# White dwarfs (2014)

Trabajo InglésUniversidad | Universidad de Barcelona (UB) |

Grado | Física - 4º curso |

Asignatura | Mecànica quàntica de N-cossos i sistemes ultrafreds |

Año del apunte | 2014 |

Páginas | 8 |

Fecha de subida | 27/07/2014 |

Descargas | 0 |

### Descripción

Quantum mechanical model for the structure of white dwarfs.

### Vista previa del texto

White dwarfs
Many-body Quantum Mechanics
icshac∗
University of Barcelona, Spring 2014
A simple model for the mechanical structure of a white dwarf is proposed, considering
classical equilibrium dynamics and non-interacting constituents. The equations are
manipulated to facilitate their computational resolution. Results are discussed and compared
to observational data.

1
Model
1.1
Mechanical structure
The escape velocity for white dwarfs (WDs) is of the order of 0.02c [1] thus we can treat
their dynamics classically. Furthermore we assume that they are cold, spherically symmetric,
non-rotating objects in hydrostatic equilibrium.

Consider a small mass element ∆m = ρ∆V located at a distance r from the center of the
star. As shown in figure 1, three forces act on it: a gravitational force, an inner pressure and
an outer pressure. From Newton’s second law of motion
P (r)∆S − F (r) − P (r + ∆r)∆S = ∆m
d2 r
.

dt2
(1)
Using
P (r + ∆r) = P (r) +
dP
∆r
dr
(2)
∆V = ∆r∆S
(3)
and Newton’s law of gravitation
F (r) =
∗
Gm(r)∆m
,
r2
(4)
icshacub@gmail.com
1
P (r + ∆r)
F (r)
dr
P (r)
r
Figure 1: Sketch of the forces acting on a small mass element located at distance r.

equation (1) becomes
−
Gm(r)ρ(r)
d2 r
dP (r)
∆V −
∆V
=
ρ(r)∆V
.

dr
r2
dt2
Finally, in hydrostatic equilibrium
(5)
d2 r
= 0, so
dt2
dP (r)
Gm(r)ρ(r)
=−
dr
r2
(6)
with P (r) and ρ(r) the pressure and the density at distance r and m(r) the mass enclosed in
the sphere of radius r.

Equation (6) has introduced three radius-dependent magnitudes; we need two additional
equations. One is obtained simply from the definition of density
dm
dm(r)
dm
=
→
= 4πr2 ρ(r) .

dV
4πr2 dr
dr
The third and last equation will be the equation of state, derived in the next section.

ρ=
1.2
(7)
Material constituents
We assume neutral matter, formed by atoms with mass number A (Z electrons, Z protons,
A−Z neutrons). Furthermore the atoms are completely ionized, forming a homogenous plasma
without any type of interaction. The nucleons provide the totality of the mass but don’t
contribute to the energy. On the other hand, the electrons don’t contribute to the mass and
provide all the energy and, consequently, pressure.

The electron gas can be modeled as a free non-interacting Fermi sea at T = 0, with singleparticle energy
ε(k) =
m2e c4 +
2 k 2 c2
(8)
obeying the fundamental relation1
ρe =
1
Ne
νk 3
= F2
Ω
6π
(9)
for an unpolarized system
2
with ρe the electron density, Ne the number of electrons, Ω the volume, ν the electron degeneracy
and kF the Fermi momentum.

The interior temperature of a WD is Tint ∼ 107 K [2]; it’s rather counterintuitive to model
the electron gas at T = 0. Nonetheless, the mean density is ρmean ∼ 108 kg m−3 [1]; taking into
account that ρ = me ρe and using (8) and (9), this corresponds to a Fermi energy εF ∼ 10−13 J
and Fermi temperature2 TF ∼ 1010 K. Thus only a tiny fraction (T /TF ∼ 10−3 ) of the electrons
located near the Fermi surface can experiment thermal excitations, and can be disregarded
when computing the pressure. However, the density is sufficiently high for the electron rest
mass and kinetic energy to be comparable3 : no non-relativistic or ultra-relativistic limits may
be considered.

1.3
Degeneracy pressure
To calculate the electron gas’ pressure, we need its energy
kF
Ne
E=
ε(k) .

εi = ν
i=1
(10)
k
Applying the continuous limit4
∞
Ω
8π 3
Ω
=ν 2
2π
dk k 2 ε(k)Θ(kF − k)
dω
E=ν
0
(11)
kF
dk k
2
m2e c4 +
2 k 2 c2
0
with ω the solid angle. Introducing the dimensionless variable x = k/me c we compute the
bulk energy
E
ν m4e c5
e=
= 2 3
Ω
2π
= a¯
e(xF )
xF
√
dx x2 1 + x2
with xF = kF /me c, a = νm4e c5 /2π 2
xF
e¯(xF ) =
(12)
0
3
and
√
dx x2 1 + x2
0
1
=
(2x3F + xF ) 1 + x2F − sinh−1 (xF )
8
2
3
εF = kB TF
me c2 ∼ kF c ∼ 10−13 J
kF
4
→
k
Ω
(2π)3
d3 k Θ(kF − |k|)
3
(13)
.

Also, equation (9) becomes
ρe =
νm3e c3 x3F
.

6π 2 3
(14)
To compute the pressure we use the relation
∂(eΩ)
∂Ω
Ne
∂e
∂e ∂ρe
= −e − Ω
= −e − Ω
∂Ω
∂ρe ∂Ω
Ne
∂e
∂e
− 2 = −e + ρe
.

= −e − Ω
∂ρe
Ω
∂ρe
P =−
∂E
∂Ω
=−
(15)
We’ve effectively found the equation of state: e can be expressed in terms of ρe using (14) and
ρe is related to ρ. However, it’s more convenient to re-write equations (6-7) and (15) in terms
of xF , which is done in the next section.

2
2.1
Computation
Dimensionless equations
The density is locally dependent on the radius, ρ = ρ(r), thus also xF = xF (r). For simplicity,
hereon forward we drop the subindex F : xF → x.

For starters we express (15) in terms of x
P (x) = −a¯
e + ρe
∂(a¯
e)
∂¯
e ∂x
= a ρe
− e¯
∂ρe
∂x ∂ρe
(16)
= aP¯ (x)
and after some algebra
√
1
P¯ (x) =
(2x3 − 3x) 1 + x2 + 3 sinh−1 (x) .

24
(17)
Now we transform the differential equation for P (r) into a differential equation for x(r)
dP
dP dx
=
dr
dx dr
(18)
using the equation of state (16-17)
dP
P¯ (x)
a x4
=a
= √
.

dx
dx
3 1 + x2
(19)
4
Substituting in (6) yields
dx
Gmρ
3G mρ
=− 2
=−
dr
r (dP/dx)
a r2
√
1 + x2
x4
(20)
with m = m(r) and ρ = ρ(r).

Next we express ρ in terms of x, starting off with the electron density ρe = Ne /Ω = ZNA /Ω,
where NA is the number of atoms forming the star. The total mass of the star is
M = Nn mn + Np mp = NA (A − Z)mn + NA Zmp
= NA Z((A/Z − 1)mn + mp ) = NA Z m
˜N
(21)
where we’ve defined an effective nucleon mass m
˜ N = (y −1 − 1)mn + mp , with y = Z/A the usual
symmetry factor. Since ρ = M/Ω → ρe = ρ/m
˜ N ; the local density will have the same form.

3
Equating to (14) yields ρ(r) = bx with b = ν m
˜ N m3e c3 /6π 2 3 . Thus, the differential equations
to be solved are
dm
= 4bπr2 x3
(22)
dr
√
3Gb m 1 + x2
dx
=−
(23)
dr
a r2
x
Finally we introduce two dimensionless variables: m
¯ = m/m0 and r¯ = r/r0 . Setting
m0 = m
˜N
3 π
3
2ν αG
r0 =
me c
with αG = Gm
˜ 2N / c, we obtain two very easy equations
dm
¯
= r¯2 x3
d¯
r
2.2
m
¯
dx
=−
d¯
r
r¯
√
3 π
2ν αG
1 + x3
.

x
(24)
(25)
Numerical resolution
To solve the two differential equations given in (25) we apply the fourth-order Runge-Kutta
method. The appropriate boundary conditions are m(¯
¯ r = 0) = 0 and x(¯
r = 0) = xC ; different
values of xC will give rise to stars with different (total) mass and radius.

The radius of the star R is defined as the distance where the pressure vanishes:
P (r = R) = 0; this value also defines the star’s mass: M = m(r = R).

We’re not computing the pressure, thus we must find a condition for x. Since x is related to
the density5 , which is defined positive, we have the condition x > 0. Furthermore, x decreases
monotonically with r¯; implying that the algorithm must stop when x ≤ 06 .

Care must be taken at the first iteration as dx/d¯
r → ∞ for r¯ → 0. However, if we consider
3
m(¯
¯ r → 0) ∼ r¯ , the divergence disappears and dx/d¯
r = 0 at r¯ = 0. This approximation
is supported by the fact that the pressure and the density (and consequently x) present a
maximum at the center of the star.

5
6
√
x∼ 3ρ
This is backed by the fact that (17) has only one root (x = 0) and increases monotonically with x [3].

5
3
Results
The values of the various physical constants and parameters7 involved in our model have been
retrieved from [4] and [5].

3.1
Plots and discussion
Figure 2 shows the computed results of our model, for different values of the symmetry factor.

We’ve computed masses as low as 0.08 M , which is the minimum mass for a protostar to enter
the main sequence [6] and, thus, the theoretical lower bound for a WD’s mass.

Most WDs’ composition is mono-elemental, commonly carbon or oxygen, although some
helium WDs have been discovered [7]. The different plots would correspond, respectively8 , to
4
He/12 C/16 O, 3 He, 13 C, 17 O and 18 O. Clearly symmetric matter (y = 1/2) adjusts best to the
Chandrasekhar limit mass (MCL = 1.44 M [8]). Odd A values can be discarded as additional
nuclear isospin degeneracy should be considered. This is probably also the case for 18 O, which,
on the other hand, has a very low natural abundance (∼ 0.2% [9]).

In all cases, however, the total radius decreases with the total mass. This is explained by
the equation of state being “soft”, a direct consequence of the Pauli exclusion principle. Compressing a WD would increase the electron density and consequently the electrons’ energy. This
would augment the pressure and, in hydrostatic equilibrium, also the star’s mass. Nonetheless,
since electrons are massive particles, their energy is upper-bound by special relativity. There
comes a moment, thus, when the degeneracy pressure is unable to balance the gravitational
pull, explaining the existence of the Chandrasekhar limit mass. An additional interpretation
is that at very high densities, electrons and nuclei are forced very closely together and reverse
beta decay sets in, causing the star to be unstable.

Figure 3 shows the interior mass profile of three symmetric WDs and figure 4 the interior
density profile for the same three stars.

The mass increases very quickly in massive stars and quite slowly in lighter stars. However,
all three cases present a low exponential initial trend (roughly up to 0.2R) followed by a linear
central increase and a slow final convergence (slower for lighter stars).

The density drops very quickly in the outer regions of all three stars, but the central decrease
has a higher slope for more massive stars.

3.2
Comparison to observations
For completeness, in figure 5 we compare our results to some observational data, taken from two
different sources: HOB [10] and PSHT [11]. This comparison is not intended, by any means, to
be exhaustive or statistically relevant, but serves to illustrate a rough agreement between our
model and observations, at least in the range 0.45 − 1.00 M .

7
8
ν=2
in order of appearance in the figure’s key
6
Mass profiles
1/2
2/3
6/13
8/17
8/18
1.0 1.5 2.0
M (M )
m(r) (M )
30
25
20
15
10
5
0
0.08 0.5
MCL
R (×103 km)
Mass-radius relation
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0.0
2.5
0 1 2 3 4 5 6 7 8 9 10
r (×103 km)
Figure 2: Total radius vs. total mass for WDs,
for various symmetry factors.

Figure 3: Mass profile m(r) for three
symmetric WDs (y = 1/2).

Density profiles
Mass-radius relation
25
0.5 M
1.0 M
1.4 M
R (×103 km)
ρ(r) (kg m−3 )
14
10
1012
1010
108
106
104
102
100
0.5 M
1M
1.4 M
HOB
PSHT
model
20
15
10
5
0
0 1 2 3 4 5 6 7 8 9 10
r (×103 km)
0.25 0.5
Figure 4: Density profile ρ(r) for three
symmetric WDs (y = 1/2).

0.75 1.0 1.25
M (M )
1.5
Figure 5: Comparison between our model and
observational data.

References
[1] http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit3/extreme.html,
retrieved 2014-04-29.

[2] Kutner, M. L. (2003). Astronomy: A physical perspective. Cambridge University Press.

p. 189.

[3] http://www.wolframalpha.com/input/?i=%282x%5E3-3x%29*sqrt%281%2Bx%5E2%29%
2B3*asinh%28x%29, retrieved 2014-05-01.

[4] http://physics.nist.gov/constants, retrieved 2014-04-30.

[5] http://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html,
retrieved 2014-05-02.

7
[6] http://www.thphys.may.ie/staff/mshadmehri/AstroNotes/MINIMUM.pdf,
retrieved 2014-05-02.

[7] http://research.amnh.org/~bro/WD/faq.html, retrieved 2014-05-02.

[8] http://hyperphysics.phy-astr.gsu.edu/hbase/astro/whdwar.html#c5,
retrieved 2014-05-02.

[9] http://environmentalchemistry.com/yogi/periodic/O-pg2.html,
retrieved 2014-05-02.

[10] Holberg, J. B.; Oswalt, T. D.; Barstow, M. A. (2012). “Observational Constraints on the
Degenerate Mass-Radius Relation”. The Astronomical Journal 143:68. DOI: 10.1088/00046256/143/3/68.

[11] Provencal, J. L.; Shipman, H. L.; Høg, E.; Thejll, P. (1998). “Testing the White
Dwarf Mass-Radius Relation with Hipparcos”. The Astrophysical Journal 494:759. DOI:
10.1086/305238.

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