10 corresponding to 6 A simple calculation, similar to the one resulting in Eq. (8), but for the total torque rather than the total force, shows that B is just the center of mass of the submerged volume V filled with any uniform material.
7 It is easy (and hence is left for the reader) to prove that a small tilt of the body leads to a small lateral displacement of point B, but does not affect the position of the metacenter M.
8 A simple example of such a problem is given by the fluid equilibrium in a container rotating with a constant angular velocity . If we solve such a problem in a reference frame rotating together with the container, the real bulk forces should be complemented by the centrifugal “force” (4.93), depending on r.
9 Indeed, if the of the interface of certain two fluids is negative, it self-reconfigures to decrease U i, i.e. to increase U i, by increasing the interface area, i.e. fragments the system into a macroscopically-uniform solution.
10 For a better feeling of this number, one should remember that 1 J/m2 1 N/m.
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the potential energy U i of a few 10-2 eV per surface molecule – i.e. just a fraction of the full binding (or
“cohesive”) energy of the same liquid, which is typically of the order of 10-1 eV per molecule.
In the absence of other forces, the surface tension makes a liquid drop spherical to minimize its surface area A at a fixed volume. For the analysis of the surface tension effects for more complex geometries, and in the presence of other forces, it is convenient to reduce them to a certain additional effective pressure drop P ef at the interface. To calculate P ef, let us consider the condition of equilibrium of a small part dA of a smooth interface between two fluids (Fig. 2), in the absence of bulk forces.
P 2
P 1
dA dA
R
dA
,
1 2
r
Fig. 8.4. Deriving the Young-Laplace
formula (13).
If the pressures P 1,2 on the two sides of the interface are different, the work of stress forces on fluid 1 at a small virtual displacement r = n r of the interface (where n = dA/ dA is the unit vector normal to the interface) equals11
W dA r P
.
(8.11)
1
P 2
For equilibrium, this work has to be compensated by an equal change of the interface energy, U i =
( dA). Differential geometry tells us that in the linear approximation in r, the relative change of the elementary surface area, corresponding to a fixed solid angle d, may be expressed as
dA r r
,
(8.12)
dA
R
R
1
2
where R 1,2 are the so-called principal radii of the interface curvature.12 Combining Eqs. (10)-(12), we get the following Young-Laplace formula:13
Young-
1
1
Laplace
P P Δ P
.
(8.13)
1
2
ef
formula
R
R
1
2
11 This equality follows from the general relation (7.30), with the stress tensor elements expressed by Eq. (2), but in this simple case of the net stress force dF = ( P 1 – P 2) dA parallel to the interface element vector dA, it may be even more simply obtained just from the definition of work: W = dFr at the virtual displacement r = n r.
12 This general formula may be readily verified for a sphere of radius r (for which R 1 = R 2 = r and dA = r 2 d, so that ( dA)/ dA = ( r 2)/ r 2 = 2 r/ r), and for a round cylindrical interface of radius R (for which R 1 = r, R 2 = , and dA = rd dz, so that ( dA)/ dA = r/ r). For more on curvature, see, for example, M. do Camo, Differential Geometry of Curves and Surfaces, 2nd ed., Dover, 2016.
13 This result (not to be confused with Eq. (15), called Young’s equation) was derived in 1806 by Pierre-Simon Laplace (of the Laplace operator/equation fame) on the basis of the first analysis of the surface tension effects by Thomas Young (yes, the same Young who performed the famous two-slit experiment with light!) a year earlier.
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In particular, this formula shows that the additional pressure created by surface tension inside a spherical drop of a liquid, of radius R, equals 2/ R, i.e. decreases with R. In contrast, according to Eqs.
(5)-(6), the pressure effects of bulk forces, for example gravity, grow as gR. The comparison of these two pressure components shows that if the drop radius (or more generally, the characteristic linear size of a liquid’s sample) is much larger than the so-called capillary length
1/ 2
2
a
,
(8.14) Capillary
c
g
length
the surface tension may be safely ignored – as will be done in all following sections of this chapter, besides a brief discussion at the end of Sec. 4. For the water surface, or more exactly its interface with air at ambient conditions, 0.073 J/m2, while 1,000 kg/m3, so that a c 4 mm.
On the other hand, in very narrow tubes, such as blood capillary vessels with radius a ~ 1 m, i.e. a << a c, the surface tension effects are very important. The key notion for the analysis of these effects is the contact angle c (also called the “wetting angle”) at an equilibrium edge of a liquid wetting a solid – see Fig. 5.
(a)
(b)
lg
lg
gas
gas
liquid
liquid
c
c
sl
sg
sg
Fig. 8.5. Contact angles
for (a) hydrophilic and
solid
sl
solid
(b) hydrophobic surfaces.
According to its definition (10), the constant may be interpreted as a force (per unit length of the interface boundary) directed normally to the boundary, and “trying” to reduce the interface area. As a result, the balance of horizontal components of the three such forces, shown in Fig. 5a, immediately yields the Young’s equation
cos ,
(8.15) Young’s
sl
lg
c
sg
equation
where the indices of the three constants correspond to three possible interfaces between the liquid, solid, and gas. For the so-called hydrophilic surfaces that “like to be wet” by a particular liquid (not necessarily water), meaning that sl < sg, this relation yields cosc > 0, i.e. c < /2 – the situation shown in Fig. 5a. On the other hand, for hydrophobic surfaces with sl > sg, Eq. (15) yields larger contact angles, c > /2 – see Fig. 5b.
Let us use this notion to solve the simplest and perhaps the most practically important problem of this field – find the height h of the fluid column lifted by the surface tension forces in a narrow vertical tube made of a hydrophilic material, assuming its internal surface to be a round cylinder of radius a – see Fig. 6. Inside an incompressible fluid, pressure drops with height according to the Pascal equation (6), so that just below the surface, P P 0 – gh, where P 0 is the background (e.g., atmospheric) pressure. This means that at a << h, the pressure variation along the concave surface (called the meniscus) of the liquid is negligible, so that according to the Young-Poisson equation (13), the sum (1/ R 1 + 1/ R 2) has to be virtually constant along the surface. Due to the axial symmetry of the problem, this means that the surface has to be a part of a sphere. From the contact angle definition, the radius R of Chapter 8
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the sphere is equal to a/cosc – see Fig. 6. Plugging this relation into Eq. (3) with P 1 – P 2 = gh, we get the following result for h:
2 cos
gh
c
.
(8.16a)
a
In hindsight, this result might be obtained more directly – by requiring the total weight gV =
g( a 2 h) of the lifted liquid’s column to be equal to the vertical component F cosc of the full surface tension force F = p, acting on the perimeter p = 2 a of the meniscus. Using the definition (11) of the capillary length a c, Eq. (16a) may be represented as the so-called Jurin rule: Jurin
a 2
a 2
rule
h
c
c
cos
;
(8.16b)
a
c
a
according to our initial assumption h >> a, Eq. (16) is only valid for narrow tubes, with radius a << a c.
a
R
h
c
g
P 0
Fig. 8.6. Liquid’s rise in a vertical capillary tube.
This capillary rise is the basic mechanism of lifting water with nutrients from roots to the branches and leaves of plants, so that the tallest tree heights correspond to the Jurin rule (16), with cosc
1, and the pore radius a limited from below by a few microns, because of the viscosity effects restricting the fluid discharge – see Sec. 5 below.
8.3. Kinematics
In contrast to the stress tensor, which is frequently very simple – see Eq. (2), the strain tensor is not a very useful notion in fluid mechanics. Indeed, besides a very few situations,14 typical problems of this field involve fluid flow, i.e. a state when the velocity of fluid particles has some nonzero time average. This means that the trajectory of each particle is a long line, and the very notion of its displacement q from the initial position becomes impracticable. However, the particle’s velocity v
dq/ dt remains a very useful notion, especially if it is considered as a function of the observation point r and (generally) time t. In an important class of fluid dynamics problems, the so-called stationary (or
“ steady ” , or “static”) flow, the velocity defined in this way does not depend on time, v = v(r).
14 One of them is sound propagation, where the particle displacements q are typically small, so that the results of Sec. 7.7 are applicable. As a reminder, they show that in fluids, with = 0, the transverse sound cannot propagate, while the longitudinal sound can – see Eq. (7.114).
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There is, however, a price to pay for the convenience of this notion: namely, due to the difference between the vectors q and r, the particle’s acceleration a = d 2q/ dt 2 (that participates, in particular, in the 2nd Newton law) cannot be calculated just as the time derivative of the velocity v(r, t).
This fact is evident, for example, for the static flow case, in which the acceleration of individual fluid particles may be very significant even if v(r) does not depend on time – just think about the acceleration of a drop of water flowing over the Niagara Falls’ rim, first accelerating fast and then virtually stopping below, while the water velocity v at every particular point, as measured from a bank-based reference frame, is nearly constant. Thus the primary task of fluid kinematics is to express a via v; let us do this.
Since each Cartesian component vj of the velocity v has to be considered as a function of four independent scalar variables: three Cartesian components rj of the vector r and time t, its full time derivative may be represented as
dv
v
3
v
dr
j
j
j
j'
.
(8.17)
dt
t
j' 1
r dt
j'
Let us apply this general relation to a specific set of infinitesimal changes { dr 1, dr 2, dr 3} that follows a small displacement dq of a certain particle of the fluid: dr = dq = v dt, i.e.
dr v dt .
(8.18)
j
j
In this case, dvj/ dt is the j th component aj of the particle’s acceleration a, so that Eq. (17) yields the following key relation of fluid kinematics:
3
v
v
a
j
v
(8.19a)
j
j .
j'
t
j' 1
rj'
Fluid
Using the del operator , this result may be rewritten in the following compact vector form:15
particle’s
acceleration
v
a
(v )v .
(8.19b)
t
This relation already signals the main technical problem of the fluid dynamics: many equations involving particle’s acceleration are nonlinear in velocity, excluding such a powerful tool as the linear superposition principle (which was used so frequently in the previous chapters of this course) from the applicable mathematical arsenal.
One more basic relation of fluid kinematics is the so-called continuity equation, which is essentially just the differential version of the mass conservation law. Let us mark, inside a fluid flow, an arbitrary volume V limited by a stationary (time-independent) surface S. The total mass of the fluid inside the volume may change only due to its flow through the boundary:
dM
d
d 3 r v d 2 r v dA, (8.20a)
n
dt
dt V
S
S
15 Note that the operator relation d/ dt = / t + (v) is applicable to an arbitrary (scalar or vector) function; it is frequently called the convective derivative. (Alternative adjectives, such as “Lagrangian”, “substantial”, or
“Stokes”, are sometimes used for this derivative as well.) The relation has numerous applications well beyond the fluid dynamics – see, e.g., EM Chapter 9 and QM Chapter 1.
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where the elementary area vector dA is defined just as in Sec. 7.2 – see Fig. 7.
v
A
d
S
V
Fig. 8.7. Deriving the continuity equation.
Now using the same divergence theorem that has been used several times in this course,16 the surface integral in Eq. (20a) may be transformed into the integral of (v) over the volume V, so that relation may be rewritten as
3
j d 0
r
,
(8.20b)
t
V
where the vector j v is called either the mass flux density (or the “mass current ” ). Since Eq. (20b) is valid for an arbitrary stationary volume V, the function under the integral has to vanish at any point: Continuity
equation
j 0.
(8.21)
t
Note that similar continuity equations are valid not only for mass but also for other conserved physics quantities (e.g., the electric charge, probability, etc.), with the proper re-definitions of and j.17
8.4. Dynamics: Ideal fluids
Let us start our discussion of fluid dynamics from the simplest case when the stress tensor obeys Eq. (2) even in motion. Physically, this means that the fluid viscosity effects, leading to mechanical energy loss, are negligible. (The conditions of this assumption will be discussed in the next section.) Then the equation of motion of such an ideal fluid (essentially the 2nd Newton law for its unit volume) may be obtained from Eq. (7.25) using the simplifications of its right-hand side, discussed in Sec. 1:
