Context Navigation

← Previous Post
Next Post →

Adiabatic BE sphere

Thinking about the adiabatic BE sphere, I went through some calculations to check if we should change anything in the BE problem module. To begin, I refreshed my memory on deriving the Lane Emden equation, and then considered how the algebra would differ in an adiabatic case.

Isothermal case -

If we start with an ideal EOS,

= \rho K T /m

and assume isothermality so we can use the isothermal sound speed,

_{T} = (KT/m)^{1/2}

we can rewrite the ideal gas equation to get,

=\rho a_{T}^2

Plugging this into the equation for HSE,

\triangledown P / \rho - \triangledown \phi_{g} = 0

and integrating, gives

rho(r) = \rho_{c} exp(-\phi_{g}/a_{T}^2)

which plugging into

triangledown^2 \phi = 4 \pi G \rho

gives

frac{1}{r^2}\frac{d}{dr}r^2\frac{d\phi}{dr} = 4 \pi G \rho_{c} exp(-\phi/a_{T}^2)

Letting

psi = \phi/a_{T}^2 ~and ~\xi=(4\pi G \rho_{c}/a_{T}^2)^{1/2}r

allows one to scale the ODE into the following form called the Lane Emden equation:

frac{1}{\xi^2}\frac{d}{d\xi}\xi^2\frac{d\psi}{d\xi} = exp(-\psi)

Adiabatic case -

So, how does this change if we wish to use an adiabatic sound speed Cs, where

_{s}^2 = \gamma K T/m

The ideal gas eqn when combined with the adiabatic sound speed is:

= \rho C_{s}^2/\gamma

Plugging into HSE gives

frac{-C_{s}^2}{\gamma \rho} \triangledown \rho - \triangledown\phi = 0

implying,

rho = \rho_{c}exp(-\phi\gamma/C_{s}^2)

This gives ~ the same ODE as above, but in order to scale it this time so that we get the same form as the Lane Emden equation above, one must make following modified variable substitutions:

psi = \phi\gamma/C_{s}^2 ~and ~\xi = (4\pi G \rho_{c} \gamma/C_{s}^2)^{1/2}r.

Note the additional factor of gamma coming into xi.

Now the question is, how does this affect the BE module?

Well, the BE module reads in rhoc, r, xi, from which it computes an isothermal sound speed, aT, to get the temperature to set the pressure inside the clump.

The module uses the definition of xi above for the isothermal case to get aT:

_{T} = r(4 \pi G \rho_{c})^{1/2}/\xi

NOTE that for the adiabatic case, the sound speed should have a factor of gamma coming in the numerator under the square root.

Now, considering the adiabatic sound speed,

_{s} = (\gamma K T/m)^{1/2}

gives

= C_{s}^2 m/ \gamma K = r(4 \pi G \rho_{c} \gamma )m / \xi^2 \gamma K T = a_{T}^2 m / K

(The gamma factor cancels out anyway, leaving T unaffected as we should expect!!).

Results

T should be the same in an adiabatic BE sphere cloud (at least initially).
Xi, r, and rho_c are given in problem.data and together they fix the sound speed. Thus, the sound speed should be effectively different between the isothermal and adiabatic cases. However, as shown above, T is independent of gamma. Since T is used to set P, gamma doesn't effect the problem set up. Therefore, no changes should be made to the problem module.
Further, rho= rho(xi). Since xi has gamma in it for the adiabatic case, the stability criteria for the adiabatic case is going to be different. It would no longer be the case that for an adiabatic BE sphere, xi_crit = 6.451. Considering dp/drho shows that a factor of gamma would come into the P given by the adiabatic ideal gas law.

astrobear

Context Navigation

Adiabatic BE sphere

Comments

Download in other formats: