Nonstandard general relativity III: The Kerr class

0.09 MB·

by  Günter Scharf

Nonstandard general relativity III: The Kerr class
3 Gu¨nter Scharf ∗
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0 Institut fu¨r Theoretische Physik,
2 Universit¨at Zu¨rich,

n Winterthurerstr. 190 , CH-8057 Zu¨rich, Switzerland
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1 Abstract
v
0 WerevisittheKerrmetricinBoyer-Lindquistcoordinatesandcon-
4 struct the corresponding class of nonstandard solutions of Einstein’s
1 equations. These solutions can be used to describe the outer part of
2
spiral galaxies without assuming dark matter.
.
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0
3
1
: PACS numbers: 04.20 Cv; 04.20 Jb
v
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∗e-mail: [email protected]
1
1 Introduction
Standard general relativity is the marriage between geometry and gravity.
Indeed, geometry is such a pretty woman that it is no surprise that almost
all relativists fall in love with her. I am old enough not to do so. Besides
divorces are quite common today. So we are in accordance with the spirit of
the age when we leave geometry single as a convention. The gravitational
field on the other hand is represented by the Christoffel symbols satisfying
Einstein’s equations. The Christoffel symbols come from a metric, but the
latter has no geometric meaning. In previous papers [1] [2] we have studied
staticsphericallysymmetricsolutionsofEinstein’sequationsonthisground.
In contrast to the standard theory we have found solutions with arbitrary
circular velocity dependence V(r) which are of interest in connection with
the dark matter problem.
Sincemostgalaxies arenotsphericallysymmetricwenowinvestigate the
axial symmetric situation. The corresponding standard vacuum solution is
the Kerr metric. We use Boyer-Lindquist coordinates (t,r,ϑ,φ) which are
identical with the spherical coordinates in the Schwarzschild case and have
a clear operational physical definition [1]. We assume the line element in
the form of Chandrasekhar [3], section 52
ds2 = e2νdt2 e2ψ(dφ ωdt)2 e2µ2dr2 e2µ3dϑ2 (1.1)
− − − −
with five metric functions ν,ψ,ω,µ ,µ which depend on r,ϑ, only. In the
2 3
nonstandard theory it is important to choose the metric as general as pos-
sible. For this reason we give preference to the ansatz (1.1) compared with
other ones in the literature which contain less than five functions.
Chandrasekhar gives the Riemann tensor for (1.1) but he does not give
the Ricci tensor R . Instead he gives suitable combinations of the vacuum
µν
field equations R = 0, which allow integration of the resulting differential
µν
equations. We fully agree with all results of Chandrasekhar which finally
lead to the Kerr metric. However, in his combination of the field equations
some information gets lost; in particular the second derivatives of the func-
tions µ and µ are eliminated. Then it must be checked whether the Kerr
2 3
metric really satisfies all field equations. Such a check is done in sect.4. Our
results for the Ricci tensor in sect.3 are also needed for future studies of the
problemwithmatter. Insect.5weconstructaclassofvacuumsolutionswith
arbitrary angular velocity dependence Ω(r) in the equatorial plane. These
solutions can be used to describe the outer part of spiral galaxies where
normal matter can be neglected.
2
2 Christoffel symbols
Chandrasekhar has used differential forms to calculate the Riemann tensor.
We find the route via Christoffel symbols most simple, and we need them
anyway for the geodesic equations. According to (1.1) the non-vanishing
components of the metric tensor are
g = e2ν ω2e2ψ, g = e2ψ, g = ωe2ψ
tt φφ tφ
− −
g = e2µ2, g = e2µ3. (2.1)
rr ϑϑ
− −
Then the inverse metric is equal to
gtt = e−2ν, gφφ = e−2ψ +w2e−2ν, etφ = ωe−2ν
−
grr = e−2µ2, gϑϑ = e−2µ3. (2.2)
− −
The Christoffel symbols are given by
1
Γα = gαµ(g , +g , g , ), (2.3)
βγ 2 βµ γ γµ β− βγ µ
thecommasalwaysmeanpartialderivatives. Thenon-vanishingcomponents
with upper index t are
ω
Γt = ν, ω, e2(ψ−ν)
rt r− r 2
ω
Γt =ν, ω, e2(ψ−ν)
ϑt ϑ− ϑ 2
Γt = 1e2(ψ−ν)ω, (2.3)
rφ 2 r
Γt = 1e2(ψ−ν)ω, .
ϑφ 2 ϑ
With upper index r we have
Γr = 1e−2µ2(e2ν ω2e2ψ),
tt 2 − r
Γr = 1e−2µ2(ωe2ψ),
φt 2 r
Γr = µ , , Γr = µ ,
rr 2 r rϑ 2 ϑ
Γr = 1e−2µ2(e2µ3), (2.4)
ϑϑ −2 r
Γr = 1e−2µ2(e2ψ), .
φφ −2 r
3
With upper index ϑ we similarly get
Γϑ = 1e−2µ3(e2ν ω2e2ψ),
tt 2 − ϑ
Γϑ = 1e−2µ3(ωe2ψ),
φt 2 ϑ
Γϑ = µ , , Γϑ = µ ,
rϑ 3 r ϑϑ 3 ϑ
Γϑ = 1e−2µ3(e2µ2), (2.5)
rr −2 ϑ
Γϑ = 1e−2µ3(e2ψ), .
φφ −2 ϑ
Finally with upper index φ we have
ω,
Γφ = ω(ν ψ), r(1+ω2e2(ψ−ν))
rt − r− 2
ω,
Γφ = ω(ν ψ), ϑ(1+ω2e2(ψ−ν))
ϑt − ϑ− 2
ω
Γφ = ψ, + ω, e2(ψ−ν) (2.6)
φr r 2 r
ω
Γφ = ψ, + ω, e2(ψ−ν).
φϑ ϑ 2 ϑ
As a consequence of axial symmetry we note the symmetry under simulta-
neous exchange of r ϑ and µ µ .
2 3
↔ ↔
3 Ricci tensor
Next we calculate the Ricci tensor according to
R = ∂ Γα ∂ Γα +Γα Γβ Γα Γβ . (3.1)
µν α µν − ν µα αβ µν − νβ µα
Weperformthiscalculationinasuitableordertoreconstructtheelimination
of second derivatives done by Chandrasekhar.
R = ∂ Γr ∂ (Γt +Γr )+Γr (Γt +Γr +Γφ )
rϑ r rϑ− ϑ rt rr rϑ tr rr rφ
+Γϑ (Γt +Γr +Γφ ) Γt Γt Γt Γφ Γr Γr Γr Γϑ Γφ Γt Γφ Γφ
rϑ tϑ rϑ φϑ − tϑ rt− ϑφ rt− rϑ rr− ϑϑ rr− tϑ rφ− ϑφ rφ
= (ψ+ν), , +µ , (ψ+ν), +µ , (ψ+ν),
r ϑ 2 ϑ r 3 ϑ ϑ
− −
1
ψ, ψ, ν, ν, + ω, ω, e2(ψ−ν). (3.2)
r ϑ r ϑ r ϑ
− − 2
This agrees with equation (8) (p.274) of Chandrasekhar.
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R = ∂ Γr +∂ Γϑ +Γr (Γt +Γr +Γϑ )
φφ r φφ ϑ φφ φφ tr rr rϑ
+Γϑ (Γt +Γr +Γϑ ) 2Γt Γr 2Γϑ Γt
φφ tϑ rϑ ϑϑ − rφ tφ− tφ φϑ
1
= e2(ψ−µ2)[ψ, , +ψ, (ψ+ν µ +µ ), ] ω,2e2(2ψ−ν−µ2)
− r r r − 2 3 r − 2 r
+[r ϑ,µ µ ]. (3.3)
2 3
↔ ↔
This agrees with equation (6) of Chandrasekhar. The last square bracket
denotes the corresponding ϑ-terms which are present because of axial sym-
metry. If we set R = 0 we get
φφ
ω,2
ψ, , +ψ, , = ψ, (ψ+ν µ +µ ), re2(ψ−ν)
r r ϑ ϑ r 2 3 r
− − − 2
+[r ϑ,µ µ ] (3.4)
2 3
↔ ↔
R = ∂ Γr +∂ Γϑ +Γr (Γr +Γϑ )+Γϑ (Γr +Γϑ )
tφ r tφ ϑ tφ tφ rr ϑr tΦ rϑ ϑϑ
ΓrΓt ΓϑΓt ΓφΓr ΓΦΓϑ .
− tt rφ− tt φϑ− tr φφ− tϑ φφ
1
= e2(ψ−µ2)[ω, , +2ωψ, , +ω, (3ψ ν µ +µ ), +
r r r r r 2 3 r
2 − −
ω
+2ωψ, (ψ+ν µ +µ ), ]+ ω,2e2(2ψ−ν−µ3)
r − 2 3 r 2 r
+[r ϑ,µ µ ]. (3.5)
2 3
↔ ↔
Substituting (3.4) inhere we obtain equation (7) of Chandrasekhar:
e−2µ2ω, , +e−2µ3ω, , = e−2µ2ω, (3ψ ν µ +µ ),
r r ϑ ϑ r 2 3 r
− − −
+[r ϑ,µ µ ] (3.6)
2 3
↔ ↔
Next we calculate
R = ∂ Γr +∂ Γϑ +Γr(Γr +Γϑ +Γφ )
tt r tt ϑ tt tt rr rϑ rφ
+Γϑ(Γr +Γϑ +Γφ ) Γt Γr 2Γr Γφ ΓϑΓt 2Γϑ Γφ
tt rϑ ϑϑ φϑ − tr tt − tφ tr − tt tϑ− tφ tϑ
= e−2µ2 (ν, , +2ν,2)e2ν e2ψ(ωω, , +ω,2+4ωω, ψ, +ω2ψ, , +2ω2ψ,2)
h r r r − r r r r r r r r i
ω,
+e2(ψ−µ2)(ω, +2ωψ, ) r(1+ω2e2(ψ−ν))+ω(ψ ν),
r r h 2 − ri
5
+e−2µ2 ν, e2ν ωe2ψ(ω, +ωψ, ) (ψ ν +µ µ ), +ωω, e2(ψ−ν)
h r r r ih 3 2 r r i
− − −
+[r ϑ,µ µ ] (3.7)
2 3
↔ ↔
Using (3.4) and (3.6) we can eliminate the second derivatives of ω and ψ.
Then Einstein’s equation R = 0 implies
tt
ω,2
0= e−2µ2 ν, , +ν, (ψ+ν +µ µ ), re2(ψ−ν−µ2)
h r r r 3− 2 ri− 2
+[r ϑ,µ µ ]. (3.8)
2 3
↔ ↔
This is equation (5) of Chandrasekhar.
Theremainingnon-vanishingcomponents areR andR andtheseare
rr ϑϑ
most subtle because they contain second derivatives of four functions. We
have
R = ∂ Γϑ ∂ (Γt +Γϑ +Γφ )+Γr (Γt +Γϑ +Γφ )
rr ϑ rr − r rt rϑ rφ rr tr rϑ rφ
+Γϑ (Γt +Γϑ +Γφ ) (Γt )2 2Γt Γφ
rr tϑ ϑϑ φϑ − rt − rφ tr
Γr Γϑ (Γϑ )2 (Γφ )2
− rϑ rr − rϑ − rφ
= e2(µ2−µ3) µ , , +2µ , (µ µ ), µ , , ∂2(ψ+ν)
− h 2 ϑ ϑ 2 ϑ 2− 3 ϑi− 3 r r− r
µ2, ψ,2 ν,2+µ , (ψ+ν +µ ),
− 3 r− r− r 2 r 3 r−
1
e2(µ2−µ3)µ , (ψ+ν +µ µ ), + ω,2e2(ψ−ν). (3.9)
− 2 ϑ 3− 2 ϑ 2 r
R is obtained from R by the substitutions r ϑ µ µ as before.
ϑϑ rr 2 3
↔ ↔
Multiplying R by exp[2(µ µ )] and subtracting it from (3.9) the second
ϑϑ 2 3
−
derivatives of µ and µ drop out. Using Einstein’s equations we arrive at
2 3
the following simpler equation
ω,2
e−2µ3 ∂2(ψ+ν)+ψ,2 +ν,2 (ψ+ν), (µ +µ ), ϑe2(ψ−ν−µ3) =
h ϑ ϑ ϑ− ϑ 2 3 ϑi− 2
ω,2
= e−2µ2 ∂2(ψ+ν)+ψ,2+ν,2 (ψ+ν), (µ +µ ), + re2(ψ−ν−µ2). (3.10)
h r r r− r 3 2 ϑi 2
On the other hand adding (3.3) and (3.8) we get
e−2µ3∂2(ψ+ν)+e−2µ3(ψ+ν), (ψ+ν +µ µ ), =
ϑ ϑ 2− 3 ϑ
= e−2µ2∂2(ψ+ν) e−2µ2(ψ+ν), (ψ+ν +µ µ ), . (3.11).
− r − r 3− 2 r
6
Combining this with (3.10) we find Chandrasekhar’s last two equations (9)
and(10). Butitis clear fromthederivation of(3.10) thatthesix differential
equations of Chandrasekhar contain less information than the six Einstein’s
equations. Therefore we must check at the end that all Einstein’s equations
are satisfied, in particular R = 0.
rr
4 Calculation of R
rr
ThesixdifferentialequationsofChandrasekharcanbeintegratedandfinally
lead to the Kerr metric [1]. Let M and a be mass and angular momentum
of the Kerr black hole, then the five metric functions in (1.1) are given by
̺2 = r2+a2cos2ϑ, ∆ = r2 2Mr+a2 (4.1)
−
Σ2 = (r2+a2)2 a2∆sin2ϑ (4.2)
−
2aMr
ω = (4.3)
Σ2
Σ2
e2ψ = sin2ϑ (4.4)
̺2
̺2∆
e2ν = (4.5)
Σ2
̺2
e2µ2 = , e2µ3 = ̺2. (4.6)
∆
It is straightforward butquite cumbersometo calculate R (3.9) with these
rr
results. The first derivatives of the 5 functions have been given by Chan-
drasekhar on p.290. We need the following second derivatives
1 (r M)2
∂2(ψ+ν) = 2 −
r ∆ − ∆2
a2 a4
∂2µ = cos2ϑ sin22ϑ (4.7)
ϑ 2 −̺2 − 2̺4
1
∂2µ = (a2cos2ϑ r2).
r 3 ̺4 −
Substituting everything into (3.9) we obtain
1 a4 1 1 (r M)2
e2µ3R = sin22ϑ+a2cos2ϑ ̺2 (a2cos2ϑ r2)+ 2 − +
rr ∆(cid:16)2̺2 (cid:17)− h̺4 − ∆− ∆2
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r2 r r M r M r
+ +ν2, +ψ2, − − + +
̺4 r r−(cid:16)̺2 − ∆ (cid:17)(cid:16) ∆ ̺2(cid:17)i
a2 2a2M2
+ sin2ϑcotϑ+ [(r2+a2)(3r2 a2) a2(r2 a2)sin2ϑ]2
2∆ Σ8 − − − ×
sin2ϑ Σ2
sin2ϑ r2+a2+2a2Mr . (4.8)
× (cid:16) ̺2 (cid:17)∆
We use
ψ2, +ν2, = (ψ, +ν, )2 2ψ, ν,
r r r r r r
−
and
̺2
2̺2ψ, ν, = 2 [2r(r2+a2) a2(r M)sin2ϑ] r
r r nΣ2 − − − o×
1 r r M
[2r(r2+a2) a2(r M)sin2ϑ]+ + − .
×n−Σ2 − − ̺2 ∆ o
Then we collect all terms on the common denominator Σ6∆2̺2:
e2µ3R Σ6∆2̺2 = a2∆Σ6(1a2sin22ϑ+̺2cos2ϑ)+Σ6∆2(r2 a2cos2ϑ)+
rr 2 −
+̺4Σ6(r2 2Mr+2M2 a2) 2Σ6̺4(r M)2+
− − − −
+2 Σ2∆̺2[2r(r2+a2) a2(r M)sin2ϑ] rΣ4∆
{ − − − }×
∆̺2[2r(r2+a2) a2(r M)sin2ϑ]+rΣ2∆+(r M)Σ2̺2 +
×{− − − − }
+a2Σ6∆̺2cos2ϑ+2a2M2∆sin2ϑ[(r2+a2)(3r2 a2) a2(r2 a2)sin2ϑ]2
− − − ×
[̺2(r2+a2)+2a2Mrsin2ϑ]. (4.9)
×
To minimize the probability of a computational error we have computed
the polynomial on the r.h.s. of (4.9) by means of the algebraic computer
program FORM [4]. There is a huge cancellation and the r.h.s. is indeed
zero. Then by symmetry R vanishes as well. Hence the Kerr metric
ϑϑ
satisfies all Einstein’s equations. This closes a loop-hole in the treatment by
Chandrasekhar.
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5 The Schwarzschild and Kerr classes
Chandrasekhar in his solution of the differential equations makes some sim-
plifications of theintegration procedurebychoosing asuitablegauge. Inthe
nonstandard theory we want a physical gauge fixing, for example by giving
the rotation curve V(r) in the equatorial plane. To integrate the field equa-
tions with such a constraint is a formidable task. Fortunately this can be
completely avoided by usinggauge (or diffeomorphism)invariance. In[1] we
have found in the Schwarschild case that the solutions found by integrating
the field equations can also be obtained by a suitable gauge transformation
from the Schwarzschild solution. For better understanding of the following
we repeat the argument here. We start from the Schwarzschild metric
ds2 = 1 rs dt¯2 1 rs −1dr¯2 r¯2(dϑ2+sin2ϑdφ2) (5.1)
(cid:16) − r¯(cid:17) −(cid:16) − r¯(cid:17) −
with a fictitious unphysical radial coordinate r¯. We set our physical, given
circular velocity squaredu(r) = V2(r)equaltotheSchwarzschild expression
r
s
u(r) = . (5.2)
2(r¯ r )
s
−
where r = 2M is the Schwarschild radius. Then solving for r¯ gives the
s
desired diffeomorphism
2u(r)+1
r¯= r . (5.3)
s
2u(r)
Indeed we get
r 1
s
g = 1 =
tt
− r¯ 2u+1
2u+1 2
g = r¯2 = r2
ϑϑ 2(cid:16) 2u (cid:17)
rs −1 dr¯ 2 rs 2 u′ 2
g = 1 = (2u+1) (5.4)
rr (cid:16) − r¯(cid:17) (cid:16)dr(cid:17) (cid:16) 2 (cid:17) (cid:16)u2(cid:17)
which is the class of nonstandard solutions found in [1].
Instead of using the circular velocity we can work with the angular ve-
locity
dφ √M
Ω¯(r¯)= = . (5.5)
dt r¯3/2
Then the corresponding diffeomorphism is
M 1/3
r¯= (5.6)
(cid:16)Ω2(cid:17)
9
and the Schwarzschild class gets the following form
g = 1 (8M2Ω2)1/3
tt
−
M 2/3
g =
ϑϑ (cid:16)Ω2(cid:17)
g = 4[1 (8M2Ω2)1/3]−1 M 2/3(Ω′)2. (5.6)
rr 9 − (cid:16)Ω5(cid:17)
Here Ω(r) is the prescribed angular velocity.
In the Kerr case the angular velocity in the equatorial plane is equal to
([3], p.336)
√M
Ω¯ = ± . (5.7)
r¯3/2 a√M
±
This depends on r¯ only, so that with this observable the Kerr class be-
comes as simple as the Schwarzschild class. Solving for r¯gives the required
diffeomorphism
1 2/3
r¯= M1/3 a (5.8)
(cid:16)Ω(r) − (cid:17)
and
dr¯ 2 1 −1/3Ω′
= M1/3 a . (5.9)
dr −3 (cid:16)Ω − (cid:17) Ω2
Then the quantities (4.1) (4.2) giving the Kerr metric must be changed
according to
1 4/3
̺¯2 = M2/3 a +a2cos2ϑ (5.10)
(cid:16)Ω − (cid:17)
1 4/3 1 2/3
∆¯ = M2/3 a 2M4/3 a +a2 (5.11)
(cid:16) (cid:17) (cid:16) (cid:17)
Ω − − Ω −
1 4/3 2
Σ¯2 = M2/3 a +a2 a2∆¯ sin2ϑ. (5.12)
h (cid:16)Ω − (cid:17) i −
The quantities (4.3-6) must now be calculated with these new expressions
with bar and with r¯. Finally the modified axial symmetric metrics follow
from (2.1) where g gets the factor (5.9) squared. This gives the Kerr
rr
class. These metrics are physically different from the Kerr metric and have
a better chance to describe the outer part of spiral galaxies where matter
can be neglected.
10