Re: [Phys-l] models of radioactivity

[John Denker <jsd@av8n.com>]

On 04/20/2009 02:14 PM, I wrote:

> 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.

Continuing down that road:  The decay makes a straight line on semi-log
paper.  To say the same thing in more formal terms, once you take the 
logarithm, the problem is linear.  Therefore

                 -d(ln x)  
       lambda =  --------                                  (14a)
                    dt     

                 -Δ(ln x)
              =  ---------                                 (14b)
                    Δt

even for a non-infinitesimal Δ.

Let's consider the case where Δt spans the time interval from 0 to t.
Writing x in terms of P we get
 
                  -ln(1-P(t))
       lambda  =  -----------                                  (15)
                       t

To first order, in the limit where P(t) is small compared to unity, that 
reduces to P/t which is just p (the probability density).

The dice model has P(1) = 1/6, which is not super-small.  So if you
want accuracy to 9% or better, you need to use equation (15), not some
first-order approximation thereto.

In some circles it is traditional to "define" lambda as dP/dt, but that 
is not the most helpful definition, especially as applied to the dice
model where t is not a continuous variable.  Equation (15) or (14b) is 
in all ways preferable.

================

The larger lesson is that formulas such as 
  lambda = P/t 
and
  halflife = ln(2)/(P/t)
always come with provisos.  In this case these formulas work _provided_ 
P is small compared to unity.

It pays to be skeptical.  If you see "lambda = P/t" without any provisos,
check to see whether it makes sense.  Check the case of P near zero, and
then check the case of P near unity.  If you're in practice, it takes
only a couple of seconds to make such checks.

The hard part comes later;  once you know that "lambda = P/t" is not the
whole story, you might need to hunt around to find something usable.  You
might find provisos sufficient to make the formula usable, or you might
find a more general expression (such as equation 15).

Conversely, once you have the general expression, it pays to check that it
upholds the correspondence principle:  It should reproduce the less-general
expression in the appropriate limit.