Editing Apps/SMS (section)

===Expression for Adiabatic Exponent===
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<tr><th align="center">Three Equivalent Expressions</th>
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Eq. (24) from {{ Eddington18 }}
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<math>~\Gamma_1 - \frac{4}{3}</math>
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<math>~=</math>
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<math>~\frac{(\gamma-\tfrac{4}{3})(4-3\beta)}{1 + 12(\gamma-1) (1 - \beta)/\beta} </math>
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----
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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>]]
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<math>~\Gamma_1</math>
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<math>~=</math>
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<math>~\beta + \frac{(\gamma-1)(4-3\beta)^2}{\beta + 12(\gamma-1) (1 - \beta)} </math>
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----
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[[#GeneralizingBAC84|Derived below]] &hellip; noting that <math>~\beta = (1+\sigma)^{-1}</math>
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<math>~\Gamma_1 - \frac{4}{3}</math>
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<math>~=</math>
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<math>~\frac{(3\gamma -4)(1+4\sigma)}{3(1+\sigma)[1 + 12(\gamma-1)\sigma]} </math>
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====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,
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<math>\Gamma_1 \equiv \frac{d\ln P}{d\ln\rho} \, ,</math>
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is,
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<math>~\Gamma_1</math>
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<math>~=</math>
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<math>~\beta + \frac{(\gamma-1)(4-3\beta)^2}{\beta + 12(\gamma-1) (1 - \beta)} \, ,</math>
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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,
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<math>~\Gamma_1\biggr|_{\beta=1}</math>
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<math>~=</math>
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<math>~\gamma \, ,</math>
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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,
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<math>~\Gamma_1\biggr|_{\beta=0}</math>
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<math>~=</math>
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<math>~\frac{16(\gamma-1)}{12(\gamma-1) } = \frac{4}{3} \, .</math>
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====Used by BAC84====
On the other hand, without derivation {{ BAC84hereafter }} state (see their equation 4) that the adiabatic exponent is,
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<math>~\Gamma_1</math>
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<math>~=</math>
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<math>~\frac{4}{3} + \frac{4\sigma + 1}{3(\sigma+1)(8\sigma + 1)} \, .</math>
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<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>
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<math>~\Gamma_1\biggr|_{\sigma \gg 1}</math>
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<math>~\approx</math>
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<math>~\frac{4}{3} + \frac{1}{6\sigma} \, .</math>
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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,
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<math>~\Gamma_1\biggr|_{\sigma = 0}</math>
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<math>~\approx</math>
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<math>~\frac{4}{3} + \frac{1}{3} = \frac{5}{3} \, .</math>
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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,
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<math>\sigma = \frac{1-\beta }{\beta} \, ,</math>
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the {{ LP41 }} expression for the adiabatic exponent becomes,
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<math>~\Gamma_1</math>
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<math>~=</math>
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<math>~\beta + \frac{\tfrac{2}{3}(4-3\beta)^2}{\beta + 8(1 - \beta)} </math>
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&nbsp;
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<math>~=</math>
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<math>~\frac{1}{3\beta} \biggl[ \frac{2(4-3\beta)^2}{1 + 8\sigma} \biggr] + \beta</math>
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&nbsp;
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<math>~=</math>
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<math>~\frac{1}{3\beta(1 + 8\sigma)} \biggl[ 2(4-3\beta)^2  + 3\beta^2(1 + 8\sigma) \biggr]</math>
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&nbsp;
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<math>~=</math>
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<math>~\frac{1}{3\beta(1 + 8\sigma)} \biggl[ 32 -48\beta + 18\beta^2  + 3\beta^2 + 24\beta^2\sigma \biggr]</math>
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&nbsp;
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<math>~=</math>
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<math>~\frac{1}{3\beta(1 + 8\sigma)} \biggl[ 32 -48\beta + \beta^2(  21 + 24\sigma) \biggr]</math>
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<math>~\Rightarrow ~~~~\Gamma_1 - \frac{4}{3}</math>
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<math>~=</math>
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<math>~\frac{1}{3\beta(1 + 8\sigma)} \biggl[ 32 -48\beta + \beta^2(  21 + 24\sigma) -4\beta(1+8\sigma)\biggr]</math>
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&nbsp;
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<math>~=</math>
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<math>~\frac{1}{3\beta(1 + 8\sigma)} \biggl[ 32 -4\beta(13+8\sigma) + 3\beta^2(  7 + 8\sigma) \biggr]</math>
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&nbsp;
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<math>~=</math>
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<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>
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&nbsp;
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<math>~=</math>
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<math>~\frac{1}{3\beta(1 + 8\sigma)} \biggl[ 32 -32 - 20\beta + 24\beta -3\beta^2\biggr]</math>
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&nbsp;
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<math>~=</math>
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<math>~\frac{(4-3\beta)}{3(1 + 8\sigma)} = \frac{\tfrac{4}{\beta}-3}{\tfrac{3}{\beta}(1 + 8\sigma)} </math>
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&nbsp;
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<math>~=</math>
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<math>~\frac{4(1+\sigma) - 3}{3(1+\sigma)(1 + 8\sigma)} </math>
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&nbsp;
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<math>~=</math>
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<math>~\frac{1+4\sigma}{3(1+\sigma)(1 + 8\sigma)} \, .</math>
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<span id="GeneralizingBAC84">Or, even more generally, we can show that,</span>
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<math>~\Gamma_1 - \frac{4}{3}</math>
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<math>~=</math>
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<math>~\frac{(3\gamma -4)(1+4\sigma)}{3(1+\sigma)[1 + 12(\gamma-1)\sigma]} \, .</math>
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<math>~\Gamma_1</math>
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<math>~=</math>
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<math>~\beta + \frac{g(4-3\beta)^2}{\beta + 12g (1 - \beta)} </math>
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<math>~=</math>
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<math>~\beta + \frac{g}{\beta}\biggl[4-3\beta\biggr]^2\biggl[1 + \frac{12g (1 - \beta)}{\beta}\biggr]^{-1} \, ,</math>
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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,
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<math>~\Gamma_1</math>
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<math>~=</math>
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<math>~\frac{1}{(1+\sigma)} + g(1+\sigma)\biggl[4-\frac{3}{(1+\sigma)}\biggr]^2\biggl[1 + 12g \sigma\biggr]^{-1} </math>
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&nbsp;
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<math>~=</math>
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<math>~\frac{1}{(1+\sigma)} + \frac{g[4(1+\sigma)-3]^2 }{(1 + 12g \sigma)} </math>
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