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,

$\frac{(1+\lambda )L_{*}^6}{\lambda (c_2{)}_{\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{y}}}\left[\frac{{\overline{a}}_{\mathrm{C}\mathrm{O}}^3{\omega }_f^2}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right]$	$=$	$\left\{\right[\frac{\lambda }{(1+\lambda {)}^2}\left]\right(\frac{d}{R}{)}^{1/2}+\frac{2}{5}\left(\frac{1}{1+\lambda }\right)[\lambda +(\frac{R^{\text{'}}}{R}{)}^2\left]\right(\frac{d}{R}{)}^{-3/2}{\}}^6$
$\Rightarrow \left[\frac{{\overline{a}}_{\mathrm{C}\mathrm{O}}^3{\omega }_f^2}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right]$	$=$	$\frac{\lambda (c_2{)}_{\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{y}}}{(1+\lambda )L_{*}^6}\left\{\right[\frac{\lambda }{(1+\lambda {)}^2}\left]\right[\frac{(1+\lambda )}{\lambda (c_2{)}_{\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{y}}}{]}^{1/6}+\frac{2}{5}\left(\frac{1}{1+\lambda }\right)[\lambda +(\frac{(c_2{)}_{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{a}\mathrm{r}\mathrm{y}}}{\lambda (c_2{)}_{\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{y}}}{)}^{2/3}\left]\right[\frac{\lambda (c_2{)}_{\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{y}}}{(1+\lambda )}{]}^{1/2}{\}}^6$
$\Rightarrow {\omega }_f^2$	$=$	$\frac{\lambda (c_2{)}_{\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{y}}}{(1+\lambda )L_{*}^6}\left[\frac{3(c_0{)}_{\mathrm{C}\mathrm{O}}{P}^2}{(4\pi {)}^2}\right]\left\{\right[\frac{\lambda }{(1+\lambda {)}^2}\left]\right[\frac{(1+\lambda )}{\lambda (c_2{)}_{\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{y}}}{]}^{1/6}+\frac{2}{5}\left(\frac{1}{1+\lambda }\right)[\lambda +(\frac{(c_2{)}_{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{a}\mathrm{r}\mathrm{y}}}{\lambda (c_2{)}_{\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{y}}}{)}^{2/3}\left]\right[\frac{\lambda (c_2{)}_{\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{r}\mathrm{y}}}{(1+\lambda )}{]}^{1/2}{\}}^6.$

Critique

Background

Let configuration #1 be a Maclaurin spheroid. Once the spheroid's eccentricity (e) has been specified, we know its dimensionless spin frequency [see note 1, below] and we know its dimensionless angular momentum [see note 2, below]. The configuration's spin frequency (Omega_0) and angular momentum (J_tot) can be expressed in physical units if, furthermore, the configuration's mass and spin period are specified.

Let configuration #2 be a binary system that contains two uniform-density spheres in circular orbit about one another, and that both have a spin frequency that is the same as (is synchronized with) the orbital frequency. Darwin (1906) provides an expression for this system's dimensionless total angular momentum, in terms of the binary's mass ratio and the ratio of the radius of each star to the orbital separation [see note 3, below].

Manuscript Review

The authors begin (their section 2) by supposing that (the pre-collapse core of) a CO star is represented by a Maclaurin spheroid, that is, represented by "configuration #1" as we have described above. They assume that the configuration's eccentricity, e = 0.812670; therefore its dimensionless spin frequency is 0.37423 (their Eq. 1) and its dimensionless angular momentum is 0.30375 (their Eq. 5). As their abstract indicates, their ultimate aim is to understand long GRBs, so they assume that the CO_star's mass is 10 solar masses and that its spin period, P, is somewhere between 5 minutes to 110 minutes (see their Table 1). Given these configuration specifications, they are able to determine the values of a variety of physical parameters (see their Table 1); of particular interest is the configuration's total angular momentum, J_CO.

Next (their section 3), the authors ask what the final system would look like if the rotating CO_star were to undergo fission while conserving its total mass and total angular momentum. As they have envisioned it, the final state is essentially "configuration #2," as we have described above.

Notes

[1] See, for example, Eq. (6) on p. 78 of Chandrasekhar's EFE, or much earlier, Eq. (1) on p. 613 of Thomas & Tait (1867).

[2] See, for example, Eq. (4.2) on p. 591 of Marcus, Press, & Teukolsky (1977).

[3] See, for example, the expression for L_1 immediately following Eq. (1) on p. 165 of Darwin (1906).

Overview

Start with a (10 solar-mass) "CO star" in orbit about a NS with orbital period, P = 2pi/Omega, and assume the CO star also spins with period P (synchronous). For the sake of simplicity, the authors treat the CO star as a rotationally flattened, uniform-density (Maclaurin) spheroid, in which case by simply **specifying** the spheroid's eccentricity, e_CO, they know the star's density and its average radius, and also ...

• Its dimensionless angular momentum; see, for example, Eq. (4.2) on p. 591 of [1];
• Its ratio of physically relevant time-scales, Omega^2/(pi G \rho); see, for example, Eq. (6) on p. 78 of [2], or much earlier, Eq. (1) on p. 613 of [3].

The numerical values -- 0.30375 and 0.37423, respectively -- that the authors have provided for these two physically interesting quantities (see their Eqs. 5 and 1, respectively) come from their assumption that e_CO = 0.812760.

Now assume that this CO star "fissions" into two component stars in orbit about one another, and that it does so while conserving total mass and total angular momentum.

Additional Minor Concerns

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