Molecular Imaging. Markus Rudin. Читать онлайн. Newlib. NEWLIB.NET

Автор: Markus Rudin
Издательство: Ingram
Серия:
Жанр произведения: Медицина
Год издания: 0
isbn: 9781786346865
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are not independent because E is the negative gradient of a potential dependent on the charges but also on the polarization Pu, as we have seen in Eq. (2.11b). In [2] it is demonstrated that Eq. (2.29) can be expressed as:

      Putting δF = 0 we have:

image

      Note that if there were no constraints on δPu(r), i.e. if its arbitrary variations would not affect E, Eq. (2.24) would be satisfied for arbitrary variations δPu(r) only if

image

      that is, if

image

      which is the usual electrostatic relation between Pu and E for systems with equilibrium U-type polarization. In our nonequilibrium system, the variations δPu must satisfy the equation of constraint (2.24) where from we see that:

image

      The equation for the variation δF is (see Eq. (2.30a)):

      Subtracting Eqs. (2.30b) from (2.30a) one has:

      Equations (2.30a) and (2.31b) are to be satisfied simultaneously. Using the method of Lagrangian multipliers (see, e.g., Refs. [2025]), the second condition multiplied by a Lagrangian multiplier m is added to the first one to have:

image

      This is an identity valid for every arbitrary variation of Pu in each volume element. It can be equal to zero only if the term in braces is equal to zero and this condition gives:

image

      So this is the Pu we were looking for. The first term is the usual electrostatic equilibrium relation between Pu and E. A physical interpretation of Eq. (2.26) was given in the preceding section.

      M. expresses now E and E in terms of image and Ec which can be easily calculated from the known charge distribution. The formulas for E and E are [1]:

image

      With the aid of these equations, Eq. (2.26) for Pu becomes:

image

      Introducing the equations for E, E, and Pu into Eq. (2.13) for the electrostatic free energy of state X, we obtain:

      We also have:

      where from m can be calculated.

      We designate now the charges of the reactants 1 and 2 in state X by image and image The corresponding charges in X will be e1 and e2. The radii a1 and a2 are assumed not to change essentially during the course of the reaction.

      The vector image is the negative gradient of the potential which the reacting ions in state X would exert if they were in a vacuum rather than in a polarized medium. The potential exerted at point r by two ions of charges e1 and e2 is:

image

      where r1 and r2 are the distances of the field point r from the centers of the ions. The vector Ec is −∇ψ, ψ being given by the earlier equations. We thus have:

      The expressions for Ec and image are introduced into Eq. (2.32) and into Eq. (2.33) and the integrations are performed. The following integrals are used for this purpose:

      where i can be 1 or 2, R is the distance between the centers of the ions and the integration volume excludes the volume physically occupied by the two ionic spheres, that is, r1a1 and r2a2 simultaneously.

      With the aid of Eqs. (2.24), (2.32), (2.34), (2.35), (2.25),