Editing SSC/Stability/InstabilityOnsetOverview (section)

====Properties of the Marginally Unstable, Fundamental-Mode Eigenfunctions====

=====Radial Displacement Function=====
The right panel of Figure 4 shows how the displacement function, <math>~x_P(\xi)</math>, varies with the radial coordinate, <math>~\xi</math>, for five different values of the polytropic index; specifically, as labeled, for n = 3, 3.05, 3.5, 5, and 6.  These are the same index values for which mass-radius equilibrium sequences have been displayed, above, in [[#Turning_Points_along_Sequences_of_Pressure-Truncated_Polytropes|Figure 3]].  The Yabushita displacement function, which is relevant to isothermal <math>~(n=\infty)</math> configurations, is also shown, for comparison.  The same set of curves (unlabeled, but having the same colors) have been redrawn on a semi-log plot in the left panel of Figure 4; in this panel the "isothermal" curve is identical to the curve that appears in the top panel of [[#Elaboration|Figure 2]].

<table border="1" cellpadding="8" align="center">
<tr>
  <th align="center"><br />Figure 4: &nbsp; Fundamental-Mode Eigenfunctions[[File:DataFileButton02.png|right|60px|file = Dropbox/WorkFolder/Wiki edits/EmbeddedPolytropes/CombinedSequences.xlsx --- worksheets = YabushitaCombined & YabushitaLinearPlot]]</th>
</tr>
<tr>
  <td align="center">
[[File:YabushitaMontageB.png|750px|Yabushita Analytic Eigenfunction]]
</td></tr>
</table>

In both panels of Figure 4, the curve associated with each polytropic index, <math>~n > 3</math>, has been displayed in two segments:  &nbsp; (a) A solid, colored segment that extends from the center of the configuration out to the radial location where the logarithmic derivative of the relevant displacement function first presents the value, <math>~d\ln x/d\ln\xi = -3</math>; and (b) a dashed, black segment that, for <math>~n < 5</math>, continues on to the natural edge of the isolated polytrope having the same index or, for <math>~n \ge 5</math>, extends to infinity.  (In all cases, the outer coordinate edge of this dashed, black curve segment lies beyond the edge of the plot and is, therefore, not actually shown.)  As we have [[#Analyses_of_Radial_Oscillations|just discussed]], for the special case of n = 3, it is the ''isolated'' polytrope &#8212; not a pressure-truncated configuration &#8212; and the more conventional surface boundary condition that are relevant, so in Figure 4 this particular displacement function has been drawn as a solid (black) curve that extends all the way out to the natural edge of an isolated n = 3 polytrope, which &#8212; see, for example, [[SSC/Structure/Polytropes#Horedt2004|p. 77 of Horedt (2004)]] &#8212; is located at <math>~\xi_\mathrm{surf} \approx 6.89684862</math>.

Drawing direct parallels with our detailed discussion, [[#Elaboration|above]], of Yabushita's displacement function in the context of isothermal spheres, we recognize that the solid curve segments displayed in Figure 4 each:
* Represents a true eigenfunction because its associated (truncated) displacement function satisfies, both, the polytropic LAWE and the appropriate surface boundary condition;
* Identifies the radial profile of the underlying equilibrium configuration's  ''fundamental mode'' of radial oscillation because it exhibits no radial nodes;
* Is associated with a configuration that is marginally [dynamically] unstable because the value of the associated (square of the) oscillation frequency is, <math>~\sigma_c^2 = 0</math>.

On each Figure 4 curve, a small, green circular marker has been placed along the displacement function at the radial coordinate location that is associated with the maximum-mass turning point of the corresponding equilibrium sequence shown in Figure 3.  In every case, this green marker sits at the same location where the transition is made from the solid segment to the dashed segment of the curve.  In each case, this is a graphical illustration of the key point made earlier:  A precise association can be made between the configuration at the maximum-mass (and <math>~P_e</math>-max) turning point and the configuration along the equilibrium sequence whose fundamental, radial mode of oscillation has an oscillation frequency of zero and, therefore, is marginally [dynamically] unstable.

=====Pressure and Density Displacement Functions=====

Referring back to the [[SSC/Perturbations#Summary_Set_of_Linearized_Equations|summary set of linearized governing equations]], we can derive analytic expressions for the pressure, <math>p_P(\xi)</math>, and/or density, <math>d_P(\xi)</math>, displacement functions that correspond to our expression for <math>x_P(\xi)</math>.  More specifically, from the linearized equation of continuity, we appreciate that,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>d_P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>-3x_P - r_0\frac{dx_P}{dr_0}</math></td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>-3x_P - \xi \cdot \frac{dx_P}{d\xi} \, .</math></td>
</tr>
</table>
Now, from our above-derived,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="1"><font color="maroon"><b>Exact Solution to the <math>~(3 \le n < \infty)</math> Polytropic LAWE</b></font></td>
</tr>
<tr>
  <td align="left">
<math>x_P \equiv \frac{3(n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{1}{\xi \theta^{n}}\biggr) \frac{d\theta}{d\xi}\biggr] \, ,</math>
  </td>
</tr>
</table>
we see that,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>\frac{dx_P}{d\xi}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{3(n-3)}{2n}\biggl[ 
-\xi^{-2}\theta^{-n} \frac{d\theta}{d\xi}
-
n\xi^{-1}\theta^{-n-1} \biggl(\frac{d\theta}{d\xi}\biggr)^2
+
\xi^{-1}\theta^{-n} \frac{d^2\theta}{d\xi^2}
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{3(n-3)}{2n}\biggl[ 
-\xi^{-1}\theta^{-n} \frac{d\theta}{d\xi}
-
n\theta^{-n-1} \biggl(\frac{d\theta}{d\xi}\biggr)^2
+
\theta^{-n} \frac{d^2\theta}{d\xi^2}
\biggr]\frac{1}{\xi} \, .
</math>
  </td>
</tr>
</table>
And, from the [[SSC/Structure/Polytropes#Lane-Emden_Equation|Lane-Emden equation]], we appreciate that,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>-\theta^n</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{\xi^2} \frac{d}{d\xi}\biggl(\xi^2 \frac{d\theta}{d\xi}\biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{2}{\xi}\biggr) \frac{d\theta}{d\xi} 
+
\frac{d^2\theta}{d\xi^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{d^2\theta}{d\xi^2}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
-\biggl(\frac{2}{\xi}\biggr) \frac{d\theta}{d\xi} 
-\theta^n
 
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \theta^{-n}\frac{d^2\theta}{d\xi^2}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
-2\xi^{-1}\theta^{-n}\frac{d\theta}{d\xi} -1 
\, .
</math>
  </td>
</tr>
</table>

As a result,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>\frac{dx_P}{d\xi}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{3(n-3)}{2n}\biggl[ 
-3\xi^{-1}\theta^{-n} \frac{d\theta}{d\xi}
-
n\theta^{-n-1} \biggl( \frac{d\theta}{d\xi} \biggr)^2
-1 
\biggr]\frac{1}{\xi} \, ;
</math>
  </td>
</tr>
</table>
and,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>d_P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>-3\biggl\{ \frac{3(n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{1}{\xi \theta^{n}}\biggr) \frac{d\theta}{d\xi}\biggr] \biggr\} 
- 
\biggl\{ \frac{3(n-3)}{2n}\biggl[ -3\xi^{-1}\theta^{-n} \frac{d\theta}{d\xi} - n\theta^{-n-1} \biggl(\frac{d\theta}{d\xi}\biggr)^2 - 1 
\biggr]\biggr\} </math></td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>-  \biggl[\frac{9(n-1)}{2n} 
+ \frac{9(n-3)}{2n} \biggl( \frac{1}{\xi \theta^{n}}\biggr) \frac{d\theta}{d\xi}\biggr]  
+ 
\biggl\{ \frac{3(n-3)}{2n}\biggl[ 3\xi^{-1}\theta^{-n} \frac{d\theta}{d\xi} + n\theta^{-n-1} \biggl(\frac{d\theta}{d\xi}\biggr)^2 + 1 
\biggr]\biggr\} </math></td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>-  \biggl[\frac{9(n-1)}{2n} 
\biggr]  
+ 
\frac{3(n-3)}{2n}\biggl[ \frac{n}{\theta^{n+1}} \biggl(\frac{d\theta}{d\xi}\biggr)^2 + 1 
\biggr] </math></td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
-3 \biggl\{1   
- 
\frac{(n-3)}{2}\biggl[ \frac{1}{\theta^{n+1}} \biggl( \frac{d\theta}{d\xi}\biggr)^2 \biggr] \biggr\}
\, .
</math></td>
</tr>
</table>
Finally, according to the [[SSC/Perturbations#Summary_Set_of_Linearized_Equations|linearized (adiabatic form of the) first law of thermodynamics]], <math>p = \gamma_g d</math>.  That is, when we set <math>\gamma_g = (n+1)/n</math> along with <math>\sigma_c^2 = 0</math>, we find that the pressure displacement function is given by the expression,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>p_P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
-\frac{3(n+1)}{n} \biggl\{1   
- 
\frac{(n-3)}{2}\biggl[ \frac{1}{\theta^{n+1}} \biggl( \frac{d\theta}{d\xi}\biggr)^2 \biggr] \biggr\}
\, .
</math></td>
</tr>
</table>

For later use ([[#Derivative|immediately below]]) we note as well that,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>\frac{d(p_P)}{d\xi}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
\frac{3(n+1)(n-3)}{2n}   \frac{d}{d\xi}
\biggl[ \theta^{-(n+1)} \biggl( \frac{d\theta}{d\xi}\biggr)^2 \biggr] 
</math></td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
\frac{3(n+1)(n-3)}{2n}   
\biggl[ -\frac{(n+1)}{\theta^{n+2}}\biggl( \frac{d\theta}{d\xi}\biggr)^2
+ 
2\theta^{-(n+1)} \frac{d^2\theta}{d\xi^2} 
\biggr] \biggl(\frac{d\theta}{d\xi}\biggr) 
</math></td>
</tr>
</table>

Again, from the [[SSC/Structure/Polytropes#Lane-Emden_Equation|Lane-Emden equation]], we appreciate that,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>\theta^{-n}\frac{d^2\theta}{d\xi^2}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
-2\xi^{-1}\theta^{-n}\frac{d\theta}{d\xi} -1 
\, .
</math>
  </td>
</tr>
</table>
Hence,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>\frac{d(p_P)}{d\xi}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
\frac{3(n+1)(n-3)}{2n}   
\biggl\{ -\frac{(n+1)}{\theta^{n+2}}\biggl( \frac{d\theta}{d\xi}\biggr)^2
+ 
\frac{2}{\theta} \biggl[-2\xi^{-1}\theta^{-n}\frac{d\theta}{d\xi} -1\biggr]
\biggr\} \biggl(\frac{d\theta}{d\xi}\biggr) 
</math></td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
-\frac{3(n+1)(n-3)}{n}   
\biggl\{ \frac{(n+1)}{2\theta^{n+1}}\biggl( \frac{d\theta}{d\xi}\biggr)^2
+ 
2\xi^{-1}\theta^{-n}\biggl(\frac{d\theta}{d\xi}\biggr) + 1
\biggr\} \biggl(\frac{d\theta}{d\xi}\biggr) \frac{1}{\theta} 
</math></td>
</tr>
</table>

=====Cross-Check=====

Can you determine <math>x_P</math>, given this expression for <math>p_P</math>?

When the square of the radial-oscillation frequency is set to zero, the [[SSC/Perturbations#Summary_Set_of_Linearized_Equations|linearized Euler + Poisson equations]] states,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>\biggl[ \frac{P_0}{\rho_0 r_0} \biggr] \frac{d(p)}{d\ln r_0}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math> 
(4x + p)g_0 + \cancelto{0}{\omega^2 r_0 x} 
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ 4x_P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math> 
\biggl[ \frac{P_0}{\rho_0 r_0 g_0} \biggr] \frac{d(p_P)}{d\ln \xi}
-
p_P
</math>
  </td>
</tr>
</table>
Now, according to our [[SSC/Stability/Polytropes#Groundwork|accompanying discussion of the equilibrium structure of isolated polytopes]],
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~r_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_n \xi \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\rho_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\rho_c \theta^{n} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~P_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>K\rho_0^{(n+1)/n} = K\rho_c^{(n+1)/n} \theta^{n+1} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~g_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{GM(r_0)}{r_0^2} = \frac{G}{r_0^2} \biggl[ 4\pi a_n^3 \rho_c \biggl(-\xi^2 \frac{d\theta}{d\xi}\biggr) \biggr]
\, ,</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>a_n</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[\frac{(n+1)K}{4\pi G} \cdot \rho_c^{(1-n)/n} \biggr]^{1/2} </math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; <math>\Rightarrow</math>&nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>K</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[\frac{4\pi G a_n^2}{(n+1)} \cdot \rho_c^{(n-1)/n} \biggr] \, .</math>
  </td>
</tr>
</table>
</div>
Therefore,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{P_0}{\rho_0 r_0 g_0}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[K\rho_c^{(n+1)/n} \theta^{n+1}\biggr]
\cdot \biggl[ \frac{1}{\rho_c \theta^n} \biggr]
\cdot \frac{1}{\xi a_n}  
\cdot \frac{a_n^2 \xi^2}{G} \biggl[ 4\pi \rho_c a_n^3 \biggl(-\xi^2 \frac{d\theta}{d\xi}\biggr) \biggr]^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{4\pi G a_n^2}{(n+1)} \cdot \rho_c^{(n-1)/n} \biggr]
\biggl[\rho_c^{(n+1)/n} \theta^{n+1}\biggr]
\cdot \biggl[ \frac{1}{\rho_c \theta^n} \biggr]  
\cdot \frac{\xi}{4\pi \rho_c a_n^2 G} \biggl(-\xi^2 \frac{d\theta}{d\xi}\biggr)^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{1}{(n+1)}  \biggr]
\cdot \xi \theta \biggl(-\xi^2 \frac{d\theta}{d\xi}\biggr)^{-1} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-\frac{1}{(n+1)}
\cdot \frac{\theta}{\xi} \biggl(\frac{d\theta}{d\xi}\biggr)^{-1} 
\, .
</math>
  </td>
</tr>
</table>

<span id="Derivative">As a result</span>, we find that,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>\biggl[ \frac{P_0}{\rho_0 r_0 g_0} \biggr] \frac{d(p_P)}{d\ln \xi}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
\biggl\{
\frac{1}{(n+1)}
\cdot \frac{\theta}{\xi} \biggl(\frac{d\theta}{d\xi}\biggr)^{-1} 
\biggr\}
\frac{3(n+1)(n-3)}{n}   
\biggl\{ \frac{(n+1)}{2\theta^{n+1}}\biggl( \frac{d\theta}{d\xi}\biggr)^2
+ 
2\xi^{-1}\theta^{-n}\biggl(\frac{d\theta}{d\xi}\biggr) + 1
\biggr\} \biggl(\frac{d\theta}{d\xi}\biggr) \frac{\xi}{\theta} 
</math></td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
\frac{3(n-3)}{n}   
\biggl\{ \frac{(n+1)}{2\theta^{n+1}}\biggl( \frac{d\theta}{d\xi}\biggr)^2
+ 
2\xi^{-1}\theta^{-n}\biggl(\frac{d\theta}{d\xi}\biggr) + 1
\biggr\} \, ;
</math></td>
</tr>
</table>
and

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>4x_P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math> 
\biggl[ \frac{P_0}{\rho_0 r_0 g_0} \biggr] \frac{d(p_P)}{d\ln \xi}
-
p_P
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math> 
\frac{3(n-3)}{n}   
\biggl\{ \frac{(n+1)}{2\theta^{n+1}}\biggl( \frac{d\theta}{d\xi}\biggr)^2
+ 
2\xi^{-1}\theta^{-n}\biggl(\frac{d\theta}{d\xi}\biggr) + 1
\biggr\} 
+ \frac{3(n+1)}{n} \biggl\{1   
- 
\frac{(n-3)}{2}\biggl[ \frac{1}{\theta^{n+1}} \biggl( \frac{d\theta}{d\xi}\biggr)^2 \biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math> 
\frac{3(n-3)}{n}\biggl\{   
2\xi^{-1}\theta^{-n}\biggl(\frac{d\theta}{d\xi}\biggr) + 1
\biggr\} 
+ \frac{3(n+1)}{n} 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math> 
\frac{3(n-3)}{n}\biggl[   
2\xi^{-1}\theta^{-n}\biggl(\frac{d\theta}{d\xi}\biggr)
\biggr] 
+ \frac{6(n-1)}{n} 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{6(n-1)}{n} \biggl\{ 
1 + 
\frac{n}{6(n-1)} \cdot \frac{3(n-3)}{n}\biggl[   
2\xi^{-1}\theta^{-n}\biggl(\frac{d\theta}{d\xi}\biggr)
\biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{6(n-1)}{n} \biggl\{ 
1 + 
\frac{(n-3)}{(n-1)}\biggl[   
\xi^{-1}\theta^{-n}\biggl(\frac{d\theta}{d\xi}\biggr)
\biggr] 
\biggr\} \, .
</math>
  </td>
</tr>
</table>

Q. E. D.

=====Displacement Functions Summary=====

<table border="1" align="center" width="60%" cellpadding="8"><tr><td align="left">

<div align="center"><b>Summary &hellip;</b></div>

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right"><math>x_P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
\frac{3(n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{1}{\xi \theta^{n}}\biggr) 
\frac{d\theta}{d\xi}\biggr] \, ,</math>
  </td>
</tr>

<tr>
  <td align="right"><math>d_P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
-3 \biggl\{1   
- 
\frac{(n-3)}{2}\biggl[ \frac{1}{\theta^{n+1}} \biggl( \frac{d\theta}{d\xi}\biggr)^2 \biggr] \biggr\}
\, ,
</math></td>
</tr>

<tr>
  <td align="right"><math>p_P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math> 
\biggl( \frac{n+1}{n} \biggr) d_P
\, .
</math></td>
</tr>
</table>

</td></tr></table>