black hole and neutron stars
TRANSCRIPT
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lack Holes and eutron Stars
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IntroductionThis is a course about black holes and neutron stars.My main goal is to give you facility in computing physicalprocesses in black hole and neutron star spacetimes. I will emphasize general relativistic processes that arerelevant to directly observable effects.I will deemphasizeobservations and the associated astrophysical
details,giving only enough of the former to motivate a calculation and of the latter to carry one out. I hope thatwe will build a toolkit useful for exploring the theoretical properties of compact objects and,perhaps forproviding a framework for interpreting the observations. This is not a course about the foundations of relativityor about many of the fundamental theoretical physics questions that have been impacted by carrying outgedanken experiments in a black hole context.I will eschew charged holes, Hawking radiation, singularities,string entropy, two dimensional black holes and so on. It is part of the current excitement of this field that wewill have plenty to occupy ourselves dealing with astrophysicsinspiredmatters. I will also deemphasizegravitational radiation as this has been covered extensively in several recent courses.
In order accomplish my goal without imposing an unreasonable calculational burden on you,I shall makeheavy use of Mathematica.I have chosen this program simply because I am most familiar with it.There are
alternative programs with which you may be familiar and which may be superior.The course will be structuredso that they can be substituted.What I want is for you to finish the course with a collection of routines that youunderstand and which you can use with confidence.The homeworks will often entail performing calculations inthis manner and producing suitable graphical output.
I will start by discussing the Schwarzschild spacetime dealing with familiar material and allowing you toestablish your own framework for performing calculations.I shall then go on to discuss the Kerrspacetime,exploring the geometry using alternative coordinate systems and navigating them with null geodesic-s.I will next generalize these to discuss particle orbits,the geometrical optics of plasma modes and gravitationallensing.In the fourth week,I plan to discuss thin accretion disks using relatively unsophisticated prescriptionsfor their angular momentum transport.I hope that we will be able to compute their optical appearance.Thick
accretion disks introduce us to relativistic fluid dynamics.I will show you how to compute their structure,againmaking simple assumptions about their internal thermodynamics etc.This will then allow us to consider twoadditional fluid problems,wave modes in disks and tidal effects on passing stars.In week 7, I will explain howMaxwells equations are solved in a Kerr spacetime under the forcefreeapproximation.This will allow us tosee how the rotational energy stored in the spacetime can be extracted and put to use.I will next discuss thestructure and appearance of spherical neutron stars and outline how to tackle the exterior spacetime of rapidlyspinning stars.The course will finish with a relatively unsophisticated discussion of the calculation of gravita-tional radiation by orbiting stars and compact objects.
This, at least, is my plan. In practice, we will see how we go and I am open to suggestions for other material.
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1. Spherical Holes
Schwarzschild Metric
Metric tensor
Use units in which GM = c = 1.ds2 = 1 2 r dt2 + 1 2 r 1 dr2 r 2 d 2 sin 2 d2 from which we read off the metric tensor
Clear coord, metric, inversemetric, affine, riemann, co2on, on2co ;coord t, r, , ; metric 1 2 r, 0, 0, 0 ,
0, 1 1 2 r , 0, 0 , 0, 0, r 2 , 0 , 0, 0, 0, r 2 Sin 2 ;Print MatrixForm metric ;
12r 0 0 0
0 11 2
r
0 0
0 0 r 2 0
0 0 0 r 2 Sin 2
Note that time has index 1.
Bases
We will mostly use a coordinate basis for calculationet t etc
wheree e = g= etc.This has a dual basiset t etcdefined bye e =
We transform between different bases usinge = e L etcwhere we use the summation convention and
L = x
x
for coordinate bases.We will mostly interpret our results in orthonormal bases, e satisfyinge e For the Schwarzschild metric,e
t 1 2 r 12 e t etc.
So, transforming from a Schwarzschild coordinate basis to a locally orthonormal basis involves L
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co2on 1 2 r12 , 0, 0, 0 ,
0, 1 2 r12 , 0, 0 , 0, 0, 1 r, 0 , 0, 0, 0, Csc r ;
Print MatrixForm co2on ;
1
1 2r
0 0 0
0 1 2r
0 0
0 0 1r
0
0 0 0 Csc r
and its inverse L .
on2co : on2co Simplify Inverse co2on ;Print MatrixForm on2co ;
2 rr
0 0 0
0 12 rr
0 0
0 0 r 0
0 0 0 r Sin
Vectors
Vectors live in the tangent space at a point and can be expanded in componentsA = A e = A e etc
which transform according to A = L A ; A = A L etcTensors are real functions of vectorsT(e ,e ) T etcand their components transform according to the natural generalization of these rules.
Differentiation
The directional derivative of a vector along e is
e A A A
, A e = A ; e A , A
e = A ; e
where A , e
A .The connection for a coordinate basis is
e
e e =12 g
g, g , g ,and we have used the inverse of the metric tensor. The derivative for a noncoordinatebasis contains extraterms coming from the nonzerocommutator.
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inversemetric : inversemetric Simplify Inverse metric ;affine : affine Simplify Table 1 2 Sum inversemetric i, s
D metric s, j , coord k D metric s, k , coord j D metric j, k , coord s ,s, 1, 4 ,i, 1, 4 , j, 1, 4 , k, 1, 4
;
Print MatrixForm affine
0
12 r r 2
0
0
12 r r 2
0
0
0
0
0
0
0
0
0
0
0
2 rr 3
0
0
0
0
12 r r 2
0
0
0
0
2 r
0
0
0
0
2 r Sin 2
0
0
0
0
0
0
1r
0
01
r
0
0
0
0
0
Cos Sin
0
0
0
0
0
0
0
1r
0
0
0
Cot
0
1r
Cot 0
We can check that the covariant derivative of the metric tensor vanishes, as it should.
delmetric : delmetricSimplify Table D metric i, j , coord k
Sum metric s, j affine s, i, k metric i, s affine s, j, k ,s, 1, 4 , i, 1, 4 , j, 1, 4 , k, 1, 4
Print MatrixForm delmetric ;
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
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Geodesic Motion
Stationarity of the proper time implies
ds L = ds g dx
dsdx
ds
12
=0where L is the Lagrangian. Performing the variation, we obtain the equation of motion of a freelyfallingparticle moving with velocity u
u u = 0where u = dx / d with u u = 1and so we get the geodesic equationd 2 x
d2
dx
ddx
d =0For photons is replace by an affine parameter and u u = 0
Killings Equation
If there is a symmetry, a direction along which the metric does not change, then it satisfies ; + ; = 0.Lagranges equations then imply that u is conserved along the geodesic. This gives us constants of themotion. For motion in the equatorial plane, t and symmetries imply E ut and L u are constant.
Freefall
As an example, consider the geodesic motion of a particle falling from rest at infinity. We have (12/r)dt/d E=1. We also know that
d2 = ds2 1 2 r dt2 1 2 r 1 dr2
from which we obtain ur dr/d = 2 r 1 2 . The full velocity is then given by ensuring that u u = 1.u = [ 1 2 r 1 , 2 r 1 2 , 0, 0]The speed is = 2 r 1 2 similar to the Newtonian expression and the associated Lorentz factor is = (12/ r) 1 2 .
Riemann and Ricci tensorsThe law of geodesic deviation for a connection vector n is
u u n +R[,u,n,u ]=0or in components,
u u n + R u n u =0
The components of the Riemann tensor R are defined in a coordinate basis by R = , , +
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riemann : riemannSimplify Table
D affine i, j, l , coord k D affine i, j, k , coord l Sum affine i, s, k affine s, j, l affine i, s, l affine s, j, k ,
s, 1, 4 , i, 1, 4 , j, 1, 4 , k, 1, 4 , l, 1, 4Print MatrixForm riemann ;
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 22 r r 2
0 0
22 r r 2
0 0 0
0 0 0 0
0 0 0 0
0 0 1r
0
0 0 0 0
1r
0 0 0
0 0 0 0
0 0 0
0 0 0
0 0 0
Sin 2
r0 0
0 2 2 rr 4
0 0
4 2 rr 4
0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 1r
0
0 1r
0 0
0 0 0 0
0 0 0
0 0 0
0 0 0
0 Sin 2
r0
0 0 2 rr 4
0
0 0 0 0
2 rr 4
0 0 0
0 0 0 0
0 0 0 0
0 0 12 r r 2 0
0 12 r r 2
0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0
0 0 0
0 0 0
0 0 2 Sin r
0 0 0 2 rr 4
0 0 0 0
0 0 0 0
2 rr 4
0 0 0
0 0 0 0
0 0 0 12 r r 2
0 0 0 0
0 12 r r 2
0 0
0 0 0 0
0 0 0 0
0 0 0 2r
0 0 2r
0
0 0
0 0
0 0
0 0
The Ricci tensor is defined by R = R
All elements vanish as they must.
ricci : ricciSimplify Table Sum riemann k, i, k, j , k, 1, 4 , i, 1, 4 , j, 1, 4
Print MatrixForm ricci ;
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
Riemann tensor in freefalling frame
First transform the Riemann tensor into an orthonormal basis, at rest in the freefalling frame R L
L L
L
R
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onriemann : onriemannTable Sum on2co i, m co2on j, n co2on k, o
co2on l, p riemann m, n, o, p , m, 1, 4 , n, 1, 4 ,o, 1, 4 , p, 1, 4 , i, 1, 4 , j, 1, 4 , k, 1, 4 , l, 1, 4
Print MatrixForm onriemann
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
02 1 2r
2 rr
2 r r 20 0
2 1 2r
2 rr
2 r r 20 0 0
0 0 0 0
0 0 0 0
0
0
2 rr
1 2r
r 3
0
0 2 2 r
1 2r
2 rr
r 40 0
4 2 r
1 2r
2 rr
r 40 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0
0 0
01 2
r
2 rr
0 0
0 0 2 r1 2
rr 4
0
0 0 0 0
2 r
1 2r
r 40 0 0
0 0 0 0
0 0 0 0
0 01 2
r
2 r r 20
01
2r
2 r r 20 0
0 0 0 0
0
0
0
0
0 0 0 2 r1 2
rr 4
0 0 0 0
0 0 0 02 r
1 2r
r 40 0 0
0 0 0 0
0 0 01 2
r
2 r r 2
0 0 0 0
01 2r
2 r r 20 0
0
0
0
0
Next, boost into the freefalling frame with infall speed by introducing the transformation matrices L and
its inverse L where the ~ denotes the freefall coordinates.
on2ff gammaff, gammaff betaff, 0, 0 ,gammaff betaff, gammaff, 0, 0 , 0, 0, 1, 0 , 0, 0, 0, 1 . betaff 2 r ^ 1 2 , gammaff 1 2 r ^ 1 2 ;
ff2on Simplify Inverse on2ff ;Print MatrixForm on2ff , MatrixForm ff2on ;
1
1 2r
2 1r
1 2r
0 0
2 1r
1 2r
1
1 2r
0 0
0 0 1 0
0 0 0 1
1
2 rr
2 1r
2 rr
0 0
2 1r
2 rr
1
2 rr
0 0
0 0 1 0
0 0 0 1
Finally, we boost from the rest orthonormal basis to the freefall orthonormal basis
R L
R
L
L
L
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ffriemann :ffriemann Simplify Table Sum ff2on i, m on2ff j, n on2ff k, o
on2ff l, p onriemann m, n, o, p , m, 1, 4 , n, 1, 4 ,o, 1, 4 , p, 1, 4 , i, 1, 4 , j, 1, 4 , k, 1, 4 , l, 1, 4 ;
Print MatrixForm ffriemann
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 2r 3
0 0
2r 3
0 0 0
0 0 0 0
0 0 0 0
0 0 1r 3
0
0 0 0 0
1r 3
0 0 0
0 0 0 0
0 0 0 1r 3
0 0 0 0
0 0 0 0
1r 3
0 0 0
0 2r 3
0 0
2r 3
0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 1r 3
0
0 1r 3
0 0
0 0 0 0
0 0 0 0
0 0 0 1r 3
0 0 0 0
0 1r 3
0 0
0 0 1r 3
0
0 0 0 01
r 3 0 0 00 0 0 0
0 0 0 0
0 0 1r 3
0
01
r 3 0 00 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 2
r3
0 0 2r 3
0
0 0 0 1r 3
0 0 0 0
0 0 0 0
1r 3
0 0 0
0 0 0 0
0 0 0 1r 3
0 0 0 0
0 1r 3
0 0
0 0 0 0
0 0 0 0
0 0 0 2r 3
0 0 2r 3
0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
This is nonsingularat the horizon.
Photon Orbits
Motivation
The flux that we measure from a neutron star or a disk will be influenced by the path that the photons havefollowed from the sources to us. Let us first consider this problem in a Schwarzschild spacetime as a prelude torepeating the exercise in a Kerr spacetime. We have not yet been able to image the spacetime directly, thoughthere are futuristic plans to do so using submm VLBI and Xrayinterferometry.
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Gravitational Redshift
From symmetry, orbits will be confined to a plane and there will be two nontrivial constants of the motionassociated with the t and symmetries. (For photons, we use the momentum p instead of the velocity with nochange in the argument.)
Now the photon energy measured by an observer moving with velocity u is = up . For an observer at rest onthe surface of a star u = [ 1 2 r 1 2 ,0,0,0]. Therefore, u = 1 2 r 1 2 t where t is the time Killingvector. The energy of a photon on the surface of a star is = 1 2 r 1 2 t p . However, along a geode-sic, E t p is conserved. Therefore = 1 2 r 1 2 .We can also use this relation to transform the intensity using the fact that the occupation number is invariant I 1 2 r
3 2 I .Note that this intensity must be computed at the transformed frequency.
Null Geodesic Equation
For nonradialmotion, there is a Killing vector and a second constant L = p . We can therefore writedown two equations of motion assuming = /2:
dtd = 1 2 r
1 Edd = r
2 Lwhere is the affine parameter. We obtain the radial equation from the interval gd x d x = 0. This gives:
drd
2E 2 1 2 r L2 r 2
Now change L ; L/E b to getdtd = b
1 1 2 r 1dd = r
2
drd
2= 1 b2 V(r)
where V(r) = r 2 2 r 3 . Note that far from the star r 2 d /dt b, which identifies b as the impact parameter.
The effective potential V(r) has the form:
veff r_ : 1 r ^ 2 2 r^3;Plot veff r , r, 2, 10 , Frame True, PlotRange 0, 10 , 0, .05 ,
FrameLabel "r", "V r " , RotateLabel False ;
2 4 6 8 10r
0.01
0.02
0.03
0.04
0.05
V r
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From this we can deduce that the cross section for capture of an infalling photon is V r . (Note that thecapture crosssectionfor a hole is 27 .)We want to map the stellar surface onto the observers sky. To do this introduce u=1/r and observe that
dud
2= 1 b2 u2 (12u)
and we can obtain the longitude of emission by quadrature
rstar 5;long b_ : NIntegrate 1 Sqrt 1 b^2 x ^ 2 1 2 x , x, 0, 1 rstarPlot long b , b, .001, 1 Sqrt veff rstar , Frame True,
PlotRange 0, 8 , 0, 3 , FrameLabel "b", " " , RotateLabel False ;
1 2 3 4 5 6 7 8b
0.5
1
1.5
2
2.5
3
Ingoing EddingtonFinkelsteinCoordinates
This coordinate system is generally introduced to clarify the nature of the Schwarzschild nonsingularity.Wewill use it as a convenience for numerical computations which we dont want to blow up near the horizon. Thecoordinates are Schwarzschild except that t v = t+r+2 log(r2).The metric becomesds2 = 1 2 r dv2 + 2dvdr r 2 d 2 sin 2 d2 Radial light rays satisfy
1 2 r dv2 + 2dv dr=0Ingoing photons follow v= v(0); outgoing photons v = 2[r r 0 + 2 log r 2 r 0 2 ]+v(0).In order to see its significance, let us first consider the trajectories of two photons, one moving radially inwardsfrom r=3 at t=0, the other moving outward in Schwarzschild coordinates.
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tin NDSolve t r r r 2 , t 3 0 , t, r, 3, 2.01 ;tout NDSolve t r r r 2 , t 3 0 , t, r, 3, 10 ;inplot Plot t r . tin, r, 3, 2.01 , Frame True, AspectRatio Automatic,PlotRange 0, 10 , 1, 10 , FrameLabel "r", "t", "Schwarzschild", ""
RotateLabel False, DisplayFunction Identity ;outplot Plot t r
. tout, r, 3, 10 , DisplayFunction Identity ;
Show inplot, outplot, DisplayFunction $DisplayFunction ;
2 4 6 8 10
r
0
2
4
6
8
10
t
Schwarzschild
The ingoing photon never reaches the horizon.Now repeat the exercise in EddingtonFinkelsteincoordinates.
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vin NDSolve v r 0, v 3 0 , v, r, 3, 2 ;vout NDSolve v r 2 r r 2 , v 3 0 , v, r, 3, 10 ;inplot Plot v r . vin, r, 3, 2 , Frame True, AspectRatio Automatic, PlotRange 0, 10 , 1, 10 ,
FrameLabel "r", "v", "Eddington Finkelstein", "" ,RotateLabel False, DisplayFunction Identity ;
outplot Plot v r . vout, r, 3, 10 , DisplayFunction Identity ;Show inplot, outplot, DisplayFunction $DisplayFunction ;
2 4 6 8 10r
0
2
4
6
8
10
v
Eddington Finkelstein
The ingoing photon passes straight through the horizon. It is conventional to replot this using v = vras theordinate
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inplot Plot v r 3 r . vin, r, 3, 2 , Frame True, AspectRatio Automatic, PlotRange 0, 10 , 1, 10 ,FrameLabel "r", "v ", "Eddington Finkelstein", "" ,RotateLabel False, DisplayFunction Identity ;
outplot Plot v r 3 r . vout, r, 3, 10 , DisplayFunction Identity ;Show inplot, outplot, DisplayFunction $DisplayFunction ;
2 4 6 8 10r
0
2
4
6
8
10
v
Eddington Finkelstein
Homework 1 (due before class Lecture 4)
1. Fun? with Riemann tensors
This is an exercise to coerce you to start writing your own software and develop your own conventions forcalculating. It also brings out an important point. We computed the Riemann tensor in the freefall frame by atwo stage process first transform to the orthonormal basis at rest in the Schwarzschild coordinates, thenLorentz boost to the orthonormal freefall frame. We can actually do it in one step.a. Write your own software for computing R in Schwarzschild coordinates starting from the metric tensorb. Observe that the velocity u is (1,0,0,0) in the free fall orthonormal basis. It is therefore the time componentof the freefall basis vector set, e 0 .What are the components of the vector e 0 in Schwarzschild coordinates.c. Use the fact that the other three freefall basis vectors must be normalized and orthogonal to e 0 to determinetheir components in Schwarzschild coordinates. Hint: Use symmetries to construct this basisd. Now construct the freefall components of the Riemann tensor directly by evaluating
R = e e
e
e R.
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2.Appearance of a Neutron Star
EXO 0748676is a newly observed neutron star. The Fe line normally seen at 1.27keV is now observed at0.94keV. The observed continuum spectrum may be crudely modeled as a black body with temperature 1.6keV. (See astroph0303147)
a. Compute the radius, assuming that the neutron star has a "standard" mass of 1.4 times that of the sun.b. Compute the intensity measured on the stellar surface assuming that it is isotropic and uniform. In otherwords there is no limbbrighteningor darkening and no "hot spots" on the surface.c. Compute the total luminosity emitted by this star as measured at large distanced. Assume that a local geographer has painted lines of latitude and longitude on the surface of the star anda future Xrayinterferometer can resolve the surface. Compute the image you would see.
2. Spinning Holes
Kerr metric
BoyerLindquistcoordinates
The Kerr metric in BoyerLindquistcoordinates isds2 = 1 2 r 2 dt2 (4 a r sin 2 / 2 ) ddt +( 2 / ) dr 2 + 2 d 2 + ( 2 sin 2 2 )d2
where 2 = r 2 a 2 cos 2 , = r 2 2r+ a 2 , 2 r 2 a 22
a 2 sin 2 .For derivations and discussions of its uniqueness, see Chandrasekhar or Isham. The parameter a measures thespecific angular momentum of the hole and lies in [0,1] assuming cosmic censorship.
Clear coord, metric, inversemetric, affine, riemann, ricci ;coord t, r, , ;
metricr 2 2 r a 2 Cos 2
r 2 a 2 Cos 2, 0, 0,
2 a r Sin 2
r 2 a 2 Cos 2,
0,r 2 a 2 Cos 2
r 2 2 r a 2, 0, 0 , 0, 0, r 2 a 2 Cos 2 , 0 ,
2 a r Sin 2
r 2 a 2 Cos 2, 0, 0,
r 2 a 2 r 2 a 2 Cos 2 2 r a 2 Sin 2
r 2 a 2 Cos 2Sin 2 ;
r 2 a 2 Cos 2 ;
r 2 2 r a 2 ;
r 2 a 2 2 a 2 Sin 2 ;Print MatrixForm metric ;
2 r r 2 a 2 Cos 2
r 2 a 2 Cos 20 0 2 a r Sin
2
r 2 a 2 Cos 2
0 r2 a 2 Cos 2
a 2 2 r r 2 0 0
0 0 r 2 a 2 Cos 2 02 a r Sin 2
r 2 a 2 Cos 20 0 Sin
2 a 2 r 2 r 2 a 2 Cos 2 2 a 2 r Sin 2r 2 a 2 Cos 2
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Riemann and Ricci
We will need the inverse metric coefficients,
gtt 2
2 , gt 2 ar
2, grr 2 , g
= 1 2 , g 2 2 r
2 sin2
the connection and the Riemann tensor.
inversemetric FullSimplify Inverse metric ;affine FullSimplify Table Sum inversemetric i, s
D metric s, j , coord kD metric s, k , coord j D metric j, k , coord s ,
s, 1, 4 2, i, 1, 4 , j, 1, 4 , k, 1, 4 ;Print MatrixForm affine ;riemann FullSimplify
Table D affine i, j, l , coord k D affine i, j, k , coord lSum affine i, s, k affine s, j, l affine i, s, l affine s, j, k
s, 1, 4 , i, 1, 4 , j, 1, 4 , k, 1, 4 , l, 1, 4 ;Print MatrixForm riemann ;
FullSimplify 2 2 r a 2 r 2 r 2 a 2 2
a 2 a 2 2 r r Sin 2
Again the vanishing of the Ricci tensor is a sanity check.
ricciSimplify Table Sum riemann k, i, k, j , k, 1, 4 , i, 1, 4 , j, 1, 4 ;
Print MatrixForm ricci ;
0 0 0 0
0 0 0 00 0 0 00 0 0 0
Why not KerrNewman?
Gravitationally significant charges discharge immediately.
Event Horizon
Location
The event horizon, located where grr 0,
f x_ : . r xsoln Solve f x 0, x ;
rh a_ : 1 1 a 2
at r = r = 1 + (1 a 21 2
,
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coord , ;inverse metric Simplify Inverse metric ;affine Simplify Table Sum inverse metric i, s
D metric s, j , coord kD metric s, k , coord j D metric j, k , coord s
s, 1, 2 2, i, 1, 2 , j, 1, 2 , k, 1, 2 ;
riemann Simplify Table D affine i, j, l , coord kD affine i, j, k , coord l Sum affine i, s, kaffine s, j, l affine i, s, l affine s, j, k , s, 1, 2 ,
i, 1, 2 , j, 1, 2 , k, 1, 2 , l, 1, 2 ;ricci Simplify Table Sum riemann k, i, k, j , k, 1, 2 ,
i, 1, 2 , j, 1, 2 ;Print MatrixForm ricci . a Sqrt 3 2, 0 ;
0 00 0
Area
We can compute the area A H = 8 r H from
area FullSimplify4 Pi Integrate Sqrt metric 1, 1 metric 2, 2 , , 0, Pi 2 ;
ah a_ : 8 Pi rh a
Ergosphere
Photon Orbits
Consider photons moving instantaneously azimuthally outside the horizon. Let =d /dt. The interval gives us
g 2 2 g0 g00 =0. Solving for the two angular velocities in the equatorial plane,
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1 Simplify metric 1, 4 Sqrt metric 1, 4 ^ 2 metric 1, 1 metric 4, 4 metric 4, 4 . Pi 22 Simplify metric 1, 4 Sqrt metric 1, 4 ^ 2
metric 1, 1 metric 4, 4 metric 4, 4 . Pi 2omegamin r1_, a1_ : 1 . r r1, a a1omegamax r1_, a1_ : 2 . r r1, a a1omegaplt a_ :
Plot omegamin r, a , omegamax r, a , omegah a , r, rh a , 6 ,Frame True, PlotRange rh a , 6 , .2, .5 , FrameLabel "r", " "
omegaplt .9 ;
2 a r a 2 2 r rr 3 a 2 2 r
2 a r a 2 2 r rr 3 a 2 2 r
2 3 4 5 6r
-0.1
0
0.1
0.2
0.3
0.4
0.5
These two angular frequencies define a range in which azimuthally moving timelike geodesics exist.
Static Limit
The boundary of the region within which there are no stationary, azimuthal timelike orbits is defined byg00 (r, ;a)=0. Sometimes this is called the static limit.
GraphicsGraphics;
rergo _, a_ : 1 1 a 2 Cos 2
ergoplt a_ : PolarPlot rergo Pi 2 , a , , 0, 2 Pi ,Frame True, AspectRatio Automatic, DisplayFunction Identity
holeplt a_ : Graphics Disk 0, 0 , rh aergoholeplt a_ :
Show ergoplt a , holeplt a , DisplayFunction $DisplayFunctionergoholeplt .9 ;
The region within the static limit is the ergosphere.
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Hawking Area Theorem
The area of the horizon must increase. Introduce irreducible mass mi = A H 16 = a 4 H . Reinstating the
gravitational mass m we have that mi 2 =m ( m+ m2 a 2 )/2. Equivalently, m2 mi 2 S2
4 mi 2 , where S=ma is the spin
or m mi 1 4 H 2 mi 2
1 2. The extractable mass is given by mr m mi .
ParametricPlot omegah a , 1 Sqrt a 4 omegah a ,a, 0, 1 , Frame True, PlotRange 0, .5 , 0, .3 ,
FrameLabel , m r m , RotateLabel False ;
0.1 0.2 0.3 0.4 0.5
0.05
0.1
0.15
0.2
0.25
m r
m
Up to 0.29 of the gravitational mass is extractable.
ZAMO Frame
Angular Velocity
Physical observers must rotate with respect to infinity close to the hole. A particular family of rotating frameshave zero angular momentum (also known as FIDucial ObserverS). Consider nongeodesicmotion with
u = [ gtt1 2 ,0,0,0]
The vanishing of the fourth component is the ZAMO condition. Compute its angular velocity
z =u
ut =2 ar
r 2 a2 2 a2 sin2
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ublcov FullSimplify 1 2 , 0, 0, 0 ;ublcon FullSimplify 1 2 , 0, 0, 2 a r 1 2 ;FullSimplify
Sum ublcon i metric i, j ublcon j , i, 1, 4 , j, 1, 4 ;
omegaz r_, _, a_ :2 a r
r 2 a 2 2 r 2 2 r a 2 a 2 Sin 2 plt1 Plot omegaz r, 2, 1 , r, rh 1 , 6 , Frame True,
FrameLabel "r", " z " , RotateLabel False, DisplayFunction Identity ; plt2 Plot omegaz 3, 2, a , a, 0, 1 , Frame True, FrameLabel "a", " z " ,
RotateLabel False, DisplayFunction Identity ; plt3 Plot omegaz 3, , 1 , , 0, 2 , Frame True, FrameLabel " ", " z " ,
RotateLabel False, DisplayFunction Identity ;Show GraphicsArray plt1, plt2, plt3 ;
1 2 3 4 5 6r
0
0.1
0.2
0.3
0.4
0.5
z
0 0.2 0.4 0.6 0.8a
0
0.01
0.02
0.03
0.04
0.05
0.06
z
It turns out that z =12 1 2 . Another useful property is that if we imagine a set of ZAMOs stringing a
loop of optical fiber (along which photons travel at the speed of light) around the hole at fixed r, , then theproand retrograde circuit times are equal. For this reason, ZAMOs are sometimes also called locally nonrotating observers.
Transformation to ZAMO
We transform the components using,
u = L u ,
u = L u ,u = u L
,
u = u L
,
L L = If the velocity of the ZAMO in BLcoordinates is u , then the timelike orthonormal basis vector in the ZAMOframe satisfies Lt = et
u and we can construct the remaining basis vectors by imposing orthonormality.
bl2z 1 2 , 0, 0, 0 , 0, 1 2 , 0, 0 ,0, 0, 1 , 0 , 2 a r 1 2 , 0, 0, Sin ;
z2bl FullSimplify Inverse bl2z ;
The sanity check comprises computing the components of the velocity in the ZAMO rest frame.
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uzcov FullSimplify Table Sum ublcov j bl2z j, i , j, 1, 4 , i, 1, 4uzcon FullSimplify Table Sum z2bl i, j ublcon j , j, 1, 4 , i, 1, 4
1, 0, 0, 0
1, 0, 0, 0
3+1 Split
It is possible to rewrite the metric in the formds2 = ( dt 2 + g jk dx
j + j dt dxk k dt)where the lapse = d / dt = 1 2 / , x j = (r, , ) and the shift j = (0, 0, H ). The spatial metric is diagonalwith grr 2 / , g = 2 , g= ( sin
2 . g jk is the metric of a surface of constant t inhabited by ZAMOs,moving on worldlines that are orthogonal to it. This is often a convenient way to develop physics so as tomake look as much as possible like flat space.
Gravity
ZAMOs have to accelerate and their acceleration can be shown to be g = du / d = ln along er . Thisdiverges as thehorizon is approached. A finite surface gravity can be defined for the horizon by taking the limit of du/dt = gas r r H .Calculating this gives g H =
12
12 r H
.
gh a_ : 1 2 1 a 2 1 2 1Plot gh a , a, 0, .9999 , Frame True,
FrameLabel "a", "g H" , RotateLabel False ;
0 0.2 0.4 0.6 0.8 1a
0
0.05
0.1
0.15
0.2
0.25
g H
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Note that grr no longer vanishes and that we have not introduced any off diagonal elements involving .
Ricci Tensor
As a sanity check,compute the Ricci tensor.
inverseksmetric FullSimplify Inverse ksmetric ;ksaffine FullSimplify Table Sum inverseksmetric i, s
D ksmetric s, j , kscoord kD ksmetric s, k , kscoord j D ksmetric j, k , kscoord s
s, 1, 4 2, i, 1, 4 , j, 1, 4 , k, 1, 4 ;Print MatrixForm ksaffine ;ksriemann FullSimplify
Table D ksaffine i, j, l , kscoord k D ksaffine i, j, k ,kscoord l Sum ksaffine i, s, k ksaffine s, j, l
ksaffine i, s, l ksaffine s, j, k , s, 1, 4 ,i, 1, 4 , j, 1, 4 , k, 1, 4 , l, 1, 4 ;
Print MatrixForm ksriemann ;ksricci
Simplify Table Sum ksriemann k, i, k, j , k, 1, 4 , i, 1, 4 , j, 1, 4 ;Print MatrixForm ksricci ;
0 0 0 00 0 0 00 0 0 00 0 0 0
Homework 2 (due before class Lecture 6)
1. Meaning of Coordinates.
In one concession to thought experiments, imagine that you command a platoon of trained and personallybrave surveyors, equipped with clocks, synchronized far from a hole (with specific angular momentum a =0.95), short strands of optical fiber and maneuverable vehicles.a. Explain how you could position them at rest on a grid of colatitude and longitude at the same r =6, beyondthe static limit.b. Compute the time it takes light to complete a polar orbitc. Plot the time it takes light to traverse an orbit with fixed colatitude as a function of .e. Explain how you could get them to synchronize their clocks among themselves and with you, (who havewiselychosen to remain far from the horizon).f. After your time has advanced by 10, resynchronise. Plot the associated change in proper time of yourassociates as a function of g. Now have a very few of your associates venture inside the ergosphere to r = 1.5, = /2. Have them becomeZAMOobservers by calculating what their proper orbital periods should be. Compute the time it will take light tocomplete anequatorial orbit and return to them and show that it is the same for forward and backward directed orbits
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2. Black Hole Thermodynamics
a. Use the area theorem to verify that m mi 1 2 1 2 where = 2 H mi .
b. Plot the reducible mass m mi as a function of a and H on the same graph for m = 1.c. Verify the computation of surface gravity and the Smarr formula.
3. Particle Orbit.
Compute the trajectory of a particle falling into a spinning hole with (a=0.9) in KerrSchildcoordinatesassuming itstarts from infinity at rest with zero angular momentum.
3. Orbits and Rays
Motivation
Stellar orbits
Stars are observed in orbit around the 2.8 million solar mass hole at our Galactic center (Sgr A*). Theyapproach as closeas 60 AU ~ 2000m and follow Keplerian orbits. There is every reason to believe (and some evidence to show)that most massive black holes are orbited by star clusters extending in close to the horizon.
Particle Orbits
BL Coordinates
Clear coord, metric, inversemetric, affine ;coord t, r, , ;
metricr 2 2 r a 2 Cos 2
r 2 a 2 Cos 2, 0, 0,
2 a r Sin 2
r 2 a 2 Cos 2,
0,r 2 a 2 Cos 2
r 2 2 r a 2, 0, 0 , 0, 0, r 2 a 2 Cos 2 , 0 ,
2 a r Sin 2
r 2 a 2 Cos 2, 0, 0,
r 2 a 2 r 2 a 2 Cos 2 2 r a 2 Sin 2
r 2 a 2 Cos 2Sin 2 ;
r 2 a 2 Cos 2 ;
r2
2 r a2
; r 2 a 2 2 a 2 Sin 2 ;
inversemetric FullSimplify Inverse metric ;affine FullSimplify Table Sum inversemetric i, s
D metric s, j , coord kD metric s, k , coord j D metric j, k , coord s ,
s, 1, 4 2, i, 1, 4 , j, 1, 4 , k, 1, 4 ;
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Horizon and Ergosphere
rh a_ : 1 1 a 2
rergo _, a_ : 1 1 a 2 Cos 2
Killing Vectors
Killings equation assures us that ; ; = g g = 0
in BL coordinates. Verifying for the temporal and azimuthal Killing vectors,
Print MatrixForm FullSimplifyTable Sum metric i, s affine s, j, 3 metric j, s affine s, i, 3
s, 1, 4 , i, 1, 4 , j, 1, 4 ;
Killing Tensor
There is also a Killing tensor that we construct from the ingoing and outgoing principal null vectorsl = [ r 2 + a 2 , , 0, a ] / n = [ r 2 + a2 , ,0, a ] / 2 2
where l.l = n.n = 0; l.n = 1.
lnull r 2 a 2 , 1, 0, a ;nnull r 2 a 2 2 2 , 2 2 , 0, a 2 2 ;FullSimplify lnull.metric.lnullFullSimplify nnull.metric.nnullFullSimplify lnull.metric.nnull
0
0
1
The covariant form of the Killing tensor is given by = 2 l n 2 + r 2 g
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fac1 FullSimplify Sum lnull i ucov i , i, 1, 4fac2 FullSimplify Sum nnull i ucov i , i, 1, 4
ea l 2 e r
a 2 2 r ru r
a 2 e a l e r 2 a 2 2 r r u r2 r 2 a 2 Cos 2
We use the invariant length of the velocity to eliminate ur
ur FullSimplify Solve ucov.inversemetric.ucov 1, u r 2fac3 FullSimplify fac1 . ur ;fac4 FullSimplify fac2 . ur ;fac5 FullSimplify fac3 fac4
u r1
a 2 2 r r
r 2 a 2 Cos 2 1 e 24 r a 2 e a l e r 2
2
a 2 2 r r a 2 r 2 a 2 2 r 2 a 2 Cos 2 l 2 Csc 2
a 2 r 2u
2
r 2 a 2 Cos 2
a a a e 2 4 e l 2 r 2 a 2 1 e 2 Cos 2 2 l 2 Csc 2 2 u 2
2 a 2 2 r 2 a 2 Cos 2
K is then given by
int3 FullSimplify 2 fac5 2 r 2
12
a a a e 2 4 e l a 2 1 e 2 Cos 2 2 l 2 Csc 2 2 u 2
It turns out to be more convenient to use the "Carter constant"Q = K(La E 2 = u 2 + a 2 (1 E 2 ) cos 2 L2 cot 2 Q provides an energy equation for vertical epicyclic motion about the equatorial plane.
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Equations of Motion
We can now solve for the equations of motion directly, without having to appeal to Lagranges equations.Taking them in turn,
dtd u
t = gt u =r 2 a2 r 2 a 2 E a L
2a a E sin
2 L 2
It turns out to be convenient to replace the proper time with a new parameter x defined byddx
2
We end up with
t dtdx r 2 a2 r 2 a 2 E a L a a E sin2 L
ucov e, u r , u , l ;FullSimplify 2 inversemetric .ucov 1 ;
ft r_, _ :r 2 a 2 r 2 a 2 e a l
r 2 2 r a 2a a e Sin 2 l
f r_, _ : r 2 a 2 Cos 2
Next, turn to the radial equation
r 2 2 ur 2
2 grr ur 2
=R(r)= r 2 a 2 E aL2
r 2 aE L 2 Q
eq1 FullSimplify ucov.inversemetric.ucov 1eq2 FullSimplify q u 2 a 2 1 e 2 Cos 2 l 2 Cot 2
FullSimplify 4 inversemetric 2, 2 2 u r 2 . FullSimplify Solve eq1, eq2 , u r , u 2
fr r_ : r2
a2
e a l2
r2
2 r a2
r2
a e l2
q
The radial equation has turning points and it is simpler, though slower, to differentiate and deal with the secondorder equationr= R(r)/2 = 2r E r 2 a 2 E aL (r1) r 2 aE L 2 Q r
Simplify D fr r , r 2dfr r_ : 2 r e r 2 a 2 e a l r 1 r 2 a e l 2 q r r 2 2 r a 2
12
2 r a 2 2 r r 2 1 r a e l 2 q r 2 4 e r a l e a 2 r 2Treat the equation similarly.
2 2 u 2
2 g u 2
= Q a 2 1 E 2 cos 2 L2 cot 2
FullSimplify 4 inversemetric 3, 3 2 u 2 . FullSimplify Solve eq2, u 2
f _ : q a 2 1 e 2 Cos 2 l 2 Cot 2
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Orbit Calculation
run1 orbit;
Orbit Plots
orbitplt : Show GraphicsArrayParametricPlot Evaluate x , t x . run1 , x, 0, xend , Frame TrPlotRange All, FrameLabel " ", "t" , DisplayFunction Identity ,
ParametricPlot Evaluate x , r x . run1 , x, 0, xend , Frame TruePlotRange All, FrameLabel " ", "r" , DisplayFunction Identity ,ParametricPlot Evaluate x , x . run1 ,x, 0, xend , Frame True, PlotRange All,
FrameLabel " ", " " , DisplayFunction Identity ,ParametricPlot Evaluate x , x . run1 ,x, 0, xend , Frame True, PlotRange All,FrameLabel " ", " " , DisplayFunction Identity ;
orbitplt;
0 250 500 750 1000 1250 1500
1.4
1.5
1.6
1.7
1.8
0 250 500 750 10
0
2
4
6
8
10
12
0 250 500 750 1000 1250 1500
0
250
500
750
1000
1250
1500
t
0 250 500 750 10
10
15
20
25
30
35
r
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2D Representation
scale 40;hole2D
Graphics PointSize rh a scale , Point 0, 0 , AspectRatio Automatic ;xyplt Show ParametricPlot
Evaluate r x Sin x Cos x , r x Sin x Sin x . run1 ,x, 0, xend , Frame True, FrameLabel "x", "y" ,PlotRange scale, scale , scale, scale , AspectRatio Automatic,DisplayFunction Identity , hole2D, DisplayFunction Identity ;
xzplt Show ParametricPlot Evaluater x Sin x Cos x , r x Cos x . run1 ,x, 0, xend , Frame True, FrameLabel "x", "z" ,
PlotRange scale, scale , scale, scale , AspectRatio Automatic,DisplayFunction Identity , hole2D, DisplayFunction Identity ;
yzplt Show ParametricPlot Evaluater x Sin x Sin x , r x Cos x . run1 ,x, 0, xend , Frame True, FrameLabel "y", "z" ,
PlotRange scale, scale , scale, scale , AspectRatio Automatic,DisplayFunction Identity , hole2D, DisplayFunction Identity ;
Show GraphicsArray xyplt, xzplt , yzplt ;
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-30 -20 -10 0 10 20 30 40y
-30
-20
-10
0
10
20
30
40
z
-30 -20 -10 0 10 20 30 40x
-30
-20
-10
0
10
20
30
40
y
-30 -20 -10 0 10 20 3x
-30
-20
-10
0
10
20
30
40
z
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3D Representation
hole3DGraphics3D PointSize rh a 40 , Point 0, 0, 0 , AspectRatio Automatic
xyzplt Show ParametricPlot3D Evaluater x Sin x Cos x , r x Sin x Sin x , r x Cos x
.
run1 , x, 0, 3 , Boxed False, AspectRatio Automatic,PlotRange 40, 40 , 40, 40 , 40, 40 , DisplayFunction Identity ,
hole3D, DisplayFunction $DisplayFunction ;
-40
-20
0
20
40
-40
-20
0
20
40
-40
-20
0
20
40
-40
-20
0
20
0
-20
0
20
Circular Equatorial Orbits
Energy
We require that r=r=0. Solving for E and choosing the relevant root, adopting the convention that a>0 is aprograde orbit, we obtain
E= 1 2 r 1 a r 3 2
1 3 r 1 2 a r 3 2 1 2
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FullSimplify Solve fr r 0, dfr r 0 . q 0, e, l 4ekep r_, a_ :
1 2 r 1 a r 3 2
1 3 r 1 2 a r 3 2 1 2
ea 2 r r 2
2 a r 3 2 3 r r 2
Photon Orbit
The energy becomes infinite at the photon orbit where the denominator of E>0
r ph1 1 2 a a 1 a 2
1 3 2
1 2 a a 1 a 21 3
rph FullSimplify Solve 1 3 r 1 2 a r 3 2 0, r ;
rphot a1_ : Chop If a1 0, r . rph 3 . a a1, r . rph 1 . a a1Marginally Bound Orbit
The marginally bound circular orbit occurs when E=1r mb = 2a+ 2 1 a
1 2
FullSimplify Solve 1 2 r 1 a r 3 22
1 3 r 1 2 a r 3 2 , r 2 ;
rkepmb a_ : 2 a 2 1 a 1 2
Angular Momentum
Similarly, solving for L
L= 1 2 a r 3 2 a2 r 2 r 1 2
1 3 r 1 2 a r 3 2 1 2
FullSimplify Solve fr r 0, dfr r 0 . q 0, l, e 4 ;lkep r_, a_ :
1 2 a r 3 2 a 2 r 2 r 1 2
1 3 r 1 2 a r 3 2 1 2
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Stability
Stability requires that the radial potential have positive curvature orR(r) < 0which leads to
r ms 3 3 a 2 3 a 1 a 1 3 1 a 2 3 1 a 2 3 1 a 1 3 3 a18 3 a 1 a 1 3 1 a 2 3 1 a 2 3 1 a 1 3 3 a
2 6 a 2 16 a2
3 a2 3 a 1 a 1 3 1 a 2 3 1 a 2 3 1 a 1 3 3 aafter choosing the relevant signs.
FullSimplify D dfr r . q 0, r . e ekep r , l lkep r ;FullSimplify Solve r r 6 3 a 2
264 a 2 r, r ;
rms a_ : Chop
3 Sign a 3 a2
3 a 1 a1 3
1 a2 3
1 a2 3
1 a1 3
3 a18 3 a 1 a 1 3 1 a 2 3
1 a 2 3 1 a 1 3 3 a 2 6 a 2 Sign a 16 a 2 3 a 2 3 a 1 a 1 3 1 a 2 3 1 a 2 3 1 a 1 3 3 a
Plot of Equatorial Radii
Let us now plot these various radii on a single graph.
rhplt Plot rh a , a, 1, 1 , Frame True, PlotRange 1, 1 , 1, 9 ,FrameLabel "a", "r" , DisplayFunction Identity ;
rergoplt Plot rergo 2, a , a, 1, 1 , DisplayFunction Identity ;rphotplt Plot rphot a , a, 1, 1 , DisplayFunction Identity ;rkepmbplt Plot rkepmb a , a, 1, 1 , DisplayFunction Identity ;rkepmsplt Plot rkepms a , a, 1, 1 , DisplayFunction Identity ;Show rhplt, rergoplt, rphotplt, rkepmbplt, rkepmsplt,
DisplayFunction $DisplayFunction ;
-0.75 -0.5 -0.25 0 0.25 0.5 0.75 1a
2
3
4
5
6
7
8
9
r
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Energy and Angular Momentum Plots
The energy of a circular orbit of radius r with a range of values of a.
ekep1pltPlot ekep r, 1 , r, rkepmb 1 , 9 , Frame True, PlotRange 1, 9 , .5, 1 ,
FrameLabel "r", "E" , DisplayFunction Identity ;ekep2plt Plot ekep r, .99 , r, rkepmb .99 , 9 , DisplayFunction Identity ;ekep3plt Plot ekep r, 0. , r, rkepmb 0. , 9 , DisplayFunction Identity ;ekep4plt Plot ekep r, 1. , r, rkepmb 1. , 9 , DisplayFunction Identity ;Show ekep1plt, ekep2plt, ekep3plt,
ekep4plt, DisplayFunction $DisplayFunction ;
2 3 4 5 6 7 8 9r
0.6
0.7
0.8
0.9
1
E
It will be useful to plot the binding energy of the marginally stable circular orbit (ISCO).
beplt1 Plot 1 ekep rkepms a , a , a, 1, 1 , Frame True,PlotRange 1, 1 , 0, .5 , FrameLabel "a", "1 E" ,RotateLabel False, DisplayFunction Identity ;
beplt2 Plot 1 ekep rkepms a , a , a, .99, 1 , Frame True,PlotRange .99, 1 , 0, .5 , FrameLabel "a", "1 E" ,RotateLabel False, DisplayFunction Identity ;
Show GraphicsArray beplt1, beplt2 ;
The corresponding angular momenta, shown as positive functions.
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lkep1pltPlot lkep r, 1 , r, rkepmb 1 , 9 , Frame True, PlotRange 1, 9 , 0, 5 ,
FrameLabel "r", "L" , DisplayFunction Identity ;lkep2plt Plot lkep r, .99 , r, rkepmb .99 , 9 , DisplayFunction Identity ;lkep3plt Plot lkep r, 0. , r, rkepmb 0. , 9 , DisplayFunction Identity ;lkep4plt Plot lkep r, 1. , r, rkepmb 1. , 9 , DisplayFunction Identity ;
Show lkep1plt, lkep2plt, lkep3plt,lkep4plt, DisplayFunction $DisplayFunction ;
2 3 4 5 6 7 8 9r
1
2
3
4
5
L
Angular Frequency
We next turn to the two equations of motion, reverting to as the independent vari-able
dtd
r 3 2 a L E r 3 a2 E 2 r a2 2 r r
d
d
2 a E L 2 r r 3
a2
2 r r from which we obtain= 1r 3 2 a
ut FullSimplify r 2 ft r, 2 ;u FullSimplify r 2 f r, 2 ;FullSimplify u ut . e ekep r, a , l lkep r, a ;
omkep r_, a_ : r 3 2 a1
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A useful relation for circular geodesics
The acceleration vanishes. Thereforea u u; = u u; u ; u u, u , 0Hence
a r ut ut ,r u u,r =0As we change the radius of the orbit,ut du t u du 0and sodEdL =
Another way to see this is to note that, if there are no other forces involved, the work done by a tangential forceequals the change in the energy.To verify,
FullSimplify D ekep r, a , r omkep r, a D lkep r, a , r
Rays
Equations of Motion
The equations for null geodesics can be obtained simply from the those for timelike geodesics by taking thelimit E, L >and choosing a suitable affine parameter. As we will mainly be interested in what a distant observer might see,we replace our impact parameter b with an impact vector b with components b1 , b2 resolved perpendicularand parallel to the projected hole spin axis. We let Ex>x, L E l b1 sin 0 , u E u b2 and
Q E 2 q
2 b22 cos 2 0 b1
2 a 2 .The orbits just depend upon l and q which are, in turn, fixed by b . I shall eschew finesse; fast routines perform
the elliptic integral quadratures substitute u=1/r and use look up tables. The equations become:t r
2 a2 r 2 a 2 a l a a sin 2 l
r 2 r 2 a 2 a l2
a l 2 q2
r= 2 r r 2 a2 a l r 1 a l 2 q2
2 q2 a 2 cos 2 l2 cot 2 = a 2 cos sin l2 cot csc 2
= a r 2 a2 a l l csc 2 a
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gt u_, _ :r 2 a 2 r 2 a 2 a l a r 2 2 r a 2 a Sin 2 l
r 2 2 r a 2. r
1
u
gu u_ : r 2 a 2 a l2
r 2 2 r a 2 a l 2 q 2 . r1
u
dgu u_ : 2 r r 2 a 2 a l r 1 a l 2 q 2 . r1
ug _ : q 2 a 2 Cos 2 l 2 Cot 2
dg _ : a 2 Sin Cos l 2 Cot Csc 2
g u_, _ :a r 2 a 2 a l r 2 2 r a 2 l Csc 2 a
r 2 2 r a 2. r
1
u
Numerical Integration
We integrate backwards along a ray from the observer plane located at a radius = u0 1 .
ray : NDSolvet x gt u x , x ,
u x2
u xu x 2 u x 2 dgu u x ,
x dg x , x g u x , x ,t 0 0,u 0 u 0 ,u 0 u 0 2 Sqrt gu u 0 , 0 0 , 0 Sqrt g 0 , 0 0,t, u, , ,x, 0, xstart
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Ray Plots
rayplt : Show GraphicsArrayPlot Evaluate t x . run2 , x, 0, xstart , Frame True,PlotRange All, FrameLabel "x", "t" , DisplayFunction Identity ,
Plot Evaluate 1 u x
. run2 , x, 0, xstart , Frame True,
PlotRange All, FrameLabel "x", "r" , DisplayFunction Identity ,Plot Evaluate x . run2 , x, 0, xstart , Frame True,PlotRange All, FrameLabel "x", " " , DisplayFunction Identity ,
Plot Evaluate x . run2 , x, 0, xstart , Frame True, PlotRange All, FrameLabel "x", " " , DisplayFunction Identity ;rayplt;
-0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0x
0.75
1
1.25
1.5
1.75
2
2.25
2.5
-0.6 -0.5 -0.4 -0
1
2
3
4
-0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0x
-350
-300
-250
-200
-150
-100
-50
0
t
-0.6 -0.5 -0.4
50
100
150
200
r
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2D Representation
scale 100;
rh a_ : 1 1 a 2hole2D
Graphics PointSize rh a scale , Point 0, 0 , AspectRatio Automatic ;xyrayplt Show ParametricPlotEvaluate Sin x Cos x u x , Sin x Sin x u x . run2 ,x, 0, xstart , Frame True, FrameLabel "x", "y" ,PlotRange scale, scale , scale, scale , AspectRatio Automatic,DisplayFunction Identity , hole2D, DisplayFunction Identity ;
xzrayplt Show ParametricPlot EvaluateSin x Cos x u x , Cos x u x . run2 ,x, 0, xstart , Frame True, FrameLabel "x", "z" ,
PlotRange scale, scale , scale, scale , AspectRatio Automatic,DisplayFunction Identity , hole2D, DisplayFunction Identity ;
yzrayplt Show ParametricPlot EvaluateSin x Sin x u x , Cos x u x . run2 ,x, 0, xstart , Frame True, FrameLabel "y", "z" ,
PlotRange scale, scale , scale, scale , AspectRatio Automatic,DisplayFunction Identity , hole2D, DisplayFunction Identity ;
Show GraphicsArray xyrayplt, xzrayplt , yzrayplt ;
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-75 -50-25 0 25 50 75 100y
-75
-50
-25
0
25
50
75
100
z
-75 -50-25 0 25 50 75 100x
-75
-50
-25
0
25
50
75
100
y
-75 -50-25 0 25 50 75 100x
-75
-50
-25
0
25
50
75
100
z
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3D Representation
hole3D Graphics3DPointSize rh a scale , Point 0, 0, 0 , AspectRatio Automatic ;
xyzrayplt Show ParametricPlot3D Evaluate Sin x Cos x u x ,
Sin x Sin x u x , Cos x u x . run2 ,x, 0, xstart , Boxed False, AspectRatio Automatic,PlotRange scale, scale , scale, scale , scale, scale , AxesLabel "x", "y", "z" , DisplayFunction Identity ,
hole3D, DisplayFunction $DisplayFunction ;
-100
-50
0
50
100
x
-100
-50
0
50
100y
-100
-50
0
50
100
z
100
-50
0
50x
0
-50
0
50y
Penrose Process
Negative Energy States
The key to this process is the realization that t becomes spacelike inside the ergosphere as gtt t . t < 0.This means that the energy E= p. can be negative as it is now a component of 3momentum. Clearly we can
be most negative with vanishing rest mass, ie with photons whose angular momentum we can set to unitywlog. It is also clear that the energy can be most negative when there is no radial or motion. Keep thingssimple by restricting to the equatorialplane. Let the interval vanish and solve for the energy, retaining only the solution with future directed energy.
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Clear a ; momen en, 0, 0, 1 ;interv momen.inversemetric.momen . 2;energy FullSimplify Solve interv 0, en 2a 1.; photen r1_ : en . energy . r r1
Plot photen r , r, rh a , rergo 2, a ,Frame True, FrameLabel "r", "E" ;Clear
a ;
en2 a r a 2 2 r r
r 3 a 2 2 r
1 1.2 1.4 1.6 1.8 2r
-0.5
-0.4
-0.3
-0.2
-0.1
0
E
As we approach the horizon, E H L
FullSimplify energy . r rh a
ena
2 2 1 a 2
This is the best we can do.
Retrograde orbits have negative energy in the ergosphere. This extends to higher latitude, of course. If we addsome ur , u , and rest mass, the phase space of negative energy orbits diminishes and E H L at the horizon.
Energy Extraction
Suppose that the photon has negative radial motion so that it approaches the horizon (in BL) and actuallycrosses it in KS. It will change the mass by E and the spin by L. If we recall the mass formuladm = ( g H 8 ) d A H + H d Swe see the the area(entropy) is destined not to decrease and the purely retrograde photon hovering just abovethe horizon corresponds to a reversible change.
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Penrose Orbits
Energy extraction is possible in principle because we can scatter a photon (or massive particle) onto a negativeenergy orbit within the ergosphere. Momentum must be conserved in this process and the remaining outgoingenergy must exceed the ingoing energy. However, the phase space for this to happen is small and this is not an
important process naturally.
Weak Field Limit
Gravitomagnetism
In the gravitomagnetic approach the field equations and geodesic equation are taken to the weak field, slowmotion limit and expressed in the language of electromagnetic theory. A gravitoelectric and gravitomagneticfield are introduced so thatE= ;= 1 g00 /2B = A ; Ai = g0 i
E = 4 ; B = 0; E =0; B = 4(4 v E
t )dvdt = E+v B
Note the sign changes and absence of a term in the third equation and th presence of a factor 4 in the fourthequation.Proceeding by direct analogy with electromagnetic theory, the gravitomagnetic field for a body with spin S is
S = 3 S r r Sr 2
r 5
A slightly inclined orbit will then precess at a rate given by integrating the torque around the orbit
p =2 Sr 3
as may be verified directly by expanding the equations for ,.This formalism can also be used to discuss the precession of spinning black holes. Specifically the gravitomag-netic
field associated with motion through the gravitoelectric field of a nonspinningmass, vE induces a spinprecessionat a rate H/ 2 = / 2r. This must be added to the geometrical precession induced by space curvature to give acombined geodetic precession of 3 /2r.Finally, there is Lense Thirring precession which arises when an orbiting spin precesses in the gravitomagneticfield. This is retrograde and of magnitude S 2 r 3 when the orbit is equatorial and progradewith twice the magnitude when the orbit is polar. We will return to these matters.
Homework 3 (due Tuesday May 6)
1. Penrose OrbitsWe discussed the nature of Penrose orbits and showed how they could not exist outside the ergosphere.Consider a single, horizoncrossing,negative energy orbit. Where does it come from?
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2. QuasiperiodicXraySignal from a Star Orbiting a Massiove Black Hole
A phenomenon which has been sought in Xrayobservations (and which may well have been detected in recentobservations of Sgr A*) is a burst of hard Xraysproduced every time a star crashes through an equatorialaccretion disk. Consider a 3 million solar mass, spinning black hole with a=0.9. Suppose that there is a half
solar mass star on a prograde orbit with r = 6, inclined at 30o
to the equator and crossing it first time at theEobservers longitude .a. Determine the BL coordinates of successive crossings of the equatorial plane.b. Compute rays which connect these events to the observers sky plane and hence determine the time intervalbetween successive, observed Xraypulses.c. Perform some sanity checks of your routines.(If you have difficulty debugging your software, generous partial credit will be given for a good description of what you are trying to do.)
4. Thin Disks
Motivation
XrayBinary and AGN Accretion Disks
We regularly observe accretion disks around stellar and massive black holes and neutron stars.It appears thatAGN holes create radiant energy per unit mass in the ratio~0.2,suggesting that the holes are spinning rapidly asthis exceeds the binding energy of the most bound,stable circular orbits.In addition,broad Fe lines are observedfrom disks around both stellar and massive holes,suggesting that the disks exist very close to the horizon,againindicating rotation.In fact the power in the line is large that it has been suggested that it must also derive fromthe ergosphere.
Simplifying Assumptions
Observed disks
Accretion disks are complex. In reality, they are nonstationary,nonaxisymmetric,thick, nonplanar,inhomoge-neous and the (magnetic) torque is nonNewtonian.The emission is nonthermal and anisotropic. The relativityis the best understood feature of their description although numerical simulation is leading to a much moresophisticated description. In order to bring out their relativistic properties I shall suppose that disks are 2D,equatorial, cold, prograde, circular stationary and axisymmetric while the emission is thermal and isotropic.
Geometry and KinematicsMetric
The 3D metric is given by
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dcoord t, r, ;dmetric
12
r, 0,
2 a
r, 0,
r 2
r 2 2 r a 2, 0 ,
2 a
r, 0,
r 2 r 2 a 2 2 a 2 r
r 2;
Print MatrixForm dmetric
r2
2 r a2
;inversedmetric FullSimplify Inverse dmetric ;daffine FullSimplify Table Sum inversedmetric i, s
D dmetric s, j , dcoord kD dmetric s, k , dcoord j D dmetric j, k , dcoord s ,
s, 1, 3 2, i, 1, 3 , j, 1, 3 , k, 1, 3 ;ddetg FullSimplify Det dmetric
1 2r
0 2 ar
0 r2
a 2 2 r r 20
2 ar
0 2 a2 r r 2 a 2 r 2
r 2
r 2
Angular Frequency, Energy, Angular Momentum and Marginally Stable Radius
From previous sections we define
omkep r_, a_ : r 3 2 a1
ekep r_, a_ :1 2 r 1 a r 3 2
1 3 r 1 2 a r 3 2 1 2
lkep r_, a_ :1 2 a r 3 2 a 2 r 2 r 1 2
1 3 r 1 2 a r 3 2 1 2
rmin a_ : Chop
3 3 a 2 3 a 1 a 1 3 1 a 2 3 1 a 2 3 1 a 1 3 3 a 183 a 1 a 1 3 1 a 2 3 1 a 2 3 1 a 1 3 3 a 2 6 a 2 16 a 2
3 a 2 3 a 1 a 1 3 1 a 2 3 1 a 2 3 1 a 1 3 3 aAngular Frequency Plot
For a=0, 1.
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omplt0 Plot omkep r, 0 , r, rmin 0 , 10 , Frame True, FrameLabel "r", " "PlotRange 0, 10 , 0, .5 , DisplayFunction Identity ;
omplt1 Plot omkep r, 1 , r, rmin 1 , 10 , DisplayFunction Identity ;Show omplt0, omplt1, DisplayFunction $DisplayFunction ;
2 4 6 8 10r
0.1
0.2
0.3
0.4
0.5
Disk Velocity
We need the velocity u of the disk in the BL frame. We express it in terms of , using its invariant length
ucov ekep r, a , 0, lkep r, aucon
FullSimplify Table Sum inversedmetric i, j ucov j , j, 1, 3 , i, 1, 3
1 ar 3 22r
1 2 ar 3 23r
, 0,1 a
2
r 22 a
r 3 2 r
1 2 ar 3 23r
a r 3 2
1 2 ar 3 23r r
3 2, 0,
1
1 2 ar 3 23r r
3 2
Coordinate Transformation
We generalize the approach we used for the ZAMO frame. Lt = et
= ue e
and we find that the transformation from BL to disk coordinates is given by
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bl2d ucon 1 , 0,ucov 3
1 2, 0,
1 2
r, 0 , ucon 3 , 0,
ucov 11 2
;
Print MatrixForm bl2d ;Print MatrixForm FullSimplify Table Sum dmetric k, l
bl2d k, i bl2d l, j , k, 1, 3 , l, 1, 3 , i, 1, 3 , j, 1, 3 ;d2bl FullSimplify Inverse bl2d ;Print MatrixForm d2bl ;
a r 3 2
1 2 ar 3 23r r 3 2
01 a
2
r 22 a
r 3 2 r
1 2 ar 3 23r a 2 2 r r 2
0 a2 2 r r 2
r 0
1
1 2 ar 3 23r r
3 20
1 ar 3 22r
1 2 ar 3 23r a
2 2 r r 2
1 0 00 1 0
0 0 1
a 2 r r
1 2 ar 3 23r r 3 2
0 a2 2 a r r 2
1 2 ar 3 23r r 3 2
0 ra 2 2 r r
0
a 2 2 r r
1 2 ar 3 23r r
3 20 a
2 2 r r a r 3 2
1 2 ar 3 23r r
3 2
Mass conservation
Mass flux vector
Introduce a proper mass surface density . The mass flux vector u has vanishing divergence. Therefore, asg =r,
2rJ = 2 r u r = M =const. Note that J < 0. The other two components of the mass flux will not be needed.
Stress energy tensor
Conservation Laws
The divergence of the full stress energy tensor must vanish. If we have a Killing vector , satisfying Killings
equation, then T ; =0.Using the temporal and axial Killing vectors and imposing stationarity and axisymmetry, we immediatelydefine a conserved energy flux vector
T t and a conserved angular momentum flux vector
T .Again, we are only interested in the radial components, = r , = r .
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Mass transfer
In absence of pressure, the matter contribution to the stress tensor is given by T u u . The contributionto the energy flux vector is = Ju t = JE and to the angular momentum flux vector is = Ju =JL.
Torque
We do not have a good understanding of the torque, T r = G. However we do know that it will do work at arate G and there will be a corresponding energy flux. Therefore the contribution to the stress energy tensor inBL coordinates must have the form T r t T
r
As a sanity check, assume that we have a Newtonian shear viscosity . The rate of shear tensor is given by thesymmetrictrace free part of u; projected perpendicular to u . (The acceleration a u u; and the expansion u; vanish.) u ; u u ; u
The viscous part of the stress energy tensor is then given by T =2 .(eg MTW pp566568). evaluating,
ducov FullSimplify Table D ucov i , dcoord jSum daffine k, i, j ucov k , k, 1, 3 , i, 1, 3 , j, 1, 3 ;
thet FullSimplify Sum inversedmetric i, j ducov i, j ,i, 1, 3 , j, 1, 3 ;
pducov FullSimplify Table ducov i, jSum ucov i ducov j, k ucon k , k, 1, 3 , i, 1, 3 , j, 1, 3 ;
sigcov FullSimplify1
2Table pducov i, j pducov j, i ,
i, 1, 3 , j, 1, 3 ;newtorque 2 Table Sum inversedmetric i, k sigcov k, j , k, 1, 3 ,
i, 1, 3 , j, 1, 3 ;Print MatrixForm newtorque
0 23 a 1 2 ar 3 2
3r a r
3 2
2 2 a 3 r r2
r 2
3 1 2 ar 3 23r a
2 2 r r 1 2r 4a 2 2 r r4 2 a 3 r r
2r
3 1 2 ar 3 23r a
2 2 r r2
2 2 a 3 r r2
r 30
0 23 a 1 2 ar 3 2
3r
2 2 a 3 r r2
r 2
3 1 2 ar 3 23r 2 r a r 3 2
4 2 a 3 r r2
r 2
The contribution to the nergy flux automatically has the form described above.
FullSimplify newtorque 2, 1 newtorque 2, 3 omkep r, a
0
We can also transform this tensor into the disk frame
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FullSimplifyTable Sum d2bl i, k newtorque k, l bl2d l, j , k, 1, 3 , l, 1, 3 ,
i, 1, 3 , j, 1, 3
0, 0, 0 , 0, 0,3 a 2 2 r r
2 2 a 3 r r r 2, 0,
3 a 2 2 r r
2 2 a 3 r r r 2, 0
Note that there is no energy flux in this frame.
Radiation
We must next consider the contribution of the escaping radiation to the stressenergybudget. We assume thatthe disk is efficiently radiative so that the energy that is dissipated is immediately radiated isotropically in thedisk frame. If we denote the radiation flux leaving one surface by F, the four dimensional stress energy tensorhas nonzerocomponents
T t = T t = F; T t t = 4F; T r r = T = T = 4 F 3 .
Note that this is tracefree,as it should be.In BL coordinates, T t =
FEr , while the vertical angular momentum flux is T
=
FLr .
Torque and Flux
Energy and Angular Momentum Conservation
The fluxes of radiative energy and angular momentum escaping from the disk can be equated to the minus thedivergence of the 3D stress tensor. In BL coordinates, this implies that1r
d dr r(G + JL) + 2FL=0
1r
d dr r(G + JE) + 2FE=0
Solution
Solving these two equations and mass conservation, using dE= dL and assuming that G and F vanish at r min gives
G = M 2 r
r
r mindL E L
E Land
F = M 2 rddr
r
r mindL E L
E L 2
which reduce to the nonrelativistic expressions. Note that these are independent of the nature of the torqueexcepting that it and the associated dissipation be local.
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a .95;eoml r_, a_ : FullSimplify ekep r, a omkep r, a lkep r, adldr r_, a_ : FullSimplify D lkep r1, a , r1 . r1 rdomdr r_, a_ : FullSimplify D omkep r1, a , r1 . r1 rtorque r_ :
NIntegrate eoml rp, a dldr rp, a , rp, rmin a , r
2 r eoml r, a
flux r_ : NIntegrate eoml rp, a dldr rp, a , rp, rmin a , r domdr r, a
2 r eoml r, a 2
Analytic expressions can be given for these functions (eg Krolik pp150152),but they are not very illuminatingand it is easier to compute them numerically.
Plot torque r , r, rmin a , 10 , Frame True, FrameLabel "r", "G" ;Plot flux r , r, rmin a , 10 , Frame True, FrameLabel "r", "F" ;Clear a
2 4 6 8 10
r
0
0.005
0.01
0.015
0.02
G
2 4 6 8 10r
0
0.0002
0.0004
0.0006
0.0008
0.001
0.0012
F
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Equilibrium Flow
Metric etc
Clear coord, metric, inversemetric ;coord t, r, , ;
metricr 2 2 r a 2 Cos 2
r 2 a 2 Cos 2, 0, 0,
2 a r Sin 2
r 2 a 2 Cos 2,
0,r 2 a 2 Cos 2
r 2 2 r a 2, 0, 0 , 0, 0, r 2 a 2 Cos 2 , 0 ,
2 a r Sin 2
r 2 a 2 Cos 2, 0, 0,
r 2 a 2 r 2 a 2 Cos 2 2 r a 2 Sin 2
r 2 a 2 Cos 2Sin 2 ;
r 2 a 2 Cos 2 ;
r 2 2 r a 2 ;
r 2 a 2 2 a 2 Sin 2 ;inversemetric FullSimplify Inverse metric ;
ekep r_, a_ :1 2 r 1 a r 3 2
1 3 r 1 2 a r 3 2 1 2
lkep r_, a_ :1 2 a r 3 2 a 2 r 2 r 1 2
1 3 r 1 2 a r 3 2 1 2
omegakep r_, a_ : r 3 2 a1
rh a_ : 1 1 a 2
rmb a_ : 2 a 2 1 a 1 2
rms a_ : Chop3 3 a 2 3 a 1 a 1 3 1 a 2 3 1 a 2 3 1 a 1 3 3 a 18
3 a 1 a 1 3 1 a 2 3 1 a 2 3 1 a 1 3 3 a 2 6 a 2 16 a 23 a 2 3 a 1 a 1 3 1 a 2 3 1 a 2 3 1 a 1 3 3 a
General::spell1 : Possible spelling error: new symbol name "lkep" is similar to existing symbol "ekep".
Fluid
Write the enthalpy asw = nm + P 1where n is the baryon density and m is the rest mass per baryon. This is good for ion tori with =5/3 andradiation tori with =4/3. The fluid stress tensor isT = wu u + Pg
It will also beuseful to define an entropy function, which is monotonically related to the true entropy
S= P1
nm .
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Euler equation
Ignore inflow and assume axisymmetry. Project the stress energy conservation law perpendicular tou ut , 0, 0, u . Introduce a, b, c for r, , i, j, k, for t, .P T ;
=> wa a P ,a 0where a u u; is the acceleration.Using u u ; 0, it is helpful to writea u u, u ,Hence
,a
w = ut ut ,a u u,a
which impliesP
w = ut ut u u
where here represents the two dimensional derivative r ,1r and we can map the meridional structure
onto the Euclidean plane.
Fluid Angular MomentumIt is convenient to replace the dynamical angular momentum L with the fluid angular momentumH = L E
uut
Usingut = 1 E 1 Hwe obtain Pw = ln E
H 1 H
hkep r_, a_ :lkep r, a
ekep r, a
General::spell : Possible spelling error: new symbol name "hkep" is similar to existing symbols ekep, lkep .
Barytropic Disks
von Zeipels theorem
To proceed we need to make an assumption about the equation of state. A convenient (and possibly relevant)choice is that it is barytropic, ie w=w(P). This implies that = (H). This is the relativistic von Zeipel theorem.If we use the relation
= gt g H
gtt gt H
then we see that this defines von Zeipel surfaces on which , H are simultaneously constant, independent of the equation of state.
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Isobars
Given a relation (H), we can define= ln E dH1 H
where we compute E using
E gtt 2 gt H g H 21 2
The isobars are then surfaces of constant obtained by solvingdPd w P
Equatorial Angular Momentum
Consider the plot of the two definitions of angular momentum with radius for equatorial Keplerian orbits.
a .9;Plot lkep r, a , hkep r, a , r, rmb a , 6 ,
Frame True, FrameLabel "r", "L,H" , RotateLabel False ;
2 3 4 5 6r
2.2
2.4
2.6
2.8
3
L,H
Clearly H mb Lmb . The minima also coincide in radius (because dE = dL for a circular geodesic). Now thisradius is simply r ms . To see this recall that the stable circular orbits for given L are located at the minima of R(r); the unstable orbits at the maxima and as we reduce L, the two solutions join at a point of inflexion in R(r)and then disappear. Therefore, for H ms H H mb (or Lms L Lmb ), there are two bound (E
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h0 .5 hkep rmb a , a hkep rms a , a ;hxy x_, y_ : h0;rmax rms a ; emax ekep rmax, a ; omegamax omegakep rmax, a ;rcusp r . FindRoot hkep r, a h0, r, rmb a , rmb a .001 ;ecusp ekep rcusp, a ; omegacusp omegakep rcusp, a ;rmax r . FindRoot hkep r, a h0, r, 2 rms a , 2 rms a .001 ;
emax ekep rmax, a ; omegamax omegakep rmax, a ;Print "H ", h0 ;Print "r cusp ", rcusp, " E cusp ", ecusp, " cusp ", omegacusp ;Print "r max ", rmax, " E max ", emax, " max ", omegamax ;
H 2.55981
r cusp 1.85884 E cusp 0.899481 cusp 0.291177
r max 3.12744 E max 0.864627 max 0.155503
Von Zeipel surfaces
The von Zeipel surfaces are the isorotational surfaces, = constant or, equivalently, H=constant.
omegaxy x_, y_ :
If Sqrt x 2 y 2 rh a ,inversemetric 1, 4 h inversemetric 4, 4
inversemetric 1, 1 h inversemetric 1, 4.
r Sqrt x 2 y 2 , ArcTan y, x , h hxy x, y , 10 ;rout 2 rmax;vonzeipel
ContourPlot omegaxy x, y , x, .001, rout , y, .5 rout, .5 rout , ContoursTable omegaxy rout, 0 .2 omegacusp omegaxy rout, 0 i, i, 0, 5 ,
ContourShading False, PlotPoints 100, DisplayFunction Identity ;
Isobars
The isobars are surfaces of constant =lnE.
energy x_, y_, h_ : Max inversemetric 1, 1 2 inversemetric 1, 4 h inversemetric 4, 4
h 2 , .01.5
. r Sqrt x 2 y 2 , ArcTan y, x
phixy x_, y_ : If x 2 y 2 1 Sqrt 1 a 22
,
Logenergy x, y, h
ecusp . h hxy x, y , 10
isobars ContourPlot phixy x, y , x, .001, rout , y, .5 rout, .5 rout ,Contours Table 0.2 i phixy rmax, 0 , i, 0, 5 ,ContourShading False, PlotPoints 100, DisplayFunction Identity ;
bhns.nb 60
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Plot
holeplt
ContourPlot x 2 y 2 1 Sqrt 1 a 22
, x, .001, rout , y, .5 rout, .5 rout ,
Contours 0 , PlotPoints 100, DisplayFunction Identity ;Show holeplt, vonzeipel, isobars, DisplayFunction $DisplayFunction ;
0 1 2 3 4 5 6
-3
-2
-1
0
1
2
3
As H is increased, and the cusp approaches r mb , and the binding energy (1H)of the gas that flows over intothe hole decreases. If r cusp r mb , it is possible to accrete without liberating any binding energy.
Pressure Variation
The density and entropy are also constant on the isobars. The simplest choice is to suppose that the entropyfunction, S, is constant and the equation of hydrostatic equilibrium is integrable.
1+ 1 P 1
S E cusp E Set S=1 wlog and specialize to a radiationdominatedtorus.
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press en_ : 1
ecusp
en1
1
. 4 3
Plot press en , en, emax, ecusp ,PlotRange All, Frame True, FrameLabel "E", "P" ;
0.865 0.87 0.875 0.88 0.885 0.89 0.895 0.9E
0
1 10-7
2 10-7
3 10-7
4 10-7
5 10-7
6 10-7
P
General angular momentum variation
The angular velocity must have the Keplerian value at the cusp and the pressure maximum,r max . It must also rise eventually as H H 0 r 1 2 , H 0 1. We can interpolate a simple, radial variation H(r)with these properties.
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omegaxyh x_, y_, h_ :
If h hcusp, FullSimplifyinversemetric 1, 4 h inversemetric 4, 4
inversemetric 1, 1 h inversemetric 1, 4.
r Sqrt x 2 y 2 , ArcTan y, x , 10hxy x_, y_ : If x rcusp, h . FindRoot omegaxyh x, y, h omegah h ,
h, heq x , heq x .1 , 0vonzeipel ContourPlot hxy x, y , x, .001, rout , y, .5 rout, .5 rout ,
Contours hcusp, heq 5 , heq 10 , heq 20 ,ContourShading False, DisplayFunction Identity ;
We next compute the isobars (x,y)= ln E(x,y,H) H
H cuspdH
1 H =constant.
energy x_, y_, h_ : Max inversemetric 1, 1 2 inversemetric 1, 4 h inversemetric 4, 4
h 2 , .01.5
. r Sqrt x 2 y 2 , ArcTan y, x
ndsg NDSolve g homegah h
1 h omegah h, g hcusp 0 , g, h, hcusp, hout ;
g1 h1_ : g h . ndsg 1 . h h1 phixy x_, y_ : If x rcusp && hxy x, y hcusp,Log
energy x, y, h
ecuspg1 h . h hxy x, y , 1
isobars ContourPlot phixy x, y , x, 0.001, rout ,y, .5 rout, .5 rout , Contours 0, .04, .07, .09, .1, .105 ,
ContourShading False, PlotPoints 100, DisplayFunction Identity ;
Isobars and von Zeipel surfaces
The plot of the isobars exhibits a very steep funnel with this choice of parameters.
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holeplt
ContourPlot x 2 y 2 1 Sqrt 1 a 22
, x, .001, rout , y, .5 rout, .5 rout ,
Contours 0 , PlotPoints 100, DisplayFunction Identity ;Show holeplt, vonzeipel, isobars, DisplayFunction $DisplayFunction ;
0 5 10 15 20 25 30-15
-10
-5
0
5
10
15
Gyrentropic Disks
An alternative prescription is that the disks are gyrentropic. This means that L = L(S). This may be morephysical for fluid disks. The reason is that this is the condition for marginal stability to axisymmetric convec-tion.It is convenient to introduce the relativistic Bernoulli functionB= wnmand it is then straightforward to show that
ln B w nmw ln S H
1 H
so that the "gyrentropes" are also "isoberns". We use a similar angular momentum variation to before
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Clear a, h, rmax ; a 0.9; h0 0.75; 1.67; n 0.58;
agam 3 5 2 n 2 n
5 2 n 2 n;
rcusp rmb a .1;hcusp hkep rcusp, a ; ecusp ekep rcusp, a ;rmax 4.5; hmax hkep rmax, a ; emax ekep rmax, a ;coeff Solve c1 rcusp .5 c2 h0 rcusp .5 hcusp,
c1 rmax .5 c2 h0 rmax .5 hmax , c1, c2 1 ;heq r_ : If r rcusp, hcusp, c1 r .5 c2 h0 r .5 . coeffrout 30; hout heq rout ;lnseq r_ :
agam
1 agam rgtt r1_ : inversemetric 1, 1 . r r1, 2gt r1_ : inversemetric 1, 4 . r r1, 2g r1_ : inversemetric 4, 4 . r r1, 2energyeq r_ : Max gtt r 2 gt r heq r g r heq r 2 , .01
.5
omegaeq r_ :gt r heq r g r
gtt r heq r gt r
dheq r : If r rcusp, 0, .5 c1 r 1.5 .5 h0 r .5 . coeffreq h_ : If h hcusp, rcusp,c2 2 2 c2 h h 2 2 c1 h0 4 c1 2 h0 2 c2 2 2 c2 h h 2 2 c1 h0
2
2 h0 2 . coeffnds1 NDSolve bern r bern r energyeq r lnseq r
omegaeq r bern r dheq r
1 omegaeq r heq r,
bern rcusp energyeq rcusp , bern, r, rcusp, rout ;fig7a Plot hkep r, a , heq r , r, rcusp, 10 ,
Frame True, PlotRange 0, 10 , All , FrameLabel "r", "L " ,RotateLabel False, DisplayFunction Identity ;
fig7b Plot Exp lnseq r , bern r . nds1 , r, rcusp, 10 ,Frame True, PlotRange 0, 10 , All , FrameLabel "r", "S ,B " ,
RotateLabel False, DisplayFunction Identity ;Fig7 Show GraphicsArray fig7a, fig7b ;
2 4 6 8 10r
2.6
2.8
3
3.2
3.4
3.6
L
2 40.88
0.9
0.92
0.94
0.96
0.98
S ,B
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bernh h_ : bern r . nds1 1 . r req hdbernh h_ : 10000 bernh h .0001 bernh hlnsh h_ : lnseq req hdlnsh h_ : 10000 lnsh h .0001 lnsh homegaxyh x_, y_, h_ : If h hcusp, omegaeq rcusp ,
FullSimplifyinversemetric 1, 4 h inversemetric 4, 4
inversemetric 1, 1 h inversemetric 1, 4 .
r Sqrt x 2 y 2 , ArcTan y, x
energyxyh x_, y_, h_ : If h hcusp, energyeq rcusp , Max inversemetric 1, 1 2 inversemetric 1, 4 h
inversemetric 4, 4 h 2 , .01.5
. r Sqrt x 2 y 2 , ArcTan y, x
hxy x_, y_ : h .FindRoot dbernh h bernh h energyxyh x, y, h dlnsh h omegaxyh x, y,
bernh h 1 omegaxyh x, y, h h , h, heq x , heq x .1
hxy 4.5, 0hkep 4.5, a
2.71917
2.77943
energyxy x_, y_ : Max inversemetric 1, 1 2 inversemetric 1, 4 h inversemetric 4, 4
h 2 , .01.5
. r Sqrt x 2 y 2 , ArcTan y, x . h hxy x, yomegaxy x_, y_ :
inversemetric 1, 4 h inversemetric 4, 4
inversemetric 1, 1 h inversemetric 1, 4 .
r Sqrt x 2 y 2 , ArcTan y, x . h hxy x, y bernxy x_, y_ : bernh hxy x, ysxy x_, y_ : Exp lnsh hxy x, y
pxy x_, y_ : 1
sxy x, y
bernxy x, y
energyxy x, y1
1
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ContourPlot pxy x, y , x, 2, 10 , y, 5, 5 , ContourShading False,Contours , DisplayFunction $DisplayFunction ;
2 4 6 8 100
2
4
6
8
10
holeplt
ContourPlot x 2 y 2 1 Sqrt 1 a 22
, x, .001, rout , y, .5 rout, .5 rout ,
Contours 0 , PlotPoints 100, DisplayFunction Identity ;Show holeplt, gyrentropes, isobars, DisplayFunction $DisplayFunction ;
6. Tides and Oscillations
Tidal Effects
Motivation
The discovery of large numbers of young stars in close orbit around our central black hole suggests strongly
that the supply of stars in galactic nuclei is much greater than suspected. It further raises the possibility thatstars can be capturedonto relativistic orbits quite readily and that tidal disruption events might be quite common in nearby galaxies.
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Metric etc
Clear coord, metric, inversemetric, affine, a ;coord t, r, , ;
metric
r 2 2 r a 2 Cos 2
r 2 a 2 Cos 2 , 0, 0,
2 a r Sin 2
r 2 a 2 Cos 2 ,
0,r 2 a 2 Cos 2
r 2 2 r a 2, 0, 0 , 0, 0, r 2 a 2 Cos 2 , 0 ,
2 a r Sin 2
r 2 a 2 Cos 2, 0, 0,
r 2 a 2 r 2 a 2 Cos 2 2 r a 2 Sin 2
r 2 a 2 Cos 2Sin 2 ;
r 2 a 2 Cos 2 ;
r 2 2 r a 2 ;
r 2 a 2 2 a 2 Sin 2 ;inversemetric FullSimplify Inverse metric ;
affine FullSimplify Table Sum inversemetric i, sD metric s, j , coord kD metric s, k , coord j D metric j, k , coord s ,
s, 1, 4 2, i, 1, 4 , j, 1, 4 , k, 1, 4 ;eqaffine affine . 2;riemann FullSimplify
Table D affine i, j, l , coord k D affine i, j, k , coord lSum affine i, s, k affine s, j, l affine i, s, l affine s, j, k
s, 1, 4 , i, 1, 4 , j, 1, 4 , k, 1, 4 , l, 1, 4 ;
omkep r_, a_ : r 3 2 a1
ekep r_, a_ :1 2 r 1 a r 3 2
1 3 r 1 2 a r 3 2 1 2
lkep r_, a_ :1 2 a r 3 2 a 2 r 2 r 1 2
1 3 r 1 2 a r 3 2 1 2
rms a_ : Chop
3 Sign a 3 a 2 3 a 1 a 1 3 1 a 2 3 1 a 2 3 1 a 1 3 3 a18 3 a 1 a 1 3 1 a 2 3
1 a 2 3 1 a 1 3 3 a 2 6 a 2 Sign a 16 a 2 3 a 2 3 a 1 a 1 3 1 a 2 3 1 a 2 3 1 a 1 3 3 a
Equatorial Riemann tensor
We confine our attention to stars on circular orbits. These might be formed in an accretion disk and migrateslowly inwards or be captured and dragged into the equatorial plane by interaction with a disk. The Riemanntensor becomes
eriemann riemann . 2 ;
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Clear a ;tidalplt Table
Plot tidal 1, 1 . a .999 i, tidal 2, 2 . a .999 i,tidal 3, 3 . a .999 i , r, rms .999 i , 10 , Frame True,FrameLabel "r", "T" , DisplayFunction Identity , i, 1, 1 ;Show GraphicsArray tidalplt ;
9 9.2 9.4 9.6 9.8 10r
-0.004-0.003-0.002-0.001
00.0010.002
T
6 7 8 9 10r
-0.01
-0.005
0
0.005
T
0
-0.2
0
0.2
0.4
T
Numerical Values
A star will be ripped apart outside r ms if its density < 70 T 11 M 8
2 g cm 3 .
Rotation of Inertial Frame
The basis vectors associated with an inertial frame tied to a freefallingstar will be parallel propagated and willrotate with respect to our BL basis and our choice of local orthonormal basis used above. Consider one spacelike basis vector, e chosen to lie in the equatorial plane. Writee = e, er , 0, ein BL, ensuring that ue = 0, and ignoring normalization. The law of parallel propagation is u e =0 orded e
u
from which we obtaind 2 ed2
u u e Evaluating the rhs,
evec omkep r, a e , e r , 0, e ;rhs FullSimplify Table
Sum eqaffine j, l, m eqaffine l, i, k evec j ucon k ucon m ,j, 1, 4 , k, 1, 4 , l, 1, 4 , m, 1, 4 , i, 1, 4
e a r 3 r 9 2
,e rr 3
, 0,e r 3
so thatd 2 ed2
= er 3independent of a! The dragging of inertial frames is really a consequence of the choice of coordinates .The rate of geodetic (plus LenseThirring)precession (in the BL frame) is given by
geo 1
r 3 2 ut 1 1 3r
2 ar 3 2
1 2
r 3 2 a3
2 r 5 2ar 3 +
in agreement with the weak field result.
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omgeo r1_, a1_ : omkep r1, a1r1 3 2
ucon 1 . r r1, a a1geoplt1 Plot omgeo r, 1 , r, rms 1 , 10 , Frame True,
FrameLabel "r", " geo " , RotateLabel False, DisplayFunction Identity ;geoplt2 Plot omgeo r, 0 , r, rms 0 , 10 , Frame True,
FrameLabel "r", " geo " , RotateLabel False, DisplayFunction Identity ;geoplt3 Plot omgeo r, 1 , r, rms 1 , 10 , Frame True,FrameLabel "r", " geo " , RotateLabel False, DisplayFunction Identity ;
Show GraphicsArray geoplt1, geoplt2, geoplt3 ;
9 9.2 9.4 9.6 9.8 10
r
0.007
0.0075
0.008
0.0085
geo
6 7 8 9 10r
0.0060.008
0.010.0120.0140.0160.018
0.02
geo
0
0.1
0.2
0.3
0.4
0.5
geo
This might be observable if there is a transfer of energy from the orbit to the star as it passes through a reso-nance with an internal oscillation frequency. The generalization to an inclined, eccentric orbit is given byDiener et al ApJ 479 164.
QPOs and Diskoseismology
Motivation
Many accretion disks exhibit quasiperioidcoscillations modulations of the hard Xrayemission with ampli-tude of a few percent and Q of order 10100.Both BH and NS show these oscillations which are mostlysupposed to be disk modes. Specific frequencies are associated with individual sources. NS often show pairs of frequencies which drift together with a spacing related to the spin frequency, or half this value. They do notdrift much in frequency as the luminosity changes. If these are disk modes then the disk is only the clock as itraidates soft Xrays;presumably the emission comes from a hot, Comptonizing corona which is modulated bya variable soft photon spectrum from the disk or a variable heating
Small perturbations for cold disks
When the disk is cold and thin we can ignore pressure. The equations of motion contain the quantities R(r),( ).
ft r_, _ :r 2 a 2 r 2 a 2 e a l
r 2 2 r a 2a a e Sin 2 l
dfr r_ : 2 r e r 2 a 2 e a l r 1 r 2 a e l 2 q r r 2 2 r a 2
df _ : a 2 1 e 2 Sin Cos l 2 Cot Csc 2
Specializing to small departures from circular equatorial orbits with = /2 gives
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ft r_ :r 2 a 2 r 2 a 2 e a l
r 2 2 r a 2a a e l
dfr r_ : 2 r e r 2 a 2 e a l r 1 r 2 a e l 2 q r r 2 2 r a 2
df _ : a 2 1 e 2 l 2
Vertical epicyclic motion
We see that the vertical equation of motion isd 2
dx2 a 2 1 e2 l2
Evaluating for circular orbits, we find
ddpsidxx FullSimplify a 2 1 e 2 l 2 . e ekep r, a , l lkep r, a
3 a 2 r 4 a r r 5 2
2 a 3 r r
We also need dtdx
dtdx FullSimplify
r 2 a 2 r 2 a 2 e a l
r 2 2 r a 2a a e l . e ekep r, a , l lkep r, a
a r r 2
1 2 ar 3 23r
so that the square of the vertical epicyclic frequency is
FullSimplify ddpsidxx dtdx 2
3 a 2 4 a r r 2
r 2 a r 3 2 2
v 2 2 14 ar 3 2 3
a2
r 2
(We are interested in freqyuencies measured from infinity.)Expanding at large radius, we obtain v
2 ar 3
This corresponds to prograde, orbital LenseThirringprecession as measured from infinity.
LT2 ar 3
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Radial epicyclic motion
Setting Q=0 in the linear approximation and usingr= R r 2 r
ddrdxx FullSimplify D2 r e r 2 a 2 e a l r 1 r 2 a e l 2 r r 2 2 r a 2 , r .e ekep r, a , l lkep r, a
r 3 a 2 8 a r 6 r r
2 a 3 r r
and so the radial epicyclic frequency is given by
FullSimplifyddrdxx
dtdx2
3 a 2 8 a r 6 r r
r 2 a r 3 2 2
2 2 1 6r 8 ar 3 2
3 a2
r 2
Again, expanding at large radius, 3r 5 2corresponding to prograde apsidal motion at a rate
aps3
r 5 2
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