## Determining Code Units

### Logic Used by Dominic Marcello

At our group meeting on 21 July 2010, Dominic explained how he had established the values of various coupling constants in his first, long, q0 = 0.7 simulations. In place of the physical constants, $~G$, $~c$, $~\Re$, and $~a_\mathrm{rad}$, Dominic used the following code-unit values — hereafter referred to as Case A:

• $\tilde{g} = 1$
• $\tilde{c} = 198$
• $\tilde{r} = 0.44$
• $\tilde{a} = 0.044$

This means that any temperature in the simulation that has a value Tcode in code units must represent an actual physical temperature Tcgs in cgs units (i.e., measured in Kelvins) of,

$T_\mathrm{cgs} = \biggl[ \biggl(\frac{c^2}{\Re}\biggr)\biggl(\frac{\tilde{c}^2}{\tilde{r}}\biggr)^{-1} \biggr] T_\mathrm{code} ;$

any length-scale in the simulation that has a value $\ell_\mathrm{code}$ must represent an actual physical length $\ell_\mathrm{cgs}$ in cgs units of,

$\ell_\mathrm{cgs} = \biggl[\biggl(\frac{\Re^4}{c^4 G a_\mathrm{rad}}\biggr)\biggl(\frac{\tilde{r}^4}{\tilde{c}^4 \tilde{g}\tilde{a}}\biggr)^{-1}\biggr]^{1/2} \ell_\mathrm{code} ;$

any time in the simulation that has a value tcode must represent an actual physical time tcgs in cgs units of,

$t_\mathrm{cgs} = \biggl[\biggl(\frac{\Re^4 }{c^6 G a_\mathrm{rad}}\biggr)\biggl(\frac{\tilde{r}^4}{\tilde{c}^6 \tilde{g}\tilde{a}}\biggr)^{-1}\biggr]^{1/2} t_\mathrm{code} ;$

and, finally, any mass in the simulation that has a value mcode must represent an actual physical mass mcgs in cgs units of,

$m_\mathrm{cgs} = \biggl[\biggl(\frac{\Re^4}{G^3 a_\mathrm{rad}}\biggr)\biggl(\frac{\tilde{r}^4}{\tilde{g}^3 \tilde{a}}\biggr)^{-1}\biggr]^{1/2} m_\mathrm{code} .$

Now, the SCF-code-generated polytropic binary that Wes Even gave to Dominic had the following properties, in dimensionless code units:

• [Mtotal]code = 0.85;
• [RAccretor]code = 0.4; and
• [Porbit]code = 31.

According to Dominic's calculations this means that his simulation represents a real binary system with the following properties:

• $[M_\mathrm{total}]_\mathrm{cgs} = 0.1 M_\odot$;
• $[R_\mathrm{Accretor}]_\mathrm{cgs} = 0.56 R_\odot$; and
• $[P_\mathrm{orbit}]_\mathrm{cgs} = 28~\mathrm{minutes}$.

Conversely — assuming pure helium, that is, a mean molecular weight $~\bar{\mu}$ of 2 — since the Thompson cross-section is $[\sigma_T]_\mathrm{cgs} = 0.2~\mathrm{cm}^2~\mathrm{g}^{-1}$, Dominic determined that, in the code, he needed to set the Thompson cross-section value to $[\sigma_T]_\mathrm{code} = 8\times 10^{12}$. Finally, Dominic pointed out that the characteristic size of a grid cell in the code is z]code = 0.025. Hence, if only the Thompson cross-section is relevant, the mean-free-path of a photon will equal the size of one grid cell if,

$\biggl[\frac{1}{\sigma_T\rho}\biggr]_\mathrm{code} = [\Delta z]_\mathrm{code}$

$\Rightarrow ~~~~~ [\rho]_\mathrm{code} = \biggl[\frac{1}{\sigma_T(\Delta z)}\biggr]_\mathrm{code} = \frac{1}{2\times 10^{11}} .$

### Joel's Check of Dominic's Logic and Numbers

Let's plug in values of the physical units that we have tabulated in a Variables Appendix to see if we agree with Dominic's conversions.

 $\frac{c^2}{\Re}$ = $\frac{(3\times 10^{10})^2}{8.314\times 10^7}~\mathrm{cgs}$ = $1.083\times 10^{13}~\mathrm{K}$ $\biggl(\frac{\Re^4}{c^4 G a_\mathrm{rad}}\biggr)^{1/2}$ = $\frac{(8.314\times 10^7)^2}{(3\times 10^{10})^2 (6.674\times 10^{-8})^{1/2}(7.566\times 10^{-15})^{1/2}}~\mathrm{cgs}$ = $3.418\times 10^{5}~\mathrm{cm}$ $\biggl(\frac{\Re^4}{c^6 G a_\mathrm{rad}}\biggr)^{1/2}$ = $\frac{(8.314\times 10^7)^2}{(3\times 10^{10})^3 (6.674\times 10^{-8})^{1/2}(7.566\times 10^{-15})^{1/2}}~\mathrm{cgs}$ = $1.139\times 10^{-5}~\mathrm{s}$ $\biggl(\frac{\Re^4}{G^3 a_\mathrm{rad}}\biggr)^{1/2}$ = $\frac{(8.314\times 10^7)^2}{(6.674\times 10^{-8})^{3/2}(7.566\times 10^{-15})^{1/2}}~\mathrm{cgs}$ = $4.609\times 10^{33}~\mathrm{g}$

Hence,

 General Relations $\frac{T_\mathrm{cgs}}{T_\mathrm{code}}$ = $1.083\times 10^{13}~\mathrm{K} \biggl( \frac{\tilde{r}}{\tilde{c}^2} \biggr)$ $\frac{\ell_\mathrm{cgs}}{\ell_\mathrm{code}}$ = $3.418\times 10^{5}~\mathrm{cm} \biggl( \frac{\tilde{c}^4 \tilde{g} \tilde{a} }{\tilde{r}^4} \biggr)^{1/2}$ $\frac{t_\mathrm{cgs}}{t_\mathrm{code}}$ = $1.139\times 10^{-5}~\mathrm{s} \biggl( \frac{\tilde{c}^6 \tilde{g} \tilde{a} }{\tilde{r}^4} \biggr)^{1/2}$ $\frac{m_\mathrm{cgs}}{m_\mathrm{code}}$ = $4.609\times 10^{33}~\mathrm{g} \biggl( \frac{\tilde{g}^3 \tilde{a} }{\tilde{r}^4} \biggr)^{1/2}$

For the Case A parameter values adopted by Dominic, above, and for the particular SCF-code-generated model provided by Wes, I derive,

 Case A RAccretor = $3.418\times 10^{5}~\mathrm{cm} \biggl[ \frac{198^4 \times 0.044 }{(0.44)^4 } \biggr]^{1/2} \times 0.40$ = $5.8\times 10^{9}~\mathrm{cm} = 0.083~\mathrm{R}_\odot$ Porbit = $1.139\times 10^{-5}~\mathrm{s} \biggl[ \frac{198^6 \times 0.044 }{(0.44)^4} \biggr]^{1/2} \times 31$ = $2.97\times 10^{3}~\mathrm{s} = 49.5 ~\mathrm{minutes}$ Mtotal = $4.609\times 10^{33}~\mathrm{g} \biggl[ \frac{0.044 }{(0.44)^4 } \biggr]^{1/2} \times 0.85$ = $4.245\times 10^{33}~\mathrm{g} = 2.1~\mathrm{M}_\odot$

These values do not agree with the ones derived by Dominic.

 Possible Point of Confusion/Disagreement NOTE: Either Dominic wrote the wrong values on my whiteboard or I copied them down incorrectly, but based on the SCF-code parameters that were given to me by Wes Even, in dimensionless code units the model parameters should be: [Mtotal]code = 0.0237 and [RAccretor]code = 0.273 and [Porbit]code = 31.19; the orbital separation is [aseparation]code = 0.83938. Combining these values with Dominic's Case A parameter values gives: $[M_\mathrm{total}]_\mathrm{cgs} = 0.059 M_\odot$; $[R_\mathrm{Accretor}]_\mathrm{cgs} = 0.057 R_\odot$; $[P_\mathrm{orbit}]_\mathrm{cgs} = 50~\mathrm{minutes}$; and $[a_\mathrm{separation}]_\mathrm{cgs} = 0.174 R_\odot$. On 7/24/2010, Joel checked this boxed-in group of numbers against a "polytropic unit conversion spreadsheet" that he developed while at the Lorentz Institute in the Fall of 2010. They are all consistent with Wes Even's SCF-generated Q07 model.

### Response from Dominic

What he wrote on my whiteboard contained some mistakes. For example, the correct code units for various quantities are:

• max]code = 1.000;
• [MAccretor]code = 0.403;
• [RAccretor]code = 0.850;
• [Porbit]code = 31.2;
• [Mtotal]code = 0.685; and
• [aseparation]code = 2.58.

And when he applies the unit conversions, he gets:

• $[\rho_\mathrm{max}]_\mathrm{cgs} = 5.12\times 10^3~\mathrm{g}~\mathrm{cm}^{-3}$;
• $[M_\mathrm{Accretor}]_\mathrm{cgs} = 0.569 M_\odot$;
• $[R_\mathrm{Accretor}]_\mathrm{cgs} = 0.100 R_\odot$;
• $[P_\mathrm{orbit}]_\mathrm{cgs} = 28.12~\mathrm{minutes}$;
• $[M_\mathrm{total}]_\mathrm{cgs} = 0.968 M_\odot$; and
• $[a_\mathrm{separation}]_\mathrm{cgs} = 0.302 R_\odot$.

Two other pieces of information are needed in order to reconcile our numbers. First, Dominic has included a value of $~\bar{\mu}$ = 4 / 3 in his cgs value of $~\Re$, that is, he has set $\Re/\bar{\mu} = 6.236\times 10^7~\mathrm{cgs}$. Second, the length-scale he has adopted in his rad-hydro code is different from the one that Wes provided straight from the SCF code. In particular, Dominic thinks Wes sets,

$[\Delta R]_\mathrm{Wes\_code} = \frac{1}{128-3} = 8\times 10^{-3} ,$

whereas, in order to conform to the constraints imposed by HAD, Dominic sets,

$[\Delta R]_\mathrm{Nic\_code} = \frac{\pi}{128} = 2.454\times 10^{-2} .$

Hence, in order to transform from the code units used by Wes (and the SCF code) to code units used by Dominic, every quantity that includes a unit of length must be multiplied by,

$\biggl[ \frac{\ell_\mathrm{Nic}}{\ell_\mathrm{Wes}}\biggr]_\mathrm{code} = \frac{\pi (128-3)}{128} = 3.067962.$

### Other Thoughts

Notice that Dominic's method for converting from code units to cgs units frequently involves the following ratio of physical constants:

$\Lambda \equiv \biggl( \frac{\Re^4}{G a_\mathrm{rad}} \biggr)^{1/2} = 3.076 \times 10^{26}~\mathrm{cm}^3~\mathrm{s}^{-2}.$

In terms of this new physical constant,

 $\ell \sim \frac{\Lambda}{ c^2 \bar{\mu}^2} ;$ $t \sim \frac{\Lambda}{ c^3 \bar{\mu}^2} ;$ $\mathrm{and} ~~~~~ m \sim \frac{\Lambda}{ G \bar{\mu}^2} .$

### Corrected Logic

Taking all of the above into consideration, the expressions that should be used to convert from Dominic's code units to real units are the following:

General Relations (taking $~\bar{\mu}$ into account)

 $\frac{T_\mathrm{cgs}}{T_\mathrm{code}}$ = $\frac{c^2}{(\Re/\bar{\mu})} \biggl( \frac{\tilde{r}}{\tilde{c}^2} \biggr)$ = $1.083\times 10^{13}~\mathrm{K} \biggl( \frac{\tilde{r} \bar{\mu}}{\tilde{c}^2} \biggr)$ $\frac{\ell_\mathrm{cgs}}{\ell_\mathrm{code}}$ = $\frac{\Lambda}{ c^2 \bar{\mu}^2} \biggl( \frac{\tilde{c}^4 \tilde{g} \tilde{a} }{\tilde{r}^4} \biggr)^{1/2}$ = $3.418\times 10^{5}~\mathrm{cm} \biggl( \frac{\tilde{c}^4 \tilde{g} \tilde{a} }{\tilde{r}^4 \bar{\mu}^4} \biggr)^{1/2}$ $\frac{t_\mathrm{cgs}}{t_\mathrm{code}}$ = $\frac{\Lambda}{ c^3 \bar{\mu}^2} \biggl( \frac{\tilde{c}^6 \tilde{g} \tilde{a} }{\tilde{r}^4} \biggr)^{1/2}$ = $1.139\times 10^{-5}~\mathrm{s} \biggl( \frac{\tilde{c}^6 \tilde{g} \tilde{a} }{\tilde{r}^4 \bar{\mu}^4} \biggr)^{1/2}$ $\frac{m_\mathrm{cgs}}{m_\mathrm{code}}$ = $\frac{\Lambda}{ G \bar{\mu}^2} \biggl( \frac{\tilde{g}^3 \tilde{a} }{\tilde{r}^4} \biggr)^{1/2}$ = $4.609\times 10^{33}~\mathrm{g} \biggl( \frac{\tilde{g}^3 \tilde{a} }{\tilde{r}^4 \bar{\mu}^4} \biggr)^{1/2}$

Hence, for Dominic's first simulation (Case A), the following conversions apply.

Case A:
$\tilde{g} = 1$; $\tilde{c} = 198$; $\tilde{r} = 0.44$; $\tilde{a} = 0.044$; $\bar\mu = 4/3$

 $\frac{T_\mathrm{cgs}}{T_\mathrm{code}}$ = $1.083\times 10^{13}~\mathrm{K} \biggl[ \frac{0.44 (4/3)}{198^2} \biggr]$ $= 1.621\times 10^8~\mathrm{K}$ $\frac{\ell_\mathrm{cgs}}{\ell_\mathrm{code}}$ = $3.418\times 10^{5}~\mathrm{cm} \biggl[ \frac{198^4 (0.044) }{(0.44)^4 (4/3)^4} \biggr]^{1/2}$ $= 8.167\times 10^9~\mathrm{cm}$ $\frac{t_\mathrm{cgs}}{t_\mathrm{code}}$ = $1.139\times 10^{-5}~\mathrm{s} \biggl[ \frac{198^6 (0.044) }{(0.44)^4 (4/3)^4} \biggr]^{1/2}$ $= 5.334\times 10^1~\mathrm{s}$ $\frac{m_\mathrm{cgs}}{m_\mathrm{code}}$ = $4.609\times 10^{33}~\mathrm{g} \biggl[ \frac{ 0.044 }{(0.44)^4 (4/3)^4} \biggr]^{1/2}$ $= 2.809\times 10^{33}~\mathrm{g}$
 $\frac{\rho_\mathrm{cgs}}{\rho_\mathrm{code}}$ = $\biggl( \frac{m_\mathrm{cgs}}{m_\mathrm{code}} \biggr) \biggl( \frac{\ell_\mathrm{cgs}}{\ell_\mathrm{code}} \biggr)^{-3}$ $= 5.157\times 10^{3}~\mathrm{g}~\mathrm{cm}^{-3}$ $\frac{\kappa_\mathrm{cgs}}{\kappa_\mathrm{code}}$ = $\biggl( \frac{\ell_\mathrm{cgs}}{\ell_\mathrm{code}} \biggr)^{2} \biggl( \frac{m_\mathrm{cgs}}{m_\mathrm{code}} \biggr)^{-1}$ $= 2.375\times 10^{-14}~\mathrm{cm}^2~\mathrm{g}^{-1}$

When using the above tabulated Case A conversion units, it must be understood that the "code unit" values refer to units used in Dominic's rad-hydro code. But it should also be appreciated, as discussed above, that the initial model provided to Dominic by Wes — which had been generated by the SCF code — used a different unit of length from Dominic. The conversion factor from SCF-code lengths to the length's used in Dominic's code is:

$\frac{\ell_\mathrm{code}}{\ell_\mathrm{SCF}} = \biggl[ \frac{\pi (128-3)}{128} \biggr] = 3.068 .$

Hence, beginning with the values of various binary system parameters as generated by the SCF code, we conclude that the initial model used by Dominic in his Case A rad-hydro simulations has the following properties:

 Properties of Initial Q0.7 Polytropic Binary Quantity SCF-code Value Conversion Factor RadHydro-code Value Case A physical units MAccretor 0.01394 $\biggl( \frac{\ell_\mathrm{code}}{\ell_\mathrm{SCF}} \biggr)^{3}$ 0.4025 $0.565~M_\odot$ MDonor 0.009761 $\biggl( \frac{\ell_\mathrm{code}}{\ell_\mathrm{SCF}} \biggr)^{3}$ 0.2819 $0.396~M_\odot$ ρAccretor 1.000 1 1.000 $5.16\times 10^{3}~\mathrm{g}~\mathrm{cm}^{-3}$ aseparation 0.8394 $\biggl( \frac{\ell_\mathrm{code}}{\ell_\mathrm{SCF}} \biggr)$ 2.575 $0.300~R_\odot$ Porbit 31.19 1 31.19 $27.7~\mathrm{min}$

#### Review

The characteristic mass, length, and time scales that are associated with a self-gravitating, degenerate-electron gas are identified in an accompanying Wiki page in the context of our discussion of the structure of spherically symmetric white dwarfs and the Chandrasekhar mass. All three of these scales depend on the characteristic Fermi pressure, $~A_\mathrm{F}$, and characteristic Fermi density, $~B_\mathrm{F}$, that are familiar to the condensed-matter community. As recorded in our accompanying variables appendix, the definition of these two condensed-matter relevant quantities is, respectively,

$A_\mathrm{F} \equiv \frac{\pi m_e^4 c^5}{3 h^3} = 6.00233\times 10^{22}~\mathrm{erg}~\mathrm{cm}^{-3},$

and,

$\frac{B_\mathrm{F}}{\mu_e} \equiv \frac{8\pi m_p}{3} \biggl( \frac{m_e c}{h} \biggr)^3 = 9.81019\times 10^{5}~\mathrm{g}~\mathrm{cm}^{-3};$

and the characteristic, astrophysically relevant mass (MCh) and length $(\ell_\mathrm{Ch})$ scales identified by Chandrasekhar are,

$\mu_e^2 M_\mathrm{Ch} = 4\pi m_3 \biggl(\frac{2 A_\mathrm{F}}{\pi G}\biggr)^{3/2} \frac{\mu_e^2}{B_\mathrm{F}^2} = 1.14169\times 10^{34}~\mathrm{g},$

and,

$\mu_e \ell_\mathrm{Ch} \equiv \biggl( \frac{2 A_\mathrm{F}}{\pi G} \biggr)^{1/2} \frac{\mu_e}{B_\mathrm{F}} = 7.71311\times 10^8~\mathrm{cm} ,$

where the dimensionless coefficient m3 = 2.01824. We could just as well define a characteristic dynamical timescale associated with white dwarfs as,

$\mu_e^{1/2} t_\mathrm{Ch} \equiv \biggl[\frac{\mu_e}{GB_\mathrm{F}} \biggr]^{1/2} = \biggl[ \frac{3h^3}{8\pi G m_p m_e^3 c^3} \biggr]^{1/2} = 3.90812~\mathrm{s}$ .

#### Application to Unit Conversion Expressions

Rewriting MCh only in terms of the fundamental physical constants, we obtain,

$\mu_e^2 M_\mathrm{Ch} = \biggl[ \frac{3 m_3^2}{2^5 \pi^2} \cdot \frac{c^3 h^3}{G^3 m_p^4} \biggr]^{1/2} .$

But also note that,

$\frac{\Lambda}{G} = \biggl[ \frac{3\cdot 5}{2^3 \pi^5} \biggl( \frac{m_p}{m_u} \biggr)^4 \cdot \frac{c^3 h^3}{G^3 m_p^4} \biggr]^{1/2} .$

Hence, we can also write,

 $\frac{m_\mathrm{cgs}}{m_\mathrm{code}}$ = $M_\mathrm{Ch} \biggl[ \frac{3\cdot 5}{2^3 \pi^5} \biggl( \frac{m_p}{m_u} \biggr)^4 \cdot \frac{2^5 \pi^2 }{3 m_3^2 } \biggr]^{1/2} \frac{\mu_e^2}{\bar{\mu}^2} \biggl( \frac{\tilde{g}^3 \tilde{a}}{\tilde{r}^4 } \biggr)^{1/2}$ = $\mu_e^2 M_\mathrm{Ch} \biggl[ \frac{2^2 \cdot 5}{\pi^3 m_3^2 } \biggr]^{1/2} \frac{m_p^2}{m_u^2} \biggl( \frac{\tilde{g}^3 \tilde{a}}{\tilde{r}^4 \bar{\mu}^4} \biggr)^{1/2}$ = $0.4038 ~\mu_e^2 M_\mathrm{Ch} \biggl( \frac{\tilde{g}^3 \tilde{a}}{\tilde{r}^4 \bar{\mu}^4} \biggr)^{1/2} .$

Similarly,

$\frac{\Lambda}{c^2} = \biggl[ \frac{3\cdot 5}{2^3 \pi^5} \biggl( \frac{m_p}{m_u} \biggr)^4 \cdot \frac{ h^3}{m_p^4 c G} \biggr]^{1/2} ,$

so,

 $\frac{\ell_\mathrm{cgs}}{\ell_\mathrm{code}}$ = $\mu_e \ell_\mathrm{Ch} \biggl[ \frac{2^2 \cdot 5}{\pi^3 } \biggr]^{1/2} \frac{m_e m_p}{m_u^2} \biggl( \frac{\tilde{c}^4 \tilde{g} \tilde{a}}{\tilde{r}^4 \bar{\mu}^4} \biggr)^{1/2}$ = $4.438\times 10^{-4} ~\mu_e \ell_\mathrm{Ch} \biggl( \frac{\tilde{c}^4 \tilde{g} \tilde{a}}{\tilde{r}^4 \bar{\mu}^4} \biggr)^{1/2} .$

And,

$\frac{\Lambda}{c^3} = \biggl[ \frac{3\cdot 5}{2^3 \pi^5} \biggl( \frac{m_p}{m_u} \biggr)^4 \cdot \frac{ h^3}{m_p^4 c^3 G} \biggr]^{1/2} ,$

so,

 $\frac{t_\mathrm{cgs}}{t_\mathrm{code}}$ = $\mu_e^{1/2} t_\mathrm{Ch}\cdot \frac{5^{1/2}}{\pi^2 } \biggl( \frac{m_e^3 m_p}{m_u^4}\biggr)^{1/2} \biggl( \frac{\tilde{c}^6 \tilde{g} \tilde{a}}{\tilde{r}^4 \bar{\mu}^4} \biggr)^{1/2}$ = $2.9261\times 10^{-6} ~\mu_e^{1/2} t_\mathrm{Ch} \biggl( \frac{\tilde{c}^6 \tilde{g} \tilde{a}}{\tilde{r}^4 \bar{\mu}^4} \biggr)^{1/2} .$

## Opacities

How should an opacity coefficient be introduced into Dominic's rad-hydrocode? Let's examine the simplest case of free-free absorption, κT (i.e., Thompson scattering):

$\kappa_\mathrm{T}|_\mathrm{cgs} = \frac{\sigma_\mathrm{T}}{m_p} \biggl[\frac{1}{2}(1+X) \biggl(\frac{m_p}{m_u}\biggr)\biggr] = 0.2003101 (1+X)~\mathrm{cm}^2~\mathrm{g}^{-1},$

where, $~\sigma_T$, $~m_u$, and $~m_p$ are all physical constants defined in an accompanying appendix. Therefore, for Case A (which assumes pure helium, so X = 0), the value of this free-free (Thompson) opacity in code units is,

$\kappa_\mathrm{T}|_\mathrm{code} = 0.2003101~\mathrm{cm}^2~\mathrm{g}^{-1} \biggl[ \frac{\kappa_\mathrm{cgs}}{\kappa_\mathrm{code}} \biggr]^{-1} = 8.434\times 10^{12}$ .

When Thompson scattering dominates the opacity, the mean-free-path of a photon is,

$\ell_\mathrm{mfp} = \frac{1}{\kappa_\mathrm{T}\rho}.$

This means that, in Dominic's rad-hydrocode, $\ell_\mathrm{mfp}$ will be less than or equal to the size of one radial grid zone, R)Nic_code, whenever,

$[\kappa_\mathrm{T}]_\mathrm{code} ~\rho_\mathrm{code} \ge \frac{1}{(\Delta R)_\mathrm{Nic\_code}} = \frac{128}{\pi}$

$\Rightarrow ~~~~~\rho_\mathrm{code} \ge 4.83\times 10^{-12} .$

It is perhaps more instructive to write this last expression in a form that will permit us to determine how this threshold value of ρcode depends on the chosen set of scaling parameters. Specifically, we can write,

$\rho_\mathrm{code} \ge \rho_\mathrm{threshold} \equiv \frac{128}{\pi (0.200 ~\mathrm{cm}^2~\mathrm{g}^{-1}) } \biggl[ \biggl(\frac{\ell_\mathrm{cgs}}{\ell_\mathrm{code}} \biggr)^2 \biggl(\frac{m_\mathrm{cgs}}{m_\mathrm{code}} \biggr)^{-1} \biggr] = 5.164\times 10^{-21} \biggl[ \frac{\tilde{c}^4 \tilde{a}^{1/2}}{\tilde{r}^2 \bar{\mu}^2 \tilde{g}^{1/2}} \biggr] .$

To check this relation, note that when Case A parameter values are used, the combination of factors inside the last set of square brackets gives $9.367\times 10^{8}$, which produces the same value for ρthreshold (in code units) as before.

## Ratio of Gas Pressure to Radiation Pressure

Let's define the following pressure ratios:

$\Gamma \equiv \frac{P_\mathrm{gas}}{P_\mathrm{rad}} = \frac{3 (\Re/\bar\mu)}{a_\mathrm{rad}} \biggl[ \frac{\rho}{T^3} \biggr]_\mathrm{cgs} ,$

and,

$\beta \equiv \frac{P_\mathrm{gas}}{(P_\mathrm{gas}+P_\mathrm{rad})} = \frac{\Gamma}{1 + \Gamma} .$

Following Dominic's definition of code units, above: T3 should be normalized by $[ c^6/(\Re/\bar\mu )^3]$; the mass density should be normalized by the quantity $[a_\mathrm{rad} c^6/(\Re/\bar\mu)^4]$; and $(\Re/\bar\mu)$ should be replaced by $\tilde{r}$. Hence, the ratio of gas pressure to radiation pressure can be written as,

$\Gamma = \frac{3(\Re/\bar\mu)}{a_\mathrm{rad}} \biggl[\frac{\rho}{T^3}\biggr]_\mathrm{code} \biggl[\frac{c^6 a_\mathrm{rad}}{(\Re/\bar\mu)^4} \biggl( \frac{\tilde{r}^4}{\tilde{c}^6 \tilde{a}} \biggr)\biggr] \biggl[\frac{(\Re/\bar\mu)^3}{c^6} \biggl( \frac{\tilde{c}^6 }{\tilde{r}^3} \biggr)\biggr] = 3\biggl[ \frac{\rho_\mathrm{code}}{T_\mathrm{code}^3} \biggr] \frac{\tilde{r}}{\tilde{a}} .$

We might, in addition, ask what the central temperature is in an n = 3 / 2 polytrope. Well, if the gas pressure dominates (i.e., if $\Gamma \gg 1$),

$P_\mathrm{cgs} \approx \frac{\Re}{\bar\mu} \rho_\mathrm{cgs} T_\mathrm{cgs};$

and in code units,

$P_\mathrm{code} = \kappa_\mathrm{code} \rho_\mathrm{code}^{5/3} .$

Hence,

 Tcode = $\biggl( \frac{T_\mathrm{cgs}}{T_\mathrm{code}} \biggr)^{-1} T_\mathrm{cgs} \approx \biggl( \frac{T_\mathrm{cgs}}{T_\mathrm{code}} \biggr)^{-1} \frac{1}{(\Re/\bar\mu)} \biggl(\frac{P_\mathrm{cgs}}{\rho_\mathrm{cgs}}\biggr)$ = $\frac{1}{(\Re/\bar\mu)} \biggl( \frac{T_\mathrm{cgs}}{T_\mathrm{code}} \biggr)^{-1} \biggl(\frac{P_\mathrm{cgs}/P_\mathrm{code}}{\rho_\mathrm{cgs}/\rho_\mathrm{code}}\biggr) \biggl(\frac{P_\mathrm{code}}{\rho_\mathrm{code}}\biggr)$ = $\frac{1}{(\Re/\bar\mu)} \biggl( \frac{T_\mathrm{cgs}}{T_\mathrm{code}} \biggr)^{-1} \biggl( \frac{\ell_\mathrm{cgs}}{\ell_\mathrm{code}} \biggr)^{2}\biggl( \frac{t_\mathrm{cgs}}{t_\mathrm{code}} \biggr)^{-2} \kappa_\mathrm{code}\rho_\mathrm{code}^{2/3}$ = $\frac{1}{\tilde{r}} ~\kappa_\mathrm{code}\rho_\mathrm{code}^{2/3} .$

This, in turn, tells us that at the center of a polytropic star,

$\Gamma \approx \frac{3\tilde{r}^4}{\tilde{a}}~\kappa_\mathrm{code}^{-3} \rho_\mathrm{code}^{-1} .$

This derivation will need to be modified to handle the more general case when Γ is not necessarily large.

## Super-Eddington Accretion

In the simplest case of spherically symmetric accretion, the Eddington luminosity and accretion luminosity are defined, respectively, as

 LEdd $\equiv$ $\frac{4\pi G M_\mathrm{a} m_p c}{\sigma_\mathrm{T}} ,$ Lacc $\equiv$ $\frac{G M_\mathrm{a} \dot{M}}{R_\mathrm{a}} ,$

where, $~G$, $~c$, $~m_p$, and $~\sigma_T$ are physical constants defined in an accompanying appendix; Ma and Ra are the mass and radius of the accreting star; and $\dot{M}$ is the mass accretion rate. Expressed in code units, these two expressions become,

 $L_\mathrm{Edd}\biggr|_\mathrm{code}$ = $4\pi \tilde{g} \tilde{c} \biggl[ \biggl(\frac{m_p}{\sigma_\mathrm{T}}\biggr) \biggl( \frac{\ell_\mathrm{cgs}}{\ell_\mathrm{code}} \biggr)^2 \biggl( \frac{m_\mathrm{cgs}}{m_\mathrm{code}} \biggr)^{-1} \biggr] [M_a]_\mathrm{code}$ = $4\pi \tilde{g} \tilde{c} \biggl[ 6.373\times 10^{-23} \biggl( \frac{\tilde{c}^4 \tilde{g} \tilde{a}}{\tilde{r}^4 \bar{\mu}^4} \biggr) \biggl( \frac{\tilde{r}^2 \bar{\mu}^2}{\tilde{g}^{3/2} \tilde{a}^{1/2}} \biggr) \biggr] [M_a]_\mathrm{code}$ = $8.009\times 10^{-22} \biggl( \frac{\tilde{c}^5 \tilde{g}^{1/2} \tilde{a}^{1/2}}{\tilde{r}^2 \bar{\mu}^2} \biggr) [M_a]_\mathrm{code} ,$ $L_\mathrm{acc}\biggr|_\mathrm{code}$ = $\tilde{g} \biggl[\frac{M_\mathrm{a} \dot{M}}{R_\mathrm{a}} \biggr]_\mathrm{code} .$

Hence,

$f_\mathrm{Edd} \equiv \frac{L_\mathrm{acc}}{L_\mathrm{Edd}} = 1.25\times 10^{21} \biggl( \frac{\tilde{g}^{1/2} \tilde{r}^2 \bar{\mu}^2}{\tilde{c}^5 \tilde{a}^{1/2}} \biggr) \biggl[\frac{\dot{M}}{R_\mathrm{a}} \biggr]_\mathrm{code} ,$

that is, the accretion will be super-Eddington (facc > 1) if,

$[ \dot{M}]_\mathrm{code} > 8\times 10^{-22}\biggl( \frac{\tilde{c}^5 \tilde{a}^{1/2}}{\tilde{g}^{1/2} \tilde{r}^2 \bar{\mu}^2} \biggr) [R_\mathrm{a}]_\mathrm{code} .$

For Case A parameters and for the Q0.7 polytropic binary model in which [Ra]code = 0.85, the condition for super-Eddington accretion is,

$[ \dot{M}]_\mathrm{code} > 8\times 10^{-22} \biggl[ \frac{198^5 (0.044)^{1/2}}{ (0.44)^2 (4/3)^2} \biggr] [0.85] = 1.26 \times 10^{-10} .$

For purposes of comparison with results from some previously published mass-transfer simulations, we should normalize $[ \dot{M}]_\mathrm{code}$ to $[M_\mathrm{Donor}/ P_\mathrm{orbit}]^\mathrm{code}_0$, where the subscript "0" means initial values. Doing this gives the following condition for super-Eddington accretion:

$[ \dot{m}]^\mathrm{code}_\mathrm{norm} > 1.26 \times 10^{-10} \biggl[ \frac{P_\mathrm{orbit}}{M_\mathrm{Donor}} \biggr]^\mathrm{code}_0 = 1.26 \times 10^{-10} \biggl[ \frac{31.19}{0.2819} \biggr] = 1.4\times 10^{-8}.$

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