Editing Appendix/Ramblings/Photosphere (section)

==Locating the Photosphere of Stably Accreting Double White Dwarf Binaries==
===Context===
At our regularly scheduled astrophysics group meeting on Monday, 2 May 2011, Juhan Frank and I started debating the answer to the following question:  What should the photospheric radius be of the common envelope that surrounds a stably accreting, double white dwarf (DWD) binary?  That is, does an accreting DWD binary that is destined to be an AM CVn system look like a single bloated star?  The various mathematical relations that we think are relevant to this question were sketched on the whiteboard in room 218 Johnston Hall (CCT).  Here is a photo of that whiteboard discussion and derivation.

[[Image:May2011WhiteBoard.JPG|400px|center]]<br />

===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,
<div align="center">
<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,
<div align="center">
<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,
<div align="center">
<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,
<div align="center">
<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>.

===Better Temperature Estimate===

Now let's assume that the common envelope puffs up adiabatically, in which case the temperature of the photosphere (and envelope) should be,
<div align="center">
<math>T_{ph} \approx T_{sh} \biggl( \frac{\rho_{ph}} {\rho_{sh}} \biggr)^{\gamma - 1}
\approx T_{sh} \biggl( \frac{R_{ph}} {R_a} \biggr)^{3(1-\gamma)}</math> .
</div>
So, by analogy with the above initial derivation, if we demand that,
<div align="center">
<math>
4\pi R_{ph}^2 \sigma T_{ph}^4 = L_{acc} \, ,
</math>
</div>
then,
<div align="center">
<math>
\biggl[ \frac{R_{ph}}{R_a} \biggr]^2 = \frac{L_{acc}}{\pi a_{rad}c R_a^2 T^4_{sh}} \biggl[\frac{R_{ph}}{R_a} \biggr]^{12(\gamma-1)} 
\approx \biggl[ \frac{20}{\pi^4} \biggr] \biggl[  \frac{\mu_e^2 M_{Ch}}{ m_3 M_a} \biggr]^2 \biggl[ \frac{\dot{M}}{M_a} \cdot \frac{R_a}{c }  \biggr] f^{-4} \biggl[\frac{R_{ph}}{R_a} \biggr]^{12(\gamma-1)}
</math>
</div>
<div align="center">
<math>
\Rightarrow~~~~~ \biggl[ \frac{R_{ph}}{R_a} \biggr]^{2(6\gamma - 7)} \approx 
\biggl[ \frac{20}{\pi^4} \biggr]^{-1} \biggl[  \frac{\mu_e^2 M_{Ch}}{ m_3 M_a} \biggr]^{-2} \biggl[ \frac{\dot{M}}{M_a} \cdot \frac{R_a}{c }  \biggr]^{-1} f^{4}
</math>
</div>
<div align="center">
<math>
\Rightarrow~~~~~ \frac{R_{ph}}{R_a}  \approx \biggl\{
\biggl[ \frac{20}{\pi^4} \biggr]^{-1} \biggl[  \frac{\mu_e^2 M_{Ch}}{ m_3 M_a} \biggr]^{-2} \biggl[ \frac{\dot{M}}{M_a} \cdot \frac{R_a}{c }  \biggr]^{-1} f^{4}
\biggr\}^{1/[2(6\gamma-7)]} \, .
</math>
</div>
Hence, for a <math>\gamma=5/3</math> envelope,
<div align="center">
<math>
\frac{R_{ph}}{R_a}  \approx \biggl\{
\biggl[ \frac{20}{\pi^4} \biggr]^{-1} \biggl[  \frac{\mu_e^2 M_{Ch}}{ m_3 M_a} \biggr]^{-2} \biggl[ \frac{\dot{M}}{M_a} \cdot \frac{R_a}{c }  \biggr]^{-1} f^{4}
\biggr\}^{1/6} \, .
</math>
</div>
Now, based on the accompanying discussion of the [[SSC/Structure/WhiteDwarfs#Chandrasekhar_mass|Chandrasekhar mass]],
<div align="center">
<math>\frac{ m_3 M_\odot}{\mu_e^2 M_{Ch}}=0.353 \, .</math>
</div>
Hence,
<div align="center">
<math>
\frac{R_{ph}}{R_a}  \approx 0.920 \biggl[ \frac{\dot{M}}{M_a} \cdot \frac{R_a}{c }  \biggr]^{-1/6} f^{2/3}  \approx \biggl[ \frac{\dot{M}}{M_a} \cdot \frac{R_a}{c }  \biggr]^{-1/6}  \, .
</math>
</div>
This relation behaves in a very different way from the initially derived relation.  For small mass-transfer rates, this relation predicts that the photospheric radius will be quite large relative to the radius of the accretor.

===Derivation from Juhan Frank===
In his independent derivation, Juhan substitutes the following dimensionless mass-transfer rate:
<div align="center">
<math>
\dot{m} \equiv \frac{\dot{M}}{\dot{M}_{int}} = \frac{\kappa\dot{M}}{4\pi c R_a} \, ,
</math>
</div>
where, <math>\kappa = \sigma_T/m_p</math>.  Hence,
<div align="center">
<math>
\frac{\dot{M}}{M_a} \cdot \frac{R_a}{c} = \dot{m} \biggl[ \frac{4\pi R_a^2}{\kappa M_a} \biggr] \, ,
</math>
</div>
and,
<div align="center">
<math>
\frac{R_{ph}}{R_a}  \approx \biggl\{
\biggl[ \frac{20}{\pi^4} \biggr]^{-1} \biggl[  \frac{\mu_e^2 M_{Ch}}{ m_3 M_a} \biggr]^{-2} \biggl[ \frac{4\pi R_a^2}{\kappa M_a} \biggr]^{-1} {\dot{m}}^{-1}  f^{4}
\biggr\}^{1/6} =
 \biggl\{
\biggl[ \frac{\pi^4}{20} \biggr] \biggl[  \frac{ m_3 M_\odot}{\mu_e^2 M_{Ch}} \biggr]^{2} \biggl[ \frac{\kappa M_\odot}{4\pi R_9^2 } \biggr] 
\biggr\}^{1/6} \biggl[ \frac{f^{2/3}}{\dot{m}^{1/6}} \biggl( \frac{M_a}{M_\odot} \biggr)^{1/2} \biggl( \frac{R_9 }{R_a} \biggr) ^{1/3}\biggr]  \, ,
</math>
</div>
where <math>R_9 \equiv 10^9~\mathrm{cm}</math>.  Again, since,
<div align="center">
<math>\frac{ m_3 M_\odot}{\mu_e^2 M_{Ch}}=0.353 \, ,</math>
</div>
we obtain,
<div align="center">
<math>
\frac{R_{ph}}{R_a} = 183.9 \biggl[ \frac{f^{2/3}}{\dot{m}^{1/6}} \biggl( \frac{M_a}{M_\odot} \biggr)^{1/2} \biggl( \frac{R_9 }{R_a} \biggr) ^{1/3}\biggr]  \, .
</math>
</div>

===Mass-Transfer Rates===
Juhan chose the situation where everything inside the final square brackets is unity, in which case,
<div align="center">
<math>
f = 1; ~ \dot{m} = 1; ~ M_a = 1M_\odot; ~ R_a = 10^9\mathrm{cm}; ~ \mathrm{and}~ R_{ph} \approx 2.63 R_\odot \, .
</math>
</div>
If I use the same values, I get,
<div align="center">
<math>
\frac{\dot{M}}{M_a} \cdot \frac{R_a}{c} = \dot{m} \biggl[ \frac{4\pi R_a^2}{\kappa M_a} \biggr] = 1.57\times 10^{-14} \, .
</math>
</div>
This means,
<div align="center">
<math>
\frac{\dot{M}}{M_a} = 4.7\times 10^{-13} ~\mathrm{s}^{-1} \approx 1.5\times 10^{-5} ~\mathrm{yr}^{-1} \, .
</math>
</div>
According to [http://iopscience.iop.org/0004-637X/655/2/1025/ Kopparapu &amp; Tohline (2007)] (hereafter, KT07), the mass-transfer rate for stable, mass-transferring double-white dwarfs is,
<div align="center">
<math>
\frac{\dot{M}}{M_a} \approx \tau_\mathrm{chirp}^{-1} \, .
</math>
</div>
Combining these last two expressions means,
<div align="center">
<math>
\tau_\mathrm{chirp}  \approx 2\times 10^{12}~ \mathrm{s} = 6.7\times 10^4~\mathrm{yr} \, .
</math>
</div>

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