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===Nonrotating, Spherically Symmetric Collapse=== When describing the [[SSC/Dynamics/FreeFall#Free-Fall_Collapse|free-fall (pressure-free) collapse]] from rest of a uniform-density sphere of mass, <math>~M</math>, and initial radius, <math>~r_0</math> — hence, initial density <math>~\rho_0 = 3M/(4\pi r_0^3)</math> — it is convenient to use the Lagrangian radial coordinate, <math>~r(t)</math>, which tracks the radius of the sphere at any time, <math>~t</math>. The relevant equation of motion is (see [http://adsabs.harvard.edu/abs/1965ApJ...142.1431L Lin, Mestel & Shu 1965] or our [[SSC/Dynamics/FreeFall#Uniform-Density_Sphere|accompanying discussion]]), <div align="center"> <table border="0", cellpadding="5"> <tr> <td align="right"> <math>~\frac{d^2r}{dt^2} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- \frac{GM}{r^2} = - \frac{G}{r^2} \biggl[ \frac{4\pi}{3} ~r_0^3 \rho_0 \biggr]</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~~\frac{1}{r_0} \frac{d^2r}{dt^2} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- \frac{4\pi G \rho_0}{3} \biggl( \frac{r_0}{r} \biggr)^2 \, ,</math> </td> </tr> </table> </div> which integrates to give, <div align="center"> <math>~r = r_0 \cos^2 \zeta \, ,</math> </div> where, <div align="center"> <table border="0", cellpadding="5"> <tr> <td align="right"> <math>~\frac{2}{\pi} \biggl[ \zeta + \frac{1}{2} \sin(2\zeta) \biggr]</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{t}{\tau_\mathrm{ff}} \, ,</math> </td> </tr> <tr> <td align="right"> <math>~\tau_\mathrm{ff}</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\biggl[ \frac{3\pi}{32 G \rho_0} \biggr]^{1/2} \, .</math> </td> </tr> </table> </div> (Detailed steps through this derivation, as well as an in-depth discussion of the free-fall collapse problem, are provided [[SSC/Dynamics/FreeFall#Free-Fall_Collapse|in an accompanying chapter of this H_Book]].) A set of dimensionless expressions drawn from this analytic solution that will prove useful to our discussion of gravitational-wave signals from core-collapse supernovae is provided in the following table. All lengths have been normalized to <math>~r_0</math>, and times have been normalized to the free-fall time, <math>~\tau_\mathrm{ff}</math>. In the lower half of the table, these analytic functions have been evaluated at six different times during the free-fall collapse. <div align="center" id="FreeFallSpherical"> <table border="1" cellpadding="5" width="75%"> <tr> <th align="center" colspan="7"> Analytic Solution to Spherical Free-Fall Collapse </th> </tr> <tr><td align="center" colspan="7"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~x(\zeta) \equiv \frac{r}{r_0}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\cos^2 \zeta \, ,</math> </td> </tr> <tr> <td align="right"> <math>~x'(\zeta) \equiv \biggl( \frac{\tau_\mathrm{ff}}{r_0} \biggr) \frac{dr}{dt}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~\frac{\pi}{2} \tan \zeta \, ,</math> </td> </tr> <tr> <td align="right"> <math>~x''(\zeta) \equiv \biggl(\frac{\tau^2_\mathrm{ff}}{r_0} \biggr) \frac{d^2r}{dt^2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-~\frac{\pi^2}{8} \cdot \frac{1}{\cos^4 \zeta} = ~-~\frac{\pi^2}{8} \cdot \frac{1}{x^2} \, ,</math> </td> </tr> <tr> <td align="right"> <math>~\frac{1}{2} \ddot{I} \equiv \frac{1}{2} \biggl(\frac{\tau_\mathrm{ff}}{r_0} \biggr)^2 \frac{d^2(r^2)}{dt^2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~+~\frac{\pi^2}{8} \cdot (\tan^2\zeta - 1) \, .</math> </td> </tr> </table> </td></tr> <tr> <td align="center"> <math>~\zeta</math> </td> <td align="center"> <math>~\frac{t}{\tau_\mathrm{ff}}</math> </td> <td align="center"> <math>~x</math> </td> <td align="center"> <math>~x'</math> </td> <td align="center"> <math>~x''</math> </td> <td align="center"> <math>~\frac{1}{2}\ddot{I}</math> </td> </tr> <tr> <td align="center"> <math>~0</math> </td> <td align="center"> <math>~0.0</math> </td> <td align="center"> <math>~1.0</math> </td> <td align="center"> <math>~0.0</math> </td> <td align="center"> <math>~-1.2337</math> </td> <td align="center"> <math>~-1.2337</math> </td> </tr> <tr> <td align="center"> <math>~\frac{\pi}{8}</math> </td> <td align="center"> <math>~0.47508</math> </td> <td align="center"> <math>~0.85355</math> </td> <td align="center"> <math>~-0.65065</math> </td> <td align="center"> <math>~-1.69336</math> </td> <td align="center"> <math>~-1.02203</math> </td> </tr> <tr> <td align="center"> <math>~\frac{\pi}{6}</math> </td> <td align="center"> <math>~0.60900</math> </td> <td align="center"> <math>~0.75000</math> </td> <td align="center"> <math>~-0.90690</math> </td> <td align="center"> <math>~-2.19325</math> </td> <td align="center"> <math>~-0.82247</math> </td> </tr> <tr> <td align="center"> <math>~\frac{\pi}{4}</math> </td> <td align="center"> <math>~0.81831</math> </td> <td align="center"> <math>~0.50000</math> </td> <td align="center"> <math>~-1.57080</math> </td> <td align="center"> <math>~-4.93480</math> </td> <td align="center"> <math>~0.0000</math> </td> </tr> <tr> <td align="center"> <math>~\frac{\pi}{3}</math> </td> <td align="center"> <math>~0.94233</math> </td> <td align="center"> <math>~0.25000</math> </td> <td align="center"> <math>~-2.72070</math> </td> <td align="center"> <math>~-19.73921</math> </td> <td align="center"> <math>~+2.46740</math> </td> </tr> <tr> <td align="center"> <math>~\frac{\pi}{2} - 0.1</math> </td> <td align="center"> <math>~0.99958</math> </td> <td align="center"> <math>~0.00997</math> </td> <td align="center"> <math>~-15.6556</math> </td> <td align="center"> <math>~-1.24196\times 10^4</math> </td> <td align="center"> <math>~+121.315</math> </td> </tr> </table> </div> Note that the second time-derivative of the moment of inertia, <math>~\ddot{I}</math>, goes through zero when <math>~\zeta = \pi/4</math>, that is, at time, <div align="center"> <math>~\frac{t}{\tau_\mathrm{ff}} = \frac{2}{\pi}\biggl[ \zeta + \frac{1}{2}\sin(2\zeta) \biggr] = \biggl( \frac{1}{2} + \frac{1}{\pi} \biggr) \, .</math> </div> <!-- OPTIONAL INTRODUCTION OF POWER SPECTRUM CONCEPT Because observers tend to think in terms of the power spectrum of a source's signal, it is worth pointing out that for the free-fall collapse the Fourier transform of the strain — that is, of the function <math>~f(t) = \ddot{I}</math> — is, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\hat{f}(\xi)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\int_{-\infty}^{\infty} f(t) e^{-2\pi i t \xi} dt \, .</math> </td> </tr> </table> </div> Then, because, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~t</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{2\tau_\mathrm{ff}}{\pi} \biggl[ \zeta + \frac{1}{2} \sin(2\zeta) \biggr]</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~~ dt</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{2\tau_\mathrm{ff}}{\pi} \biggl[ 1 + \cos(2\zeta)\biggr]d\zeta \, ,</math> </td> </tr> </table> </div> we may write, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\hat{f}(\xi)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{\pi^2}{4} \biggl( \frac{2\tau_\mathrm{ff}}{\pi} \biggr) \int_{-\infty}^{\infty} \biggl[ \tan^2\zeta - 1 \biggr] \biggl[1 + \cos(2\zeta) \biggr] \exp\biggl\{ -2\pi i \xi \cdot \frac{2\tau_\mathrm{ff}}{\pi} \biggl[ \zeta + \frac{1}{2}\sin(2\zeta) \biggr] \biggr\} d\zeta </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{\pi \tau_\mathrm{ff}}{2} \int_{-\infty}^{\infty} \biggl[ \tan^2\zeta - 1 \biggr] \biggl[1 + \cos(2\zeta) \biggr] \exp\biggl\{ -i \xi \cdot 4\tau_\mathrm{ff} \biggl[ \zeta + \frac{1}{2}\sin(2\zeta) \biggr] \biggr\} d\zeta \, .</math> </td> </tr> </table> </div> END OF OPTIONAL DISCUSSION -->
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