Last updated: Feb. 15, 2003
Molecular Dynamical Estimates of Vibrational
Frequencies of a Cubic Crystalline Sphere
Vibrational frequencies of a crystal sphere are calculated in terms of the elastic constants for the case where C12=C44, based on molecular dynamics simulations of a simple cubic lattice with first and second neighbour coupling. The phonon spectra of spheres of NaCl, CsBr and RbI are obtained.

Introduction
   Current interest in the phonon spectrum and density of states of spherical nanoparticles [Cerullo, De Silvestri and Banin 1999 pdf ::] [Geller et al. 2001 pdf ::] [Patton and Geller 2001 abstract, preprint pdf ::] [Portales et al. 2002 pdf ::] [Simon and Geller 2002 PRB preprint :: preprint pdf ::] has led to a need for more precise theoretical calculation of the frequencies of the vibrational modes. These particles, ranging from 0.5 nm to 50 nm in size, are small enough that they may be single crystals, without grain boundaries. In this case, their elastic properties are not isotropic. Rather, their Young's modulus and other elastic properties depend on the direction along which the particle is stretched.
   The necessity to interpret experimental data has often led to the employment of isotropic continuum elasticity theory in situations where it does not apply. Furthermore, the observed density of states sometimes exhibits broadening which is not explainable from isotropic theory [Simon and Geller 2002 :: ::].
   In what follows, calculations of the vibrational spectrum of a sphere are calculated using a method that should be a very good approximation for NaCl and CsBr, and a fairly good approximation for RbI and RbBr. In addition, this provides some qualitative insight that should be of use for other materials.

Elasticity Theory
   An isotropic elastic material has two independent elastic constants, which are any two of the Young's modulus (Y), Bulk modulus (B), Shear modulus (μ) and Poisson ratio (ν). A general crystalline material is anisotropic and has a number of elastic constants according to its symmetry, summarized in table I:

Table I: Crystal Elastic Constants
Source: Ashcroft & Mermin, "Solid State Physics" Table 22.1, page 445
Kittel, "Introduction to Solid State Physics" Table 4, page 38
Crystal
symmetry		No. of elastic
 constants		Example Crystal structures
Isotropic 2 polycrystalline, glass, etc.
Cubic 3 sc, fcc, bcc, diamond, NaCl, CsCl
Hexagonal 5 hcp
Rhombohedral 6 B, As, Sb, Bi, Hg
Tetragonal 6 Pa
Orthorhombic 9  
Monoclinic 12  
Triclinic 18  

   The stress tensor σ has components
σxxσxyσxz
σyxσyyσyz
σzxσzyσzz
while the strain tensor e has components
exxexyexz
eyxeyyeyz
ezxezyezz
   Both σ and e are symmetric and only have six independent components. For a linear, homogenous material these are linearly related through the matrix equation

σxx  C11 C12 C13 C14 C15 C16   exx
σyy  C21 C22 C23 C24 C25 C26   eyy
σzz= C31 C32 C33 C34 C35 C36   ezz
σyz  C41 C42 C43 C44 C45 C46   eyz
σxz  C51 C52 C53 C54 C55 C56   exz
σxy  C61 C62 C63 C64 C65 C66   exy

   If the material possesses cubic symmetry, then the Cij are not independent, and the relation simplifies to
σxx  C11 C12 C12   exx
σyy  C12 C11 C12   eyy
σzz= C12 C12 C11   ezz
σyz  C44   eyz
σxz  C44   exz
σxy  C44   exy

so that all elastic properties of a cubic crystal require the knowledge of just three constants, C11,C12 and C44. Furthermore, if the material is isotropic, then

C12 = C11 - 2 C44
so that just two elastic constants, C11 and C44 are sufficient to specific the behavior of an isotropic material.
   For either case (isotropic or cubic), these may be used to obtain bulk modulus

B = (C11 + 2 C12) / 3

and shear modulus
μ = (C11-C12)/2

   Plane wave vibrations travelling in an elastic solid (whether isotropic or cubic) can be longitudinally (L) or transversely (T) polarized. Longitudinal waves are associated with compression of the material. Transverse waves create shear strains while keeping the density constant. In an isotropic elastic material with density ρ, the speed of longitudinal waves is CL = sqrt(C11/ρ) and transverse ("shear") waves move at speed CS = sqrt(C44/ρ). In a cubic crystal, the speed of plane waves also depends on the direction of their propagation, as given in table II:

Table II: Speed of Waves
Source: Kittel "Introduction to Solid State Physics" Figure 9, page 147
DirectionPolarization(Speed2) ρ
[100]longitudinal C11
[100]transverse C44
[110]longitudinal (1/2)(C11 + C12 + 2 C44)
[110]transverse C44
[110]transverse (1/2)(C11 - C12)
[111]longitudinal (1/3)(C11 + 2 C12 + 4 C44)
[111]transverse (1/3)(C11 + C44 - C12 )

Crystal Lattice Simulation
   The frequencies of the vibrational modes of a crystal sphere are determined by constructing a crystal lattice from point masses connected by springs. The resulting structure is given an initial velocity or initial strain, and the subsequent motion is simulated. Motions of individual atoms are Fourier analyzed to yield the spectrum of vibrational frequencies.
   The mass of the sphere is m. The mass of one atom in the sphere is matom. N atoms are used. The radius of the sphere is R. The atoms are arranged on a simple cubic lattice with side a. The density of the material is ρ = matoma-3. Memory limitations restricted N to 10000 or less.
   Each atom has six nearest neighbours, each a distance a away. A spring of force constant ksp1 connects nearest neighbour atoms. A spring of force constant ksp2 connects each atom to its 12 second neighbours, each at distance sqrt(2) a. Finally, a spring of force constant ksp3 connects each atom to its 8 third neighbours at distance sqrt(3) a.
   It is necessary to relate the spring constants to the bulk elastic constants, C11, C12 and C44. This is done in three steps:
Step 1. First, recall that the bulk modulus

B = (C11+2C12)/3 = V0 dP/dV.

Suppose the crystal is compressed by δ = dV/V. The energy required is obtained by integrating the work dW = PdV. For the uncompressed crystal, P=0. P increases linearly with dV and dP/dV = B/V. Integrating, U = (1/2)V0 B δ2. The strain of the crystal is exx = eyy = ezz = δ/3, and eyz = exz = exy = 0. Each nearest neighbour spring of length a is shortened by distance δ a/3, resulting in the storage of energy

(1/2) ksp1 (δ a/3)2.

   Considering first, second, third, fourth and fifth neighbours,

B = (1)(6) ksp1
-------------
(2) (9) a		 + (2)(12) ksp2
--------------
(2) (9) a		 + (3)(8) ksp3
-------------
(2) (9) a		 + (4)(6) ksp4
-------------
(2) (9) a		 + (5)(24) ksp5
--------------
(2) (9) a
so that
a (C11+2 C12) = ksp1 + 4 ksp2 + 4 ksp3 + 4 ksp4 + 20 ksp5

Step 2. Consider a longitudinal wave moving in the [100] direction. The speed (phase velocity) of the wave is sqrt(C11/ρ). The wavelength of the wave is λ = 2 π / kx, where kx is the wavevector. ux(nx,ny,nz) is the x displacement of the atom whose lattice position is (nx,ny,nz). All y and z displacements are zero.

ux(nx,ny,nz) = ux(0,0,0) exp(i nx kx a )

F(0,0,0,nx,ny,nz) is the magnitude of the force on atom (0,0,0) from atom (nx,ny,nz).
F(0,0,0,nx,ny,nz) = ksp.. ( r - a sqrt((nx)2+(ny)2+(nz)2) )
and
r = sqrt( x2 +y2 +z2 )

x = nx a + ux(nx,ny,nz) - ux(0,0,0)

y = ny a

z = nz a

Fx(0,0,0,nx,ny,nz) is the x component of the force on (0,0,0) from (nx,ny,nz).

Fx(0,0,0,nx,ny,nz) = (x/r) F(0,0,0,nx,ny,nz)
To linear order in ux(0,0,0),
Fx(0,0,0,nx,ny,nz) = ((nx)2/((nx)2+(ny)2+(nz)2) ksp.. (exp(i nxkx a) - 1 ) ux(0,0,0)

The next step is to expand the complex exponential exp(i nxkx a). To linear order it is i nxkx a. Because of the mirror symmetry of the lattice, this will add up to zero when all neighbours are summed over. The lowest nonvanishing term is -(1/2)(nxkx a)2. The total force on atom (0,0,0) is the sum of this. Force divided by ux(0,0,0) equals (- matom ω2) where ω is the angular frequency of the wave. Therefore,

matom ω2 = Σ ((nx)2/((nx)2+(ny)2+(nz)2) ksp.. (0.5)(nxkx a)2

The speed of the waves is sqrt(C11/ρ) = ω/kx. Thus, C11 = ρ (ω/kx)2. Recall also that ρ = matom a-3. Finally, C11 is obtained by summing over all neighbours the expression

C11 = Σ (ksp../(2a)) ((nx)4 /((nx)2+(ny)2+(nz)2)

   Considering up to fifth neighbours,

a C11 = ksp1 + 2 ksp2 + (4/3) ksp3 + 4 ksp4 + 13.6 ksp5

Step 3. Consider a transverse wave moving in the [100] direction. The speed (phase velocity) of the wave is sqrt(C44/ρ). The wavelength of the wave is λ = 2 π / kx, where kx is the wavevector. uy(nx,ny,nz) is the y displacement of the atom whose lattice position is (nx,ny,nz). All x and z displacements are zero.

uy(nx,ny,nz) = uy(0,0,0) exp(i nx kx a )

F(0,0,0,nx,ny,nz) is the magnitude of the force on atom (0,0,0) from atom (nx,ny,nz).
F(0,0,0,nx,ny,nz) = ksp.. ( r - a sqrt((nx)2+(ny)2+(nz)2) )
and
r = sqrt( x2 +y2 +z2 )

x = nx a

y = ny a + uy(nx,ny,nz) - uy(0,0,0)

z = nz a

Fy(0,0,0,nx,ny,nz) is the y component of the force on (0,0,0) from (nx,ny,nz).

Fy(0,0,0,nx,ny,nz) = (y/r) F(0,0,0,nx,ny,nz)
To linear order in uy(0,0,0),
Fy(0,0,0,nx,ny,nz) = [(ny)2/((nx)2+(ny)2+(nz)2)] ksp.. (exp(i nxkx a) - 1 ) uy(0,0,0)

The next step is to expand the complex exponential exp(i nxkx a). To linear order it is i nxkx a. Because of the mirror symmetry of the lattice, this will add up to zero when all neighbours are summed over. The lowest nonvanishing term is -(1/2)(nxkx a)2. The total y force on atom (0,0,0) is the sum of this. Force divided by uy(0,0,0) equals (- matom ω2) where ω is the angular frequency of the wave. Therefore,

matom ω2 = Σ (ny)2/((nx)2+(ny)2+(nz)2) ksp.. (0.5)(nxkx a)2

The speed of the waves is sqrt(C44/ρ) = ω/kx. Thus, C44 = ρ (ω/kx)2. Recall also that ρ = matom a-3. Finally, C44 is obtained by summing over all neighbours the expression

C44 = Σ (ksp../(2a)) (nx)2(ny)2 /((nx)2+(ny)2+(nz)2)

   Considering up to fifth neighbours,

a C44 = ksp2 + (4/3) ksp3 + 3.2 ksp5

Couplings Beyond Second Neighbours:
   Combining previous results,

a C12 = ksp2 + (4/3) ksp3 + 3.2 ksp5

In other words, C12 and C44 are always exactly the same, no matter what values of the spring constants are chosen. This equivalence has been verified numerically up to tenth neighbours.

First and Second Neighbour Model:
   Incorporation of springs connecting higher order nearest neighbours has two practical disadantages. First, there is a larger number of interactions that must be calculated, which will have a cost for calculation time. Storage space will be strained since larger lookup tables must be maintained. Second, since the distance over which the interaction is non-local is larger if more neighbours are considered, it would be expected that the continuum limit would be approached more slowly.
   It is thus desirable to test the lattice calculation in the situation in which it can be expected to perform the best, and that is when only first and second neighbour interactions are kept. In this situation,

a (C11+2 C12) = ksp1 + 4 ksp2

a C11 = ksp1 + 2 ksp2
and
a C44 = ksp2

Combining these, it turns out that C12 = C44. Even if more distant neighbour interations are included this still remains the case. Therefore, there is no value in including springs beyond second neighbours.    Analytical solutions for a sphere are available for an isotropic elastic medium, for which C12 = C11 - 2 C44. Combining these conditions, the result is that 3 C44 = C11. This corresponds to an isotropic elastic material whose Poisson ratio is exactly 0.25. Thus, the model can do two things: (1) It can simulate an isotropic material with Poisson ratio 0.25; (2) It can simulate a cubic crystal for which C12 = C44. A number of materials do have Poisson ratio close to 0.25 such as some iron alloys, Cerium (0.248), Yttrium (0.243), Gadolinium (0.259), Iridium (0.26), Lutetium (0.261) and Silicon Carbide (0.24). There are also some cubic crystals with C12 close to C44 such as NaCl, CsBr, RbI and RbBr.

Comparison With Exact Analytic Solution:
   The classic problem of the vibrational modes of an isotropic elastic sphere has been extensively studied. [Lamb 1882] [Love 1944] [Sato and Usami 1962] [Auld 1973] [Nishiguchi and Sakuma 1981] [Bastrukov 1994] [Graff 1994] [Bastrukov et al. 1998] [McDaniel and Holt 2000 pdf ::] [Ye 2000 pdf] [Patton and Geller 2001 pdf ::]. This problem finds diverse application in geophysics [geophysics introduction], seismology [simple introduction to earthquake waves], solid stars [Bastrukov 1996], silicon nanoparticles [Patton and Geller 2001 pdf ::] and aqueous foam spheres [McDaniel and Holt 2000 pdf ::], and an accurate list of the exact frequencies has recently been reported [Murray 2002].
   Consider a freely suspended isotropic elastic sphere not subject to body forces. The sphere has radius R, uniform density ρ, shear modulus μ and Poisson ratio ν. These parameters fully describe the problem. If the sphere is disturbed, it will begin to vibrate. A given eigenfunction has frequency ω (in radians per second). The speed of transverse waves (shear waves) is

Cs = sqrt(μ/ρ)

   It is convenient to express normal mode frequencies in terms of the dimensionless variable

η = ω R / Cs

   Modes are of two types: spheroidal and torsional. In spheroidal modes, the velocity field is of the form

u(r,θ,φ,t) exp(-i η Cs t / R ) = An ∇ [ j(n, kl r ) P(n,cos(θ)) ]
+ Bn ∇× ∇× [ r j(n, ks r ) P(n,cos(θ)) ]

where r is the position vector, j(n,x) are spherical Bessel functions of the first kind and P(n,x) are Legendre polynomials. The ratio Bn/An must be chosen so as to satisfy the boundary conditions [Murray 2000].
   The expression for the displacement field u for torsional modes is
u(x,y,z,t) exp(-i η Cs t / R ) = Cn ∇× [ (x,y,z) j(n, ks r ) P(n,cos(θ)) ]
    In the calculation, distance units are normally chosen so that the center to center distance between two atoms is 1.
   If the velocities of the atoms became too large, the forces would become nonlinear. Since we are interested in linear elastic behavior, the velocities of the atoms are kept small enough that the forces are linear. Normally, in these simulation, the peak speed of an atom is on the order of 1x10-5 Cs. If the speed is made too small, there are numerical precision problems.
   The sharpness of the Fourier spectrum peaks depends on the amount of time over which the integrations are made.

Figure 1. Dimensionless frequency parameter η for vibrational modes of a cubic crystal sphere are shown. The vertical axis is the reciprocal of the sphere's radius. The upper edge of the chart (above the black band) corresponds to infinite R. The bottom edge of the chart corresponds to R=2. The maximum R and maximum number of atoms used are plotted at the lower right. (cc2mod.cpp cc2sph0.gif cc2sph1.gif cc2sph2.gif cc2sph3.gif cc2tor2.gif cc2tor3.gif)

Table III: Test of Lattice Calculation
Typel η
(simul.)		 η
(exact)		 % diff.
sph 0 4.45 4.440 0.2%
sph 1 3.43 3.425 0.1%
sph 2 2.64 2.640 0%
sph 3 3.95 3.917 0.8%
tor 2 2.50 2.502 -0.1%
tor 3 3.87 3.865 0.1%

   The values of η from the simulations as shown in Table III were obtained by simple interpolation of Figures 1(a-f). The results in Table III show that the lattice calculation is able to accurately calculate the vibrational frequencies of an isotropic elastic sphere. This provides an independent check on the analysis as well as on the correctness of the computer program. The rapidity of the convergence as N increases demonstrates that reasonably accurate frequencies could have been obtained even if the simulation been limited to a much smaller number of atoms.

Vibrational Frequencies of Salt Spheres:
   The requirement that C12 = C44 restricts the applicability of the simulations. Table IV shows elastic constants of some salt crystals at 300K or room temperature.

Table IV. Elastic Constants of Selected Cubic Crystals
Source: Ashcroft & Mermin "Solid State Physics" Table 22.2 page 447
C. Kittel "Introduction to Solid State Physics" Table 3, page 150
Substance C11 C12 C44 C12 -C44
  (GPa) (GPa) (GPa) (%)
NaCl 48.7 12.4 12.6 -1.6%
CsBr 30.0 7.8 7.6 2.6%
RbI 25.6 3.1 2.9 6.7%
GaSb 88.5 40.4 43.3 -7.0%
RbBr 31.7 4.2 3.9 7.3%
LiCl 49.4 22.8 24.6 -7.6%
CsI 24.6 6.7 6.2 7.8%
Ni 245 140 125 11.3%
Fe 234 136 118 14.2%

   In order to make an approximate calculation of the vibrational frequencies, I use the average of C12 and C44. The results are shown in Figures 2, 3 and 4.

Figure 2. Dimensionless frequency parameter η for vibrational modes of a sphere of Sodium Chloride (NaCl) are shown. The vertical axis is the reciprocal of the sphere's radius. The upper edge of the chart (above the black band) corresponds to infinite R. The bottom edge of the chart corresponds to R=2. The maximum R and maximum number of atoms used are plotted at the lower right. (cc2mod.cpp cc2nacl1.gif cc2nacl2.gif cc2nacl3.gif cc2nacl4.gif)

   For NaCl at 300 K Cs[100] = 2410 m/s. The lowest few values of η are 2.48, 2.64, 2.74, 3.12, 3.60 and its breathing mode is at 4.77. For example, for a NaCl sphere of diameter 30 nm, the frequency of the lowest vibrational mode is ω = (2.48)(2410 m/s)/(1.5e- m) = 3.983e11 rad/s, or 6.34e10 Hz or 2.11 cm-1.


Figure 3. Dimensionless frequency parameter η for vibrational modes of a sphere of Cesium Bromide (CsBr) are shown. The vertical axis is the reciprocal of the sphere's radius. The upper edge of the chart (above the black band) corresponds to infinite R. The bottom edge of the chart corresponds to R=2. The maximum R and maximum number of atoms used are plotted at the lower right. (cc2mod.cpp cc2csbr1.gif cc2csbr2.gif cc2csbr3.gif cc2csbr4.gif)

   For CsBr at 300 K Cs[100] = 1310 m/s. The lowest few values of η are 2.54, 2.67, 2.80 and 3.17. The breathing mode is at η = 4.79. For a 30 nm diameter sphere of CsBr, the frequency of the lowest vibrational mode is 1.18 cm-1.

Figure 4. Dimensionless frequency parameter η for vibrational modes of a sphere of Rubidium Iodide (RbI) are shown. The vertical axis is the reciprocal of the sphere's radius. The upper edge of the chart (above the black band) corresponds to infinite R. The bottom edge of the chart corresponds to R=2. The maximum R and maximum number of atoms used are plotted at the lower right. (cc2mod.cpp cc2rbi1.gif cc2rbi2.gif cc2rbi3.gif)

   For RbI at 300 K Cs[100] = 904 m/s. The lowest few values of η are 2.58, 2.73 and 3.25. For the breathing mode, η = 6.27. For a 30 nm diameter sphere of RbI, the lowest frequency vibrational mode is 0.82 cm-1.

Finite Size Dependence
    As radius varies, a given mode appears as a sloped but approximately straight line on the graph. In every case, η increases as R gets bigger. This is likely a result of decreased phase velocity of phonons as the boundary of the Brillouin zone is approached, in other words, when the wavelengths are comparable to atomic distances. Thus, the finite size dependence of the mode frequencies is of the form

η(R) = β(∞)(1 - D / R )

ω(R) = (η(∞) Cs / R)( 1 - D / R )

where constant D depends on the mode, and Cs is the speed of transverse waves. Some modes have stronger finite size dependence than others. Apparently, D is always positive. For most modes, D is on the order of 1 or less. But remember that I am using microscopic distance units. If R was in metres, then D would be on the order of 0.3 nm or so.
   Researchers studying nanoparticles normally adapt results from continuum elasticity theory without considering finite size effects [Cerullo, De Silvestri and Banin 1999 pdf ::] [Geller et al. 2001 pdf ::] [Patton and Geller 2001 abstract, preprint pdf ::] [Portales et al. 2002 pdf ::].
   Finite size dependence of vibrational modes has been seen in nanospheres [ Cerullo, De Silvestri and Banin 1999 pdf :: ].

References:

H. Lamb, "On the vibrations of an elastic sphere" Proc. London Math. Soc. 13, 189 (1881-1882).
A. E. H. Love, A Treatise on the Mathematical Theory of Elasticity, (Dover, New York, 1944).
Y. Sato and T. Usami, Geophys. Mag. 31, 15 (1962)
C. Kittel, "Introduction to Solid State Physics" Wiley 4th Ed. 1971, see table 4 on page 38
B. A. Auld, Acoustic Fields and Waves in Solids, (John Wiley & Sons, New York, 1973).
N. Nishiguchi and T. Sakuma, Sol. Stat. Comm. 38, 1073 (1981).
Sergey I Bastrukov, Phys. Rev. E 49, 3166 (1994).
F. K. Graff, Wave Motion in Elastic Solid, (Ohio University Press, Ohio, 1994).
S. Bastrukov, Phys. Rev. E (1996)
Bastrukov et al., Physica A 250, 435 (1998).
Zhen Ye, "On the Low Frequency Elastic Response of a Spherical Particle" Chinese Journal of Physics Vol. 38, pages 103-110, (2000). pdf
J. Gregory McDaniel and R. Glynn Holt "Measurement of Aqueous foam Rheology by acoustic levitation" Phys. Rev. E Rapid Communications volume 61 R2204 (2000). pdf ::
Kelly Patton & Michael Geller "Phonon Spectrum in a Nanoparticle mechanically coupled to a substrate" Journal of Luminescence 94-95 (2001) 747-570 pdf ::
Michael R. Geller et al. "Theory of electron-phonon dynamics in insulating nanoparticles" Preprint 11 July 2001 t / d where vt is the bulk transverse sound velocity." > pdf ::
Kelly R. Patton, Michael R. Geller, "Phonon Spectrum in a Nanoparticle Mechanically Coupled to a Substrate" (Preprint Tues, June 26, 2001) abstract, preprint pdf ::
G. Cerullo, S. De Silvestri and U. Banin "Size-dependent dynamics of coherent acoustic phonons in nanocrystal quantum dots" Phys. Rev. B volume 60 (July 15, 1999) pdf ::
H. Portales, L. Saviot, E. Duval, M. Gaudry, E. Cottancin, M. Pellarin, J. Lermé and M. Broyer "Resonant Raman Scattering by Quadrupolar Vibrations of Ni-Ag Core-shell nanoparticles" (Preprint Mar 22 2002) pdf ::
Daniel T. Simon and M. Geller "Electron Phonon Dynamics in an Ensemble of Nearly Isolated Nanoparticles" (Preprint June 8, 2002) To appear in Phys. Rev. B volume 64 PRB preprint :: preprint pdf ::
Daniel B. Murray, "Molecular Dynamics Simulation of an Elastic Material" 2002 (link to article)
Daniel B. Murray, "Vibrational Frequencies of an Elastic Sphere" 2002 (link to article)

Byeong-Joo Lee "A Semi-Empirical Atomistic Approach In Materials Science and Engineering" pdf ::

also
http://www.postech.ac.kr/~calphad/files/kissme.pdf

http://www.ees4.lanl.gov/nonlinear/iwnem/abstracts.html TJ Ulrich, KR McCall, PA Johnson, TW Darling, A Migliori "Application of Resonant Ultrasound Spectroscopy (RUS) to Determine the Elastic Properties of Rock Samples " Summer Workshop 1999 CdSe nanocrystals with colour
http://home.uchicago.edu/~cwang2/research.html

Physical Properties of semiconductors
http://newton.ex.ac.uk/useful/scp.html

T. Iitaka and T. Ebisuzaki "First principles calculation of elastic properties of solid argon at high pressures" pdf ::

Albert Brown "..Ultrasonics.." ::
Daniel Murray
Associate Professor
Math, Stats & Physics Unit
University of British Columbia - Okanagan
Kelowna, BC, Canada
daniel "dot" murray "at" ubc "dot" ca

For a list of related articles click on the link.
Hosted by www.Geocities.ws

1