content_image/337989.jpg content_image/337975.jpg content_image/337976.jpg content_image/337977.jpg content_image/337978.jpg content_image/337979.jpg
where m$_{0}$ = ∑ n j $_{=1}$m$_{j}$ . We rewrite the full integral as k 1 1 c + i ∞ n m$_{j}$ 

<img src='content_image/337989.jpg'>

<img src='content_image/337975.jpg'>

Combining the factors resulting from the momentum integrals with those due to the fragment multiplicities, we define the translational Helmholtz free energies as 

<img src='content_image/337976.jpg'>

where we recognize that m$_{l}$ = m$_{N}$A$_{l}$ and m$_{0}$ = m$_{N}$A$_{0}$ , with m$_{N}$ the nucleon mass. We can then define the total Helmholtz free energy as 

<img src='content_image/337977.jpg'>

where we have replaced the sum over the fragments with a sum over distinct fragments times their multiplicities, and write the density of final states as 

where we have defined 

<img src='content_image/337978.jpg'>

<img src='content_image/337979.jpg'>

Up to this point, we have made several definitions but no approximations. This expression is exactly equivalent to that of Eq. (9). 

[1] E. Fermi, Prog. Theor. Phys. 5 (1950) 570. 

[2] E. Fermi, Phys. Rev. 81 (1951) 683. 

[3] J. J. Lord, J. Fainberg, and M. Schein, Phys. Rev. 80 (1950) 970. 

