Standing Waves Between Singularities In An Elastic Continuum Of Infinite Extension

STANDING WAVES BETWEEN SINGULARITIES IN AN ELASTIC CONTINUUM OF INFINITE EXTENSION BJØRN URSIN KARLSEN To Elizabeth Abstract. Waves in an elastic continuum are mainly of two different types, longitudinal and transversal waves, which are well documented in various textbooks. In this paper I will look into the possibility that there might be a third alternative, namely standing waves between oscillating nodes in the form of singularities in an elastic continuum of infinite extension. First I will show that a standing wave can form between a hypothetical rigid sphere embedded in the spatial continuum and the center node, thus establishing that a singularity may form the one endpoint in such oscillations. Next I will show that if the spatial continuum initially is agitated to an extent that it contains a plethora of such oscillating nodes, they will tend to organize along endless strings. Finally I will try to look into the possibility that there might be a coupling between the two field components in question, irrotational and solenoidal fields.

1. The Navier-Cauchy Equation In order to find how scalar waves propagate in the spatial continuum we start with recalling the Navier-Cauchy equation (1.1)

(λs + 2µs )∇ div u − µs curl curl u + b = ρs u ¨.

or equivalently by the mathematical identity curl curl u = ∇ div u − ∇2 u (1.2)

µs ∇2 u + (λs + µs )∇ div u + b = ρs u ¨,

At this point it may be appropriate to stress the point that the Navier-Cauchy equation only treats the limit where deformations can be considered infinitesimal, and it must not be mixed up with Navier-Stokes equation, which also incorporates viscosity and takes into account the hydrodynamic property that v˙ may be different from ∂v/∂t [i.e. v˙ = ∂v/∂t + (v · ∇)v]. According to Helmholtz’s Theorem any vector field satisfying [∇ · v]∞ = 0, [∇ × v]∞ = 0, (no velocities at infinite distance from considered area) may be written as the sum of an irrotational part and a solenoidal part, v = −∇φ + ∇ × A, Date: 20-08-08. 1

2

BJØRN URSIN KARLSEN

where

Z

∇·v d3 r0 , 0 V 4π|r − r| Z ∇×v 3 0 A= d r, 0 − r| 4π|r V

φ=−

(see http://mathworld.wolfram.com/HelmholtzsTheorem.html). As the spatial continuum is of infinite extension, or nearly so, any deformations have to be confined to a finite part of space, so this theorem will be applicable on all deformations. Hence the displacement field can be decomposed into two properties u = u1 + u2 , where u1 = −∇φ = − grad φ, u2 = ∇ × Ψ = curl Ψ,

div Ψ = 0.

Since curl grad φ ≡ 0, and div curl Ψ ≡ 0, the Navier-Cauchy equation (1.1) can be divided into two independent equations, one for an irrotational field b1 ρs (1.3) u ¨1 − , ∇ div u1 = (λs + 2µs ) λs + 2µs and the other for a solenoidal field ρs b2 (1.4) − curl curl u2 = u ¨2 − . µs µs By defining two new constants s r λs + 2µs µs c1 = (1.5) , c2 = , ρs ρs the N-C equation takes the form b =u ¨. ρs Operating on Equation (1.1) with the div operator and on Equation (1.2) with the curl operator yields respectively

(1.6)

c12 ∇ div u − c22 curl curl u +

1 ∂ 2 ( div u) div b =− , 2 2 c1 ∂t λs + 2µs

(1.7)

∇2 ( div u) −

(1.8)

∇2 ( curl u) −

1 ∂ 2 ( curl u) curl b =− . c22 ∂t2 µs

With the outer force, b, set to zero we have two wave equations where the dilatation, div u, satisfies a wave moving with the speed c1 , while the rotational component curl u, satisfies a wave moving with the speed c2 . In fact the Propagation theorem for isotropic bodies states that if a body is isotropic, then a wave is either longitudinal, in which case c = c1 , or transversal, in which case c = c2 [1, page 256]. This splitting of the Navier-Cauchy equation into one irrotational and one solenoidal part, allows us to examine these two parts separately and thereby simplifies the strain-stress relation immensely by reducing the elastic constants to only one single constant (the wave speed) in each equation (c1 6= c2 ). We see from Equation (1.5) that the two wave speeds are related to each other with a fixed constant given by

3

p the relation c1 = 2 + λs /µs · c2 , where c1 might be about the double of c2 (also dubbed c without the index in the text to follow). Notice also that all information of curl u is lost in Equation (1.7) and all information of div u in Equation (1.8). The energy in a deformation field with no surface trajectories is given by Kelvin’s theorem [1, page 208]: Z h i 1 (1.9) E= ˙ 2 + 21 (λs + 2µ)(div u)2 + 21 µ(curl u)2 dv. 2 ρs u B

The theorem states that to a curl u, a div u, and a velocity field u˙ there always corresponds an energy equal to E, but it does not tell exactly where in the field the energy is to be found. With this restriction in mind, the local energy density, e, in a spatial continuum of infinite extension can all the same be define as1 (1.10)

e=

1 2

ρs u˙ 2 +

1 2

(λs + 2µs )( div u)2 +

1 2

µs ( curl u)2 .

2. Scalar longitudinal waves in the spatial continuum By setting φ = div u in Equation (1.7) and consider a field without any outer forces, we get the wave equation (2.1)

∇2 φ =

1 ∂2φ , c12 ∂t2

or the even simpler wave equation for plane, longitudinal waves (2.2)

∂2φ 1 ∂2φ = . ∂x2 c12 ∂t2

It can be solved in several ways, but here I will apply d’Alembert’s solution. Let ξ ≡ c1 t − x η ≡ c1 t + x. Since x = f1 (ξ, η) and t = f2 (ξ, η) we have by the chain rule ∂ ∂ ∂ξ ∂ ∂η ∂ ∂ = · + · =− + , ∂x ∂ξ ∂x ∂η ∂x ∂ξ ∂η ∂ ∂ξ ∂ ∂η ∂ ∂ ∂ = · + · = c1 + c1 . ∂t ∂ξ ∂t ∂η ∂t ∂ξ ∂η By applying these operators on Equation (2.2) it reduces to ∂2φ = 0. ∂ξ∂η This partial differential equation can be integrated in two steps, and the general solution is φ = f (ξ) + g(η) or

φ = f (c1 t − x) + g(c1 t + x).

The two functions f (c1 t − x) and g(c1 t + x) represent by a certain time t = t0 a disturbance, or wave-formation, which when it first is started, propagates with velocity c1 in the x-axis’s positive or negative direction respectively. 1The corresponding expression for the energy density in an electromagnetic field has the same limitation, but nonetheless it is usually interpreted as the local energy density.

4

BJØRN URSIN KARLSEN

In order to find how spherical waves propagate from a disturbance center, I will write Equation (2.1) in polar coordinates φ = φ(r, v, w): 1 ∂ ³ 2 ∂φ ´ 1 ∂2φ 1 ∂ ³ ∂φ ´ r + + sin w r2 ∂r ∂r r2 sin w ∂w ∂w r2 sin2 w ∂v 2 2 1 ∂ φ = 2 2. c1 ∂t In this connection I will only consider the spherical symmetric case so ∂φ/∂v = ∂φ/∂w = 0, and we acquire 1 ∂ ³ 2 ∂φ ´ 1 ∂2φ , r = r2 ∂r ∂r c12 ∂t2 which by the substitution φ = ψ/r goes over in the simpler ∂2ψ 1 ∂2ψ = . ∂r2 c12 ∂t2 This equation is formally equivalent to Equation (2.2) and has a solution corresponding to what is the case for plane waves: ψ = f (c1 t − r) + g(c1 t + r). Hence the general expression for a spherical wave spreading from a given wavecenter is given by f (c1 t − r) . r The other solution is for a spherical wave converging towards a focal point in space:

(2.3)

div u =

(2.4)

div u =

g(c1 t + r) . r

3. Irrotational standing waves In this section I will investigate if standing waves can occur between hypothetical concentric rigid shells embedded in the spatial continuum, especially in order to see if standing waves can form between the inside of a spherical shell and the center node. I perform the thought experiment that imbedded in the spatial continuum is a hypothetical rigid, undeformable sphere inside which there might be a standing compression wave bouncing back and forth between the center and the firm shell. In order to see if such a standing wave is possible, I will set up the Navier-Cauchy equation and try to solve it with these border conditions. I must, however emphasis that the N-C equation only is valid for small deformations, in fact only when the deformations are infinitesimal. When the deformations are significant, we have got to bring in the full power of the Navier-Stokes equation. The main difference from the N-C equation is that the time derivative has got to be modified to2 ∂ d (·) = (·) + (v · ∇)(·), dt ∂t e.g. u ¨ = ∂ u/∂t ˙ + (u∇) ˙ u. ˙ Hence the result may at best be suggestive.

(3.1)

2The N-S equation a also incorporates viscosity, which is not present in the spatial continuum.

5

First we write the Navier-Cauchy equation for an irrotational field with no external forces (Eq. 1.3): 1 grad div u = 2 u ¨, c1 where c1 is the speed of longitudinal waves, and then let us consider a central symmetric deformation where u = f (r)ˆ r. From the identities (3.2)

div (fˆr) = 2

f + f 0, r

and grad f = f 0 ˆr,

(3.3) we obtain the identity

¡ 2f 0 2f ¢ grad div f = f 00 + − 2 ˆr. r r With these properties inserted, the N-C equation takes the form ¡ 00 u0 u¢ u ¨ (3.5) u +2 −2 2 ˆ r = 2 ˆr. r r c1

(3.4)

By the product method, u(r, t) = F (r) · G(t) · ˆ r, we have 2 2 1 ¨ F 00 G + F 0 G − 2 F G = 2 F G, r r c1 ¨ F 00 2 F0 2 1 G + − 2 = 2 . F r F r c1 G Since the left and right side of this expression only depend on the arguments r and t respectively, they can both be set to the same constant −p2 , and hence they can be separated into the two equations: (3.6) (3.7)

2 (−2 + p2 r2 ) F 00 + F 0 + F =0 r r2 ¨ + c21 p2 G = 0 G

Equation (3.7) has the general solution (3.8)

G(t) = C1 sin(c1 pt) − C1 cos(c1 pt).

Equation (3.6) may be solved by the Frobenius method (see e.g. [2, Sec. 4.4]): Any differential equation of the form b(x) 0 c(x) y + 2 y=0 x x where the functions b(x) and c(x) are analytic at x=0, has at least one solution that can be represented in the form y 00 +

(3.9)

y(x) = xr (a0 + a1 x + a2 x2 + · · · )

where the exponent r may be any (real or complex) number (and r is chosen so that a0 6= 0). The equation also has a second solution (such that these two solutions are linearly independent) that may be similar to (3.9) (with a different r and different coefficients) or may contain a logarithmic term.

6

BJØRN URSIN KARLSEN

Temporary replacing r with x and F with y in Equation (3.6) yields 2 0 −2 + p2 x2 y + y = 0. x x2 This equation has at least one solution of the form y 00 +

(3.10)

∞ X

y(x) = xr

am x m =

m=0

∞ X

am rm+r .

m=0

The first and second derivative becomes ∞ X y 0 (x) = am (m + r)xm+r−1 , m=0 ∞ X

y 00 (x) =

am (m + r)(m + r − 1)xm+r−2 .

m=0

We expand b(x) and c(x) in power series with coefficients b0 = 2, b1,2,··· = 0, c0 = −2, c1 = 0, c2 = p2 , c3,4,··· = 0. The indicial equation r(r − 1) + b0 r + c0 = 0 takes the form r(r − 1) + 2r − 2 = 0 (3.11)

r1 = 1,

r2 = −2.

Here we have a case 3 situation (roots differing by an integer, r1 − r2 = 3). We also note that r1 − r2 > 0 so we have the two solutions (3.12)

y1 (x) = xr1 (a0 + a1 x + a2 x2 + · · · ),

(3.13)

y2 (x) = ky1 (x) ln x + xr2 (A0 + A1 x + A2 x2 + · · · ),

where k may or may not be equal to zero. The solution and derivatives inserted into the original equation x2

∞ X

am (m + r)(m + r − 1)xm+r−2

m=0

+ 2x

∞ X

am (m + r)xm+r−1

m=0

−2

∞ X

m+r

am x

2 2

+p x

m=0

∞ X

am xm+r = 0.

m=0

From the indicial equation (3.11) we choose r = 1 and collect the parts with the same power of x as (s + 1)sxs+1 + 2as (s + 1)xs+1 − 2as xs+1 + p2 as−2 xs+1 = 0, as (s2 + s + 2s + 2 − 2) + as−2 p2 = 0, as · s(s + 3) = −as−2 p2 , as = as−2

−p2 , s(s + 3)

7

and develop the even coefficients of an a0 · 0(0 + 3) = 0, a0 = a, (−p2 ) (−p2 ) =a , 2·5 2·5 (−p2 )2 (−p2 ) =a , a4 = a2 4·7 2·5·4·7 (−p2 ) (−p2 )3 a6 = a4 =a , 6·9 2·5·4·7·6·9 ····················· a2 = a0

3 · (m + 2) , (m + 3)! h i m 3 3 = a(−1) 2 · pm − , (m + 2)! (m + 3)! i h 3 3 − , a2n = a(−1)n p2n (2n + 2)! (2n + 3)! m

am = a(−1) 2 · pm

(3.14)

m = 0, 2, 4, 6, · · · , n = 0, 1, 2, 3, · · · .

The odd coefficients of an are all zero: a1 · 1(1 + 3) = 0, a1 = 0, (−p2 ) = 0, 3·6 a5 = a7 = a9 = · · · = 0. a3 = a1

According to (3.12) the first solution is: y1 =xr1

∞ X

a2n x2n

n=0

=x

∞ X n=0

h

i 3 3 − x2n (2n + 2)! (2n + 3)!

´ p2 x3 p4 x5 x p2 x3 p4 x5 − + − ··· − + − + ··· 2! 4! 6! 3! 5! 7! ³ ´i 3a0 h (px)2 (px)4 (px)6 = 2 1− 1− + − + ··· p x 2! 4! 6! ³ ´i 3a0 h (px)3 (px)5 (px)7 − px + px − + − + · · · , p3 x2 3! 5! 7! 3a0 3a0 y1 = 3 2 sin (px) − 2 cos (px). p x p x =3a0

(3.15)

a0 (−1)n p2n

³x

There may be another solution (3.13) of the form y2 = ky1 ln(x) + (A0 + A1 x + A2 x2 + · · · ), where k might or might not be zero. First I try the solution where k is set to zero. From the indicial Equation (3.11) we choose r2 = −2 and collect the parts with the

8

BJØRN URSIN KARLSEN

same power of x. Am (m − 2)(m − 3)xm−2 + 2Am (m − 2)xm−2 − 2Am xm−2 + p2 Am xm = 0, As [(s − 2)(s − 3) + 2(s − 2) − 2] = −p2 As−2 , (3.16)

As (s − 3)s = −p2 As−2 , As = As−2

−p2 , s(s − 3)

(s 6= 0, 3),

and develop the coefficients of An A0 (0 − 3)0 = 0, A0 = A0 , (−p2 ) , 2·1 2 −p (−p2 )2 A4 = −A2 = A0 , 4·1 1·2·4 (−p2 )3 −p2 = A0 , A6 = −A4 3·6 1·2·3·4·6 ······ A2 = −A0

n n p (n − 1) An = −A0 (−1) 2 , n = 0, 2, 4, · · · , n! h pn n pn i − , n = 2, 4, 6, · · · , An = A0 (−1) 2 n! (n − 1)! h p2s p2s i A2s = A0 (−1)s − , s = 1, 2, 3, · · · . (2s)! (2s − 1)!

We see from (3.16) that A1 = 0, but that A3 may still be 6= 0. We get the odd coefficients of An : A1 (1 − 3)1 = 0, A1 = 0, A3 (3 − 3)3 = A1 (−p2 ) = 0, A3 = A3 , (−p2 ) , 5·2 (−p2 )2 (−p2 ) = A3 , A7 = A5 7·4 2·4·5·7 ······ A5 = A3

pn−3 · 3 · (n − 1) , n = 3, 5, 7, · · · , n! h p2s p2s i = 3A3 (−1)s − , s = 0, 1, 2, · · · . (2s + 2)! (2s + 3)!

An = A3 (−1) A2s+3

n−3 2

·

9

We first seek the partial solution of (3.13) with the odd coefficients of An set to zero: ∞ h ³ p2s X p2s ´ 2s i y21 = x−2 A0 + − x A0 (−1)s (2s)! (2s − 1)! s=1 ´ ³ (px)2 (px)4 (px)6 + − + ··· = A0 x−2 1 − 2! 4! 6! ³ (px)1 ´ 3 5 (px) (px) + A0 px−1 − + − ··· , 1! 3! 5! A0 p A0 y 21 = sin(px) + 2 cos(px). x x

Next we seek the partial solution of (3.13) with the even coefficients of An set to zero: y22 =x−2

∞ X

A2s+3 x2s+3

s=0

=x−2

∞ X

3A3 (−1)s p2s

s=0

i 1 1 − x2s+3 (2s + 2)! (2s + 3)!

p2 x3 p4 x5 + − ··· 2! 4! 6! ´ p2 x3 p4 x5 x − + ··· ··· − + 3! 5! 7! ³ ´i 3A3 h (px)2 (px)4 (px)6 = 2 1− 1− + − + ··· p x 2! 4! 6! ³ ´i 3A3 h (px)3 (px)5 (px)7 − px + px − + − + · · · , p3 x2 3! 5! 7! 3A3 3A3 = 3 2 sin (px) − 2 cos (px). p x p x =3A3

y 22

³x

h

−

We notice that the partial solution y22 is identical to the partial solution y1 , hence the complete solution has got to be of the form y = ky1 ln x + y21 + y22 . We can test this equation in order to see if k 6= 0. y = [y1 ] · k ln x + [y21 + y22 ] 1 y 0 = [y10 ] · k ln x + y1 + [y20 1 + y20 2 ] x 2y 0 1 y 00 = [y100 ] · k ln x + 1 − y1 2 + [y2001 + y2002 ]. x x We already know that the expressions in square brackets are partial solutions to Equation (3.6) and zeroes out, hence it only remains to see if the rest of the terms

10

BJØRN URSIN KARLSEN

may be another partial solution with k 6= 0 or if k has got to be zero. We have 2 1 1 − y1 2 ) + 2x(y1 ) x x x =2xy10 − y1 + 2y1

x2 (y10

=2xy10 + y1 2a 2a ap =2x[− 3 sin(px) + 2 cos(px) + sin(px)] px x x a a + 2 sin(px) − cos(px) px x 3a 3a = cos(px) − 2 sin (px) + 2ap sin(px). x px This expression cannot be zero for any range of r, so k has got to be zero, and the solution of the differential equation is given by i 1h 3A3 y = A0 p sin(px) − 2 cos(px) x p i 1 h 3A3 + 2 sin(px) + A cos(px) , 0 x p3 and by reentering F = y and r = x we finally acquire the solution of Equation (3.6): i 1h 3A3 (3.17) F = A0 p sin(pr) − 2 cos(pr) r p i 1 h 3A3 + 2 sin(pr) + A cos(pr) . 0 r p3 By (3.8) and (3.17) the displacement vector, u(r, t) = F G · ˆ r, becomes (3.18)

u(r, t) =[C1 sin(c1 pt) − C2 cos(c1 pt)] n1h i 3A3 · A0 p sin(pr) − 2 cos(pr) r p io 1 h 3A3 + 2 sin(pr) + A cos(pr) ˆ r. 0 r p3

Equation (3.18) is the complete solution for concentric waves in the spatial continuum. By a suitable choice of constants we can investigate how standing waves may occur between imaginary fixed spheres in the spatial continuum, but more important, we can see if a standing wave may occur between the inside of a fixed sphere and the center node. There might be a possibility as we can see from the graph in Figure 1. A0 has got to be zero (else |u| would raise beyond any limits when r approaches zero), the radius of the sphere is given by the choice of p, and A3 dictates the amplitude of the standing wave. Under these conditions there might be established a standing wave between r → 0 and r = R1 , which is oscillating with a frequency, f , given by c1 p = 2πf hence (3.19)

f=

c1 p , 2π

and p =

2πf . c1

The displacement function u(r) approaches zero as r → 0 and r = Rn , which is the necessary condition that there might be a standing wave between these two

11

div u

u A>0

A=0 0 A<0

0

R1

r

Figure 1. Standing wave between the inside of a rigid shell with radius R1 and the center node.

borders. This situation is covered by the expression (3.20)

u(r, t) = A cos(c1 pt) ·

h sin(pr) p3 r 2

−

cos(pr) i ˆ r p2 r

when C1 and A0 are set to zero, and A = −3A3 C2 . In order to find the radii of the possible spheres, we set u(r, t) = 0 which gives (3.21)

tan(pRn ) = pRn ,

where Rn are possible radii of the firm shell. There is a solution for every tan(pRn ) = pRn , and we also notice that the first radius to be considered is when pR1 ≈ 4.49341 radians. By deriving Equation (3.20) one and two times with respect on t we acquire the velocity and acceleration fields (3.22)

u(r, ˙ t) = −Ac1 sin(c1 pt) ·

h sin(pr) p2 r 2

−

cos(pr) i ˆ r, pr

−

cos(pr) i ˆ r. r

and (3.23)

u ¨ (r, t) = −Ac1 2 cos(c1 pt) ·

h sin(pr) pr2

12

BJØRN URSIN KARLSEN

˙ and u ¨ really approach zero when r → 0, we can apply In order to see that u, u, l’Hˆopital’s rule: sin(pr) − pr cos(pr) [sin(pr) − pr cos(pr)]0 = lim r→0 r→0 (pr)2 [(pr)2 ]0 1 = lim sin(pr) = 0. r→0 2 lim

(3.24)

By applying the identity, div (f (r)ˆ r) = 2f (r)/r + f 0 (r), on Equation (3.20), we acquire div u =

2u ∂u + r ∂r

(3.25)

h 2 sin(pr)

2 cos(pr) p2 r2 2 sin(pr) cos(pr) cos(pr) p sin(pr) i − + 2 2 + 2 2 + p3 r3 p r p r p2 r sin(pr) div u = A cos(c1 pt) , pr = A cos(c1 pt)

p3 r 3

−

and by the identity, grad (f (r)) = f 0 (r)ˆ r (3.26)

h sin(pr) cos(pr) i grad div u = −A cos(c1 pt) − ˆ r. pr2 r

It is noteworthy that even if the node itself is a mathematical singularity such that r 6= 0, the limit of both the displacement u and the divergence of u have finite limits when r is approaching zero3. We easily verify from Equation (3.26) and (3.23) that the N-C equation for a solenoidal field (1.3) is satisfied. The amplitude of the displacement from inside a sphere of radius r = π/p where div u is positive is given by Z π/p sin(pr) D= A 4πr2 dr pr 0 4πA h sin(pr) r cos(pr) iπ/p = − p p2 p 0 2 4π A (3.27) = . p3 This thought experiment shows that standing waves may be established between a single node and a fixed concentric shell, but it also suggests that individual points in space may be nodes in a more complicated system of standing waves. 4. Solenoidal standing waves The Navier-Cauchy equation for solenoidal deformations is (4.1) sin(x) 3Note that lim = 1. x→0 x

curl curl u = −

1 u ¨. c2

13

Solving this equation for a deformation around a singularity may be a bit tricky, so first I will try to guess a solution of the form: (4.2)

u(r, t) = g(r, t)(m ˆ ׈ r),

where m ˆ is a fixed direction in space. By the identities curl (φA) = φ curl A+ grad φ×A, curl (c×r) = 2c, curl r = 0, grad (cr) = c, and A × (B × C) = (AC)B − (AB)C where c is a constant vector, r a radius vector, and φ a scalar field, we develop g curl [g(m ˆ ׈ r)] = curl [ (m ˆ × r)] r g g ˆ × r) + grad × (m ˆ × r) = curl (m r r 0 2g g g = m ˆ + ( − 2 )ˆ r × (m ˆ × r) r r r 2g g ˆ + (g 0 − )[(ˆ r ·ˆ r)m ˆ − (ˆ r · m)ˆ ˆ r = m r r 0 ¡g ¢ ¡g g¢ = + g0 m ˆ − − 2 (r · m)ˆ ˆ r, r r r and further g g curl curl [g(m ˆ ׈ r)] =( + g 0 ) curl m ˆ + grad ( + g 0 ) × m ˆ r r 0 ¡ g0 ¡g g¢ g¢ − 2 (r · m) ˆ curl ˆ r − grad [ − 2 (r · m)] ˆ ׈ r − r r r r ¡ g0 g0 g g ¢0 =( − 2 + g 00 )(ˆ r × m) ˆ − − 2 ˆ r(r · m)] ˆ ׈ r r r r r ¡ g0 g¢ − − 2 grad (r · m) ˆ ׈ r r r ¡ g0 g0 g g¢ =( − 2 + g 00 )(ˆ r × m) ˆ − − 2 (m ˆ ׈ r) r r r r 0 2g 2g =(g 00 + − 2 )(ˆ r × m). ˆ r r With these terms inserted into Equation (4.1) we acquire (4.3)

(g 00 +

2g 0 2g 1 − 2 )(ˆ r × m) ˆ = 2 g¨(ˆ r × m), ˆ r r c

or the scalar differential equation 2g 1 2g 0 − 2 = 2 g¨, r r c which is exactly like Equation (3.5) except for the value of c, and it gives the same solution, hence h sin(qr) cos(qr) i (4.5) − (ˆ r × m), ˆ u = M cos(cqt) q3 r2 q2 r (4.4)

g 00 +

By deriving Equation (4.5) one and two times with respect on t we acquire the velocity and acceleration fields h sin(qr) cos(qr) i (4.6) − (m ˆ ׈ r), , u˙ = −M c sin(cqt) · q 2 r2 qr

14

BJØRN URSIN KARLSEN

and (4.7)

u ¨ = −M c2 cos(cpt) ·

h sin(qr) qr2

−

cos(qr) i (m ˆ ׈ r).. r

It follows from the above equations and some straightforward work that h sin(qr) cos(qr) i (4.8) − (m ˆ ׈ r), curl curl u =M cos(cqt) qr2 r which together with Equation (4.7) inserted into the Navier-Cauchy equation confirms that Equation (4.5) is a solution. We notice that the irrotational and solenoidal components of u are orthogonal to each other as expected. It might be of interest to see how curl u behave when r approaches zero. By applying l’Hˆopital’s rule we find in a straight forward way that lim curl u = 32 M cos(cqt)m. ˆ

(4.9)

r→0

This interesting result confirms that curl u - and hence also curl u˙ - has a finite value in the vicinity of an oscillating singularity. 5. Standing waves between singularities In the preceding section I discussed the possibility that there might be a standing wave between a node and a concentric firm shell. The most important result was that a singularity may form the one endpoint in a standing wave. An imaginary firm shell is of course not realizable in a spatial continuum of infinite extension, but singularities are indeed possible entities. The question therefore naturally arises if standing waves can still be formed in the spatial continuum without the participation of any firm shells. By the identity (5.1)

sin x cos y = 12 [sin(x + y) + sin(x − y)],

Equation (3.25) can be rewritten into: A cos(c1 pt) sin(pr) pr ¤ A £ = sin(pr + c1 pt) + sin(pr − c1 pt) , 2pr

div u = (5.2)

which can be interpreted as the sum of two spherical waves bouncing back and forth in opposite directions between the firm shell and the center node. The two waves can be said to be reflected successively from the firm shell and the singularity at the center. We realize that standing waves can be treated as the superposition of two or more progressive free waves. This makes the mathematics simpler, and we can investigate more complex systems of standing waves. In this case the waves move with the speed, c1 . The frequency is f = c1 p/2π, and the wavelength is λ = 2π/p. Next let us have a look at the energy which is involved in the standing wave between the center node and the shell. At the time t = 0 the whole energy is in the div u-field because the u-field ˙ equals zero all over the considered volume. By Equation (1.10) the energy inside a thin shell at the distance r from the center is de =

λ + 2µ ( div u)2 · 4πr2 dr, 2

15

Figure 2. Considering a point in spherical coordinates

and inside a shell with radius Rn Z 2π(λ + 2µ)A2 Rn 2 sin (pr)dr p2 0 sin(2pRn ) i 2π(λ + 2µ)A2 h Rn − . = 2 p 2 4p

ERn =

We notice that the total energy inside the sphere approaches infinity when Rn grows without any limit. Hence a single oscillating node cannot exist in the spatial continuum. Let us, however, make a new thought experiment. Inside a sphere with a great radius and many wavelengths of standing waves, there are two oscillating nodes oppositely placed at a short distance, z1 = +d and z2 = −d, from the center, and let the the two nodes oscillate in opposite phase with each other (see Figure 2). Then by adjusting Rn a little, it should be possible to let two progressive waves move from one node via the firm shell and encounter the other node in tune with the oscillation of that node, where it is reflected. We then would have a system of two separate waves bouncing back and forth between the two nodes via the firm shell. The superposition of the free waves would form a standing wave, and the problem is now to find the total energy in the new system. If it is finite and the energy is kept in the surrounding of the singularities, it may indicate that we could get rid of the reflecting shell and end up with a pure oscillating dipole. First I will only consider the energy in a volume element dV at a distance r from the center that is great in comparison with the distance 2d between the nodes. Then the angle between the axis through the two nodes and the direction towards the volume element can be considered to be the same, namely φ, from both nodes and the center point. The divergence in the volume element at P is the superposition

16

BJØRN URSIN KARLSEN

of the divergence from the two nodes at the time t = 0 h sin p(r + d cos φ)

div u = A

−

sin p(r − d cos φ) i p(r − d cos φ)

p(r + d cos φ) ¤ A£ ≈ sin(pr + pd cos φ) − sin(pr − pd cos φ) pr A = cos(pr) · sin(pd cos φ). pr

The energy in a zone between Rm À d and Rn when n À m in the two half spheres is ZZZ λs + 2µs E= ( div u)2 dV 2 Z Z Z o λs + 2µs Rn π π A2 n =2 cos2 (pr) · sin2 (pd cos φ)r2 sin φ dφ dθ dr 2 2 2 Rm 0 0 p r Z π Z π 2 Z Rn (λs + 2µs )A 2 2 = cos (pr)dr · sin (pd cos φ) sin φ dφ dθ p2 Rm 0 0 Z ¯ cos φ sin(pd cos φ) cos(pd cos φ) ¯π ¯ ¯π (λs + 2µs )A2 Rn ¯ ¯ ¯ ¯ + = cos2 (pr)dr¯ − ¯ ¯θ ¯ 2 p 2 2pd 0 0 Rm Z Rn i 2h sin(2pd) π(λs + 2µs )A 1− cos2 (pr)dr. E= 2 p 2pd Rm The property sin(2pd)/2pd has a maximum value of unity when 2dp approaches zero, hence the total energy has no upper bounds except when either d or p approaches zero. Hence we can conclude that a single isolated dipole cannot exist because it would take an unlimited amount of energy to keep it up. Notice by the way that E is vanishing in the equatorial direction from the origin and increases when φ shrinks towards zero (i.e. it has its maximum value in the z-direction). The next possibility I will examine is whether a chain of oscillating nodes may exist. So let an infinite chain of oscillating nodes with the same strength be organized along a straight line separated with a distance d and organized such that one node always oscillate in opposite phase with its two adjacent nodes. Hence the next node that oscillates in the same phase is at a distance λ1 = 2d. The divergence in a point P at a distance r from the chain and an offset z along the chain from node #0, is given by the superposition of the effects from all the nodes in the chain ¯ ¡ p ¢¯ ∞ ¯ 2 2 ¯ X ¯ n ¯ sin p r + (nd − z) p (−1) ¯ div u = A cos(c1 pt) ¯. ¯ p r2 + (nd − z)2 ¯ n=−∞ Let pd = π, and 0 < z < d, then we acquire c1 t) 2d ¯ ¯ ∞ ¯ sin ¡p( πr )2 + (πn − πz )2 ¢ ¯ X ¯ ¯ d (−1)n ¯ p πrd 2 · ¯. πz 2 ¯ ¯ ( d ) + (πn − d ) n=−∞

div u =A cos(2π

17

This formula should be valid for all of space, but let us see what div u amounts to strictly along the chain axis where r = 0. ¯ ∞ πz ¯¯ X ¯ c1 n ¯ sin(πn − d ) ¯ div u = A cos(2π t) (−1) ¯ ¯ 2d n=−∞ πn − πz d ¯ ¯ ∞ πz n¯ X ¯ c1 n ¯ sin(− d )(−1) ¯ (−1) ¯ = A cos(2π t) ¯ 2d n=−∞ πn − πz d = A cos(2π

∞ ¡ πz ¢¯¯ X c1 ¯¯ (−1)n t)¯ sin , ¯ d d n=−∞ |nπ − zπ d |

which is a alternating harmonic series that can be shown to be convergent. As stated above, the result is only valid strictly along the chain axis, but a simple consideration shows that the result is valid also when P is outside the axis. First take the sum when −m < n < m. It is of course finite. Then let m be so great that md À r. Then we can set r = 0 for the rest of the series, and the reminder will also be finite making the whole series finite. Even when r grows beyond all limits, the effect from one node will be canceled out by the effect from its neighbor nodes. This proves that a string of oscillating nodes may have a finite energy per unit length. Say that an extreme situation in the spatial continuum, with release of an enormous amount of energy, initiated a boiling phase where nodes of the type described above where formed. Then the nodes would have got to organize in a way that required the least amount of energy, i.e. along strings. Hence if oscillating nodes at all can be present in the spatial continuum, they will tend to organize along strings. The important result, that standing waves may form between nodes in long strings, opens for the possibility that space can be filled with a plethora of faint oscillating nodes that organize and reorganize along strings - not necessarily formed as straight lines - in an ever changing pattern. 6. Chains of irrotational and solenoidal oscillating nodes Since div u has a finite value at every point in space, we could express it as a function of type c1 πz div u = A(r, z) cos(2π t) sin( ), 2d d where A(r, z) is the amplitude at the point P . The oscillatory frequency is c1 (6.1) . f1 = 2d By the identity, cos a sin b = 1/2 sin(a + b) − 1/2 sin(a − b), we can decompose the function above into n £π ¤ £π ¤o 1 div u = A(r, z) sin (c1 t + z) − sin (c1 t − z) , 2 d d which can be interpreted as the superposition of two waves moving in opposite directions along the chain of nodes with velocity c1 . In much the same way as considered above, it should even be possible to assume a solenoidal standing wave in another chain of nodes that is given by a function of

18

BJØRN URSIN KARLSEN

the type n £ 2π ¤ £ 2π ¤o 1 curl u = A(r, θ, z) sin (ct + z) − sin (ct − z) . 2 d d The main difference between these two set of waves, is that the latter moves with the velocity c, which is probably about half of the velocity c1 , and the frequency c (6.2) f2 = . 2d Both sets of deformation fields can, however, be expressed as two progressive waves moving in opposite directions in space, and the resultant of these waves are standing waves along separate strings in space. So far all the deductions are done on the basis of the Navier-Cauchy equation, but in the N-C representation there is no coupling between the two types of oscillations. Hence a chain with both types of oscillation represented in the same set of nodes is not feasible in the N-C representation, but that doesn’t mean that a more thorough examination rules out this possibility. 7. The Navier-Stokes equation and Coupled oscillations In this section I will discuss the possibility that the two basic types of oscillation can occur in the same set of nodes. In the Navier-Cauchy representation the two wave equations are independent of each other and cannot interact, but not so in the Navier-Stokes representation of a continuous medium. Disregarding viscosity and the influence of any outer forces, the N-S equation takes the form h ∂ u˙ i (λs + 2µs ) grad div u − µs curl curl u = ρs + (u˙ · ∇)u˙ , ∂t or by the identity, grad (A · A) = 2[A × curl A + (A · ∇)A]: i h ∂ u˙ 1 + grad u˙ 2 − u˙ × curl u˙ (λs + 2µs ) grad div u − µs curl curl u = ρs ∂t 2 where the additional term is the directional derivative of u˙ with respect on t along the tangent vector to the velocity field. We now turn back to the central case with standing waves bouncing back and forth between a center node and an imaginary firm concentric sphere. The displacement vector u can as seen earlier be split into a radial component u1 = f (r, t)ˆ r and a tangential component u2 = g(r, t)(m ˆ ׈ r), which are orthogonal to each other. We see that the equation no longer can be split into two independent equations. We have at our hands two different oscillation systems which have a mutual coupling between them. I think that the NavierStokes equation is too complicated to be solved by conventional methods, so instead of solving the equation, I will try to figure out what the new terms will do to the partial solutions of the Navier-Cauchy equation. Let us first single out the term ρ(u˙ × curl u) ˙ and see what impact it may have on the oscillatory system. The effect is only significant with relatively great deformations in the near vicinity of the node. Here the N-S equation takes the form (note that u˙ 1 ⊥ u˙ 2 , div u2 = 0, curl u1 = 0): (λ + 2µ) grad div u1 − µ curl curl u2 ¡ ∂ u˙ 1 ¢ ∂ u˙ 2 1 1 =ρ + + grad u˙ 12 + grad u˙ 22 − u˙ 1 × curl u˙ 2 − u˙ 2 × curl u˙ 2 . ∂t ∂t 2 2

19

The terms in this formula are representing different components of force. If we multiply them with the velocity at a site, we get the power, or energy transfer per units of time and volume, that are transferred between the different fields. For example the formula ∂ u˙ 1 u˙ 1 · (λ + 2µ) grad div u1 = u˙ 1 · ρ , ∂t represent the energy transferred between the local potential and kinetic energies in an irrotational oscillation per time unit at a given moment. If both sides are negative, the energy transfer is from potential to kinetic energy, and if they are positive, the energy transfer is from kinetic to potential energy. By the way, this component of the oscillation is between compression and rarefaction and will not result in any net displacement of spatial mass. Now let us multiply the N-S equation with the velocity u˙ = u˙ 1 + u˙ 2 , and neglect all terms that are orthogonal to each other (note that the vector product of two vectors is orthogonal to both of them). We obtain c12 u˙ 1 · grad div u1 − c22 u˙ 2 · curl curl u2 = u˙ 1 ·

∂ u˙ 1 ∂ u˙ 2 + u˙ 2 · ∂t ∂t

1 + u˙ · grad u˙ 2 − u˙ 2 · (u˙ 1 × curl u˙ 2 ) − u˙ 1 · (u˙ 2 × curl u˙ 2 ). 2 The two last terms representing components of the power exerted by the two fields can be set like w1 = u˙ 1 · (u˙ 2 × curl u˙ 2 ), w2 = u˙ 2 · (u˙ 1 × curl u˙ 2 ). We see by the sequence in which the three terms occur that when one of the properties is positive, then the other is negative and vice versa. Clearly this shows that there may be an energy transfer back and forth between the irrotational and the solenoidal field in the vicinity of an oscillating node. Hence the two fields are coupled and there might be some normal modes of oscillation where the two fields can oscillate within the same firm shell with a common frequency. The three vectors u˙ 1 , u˙ 2 , and curl u˙ 2 make up a parallelepiped with volume w1 = w2 . Hence the energy that is given up from one field is exactly like the energy received by the other field. It is also implicit that the two velocity fields has got to be a quarter out of phase with each other such that the maximum displacement coincides with the maximum rotation. By (3.22) and (4.6) we could end up with two equations of the form u˙ 1 = A1 sin 2πνf (r)ˆ r (7.1) u˙ 2 = A2 sin (2πν + ψ)f (r)(ˆ r × m), ˆ where ψ is the phase shift between the two fields. According to the above assumption it might be like π/4. In Figure (3) I have tried to visualize some possible standard modes of oscillation which could occur depending of the initial condition. In the left columns the energy is shifted from displacement to rotation, and in the right columns the energy is shifted from rotation to displacement. In the upper row the oscillation remains symmetric, while in the lower row the oscillation becomes asymmetric, meaning that the inflation (or rarefaction) phase is stronger than the rarefaction (or inflation) phase, or the right/left orientation is stronger than the corresponding left/right

20

BJØRN URSIN KARLSEN

Rotation enhanced mode

Displacement enhanced mode

Asymmetric rotation enhanced mode

Asymmetric displacement enhanced mode

Figure 3. Coupled oscillations. The blue solid lines represent the div u-field and the red solid lines the curl u-lines. ˙ The smaller graphs in the middle represent the transfer of movement between the two main oscillations, and the thick solid lines are the resulting amplitudes. orientation. I also suspect that if these modes of oscillation (or something like them) at all can occur in a string of nodes, the lower ones would curl up along an interwound line while the two upper ones could occur along a straight line. Well, so much for speculations. Finally I will try to estimate what effect the term grad ( 12 ρ u˙ 2 ) can have on the displacement of spatial mass. The gradient of the kinetic energy density, 12 ρ u˙ 2 , can be taken up by the force (λs + 2µs ) grad div u1 , hence we can set up the equation ρs (λs + 2µs ) grad div u = grad u˙ 2 , 2 and because u = u1 + u2 , u1 ⊥ u2 , and div u2 = 0 we get (λs + 2µs ) grad div u =

ρs grad (u˙ 12 + u˙ 22 ), 2

and further, since the deformation is zero at infinity, we obtain div u =

ρs (u˙ 2 + u˙ 22 ). 2(λs + 2µs ) 1

The displacement D from a volume V is given by Z D= div u dV. V

21

By Equation (7.1) we acquire the displacement from inside a sphere with radius R nZ R £ ¤2 ρs D= A1 sin(2πνt)f (r) 4πr2 dr 2(λ + 2µ) 0 Z RZ π o £ ¤2 A2 sin(2πνt + ψ)f (r) sin φ 2πr sin φ r dφ dr + 0

=

0

£ 2 2 ¤ ρs A1 sin (2πνt) + 23 A22 sin2 (2πνt + ψ) · 2(λ + 2µ)

Z

R

f (r)2 4πr2 dr.

0

If we set ψ = ±π/2, the two oscillations are shifted one quarter of a period in relation to each other, and moreover if A12 = 2/3A22

(7.2)

the displacement is independent of time and we get Z Z R ρs 0 2 ( div u1 )dV = A f (r)2 4πr2 dr, 2(λ + 2µ) 1 0 V Z 1 ρs 2 u˙ dV, = (λ + 2µ) V 2 1 or E0 , λ + 2µ where E 0 is the kinetic energy inside the sphere. In the Navier-Cauchy representation of a string of oscillating node there is no net displacement of spatial mass from the system - the compression into one node is like the displacement from another node - but the above equation shows that the presence of kinetic energy will inevitably lead to a net displacement of spatial mass from the entire string. The energy component E1 is only a part of the total energy, E, in the system, so if it should happen to be like one third of the total energy, the displacement amounts to E D= , 3(λ + 2µ) which corresponds to the energy-displacement relation that I found in the paper Elastodynamics in a continuum of infinite extension. D=

References 1. S. Fl˝ ugge (ed.), Mechanics of solids ii, Encyclopedia of Physics, vol. VIa/2, Springer, 1972. 2. Erwin Kreyszig, Advanced engineering mathematics, 8 ed., JOHN WILEY and SONS, INC, 1999. E-mail address: [email protected]