Pressure, Incompressibility and Asymmetry Coefficients

At the saturation point, all first derivatives of the energy per nucleon have to vanish and all second derivatives have to be positive. The first derivative with respect to $\rho_0$ is related to the pressure, the second derivative with respect to $\rho_0$ is related to the incompressibility, and the second derivatives with respect to the I_i is related to the asymmetry coefficients. For symmetric matter, the first derivatives with respect to the I_i vanish because the energy per nucleon is an even function of all I_is. The pressure is given by

$$P = - \frac{\partial E}{\partial V} \bigg\vert _A = \rho_0^2 \frac{\partial {\cal H}/\rho_0}{\partial \rho_0} \quad ,$$ (48)

which gives

$${\textstyle\frac{{2}}{{5}}} {\textstyle\frac{{\hbar^2}}{{2m}}} \, \beta \, \rho_0^{5/3} F_{5/3}^{(0)}$$ P =

$$\quad + \big( C_0^{\rho} + C_1^{\rho} I_\tau^2 + C_0^{s} I_\sigma^2 + C_1^{s} I_{\sigma \tau}^2 \big) \, \rho_0^2$$

$$\quad + 3 \rho_0^2 \; \frac{\partial}{\partial \rho_0} \, \big( C... ...o} + C_1^{\rho} I_\tau^2 + C_0^{s} I_\sigma^2 + C_1^{s} I_{\sigma \tau}^2 \big)$$

$$\quad + \beta \Big( C_0^{\tau} \, F^{(0)}_{5/3} + C_1^{\tau} \, F^{(\tau)}_{5/3} \, I_\tau$$

$$\qquad + C_0^{T} \, F^{(\sigma)}_{5/3} \, I_\sigma + C_1^{T} \, F^{(\sigma\tau)}_{5/3} \, I_{\sigma \tau} \Big) \; \rho^{8/3} \quad .$$ (49)

The incompressibility is defined as

$$K = \frac{18 P}{\rho_0} + 9 \rho_0^2 \frac{\partial^2 {\cal H}/\rho_0}{\partial \rho_0^2} ,$$ (50)

which, for the Skyrme energy functional (49) at the saturation point ( $\rho_0=\rho_{\rm n.m.}$, $I_\tau = I_\sigma = I_{\sigma \tau} = 0$) gives

$$K_\infty - {\textstyle\frac{{6}}{{5}}} \left( {\textstyle\frac{{\hbar^2}}{{2m}}} - 5 C_0^\tau \, \rho_0 \right) \beta \, \rho_0^{2/3} F_{5/3}^{(0)}$$ =

$$+ 2 \rho_0^2 \frac{\partial C_0^\rho}{\partial \rho_0} + \rho_0^3 \frac{\partial^2 C_0^\rho}{\partial^2 \rho_0} .$$ (51)

The asymmetry coefficients are:

$$a_\tau \frac{1}{2} \frac{\partial^2 {\cal H}/\rho_0}{\partial I_\tau^2} \bigg\vert _{I_\tau = I_\sigma = I_{\sigma \tau} = 0}$$ =

$${\textstyle\frac{{1}}{{3}}} \left[ {\textstyle\frac{{\hbar^2}}{{2... ...3 C_1^{\tau} ) \, \rho_0 \right] \beta \, \rho_0^{2/3} + C_1^{\rho} \, \rho_0 ,$$ = (52)

$$a_\sigma \frac{1}{2} \frac{\partial^2 {\cal H}/\rho_0}{\partial I_\sigma^2} \bigg\vert _{I_\tau = I_\sigma = I_{\sigma \tau} = 0}$$ =

$${\textstyle\frac{{1}}{{3}}} \left[ {\textstyle\frac{{\hbar^2}}{{2... ...{\tau} + 3 C_0^T) \, \rho_0 \right] \beta \, \rho_0^{2/3} + C_0^{s} \, \rho_0 ,$$ = (53)

$$a_{\sigma \tau} \frac{1}{2} \frac{\partial^2 {\cal H}/\rho_0}{\partial I_{\sigma \tau}^2} \bigg\vert _{I_\tau = I_\sigma = I_{\sigma \tau} = 0}$$ =

$${\textstyle\frac{{1}}{{3}}} \left[ {\textstyle\frac{{\hbar^2}}{{2... ...\tau} + 3 C_1^T ) \, \rho_0 \right] \beta \, \rho_0^{2/3} + C_1^{s} \, \rho_0 .$$ = (54)

Here, $a_\tau$ is the well-known volume asymmetry coefficient of the liquid-drop model, and $a_\sigma$ and $a_{\sigma \tau}$ are its generalizations to the spin and spin-isospin channels of the interaction. At the saturation point, all asymmetry coefficients have to be positive.