Updated: January 23, 2008
l = 1 and l = 2 Spheroidal Decaying Modes of an
Infinite Elastic Matrix Surrounding a Fixed Rigid Sphere
The problem of the vibrational motion of an elastic sphere
surrounded by an elastic matrix can be solved to obtain different kinds of
motion.
Even though the solution of the more general problem can
readily be done, it is useful to consider the limiting case of a sphere that
is extremely heavy and rigid. In this case, the elastic behavior of the sphere
can be ignored, and the problem is solved by imposing boundary conditions
of zero displacement on the surface of the sphere.
Suppose the radius of the sphere is R.
The form of the displacement u(r,t)
outside the sphere is given
in terms of two components: which I will call "longitudinal" and "transverse".
The "longitudinal" part has zero curl (i.e. ∇×u = 0).
The "transverse" part has zero divergence (i.e. ∇.u = 0).
The wave is
assumed to be purely outgoing. This corresponds to an initial disturbance
in the vicinity of the sphere which then travels away in all directions.
The "longitudinal" (zero curl) part is:
u(r,t) = eiωt ∇ (A h(2)l(klm r) Pl(cos(θ)) )
where
A is a complex constant for the amplitude of the wave
h(2)l(x) is the spherical Hankel function of the second kind of order l
h(2)l(x) = jl(x) - i nl(x)
jl(x) = spherical Bessel function of the first kind (nonsingular at r=0)
nl(x) = spherical Bessel function of the second kind
( h(2)l(x) is an outgoing travelling wave; conversely, jl(x) + i nl(x) is an inward travelling wave )
klm = ω / clm (Note that klm is complex valued)
clm is the longitudinal speed of sound in the matrix material (in m/s)
ω is the (complex valued) frequency in radians/second
Pl(.) is the Legendre polynomial of order l
(l = 0, 1, 2, ...)
θ is the altitude angle in the range 0 < θ < π. (azimuthal angle φ goes from 0 to 2 π )
The "transverse" (zero divergence) part (which exists only for l≥1) is:
u = eiωt ∇ × ∇ × (r B h(2)l(ktm r) Pl(cos(θ)) )
where
r is the vector field whose components in spherical coordinates
are (r,0,0), and whose cartesian components are (x,y,z).
B is a complex constant for the amplitude of the wave
ktm = ω / ctm
ctm is the transverse speed of sound in the matrix material= sqrt(μmat/ρmat) (in m/s)
The boundary condition is u = 0 for r = R, for all θ and φ.
A and B are obtained subject to this condition.
Normally, the boundary conditions sufficiently constrain the
problem so that the only solution is A = B = 0. However, if ω is
allowed to be a complex number, there is a solution with nonzero A and B.
Since the result scales in a direct way with the radius of the sphere
and the speeds of sound, it is convenient to introduce a dimensionless (complex)
frequency, S.
ν = 0.01 S clm / ( d c )
where
ν (the Greek letter "nu") is the frequency expressed in "wavenumbers" (cm-1).
clm is the longitudinal speed of sound in the matrix material (in m/s)
d is the diameter of the sphere (metres)
c is the speed of light in a vacuum (2.9979×108 m/s)
I solved this problem for the particular case of an SiO2
matrix, for which clm = 5720 m/s and ctm = 3750 m/s.
If the matrix is glassy then it may be appropriate to assume that its elastic
properties are isotropic. The above solutions apply assuming isotropic elasticity.
Also, remember that for this approximate calculation, the nanoparticle itself is assumed to be infinitely
dense and hard.
Using the C++ program scp76b.cpp, I solve the problem
for the l = 1 case, and find just one solution in the entire range,
S = 0.2599 + 0.4227 i. The program scp76c.cpp was used
to precisely locate the root of the determinant. For a 3.5 nm sphere, this corresponds
to 14.17 cm-1 and a decay time of 0.2304 ps.
Using the C++ program scp76.cpp, I solve the problem
for the l = 2 case, and find just one solution in the entire range,
S = 0.5214 + 0.5026 i. Based on the real part of S, the real frequency of
the oscillations for a 3.5 nm diameter sphere would be 28.42 cm-1.
That is equivalent to f = 8.55×1011 Hz, and the period
of of the oscillation is T = 1/f = 1.17 ps.
Using the imaginary part of S, the damping time is 0.1938 ps.
In other words, the oscillations decay with
time as e-t/(0.1938 ps).
Thus, the oscillation is completely damped out
before one full oscillation is completed.
The correctness of these results has been checked against other
calculations which include the elastic response of the sphere. The limit in which
the sphere has high density and high speed of sound was then checked to see
if the results agreed. However, in making the comparison it is important to note
that these dimensionless frequencies (S) are given relative to clm,
whereas in the other calculations S is given relative to clp (longitudinal
speed of sound in the nanoparticle itself).
In the l = 2 case, the symmetry of the vibrations means
that the net force and net torque on the nanoparticle from the surrounding matrix are
always zero. Therefore, the nanoparticle will not move or turn as a result of these
vibrations.
Daniel Murray
Associate Professor
Math, Stats & Physics Unit
University of British Columbia - Okanagan
Kelowna, BC, Canada
daniel "dot" murray "at" ubc "dot" ca
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