[Wien] alakali anisotropy

Saeid Jalali sjalali at phys.ui.ac.ir
Fri Mar 19 09:10:19 CET 2004


Hi Dominik,
Larger A corresponds to smaller C_11-C_12 (compared to C_44) based on your
definition of elastic anisotropy ratio in cubic symmetry. Smaller C_11-C_12
corresponds to more isotropic bonds along 11 and 12. So the trend of your
results does not appear to be wrong based on your definition of A.
Your,
Saeid.
PS: To ensure about any numerical calculations physical
test/interpretation/insight is essential.

> Dear colleagues,
>
> recently, I came across the elastic anisotropy
> ratio in cubic metals defined as a ratio of the
> trigonal shear modulus C_44 and the tetragonal shear
> modulus C'=(1/2)(C_11-C_12), i.e. A = C_44/C'.
> There is one striking fact I would like to understand:
>
> For tungsten, a 5d metal, A = 1, i.e. tungsten is
> elastically isotropic! On the other hand, the alkali
> metals, the most free-electron metals, exhibit the
> highest elastic anisotropy factor among the elements:
>
> Element  A
> ----------
> Li   9.391
> Na   8.286
> Rb   7.435
> K    7.250
> Cs   7.220
> ----------
>
> One would expect that alkali metals, as nearly free-electron
> metals, would be also nearly elastically isotropic. On the
> other hand, a high elastic anisotropy could be expected in
> d-metals, such as tungsten, exhibiting directional bonds. As I show
> above, just the opposite is true. Does anybody of you know what
> is the origin of such a high elastic anisotropy in alkali
> metals? And why is tungsten elastically isotropic? I will be
> grateful for any hint or reference!
>
> With Best Regards,
>
> Sincerely,
>
> Dominik Legut
>
> P.S. I know that these questions are not related to the Wien code, but
> still this is a physical community...
>
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