Editing Apps/SMS

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=Rotating, Supermassive Stars=
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<font size="-1">[[H_BookTiledMenu#Context|Bond, Arnett, &amp; Carr<br />(1984)]]</font>
|}

Here we draw upon the work of {{ BAC84full }} &#8212; hereafter {{ BAC84hereafter }} &#8212; who were among the first to seriously address the question of the fate of very massive (stellar) objects.  
&nbsp;<br />

==Equation of State==
Our discussion of the equation of state (EOS) that was used by {{ BAC84hereafter }} draws on the terminology that has already been adopted in our [[SR#Equation_of_State|introductory discussion of ''supplemental relations'']] and closely parallels [[SSC/Structure/BiPolytropes/Analytic1.53#Envelope|our review of the properties of the envelope]] that {{ Milne30full }} used to construct a bipolytropic sphere.

===Expression for Total Pressure===
Ignoring the component due to a degenerate electron gas, <math>P_\mathrm{deg}</math>, the total gas pressure can be expressed as the sum of two separate components:  a component of ideal gas pressure, and a component of radiation pressure.  That is, in {{ BAC84hereafter }} the total pressure is given by the expression,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>P</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>P_\mathrm{gas} + P_\mathrm{rad} \, ,</math>
  </td>
</tr>
</table>
</div>
where,

<table width="60%" align="center" border=1 cellpadding=5>
<tr>
<td align="center" width="50%"><font color="darkblue">Ideal Gas</font></td>
<td align="center"><font color="darkblue">Radiation</font></td>
</tr>
<tr>
<td align="center">
{{Template:Math/EQ_EOSideal0A}}
</td>
<td align="center">
{{Template:Math/EQ_EOSradiation01}}
</td>
</tr>

</table>

Now, {{ BAC84hereafter }} define the rest-mass density in terms of the mean baryon mass, <math>m_B</math>, via the expression, <math>\rho = m_B n</math>, and write (see their equation 1),
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>P</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>Y_T n k T + \frac{1}{3}a_\mathrm{rad} T^4 \, .</math>
  </td>
</tr>
</table>
</div>
In converting from our notation to theirs we conclude, therefore, that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{\Re}{\bar{\mu}} (m_B n) T</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>Y_T n k T </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ Y_T </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{\Re}{k} \cdot \frac{m_B}{\bar{\mu}}  \, .</math>
  </td>
</tr>
</table>
</div>

===Ratio of Radiation Pressure to Gas Pressure ===
Following {{ Milne30 }}, we have defined the parameter, <math>\beta</math>, as the ratio of gas pressure to total pressure.  That is, in the context of {{ BAC84hereafter }}, we have,
<div align="center">
<math>\beta \equiv \frac{P_\mathrm{gas} }{P}  \, ,</math>
</div>
in which case, also,
<div align="center">
<math>\frac{P_\mathrm{rad}}{P}  = 1-\beta </math>
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp; 
<math>\frac{P_\mathrm{gas}}{P_\mathrm{rad}}  = \frac{\beta}{1-\beta} \, .</math>
</div>
 
Using a different notation, {{ BAC84hereafter }} (see their equation 5) define <span title="Ratio of radiation pressure to gas pressure"><math>\sigma</math></span> as the ratio of the radiation pressure to the gas pressure.  Therefore, in converting from our notation to theirs we have,
<div align="center">
<math>\sigma = \frac{1-\beta}{\beta} ~~~~\Rightarrow ~~~~ \beta = (1 + \sigma)^{-1} \, , </math>
</div>

as well as,

<div align="center">
<math>\sigma = \frac{P_\mathrm{rad}}{P_\mathrm{gas}} 
= \frac{a_\mathrm{rad} T^3}{3} \cdot \frac{\bar\mu}{(\Re m_B)n} 
= \frac{a_\mathrm{rad} T^3}{3Y_T n k} \, ,</math>
</div>
which is precisely the definition provided in equation (5) of {{ BAC84hereafter }}.

===Expression for Adiabatic Exponent===
<div align="center">
<table border="1" cellpadding="8" align="center">
<tr><th align="center">Three Equivalent Expressions</th>
</tr>
<tr><td align="center">

<!-- BEGIN TABLE LAYOUT -->
<table border="0" cellpadding="5" align="center">

<tr><td align="center" colspan="3">
Eq. (24) from {{ Eddington18 }}
</td></tr>
<tr>
  <td align="right">
<math>~\Gamma_1 - \frac{4}{3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(\gamma-\tfrac{4}{3})(4-3\beta)}{1 + 12(\gamma-1) (1 - \beta)/\beta} </math>
  </td>
</tr>

<tr><td align="center" colspan="3">
----
</td></tr>
<tr><td align="center" colspan="3">
Eq. (131) from Chapter II of [[Appendix/References#C67|<b>[<font color="red">C67</font>]</b>]]<br />
Eq. (74) from &sect;3.4 of [[Appendix/References#T78|<b>[<font color="red">T78</font>]</b>]]
</td></tr>
<tr>
  <td align="right">
<math>~\Gamma_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\beta + \frac{(\gamma-1)(4-3\beta)^2}{\beta + 12(\gamma-1) (1 - \beta)} </math>
  </td>
</tr>
<tr><td align="center" colspan="3">
----
</td></tr>
<tr><td align="center" colspan="3">
[[#GeneralizingBAC84|Derived below]] &hellip; noting that <math>~\beta = (1+\sigma)^{-1}</math>
</td></tr>

<tr>
  <td align="right">
<math>~\Gamma_1 - \frac{4}{3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(3\gamma -4)(1+4\sigma)}{3(1+\sigma)[1 + 12(\gamma-1)\sigma]} </math>
  </td>
</tr>
</table>

<!-- END TABLE LAYOUT -->
</td></tr>
</table>
</div>


====For any value of the ratio of specific heats====
From equation (2) of {{ LP41full }} &#8212; see, for example, [[SSC/Perturbations#Ledoux_and_Pekeris_.281941.29|our brief summary of this work]] &#8212; or, equally well, from equation (131) in Chapter II of [[Appendix/References#C67|<b>[<font color="red">C67</font>]</b>]], we see that when the total pressure is of the form being considered here, a general expression for the adiabatic exponent,
<div align="center">
<math>\Gamma_1 \equiv \frac{d\ln P}{d\ln\rho} \, ,</math>
</div>
is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Gamma_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\beta + \frac{(\gamma-1)(4-3\beta)^2}{\beta + 12(\gamma-1) (1 - \beta)} \, ,</math>
  </td>
</tr>
</table>
</div>
where, <math>~\gamma</math> is the ratio of specific heats associated with the ideal-gas component of the equation of state.  Notice that <math>~\beta = 1 </math> represents a situation where there is no radiation pressure.  In this limit the expression simplifies to,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Gamma_1\biggr|_{\beta=1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\gamma \, ,</math>
  </td>
</tr>
</table>
</div>
which makes sense.  On the other hand, setting <math>~\beta = 0</math> represents the other extreme, where there is no ideal-gas contribution to the pressure.  In this case, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Gamma_1\biggr|_{\beta=0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{16(\gamma-1)}{12(\gamma-1) } = \frac{4}{3} \, .</math>
  </td>
</tr>
</table>
</div>

====Used by BAC84====
On the other hand, without derivation {{ BAC84hereafter }} state (see their equation 4) that the adiabatic exponent is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Gamma_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4}{3} + \frac{4\sigma + 1}{3(\sigma+1)(8\sigma + 1)} \, .</math>
  </td>
</tr>
</table>
</div>
<span id="GammaApprox">They also point out that, in the case where a system is dominated by radiation pressure <math>~(\sigma \gg 1)</math>, this expression becomes,</span>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Gamma_1\biggr|_{\sigma \gg 1}</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\frac{4}{3} + \frac{1}{6\sigma} \, .</math>
  </td>
</tr>
</table>
</div>
Clearly, in the limit <math>~\sigma \rightarrow \infty</math>, this gives <math>~\Gamma_1 \rightarrow 4/3</math>, which, as it should, matches the limiting value obtained from the {{ LP41 }} expression when <math>~\beta = 0</math>.  

{{ BAC84hereafter }} do not explicitly state what value they used for the ratio of specific heats when deriving their expression for the adiabatic exponent.  But this can be deduced by examining how their expression behaves in the limit of no radiation pressure, that is, for <math>~\sigma = 0</math>.  In this limit, the {{ BAC84hereafter }} expression gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Gamma_1\biggr|_{\sigma = 0}</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\frac{4}{3} + \frac{1}{3} = \frac{5}{3} \, .</math>
  </td>
</tr>
</table>
</div>

The general {{ BAC84hereafter }} expression should therefore match the (even more) general {{ LP41 }} expression if we set <math>~\gamma = \tfrac{5}{3}</math>.  Let's check this out.  Inserting this specific value of <math>~\gamma</math>, and remembering (from above) that,
<div align="center">
<math>\sigma = \frac{1-\beta }{\beta} \, ,</math>
</div>
the {{ LP41 }} expression for the adiabatic exponent becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Gamma_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\beta + \frac{\tfrac{2}{3}(4-3\beta)^2}{\beta + 8(1 - \beta)} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{3\beta} \biggl[ \frac{2(4-3\beta)^2}{1 + 8\sigma} \biggr] + \beta</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{3\beta(1 + 8\sigma)} \biggl[ 2(4-3\beta)^2  + 3\beta^2(1 + 8\sigma) \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{3\beta(1 + 8\sigma)} \biggl[ 32 -48\beta + 18\beta^2  + 3\beta^2 + 24\beta^2\sigma \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{3\beta(1 + 8\sigma)} \biggl[ 32 -48\beta + \beta^2(  21 + 24\sigma) \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~\Gamma_1 - \frac{4}{3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{3\beta(1 + 8\sigma)} \biggl[ 32 -48\beta + \beta^2(  21 + 24\sigma) -4\beta(1+8\sigma)\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{3\beta(1 + 8\sigma)} \biggl[ 32 -4\beta(13+8\sigma) + 3\beta^2(  7 + 8\sigma) \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{3\beta(1 + 8\sigma)} \biggl[ 32 -4\beta(8+8\sigma) - 20\beta + 3\beta^2(  8 + 8\sigma) -3\beta^2\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{3\beta(1 + 8\sigma)} \biggl[ 32 -32 - 20\beta + 24\beta -3\beta^2\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(4-3\beta)}{3(1 + 8\sigma)} = \frac{\tfrac{4}{\beta}-3}{\tfrac{3}{\beta}(1 + 8\sigma)} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4(1+\sigma) - 3}{3(1+\sigma)(1 + 8\sigma)} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1+4\sigma}{3(1+\sigma)(1 + 8\sigma)} \, .</math>
  </td>
</tr>
</table>
</div>

<span id="GeneralizingBAC84">Or, even more generally, we can show that,</span>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Gamma_1 - \frac{4}{3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(3\gamma -4)(1+4\sigma)}{3(1+\sigma)[1 + 12(\gamma-1)\sigma]} \, .</math>
  </td>
</tr>
</table>
</div>



<!-- COMMENT OUT BRIEF SECTION
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Gamma_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\beta + \frac{g(4-3\beta)^2}{\beta + 12g (1 - \beta)} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\beta + \frac{g}{\beta}\biggl[4-3\beta\biggr]^2\biggl[1 + \frac{12g (1 - \beta)}{\beta}\biggr]^{-1} \, ,</math>
  </td>
</tr>
</table>
</div>
where we are using the temporary shorthand notation, <math>g \equiv (\gamma - 1)</math>.  Replacing <math>~\beta</math> with its expression in terms of <math>~\sigma</math> gives us,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Gamma_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{(1+\sigma)} + g(1+\sigma)\biggl[4-\frac{3}{(1+\sigma)}\biggr]^2\biggl[1 + 12g \sigma\biggr]^{-1} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{(1+\sigma)} + \frac{g[4(1+\sigma)-3]^2 }{(1 + 12g \sigma)} </math>
  </td>
</tr>
</table>
</div>
END OF COMMENTED OUT SECTION -->

===Mass Normalization===
Now, according to {{ BAC84hereafter }} (see their equation 8), when the total pressure is written in polytropic form &#8212; specifically, if we set,
<div align="center">
<math>P = K\rho^{(1+1/n_p)} </math>
</div>

&#8212; the mass-scaling for relativistic configurations will depend on <math>~G</math>, <math>~c</math>, <math>~K</math>, and <math>~n_p</math> via the expression,

<div align="center">
<math>~M_u = K^{n_p/2} G^{-3/2} c^{3-n_p} = \biggl( \frac{K}{G}\biggr)^{3/2} \biggl(\frac{K}{c^2}\biggr)^{(n_p-3)/2} \, .</math>
</div>

<!-- COMMENT OUT SMALL SECTION
It is convenient to rewrite this expression in the form,

<div align="center">
<math>~M_u = M_\mathrm{norm} \biggl(\frac{K}{c^2}\biggr)^{(n_p-3)/2} \, ,</math>
</div>

and to determine, first, an expression for the mass-normalization when <math>~n_p = 3</math>, namely,
<div align="center">
<math>~M_\mathrm{norm} \equiv \biggl( \frac{K}{G}\biggr)^{3/2} .</math>
</div>
-->

====Polytropic Index Equals 3====
Referencing our [[SSC/Structure/BiPolytropes/Analytic1.53#HighlightedExpressions|separate discussion of Milne's (1930) work]], when <math>~n_p = 3</math>, the polytropic constant is related to the relevant set of physical parameters via the relation,

<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~K_{3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \biggl( \frac{\Re}{\bar\mu}\biggr)^4 \biggl(\frac{1-\beta}{\beta^4}\biggr) \frac{3}{a_\mathrm{rad}} \biggr]^{1/3} \, .</math>
  </td>
</tr>
</table>
</div>
Adopting the {{ BAC84hereafter }} terminology, this means that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl(\frac{K_{3}}{G}\biggr)^3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{\Re}{\bar\mu}\biggr)^4 \biggl[\frac{1-\beta}{\beta^4}\biggr] \frac{3}{G^3 a_\mathrm{rad}} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{k Y_T}{m_B}\biggr)^4 \biggl[\sigma^4(1+\sigma^{-1})^3\biggr] \frac{3}{G^3 a_\mathrm{rad}} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ M_\mathrm{norm} \equiv \biggl(\frac{K_{3}}{G}\biggr)^{3/2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(1+\sigma^{-1})^{3/2} \biggl( \frac{k Y_T}{m_B}\biggr)^2  \biggl(\frac{3}{G^3 a_\mathrm{rad}}\biggr)^{1/2} \sigma^2 </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\biggl(1+\frac{3}{2\sigma} \biggr) \biggl( \frac{k Y_T}{m_B}\biggr)^2  \biggl(\frac{3}{G^3 a_\mathrm{rad}}\biggr)^{1/2} \sigma^2 \, .</math>
  </td>
</tr>
</table>
</div>
When radiation pressure significantly dominates over gas pressure &#8212; that is, in the limit <math>~\sigma \gg 1</math> &#8212; the leading factor is approximately unity, in which case we see that this expression for <math>~M_\mathrm{norm}</math> exactly matches the expression for <math>~M_{u,3}</math> given by equation (10) of {{ BAC84hereafter }}.

====Polytropic Index Slightly Less Than 3====
More generally, equating the two expressions for the total pressure and drawing (twice) on the expression for <math>~\sigma</math> [[#Ratio_of_Radiation_Pressure_to_Gas_Pressure|provided above]], we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~K\rho^{(1 + 1/n_p)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~Y_T n k T + \frac{a_\mathrm{rad}}{3} T^4 </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{a_\mathrm{rad}}{3} (1+\sigma^{-1})T^4 </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{a_\mathrm{rad}}{3} (1+\sigma^{-1})\biggl[ \frac{3Y_T n k \sigma}{a_\mathrm{rad}} \biggr]^{4/3} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{3}{a_\mathrm{rad}} \biggr)^{1/3}(1+\sigma^{-1})\biggl[ Y_T n k \sigma \biggr]^{4/3} \, .</math>
  </td>
</tr>
</table>
</div>

Now, from [[#GammaApprox|above]] we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~1 + \frac{1}{n_p} = \Gamma</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\frac{4}{3} + \frac{1}{6\sigma} \, ,</math>
  </td>
</tr>
</table>
</div>

so the lefthand-side of this last expression can be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~K\rho^{(1+1/n_p)}</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~K\rho^{(4/3+1/6\sigma)} = K(m_B n)^{4/3} \rho^{1/6\sigma} \, .</math>
  </td>
</tr>
</table>
</div>
This means that, for any <math>~\sigma \gg 1</math>,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~K </math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{3}{a_\mathrm{rad}} \biggr)^{1/3} \biggl( \frac{Y_T  k \sigma}{m_B} \biggr)^{4/3} (1+\sigma^{-1}) \rho^{-1/6\sigma} \, .</math>
  </td>
</tr>
</table>
</div>
This matches exactly expression (7) in {{ BAC84hereafter }}.  Again from [[#GammaApprox|above]] &#8212; and continuing to assume <math>~\sigma \gg 1</math> &#8212; we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~1 + \frac{1}{n_p} \approx \frac{4}{3} + \frac{1}{6\sigma} </math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~~~~\Rightarrow ~~~~</math>&nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\frac{1}{n_p} \approx \frac{1}{3}\biggl(1 + \frac{1}{2\sigma}\biggr) </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~~~~\Rightarrow ~~~~</math>&nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~n_p \approx 3\biggl(1 + \frac{1}{2\sigma}\biggr)^{-1} \approx 3\biggl(1 - \frac{1}{2\sigma}\biggr)</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; <math>~~~~\Rightarrow ~~~~</math>&nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\frac{(n_p-3)}{2} \approx - \frac{3}{4\sigma} \, .</math>
  </td>
</tr>
</table>
</div>

Hence, when the polytropic index is slightly less than 3, the mass normalization is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~M_u</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{K}{G}\biggr)^{3/2} \biggl(\frac{K}{c^2}\biggr)^{-3/4\sigma}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{1}{G} \biggl( \frac{3}{a_\mathrm{rad}} \biggr)^{1/3}
\biggl( \frac{Y_T  k \sigma}{m_B} \biggr)^{4/3} (1+\sigma^{-1}) \rho^{-1/6\sigma} \biggr]^{3/2} 
\biggl[\frac{1}{c^2} \biggl( \frac{3}{a_\mathrm{rad}} \biggr)^{1/3}
\biggl( \frac{Y_T  k \sigma}{m_B} \biggr)^{4/3} (1+\sigma^{-1}) \rho^{-1/6\sigma} \biggr]^{-3/4\sigma} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\biggl( \frac{3}{G^3 a_\mathrm{rad}} \biggr)^{1/2}
\biggl( \frac{Y_T  k }{m_B} \biggr)^{2} (1+\sigma^{-1})^{3/2} \sigma^2 \biggr] \rho^{-1/4\sigma} 
\biggl[\frac{1}{c^2} \biggl( \frac{3}{a_\mathrm{rad}} \biggr)^{1/3}
\biggl( \frac{Y_T  k \sigma}{m_B} \biggr)^{4/3} (1+\sigma^{-1}) \rho^{-1/6\sigma} \biggr]^{-3/4\sigma} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[M_\mathrm{norm} \biggr]  
\biggl\{\frac{1}{c^2} \biggl( \frac{3}{a_\mathrm{rad}} \biggr)^{1/3}
\biggl( \frac{Y_T  k \sigma}{m_B} \biggr)^{4/3} (m_B n)^{1/3} \biggl[(1+\sigma^{-1}) \rho^{-1/6\sigma} \biggr] \biggr\}^{-3/4\sigma} </math>
  </td>
</tr>
</table>
</div>
Drawing again from the definition of <math>~\sigma</math> [[#Ratio_of_Radiation_Pressure_to_Gas_Pressure|provided above]], we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl(\frac{3}{a_\mathrm{rad}}\biggr)^{1/3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~T (\sigma Y_T n k)^{-1/3} \, ,</math>
  </td>
</tr>
</table>
</div>
so this last relation can be rewritten as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~M_u</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
M_\mathrm{norm} \cdot f \biggl[\frac{m_B c^2}{T \sigma Y_T k} \biggr]^{3/4\sigma} 
\approx f \cdot M_{u,3} \biggl(1+\frac{3}{2\sigma} \biggr) \biggl[\frac{m_B c^2}{T \sigma Y_T k} \biggr]^{3/4\sigma} 
\, ,</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>f \equiv \biggl[(1+\sigma^{-1})^{-1}\rho^{1/6\sigma} \biggr]^{3/4\sigma} 
\approx \biggl( 1 - \frac{3}{4\sigma^2}\biggr) (n m_B)^{1/8\sigma^2}\, ,</math>
</div>
which certainly is close to unity when <math>~\sigma \gg 1</math>.  After setting <math>~f=1</math>, this last expression for <math>~M_u</math> exactly matches the expression presented as equation (9) in {{ BAC84hereafter }}.

==Entropy of Radiation Field==
===Planck's Law of Black-Body Radiation===
Drawing from the [https://en.wikipedia.org/wiki/Black-body_radiation#Equations Wikipedia discussion of black-body radiation], Planck's law states that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>B_\nu(T)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
\frac{2h\nu^3}{c^2} \frac{1}{e^{h\nu/kT} - 1} \, .
</math>
  </td>
</tr>
<tr><td align="center" colspan="3">
[[Appendix/References#H87|[<b><font color="red">H87</font></b>]]], &sect;12.1 (p. 280), Eq. (12.14)<br />
[[Appendix/References#KW94|[<b><font color="red">KW94</font></b>]]], &sect;5.1 (p. 30), Eq. (5.15)<br />
[[Appendix/References#HK94|[<b><font color="red">HK94</font></b>]]], &sect;4.1 (p. 152), Eq. (4.7)<br />
[[Appendix/References#BLRY07|[<b><font color="red">BLRY07</font></b>]]], &sect;1.6.2 (p. 23), Eq. (1.103)
</td></tr>
</table>
Letting,
<div align="center"><math>x \equiv \frac{h\nu}{kT} \, ,</math></div>
and integrating over the entire frequency spectrum, we have,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\int_0^\infty B_\nu(T)d\nu</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
\frac{2h}{c^2}\biggl(\frac{kT}{h} \biggr)^4 \int_0^\infty \frac{x^3 dx}{e^{x} - 1} 
=
\frac{2k^4}{h^3c^2} \biggl(\frac{\pi^4}{15} \biggr)T^4
= \frac{\sigma_\mathrm{SB} T^4}{\pi} \, ,
</math>
  </td>
</tr>
</table>
where [with units of mass &times; (time)<sup>-3</sup> K<sup>-4</sup>],

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\sigma_\mathrm{SB} </math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math> 
\frac{2\pi^5}{15} \frac{k^4}{h^3c^2} \, .
</math>
  </td>
</tr>
</table>
In discussions of the thermodynamic properties of (photon) radiation fields, the radiation [density] constant, {{ Template:Math/C_RadiationConstant }}, appears in many physical relations.  It is related to the Steffan-Boltzmann constant via the simple expression,
<div align="center"><math>a_\mathrm{rad} = \frac{4 \sigma_\mathrm{SB}}{c} \, ,</math></div>
and has units of energy per unit volume &times; K<sup>-4</sup>.

===Photon Entropy===
Here are expressions from the published literature that &#8212; to within an additive constant &#8212; give the entropy of the (photon) radiation field.  We begin with [<b>[[Appendix/References#Clayton68|<font color="red">Clayton68</font>]]</b>], who states that, "&hellip; the entropy per gram of the photon [field]" is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>s_\mathrm{rad}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
\frac{4a_\mathrm{rad}T^3}{3\rho} \, . 
</math>
  </td>
</tr>
</table>
[<b>[[Appendix/References#Clayton68|<font color="red">Clayton68</font>]]</b>], Chapter 2 (p. 121), Eq. (2-136)<br />
[[Appendix/References#Shu92|[<b><font color="red">Shu92</font></b>]]], Vol. I, Chapter 9 (p. 82), Eq. (9.22)
</div>
Providing the same expression, [[Appendix/References#Shu92|[<b><font color="red">Shu92</font></b>]]] states that it  is a measure of "&hellip; the entropy of blackbody radiation per unit mass of fluid."  From the group of terms on the right-hand side of this expression, we conclude that <math>s_\mathrm{rad}</math> has units of specific energy per Kelvin.  This is consistent with the units that are traditionally associated with entropy; see, for example, our [[PGE/FirstLawOfThermodynamics#VariableDimensions|separate discussion of the dimensions of various thermodynamic variables]].

According to [[Appendix/References#ST83|[<b><font color="red">ST83</font></b>]]], "&hellip; the photon entropy per baryon" is,
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right">
<math>s_r = \frac{4}{3} \frac{a_\mathrm{rad}T^3}{n}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>\frac{4m_H a_\mathrm{rad} T^3}{3\rho} \, ,</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[[Appendix/References#ST83|[<b><font color="red">ST83</font></b>]]], &sect;17.2, Eq. (17.2.2) &amp;  &sect;17.3, Eq. (17.3.7)
  </td>
</tr>
</table>
where, <span title="Hydrogen mass"><math>m_H</math></span> is the mass of a hydrogen atom, and <math>n = \rho/m_H</math> is the baryon number density.  We conclude, therefore, that <math>s_\mathrm{rad} = s_r/m_H</math>.

We have also found it useful to examine the expression for the entropy of a radiation field used by {{ FWW86full }} &#8212; hereafter {{ FWW86hereafter }} &#8212; primarily because the authors provide, for comparison, a numerical evaluation of the expression's leading coefficient.

<table border="1" width="80%" align="center" cellpadding="8">
<tr><td align="left">
In terms of the variables, {{ Template:Math/VAR_Temperature01 }} and {{ Template:Math/VAR_Density01 }}, and the four physical constants, {{ Template:Math/C_PlanckConstant }}, {{ Template:Math/C_SpeedOfLight }}, {{ Template:Math/C_BoltzmannConstant }}, {{ Template:Math/C_AvogadroConstant }}, {{ FWW86hereafter }} state that the "&hellip; entropy per baryon &hellip; for photons is,"
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>s_\gamma</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
\frac{4\pi^2}{45}\biggl[ \frac{kT }{\hbar c}\biggr]^3 \frac{1}{\rho N_A}. 
</math>
  </td>
</tr>
<tr>
<td align="center" colspan="3">
{{ FWW86hereafter }} &sect;III (p. 678), Eq. (11)
</td>
</tr>
</table>

From the numerical values of the physical constants &#8212; which can be obtained by scrolling your cursor over each representative letter in our text (see also our [[Appendix/VariablesTemplates#Physical_Constants|accompanying table of physical constants]]) &#8212; and acknowledging that <math>\hbar \equiv h/(2\pi)</math>, we deduce that, to six significant digits,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>hc</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
1.986446 \times 10^{-16} ~\mathrm{erg} \cdot \mathrm{cm}
=
1.239841 \times 10^{-10} ~\mathrm{MeV} \cdot \mathrm{cm}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~s_\gamma</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
\biggl[ \frac{(2\pi)^5}{45 N_A(h c)^3}\biggr] \frac{(kT)^3}{\rho} 
= (4.610053 \times 10^{25}~\mathrm{erg}^{-3} \cdot \mathrm{cm}^{-3}) \frac{(kT)^3}{\rho}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
(1.895998 \times 10^{8}~\mathrm{MeV}^{-3} \cdot \mathrm{cm}^{-3}) \frac{(kT)^3}{\rho} \,.
</math>
  </td>
</tr>
</table>
This is consistent with the statement in {{ FWW86hereafter }} that, when energy is measured in MeV, the leading coefficient is, <math>1.905 \times 10^8</math>.

Notice that, because the quantity, {{ Template:Math/C_BoltzmannConstant }}{{ Template:Math/VAR_Temperature01 }}, has units of energy, the dimension of <math>s_\gamma</math> is inverse mass.   In contrast to this, as has been highlighted in our [[PGE/FirstLawOfThermodynamics#VariableDimensions|separate discussion of the physical units of various thermodynamic variables]], we expect the units of specific entropy <math>(s_\mathrm{rad})</math> to be specific energy per Kelvin.  We suspect that, for convenience, {{ FWW86hereafter }} have dropped a leading factor of {{ Template:Math/C_BoltzmannConstant }} on the RHS of their expression for <math>s_\gamma</math> and that it is appropriate to draw from their presentation that the specific entropy of the radiation field is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>s_\mathrm{rad}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
k s_\gamma \, . 
</math>
  </td>
</tr>
</table>
</td></tr>
</table>


[[Image:CommentButton02.png|right|100px|Note that in the {{ BAC84hereafter }} publication, the Boltzmann constant, k, does not appear in this expression because "T is the temperature in energy units (Boltzmann constant = 1)."]]Similarly, from {{ BAC84hereafter }} we find that "&hellip; the photon entropy [is],"

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>s_\gamma</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
\frac{4}{3} \biggl[ \frac{\pi^2 }{15(\hbar c)^3}\biggr] \frac{(kT)^3}{n_B} \, , 
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">{{ BAC84hereafter }}, &sect;2b (p. 827), Eq. (5)</td>
</tr>
</table>
where <math>n_B</math> is the "baryon density."  


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