next previous
Up: Conditions for shock revival


Subsections

   
5 Structure of the atmosphere

Within the supernova shock, the infalling matter is strongly decelerated to a velocity $v_{\rm {s}} = (v_{\rm {p}}-U_{\rm {s}})/\beta +
U_{\rm {s}}$. For a stalled shock, $\vert v_{\rm {s}}\vert \ll \vert v_{\rm {p}}\vert$. Compared to the internal energy and the gravitational energy, the kinetic energy behind the shock is therefore negligibly small. The gas is further slowed down as it moves inward and settles onto the nascent neutron star. Between neutrinosphere and shock front ${\rm {d}}v/{\rm {d}}t
\approx 0$ is therefore a good assumption, i.e., the stellar structure is well approximated by hydrostatic equilibrium (Chevalier 1989; Bethe 1993, 1995; Fryer et al. 1996). Combining Eqs. (2) and (3) and using Eq. (6), the equation of hydrostatic equilibrium is found to be

 \begin{displaymath}-\,{1\over \rho}\,{\partial P\over \partial r} - {G\,\widetilde{M}\over r^2}
\,=\,0\ .
\end{displaymath} (47)

In the following, the solutions of this equation in the layers between neutrinosphere $R_{\nu }$ and EoS transition radius $R_{\rm {eos}}$and between $R_{\rm {eos}}$ and shock position $R_{\rm {s}}$ will be derived.

   
5.1 Hydrostatic equilibrium between $\mathsfsl{R}_{\nu}$ and $\mathsfsl{R}_{\mathsf{eos}}$

When nonrelativistic baryons dominate the pressure and relativistic electrons contribute, but positrons and radiation can be ignored because the electrons are mildly degenerate, the pressure can be expressed as

 
$\displaystyle P\,\approx\,P_{\rm {b}} + P_{\rm e^-}$ = $\displaystyle {\rho\over m_{\rm {u}}}\,kT\,
\left( 1 + {Y_{\rm e}\over 3}\,{{\cal F}_3(\eta_{\rm e})\over {\cal F}_2(\eta_{\rm e})} \right)$  
  $\textstyle \equiv$ $\displaystyle f_{\rm {g}}\,{\rho\over m_{\rm {u}}}\,kT \ .$ (48)

Since $Y_{\rm e}$ and the electron degeneracy do not vary strongly, the factor $f_{\rm {g}}$ can be considered as constant. Between $R_{\nu }$ and $R_{\rm {eos}}$ also the temperature is a slowly changing quantity, $T(r)\approx T_{\nu} \approx T_{\nu_{\rm e}}$ (compare Fig. 2). Hydrostatic equilibrium therefore implies

 \begin{displaymath}\rho(r)\,=\,\rho_{\nu}\,\exp\left\lbrack -\,{G\,\widetilde{M}...
...u}R_{\nu}}\,\left( 1-{R_{\nu}\over r} \right) \right\rbrack\ ,
\end{displaymath} (49)

where $\rho_{\nu}$ is the density at the neutrinosphere. Near the neutrinosphere, $r\approx R_{\nu}$, this can be approximated by
 
$\displaystyle \rho(r)$ $\textstyle \approx$ $\displaystyle \rho_{\nu}\,\exp\left( -\,{x\over h}\, \right)$ (50)
$\displaystyle {\rm {with}}$   $\displaystyle x \,\equiv\, r-R_{\nu} \ \ \ {\rm {and}}\ \ \
h \,\equiv\, f_{\rm {g}}\,{kT_{\nu}\, R_{\nu}^2 \over
G\,\widetilde{M}\,m_{\rm {u}}}\ \cdot$  

Using typical numbers gives

 \begin{displaymath}h \,\approx\, 2.9~10^4\,f_{\rm {g}} R_{\nu,6}^2
\left( {kT_{\...
...\right)^{\!\! -1}
\ \ \left\lbrack {\rm {cm}} \right\rbrack\ ,
\end{displaymath} (51)

where $R_{\nu,6}$ is the radius of the neutrinosphere in units of 106 cm.

The density declines exponentially outside the neutrinosphere with a scale height $h \ll r$, forming a sharp "cliff'' (Bethe & Wilson 1985; Bethe 1990; Woosley 1993a). For this reason the effective optical depth is dominated by the immediate vicinity of the neutrinosphere. Therefore the integration in Eq. (13) can be performed, using Eq. (50) for the density in the effective opacity of Eq. (16), to derive the neutrinospheric density (normalized to $10^{10}\,$g/cm3) as

 \begin{displaymath}\rho_{\nu, 10}\,\approx\,150\,f_{\rm {g}}^{-1}R_{\nu,6}^{-2}
...
...\left( {kT_{\nu}\over 4\,{\rm {MeV}}} \right)^{\!\! -3}\ \cdot
\end{displaymath} (52)

This result confirms that the density of the transition from the baryon-dominated to the radiation-dominated regime (Eq. (35)) is significantly lower than $\rho_{\nu}$.

   
5.2 Hydrostatic equilibrium between $\mathsfsl{R_{eos}}$ and $\mathsfsl{R_{s}}$

In the radiation-dominated region a large part of the pressure is due to relativistic electron-positron pairs and photons, but also contributions from nucleons and nuclei with number fractions Yi might not be negligible, therefore

 
P = $\displaystyle P_{\gamma} + P_{\rm e^\pm} + P_{\rm {b}}$  
  = $\displaystyle {a_{\gamma}\over 3}\,(kT)^4
\left\lbrack {11\over 4} + {15\over 2...
...ight)
+ {3\,\rho kT\,\sum_i Y_i\over m_{\rm {u}}a_{\gamma}(kT)^4} \right\rbrack$  
  $\textstyle \equiv$ $\displaystyle P_{\rm {r}}\left( 1 + {4\over g_{\rm {r}}\,s_{\gamma}} \right)
\,\equiv\, f_{\rm {r}}\,P_{\rm {r}} \ ,$ (53)

where $P_{\rm r}$ is the pressure associated with relativistic particles,

 \begin{displaymath}P_{\rm {r}} = g_{\rm {r}}\,P_{\gamma}\,=\,
{1\over 3}\,a_{\gamma}\,g_{\rm {r}}\,(kT)^4 \ ,
\end{displaymath} (54)


 \begin{displaymath}{\rm {with}}\quad
g_{\rm {r}} \equiv {11\over 4} + {15\over 2...
...\rm e}^2
\left( \pi^2 + {1\over 2}\,\eta_{\rm e}^2 \right) \ ,
\end{displaymath} (55)

$s_{\gamma}$ is the entropy per nucleon carried by photons, $s_{\gamma} = 4a_{\gamma}(kT)^3/(3\rho/m_{\rm {u}})$, and $\sum_i Y_i \approx 1$ because of the nearly complete disintegration of nuclei. If both the factor $g_{\rm {r}}$ and $s_{\gamma}$ are constant (which is roughly fulfilled in the radiation-dominated region between $R_{\rm {eos}}$ and $R_{\rm {s}}$ where the electron degeneracy parameter $\eta_{\rm e}$ divided by $\pi$ is small, and, as was discussed in Sect. 2, convective processes tend to homogenize the total entropy and thus also the radiation entropy; see Bethe 1996b) then also $f_{\rm {r}}$ can be considered as constant. In this case the pressure is simply proportional to (kT)4, both for the contribution from nucleons and for the contribution from photons plus electron-positron pairs (for a detailed discussion, see Bethe 1993).

This implies that the density $\rho $ is proportional to T3, i.e.,

 \begin{displaymath}P\,=\, f_{\rm {r}}\,g_{\rm {r}}\,{a_{\gamma}\over 3}\,(kT)^4
\,=\, K\,\rho^{4/3}\ .
\end{displaymath} (56)

Note that Eq. (56) is valid more generally than for radiation-dominated conditions (Bethe 1996b). Using

 \begin{displaymath}n_{\rm e} \,=\, Y_{\rm e}\,{\rho\over m_{\rm {u}}}\,=\, {8\pi...
...,
{(kT)^3\over (hc)^3}\,\eta_{\rm e}(\pi^2+\eta_{\rm e}^2) \ ,
\end{displaymath} (57)

the coefficient K can be determined as

 \begin{displaymath}K\,=\, f_{\rm {r}}g_{\rm {r}}{a_{\gamma}\over 3}\,(hc)^4
\lef...
...}\eta_{\rm e}(\pi^2 + \eta_{\rm e}^2)}
\right)^{\! 4/3} \cdot
\end{displaymath} (58)

If K in Eq. (58) is approximately constant, which is typically fulfilled in the region between $R_{\rm {g}}$ and $R_{\rm {s}}$(Bethe 1996b), Eq. (56) is a useful representation of the equation of state.

With Eq. (47), one can now determine the density distribution between $R_{\rm {eos}}$ and $R_{\rm {s}}$ in hydrostatic equilibrium as

 \begin{displaymath}\rho(r)\,=\,\left\lbrack \rho_{\rm {s}}^{1/3} + {1\over 4}\,
...
...1\over r}-{1\over R_{\rm {s}}} \right) \right\rbrack^3 \ \cdot
\end{displaymath} (59)

Inserting this in Eq. (56) and setting $K = P_{\rm {s}}/\rho_{\rm {s}}^{4/3}$, the pressure as a function of radius is obtained,

 \begin{displaymath}P(r)\,=\,\left\lbrack P_{\rm {s}}^{1/4} + {1\over 4}\,
{G\,\w...
...t( {1\over r}-{1\over R_{\rm {s}}} \right) \right\rbrack^4 \ ,
\end{displaymath} (60)

and kT(r) can also be found from Eq. (56) as

 \begin{displaymath}kT(r)\,=\, kT_{\rm {s}} + {1\over 4}
\left( {3\over f_{\rm {r...
...M}\over K^{3/4}}\left( {1\over r}-{1\over R_{\rm {s}}} \right)
\end{displaymath} (61)

when $kT_{\rm {s}}=\left\lbrack 3P_{\rm {s}}/(f_{\rm {r}}g_{\rm {r}}\,a_{\gamma}) \right\rbrack^{1/4}$is used. If the density-pressure relation is more general than Eq. (56), namely $P = K\rho^\gamma$ with K being constant, hydrostatic equilibrium implies

 \begin{displaymath}\rho(r)\,=\,\left\lbrack \rho_{\rm {s}}^{\gamma -1} + {\gamma...
...1\over R_{\rm {s}}} \right)
\right\rbrack^{1/(\gamma -1)} \ ,
\end{displaymath} (62)

which replaces Eq. (59).

Instead of the general solutions, Eqs. (59)-(61), simple power-laws,

 
$\displaystyle \rho(r)$ = $\displaystyle \rho_{\rm {s}}\left( {R_{\rm {s}}\over r} \right)^{\! 3}\ ,\ \
kT(r)\,=\,kT_{\rm {s}}\,{R_{\rm {s}}\over r}\ ,$  
P(r) = $\displaystyle P_{\rm {s}}\left( {R_{\rm {s}}\over r} \right)^{\! 4}\ ,$ (63)

yield a good approximation for the hydrostatic atmosphere, if K fulfills the condition $K = G\widetilde{M}/(4 R_{\rm {s}}\rho_{\rm {s}}^{1/3})$. Since $K = P_{\rm {s}}/\rho_{\rm {s}}^{4/3}$, this is equivalent to the requirement that

 \begin{displaymath}{P_{\rm {s}}\over \rho_{\rm {s}}}\,=\,
{G\,\widetilde{M}\over 4\,R_{\rm {s}}}\ \cdot
\end{displaymath} (64)

On the other hand, from Eqs. (39) and (43) one gets
 
$\displaystyle {P_{\rm {s}}\over \rho_{\rm {s}}}$ $\textstyle \approx$ $\displaystyle {\beta-1\over \beta^2}\,{2\alpha^2 G\,{\widetilde M}\over R_{\rm {s}}}
\left( 1-{U_{\rm {s}}\over v_{\rm {p}}} \right)^{\! 2}$  
  $\textstyle \sim$ $\displaystyle {6\over 49}\,{G\,{\widetilde M}\over R_{\rm {s}}}
\left( 1-{U_{\rm {s}}\over v_{\rm {p}}} \right)^{\! 2}\ \cdot$ (65)

The numerical factor on the right hand side of Eq. (65) was obtained with $\alpha\sim 1/\sqrt{2}$ (Bethe 1990, 1993) and $\beta \sim 7$ (Eq. (41)). Equation (65) shows that the requirement of Eq. (64) is reasonably well, although for small shock radii not very well, fulfilled. In the following the power-laws of Eq. (63) will therefore only serve to facilitate analytical evaluation, and the use of this approximate solution of hydrostatic equilibrium will be avoided where inconsistencies might result.

With Eqs. (33) and (63) the gain radius $R_{\rm {g}}$ and the conditions at the gain radius can be expressed in terms of the properties at the shock front and the characteristic parameters ( $T_{\nu_{\rm e}}$, $L_{\nu_{\rm e}}$) of the neutrino emission. Inserting the relation $kT_{\rm {g}} = kT_{\rm {s}}(R_{\rm {s}}/R_{\rm {g}})$ into Eq. (33) yields the gain radius (in units of $10^7\,$cm),

 \begin{displaymath}R_{{\rm {g}},7}\,\approx\,0.98\,\,R_{{\rm {s}},7}^{3\over 2}
...
...langle \mu_{\nu} \rangle_{\rm {g}}} \right)^{\! -{1\over 4}} ,
\end{displaymath} (66)

and for the temperature at the gain radius one gets

 \begin{displaymath}kT_{{\rm {g}},2}\,\approx\,
1.02\,\,R_{{\rm {s}},7}^{-{1\over...
...\langle \mu_{\nu} \rangle_{\rm {g}}} \right)^{\! {1\over 4}} ,
\end{displaymath} (67)

where $kT_{{\rm {g}},2} = kT_{\rm {g}}/(2\,{\rm {MeV}})$and $kT_{\nu_{\rm e},4}$ is the neutrinospheric temperature and $kT_{{\rm {s}},2}$ the postshock temperature normalized to 4$\,$MeV and 2$\,$MeV, respectively.

The assumptions made in this section to solve the equation of hydrostatic equilibrium in the layer between $R_{\rm {eos}}$ and $R_{\rm {s}}$do not seem to be very restrictive, because two-dimensional as well as one-dimensional simulations without convection (e.g., Bruenn 1993; Janka & Müller 1996, Fig. 6; Rampp 2000) yield density and temperature profiles in the postshock region which are very close to power laws with power law indices around 3 and 1, respectively. Near $R_{\rm {eos}}$ the contributions of relativistic and nonrelativistic gas components will become equally important. Here the exponentially steep density decline just outside the neutrinosphere must change to the power-law behavior behind the shock, and both of these limiting solutions will not provide a good description. The exact structure in the intermediate layer between $R_{\rm {eos}}$ and $R_{\rm {g}}$, however, does not play an important role in the further discussion and therefore a more accurate treatment is not necessary.


next previous
Up: Conditions for shock revival

Copyright ESO 2001