Re: [Phys-l] models of radioactivity

[John Denker <jsd@av8n.com>]

On 04/20/2009 12:57 PM, Edmiston, Mike wrote:
> Perhaps a minor point, but it initially threw me off, and perhaps
> others, was Hugh's use of the letter p for the radioactive decay
> constant.  There is a standard symbol for the decay constant, and it is
> lower-case lambda.  Thus, the half life is ln(2)/lambda.

It seems to me there are at least three points in play here.

Let's suppose for the moment that the intended object of interest 
was in fact the "radioactive decay constant".  It's not 100% clear
that this was the original intention, but it is at least plausible, 
given how this object appeared in the original equation (1):

         T = (ln 2)/p		(1)

I see the issues as:
 a) What symbol is used for this object;
 b) How this object is defined in words; and
 c) How this object is used later in the calculation.

I agree with M.E. that item (a) is a minor point.  I have nothing
to add or subtract, except to say that this minor point should not
distract us from (b) and (c), which are more important points, i.e.
real conceptual points.  Concepts are always more important than
notation.

The original post defined "p" in a way that is inconsistent with
the established definition of radioactive decay constant.  It went
on to use this symbol in a way consistent with the unconventional
definition (and therefore inconsistent with conventional usage).

On 04/20/2009 01:14 PM, Polvani, Donald G. wrote:

> > I'm not a nuclear physicist, so I don't have a keen sense of the
> > conventions used by nuclear physicists.  However, my understanding of
> > radioactive decay is that, if lambda is the decay constant expressed in
> > number of nuclei decaying per time unit (e.g. in 1 s), then the
> > probability (p) of any ONE nucleus decaying in that same time unit is
> > equal to lambda, so p has "dimensions" of probability per unit time and
> > is numerically the same as lambda.

Alas, that's not the proper definition of lambda.  That might be
kinda sorta maybe OK if the halflife is very long compared to one
second ... but there are plenty of radioactive species for which
the halflife is *not* at all long compared to one second.  This is
directly on-point because the dice-based model sits in the regime 
where the halflife is not much different from the unit of time.

If you want to do it right, define 

  lambda := -(d/dt) ln x                                 (11)

where x(t) is the amount of remaining material at time t.  Because of 
the logarithm, the dimensions of x drop out, and lambda has dimensions 
of inverse time.  Because of the derivative, lambda depends on the 
amount of decay in an _infinitesimal_ time, not "unit" time.  It is 
therefore insensitive to the nonlinearities intrinsic to the problem 
... in contrast to the definition (b) and usage (c), which were needlessly 
tangled up in the nonlinearities.

One can connect lambda to probability by writing 

      x(t) = x0 (1-P(t))                                  (12)

where P (capital P) is the _cumulative_ probability of decay up to time
t.  Then we have

      lambda = - (d/dt) ln (x0 (1-P))                      (13a)

                 dP/dt
             = ----------                                  (13b)
                1 - P(t)

so in terms of the probability density p := dP/dt, we see that p is not
the same as lambda.  It's not just a new symbol, it is a whole new concept.
It reduces to lambda in some cases ... but not in all cases, in particular
not in the case of the dice-model.


If you define things in the conventional way, the halflife of the dice model
is just what you would expect.