Download - Black Hole Astrophysics Chapter 7.4
Black Hole AstrophysicsChapter 7.4
All figures extracted from online sources of from the textbook.
FlowchartBasic properties of the Schwarzschild metric
Coordinate systemsEquation of motion and conserved quantities
Letβs throw stuff in!What does it feel like to orbit a Black Hole?
General motion in Schwarzschild MetricHorizon Penetrating coordinates
The Schwarzschild Matric(πSH
Sch )Ξ±Ξ²=(βπ2 (1β
π π
π ) 0 0 0
01
(1βπ π
π )0 0
0 0 π2 00 0 0 π 2 sin2 π
)βSchβ means that this metric is describing a Schwarzschild Black Hole.
Recall: A Schwarzschild Black Hole is a solution of the Einstein Equations assuming that we put a point mass M in free space and then assume that we are in a static coordinate.βSHβ means that we are in the Schwarzschild-Hilbert coordinate system.Why bother?Remember that we are now in curved space, but we can sometimes for convenience still choose a locally flat coordinate to consider the physics. The SH coordinate is just like considering the whole surface of the Earth as a curved surface.
The Schwarzschild radius
The metric being diagonal also says that relativistic spherical gravity is still a radial r=force.
Some basic propertiesds β
dr
β1βππ π
π =π βπ (1βπ π
π )+ππ ΒΏ
Integrate[ 1
β1βπ π
r 0
, {r 0 ,ππ ,π }]]
When and it reduces to Newtonian gravity as expected
Limits at infinity(πSH
Sch )Ξ±Ξ²=(βπ2 (1β
π π
π ) 0 0 0
01
(1βπ π
π )0 0
0 0 π2 00 0 0 π 2 sin2 π
)β (βπ2 0 0 00 1 0 00 0 π 2 00 0 0 π 2 sin2 π
)If we take or Reduces to the Minkowski metric!
Passing the horizon(βπ2 (1β
π π
π ) 0 0 0
01
(1βπ π
π )0 0
0 0 π2 00 0 0 π 2 sin2 π
)β (π2 ( π π
πβ1) 0 0 0
0 β1
( π π
πβ1)
0 0
0 0 π 2 00 0 0 π2 sin2 π
)Outside the horizon Inside the horizon
Whatβs so interesting?We know that particles can only travel on timelike trajectories, that is, .Outside the horizon, is the negative term so we can be on a timelike trajectory if we have
Inside the horizon, it is that is negative! So to be on a timelike trajectory, the simplest case would be to have This means that we can only fall toward the BH once we pass the horizon!
Coordinate Systems1. The moving body frame (MOV)
2. Fixed local Lorentz frame (FIX)
3. Schwarzschild-Hilbert frame (SH)
The moving body frame (MOV)In this frame, we are moving with the object of interest. Since spacetime is locally flat, we have a Minkowski metric in this case and by definition the 4-velocity
This frame is useful for expressing microphysics, such as gas pressure, temperature, and density, but not motion.
Fixed local Lorentz frame (FIX)In this frame, we consider some locally flat part of the Schwarzschild spacetime to sit on and watch things fly past. Therefore the metric is still the Minkowski onebut now the 4-velocity of objects become which is obvious since the FIX and MOV frames
are simply related by a Lorentz Transform.
It is a convenient frame for looking at motion of particles.However, it is not unique, there is a different FIX frame for every point around the black hole. This also means that time flows differently in different frames.
Schwarzschild-Hilbert frame (SH)This is a global coordinate, so it does not have the problems in the FIX frame, there is a unique time coordinate and a single system.
For this coordinate, the metric is the one we presented earlier However, in such a case is hard to interpret.
Which frame to use?
How to go from FIX to SH frame?
(πFIXSch )Ξ±Ξ²=(
β1 0 0 00 1 0 00 0 1 00 0 0 1
)
ππΌ β² π½ β²=π¬πΌ β²πΌ π¬π½ β²
π½ πΞ±Ξ²
(πSHSch )πΌ β² π½β²=π¬SH (πΌ β² )
FIX (πΌ) π¬SH ( π½ β²)FIX (π½) (πFIX
Sch )Ξ±Ξ²
(πtt 0 0 00 πrr 0 00 0 πΞΈΞΈ 00 0 0 πΟΟ
)=π¬SH (πΌ β² )FIX (πΌ ) π¬SH (π½ β² )
FIX (π½ ) (β1 0 0 00 1 0 00 0 1 00 0 0 1
)
π¬SHFIX=(β
βπtt 0 0 0
0 βπrr 0 0
0 0 βπΞΈΞΈ 0
0 0 0 βπΟΟ
)=(π β1β
ππ π
0 0 0
01
β1βπ π
π
0 0
0 0 π 00 0 0 π sinΞΈ
)π¬FIX
SH=(1
ββπtt
0 0 0
01
βπrr
0 0
0 01
βπΞΈΞΈ
0
0 0 01
βπΟΟ
)=(1
π β1βπ π
π
0 0 0
0 β1βπ π
π0 0
0 01π
0
0 0 01
π sinΞΈ
)
Generalized Lorentz Transform
Expressing the 4-velocity in SH coordinatesπ¬FIX
SH=(1
ββπtt
0 0 0
01
βπrr
0 0
0 01
βπΞΈΞΈ
0
0 0 01
βπΟΟ
)=(1
π β1βπ π
π
0 0 0
0 β1βπ π
π0 0
0 01π
0
0 0 01
π sinΞΈ
)(π SH )πΌ β²=(
ππ‘
π π
π π
π π)=π¬FIX (πΌ )SH (πΌ β² ) (π FIX )πΌ=(
1
π β1βπ π
π
0 0 0
0 β1βπ π
π0 0
0 01π
0
0 0 01
π sinΞΈ
) (Ξ³c
Ξ³V π
Ξ³V οΏ½ΜοΏ½
Ξ³V οΏ½ΜοΏ½)=(πΎ
β1βπ π
π
β1βππ π
Ξ³V π
Ξ³V οΏ½ΜοΏ½
π
Ξ³V οΏ½ΜοΏ½
π sinΞΈ
)(π FIX )πΌ=(
Ξ³cΞ³V οΏ½ΜοΏ½
Ξ³V οΏ½ΜοΏ½
Ξ³V οΏ½ΜοΏ½)(π SH )πΌ=(ππ‘
π π
ππ
π π)
This now becomes more convenient to use and interpret. is the 4-velocity of the global frame and we write its components in terms of local frame parameters
Letβs examine the 4-velocityπ SH
πΌ=(πΎ
β1βπ π
π
β1βππ π
Ξ³V π
Ξ³V οΏ½ΜοΏ½
π
Ξ³V οΏ½ΜοΏ½
π sinΞΈ
)=(dtdΟdrdΟdΞΈdΟdΟdΟ
)We expect from our old idea of gravity that the velocity of objects should approach c as we get to the black hole, but if we check
Then when οΌ
Particles seen in the SH frame apparently are βstuckβ at the horizon and never get across it!But particles should fall into black holes!
This is simply due to the Generalized Lorentz Transform.
What happened?dt SH =
dt FIX
β1βπ π
π
Consider someone falling into a black hole, the local FIX frame observes the time of the person as , then, for a person sitting watching the BH very far away he would observe .Given that should be finite, as , It would take the far away observer infinite amount of time to watch the unfortunate person falling into the hole!This also says that any photon sent out by the falling person would be infinitely redshifted.
( π¬FIXSH )diag=( 1
π β1βπ π
π
,β1βπ π
π,1π
,1
π sinΞΈ )
In the SH frameIn the FIX frame
What happened?dr SH=dr FIX β1β
ππ π
Similarly, for finite , as , οΌNo matter how much the person moves in some instant, a far away observer would observe him as stuck!
Therefore combining and itβs obvious that the apparent velocity for an observer at infinity is zeroοΌ
( π¬FIXSH )diag=( 1
π β1βπ π
π
,β1βπ π
π,1π
,1
π sinΞΈ )
Ahhhhh Ah β¦hβ¦β¦.hβ¦β¦β¦β¦.
The Equation of motion
π π‘ (πππ
ππ‘β12
πrr ππtt
πππ π‘)+π π (πππ
ππ+ 12
πrr ππrr
ππππ)=0
Considering only radial motion, and applying the relation between Christoffel symbols and the metric (we are now working in the SH coordinate),
In general , the equation of motion expands to The equation of motion:
Hello GravityοΌNow comes the hidden trick used in the bookβ¦
πππSH
ππ SH
+12
πrrSH ππrr
SH
ππSH
ππSH=
1
βπrr
πππ
FIX
ππ SH
β12
ππ
FIX
(πrrSH )1.5
ππrr
SH
ππ SH
+( 12 1
πrrSH
ππrrSH
ππSH ) ( ππ
FIX
βπrr)
π π‘SH (ππ
πSH
ππ‘ SH
β12
πrrSH ππtt
SH
ππSH
ππ‘SH )+π π
SH (πππ
SH
ππSH
+ 12
πrrSH ππrr
SH
ππ SH
ππSH)=0
Time dilation factor
General Relativistic Term! Newtonian Gravity with relativistic mass
π π‘SH πππ
FIX
ππ‘ SH
+π πSH πππ
FIX
ππ SH
βπ π‘
SH π π‘SH
βπrr
12
ππtt
SH
ππ SH
=0
dPπFIX
dΟ=π (Ξ³m0 π π )
dΟ=β
πΎ
β1βrsπ
(πΊ π ( Ξ³m0 )π2 )
Itβs from the gradient operator!
Conserved Quantities β1-forms are usefulοΌAgain the equation of motion (π π
β
dΟ )πΌ
=[(πβ Β·π»β ) π
β ]πΌ
=π π½ (πππΌ
π π½+π€πΌ
ΞΌΞ² ππ)=πΉπΌ
Considering the direction,
Replacing in the definitions of the Christoffel symbols,
The angular momentum of a particle is conserved along the trajectory!
Henceforth, if unspecified, all the tensor/vector components are written in the SH coordinate
ππ=πΎ π0 π οΏ½ΜοΏ½ π sinΞΈFinally, we get,
Conserved Quantities β1-forms are usefulοΌπΈ=βππ‘=β1β
ππ π
πΎ π0 π2dEdΟ
=0
The energy of a particle is also conserved along the trajectory!
Similarly, we can also find that for the energy,
This constant, E, is sometimes also called energy at infinity because as , this term goes to .
However, since it is the same at any radius, we can use it to calculate hence the velocity (we will see this on the next slide).
A moving body is bound to the BH if and unbound otherwise.
Free fallπΈ=βππ‘=β1β
ππ π
πΎ π0 π2
On the last slide we mention that we can use E to calculated the Lorentz factor as a function of radial distance r, letβs now work it out.Consider a particle falling toward a black hole stating from rest at infinity.This means that
This gives us At the event horizon,
We see that if we drop something at infinity and assuming there is noting else in the universe between it and the BH, then it arrives at the BH at exactly the speed of light!
Chucking stuff directly at the BHNow you might askοΌβ Particles accelerate to c if we drop them off at infinity, what if we kick them into the BH starting from infinity?βSpecial relativity tells us that we canβt exceed c now matter what, so somehow the particle should still end up less than c even if we throw as hard as we can! Letβs consider a general case in which we donβt specify energy at infinity, thus,
Solving this gives Interestingly, no matter what E is, when , we always get οΌ
πΈβ=π0 π2
πΈβ=2π0 π2πΈβ=5π0 π2
πΈβ=10π0 π2πΈβ=100π0 π2
OrbitsConsider the simple cases of circular orbits, again using the equation of motion
πΉ π=π π‘ (πππ
ππ‘+π€π
ΞΌt ππ)+π π ( πππ
ππ+π€π
ΞΌr ππ)+π π (πππ
π π+π€π
ΞΌΞΈ ππ)+π π ( πππ
ππ+π€π
ΞΌΟ ππ)With some further reductionβ¦
π οΏ½ΜοΏ½=βπΊ ππ βπ π
We find that the orbital velocity in general is
Photon OrbitsTo find the orbital radius of photons, lets consider case
This gives us i.e. for photons, the only place they can orbit the BH is at this radius.
However, weβll see later in a more general formulism (in Schutz) that this orbit is nowhere stable, if we accidently kick the photon a bit, it will either spiral into the BH or spiral out to infinity.
Finite mass particle orbitsFor finite mass particles, we need to consider case
By definition of the Lorentz factor
Solving for the energy and angular momentum, πΏ=ππ=πΎ π0 π οΏ½ΜοΏ½ π sinΞΈ πΈ=βππ‘=β1βππ π
πΎ π0 π2
πΏorb=β π π
2 π β3 π π
π π0 π2
πΈorb=πβπ π
βπ (π β32
π π )π0 π2
-- The radius at which
Minimum point at L
E
The ISCOThe minimum for both of these two curves happen at This is commonly called the Innermost stable circular orbit for reasons we will see later.
β The radius at which photons orbit
Minimum point at L
E
However, we will also find that the ISCO is more or less a βmarginally stableβ orbit! If we accidently kick it a bit toward the black hole, it will just give up and fall in!
πΏorb=β π π
2 π β3 π π
π π0 π2πΈorb=
πβπ π
βπ (π β32
π π )π0 π2
πΏ=β3 π π π0 π πΈ=2 β23
π0 π
At this radius, L and E are
General discussion for particle motionPreviously, we have already found that we can calculate either free-fall or orbits by considering E or L respectively.For a general consideration, it is more convenient if we write both of them in the same equation so we can discuss different the properties of the different orbits more clearly π2=βπ0
2π2=πtt (ππ‘ )2+πrr (ππ )2+πΞΈΞΈ (ππ )2+πΟΟ (π π )2
πΏ=ππ=πΎ π0 π οΏ½ΜοΏ½ π πΈ=βππ‘=β1βππ π
πΎ π0 π2
For simplicity, we take , so
β1π2
1
(1βπ π
π )πΈ2+ 1
(1βπ π
π )π0
2 ( drdΟ )
2
+ πΏ2
π2=βπ0
2 π2
( drπ dΟ )
2
=( πΈπ0 π2 )
2
β(1βπ π
π ) (1+ 1π2 ( πΏπ0 π )
2
)
General discussion for particle motion( drπ dΟ )
2
=( πΈπ0 π2 )
2
β(1βπ π
π ) (1+ 1π2 ( πΏπ0 π )
2
)We can define the effective potential as
Then, very much like the classical οΌ
Remember that both E and L are constant of trajectory
Behavior in different potentials
Falls in no matter what! Simply put, the centrifugal force canβt balance with gravity!
http://www2.warwick.ac.uk/fac/sci/physics/current/teach/module_home/px436/notes/lecture16.pdf
Unstable Circular HyperbolicCapture
Elliptical
CircularCapture
Capture
Comparing with Classical Physics
Stable Circular
Double root solution
In relation to our previous analysis
1 Circular orbit solution
Capture Capture
Forwe have two solutions.
ISCO case
No real solution
Two solutionsUnstable Circular CaptureStable Circular
The marginally bound orbitπΈβ=πΈ=π0 π2
2π π
How stable is the ISCO?
1 Circular orbit solution
Capture
( drπ dΟ )
2
=( πΈπ0 π2 )
2
β(1βπ π
π ) (1+ 1π2 ( πΏπ0 π )
2
)Taking the derivating w.r.t proper time on both sides, we get the force equation which is analogous to in Classical Physics.
The ISCO is the double root solution, it is at the same time the stable and unstable circular orbit. Unfortunately for particles flying about the black hole, the result is simply that it is unstable! Any perturbation toward the black hole and the particle would have to say goodbye to the rest of the outside universe!
Observational evidence of the ISCO?Resolving the Jet-Launch Region of the M87 Supermassive Black Hole , Science 338, 355 (2012)
(πHPSch )Ξ±Ξ²=(βπ2 (1β
ππ π ) π π
ππ 0 0
π π
ππ 1+
π π
π0 0
0 0 π2 00 0 0 π 2 sin2 π
)
The Horizon-Penetrating coordinates(πSH
Sch )Ξ±Ξ²=(βπ2 (1β
π π
π ) 0 0 0
01
(1βπ π
π )0 0
0 0 π2 00 0 0 π 2 sin2 π
)(πHP
Sch )Ξ±Ξ²=(βπ2 (1βππ π ) π π
ππ 0 0
π π
ππ 1+
π π
π0 0
0 0 π2 00 0 0 π 2 sin2 π
)
dt β²=dt +ππ
π (π βππ )dr
π¬HP (πΌ β² )SH (πΌ ) β‘(1 π π
π (πβπ π )0 1
)
(1 π π
π (π βπ π )0 1
)π
.(βπ2 (1βπ π
π ) π π
ππ
ππ π
π 1+π π
π) .(1 ππ
π (π βπ π )0 1
)=(βπ2 (π βππ )
π0
0π
π βππ )
π¬HP (πΌ β² )SH (πΌ ) π¬
HP (π½ β² )SH (π½ ) (πHP
Sch )πΌ β² π½ β²=(πSHSch )Ξ±Ξ²