Appendix/Ramblings/BdHN

Binary-driven Hpernovae

The material presented here builds on our separate discussion of close binary stars.

Setup

Initial Carbon-Oxygen Core

The total angular momentum associated with a Maclaurin spheroid of eccentricity, $e$, is given by the expression,

$L_{*}^2\equiv \frac{L^2}{(G{M}^3\overline{a})}$	$=$	$\frac{6}{5^2}[(3-2e^2)(1-e^2{)}^{1/2}\cdot \frac{{\sin }^{-1}e}{e^3}-\frac{3(1-e^2)}{e^2}](1-e^2{)}^{-2/3}.$
📚 Marcus, Press, & Teukolsky (1977), §IVa, p. 591, Eq. (4.2)

According to an accompanying table …

Jacobi Sequence Bifurcation:

• $[e,{\omega }_0^2/(4\pi G\rho ),j^2]=[0.812670,0.0935574,0.00455473]\Rightarrow [e,{\omega }_0^2/(\pi G\rho ),L_{*}]=[0.812670,0.374230,0.303751]$

Dynamically Unstable 2^nd Harmonic:

• $[e,{\omega }_0^2/(4\pi G\rho ),j^2]=[0.95289,0.11006,0.012796]\Rightarrow [e,{\omega }_0^2/(\pi G\rho ),L_{*}]=[0.95289,0.44024,0.25921]$

Now, given that, ${\omega }_0^2=(2\pi /P{)}^2$, if we specify $P$, then we know ${\rho }_{\mathrm{C}\mathrm{O}}$ from the expression,

${\rho }_{\mathrm{C}\mathrm{O}}$

$=$

$(\frac{{\omega }_0^2}{\pi G\rho }{)}^{-1}\frac{(2\pi /P{)}^2}{\pi G};$

and if we specify $M_{\mathrm{C}\mathrm{O}}$, we know ${\overline{a}}_{\mathrm{C}\mathrm{O}}$ via the expression,

${\overline{a}}_{\mathrm{C}\mathrm{O}}$

$=$

$(\frac{3M_{\mathrm{C}\mathrm{O}}}{4\pi {\rho }_{\mathrm{C}\mathrm{O}}}{)}^{1/3}.$

Initial CO Core at the Jacobi Bifurcation Point with $M_{\mathrm{C}\mathrm{O}}=10M_{\odot }$ and various spin periods, $P$
$P$ (min)	$P$ (s)	${\rho }_{\mathrm{C}\mathrm{O}}=\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{1}/P^2$		${\overline{a}}_{\mathrm{C}\mathrm{O}}=\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{2}\cdot {\rho }_{\mathrm{C}\mathrm{O}}^{-1/3}$		$L_{\mathrm{C}\mathrm{O}}=\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{3}\cdot {\overline{a}}^{1/2}$
		factor1	(g cm-3)	factor2	(cm)	factor3	(g cm2 s-1)
100	6600	…	1.155 × 101	…	7.435 × 1010	…	6.001 × 1052
5	300	…	5.592 × 103	…	9.469 × 109	…	2.142 × 1052
$G=6.672\times 10^{-8}$ cm3 g-1 s-2 $M_{\odot }=1.989\times 10^{33}$ g $\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{1}=(0.37423{)}^{-1}\cdot \frac{4\pi }{G}=5.033\times 10^8$ cgs $\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{2}=1.681\times 10^{11}$ cgs $\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{3}=(0.303751)\cdot (G{M}_{\mathrm{C}\mathrm{O}}^3{)}^{1/2}=2.201\times 10^{47}$ cgs

Related Binary

Consider the simple model of two spherical stars in circular orbit about one another, as depicted here on the right. In addition to the physical parameters explicitly labeled in this diagram, we adopt the following variable notation:

The total system mass is, $M_{\mathrm{t}\mathrm{o}\mathrm{t}}\equiv M+M^{\text{'}};$ The ratio of the primary to secondary mass is, $\lambda \equiv \frac{M}{M^{\text{'}}};$ And the separation between the two centers is, $d\equiv r_{\mathrm{c}\mathrm{m}}+r_{\mathrm{c}\mathrm{m}}^{\text{'}}.$

For a circular orbit, the angular velocity is related to the the system mass and separation via the Kepler relation,

${\omega }^2{d}^3=G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}},$

and the distances, $r_{\mathrm{c}\mathrm{m}}$ and $r_{\mathrm{c}\mathrm{m}}^{\text{'}}$, between the center of each star and the center of mass (cm) of the system must be related to one another via the expression,

$\frac{r_{\mathrm{c}\mathrm{m}}^{\text{'}}}{r_{\mathrm{c}\mathrm{m}}}$

$=$

$\frac{M}{M^{\text{'}}}=\lambda .$

Note that the following relations also hold:

$M=M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left(\frac{\lambda }{1+\lambda }\right)$	and	$M^{\text{'}}=M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left(\frac{1}{1+\lambda }\right)$
$r_{\mathrm{c}\mathrm{m}}=d\left(\frac{1}{1+\lambda }\right);$	and	$r_{\mathrm{c}\mathrm{m}}^{\text{'}}=d\left(\frac{\lambda }{1+\lambda }\right).$

Hence, the orbital angular momentum is,

$L_{\mathrm{o}\mathrm{r}\mathrm{b}}$	$=$	$[M{r}_{\mathrm{c}\mathrm{m}}^2+M^{\text{'}}(r_{\mathrm{c}\mathrm{m}}^{\text{'}}{)}^2]\omega$
	$=$	$M_{\mathrm{t}\mathrm{o}\mathrm{t}}{d}^2\left[\right(\frac{\lambda }{1+\lambda }\left)\right(\frac{1}{1+\lambda }{)}^2+\left(\frac{1}{1+\lambda }\right)\left(\frac{\lambda }{1+\lambda }{)}^2\right][\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{d^3}{]}^{1/2}$
	$=$	$(G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{3}d{)}^{1/2}[\frac{\lambda }{(1+\lambda {)}^2}].$

Assuming that both stars are rotating synchronously with the orbit, their respective spin angular momenta are,

$L_M=I_M\omega$	$=$	$\frac{2}{5}M{R}^2\omega$
	$=$	$\frac{2}{5}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}\left(\frac{\lambda }{1+\lambda }\right)R^2[\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{d^3}{]}^{1/2}$
	$=$	$\frac{2}{5}(G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{3}d{)}^{1/2}(\frac{\lambda }{1+\lambda }\left)\right(\frac{R}{d}{)}^2,$
$L_{M^{\text{'}}}=I_{M^{\text{'}}}\omega$	$=$	$\frac{2}{5}{M}^{\text{'}}(R^{\text{'}}{)}^2\omega$
	$=$	$\frac{2}{5}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}\left(\frac{1}{1+\lambda }\right)(R^{\text{'}}{)}^2[\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{d^3}{]}^{1/2}$
	$=$	$\frac{2}{5}(G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{3}d{)}^{1/2}(\frac{1}{1+\lambda }\left)\right(\frac{R^{\text{'}}}{d}{)}^2.$

Hence, the total angular momentum of the system is,

$L_{\mathrm{t}\mathrm{o}\mathrm{t}}=L_{\mathrm{o}\mathrm{r}\mathrm{b}}+L_M+L_{M^{\text{'}}}$	$=$	$(G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{3}d{)}^{1/2}[\frac{\lambda }{(1+\lambda {)}^2}]+\frac{2}{5}(G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{3}d{)}^{1/2}\left(\frac{\lambda }{1+\lambda }\right)(\frac{R}{d}{)}^2$
		$+\frac{2}{5}(G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{3}d{)}^{1/2}(\frac{1}{1+\lambda }\left)\right(\frac{R^{\text{'}}}{d}{)}^2$
	$=$	$(G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{3}R{)}^{1/2}\{\left[\frac{\lambda }{(1+\lambda {)}^2}\right](\frac{d}{R}{)}^{1/2}+\frac{2}{5}(\frac{1}{1+\lambda }\left)\right[\lambda +\left(\frac{R^{\text{'}}}{R}{)}^2\right]\left(\frac{d}{R}{)}^{-3/2}\right\}$

Solutions

Here we assume that the unprimed parameters (primary star) refer to the Maclaurin spheroid component while the primed parameters (secondary star) refer to the Jacobi ellipsoid component. Also assume that the mass ratio is,

$\lambda \equiv \frac{M}{M^{\text{'}}}$

$=$

$\frac{M_{\mathrm{C}\mathrm{O}\mathrm{f}}}{M_{\mathrm{J}\mathrm{T}\mathrm{E}}}=\frac{8.5}{1.5}=\frac{17}{3}=5.6667.$

Assume that the total angular momentum and total mass are conserved, and that the three frequencies that appear in the binary system — orbital, and two stellar spins — are identical (co-rotation). This means that the orbital separation, $d$, can be expressed in terms of ${\omega }_f$ via the Keperian relation,

$d^3$

$=$

$\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{{\omega }_f^2}.$

And the radii of the binary components are given, respectively, by the relations,

$R^3$	$=$	$(c_2{)}_{\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{y}}[\frac{\lambda G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{(1+\lambda ){\omega }_f^2}];$
$(R^{\text{'}}{)}^3$	$=$	$(c_2{)}_{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{a}\mathrm{r}\mathrm{y}}[\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{(1+\lambda ){\omega }_f^2}].$

Hence, the length-ratios that appear in the expression for $L_{\mathrm{t}\mathrm{o}\mathrm{t}}$ may be re-expressed as follows:

$(\frac{R^{\text{'}}}{R}{)}^3$	$=$	$(c_2{)}_{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{a}\mathrm{r}\mathrm{y}}[\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{(1+\lambda ){\omega }_f^2}\left]\right\{(c_2{)}_{\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{y}}[\frac{\lambda G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{(1+\lambda ){\omega }_f^2}]{\}}^{-1}$
	$=$	$\frac{(c_2{)}_{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{a}\mathrm{r}\mathrm{y}}}{\lambda (c_2{)}_{\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{y}}};$
$(\frac{d}{R}{)}^3$	$=$	$\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{{\omega }_f^2}\{(c_2{)}_{\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{y}}\left[\frac{\lambda G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{(1+\lambda ){\omega }_f^2}\right]{\}}^{-1}$
	$=$	$\frac{(1+\lambda )}{\lambda (c_2{)}_{\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{y}}}.$

Also,

$\frac{L_{\mathrm{t}\mathrm{o}\mathrm{t}}^6}{(G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{3}R{)}^3}=\frac{L_{*}^6(G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^3{\overline{a}}_{\mathrm{C}\mathrm{O}}^3)}{(G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{3}R{)}^3}$	$=$	$L_{*}^6{\overline{a}}_{\mathrm{C}\mathrm{O}}^3\{(c_2{)}_{\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{y}}\left[\frac{\lambda G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{(1+\lambda ){\omega }_f^2}\right]{\}}^{-1}$
	$=$	$\frac{(1+\lambda )L_{*}^6}{\lambda (c_2{)}_{\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{y}}}\left[\frac{{\overline{a}}_{\mathrm{C}\mathrm{O}}^3{\omega }_f^2}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right].$

As a result, we find,

$\frac{(1+\lambda )L_{*}^6}{\lambda (c_2{)}_{\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{y}}}\left[\frac{{\overline{a}}_{\mathrm{C}\mathrm{O}}^3{\omega }_f^2}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right]$	$=$	$\left\{\right[\frac{\lambda }{(1+\lambda {)}^2}\left]\right(\frac{d}{R}{)}^{1/2}+\frac{2}{5}\left(\frac{1}{1+\lambda }\right)[\lambda +(\frac{R^{\text{'}}}{R}{)}^2\left]\right(\frac{d}{R}{)}^{-3/2}{\}}^6$
$\Rightarrow \left[\frac{{\overline{a}}_{\mathrm{C}\mathrm{O}}^3{\omega }_f^2}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right]$	$=$	$\frac{\lambda (c_2{)}_{\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{y}}}{(1+\lambda )L_{*}^6}\left\{\right[\frac{\lambda }{(1+\lambda {)}^2}\left]\right(\frac{d}{R}{)}^{1/2}+\frac{2}{5}\left(\frac{1}{1+\lambda }\right)[\lambda +(\frac{R^{\text{'}}}{R}{)}^2\left]\right(\frac{d}{R}{)}^{-3/2}{\}}^6$

Critique

1. Ill-advised to refer to the new NS as "νNS" because, in this context, readers might reasonably associate the greek letter, ν, with neutrinos.

See Also