Tp

Positive mass theorem and adiabatic
index below 6/5 in a static star
A. K. M. Masood-ul-Alam
abulm@math.tsinghua.edu.cn

Mathematical Sciences Center & Department of Mathematical Sciences
Tsinghua University, Beijing 100084, China

Done with Prosper – p. 1/45

Pressure, Density, Adiabatic Index
Equation of state: a function of one variable: ρ = ρ(p)


ρ + p dp
Adiabatic index: γ =
p dρ


ρ + p dp
p dρ
p, called pressure, is usually small compared to ρ,
called density or energy density or mass-energy ensity
or total mass-energy density including internal energy.


ρ + p dp
p dρ
p
We use geometrized units c = G = 1. p was 2 where c
c
is the speed of light.


ρ + p dp
p dρ
p
We use geometrized units c = G = 1. p was 2 where c
c
is the speed of light.
Ex. If pressure is 1030 dynes/cm2 corresponding to a
density 3.2 × 1011 gm/cm3 , then with
c = 3 × 1010 cm/sec2 , p = 0.0035. Recall
c2 ρ
1 dynes/cm2 = 1 g/(cm sec2 ) ≈ 0.9872 × 10−6 atm.

History of the name, Small p/ρ
If we neglect small p/ρ, then
ρ + p dp p d log p
γ= = (1 + )
p dρ ρ d log ρ

coincides with the adiabatic index of perfect gas
ρ dp
γ= .
p dρ


History of the name, Small p/ρ
If we neglect small p/ρ, then
ρ + p dp p d log p
γ= = (1 + )
p dρ ρ d log ρ

coincides with the adiabatic index of perfect gas
ρ dp
γ= .
p dρ
The name originated from Poisson adiabatic eqns
P V γ = constant, P 1−γ T γ = constant, T V γ−1 = constant,
where γ is the ratio of speciﬁc heats cp /cv . More
generally polytropic changes (quasi-static process in
which dQ/dT = c = constant during the entire process)
satisfy above equations with γ replaced polytropic
cp − c
exponent γ = .
cv − c Done with Prosper – p. 3/45

Conditions on ρ and p
Differentiability Conditions: ρ(p) is a piecewise C 1 on
(0, pc ).
We call pc the central pressure. On (0, pc )


(0, pc ).
dp
ρ > 0 and > 0.
dρ


(0, pc ).
dp
ρ > 0 and > 0.
dρ
dp
For a polytropic gas is the velocity of sound.
dρ


(0, pc ).
dp
ρ > 0 and > 0.
dρ
dp
dρ
However constant density (incompressible ﬂuid!) is
dρ
allowed: = 0. Real motivation: history based on
dp
stability considerations.


(0, pc ).
dp
ρ > 0 and > 0.
dρ
dp
dρ
However constant density (incompressible ﬂuid!) is
dρ
allowed: = 0. Real motivation: history based on
dp
stability considerations.
lim inf γ(p) = lim sup γ(p).
p↓0 +
p↓0+


log p and γ v. log ρ

(1)

(2)

Bigger


Static stellar model
Σ is a manifold diffeomorphic to R3 .


p is now a function on Σ.


p vanishes outside a compact set in Σ. This outside is
the vacuum and the inside, where p > 0 is the star or
ﬂuid region.


p vanishes outside a compact set in Σ. This outside is
the vacuum and the inside, where p > 0 is the star or
ﬂuid region.
Assume ρ = 0 outside.


Static stellar model: Einstein Equations

g is a Riemannian metric on Σ.



V is a positive function on Σ.



4
Space-time is (Σ × R, g), where



4
4
g = −V 2 dt2 + g



4
4
g = −V 2 dt2 + g
4
Since V and g are independent of time t, g is static.



4
4
g = −V 2 dt2 + g
4
Since V and g are independent of time t, g is static.
4
Space-time metric g satisﬁes Einstein equations
4 1 4 4
Ric(g)αβ − Scalar(g)g αβ = 8πTαβ
2
4
with energy-momentum tensor Tαβ = (ρ + p)uα uβ + pg αβ


Static stellar model – cont.
uα four dimensional velocity of the ﬂuid.


∂
Static ﬂuid: uα = V −1 unit timelike vector.
∂t


∂
Static ﬂuid: uα =V −1 unit timelike vector.
∂t
Σ, 3-space at any one time, is the rest frame of the ﬂuid.


On (Σ, g)
V and g then satisfy the equations
Rij = V −1 i j V + 4π(ρ − p)gij
∆V = 4πV (ρ + 3p)


On (Σ, g)
∆V = 4πV (ρ + 3p)
Rij is the Ricci curvature of g .


On (Σ, g)
∆V = 4πV (ρ + 3p)
Rij is the Ricci curvature of g .
idenotes covariant derivative and
i ij ∂2 k ∂ denotes the Laplacian
∆= i =g i ∂xj
−Γ
∂x ∂xk
relative to g .


Surface of the Star
p = 0 on the surface of the star.


Surface of the Star
Assume that the surface is smooth.


Surface of the Star
V and g are C 3 except at the surface of the star (and
possibly at several smooth V = const surfaces inside
the star).


Surface of the Star
V and g are C 3 except at the surface of the star (and
possibly at several smooth V = const surfaces inside
the star).
V and g are globally C 1,1 .


Surface of the Star
The contracted Bianchi identity for g implies that p is a
Lipschitz function on Σ and (almost everywhere)

ip = −V −1 (ρ + p) iV


Surface of the Star

ip = −V −1 (ρ + p) iV

If ρ(p) is a Lipschitz function of p this implies

dρ i
∆p + 2 + V −1 V ip = −4π(ρ + p)(ρ + 3p) ≤ 0
dp


Surface of the Star

ip = −V −1 (ρ + p) iV


dρ i
∆p + 2 + V −1 V ip = −4π(ρ + p)(ρ + 3p) ≤ 0
dp

Thus if ρ(p) is a Lipschitz function of p then p vanishes
identically.


Surface of the Star

ip = −V −1 (ρ + p) iV


dρ i
∆p + 2 + V −1 V ip = −4π(ρ + p)(ρ + 3p) ≤ 0
dp

Thus if ρ(p) is a Lipschitz function of p then p vanishes
identically.
So for such an equation of state ρ has a jump
discontinuity across the surface of the star and V and g
can only be C 1,1 across the surface of the star.

p is a Function of V
We write i p = −V −1 (ρ + p) iV in the form
dp
= −V −1 (ρ + p)
dV


dp
= −V −1 (ρ + p)
dV
Thus the boundary of the star is a level set of V
henceforth dentoted by V = VS .


dp
= −V −1 (ρ + p)
dV
Since we assumed ρ to be a piecewise C 1 function of p
for p > 0,


dp
= −V −1 (ρ + p)
dV
Since we assumed ρ to be a piecewise C 1 function of p
for p > 0,
ρ = ρ(V ) is also a piecewise C 1 function of V on [Vc , VS ].
Vc is the minimum value of V on Σ.


Asymptotic Conditions
Outside a compact set Σ is diffeomorphic to a R3 minus
a ball, and w.r.t. to the standard coordinate system in
R3 we have
M
V =1− + O(r−2 )
r
2M
gab = 1+ δab + O(r−2 )
r


R3 we have
M
V =1− + O(r−2 )
r
2M
gab = 1+ δab + O(r−2 )
r
δab is the Euclidean metric, and r is the spherical
coordinate associated with δab .


R3 we have
M
V =1− + O(r−2 )
r
2M
gab = 1+ δab + O(r−2 )
r
The constant M is the mass of the 3-metric g.


R3 we have
M
V =1− + O(r−2 )
r
2M
gab = 1+ δab + O(r−2 )
r
The constant M is the mass of the 3-metric g.
Above decay conditions follow from weaker conditions
on V and g. Physical motivation: ﬁnite energy static star.


Impossibilty of γ ≤ 6/5 Everywhere
1
Theorem [Lindblom, Masood-ul-Alam] Assume is
γ(p)
bounded as p → 0+ . Then γ must satisfy the inequality
2
6 p 6
γ> 1+ ≥
5 ρ 5

at some point inside a ﬁnite star.


Impossibilty of γ ≤ 6/5 Everywhere
1
Theorem [Lindblom, Masood-ul-Alam] Assume is
γ(p)
bounded as p → 0+ . Then γ must satisfy the inequality
2
6 p 6
γ> 1+ ≥
5 ρ 5

at some point inside a ﬁnite star.
Limits on the adiabatic index in static stellar models, in
B.L. Hu and T. Jacobson ed. Directions in General Relativity,
Cambridge University Press (1993).


1/γ(p) is bounded
1
is bounded if we exclude indeﬁnite oscillations of γ
γ(p)
near p = 0. We also assume that ρ = ρ(p) is a
non-decreasing C 1 function in some 0 < p < .


1/γ(p) is bounded
1
γ(p)
Theorem [Lindblom, Masood-ul-Alam] γ(p) > 1 at some
point in every open interval (0, ).


1/γ(p) is bounded
1
γ(p)
Theorem [Lindblom, Masood-ul-Alam] γ(p) > 1 at some
point in every open interval (0, ).
In particular if the equation of state is nice enough such
1
that lim inf γ(p) = lim sup γ(p) then is bounded on
p↓0 +
p↓0+ γ(p)
(0, pc ].


History
2
6 p
The "γ > 1+ somewhere theorem" was
5 ρ
motivated by Beig and Simon’s observation that the
2
6 p
equality everywhere γ = 1+ gives
5 ρ
1 1/5
p(ρ) = (ρ0 − ρ1/5 ) and leads to the uniqueness of
6
Buchdahl’s solutions. In these solutions and in
corresponding Newtonian theory of polytropic fluids
ρ = Apn/n+1 , n = constant ≥ 5 the fluid extends to
infinity.


History
2
6 p
The "γ > 1+ somewhere theorem" was
5 ρ
motivated by Beig and Simon’s observation that the
2
6 p
equality everywhere γ = 1+ gives
5 ρ
1 1/5
p(ρ) = (ρ0 − ρ1/5 ) and leads to the uniqueness of
6
Buchdahl’s solutions. In these solutions and in
corresponding Newtonian theory of polytropic fluids
ρ = Apn/n+1 , n = constant ≥ 5 the fluid extends to
infinity.
Beig, R., Simon, W.: On the Spherical Symmetry of
Static Perfect Fluids in General Relativity, Lett. Math.
Phys. 21: 245-250, 1991

History cont.
Beig and Simon was in turn inspired by my use of the
2
6 p
condition γ ≥ 1+ for deriving spherical
5 ρ
symmetry of static stars using the positive mass
theorem.


History cont.
2
6 p
5 ρ
theorem.
Masood-ul-Alam: A proof of the uniqueness of static
stellar models with small dρ/dp, Classical Quantum
Gravity 5, 409-491, 1988


History cont.
2
6 p
5 ρ
theorem.
Masood-ul-Alam: A proof of the uniqueness of static
stellar models with small dρ/dp, Classical Quantum
Gravity 5, 409-491, 1988
However the condition was known earlier. It occurred in
a series expansion near the center of the star in the
study of stellar stability.


History cont.
In relation to the positive mass theorem the condition
arose when we demanded


History cont.
d2 ψ
2
≥0
dV


History cont.
d2 ψ
2
≥0
dV
so that

˜ d2 ψ
Scalar(ψ 4 g) = 8ψ −5 (W − W ) 2 ≥ 0
dV


History cont.
d2 ψ
2
≥0
dV
so that

˜ d2 ψ
Scalar(ψ 4 g) = 8ψ −5 (W − W ) 2 ≥ 0
dV
˜
if W − W ≥ 0.


History cont.
d2 ψ
2
≥0
dV
so that

˜ d2 ψ
Scalar(ψ 4 g) = 8ψ −5 (W − W ) 2 ≥ 0
dV
˜
if W − W ≥ 0.
˜ ˜
Here W = | V |2 and W = W (V ) is related to the
conformal function (more later).


2
6
Proof of γ>
5
1+
p
ρ
somewhere Thm
Consider the conformal metric ψ 4 g where

 1 VS
 (1 + V )exp − VS
 ρ(v)dv
S
ψ(V ) = 2 1 + VS V v(ρ(v) + 3p(v))
 1

 (1 + V ) in vacuum
2


2
6
Proof of γ>
5
1+
p
ρ
somewhere Thm

 1 VS
 (1 + V )exp − VS
 ρ(v)dv
S
ψ(V ) = 2 1 + VS V v(ρ(v) + 3p(v))
 1

2

Scalar curvature R of Ψ4 g vanishes in the vacuum.


2
6
Proof of γ>
5
1+
p
ρ
somewhere Thm

 1 VS
 (1 + V )exp − VS
 ρ(v)dv
S
ψ(V ) = 2 1 + VS V v(ρ(v) + 3p(v))
 1

2

Scalar curvature R of Ψ4 g vanishes in the vacuum.
Inside the star


4
Proof cont.: Scalar curvature of Ψ g
The scalar curvature R of Ψ4 g satisﬁes


4
Ψ4 (1 + VS )R = 16πρ(1 − VS )+
2
8ρ2 V S| V |2 p 2 + 3VS
3 1+ −γ
γV 2 (ρ + 3p)2 ρ 1 + VS
which is non-negative whenever


4
Ψ4 (1 + VS )R = 16πρ(1 − VS )+
2
8ρ2 V S| V |2 p 2 + 3VS
3 1+ −γ
γV 2 (ρ + 3p)2 ρ 1 + VS

2
p 1 + VS
γ ≤3 1+
ρ 2 + 3VS


4
Ψ4 (1 + VS )R = 16πρ(1 − VS )+
2
8ρ2 V S| V |2 p 2 + 3VS
3 1+ −γ
γV 2 (ρ + 3p)2 ρ 1 + VS

2
p 1 + VS
γ ≤3 1+
ρ 2 + 3VS
2
1 + VS 2 6 p
For VS < 1, > . So R ≥ 0 if γ ≤ 1+
2 + 3VS 5 5 ρ


Proof cont.
2
6 p
The mass of Ψ4 gis zero. Thus if γ ≤ 1+
5 ρ
everywhere inside the star then Ψ4 g is Euclidean by the
positive mass theorem:


Proof cont.
2
6 p
The mass of Ψ4 gis zero. Thus if γ ≤ 1+
5 ρ
everywhere inside the star then Ψ4 g is Euclidean by the
positive mass theorem:
Theorem [Schoen, Yau] Let (N, η) be a complete oriented
3-dimensional Riemannian manifold. Suppose (N, η) is
asymptotically ﬂat and has non-negative scalar
curvature. Then the mass of (N.η) is non-negative and
if the mass is zero then (N, η) is isometric to R3 with the
standard euclidean metric.


Investigating where γ is small
There has been one more paper related to the
2
6 p
"γ > 1+ somewhere theorem."
5 ρ


2
6 p
5 ρ
Stricter constraint on γ involving Vc and VS has been
derived using the positive mass theorem:


2
6 p
5 ρ
Shiromizu, S., Yoshino, H.: Positive Energy Theorem
Implies Constraints on Static Stellar Models,, Prog.
Theor. Phys. 116, No. 6 (2006) 1159-1164.


2
6 p
5 ρ
Shiromizu, S., Yoshino, H.: Positive Energy Theorem
Implies Constraints on Static Stellar Models,, Prog.
Theor. Phys. 116, No. 6 (2006) 1159-1164.
However so far nobody investigated the region where γ
is small from the viewpoint of positive mass theorem.


Conformal function
To investigate the extent of the region where γ is small
and the limits on it imposed by the positive mass
theorem we need detail analysis of spherically
symmetric equations,


Conformal function
in particular analysis of ψ such that ψ 4 g is Euclidean.


Conformal function
d ˜ ˜
Let W = | V |2 and W = W (V ). Suppose ψ satisﬁes

d2 ψ 2π dψ
= ρψ − 2V (ρ + 3p)
dV 2
W˜ dV


Conformal function
d ˜ ˜
Let W = | V |2 and W = W (V ). Suppose ψ satisﬁes

d2 ψ 2π dψ
= ρψ − 2V (ρ + 3p)
dV 2
W˜ dV

Then the scalar curvature R of ψ 4 g satisﬁes
˜ d2 ψ
R = 8ψ −5 (W − W ) 2
dV

Can assume spherical symmetry
˜
Then W = W (V ). Can take W = W to make R zero.


˜
d2 ψ
Knowledge of the sign of 2
at V = k is crucial.
dV


˜
d2 ψ
Knowledge of the sign of 2
at V = k is crucial.
dV
d2 ψ
Lemma At V < VS where 2
= 0 and ψ(V ) is thrice
dV
differentiable, we have
d3 ψ 5πρ2 ψ 6 p 2
= [γ − (1 + ) ]
dV 3
γW˜ V (ρ + 3p) 5 ρ


Notes regarding the Lemma
d2 ψ dψ
When 2
= 0, 2V (ρ + 3p) = ρψ
dV dV


d2 ψ dψ
When 2
= 0, 2V (ρ + 3p) = ρψ
dV dV
6 p 2 d2 ψ
If γ > (1 + ) near the center then 2
> 0 in a
5 ρ dV
deleted neighborhood of the center. This is because
d2 ψ
2
= 0 at the center of the star. Roughly and
dV
physically speaking "soft center" is not good for the
stability of the star.


d2 ψ dψ
When 2
= 0, 2V (ρ + 3p) = ρψ
dV dV
6 p 2 d2 ψ
If γ > (1 + ) near the center then 2
> 0 in a
5 ρ dV
deleted neighborhood of the center. This is because
d2 ψ
2
= 0 at the center of the star. Roughly and
dV
physically speaking "soft center" is not good for the
stability of the star.
d2 ψ
Near the surface of the star 2
> 0. This follows
dV
p
because → 0 as p ↓ 0.
ρ


Extent of the region where γ is small
Theorem [Masood-ul-Alam] There cannot be an interval
2
6 p
j < V < k inside a ﬁnite star where γ < 1+
5 ρ


2
6 p
5 ρ
d2 ψ
if 2
≥ 0 at V = k− .
dV


2
6 p
5 ρ
d2 ψ
if 2
≥ 0 at V = k− .
dV
d2 ψ
if 2
< 0 at V = k− and ... ? ...
dV


2
6 p
5 ρ
d2 ψ
if 2
≥ 0 at V = k− .
dV
d2 ψ
if 2
< 0 at V = k− and ... ? ...
dV
k− allows ρ to be discontinuous at k.


Proof
Choose a conformal function u(V ) on [j, k] that matches
at V = k in a C 1,1 fashion and gives positive scalar
curvature on [j, k] as before. Then if


Proof

d ln u
0 < 2j < 1,
dV V =j+

we can construct portion of a new star solving the
spherically symmetric equations with appopriate initial
˜
conditions at V = j. The new pair (W , ψ) makes the
metric ψ 4 g scalar ﬂat in this portion.


Proof

d ln u
0 < 2j < 1,
dV V =j+

we can construct portion of a new star solving the
spherically symmetric equations with appopriate initial
˜
conditions at V = j. The new pair (W , ψ) makes the
metric ψ 4 g scalar ﬂat in this portion.
d2 ψ
If 2
≥ 0 at V = k− , the above condition is satisﬁed.
dV


Proof cont.: Conformal functions on [j, k]

k ρ(s)ds
−α
u(V ) = ψ(k)e V 2s(ρ(s) + 3p(s))



k ρ(s)ds
−α
u(V ) = ψ(k)e V 2s(ρ(s) + 3p(s))

3p(k) d ln ψ
with constant α = 2k 1 +
ρ(k− ) dV
V =k



k ρ(s)ds
−α
u(V ) = ψ(k)e V 2s(ρ(s) + 3p(s))

3p(k) d ln ψ
ρ(k− ) dV
V =k
u satisﬁes
d ln u αρ
2V =
dV ρ + 3p



k ρ(s)ds
−α
u(V ) = ψ(k)e V 2s(ρ(s) + 3p(s))

3p(k) d ln ψ
ρ(k− ) dV
V =k
u satisﬁes
d ln u αρ
2V =
dV ρ + 3p
u matches with ψ at V = k in a C 1,1 fashion.



k ρ(s)ds
−α
u(V ) = ψ(k)e V 2s(ρ(s) + 3p(s))

3p(k) d ln ψ
ρ(k− ) dV
V =k
u satisﬁes
d ln u αρ
2V =
dV ρ + 3p
u matches with ψ at V = k in a C 1,1 fashion.
d ln u
α ≤ 1 =⇒ 0 < 2j <1
dV V =j+


Proof cont.:Conformal functions on [j, k]

d2 ψ 2π dψ
Recall, = ρψ − 2V (ρ + 3p)
dV 2
W˜ dV
2πρψ 3p d ln ψ
= 1 − 2V 1+
W˜ ρ dV



d2 ψ 2π dψ
dV 2
W˜ dV
2πρψ 3p d ln ψ
= 1 − 2V 1+
W˜ ρ dV

d2 ψ 2π ψρ−
= (1 − α)
dV 2 W˜
V =k−



d2 ψ 2π dψ
dV 2
W˜ dV
2πρψ 3p d ln ψ
= 1 − 2V 1+
W˜ ρ dV

d2 ψ 2π ψρ−
= (1 − α)
dV 2 W˜
V =k−

d2 ψ
Thus if ≥ 0, α ≤ 1 and hence
dV 2
V =k−



d2 ψ 2π dψ
dV 2
W˜ dV
2πρψ 3p d ln ψ
= 1 − 2V 1+
W˜ ρ dV

d2 ψ 2π ψρ−
= (1 − α)
dV 2 W˜
V =k−

d2 ψ
Thus if ≥ 0, α ≤ 1 and hence
dV 2
V =k−

d ln u
0< < 1 which allows us to continue inward.
dV V =j+


Proof cont.:Scalar curvature of u g4

u4 scalar(u4 g) =
2W α p 2
6(1 + ) − 5γ + γ(1 − α) + 16πρ(1 − α)
3p 2 ρ
γV 2 (1 + )
ρ


Proof cont.
A detailed analysis of the spherically symmetric
equations are necessary to ensure
Proof goes through even if for new star W becomes
zero early?
ψ matches at V = j in C 1,1 fashion.
ψ > 0.


Proof cont.
A detailed analysis of the spherically symmetric
equations are necessary to ensure
Proof goes through even if for new star W becomes
zero early?
ψ matches at V = j in C 1,1 fashion.
ψ > 0.
On an interval [a, b] we shall denote by r = r(V ),
2
m = m(V ), W ˜ = W (V ) ≡ 1 − 2m
˜ dr
, and
r dV
ψ = ψ(V ) to be the solutions of the following equations
with initial values of the functions speciﬁed at j :


Proof cont. Spherically Symmetric Equations

dr r(r − 2m)
= r-equation
dV V (m + 4πr3 p)



dr r(r − 2m)
= r-equation
dV V (m + 4πr3 p)

dm 4πr3 (r − 2m)ρ
= m-equation
dV V (m + 4πr3 p)



dr r(r − 2m)
= r-equation
dV V (m + 4πr3 p)

= m-equation
dV V (m + 4πr3 p)

˜ V 2 (m + 4πr3 p)2 ˜
W (V ) = W -equation
r3 (r − 2m)



dr r(r − 2m)
= r-equation
dV V (m + 4πr3 p)

= m-equation
dV V (m + 4πr3 p)

˜ V 2 (m + 4πr3 p)2 ˜
W (V ) = W -equation
r3 (r − 2m)

dψ ψ 2m
= 1− 1− 1st order ψ -equation
dV 2r W˜ r



Einstein equations yield ﬁrst two equations. With the
help of i p = −V −1 (ρ + p) i V they can be written as
TOV (Tolman-Oppenheimer-Volkoff) eqns.



The solution ψ of the last equation makes ψ 4 g
Euclidean when g is spherically symmetric.



The solution ψ of the last equation makes ψ 4 g
Euclidean when g is spherically symmetric.
Differentiating this equation wherever possible and
using the other three equations we get the second
derivative of ψ(V ) we met before:

d2 ψ 2π dψ
= ρψ − 2V (ρ + 3p) 2nd order ψ -equation
dV 2
W˜ dV


Proof cont. A new system of ODEs

Theorem [Lindblom,Masood-ul-Alam] ˜
If r(V ), m(V ), W (V )
˜
is a solution of the following system, r > 2m > 0, W > 0,
and initially the values of the three functions are related
˜ ˜
by W -equation, then W -equation continues to hold
subsequently .

1/2
dr 2m ˜ −1/2
= 1− W
dV r
1/2
dm 2m ˜
2
= 4πρr 1 − W −1/2
dV r
dW˜
= 8πV (ρ + p) − 4V mr−3
dV


Proof cont. Why/when two systems are equivalent

The old system yields the new system. The new system
satisﬁes
1/2
d 2m ˜
1− W 1/2
− (m + 4πr3 p)V r−2 = 0
dV r



satisﬁes
1/2
d 2m ˜
1− W 1/2
− (m + 4πr3 p)V r−2 = 0
dV r

˜
Thus if W -equation holds at V = j, then it continues to
hold for V < j until Vc or a new center before Vc is
reached.



satisﬁes
1/2
d 2m ˜
1− W 1/2
− (m + 4πr3 p)V r−2 = 0
dV r

˜
reached.
Only two possible types of new center can occur. For
the center and the precise meaning of subsequently :



satisﬁes
1/2
d 2m ˜
1− W 1/2
− (m + 4πr3 p)V r−2 = 0
dV r

˜
reached.
Only two possible types of new center can occur. For
the center and the precise meaning of subsequently :
Lindblom, L., Masood-ul-Alam, A.K.M.: On the
Spherical Symmetry of Static Stellar Models Commun.
Math. Phys. 162, 123-145 (1994)

Proof cont.
Integrate the system for V ≤ j with initial conditions at
V =j
˜ j 2 (m + 4πr 3 p(j))2
W (j) =
r 3 (r − 2m)
j 2m d ln ψ
1− 1− = j
2r ˜
W r dV
V =j +


Proof cont.
Integrate the system for V ≤ j with initial conditions at
V =j
˜ j 2 (m + 4πr 3 p(j))2
W (j) =
r 3 (r − 2m)
j 2m d ln ψ
1− 1− = j
2r ˜
W r dV
V =j +

d ln ψ
Provided 0 < 2 j <1 it is possible to ﬁnd
dV
V =j +
˜
positive numbers r(j), m(j), W (j) such that they satisfy
the above initial conditions. For details:


Proof cont.
Masood-ul-Alam, A.K.M.: Proof that static stellar
models are spherical Gen Relativ Gravit 39,55-85
(2007)


Proof cont.
Masood-ul-Alam, A.K.M.: Proof that static stellar
models are spherical Gen Relativ Gravit 39,55-85
(2007)
Yaohua Wang (a student of Prof. X. Zhang) exactly
solved the algebraic system of initial conditions.


Neutrino Connection?
Can ψ be related to neutrino?


(A stupid analogy) Think the conformal function as a
globally compensating function in the sense that it
makes g curvature-free. Neutrino was ﬁrst introduced
"... as a desperate remedy to save the principle of
energy conservation ...."


(A profound analogy?) ψ 4 = ||ξ||2 where ξ is a ﬁxed
spinor satisfying Dirac equation (actually neutrino
equation) relative to the nonnegative scalar curvature
metric g that produces the mass in the Witten-Bartnik
mass formula.


(A profound analogy?) ψ 4 = ||ξ||2 where ξ is a ﬁxed
spinor satisfying Dirac equation (actually neutrino
equation) relative to the nonnegative scalar curvature
metric g that produces the mass in the Witten-Bartnik
mass formula.
Inverse β−decay:

electron + proton → neutron + neutrino

Open Problems
Investigate situations at higher densities from
geometrical viewpoint. Pion condensation.
Accompanied neutrino (including massive νµ ) emission.


Open Problems
Forget adiabatic index. Consider when p/ρ is not
negligible. Ex. Pressure is 1037 dynes/cm2
corresponding to a density 10 16 gm/cm3 p ≈ 3.3.
c2 ρ


Open Problems
c2 ρ
2
6 6
Generalize the inequality γ ≥ 1+ to include
5 5
rotation:


Open Problems
c2 ρ
2
6 6
5 5
rotation:
layer by layer.


Open Problems
c2 ρ
2
6 6
5 5
rotation:
layer by layer.
angular velocity depending on radius.


Open Problems cont.
Relax staticity and isentropy conditions: Isentropic
equation of state p = p(ρ). entropy depending on radius.
In the zero entropy case thermal contributions to the
pressure and density are neglected. Examples: Cold
stars such as Neutron stars, White dwarfs. Constant
entropy per baryon independent of the radius of the star
(Supermassive stars).


Open Problems cont.
Relax staticity and isentropy conditions: Isentropic
equation of state p = p(ρ). entropy depending on radius.
In the zero entropy case thermal contributions to the
pressure and density are neglected. Examples: Cold
stars such as Neutron stars, White dwarfs. Constant
entropy per baryon independent of the radius of the star
(Supermassive stars).
For our simple situation one escape from no-go result
was that the star extends to inﬁnity. What are possible
escapes when rotation, local non-spherical symmetry
are important?


Wild Guessing: Gravitation Rules.
Gravitation and Thermodynamics are the foundation of
Physics.


Physics.
Quantization, principle of least/stationary action are
mathematical properties of differential equations.


Physics.
Space-time (Einstein equations but here we have no
"real sense" of time because of the staticity
assumption) determines the energy-momentum tensor.
Solving for the space-time metric given the
energy-momentum tensor is backward thinking.


Physics.
Space-time (Einstein equations but here we have no
"real sense" of time because of the staticity
assumption) determines the energy-momentum tensor.
Solving for the space-time metric given the
energy-momentum tensor is backward thinking.
Possible causes of the "patterns" we see in the
energy-momentum tensor: its tensorial nature, coupled
with the approximate local and asymptotic symmetry of
Minkowski space-time? Stability?


Non-staticity and time
Non-staticity and "real sense" of time.


Introduce time "minimally." Manage non-conformally ﬂat
Riemannian 3-metric with non-zero Cotton tensor and
avoid giving importance of extrinsic curvature of the
embedding in 3+1 dimension as far as possible.


Can this time be "maya" (illusion)? Real time is energy.
In what sense?


In what sense?
Is any one "more basic?" Which one: linear or angular
momentum? Rotation or translation?


In what sense?
Is any one "more basic?" Which one: linear or angular
momentum? Rotation or translation?
Dual space of 3-space.


Suggested Readings on Gravity
Hu, B. L.: General Relativity as
Geometro-Hydrodynamics, arXiv:gr-qc/9607070v1
(1996).
Barut, A. O., Cruz, M. G., Sobouti, Y.: Localized
solutions of the linearized gravitational ﬁeld equations in
free space, Class. Quantum Grav. II, (1994) 2537-2543.
Finster, F., Smoller, J., Yau, S.-T.: The coupling of
gravity to spin and electromagnetism, Mod. Phys. Lett.
A 14, (1999) 1053-1057.
Penrose, R.: Chapter 30 in The Road to Reality, (2004).


References
Lindblom, L., Masood-ul-Alam, A.: Limits on the adiabatic index in
static stellar models, in B.L. Hu and T. Jacobson ed. Directions in
General Relativity, Cambridge University Press (1993).

Beig, R., Simon, W.: On the spherical symmetry of static perfect
ﬂuids in general relativity, Lett. Math. Phys. 21, 245-250 (1991).
Shiromizu, S., Yoshino, H.: Positive Energy Theorem Implies
Constraints on Static Stellar Models,, Prog. Theor. Phys. 116, No. 6
(2006) 1159-1164.
Shiromizu, T., Yamada, S., Yoshino, H.: On existence of matter
outside a static black hole, Journal of Mathematical Physics 47,
112502(1-8) (2006).


References
Carrasco, A., Mars, M., Simon, W.: On perfect ﬂuids and black holes
in static equilibrium, Journal of Physics: Conference Series 66,
012012(1-6) (2007).
Beig, R., Simon, W.: On the uniqueness of static perfect ﬂuid
solutions in general relativity, Commun. Math. Phys. 144, 373-390
(1992).


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