Editing
SSC/Virial/PolytropesSummary
(section)
Jump to navigation
Jump to search
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
====Plotting the Virial Theorem Relation==== The relevant relation is obtained by plugging <math>~n = 5</math> into the [[User:Tohline/SSC/Virial/PolytropesSummary#ConciseVirialXY|general mass-radius relation derived above]], repeated here for clarity: <table border="1" cellpadding="8" align="center"> <tr><td> <div align="center"> <math> \mathcal{X}^4 \biggl[\frac{4\pi (5-n)}{3} \biggr] - \mathcal{X}^{(n-3)/n} \mathcal{Y}^{(n+1)/n} (4\pi)^{-1/n} \mathfrak{b}_n + \mathcal{Y}^2 \biggl(\frac{n+1}{n}\biggr) = 0 </math> </div> where, <div align="center"> <math>\mathfrak{b}_n = \biggl[ (n+1) (-\tilde\theta^')^2 + \biggl( \frac{5-n}{3} \biggr)\tilde\theta^{n+1} \biggr] \biggl( \frac{\tilde\xi}{-\tilde\theta^'} \biggr)^{(n+1)/n} </math> </div> </td></tr> </table> We will begin by plugging <math>~n = 5</math> into these expressions everywhere except for the coefficient <math>~(5-n)</math>, which we will leave unresolved, for the time being, in order to better appreciate the interplay of various terms. We obtain, <div align="center"> <table border="0" cellpadding="5"> <tr> <td align="right"> <math>~ \frac{6}{5} \mathcal{Y}^2 - \biggl( \frac{\mathfrak{b}_I^5 }{4\pi} \cdot \mathcal{X}^{2} \mathcal{Y}^{6}\biggr)^{1/5} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ (n-5)\biggl[ \biggl(\frac{4\pi}{3} \biggr) \mathcal{X}^4 - \biggl( \frac{\mathfrak{b}_{II}^5}{4\pi} \cdot \mathcal{X}^{2} \mathcal{Y}^{6}\biggr)^{1/5} \biggr] \, , </math> </td> </tr> </table> </div> where, if they are to be assigned values that are actually associated with a particular detailed force-balance model having truncation radius, <math>~\tilde\xi</math>, <div align="center"> <table border="0" cellpadding="5"> <tr> <td align="right"> <math>~ \mathfrak{b}_I^5 </math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~ \biggl[ (n+1) (-\tilde\theta^')^2 \biggl( \frac{\tilde\xi}{-\tilde\theta^'} \biggr)^{(n+1)/n} \biggr]^5_{n=5} = 2^5\cdot 3^5 \biggl[ (-\tilde\theta^')^4 {\tilde\xi}^6 \biggr] = 2^5 \cdot 3 \cdot \tilde\xi^{10} (1+\tilde\xi^2/3)^{-6}\, , </math> </td> </tr> <tr> <td align="right"> <math>~ \mathfrak{b}_{II}^5 </math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~ \biggl[ \frac{1}{3} ~ \tilde\theta^{n+1} \biggl( \frac{\tilde\xi}{-\tilde\theta^'} \biggr)^{(n+1)/n} \biggr]^5_{n=5} = \frac{1}{3^5} ~ \tilde\theta^{30} \biggl( \frac{\tilde\xi}{-\tilde\theta^'} \biggr)^{6} = 3(1+\tilde\xi^2/3)^{-6} \, . </math> </td> </tr> </table> </div> Now, if we plug <math>~n=5</math> into the remaining unresolved <math>~(n-5)</math> coefficient, the righthand side goes to zero and the mass-radius relationship provided by the virial theorem becomes, <div align="center"> <table border="0" cellpadding="5"> <tr> <td align="right"> <math>~\frac{6}{5} \mathcal{Y}^2 </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>\biggl( \frac{\mathfrak{b}_I^5 }{4\pi} \cdot \mathcal{X}^{2} \mathcal{Y}^{6}\biggr)^{1/5}</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~~ \mathcal{Y}^4 </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>\mathcal{X}^{2} \biggl[ \biggl( \frac{5^5}{2^5\cdot 3^5}\biggr) \frac{\mathfrak{b}^5_{I}}{4\pi} \biggr] </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~~ \mathcal{Y} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>\mathcal{X}^{1/2} \biggl( \frac{5^5 \mathfrak{b}^5_{I}}{2^7\cdot 3^5 \pi}\biggr)^{1/4} \, . </math> </td> </tr> </table> </div> Hence, for a given value of the structural form factor(s) — which implies a specific value of the constant coefficient, <math>~\mathfrak{b}_I</math> — the scalar virial theorem defines a relationship where the normalized mass <math>~(\mathcal{Y})</math> varies as the square root of the normalized radius <math>~(\mathcal{X})</math>. On the other hand, if we demand that the expression inside the square brackets on the righthand side of the virial theorem relation go to zero on its own — without relying on the leading coefficient to knock it zero — the mass-radius relationship provided by the virial theorem becomes, <div align="center"> <table border="0" cellpadding="5"> <tr> <td align="right"> <math>~\frac{4\pi}{3} ~ \mathcal{X}^4 </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>\biggl( \frac{\mathfrak{b}_{II}^5}{4\pi} \cdot \mathcal{X}^{2} \mathcal{Y}^{6}\biggr)^{1/5} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~~ \mathcal{Y}^6 </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>\mathcal{X}^{18} \biggl[ \biggl( \frac{4\pi}{3}\biggr)^5 \frac{4\pi}{\mathfrak{b}^5_{II}} \biggr] </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~~ \mathcal{Y} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>4\pi \mathcal{X}^{3} ( 3^5\mathfrak{b}^5_{II} )^{-1/6} \, . </math> </td> </tr> </table> </div> <!-- COMMENT 1 ....... In order to graphically display the mass-radius relation derived from the virial theorem in the <math>~\mathcal{X}-\mathcal{Y}</math> plane, as desired, we must first write out the expressions for the free-energy coefficients. After setting <math>~M_\mathrm{limit}/M_\mathrm{tot} = 1</math> in the [[User:Tohline/SSC/Virial/PolytropesSummary#Structural_Form_Factors|above summary expressions]], we obtain for all polytropic indexes, <div align="center"> <table border="0" cellpadding="5"> <tr> <td align="right"> <math>~\mathcal{A}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>\frac{1}{5} \cdot \frac{\tilde\mathfrak{f}_W}{\tilde\mathfrak{f}_M^2} = \frac{1}{5-n} \, ,</math> </td> </tr> <tr> <td align="right"> <math>~\mathcal{B}</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math> \biggl( \frac{3}{4\pi}\biggr)^{1/n} \frac{\tilde\mathfrak{f}_A}{\tilde\mathfrak{f}_M^{(n+1)/n}} = \frac{1}{3(5-n) ( 4\pi )^{1/n}} \biggl[ 3(n+1) (\tilde\theta^')^2 + (5-n)\tilde\theta^{n+1} \biggr] \biggl( \frac{\tilde\xi}{\tilde\theta^'} \biggr)^{(n+1)/n} \, . </math> </td> </tr> </table> </div> In an effort not to be caught dividing by zero while investigating the specific case of <math>~n=5</math> polytropes, we will use as shorthand notation, <div align="center"> <math>\mathfrak{b}_5 \equiv \biggl[ (5-n)\mathcal{B}\biggr]_{n=5} = \biggl[ \frac{2^3\cdot 3^5}{ \pi} \cdot (-\tilde\theta^')^{4} \tilde\xi^6 \biggr]^{1/5} = \biggl[ \frac{2^3 \cdot 3}{\pi} \biggl(1 + \frac{\tilde\xi^2}{3} \biggr)^{-6} \tilde\xi^{10} \biggr]^{1/5} \, ,</math> </div> where we have inserted the definition of <math>~\tilde\theta^'</math> as provided for <math>~n=5</math> polytropic structures in another section's [[User:Tohline/SSC/Structure/PolytropesEmbedded#Tabular_Summary_.28n.3D5.29|summary table]]. For pressure-truncated <math>~n=5</math> polytropes, we therefore have, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{R_\mathrm{eq}}{R_\mathrm{mod}} = \mathcal{X} \cdot \frac{R_\mathrm{SWS}}{R_\mathrm{mod}}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~\mathcal{X} \cdot \biggl( \frac{4\pi}{3\mathcal{B}}\biggr)^{5/6} \biggl[ \frac{3^2\mathcal{A}}{2\cdot 5\pi} \biggr]^{1/2} = (5-n)^{1/3} \mathcal{X} \cdot \biggl( \frac{4\pi}{3\mathfrak{b}_5}\biggr)^{5/6} \biggl[ \frac{3^2}{2\cdot 5\pi} \biggr]^{1/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> (5-n)^{1/3} \mathcal{X} \cdot \biggl( \frac{2^{7}\cdot 3\pi^2}{5^3} \biggr)^{1/6} \mathfrak{b}_5^{-5/6} \, , </math> </td> </tr> <tr> <td align="right"> <math>~\frac{M_\mathrm{tot}}{M_\mathrm{mod}} = \mathcal{Y} \cdot \frac{M_\mathrm{SWS}}{M_\mathrm{mod}}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ~ \mathcal{Y} \cdot \biggl( \frac{4\pi}{3\mathcal{B}} \biggr)^{5/3} \biggl[ \frac{3^2 \mathcal{A}}{2\cdot 5\pi} \biggr]^{3/2} = (5-n)^{1/6}\mathcal{Y} \cdot \biggl( \frac{4\pi}{3\mathfrak{b}_5} \biggr)^{5/3} \biggl[ \frac{3^2}{2\cdot 5\pi} \biggr]^{3/2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> (5-n)^{1/6}\mathcal{Y} \cdot \biggl( \frac{2^{11}\cdot 3^8 \pi}{5^9} \biggr)^{1/6} \mathfrak{b}_5^{-5/3} \, . </math> </td> </tr> </table> </div> So, the cross term that appears in the mass-radius relation obtained from the virial theorem may be written as, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl(\frac{R_\mathrm{eq}}{R_\mathrm{mod}}\biggr)^{2/5}\biggl(\frac{M_\mathrm{tot}}{M_\mathrm{mod}}\biggr)^{6/5} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl\{\biggl[ (5-n)^{1/3} \mathcal{X} \cdot \biggl( \frac{2^{7}\cdot 3\pi^2}{5^3} \biggr)^{1/6} \mathfrak{b}_5^{-5/6} \biggr]^2 \times \biggl[ (5-n)^{1/6}\mathcal{Y} \cdot \biggl( \frac{2^{11}\cdot 3^8 \pi}{5^9} \biggr)^{1/6} \mathfrak{b}_5^{-5/3} \biggr]^6 \biggr\}^{1/5} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~(5-n)^{1/3} \mathcal{X}^{2/5} \mathcal{Y}^{6/5} \biggl[ \biggl( \frac{2^{7}\cdot 3\pi^2}{5^3} \biggr)^{1/3} \mathfrak{b}_5^{-5/3} \cdot \biggl( \frac{2^{11}\cdot 3^8 \pi}{5^9} \biggr) \mathfrak{b}_5^{-10} \biggr]^{1/5} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~(5-n)^{1/3} \mathcal{X}^{2/5} \mathcal{Y}^{6/5} \biggl( \frac{2^{8}\cdot 3^{5} \pi}{5^{6}} \biggr)^{1/3} \mathfrak{b}_5^{-7/3} \, , </math> </td> </tr> </table> </div> and, in entirety, the virial theorem relation becomes, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ (5-n)^{1/3}\biggl\{ \mathcal{Y}^2 \cdot \biggl( \frac{2^{11}\cdot 3^8 \pi}{5^9} \biggr)^{1/3} \mathfrak{b}_5^{-10/3} - \mathcal{X}^{2/5} \mathcal{Y}^{6/5} \biggl( \frac{2^{8}\cdot 3^{5} \pi}{5^{6}} \biggr)^{1/3} \mathfrak{b}_5^{-7/3} + (5-n) \mathcal{X}^4 \cdot \biggl( \frac{2^{7}\cdot 3\pi^2}{5^3} \biggr)^{2/3} \mathfrak{b}_5^{-10/3} \biggr\} \, . </math> </td> </tr> </table> </div> A nontrivial solution is obtained by requiring that the terms inside the curly braces sum to zero. Noting that the third term must be set to zero, on its own, because it retains a leading factor of <math>~(5-n)</math>, the virial theorem relation becomes, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\mathcal{Y}^2 \cdot \biggl( \frac{2^{11}\cdot 3^8 \pi}{5^9} \biggr)^{1/3} \mathfrak{b}_5^{-10/3}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\mathcal{X}^{2/5} \mathcal{Y}^{6/5} \biggl( \frac{2^{8}\cdot 3^{5} \pi}{5^{6}} \biggr)^{1/3} \mathfrak{b}_5^{-7/3} </math> </td> </tr> </table> </div> <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Rightarrow~~~~ \mathcal{Y}^{2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\mathcal{X} \biggl( \frac{5\mathfrak{b}_5}{2 \cdot 3} \biggr)^{5/2} = \mathcal{X} \biggl( \frac{5^5}{2^2 \cdot 3^4 \pi} \biggr)^{1/2} \biggl(1 + \frac{\tilde\xi^2}{3} \biggr)^{-3} \tilde\xi^{5} \, . </math> </td> </tr> </table> </div> ....... END COMMENT --> <!-- COMMENT 2 ........ Let's go back to the [[User:Tohline/SSC/Virial/PolytropesSummary#CompactStahlerVirial|earlier virial expression]] that still contains the <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math> coefficients and multiply through by <math>~5</math> so that the coefficient of the <math>~\mathcal{X}^4</math> term matches the coefficient found in Stahler's relation. We have, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ \biggl(\frac{2^2\cdot 5\pi}{3} \biggr) \mathcal{X}^4 - 5\mathcal{B} \mathcal{X}^{2/5} \mathcal{Y}^{6/5} + 6\mathcal{A} \mathcal{Y}^{2} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~0 \, . </math> </td> </tr> </table> </div> Next, let's rewrite the other two terms so that they look more like the terms found in Stahler's expression. <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ \biggl(\frac{2^2\cdot 5\pi}{3} \biggr) \mathcal{X}^4 - 5\mathcal{X}\mathcal{Y} \biggl( \frac{\mathcal{B}^5 \mathcal{Y}}{\mathcal{X}^3} \biggr)^{1/5} + 6\mathcal{A} \mathcal{Y}^{2} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~0 </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ \biggl(\frac{2^2\cdot 5\pi}{3} \biggr) \mathcal{X}^4 - 5\mathcal{X}\mathcal{Y} + \mathcal{Y}^{2} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 5\mathcal{X}\mathcal{Y} \biggl[\biggl( \frac{\mathcal{B}^5 \mathcal{Y}}{\mathcal{X}^3} \biggr)^{1/5}-1\biggr] +(1- 6\mathcal{A} )\mathcal{Y}^{2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \mathcal{Y} \biggl\{ 5\mathcal{X}\biggl[\biggl( \frac{\mathcal{B}^5 \mathcal{Y}}{\mathcal{X}^3} \biggr)^{1/5}-1\biggr] +(1- 6\mathcal{A} )\mathcal{Y} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \mathcal{Y} \biggl\{ 5 \mathcal{B} (\mathcal{Y} \mathcal{X}^2 )^{1/5} - 5 \mathcal{X} +(1- 6\mathcal{A} )\mathcal{Y} \biggr\} \, . </math> </td> </tr> </table> </div> Now, according to Stahler's relation, the lefthand side of our derived expression should be zero when the chosen <math>~(\mathcal{X}, \mathcal{Y})</math> pair identifies an equilibrium configuration. Therefore, the terms inside the curly brackets on the righthand side of our derived expression should also sum to zero in equilibrium. Let's see if, indeed, this is the case; as shorthand, we will use, <div align="center"> <math>\Lambda \equiv \frac{\tilde{\xi}^2}{3} \, .</math> </div> From the [[User:Tohline/SSC/Structure/PolytropesEmbedded#Tabular_Summary_.28n.3D5.29|tabular summary of detailed force-balanced models]], we have, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\tilde\theta</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{(1+\Lambda)^{1/2}} \, , </math> </td> </tr> <tr> <td align="right"> <math>~-~\tilde\theta'</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{3^{1/2}} \cdot \frac{\Lambda^{1/2}}{(1+\Lambda)^{3/2}} ~~~~\Rightarrow~~~~ \biggl( \frac{\tilde{\xi}}{-\tilde\theta'} \biggr) = 3(1+\Lambda)^{3/2} \, , </math> </td> </tr> <tr> <td align="right"> <math>~\mathcal{X}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl\{ \frac{3\cdot 5}{2^2\pi} \biggl[ \frac{\Lambda}{(1+\Lambda)^2} \biggr] \biggr\}^{1/2} = \biggl( \frac{3\cdot 5}{2^2\pi} \biggr)^{1/2} \frac{\Lambda^{1/2}}{(1+\Lambda)} \, , </math> </td> </tr> <tr> <td align="right"> <math>~\mathcal{Y}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl\{ \frac{3\cdot 5^3}{2^2\pi} \biggl[ \frac{\Lambda^3}{(1+\Lambda)^4} \biggr] \biggr\}^{1/2} = 5\biggl(\frac{3\cdot 5}{2^2\pi}\biggr)^{1/2} \frac{\Lambda^{3/2}}{(1+\Lambda)^2} \, , </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ \mathcal{Y}\mathcal{X}^2</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 5\biggl(\frac{3\cdot 5}{2^2\pi}\biggr)^{3/2} \frac{\Lambda^{5/2}}{(1+\Lambda)^4} \, ; </math> </td> </tr> </table> </div> and, from the general definitions given, above, <div align="center"> <table border="0" cellpadding="5"> <tr> <td align="right"> <math>~\mathcal{A}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>\frac{1}{5-n} \, ,</math> </td> </tr> <tr> <td align="right"> <math>~\mathcal{B}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> \frac{1}{3(5-n) ( 4\pi )^{1/n}} \biggl[ 3(n+1) (\tilde\theta^')^2 + (5-n)\tilde\theta^{n+1} \biggr] \biggl( \frac{\tilde\xi}{\tilde\theta^'} \biggr)^{(n+1)/n} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ( 4\pi )^{-1/5} \biggl[ \frac{2\cdot 3}{(5-n)} (\tilde\theta^')^2 + \frac{1}{3}\cdot \tilde\theta^{6} \biggr] \biggl( \frac{\tilde\xi}{\tilde\theta^'} \biggr)^{6/5} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> ( 4\pi )^{-1/5} \biggl[ \frac{2}{(5-n)} \frac{\Lambda}{(1+\Lambda)^3} + \frac{1}{3}\cdot \frac{1}{(1+\Lambda)^3} \biggr] 3^{6/5} (1+\Lambda)^{9/5} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math> \biggl(\frac{3}{4\pi }\biggr)^{1/5} \biggl[ \frac{2\cdot 3}{(5-n)} \Lambda + 1 \biggr] (1+\Lambda)^{-6/5} \, . </math> </td> </tr> </table> </div> Hence, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~5\mathcal{B} (\mathcal{Y}\mathcal{X}^2)^{1/5}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~5 \biggl(\frac{3}{4\pi }\biggr)^{1/5} \biggl[ \frac{2\cdot 3}{(5-n)} \Lambda + 1 \biggr](1+\Lambda)^{-6/5} \cdot \biggl\{ 5\biggl(\frac{3\cdot 5}{2^2\pi}\biggr)^{3/2} \frac{\Lambda^{5/2}}{(1+\Lambda)^4} \biggr\}^{1/5} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl(\frac{3\cdot 5^3}{4\pi }\biggr)^{1/2} \biggl[ \frac{2\cdot 3}{(5-n)} \Lambda + 1 \biggr]\frac{\Lambda^{1/2}}{(1+\Lambda)^2} \, ; </math> </td> </tr> </table> </div> while, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~(1-6\mathcal{A})\mathcal{Y} - 5\mathcal{X}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ 1- \frac{2\cdot 3}{(5-n)} \biggr]\biggl(\frac{3\cdot 5^3}{2^2\pi}\biggr)^{1/2} \frac{\Lambda^{3/2}}{(1+\Lambda)^2} - \biggl( \frac{3\cdot 5^3}{2^2\pi} \biggr)^{1/2} \frac{\Lambda^{1/2}}{(1+\Lambda)} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl( \frac{3\cdot 5^3}{2^2\pi} \biggr)^{1/2} \biggl[ \Lambda - \frac{2\cdot 3 }{(5-n)} \Lambda - (1+\Lambda)\biggr] \frac{\Lambda^{1/2}}{(1+\Lambda)^2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \biggl(\frac{3\cdot 5^3}{4\pi }\biggr)^{1/2} \biggl[ \frac{2\cdot 3}{(5-n)} \Lambda + 1 \biggr]\frac{\Lambda^{1/2}}{(1+\Lambda)^2} \, . </math> </td> </tr> </table> </div> So we have the desired result, namely, that these last two expressions have opposite signs but are otherwise identical and, hence, they sum to zero. ........ END COMMENT 2 -->
Summary:
Please note that all contributions to JETohlineWiki may be edited, altered, or removed by other contributors. If you do not want your writing to be edited mercilessly, then do not submit it here.
You are also promising us that you wrote this yourself, or copied it from a public domain or similar free resource (see
JETohlineWiki:Copyrights
for details).
Do not submit copyrighted work without permission!
Cancel
Editing help
(opens in new window)
Navigation menu
Personal tools
Not logged in
Talk
Contributions
Log in
Namespaces
Page
Discussion
English
Views
Read
Edit
View history
More
Search
Navigation
Main page
Tiled Menu
Table of Contents
Old (VisTrails) Cover
Appendices
Variables & Parameters
Key Equations
Special Functions
Permissions
Formats
References
lsuPhys
Ramblings
Uploaded Images
Originals
Recent changes
Random page
Help about MediaWiki
Tools
What links here
Related changes
Special pages
Page information