Valuable discussions are gratefully acknowledged with John W. Cahn, J. G. Dash, and Yu. Khait. Research supported by Technion V.P.R. fund, S. Langberg Nuclear Research Fund, BSF 88-00240, and D.O.E. grant #DE-FG06-90ER45425.
Table i: Thermodynamic properties of ``premelted" bubbles.
.
E and S are, respectively, the energy and
entropy differences between the vibrating state 1 and the excited state 2. L
is an estimate of the number of surrounding atoms participating in the
impurity hopping in the excited state 2.
Appendix
We now derive Eq. (3) for the Mössbauer spectrum of a system with
the absorbing atom having two possible states, one vibrating about
the lattice site, and the other hopping around the lattice site. We
take P to be the probability of being in the vibrating state (state
1) with distribution around the lattice site 
. The
hopping state (state 2) has probability 
and distribution
. The origin is chosen at the lattice site.
The Mössbauer spectrum is given by[15]
where
is the Mössbauer line width, and
is the correlation between the position of the same Mössbauer particle at different times.
In evaluating 
, we are interested in those times that may
affect the Mössbauer line shape. We assume that 
, the time to
equilibrate within a given state, is much smaller than both 
and 
, where 
is the Mössbauer lifetime, and 
is the timescale for the two
states to reach equilibrium. We show below that a particle starting
in either of the two states approaches equilibrium at the same rate.
The time 
is on the order of a vibrational period, and so is
clearly much less than 
, and will therefore contribute only a
broad background to the Mössbauer spectrum. For this reason we only need to
consider times 
in evaluating 
. For such times,
is dependent on the initial state in which the atom is
distributed, but is independent of the initial position 
of
the Mössbauer atom. The average can therefore be done first over the
initial position 
, and then done over 
where care must
be taken to consider the initial state of the particle. The first
average gives
which gives the equilibrium distribution of the initial state. We are left with
where
and the average is now only over 
for each of the two
possible initial states.
These two averages are
and
both of which can easily be verified to give the correct
results for t=0 and 
. (The result for
is
. At long times, the equilibrium expression of
Eq. (A3) is recovered.). Combining all these gives
Taking the fourier transform, as in Eq. (A1), we get
where 
is the usual Lorentzian defined in Eq. (4),
and 
is the fourier transform of 
.
For 
that are gaussian distributions with mean-square
disorder 
, Eq. (A9) reduces to Eq. (3).
The rate 
is related to the relaxation of states 1 and 2 by
the relations
where 
are the numbers of particles in state i and 
are the relaxation rates of each state.
For equilibrium values of 
, there is no time dependence and
In terms of 
, we have
so that
For a deviation from equilibrium, 
, Eqs. (A10) and (A11) give
where
We note that 
for 
, which can explain
why at 
there is no noticeable broadened component starting to
appear as the transition to state 2 begins. Although at one may
expect 
to be initially small, 
will be much larger
and could then be buried in the background, consistent with the
measurements.
