Editing Appendix/Ramblings/Photosphere (section)

===Initial Derivation===
As the accretion stream from the less massive white dwarf impacts the surface of the accretor supersonically, it will heat the accreted material to a post-shock temperature, 
<div align="center">
<math>T_{sh} \approx \frac{GM_a m_p}{kR_a}\cdot f</math> ,
</div>
where <math>M_a</math> and <math>R_a</math> are the mass and radius, respectively, of the accretor, and 0 < ''f'' &le; 1 is a coefficient signifying the fraction of potential energy that is converted into heat.  Assuming the post-shock material is optically thick to photon radiation, we should ask to what "photospheric" radius, <math>R_{ph}</math>, the envelope of the accretor will have to swell in order for the star (in steady state) to be able to radiate all of the accretion energy?  It seems that it will need,
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<math>
4\pi R_{ph}^2 \sigma T_{sh}^4 = L_{acc}
</math>
</div>
<div align="center">
<math>
\Rightarrow~~~~~\biggl[ \frac{R_{ph}}{R_a} \biggr]^2 = \frac{L_{acc}}{\pi a_{rad}c R_a^2 T^4_{sh}} = 
\frac{GM_a \dot{M}}{\pi a_{rad}c R_a^3 T^4_{sh}} \, ,
</math>
</div>
where we've set <math>4\sigma = a_{rad}c</math> and,
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<math>
L_{acc} = \frac{GM_a \dot{M}}{R_a} \, .
</math>
</div>
Replacing <math>T_{sh}</math> by the approximate expression shown above gives,
<div align="center">
<math>
\biggl[ \frac{R_{ph}}{R_a} \biggr]^2 \approx \frac{GM_a \dot{M}}{\pi a_{rad}c R_a^3} \biggl[  \frac{kR_a}{GM_a m_p} \biggr]^4 f^{-4}
= \biggl[ \frac{15(hc)^3}{8\pi^5 k^4} \biggr] \frac{R_a \dot{M}}{\pi c G^3 M_a^3} \biggl[  \frac{k}{ m_p} \biggr]^4  f^{-4}
=  \biggl[ \frac{15}{8\pi^5} \biggr] \frac{R_a \dot{M}}{\pi c m_p^4} \biggl[  \frac{hc}{ GM_a} \biggr]^3 f^{-4}
\, .
</math>
</div>
We recognize that the combination of physical constants in this expression resemble the [[Appendix/Ramblings/Radiation/CodeUnits#Application_to_Unit_Conversion_Expressions | Chandrasekhar Mass]], <math>M_{Ch}</math>.  Specifically,
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<math>
\frac{c^3 h^3}{G^3 m_p^4}   = \frac{2^5 \pi^2}{3 m_3^2}  [ \mu_e^2 M_\mathrm{Ch} ]^2.
</math>
</div>
Hence,
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<math>
\frac{R_{ph}}{R_a}  \approx 
\biggl[ \frac{20}{\pi^4} \biggr]^{1/2} \biggl[  \frac{\mu_e^2 M_{Ch}}{ m_3 M_a} \biggr] \biggl[ \frac{\dot{M}}{M_a} \cdot \frac{R_a}{c }  \biggr]^{1/2} f^{-2}
\, .
</math>
</div>
Because <math>R_a/c</math> is less than 1 second for white dwarfs, we see that <math>R_{ph}/R_a \ll 1</math> unless <math>\dot{M}/M_a</math> is very large or <math>f</math> is very small.  This does not make physical sense.  The result is probably screwy because we've incorrectly assumed that the temperature of the common envelope is approximately <math>T_{sh}</math>.