Apps/SMS

Rotating, Supermassive Stars

Bond, Arnett, & Carr (1984)

Here we draw upon the work of 📚 J. R. Bond, W. D. Arnett, & B. J. Carr (1984, ApJ, Vol. 280, pp. 825 - 847) — hereafter BAC84 — who were among the first to seriously address the question of the fate of very massive (stellar) objects.  

Equation of State

Our discussion of the equation of state (EOS) that was used by BAC84 draws on the terminology that has already been adopted in our introductory discussion of supplemental relations and closely parallels our review of the properties of the envelope that 📚 E. A. Milne (1930, MNRAS, Vol. 91, pp. 4 - 55) used to construct a bipolytropic sphere.

Expression for Total Pressure

Ignoring the component due to a degenerate electron gas, $P_{\mathrm{d}\mathrm{e}\mathrm{g}}$, the total gas pressure can be expressed as the sum of two separate components: a component of ideal gas pressure, and a component of radiation pressure. That is, in BAC84 the total pressure is given by the expression,

$P$

$=$

$P_{\mathrm{g}\mathrm{a}\mathrm{s}}+P_{\mathrm{r}\mathrm{a}\mathrm{d}},$

where,

Ideal Gas	Radiation
$P_{\mathrm{g}\mathrm{a}\mathrm{s}}=\frac{\mathrm{\Re }}{\overline{\mu }}\rho T$	$P_{\mathrm{r}\mathrm{a}\mathrm{d}}=\frac{1}{3}{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}{T}^4$

Now, BAC84 define the rest-mass density in terms of the mean baryon mass, $m_B$, via the expression, $\rho =m_{B}n$, and write (see their equation 1),

$P$

$=$

$Y_{T}nkT+\frac{1}{3}{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}{T}^4.$

In converting from our notation to theirs we conclude, therefore, that,

$\frac{\mathrm{\Re }}{\overline{\mu }}(m_{B}n)T$	$=$	$Y_{T}nkT$
$\Rightarrow Y_T$	$=$	$\frac{\mathrm{\Re }}{k}\cdot \frac{m_B}{\overline{\mu }}.$

Ratio of Radiation Pressure to Gas Pressure

Following 📚 Milne (1930), we have defined the parameter, $\beta$, as the ratio of gas pressure to total pressure. That is, in the context of BAC84, we have,

$\beta \equiv \frac{P_{\mathrm{g}\mathrm{a}\mathrm{s}}}{P},$

in which case, also,

$\frac{P_{\mathrm{r}\mathrm{a}\mathrm{d}}}{P}=1-\beta$         and         $\frac{P_{\mathrm{g}\mathrm{a}\mathrm{s}}}{P_{\mathrm{r}\mathrm{a}\mathrm{d}}}=\frac{\beta }{1-\beta }.$

Using a different notation, BAC84 (see their equation 5) define $\sigma$ as the ratio of the radiation pressure to the gas pressure. Therefore, in converting from our notation to theirs we have,

$\sigma =\frac{1-\beta }{\beta }\Rightarrow \beta =(1+\sigma {)}^{-1},$

as well as,

$\sigma =\frac{P_{\mathrm{r}\mathrm{a}\mathrm{d}}}{P_{\mathrm{g}\mathrm{a}\mathrm{s}}}=\frac{a_{\mathrm{r}\mathrm{a}\mathrm{d}}{T}^3}{3}\cdot \frac{\overline{\mu }}{(\mathrm{\Re }{m}_B)n}=\frac{a_{\mathrm{r}\mathrm{a}\mathrm{d}}{T}^3}{3Y_{T}nk},$

which is precisely the definition provided in equation (5) of BAC84.

Expression for Adiabatic Exponent

Three Equivalent Expressions

Eq. (24) from 📚 Eddington (1918) ${\mathrm{\Gamma }}_1-\frac{4}{3}$ $=$ $\frac{(\gamma -\frac{4}{3})(4-3\beta )}{1+12(\gamma -1)(1-\beta )/\beta }$ Eq. (131) from Chapter II of [C67] Eq. (74) from §3.4 of [T78] ${\mathrm{\Gamma }}_1$ $=$ $\beta +\frac{(\gamma -1)(4-3\beta {)}^2}{\beta +12(\gamma -1)(1-\beta )}$ Derived below … noting that $\beta =(1+\sigma {)}^{-1}$ ${\mathrm{\Gamma }}_1-\frac{4}{3}$ $=$ $\frac{(3\gamma -4)(1+4\sigma )}{3(1+\sigma )[1+12(\gamma -1)\sigma ]}$

For any value of the ratio of specific heats

From equation (2) of 📚 P. Ledoux & C. L. Pekeris (1941, ApJ, Vol. 94, pp. 124 - 135) — see, for example, our brief summary of this work — or, equally well, from equation (131) in Chapter II of [C67], we see that when the total pressure is of the form being considered here, a general expression for the adiabatic exponent,

${\mathrm{\Gamma }}_1\equiv \frac{d\ln P}{d\ln \rho },$

is,

${\mathrm{\Gamma }}_1$

$=$

$\beta +\frac{(\gamma -1)(4-3\beta {)}^2}{\beta +12(\gamma -1)(1-\beta )},$

where, $\gamma$ is the ratio of specific heats associated with the ideal-gas component of the equation of state. Notice that $\beta =1$ represents a situation where there is no radiation pressure. In this limit the expression simplifies to,

${\mathrm{\Gamma }}_1{|}_{\beta =1}$

$=$

$\gamma ,$

which makes sense. On the other hand, setting $\beta =0$ represents the other extreme, where there is no ideal-gas contribution to the pressure. In this case, we have,

${\mathrm{\Gamma }}_1{|}_{\beta =0}$

$=$

$\frac{16(\gamma -1)}{12(\gamma -1)}=\frac{4}{3}.$

Used by BAC84

On the other hand, without derivation BAC84 state (see their equation 4) that the adiabatic exponent is,

${\mathrm{\Gamma }}_1$

$=$

$\frac{4}{3}+\frac{4\sigma +1}{3(\sigma +1)(8\sigma +1)}.$

They also point out that, in the case where a system is dominated by radiation pressure $(\sigma \gg 1)$, this expression becomes,

${\mathrm{\Gamma }}_1{|}_{\sigma \gg 1}$

$\approx$

$\frac{4}{3}+\frac{1}{6\sigma }.$

Clearly, in the limit $\sigma \to \mathrm{\infty }$, this gives ${\mathrm{\Gamma }}_1\to 4/3$, which, as it should, matches the limiting value obtained from the 📚 Ledoux & Pekeris (1941) expression when $\beta =0$.

BAC84 do not explicitly state what value they used for the ratio of specific heats when deriving their expression for the adiabatic exponent. But this can be deduced by examining how their expression behaves in the limit of no radiation pressure, that is, for $\sigma =0$. In this limit, the BAC84 expression gives,

${\mathrm{\Gamma }}_1{|}_{\sigma =0}$

$\approx$

$\frac{4}{3}+\frac{1}{3}=\frac{5}{3}.$

The general BAC84 expression should therefore match the (even more) general 📚 Ledoux & Pekeris (1941) expression if we set $\gamma =\frac{5}{3}$. Let's check this out. Inserting this specific value of $\gamma$, and remembering (from above) that,

$\sigma =\frac{1-\beta }{\beta },$

the 📚 Ledoux & Pekeris (1941) expression for the adiabatic exponent becomes,

${\mathrm{\Gamma }}_1$	$=$	$\beta +\frac{\frac{2}{3}(4-3\beta {)}^2}{\beta +8(1-\beta )}$
	$=$	$\frac{1}{3\beta }\left[\frac{2(4-3\beta {)}^2}{1+8\sigma }\right]+\beta$
	$=$	$\frac{1}{3\beta (1+8\sigma )}[2(4-3\beta {)}^2+3{\beta }^2(1+8\sigma )]$
	$=$	$\frac{1}{3\beta (1+8\sigma )}[32-48\beta +18{\beta }^2+3{\beta }^2+24{\beta }^2\sigma ]$
	$=$	$\frac{1}{3\beta (1+8\sigma )}[32-48\beta +{\beta }^2(21+24\sigma )]$
$\Rightarrow {\mathrm{\Gamma }}_1-\frac{4}{3}$	$=$	$\frac{1}{3\beta (1+8\sigma )}[32-48\beta +{\beta }^2(21+24\sigma )-4\beta (1+8\sigma )]$
	$=$	$\frac{1}{3\beta (1+8\sigma )}[32-4\beta (13+8\sigma )+3{\beta }^2(7+8\sigma )]$
	$=$	$\frac{1}{3\beta (1+8\sigma )}[32-4\beta (8+8\sigma )-20\beta +3{\beta }^2(8+8\sigma )-3{\beta }^2]$
	$=$	$\frac{1}{3\beta (1+8\sigma )}[32-32-20\beta +24\beta -3{\beta }^2]$
	$=$	$\frac{(4-3\beta )}{3(1+8\sigma )}=\frac{\frac{4}{\beta }-3}{\frac{3}{\beta }(1+8\sigma )}$
	$=$	$\frac{4(1+\sigma )-3}{3(1+\sigma )(1+8\sigma )}$
	$=$	$\frac{1+4\sigma }{3(1+\sigma )(1+8\sigma )}.$

Or, even more generally, we can show that,

${\mathrm{\Gamma }}_1-\frac{4}{3}$

$=$

$\frac{(3\gamma -4)(1+4\sigma )}{3(1+\sigma )[1+12(\gamma -1)\sigma ]}.$

Mass Normalization

Now, according to BAC84 (see their equation 8), when the total pressure is written in polytropic form — specifically, if we set,

$P=K{\rho }^{(1+1/n_p)}$

— the mass-scaling for relativistic configurations will depend on $G$, $c$, $K$, and $n_p$ via the expression,

$M_u=K^{n_p/2}{G}^{-3/2}{c}^{3-n_p}=\left(\frac{K}{G}{)}^{3/2}\right(\frac{K}{c^2}{)}^{(n_p-3)/2}.$

Polytropic Index Equals 3

Referencing our separate discussion of Milne's (1930) work, when $n_p=3$, the polytropic constant is related to the relevant set of physical parameters via the relation,

$K_3$

$=$

$\left[\right(\frac{\mathrm{\Re }}{\overline{\mu }}{)}^4\left(\frac{1-\beta }{{\beta }^4}\right)\frac{3}{a_{\mathrm{r}\mathrm{a}\mathrm{d}}}{]}^{1/3}.$

Adopting the BAC84 terminology, this means that,

$(\frac{K_3}{G}{)}^3$	$=$	$\left(\frac{\mathrm{\Re }}{\overline{\mu }}{)}^4\right[\frac{1-\beta }{{\beta }^4}]\frac{3}{G^3{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}}$
	$=$	$\left(\frac{k{Y}_T}{m_B}{)}^4\right[{\sigma }^4(1+{\sigma }^{-1}{)}^3]\frac{3}{G^3{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}}$
$\Rightarrow M_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\equiv (\frac{K_3}{G}{)}^{3/2}$	$=$	$(1+{\sigma }^{-1}{)}^{3/2}(\frac{k{Y}_T}{m_B}{)}^2(\frac{3}{G^3{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}}{)}^{1/2}{\sigma }^2$
	$\approx$	$(1+\frac{3}{2\sigma })\left(\frac{k{Y}_T}{m_B}{)}^2\right(\frac{3}{G^3{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}}{)}^{1/2}{\sigma }^2.$

When radiation pressure significantly dominates over gas pressure — that is, in the limit $\sigma \gg 1$ — the leading factor is approximately unity, in which case we see that this expression for $M_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$ exactly matches the expression for $M_{u,3}$ given by equation (10) of BAC84.

Polytropic Index Slightly Less Than 3

More generally, equating the two expressions for the total pressure and drawing (twice) on the expression for $\sigma$ provided above, we have,

$K{\rho }^{(1+1/n_p)}$	$=$	$Y_{T}nkT+\frac{a_{\mathrm{r}\mathrm{a}\mathrm{d}}}{3}{T}^4$
	$=$	$\frac{a_{\mathrm{r}\mathrm{a}\mathrm{d}}}{3}(1+{\sigma }^{-1})T^4$
	$=$	$\frac{a_{\mathrm{r}\mathrm{a}\mathrm{d}}}{3}(1+{\sigma }^{-1})[\frac{3Y_{T}nk\sigma }{a_{\mathrm{r}\mathrm{a}\mathrm{d}}}{]}^{4/3}$
	$=$	$(\frac{3}{a_{\mathrm{r}\mathrm{a}\mathrm{d}}}{)}^{1/3}(1+{\sigma }^{-1})[Y_{T}nk\sigma {]}^{4/3}.$

Now, from above we have,

$1+\frac{1}{n_p}=\mathrm{\Gamma }$

$\approx$

$\frac{4}{3}+\frac{1}{6\sigma },$

so the lefthand-side of this last expression can be written as,

$K{\rho }^{(1+1/n_p)}$

$\approx$

$K{\rho }^{(4/3+1/6\sigma )}=K(m_{B}n{)}^{4/3}{\rho }^{1/6\sigma }.$

This means that, for any $\sigma \gg 1$,

$K$

$\approx$

$\left(\frac{3}{a_{\mathrm{r}\mathrm{a}\mathrm{d}}}{)}^{1/3}\right(\frac{Y_{T}k\sigma }{m_B}{)}^{4/3}(1+{\sigma }^{-1}){\rho }^{-1/6\sigma }.$

This matches exactly expression (7) in BAC84. Again from above — and continuing to assume $\sigma \gg 1$ — we have,

$1+\frac{1}{n_p}\approx \frac{4}{3}+\frac{1}{6\sigma }$	$\Rightarrow$	$\frac{1}{n_p}\approx \frac{1}{3}(1+\frac{1}{2\sigma })$
	$\Rightarrow$	$n_p\approx 3(1+\frac{1}{2\sigma }{)}^{-1}\approx 3(1-\frac{1}{2\sigma })$
	$\Rightarrow$	$\frac{(n_p-3)}{2}\approx -\frac{3}{4\sigma }.$

Hence, when the polytropic index is slightly less than 3, the mass normalization is,

$M_u$	$\approx$	$\left(\frac{K}{G}{)}^{3/2}\right(\frac{K}{c^2}{)}^{-3/4\sigma }$
	$=$	$\left[\frac{1}{G}\right(\frac{3}{a_{\mathrm{r}\mathrm{a}\mathrm{d}}}{)}^{1/3}(\frac{Y_{T}k\sigma }{m_B}{)}^{4/3}(1+{\sigma }^{-1}){\rho }^{-1/6\sigma }{]}^{3/2}[\frac{1}{c^2}\left(\frac{3}{a_{\mathrm{r}\mathrm{a}\mathrm{d}}}{)}^{1/3}\right(\frac{Y_{T}k\sigma }{m_B}{)}^{4/3}(1+{\sigma }^{-1}){\rho }^{-1/6\sigma }{]}^{-3/4\sigma }$
	$=$	$\left[\right(\frac{3}{G^3{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}}{)}^{1/2}(\frac{Y_{T}k}{m_B}{)}^2(1+{\sigma }^{-1}{)}^{3/2}{\sigma }^2\left]{\rho }^{-1/4\sigma }\right[\frac{1}{c^2}\left(\frac{3}{a_{\mathrm{r}\mathrm{a}\mathrm{d}}}{)}^{1/3}\right(\frac{Y_{T}k\sigma }{m_B}{)}^{4/3}(1+{\sigma }^{-1}){\rho }^{-1/6\sigma }{]}^{-3/4\sigma }$
	$=$	$\left[M_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\right]\left\{\frac{1}{c^2}\right(\frac{3}{a_{\mathrm{r}\mathrm{a}\mathrm{d}}}{)}^{1/3}(\frac{Y_{T}k\sigma }{m_B}{)}^{4/3}(m_{B}n{)}^{1/3}[(1+{\sigma }^{-1}){\rho }^{-1/6\sigma }]{\}}^{-3/4\sigma }$

Drawing again from the definition of $\sigma$ provided above, we have,

$(\frac{3}{a_{\mathrm{r}\mathrm{a}\mathrm{d}}}{)}^{1/3}$

$=$

$T(\sigma Y_{T}nk{)}^{-1/3},$

so this last relation can be rewritten as,

$M_u$

$\approx$

$M_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\cdot f[\frac{m_B{c}^2}{T\sigma Y_{T}k}{]}^{3/4\sigma }\approx f\cdot M_{u,3}(1+\frac{3}{2\sigma }\left)\right[\frac{m_B{c}^2}{T\sigma Y_{T}k}{]}^{3/4\sigma },$

where,

$f\equiv [(1+{\sigma }^{-1}{)}^{-1}{\rho }^{1/6\sigma }{]}^{3/4\sigma }\approx (1-\frac{3}{4{\sigma }^2})(n{m}_B{)}^{1/8{\sigma }^2},$

which certainly is close to unity when $\sigma \gg 1$. After setting $f=1$, this last expression for $M_u$ exactly matches the expression presented as equation (9) in BAC84.

Entropy of Radiation Field

Planck's Law of Black-Body Radiation

Drawing from the Wikipedia discussion of black-body radiation, Planck's law states that,

$B_{\nu }(T)$	$=$	$\frac{2h{\nu }^3}{c^2}\frac{1}{e^{h\nu /kT}-1}.$
[H87], §12.1 (p. 280), Eq. (12.14) [KW94], §5.1 (p. 30), Eq. (5.15) [HK94], §4.1 (p. 152), Eq. (4.7) [BLRY07], §1.6.2 (p. 23), Eq. (1.103)

Letting,

$x\equiv \frac{h\nu }{kT},$

and integrating over the entire frequency spectrum, we have,

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{B}_{\nu }(T)d\nu$

$=$

$\frac{2h}{c^2}(\frac{kT}{h}{)}^4\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x^{3}dx}{e^x-1}=\frac{2k^4}{h^3{c}^2}(\frac{{\pi }^4}{15})T^4=\frac{{\sigma }_{\mathrm{S}\mathrm{B}}{T}^4}{\pi },$

where [with units of mass × (time)^-3 K^-4],

${\sigma }_{\mathrm{S}\mathrm{B}}$

$\equiv$

$\frac{2{\pi }^5}{15}\frac{k^4}{h^3{c}^2}.$

In discussions of the thermodynamic properties of (photon) radiation fields, the radiation [density] constant, $a_{\mathrm{r}\mathrm{a}\mathrm{d}}$, appears in many physical relations. It is related to the Steffan-Boltzmann constant via the simple expression,

$a_{\mathrm{r}\mathrm{a}\mathrm{d}}=\frac{4{\sigma }_{\mathrm{S}\mathrm{B}}}{c},$

and has units of energy per unit volume × K^-4.

Photon Entropy

Here are expressions from the published literature that — to within an additive constant — give the entropy of the (photon) radiation field. We begin with D. D. Clayton (1968), who states that, "… the entropy per gram of the photon [field]" is,

$s_{\mathrm{r}\mathrm{a}\mathrm{d}}$

$=$

$\frac{4a_{\mathrm{r}\mathrm{a}\mathrm{d}}{T}^3}{3\rho }.$

[Clayton68], Chapter 2 (p. 121), Eq. (2-136)
[Shu92], Vol. I, Chapter 9 (p. 82), Eq. (9.22)

Providing the same expression, [Shu92] states that it is a measure of "… the entropy of blackbody radiation per unit mass of fluid." From the group of terms on the right-hand side of this expression, we conclude that $s_{\mathrm{r}\mathrm{a}\mathrm{d}}$ has units of specific energy per Kelvin. This is consistent with the units that are traditionally associated with entropy; see, for example, our separate discussion of the dimensions of various thermodynamic variables.

According to [ST83], "… the photon entropy per baryon" is,

$s_r=\frac{4}{3}\frac{a_{\mathrm{r}\mathrm{a}\mathrm{d}}{T}^3}{n}$	=	$\frac{4m_H{a}_{\mathrm{r}\mathrm{a}\mathrm{d}}{T}^3}{3\rho },$
[ST83], §17.2, Eq. (17.2.2) & §17.3, Eq. (17.3.7)

where, $m_H$ is the mass of a hydrogen atom, and $n=\rho /m_H$ is the baryon number density. We conclude, therefore, that $s_{\mathrm{r}\mathrm{a}\mathrm{d}}=s_r/m_H$.

We have also found it useful to examine the expression for the entropy of a radiation field used by 📚 G. M. Fuller, S. E. Woosley, & T. A. Weaver (1986, ApJ, Vol. 307, pp. 675 - 686) — hereafter FWW86 — primarily because the authors provide, for comparison, a numerical evaluation of the expression's leading coefficient.

In terms of the variables, $T$ and $\rho$, and the four physical constants, $h$, $c$, $k$, $N_A$, FWW86 state that the "… entropy per baryon … for photons is," $s_{\gamma }$ $=$ $\frac{4{\pi }^2}{45}[\frac{kT}{\hslash c}{]}^3\frac{1}{\rho N_A}.$ FWW86 §III (p. 678), Eq. (11) From the numerical values of the physical constants — which can be obtained by scrolling your cursor over each representative letter in our text (see also our accompanying table of physical constants) — and acknowledging that $\hslash \equiv h/(2\pi )$, we deduce that, to six significant digits, $hc$ $=$ $1.986446\times 10^{-16}\mathrm{e}\mathrm{r}\mathrm{g}\cdot \mathrm{c}\mathrm{m}=1.239841\times 10^{-10}\mathrm{M}\mathrm{e}\mathrm{V}\cdot \mathrm{c}\mathrm{m}$ $\Rightarrow s_{\gamma }$ $=$ $\left[\frac{(2\pi {)}^5}{45N_A(hc{)}^3}\right]\frac{(kT{)}^3}{\rho }=(4.610053\times 10^{25}{\mathrm{e}\mathrm{r}\mathrm{g}}^{-3}\cdot {\mathrm{c}\mathrm{m}}^{-3})\frac{(kT{)}^3}{\rho }$   $=$ $(1.895998\times 10^8{\mathrm{M}\mathrm{e}\mathrm{V}}^{-3}\cdot {\mathrm{c}\mathrm{m}}^{-3})\frac{(kT{)}^3}{\rho }.$ This is consistent with the statement in FWW86 that, when energy is measured in MeV, the leading coefficient is, $1.905\times 10^8$. Notice that, because the quantity, $k$$T$, has units of energy, the dimension of $s_{\gamma }$ is inverse mass. In contrast to this, as has been highlighted in our separate discussion of the physical units of various thermodynamic variables, we expect the units of specific entropy $(s_{\mathrm{r}\mathrm{a}\mathrm{d}})$ to be specific energy per Kelvin. We suspect that, for convenience, FWW86 have dropped a leading factor of $k$ on the RHS of their expression for $s_{\gamma }$ and that it is appropriate to draw from their presentation that the specific entropy of the radiation field is, $s_{\mathrm{r}\mathrm{a}\mathrm{d}}$ $=$ $k{s}_{\gamma }.$

Note that in the BAC84 publication, the Boltzmann constant, k, does not appear in this expression because "T is the temperature in energy units (Boltzmann constant = 1)."

Similarly, from BAC84 we find that "… the photon entropy [is],"

$s_{\gamma }$	$=$	$\frac{4}{3}\left[\frac{{\pi }^2}{15(\hslash c{)}^3}\right]\frac{(kT{)}^3}{n_B},$
BAC84, §2b (p. 827), Eq. (5)

where $n_B$ is the "baryon density."